Algebrai and dierential formsMiquel Dalmau Vilalda h

2 .

3Prefa eThis is a problem book evolved from problem le tures I gave in UPC(Universitat Politè ni a de Catalunya). There are one hundred and eightyone problems ompletely solved in detail. Some are standard, some are re re-ations and some others have been told by olleagues. If any merit an begiven to the olle tion it ould be the sele tion and grouping of the material,mainly following Polya's di tum that 'problems grow as mushrooms'.These problems ould interest students of mathemati s or theoreti alphysi s. Some students of mathemati s, some physi ists not so theoreti- al and students of engineering may prefer to browse through the ompanionproblem book ve tor al ulus.A qui k view at the table of ontents reveals that half of the problemsare in the land of linear or multilinear algebra and the other half are aboutdierential forms and its use in Stokes theorem and other matters as well.If you are in a hurry take a look at my favorite problems: 13, 42, 76,84 (with an unproved funny formula for the omputation of the sign of a anoni al permutation), 102, 104, 154, 159, 169, 176 and the presentationof Maxwell's equations in terms of dierential forms.I only assume a mild responsability as to possible errors be ause as lateProf. Wieszlaw Slen k said orre ting is an innite non onvergent pro ess.The existen e of errors an be looked at as a stimulus for the reader to be areful. In any ase if somebody wants to point out an error, he an do sowriting to miquel.dvgmail. om, but he needn't use this address only forthat reason and if he liked some part(s) of the book he an show it throughthe same hannel.Needless to say that my English is IE (International English), not SE(Shakespeare English); I apologize about that.

4 A knowledgmentsThe author would like to thank Profs. Jordi Saludes for the onden e,and Juan Jose Morales who took are of the ourse later on. Thanks as wellto Prof. Natalia Sadovskaia who pointed out errors and ontributed someproblems.Bar elona May 2013

5Notationa := b denes a in terms of a known b.vs:=ve tor spa e.og ve tors:=orthogonal ve tors.on set of ve tors:=orthonormal set of ve tors.li:=linearly independent.Matrix (m,n) :=m rows, n olumns.T Beginning of a theoreti al se tion. End of a theoreti al se tion or end of a problem. :=isomorphism at.♯ :=isomorphism sharp.i:=if and oly if.πi, dxi := i− th proje tion in Rn.T kE := k-multilinear appli ations from E to K.ΛkE := k-multilinear alternating appli ations from E to K.Ωk(U) :=dierential k-forms on an open set U .V(U) :=ve tor elds on an open set U .I = (i1, . . . , ik) ⊂ (1, . . . , n)k a multiindex.J = (j1, . . . , jk)ր, 1 ≤ j1 < · · · < jk ≤ n a stri tly in reasing multiindex.ω⊗ = ωi1 ⊗ · · · ⊗ ωik(i1,...,ik) a basis of T kE (the (ωi1, . . . , ωn) being abasis of E∗.ω∧ = ωi1 ∧ · · · ∧ ωik(i1,...,ik)ր a basis of ΛkE (the (ωi1, . . . , ωn) being abasis of E∗ .S1 :=unit ir umferen eS2 :=unit sphere

6

Contents1 Linear algebra 111.1 Basi s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.1.1 Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.1.2 Change of basis . . . . . . . . . . . . . . . . . . . . . . 121.1.3 Linear maps . . . . . . . . . . . . . . . . . . . . . . . . 131.1.4 Coordinates of a basis . . . . . . . . . . . . . . . . . . 151.1.5 Interlude . . . . . . . . . . . . . . . . . . . . . . . . . . 171.2 Dual spa e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.1 Dual spa e . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.2 Dual map . . . . . . . . . . . . . . . . . . . . . . . . . 271.3 First anoni al isomorphism . . . . . . . . . . . . . . . . . . . 281.3.1 Metri s . . . . . . . . . . . . . . . . . . . . . . . . . . 281.3.2 First anoni al isomorphism . . . . . . . . . . . . . . . 402 Alternating multilinear forms 492.1 Multilinear appli ations and forms . . . . . . . . . . . . . . . . 492.1.1 Basi s . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.1.2 Pull-ba k . . . . . . . . . . . . . . . . . . . . . . . . . 582.2 Alternating forms . . . . . . . . . . . . . . . . . . . . . . . . . 612.2.1 Permutations . . . . . . . . . . . . . . . . . . . . . . . 612.2.2 A tion on lists and fun tions . . . . . . . . . . . . . . . 642.2.3 Multilinear alternating forms . . . . . . . . . . . . . . 662.3 Exterior produ t . . . . . . . . . . . . . . . . . . . . . . . . . 732.3.1 Exterior produ t 1 . . . . . . . . . . . . . . . . . . . . 732.3.2 De omposable forms . . . . . . . . . . . . . . . . . . . 772.3.3 Exterior produ t 2 . . . . . . . . . . . . . . . . . . . . 882.3.4 Medley . . . . . . . . . . . . . . . . . . . . . . . . . . . 922.4 Orientation and volume . . . . . . . . . . . . . . . . . . . . . 1007

8 CONTENTS2.4.1 Orientation . . . . . . . . . . . . . . . . . . . . . . . . 1002.4.2 Volume and volume form . . . . . . . . . . . . . . . . . 1032.4.3 Contra tion . . . . . . . . . . . . . . . . . . . . . . . . 1092.4.4 Volume form of subspa es . . . . . . . . . . . . . . . . 1122.4.5 Cross produ t . . . . . . . . . . . . . . . . . . . . . . 1172.4.6 Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . 1232.4.7 Se ond anoni al isomorphism . . . . . . . . . . . . . . 1292.5 Hodge star appli ation . . . . . . . . . . . . . . . . . . . . . . 1322.5.1 S alar produ t of k-forms . . . . . . . . . . . . . . . . 1322.5.2 Geometri al denition . . . . . . . . . . . . . . . . . . 1362.5.3 General denition . . . . . . . . . . . . . . . . . . . . . 1412.5.4 Hodge and anoni al isomorphisms . . . . . . . . . . . 1583 Dierential forms 1673.1 Dierential forms . . . . . . . . . . . . . . . . . . . . . . . . . 1673.1.1 Basi s . . . . . . . . . . . . . . . . . . . . . . . . . . . 1673.1.2 Exterior derivative . . . . . . . . . . . . . . . . . . . . 1693.1.3 Pull-ba k for dierential forms . . . . . . . . . . . . . 1793.2 Translation isomorphisms . . . . . . . . . . . . . . . . . . . . 1893.2.1 Metri s in an open set . . . . . . . . . . . . . . . . . . 1893.2.2 Canoni al isomorphisms . . . . . . . . . . . . . . . . . 1903.3 Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1973.3.1 Coordinates and oordinate urves . . . . . . . . . . . 1973.3.2 Canoni al isomorphisms in oordinates . . . . . . . . . 2263.3.3 Dierential operators in oordinates . . . . . . . . . . . 2293.4 Hodge's operator . . . . . . . . . . . . . . . . . . . . . . . . . 2403.4.1 Hodge's operator . . . . . . . . . . . . . . . . . . . . . 2443.4.2 Hodge and dierential operators . . . . . . . . . . . . . 2454 Stokes theorem 2614.1 Closed and exa t forms . . . . . . . . . . . . . . . . . . . . . 2614.1.1 Closed and exa t forms . . . . . . . . . . . . . . . . . . 2614.1.2 Poin aré formula . . . . . . . . . . . . . . . . . . . . . 2654.2 Stokes theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 2774.2.1 Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . 2774.2.2 Integration on hains . . . . . . . . . . . . . . . . . . . 2814.2.3 Stokes theorem . . . . . . . . . . . . . . . . . . . . . . 2894.3 Maxwell equations and dierential forms . . . . . . . . . . . . 310

CONTENTS 94.3.1 The rst two equations . . . . . . . . . . . . . . . . . . 3114.3.2 The last two equations . . . . . . . . . . . . . . . . . . 3135 Graphi al bibliography 317

10 CONTENTS

Chapter 1Linear algebra1.1 Basi s1.1.1 BasisT Let E be a K-ve tor spa e (mainly K = R, only o asionally K = C)and e = (e1, . . . , en) a basis (a matrix (1,n) with the ve tors of the basis asentries). The main role of a basis is to show ea h abstra t ve tor v ∈ E as afamiliar list of n numbers (x1, . . . , xn), the list of omponents (or oordinates)of v in the basis e:

v = x1e1 + · · ·+ xnen

e2 u2

e1

u1

.

v’

v

The gures show the bases e = (e1, e2) (the anoni al one for familiarity)and u = (u1,u2) of R2. Ea h one has its own point of view, its own way ofstratifying the plane, of onverting ve tors into lists of numbers. The gure11

12 CHAPTER 1. LINEAR ALGEBRAon the right has re tangular axes in it only for referen e purposes. Theve tors v and v′ have the same numeri al oordinates, namely (2, 3), despitebeing ompletely dierent ve tors.For omputational reasons among others, we shall list the omponents ofa ve tor in a olumn, thusX =

x1...xn

Then, using the simplest matrix produ t, that of a row by a olumn, we anwrite the expression of a ve tor in a ompa t form:

v = x1e1 + . . .+ xnen = (e1, . . . , en)

x1...xn

= eX

v ve tor e basisv = eX

1.1.2 Change of basisT Let e = (e1, . . . , en) and e′ = (e′1, . . . , e

′n) be bases of E; the matrix Cwhose olumns are the omponents of the e′i in the basis e is alled the hangeof basis matrix from the basis e to the basis e′; then

e′ = (e′1, . . . , e′n) = (e1, . . . , en)

c11 . . . c1n... ... ...cn1 . . . cnn

= eCExample:Take E = R3, let e be the anoni al basis and onsider the new basis

e′1 = (1,−1, 1), e′2 = (−1, 0, 1), e′3 = (2, 1, 0).The hange of basis matrix from the anoni al basis to the basis e′ is:C =

1 −1 2−1 0 1

1 1 0

1.1. BASICS 13Let e be a basis and X the olumn of omponents of a ve tor v in thatbasis, v = eX. Let e′ be another basis related to the basis e through e′ = eCand let X ′ be the olumn of omponents of v in that basis, v = e′X ′. Thematrix C allows the omputation of X ′ from X:v = eX = e′X ′ = eCX ′

X = CX ′Being C the matrix of the new basis in terms of the old one, we ould thinkthat we should have X ′ = CX. To emphasize that it is not so, we say thatthe omponents of the ve tors are ontravariant in a hange of basis.Example:In the notation of the pre eding example if v = e

4−1

0

, its omponentsin the basis e′ areX ′ = C−1X =

1

4

1 −1 1−1 2 3

1 2 1

4−1

0

=1

4

5−6

2

Remember: (Old omponents olumn)=C(New omponents olumn)1.1.3 Linear maps

T A map between two ve tor spa es E,F over the same eld K

f : E −→ F

e 7−→ f(e)is linear if it satisesf(e + e′) = f(e) + f(e′)

f(λe) = λf(e)

14 CHAPTER 1. LINEAR ALGEBRALet e = (e1, . . . , en) be a basis of E and u = (u1, . . . ,um) a basis of F .Linear maps are then easy to handle be ause applying linearityf(v) = f(eX) = f(x1e1 + . . .+ xnen) = x1f(e1) + · · ·+ x1f(e1)and f is determined by the n ve tors f(ei) ∈ F . Expressing ea h of them inthe basis u:f(ei) = a1

iu1 + · · ·+ ami um = (u1, . . . ,um)

a1i

. . .ami

= uAiand olle ting those results,f(e) := (f(e1), . . . , f(en)) = (u1, . . . ,um)(

a1

1

. . .am1

, . . . ,

a1n

. . .amn

) = uAorf(e) = uAwhere A, the juxtaposition of the olumns Ai of omponents of the f(ei), is alled the matrix of f in the given bases. With its aid we have

f(v) = (u1, . . . ,um)(x1

a1

1

. . .am1

+ · · ·+ xn

a1n

. . .amn

) =

= (u1, . . . ,um)

a1

1 . . . . . . a1n

. . . . . .am1 . . . . . . amn

x1

. . .xn

= uAX

f(v) = uAX

Problem 1:The map f : R2 → R3 given by f(x, y) = (2x + 3y, x − y, x + y) is linear.Find its matrixa) In the anoni al bases of R2 and R3; use it to ompute f(2,−3).b) In the basis e = ((1, 1), (1− 1)) of R2 and the anoni al basis of R3.

1.1. BASICS 15Solution:a) We have f((1, 0)) = (2, 1, 1) and f((0, 1)) = (3,−1, 1) and its matrixin the given bases isA =

2 31 −11 1

Thenf((2, 3)) = u

2 31 −11 1

(

2−3

)= u

−5

5−1

b) Now f((1, 1) = (5, 0, 2) and f((1,−1)) = (−1, 2, 0) and the matrix inthe given bases isA′ =

5 −10 22 0

1.1.4 Coordinates of a basisT Let us fo us the attention on the omponents of a ve tor v in a basise1, e2 as fun tions of the ve tor:

1p

2p

e2 e

2e

1 e1

. .

v

v

Proje ting v onto the line of e1 following the dire tion of e2 we obtainp1(v), a ertain multiple x1 of e1. This x1 is the rst omponent of v in thebasis, and analogously we obtain the se ond omponent.

16 CHAPTER 1. LINEAR ALGEBRALet ϕ1 be the fun tion that sifts the rst omponent of a ve tor; it is alinear form, i.e. a linear mapϕ1 : E −→ K

e 7−→ ϕ(e) alled the rst oordinate. Analogously let ϕ2 be the linear form that siftsthe se ond omponent of a ve tor.We all the pair of linear forms (ϕ1, ϕ2) the oordinates of the basis e.Observe thatϕi(ej) =

1 if i = j0 if i 6= j

=: δij , i, j = 1, 2In general if e = (e1, . . . , en) is a basis, the oordinates of this basis arethe linear forms:

ϕi : E −→ K

e 7−→ ϕi(e)dened byϕi(ej) =

1 if i = j0 if i 6= j

=: δij, i, j = 1, . . . , nNote: From now on we shall denote the a tion of forms on ve tors by meansof straight bra kets.The a tion of the i-th oordinate on the ve tor v = eX is:

ϕi[v] = ϕi[x1e1 + · · ·+ xnen] = x1ϕi[e1] + · · ·+ xnϕi[en] = xiϕi[ei] = xi,and again the linear form ϕi sifts the i-th omponent of v in the basis e.The linear mapφ : E → Kn

v 7→ (ϕ1[v], . . . , ϕn[v]) = (x1, . . . , xn)

1.1. BASICS 17is an isomorphism (it su es to observe that its kernel is 0, for E andKn have the same dimension) alled the oordinate system of the basis e.Re ipro ally, an isomorphism

ψ : E → Knis the oordinate system of the basise1 = ψ−1(1, 0, . . . , 0), . . . , en = ψ−1(0, . . . , 1)

1.1.5 InterludeT The oordinates of a basis are not the only linear forms. To give animportant new example of linear forms, onsider R3 and the standard s alarprodu t. Fix u ∈ R3; the map

ϕu : R3 → R

v 7→ ϕu[v] = v · uis a linear form, for the properties of the s alar produ t tell us thatϕu[v + v′] = (v + v′) · u = v · u + v′ · u = ϕu[v] + ϕu[v′]

ϕu[λv] = (λv) · u = λ(v · u) = λϕu[v]This operation has a geometri al meaning; let P [v] be the signed length ofthe orthogonal proje tion of v on u:[ v ]>0

[ v ]< 0

v

u

uv

P P

θ

θ

Thenϕu[v] = v · u =| u || v | cos θ =| u | P [v]

18 CHAPTER 1. LINEAR ALGEBRAWhenever u has length 1 we see that ϕu[v] = P [v], the signed length ofthe proje tion. If u is not unitary then ϕu[v] is the signed length of theorthogonal proje tion multiplied by a weight | u |.Every linear form in a spa e with a s alar produ t is a weighted orthogonalproje tion (see p.40)1.2 Dual spa e1.2.1 Dual spa e

T We now formalize the pre eding onsiderations.The set of all linear forms on a ve tor spa e E is the dual spa e of E:E∗ = α : E → K, α linearAs we an add linear forms and multiply them by s alars, E∗ is a K-ve torspa e. Let e = (e1, . . . , en) be a basis of E; the oordinates of this basis

φ = (ϕ1, . . . , ϕn) are a basis of E∗, alled the dual basis of the basis e.If e = (e1, . . . , en) happens to be the anoni al basis of Rn, its dual basiswill usually be notated π = (π1, . . . , πn) or (dx1, . . . , dxn).Problem 2: Dual basis.Show that the oordinates φ = (ϕ1, . . . , ϕn) of a basis e = (e1, . . . , en) are abasis of E∗.Solution:We must show that they are linearly independent and that they generate thespa e.

• They are linearly independent, be ause if a linear ombination vanishesλ1ϕ1 + · · ·+ λnϕn = 0,applying it to the basis ve tors we obtain:

0 = (λ1ϕ1 + · · ·+ λnϕn)[ei] = λi, i = 1, . . . , n

1.2. DUAL SPACE 19• They generate the spa e, be ause if α ∈ E∗ we have the equality

α = α[e1]ϕ1 + . . .+ α[en]ϕnthat is seen to be true applying both members to ea h ve tor of thebasis e:(α[e1]ϕ1 + · · ·+ α[en]ϕn)[ei] = α[ei]ϕi[ei] = α[ei]Noti e that the outlined formula tells us that the i−th omponent of theform α is α[ei].

T E and E∗ have the same dimension, sin e the basis φ has the samenumber of elements as the basis e; they are isomorphi vs. The linear mapf : E → E∗

ei 7→ ϕiis su h an isomorphism (its matrix in the bases e, φ is I) but it depends onthe sele ted basis; one says that it is not anoni al.Problem 3:In R2 the ve tors u = (3, 2),v = (1, 0) are a basis; ompute its dual basis interms of π = (π1, π2), the dual of the anoni al basis.Solution:We look for two linear forms ϕ, ψ that we express in the basis π:

ϕ = aπ1 + bπ2

ψ = cπ1 + dπ2satisfying the onditionsϕ[u] = 1 : 3a+ 2b = 1ϕ[v] = 0 : a = 0ψ[u] = 0 : 3c+ 2d = 0ψ[v] = 1 : c = 1

20 CHAPTER 1. LINEAR ALGEBRAwe obtain a = 0, b = 1/2, c = 1, d = −3/2 andϕ =

1

2π2, ψ = π1 −

3

2π2Done. But what if there are more dimensions? Is there a simple pattern?It is lear that we are making a hange of basis from e, the 'old basis' (the anoni al one in this ase) to e′ the 'new basis' (the basis (u,v)) and thatthe hange of basis matrix is

C =

(3 12 0

)What does the dual basis do? Well, ea h linear form sifts its oordinate fromthe ve tor. But we know the new oordinates:X ′ = C−1X = −1

2

(0 −1−2 3

)orx′ =

1

2y

y′ = x− 3

2ythat is x′ is half what the se ond oordinate sifts, and y′ is what the rst oordinate sifts less ... Then

ϕ =1

2π2

ψ = π1 −3

2π2

T Let Y = (α[e1], . . .,α[en]) be the row of omponents of the linear formα in the basis φ = (ϕ1, . . . , ϕn) dual to the basis e; observe that Y is also thematrix of the linear map α : E → K in the basis e of E and 1 of K. That'swhy we use the rows of omponents for linear forms. Then, orresponding tothe expression v = eX for a ve tor, we now have for a linear form:

α = (y1, . . . , yn)

ϕ1...ϕn

= Y φT

1.2. DUAL SPACE 21where φT is the transposed of φ.A basis e = (e1, . . . , en) of E and a basis φ = (ϕ1, . . . , ϕn) of E∗ are dualiφT (e) =

ϕ1

. . .ϕn

(e1, . . . , en) =

ϕ1[e1] . . . ϕ1[en]. . . . . . . . .

ϕn[e1] . . . ϕn[en]

=

1 . . . 0. . . . . . . . .0 . . . 1

that is i they satisfy the duality onditionφT (e) = INoti e that if we forget about nearly everything and view that as a prod-u t, the dual basis φ is a kind of 'left inverse' of the basis e.

Problem 4:a) Let e = (e1, e2, e3) be a basis of a vs E and φ = (ϕ1, ϕ2, ϕ3) its dualbasis. Let v = e(2, 0,−1)T and α = (−3, 1, 5)φT ; ompute α[v].b) Let e1 = (1,−1, 1), e2 = (−1, 0, 1), e3 = (2, 1, 0) be a basis of R3; ompute the dual basis φ = (ϕ1, ϕ2, ϕ3) in terms of the basis π =(π1, π2, π3) dual of the anoni al basis. ) Let F = 〈e1, e2〉 ⊂ E; nd the dual basis.d) Let β ∈ E∗ be su h that β[e1] = 2, β[e2] = −1, β[e3] = 1; ompute the omponents of β in the basis π, dual of the anoni al basis.Solution:a) By the duality ondition

α[v] = (−3, 1, 5)φT (e)

20−1

= (−3, 1, 5)I

20−1

= −11

22 CHAPTER 1. LINEAR ALGEBRAor, in generalα[v] = (Y φT )[eX] = Y (φT [e])X = Y IX = Y XThe on lusion is that using olumns for the omponents of ve tors androws for the omponents of a form then, in the dual basis, to apply aform to a ve tor we simply apply the usual 'row by olumn' pro edure.b) We have seen how to do that in problem 3, but let's work it anew. Let

ϕi = a1iπ1 + a2

iπ2 + a3iπ3, i = 1, 2, 3; then

ϕi[e1] = (a1iπ1 + a2

iπ2 + a3iπ3)[i− j + k] = (a1

i , a2i , a

3i )

1−1

1

and similar expressions for the other two ve tors. Now φ will be thedual basis to e i

a1

1 a21 a3

1

a12 a2

2 a32

a13 a2

3 a33

1 −1 2−1 0 1

1 1 0

=

1 0 00 1 00 0 1

a system that we an solve inverting the matrix of the omponents ofthe ve tors e:

a1

1 a21 a3

1

a12 a2

2 a32

a13 a2

3 a33

=

1 −1 2−1 0 1

1 1 0

−1

=1

4

1 −2 1−1 2 3

1 2 1

and we have obtained for the dual basisϕ1 =

1

4(π1 − 2π2 + π3)

ϕ2 =1

4(−π1 + 2π2 + 3π3)

ϕ3 =1

4(π1 + 2π2 + π3) ) We look for two linear forms ψi satisfying ψi(ej) = δji . But the forms

ϕ1, ϕ2 do that; it su es to restri t them to F .

1.2. DUAL SPACE 23d) We know that β = 2ϕ1−ϕ2 +ϕ3 and substituting the pre eding valueswe obtainβ =

1

4[(2π1 − 4π2 + 2π3)− (−π1 + 2π2 + 3π3) + (π1 + 2π2 + π3) =

= π1 − π2

Problem 5: Change of basis in the dual.Let e and e′ be two bases of E , e′ = eC. Let φ = (ϕ1, . . . , ϕn) and φ′ =(ϕ′1, . . . , ϕ

′n) be the respe tive dual bases; let D be the hange of basis matrixin the dual, φ′ = φD.a) Compute D in terms of C.b) Compute the hange of omponents of a linear form in terms of C. ) Redo problem 4 d).Solution:a) From the duality ondition we have:

I = φ′T (e′) = DTφT (eC) = DTφT (e)C = DT IC = DTC

D = (C−1)Tb) Let α = Y φT = Y ′φ′T where Y, Y ′ are the rows of omponents of thelinear form α in the bases φ, φ′ respe tively. ThenY φT = Y ′φ′T = Y ′DTφT = Y ′C−1φTor

Y ′ = Y Cand we see that the omponents of a form ovariate under a hange ofbasis.

24 CHAPTER 1. LINEAR ALGEBRA ) In problem 4 we had C =

1 −1 2−1 0 1

1 1 0

and we know that Y ′ =

(2,−1, 1). We have:Y = Y ′C−1 = (2,−1, 1)

1

4

1 −2 1−1 2 3

1 2 1

= (1,−1, 0)

Problem 6:Let E be a vs and E∗ its dual spa e; if F ⊂ E∗ is a subspa e, its in identspa e isN(F ) = v ∈ E : ω[v] = 0, ∀ω ∈ F ⊂ E.a) Let e = (e1, e2, e3) be a basis of E and φ = (ϕ1, ϕ2, ϕ3) its dual basis.If ω1 = 2ϕ1 + ϕ2 − ϕ3, ω2 = ϕ1 − ϕ2 − ϕ3, ompute (a basis of)

N(ω1), N(ω2), N(〈ω1, ω2〉). What is N(E∗)?b) Let e = (e1, e2, e3, e4) be a basis of E and φ = (ϕ1, ϕ2, ϕ3, ϕ4) its dualbasis. If ω1 = 2ϕ1 + ϕ2 − ϕ3 + ϕ4,ω2 = ϕ1 − ϕ2 − ϕ3 + ϕ4 omputeN(〈ω1, ω2〉).Solution:a) i) If v = x1e1 + x2e2 + x3e3 then

ω1[v] = (2ϕ1 + ϕ2 − ϕ3)[x1e1 + x2e2 + x3e3] =

= 2x1 + x2 − x3andN(ω1) = (x1, x2, x3) : 2x1 + x2 − x3 = 0,a plane, and a basis ofN(ω1) is easily seen to be u1 = (1,−1, 1),u2 =

(0, 1, 1).

1.2. DUAL SPACE 25ii) Similarly N(ω2) is the planex1 − x2 − x3 = 0and a basis of N(ω2) is u′1 = (1, 1, 0),u′2 = (0, 1,−1).

ω

N (ω )

iii) Now, as the linear forms ω1, ω2 are li we have N(〈ω1, ω2〉) =N(ω1) ∩N(ω2) = 〈(0,−1, 1)〉, a 1-dimensional subspa e.

ω1

ω >2

< ,

ω1

ω2, )N (iv) N(E∗) = 0 for if v 6= 0 it has a nonvanishing omponent, say x1,and then ϕ1[v] = x1 6= 0.b) Similarly writing v = x1e1+x2e2+x3e3+x4e4, N(ω1) is the hiperplane2x1 + x2 − x3 + x4 = 0and a basis ofN(ω1) is u1 = (1,−1, 1, 0),u2 = (0, 1, 1, 0),u3 = (1, 1, 1,−2).Also N(ω2) is the hiperplanex1 − x2 − x3 + x4 = 0

26 CHAPTER 1. LINEAR ALGEBRAand a basis ofN(ω2) is u′1 = (1, 1, 0, 0),u′2 = (0, 1,−1, 0),u′3 = (1, 1, 1, 1).The ve tors in N(〈ω1, ω2〉) = N(ω1) ∩ N(ω2) are to be found solvingthe system of four equations in the six unknowns λ, µ, ν, λ′, µ′, ν ′λu1 + µu2 + νu3 = λ′u′1 + µ′u′2 + ν ′u′3We nd

λ′ = 2λ −2µµ′ = −2λ +µν ′ = −2

3λ +4

3µ

ν = 13λ −2

3µFor λ = 1, µ = 0 we obtain the ve tor f1 = (4,−2, 4,−2) and for

λ = 0, µ = −1 we obtain f2 = (−23, 1

3, 1

3,−4

3). Then

N(〈ω1, ω2〉) = 〈(2,−1, 2− 1), (−2, 1, 1,−4)〉

Problem 7: Bidual.Show that E is anoni ally (:=independent of any basis) isomorphi toE∗∗ := (E∗)∗ ( alled the bidual of E).Solution:Asso iate to ea h ve tor eǫE a linear form e on E∗:

e : E∗ → K

α 7→ e[α] := α[e]Clearly e is a linear form on E∗ and so it is a ve tor in E∗∗. Now it is easyto see thatE → E∗∗

e → eis an isomorphism: E and E∗∗ have the same dimension and if e = 0, thenα[e]=0 for every α, that informs us that e = 0. This isomorphism isbasis-independent; we say that it is a anoni al isomorphism.

1.2. DUAL SPACE 271.2.2 Dual mapT Let f : E → F be a linear map. The dual map (also alled the transposedmap) is the linear map f ∗

f : E → FE∗ ← F ∗ : f ∗

f ∗(α) ← αdened by f ∗(α)[v] = α[f(v)], v ∈ E. A diagram may be helpful:E

f−→ Ff ∗(α) ց ւ α

K

Problem 8:Let e and u be basis of E and F respe tively and A the matrix of f in thosebases. Let φ = (ϕ1, . . . , ϕn) and Ψ = (ψ1, . . . , ψn) be the orresponding dualbases. Find the matrix B of f ∗ in the bases Ψ, φ.Solution:The i olumn of B onsists of the omponents of f ∗(ψi) in the basis φ; as itis the dual basis of e, the omponent j isf ∗(ψi)[ej ] = ψi[f(ej)] = aijand we see that the i-th olumn onsists of the elements in the i-th row of

A. That isB = ATA more formal proof:

B is the matrix of f ∗: f ∗(Ψ) = φBφ is the dual basis of e: φT (e) = IΨ is the dual basis of u: ΨT (u) = Iwe obtain : f ∗(ΨT )(e) = BTφT (e) = BT I = BTby the denition of f ∗ : f ∗(ΨT )(e) = ΨT (f(e)) = ΨT (uA) = IA = Aand, as before, B = AT .

28 CHAPTER 1. LINEAR ALGEBRA1.3 First anoni al isomorphism1.3.1 Metri sT A s alar produ t (or a metri ) in an R-vs E is an appli ation

g : E × E → R

(e,v) 7→ g(e,v)satisfying:a) It is linear in the rst omponent:g(e + e′,v) = g(e,v) + g(e′,v)

g(ae,v) = ag(e,v)b) It is symmetri g(e,v) = g(v, e) ) It is non degenerate

g(e,v) = 0, ∀v⇒ e = 0Observe that a) and b) implyg(e,v + v′) = g(e,v) + g(e,v′)

g(e, av) = ag(e,v)and we say that g is bi linear symmetri (there are also multi linear appli a-tions).If we write, as it is usual to do, g(u,v) = u · v, the bilinearity expressesthat the produ t is distributive and the simmetry tells us that the produ tis ommutative.Two ve tors u and v are alled orthogonal whenever u ·v = 0. Then a metri is nondegenerate i 0 is the only ve tor orthogonal to any ve tor in the spa e.The ve tors su h that u · u = 0 are alled isotropi .If instead of ) the produ t satises ') Denite positivenessg(e, e) > 0, ∀e 6=0,

1.3. FIRST CANONICAL ISOMORPHISM 29we shall say that we have an eu lidian produ t (or an eu lidian metri ).A ve tor spa e with an eu lidian metri is an eu lidian spa e. The re-stri tion of an eu lidian metri to a subspa e remains an eu lidian metri . Aspa e with a non eu lidian metri is an orthogonal spa e. The restri tion ofa non eu lidian metri to a subspa e may well be degenerate, and then theprodu t does not satisfy ) (see p.41). If the restri tion is nondegenerate thesubspa e is alled non singular.We shall use mainly the notations u · v or g(u,v) for the s alar produ t,but other notations are in use as well:u · v = g(u,v) = (u,v) = (u|v) = 〈u,v〉 = 〈u|v〉.

Coordinate expression of a metri T If e = (e1, . . . , en) is a basis, the symmetri matrix G = eT · e is alledthe matrix of the produ t in the base e, or the matrix of the metri , or themetri matrix to be brief ; Physi s has promoted the symbol G:G = eT · e =

e1

. . .en

· (e1, . . . , en) =

e1 · e1 e1 · e2 . . . e1 · en. . . . . . . . . . . .en · e1 en · e2 . . . en · en

Let X, Y be the olumns of omponents of the ve tors u and v in the basise, that is u = eX,v = eY ; then

u · v = eX · eY = XT eT · eY = XT (eT · e)Y = XTGY

u · v = XTGYThis formula allows the omputation of the produ t of two ve tors on e weknow their omponents and the metri matrix G.Whenever the produ ts of the basis ve tors are ei · ej = ±δji we say thatthe basis is orthonormal (on basis). If we have a eu lidian produ t and thebasis is on, the metri matrix is G = I and the pre eding formula is 'rowtimes olumn', the usual way of omputing a s alar produ t:(x1, . . . , xn)I

y1

. . .yn

= x1y1 + . . .+ xnyn

30 CHAPTER 1. LINEAR ALGEBRAExamples:Let u = (u1, . . . , un) and v = (v1, . . . , vn) be two ve tors in Rn; the followingformulae dene bilinear maps:a) u · v = u1v1 + u2v2 + . . .+ unvnb) u · v = u1v1 + 2u2v2 + . . .+ nunvn ) u ·v = µ1u1v1 +µ2u2v2 + . . .+µnunvn (the µi are real numbers, µi 6= 0)d) u · v = u1v1 + u2v2 + u3v3 − u4v4e) u · v = u1v1 − u2v2 in R2f) u · v = u1v1 − u2v2 in R3a) and b) are eu lidian metri s; the rst is refered to as the standard metri in Rn. ) is eu lidian i µi > 0, i = 1, . . . , n. d) is the noneu lidian Lorentzmetri in R4. In example e) (0, 1) · (0, 1) = −1 (there are ve tors withnegative self produ t) and (1, 1) · (1, 1) = 0 (there are isotropi ve tors). Inf) the ve tor (0, 0, 1) is orthogonal to every ve tor and, being degenerate,a ording to the denition it is not a metri .Example:Let E = C([a, b]), the R-vs of ontinuous real fun tions on the interval [a, b].The appli ation

E × E → R

f , g 7→ (f, g)where (f, g) =∫ baf(x)g(x)dx, is bilinear and symmetri due to the propertiesof the integral. Have a look to see if it is nondegenerate and, the ase being,de ide if it is eu lidian.The novelty here is that the spa e is not nite dimensional. Neverthelessit is a generalization to the ontinuous ase of a) in the pre eding example.

1.3. FIRST CANONICAL ISOMORPHISM 31Problem 9: In the pre eding examples write the metri matrix in the anon-i al basis.Solution:a) G =

11

. . .1

= I

b) G =

12

. . .n

) G =

µ1

µ2

. . .µn

d) G =

11

1−1

e) G =

(1−1

)f) Despite not being a metri we an write the matrix of the produ t:G =

1−1

0

Problem 10:Find the new metri matrix when the basis undergoes a hange u = eC.Apply it to E = R3 with the standard s alar produ t and the new basisu1 = (2, 0, 1),u2 = (−1, 1, 0),u3 = (0, 0, 1).

32 CHAPTER 1. LINEAR ALGEBRASolution:In the example we an ompute the new metri matrix dire tly:G′ =

u1 · u1 u1 · u2 u1 · u3

u2 · u1 u2 · u2 u2 · u3

u3 · u1 u3 · u2 u3 · u3

=

5 −2 1−2 2 0

1 0 1

But we want to treat as well the general ase:G′ = uT · u = CT eT · eC = CTGC

G′ = CTGCApplying this to the example we have:C =

2 −1 00 1 01 0 1

Then, being I the metri matrix of the standard produ t in the anoni albasis, we have:G′ = CT IC = CTC =

2 0 1−1 1 0

0 0 1

2 −1 00 1 01 0 1

=

5 −2 1−2 2 0

1 0 1

Problem 11:Let u be the basis in the pre eding problem and let v1 = u1 − u2,v2 =2u2 + u3. Compute v1·v2.Solution:a) From the point of view of the anoni al basis the omponents of v1 and

v2 areC

1−1

0

=

3−1

1

, C

021

=

−2

21

.

1.3. FIRST CANONICAL ISOMORPHISM 33Then:v1 · v2 = (3,−1, 1)I

−2

21

= −7b) From the point of view of the basis u, using the new metri matrix G′we have:v1 · v2 = XTG′Y = (1,−1, 0)

5 −2 1−2 2 0

1 0 1

021

=

= (7,−4, 1)

021

= −7 ) We an view the last omputation thus:i) Convert the ve tor v1 into the linear form αv1 that, in the basisφ dual to u, has omponents XTG′:

(1,−1, 0)

5 −2 1−2 2 0

1 0 1

= (7,−4, 1).ii) Apply αv1 to v2 to obtain v1 · v2 = αv1[v2]:v1 · v2 = (7,−4, 1)

021

= −7Analogously we an onvert v2 into a linear form αv2 and applyit to v1; we obtain (−3, 4, 1) andv1 · v2 = (−3, 4, 1)

1−1

0

= −7For more details see the rst anoni al isomorphism in p.40.

34 CHAPTER 1. LINEAR ALGEBRAProblem 12:a) Let E be a real vs and g(x,y) = x ·y a symmetri bilinear form; showthat g is nondegenerate i detGe = det(eT · e) 6= 0 for any basis e.b) Let E be an orthogonal spa e and F ⊂ E a nonsingular subspa e.Dene the orthogonal of F as the subspa e F⊥ = e ∈ E : e · f =0, ∀f ∈ F. Prove:i) dimF⊥ = dimE − dimF .ii) (F⊥)⊥ = F .iii) F⊥ is nonsingular.iv) E = F ⊕ F⊥.Solution:a) With the usual notations we have: g is nondegenerate i x · y =XTGeY = 0, ∀Y ⇒ X = 0. But XTGeY = 0, ∀Y is the same asXTGe = 0; this is an n× n homogeneous linear system and we have:

g is nondegenerate ⇔ X = 0 is the only solution ⇔ detGe 6= 0b) i) Let u = (u1, . . . ,uk) be a basis of F and omplete it to a basis ofE:

e = (u1, . . . ,uk, ek+1, . . . , en)Let x = x1u1 + · · ·+ xkuk + xk+1ek+1 + · · ·+ xnen; then x ∈ F⊥i x · ui = 0, i = 1, . . . , k:x1u1 · u1 + · · ·+ xkuk · u1 + xk+1ek+1 · u1 + · · ·+ xnen · u1 = 0

. . . . . . . . .x1u1 · uk + · · ·+ xkuk · uk + xk+1ek+1 · uk + · · ·+ xnen · uk = 0

a system of k equations in the unknowns (x1, . . . , xk, xk+1, . . . , xn);write it asx1u1 · u1 + · · ·+ xkuk · u1 = −xk+1ek+1 · u1 − · · · − xnen · u1

. . . . . . . . .x1u1 · uk + · · ·+ xkuk · uk = −xk+1ek+1 · uk − · · · − xnen · uk

,

1.3. FIRST CANONICAL ISOMORPHISM 35take (xk+1, . . . , xn) as parameters and onsider the system of kequations in the unknowns (x1, . . . , xk) whose matrix is Gu. As Fis nonsingular we know from a) thatdetGu 6= 0; then the solutionsof the system depend then on n− k parameters, that isdimF⊥ = dimE − dimFii) Obviously F ⊂ (F⊥)⊥ and moreover dim(F⊥)⊥ = n − dimF⊥ =

n− (n− dimF ) = dimF and this proves that F = (F⊥)⊥.iii) A subspa e H is nonsingular i H ∩H⊥ = 0 and as F is non-singular we have:F⊥ ∩ (F⊥)⊥ = F⊥ ∩ F = 0and F⊥ is nonsingular.iv) The sum F + F⊥ ⊂ E is dire t be ause F⊥ ∩ F = 0 and

dim(F ⊕F⊥) = dimF +dimF⊥ = k+n−k = n. Then F ⊕F⊥ =E.

Problem 13: Existen e of orthonormal bases.Show that in an orthogonal spa e E:a) There are bases u = (u1, . . . ,un) su h that ui · uj = 0 whenever i 6= j;they are alled orthogonal bases.b) The ve tors ui in an og basis satisfy ui · ui 6= 0 (Hint: indu tion ondimE). ) There are bases u = (u1, . . . ,un) su h that ui · uj = ±δij ; they are alled orthonormal bases. What is the form of the metri matrix in anon basis? Write the expression in omponents of the produ t of twove tors.d) What are the hange of basis matri es between two on bases?Solution:

36 CHAPTER 1. LINEAR ALGEBRAa) Indu tion on dimE:i) If dimE = 1, any ve tor u 6= 0 is an og basis (and satisesu · u 6= 0 for in other ase g ≡ 0).ii) Assume the result true for dimE < n and onsider the asedimE = n.There is a ve tor u su h that u · u 6= 0 be ause if not

0 = (u + v) · (u + v) = 2u · vand again g ≡ 0. Then F = 〈u〉 is nonsingular and so is F⊥by the pre eding problem; we an apply the indu tion hypothesisto F⊥ and on lude that F⊥ has an og basis (u2, . . . ,un). Then(u,u2, . . . ,un) is an og basis of E.b) If ui · ui = 0 then ui · (λ1u1 + . . .+ λiui + . . .+ λnun) = λiui · ui = 0,and ui ∈ E⊥; but then g would be degenerated. ) If v = (v1, . . . ,vn) is an og basis, the ve tors ui = vi√

|vi·vi|satisfy

ui · ui =vi · vi

(√|vi · vi|)2

=vi · vi

|vi · vi|=

1 if vi · vi > 0−1 if vi · vi < 0and are an on basis. In su h a basis (with the wright order) the metri matrix is

G =

1(r. . .

1−1

(s. . .−1

=: Irs

One an prove that r the number of 1's and s the number of -1's areindependent of the on basis sele ted to ompute G (Sylvester's inertialaw). The number s is alled the index of the metri and t = r − s isthe signature of the metri .

1.3. FIRST CANONICAL ISOMORPHISM 37If the spa e is eu lidian the basis u is on i G = uT ·u = I. If the spa eis not eu lidian the basis u is on i G = uT ·u = Irs . Now onsider twove tors e = uX,v = uY ; then the produ t in the eu lidian ase ise · v = x1y1 + · · ·+ xnynand in the noneu lidian ase

e · v = x1y1 + · · ·+ xryr − xr+1yr+1 − · · · − xnynd) Let u, e be two on bases with u = eC; whenever the produ t is eu lidianI = uT · u = CT eT · eC = CTCand the hange of basis matrix satises CTC = I; those are alledorthogonal matri es. Let Ci be the i-th olumn of C and look at it asthe olumn of omponents of the ve tor ui, then the pre edent onditionreads:

CTC =

c11 . . . cn1. . . . . . . . .c1n . . . cnn

c11 . . . c1n. . . . . . . . .cn1 . . . cnn

=

=

C1 · C1 . . . C1 · Cn. . . . . . . . .

Cn · C1 . . . Cn · C1

=

=

1 . . .1

,and we an write the equality in the form Ci · Cj = δji that shows theorthonormality of the u basis.If the produ t is noneu lidian we have

Irs = uT · u = CT eT · eC = CT IrsCWriting as before Ci for the i-th olumn we have:C = (C1, . . . , Cn), C

T = ((CT )1, . . . , (CT )n)

38 CHAPTER 1. LINEAR ALGEBRAand thenCT Irs = ((CT )1, . . . , (C

T )r,−(CT )r+1, . . . ,−(CT )n)

CT IrsC =

=

c11 . . . cr1 −cr+1

1 . . . −cn1. . . . . . . . . . . . . . . . . .c1n . . . crn −cr+1

n . . . −cnn

c11 . . . c1r c1r+1 . . . c1n. . . . . . . . . . . . . . . . . .cn1 . . . cnr cnr+1 . . . cnn

= IrsTo have an idea of the form of this ondition we ompute:i) The produ t of the rst row and the rst olumn:(c11)

2 + · · ·+ (cr1)2 − (cr+1

1 )2 − · · · − (cn1 )2 = 1that is: C1 · C1 = 1.ii) The produ t of the rst row and the se ond olumn:c11c

12 + · · ·+ cr1c

r2 − cr+1

1 cr+12 − · · · − cn1cn2 = 0that is: C1 · C2 = 0.iii) The produ t of the row r + 1 and the olumn r + 1:

(c1r+1)2 + · · ·+ (crr+1)

2 − (cr+1r+1)

2 − · · · − (cnr+1)2 = −1that is: Cr+1 · Cr+1 = −1.These onditions show the orthonormality of the basis u.

Problem 14:Let E be a vs, E∗ its dual, e = (e1, . . . , en) a basis of E and φ = (α1, . . . , αn)its dual basis. In W = E × E∗ dene the produ t of the ve tors ω =(v, α), ω′ = (v′, α′) by

〈ω, ω′〉 =1

2(α[v′] + α′[v])a) Find a basis of W and ompute the metri matrix in this basis.b) Is this produ t nondegenerate? Is this produ t eu lidian?. ) Find an on basis of W .

1.3. FIRST CANONICAL ISOMORPHISM 39Solution:a) The 2n ve tors ofW , ui = (ei, 0), χi = (0, αi) are a basisΨ = (u1, . . . , un, χ1, . . . , χn),for ifλ1u1 + · · ·+ λnun + µ1χ1 + · · ·+ µnχn = 0then

λ1e1 + · · ·+ λnen = 0

µ1α1 + · · ·+ µnαn = 0and, being the e1, . . . , en and α1, . . . , αn bases, we obtain λi = µi = 0.The produ ts are:〈ui, uj〉 = 〈(ei, 0), (ej, 0)〉 = 0

〈χi, χj〉 = 〈(0, αi), (0, αj)〉 = 0

〈ui, χj〉 = 〈(ei, 0), (0, αj)〉 =

0 if i 6= j12

if i = jThe metri matrix in this basis is:G =

12

. . . 0

0... . . . ...0 . . . 1

212

. . . 0... . . . ... 00 . . . 1

2

=1

2

(0 II 0

)

b) The produ t is non eu lidian, for the ve tors ui, χj are isotropi ; it isnondegenerate be ause if we had a ve tor, say Ψ(x1, . . . , xn, λ1, . . . , λn)T ,orthogonal to ea h ve tor in W , we would have

(x1, . . . , xn, λ1, . . . , λn)G = (0, . . . , 0, 0, . . . , 0)and then(0, 0) = (x, λ)G = (x, λ)

1

2

(0 II 0

)=

1

2(λ, x)That is x = λ = 0 and the metri is nondegenerate.Another way to see the nondegenera y is to use the result in problem

40 CHAPTER 1. LINEAR ALGEBRA12; it su es then to show that the determinant of the metri matrixdoesn't vanish. Using Lapla e's rule we obtain| detG |= (1/2)2n 6= 0 ) We do it small rst; if dimE = 1 thenu = (e, α), χ = (−e, α)is a basis of W for if λu + µχ = 0 then (λ − µ)e = 0, (λ + µ)α = 0implies λ = µ = 0.The metri matrix is

G =

(1 00 −1

)so the basis is on.In the general ase (e1, α1), . . . , (en, αn), (−e1, α1), . . . , (−en, αn) is abasis and the metri matrix isG =

1 . . . 01−1

0. . .−1

that shows an on basis.

1.3.2 First anoni al isomorphismT In an orthogonal spa e E we an ouple a ve tor e with the metri g,obtaining a 1-form:

(eyg)[u] := g(u, e)(read e in g) that we an write also as eyg = g( . , e). Now we may look atve tors as linear forms and this hange in the viewpoint is the isomorphism

1.3. FIRST CANONICAL ISOMORPHISM 41at (it 'attens' ve tors to forms):E

→ E∗

e 7→ ewhere e = g( . , e)It is alled the rst anoni al isomorphism be ause it is basis independent.It allows the identi ation of ve tors and forms (see next problem).Can we visualize e? The linear form e a ts on u doing the weightedorthogonal proje tion of u on e. The kernel of e onsists of all the ve torsorthogonal to e. The level sets of e are hiperplanes parallel to the kernel.So to say the familiar ve tor e shades away and what remains is the way ita ts on ea h ve tor u, that is as the 1-form eyg. This form forgets aboutanything not in its dire tion. By the way, in many situations you have twoworlds: the tangent one and the outside one. Here we are de idedly lookingat what happens in the 'tangent' dire tion:

bvbv[v’]v

v’

The inverse isomorphism is alled sharp (it 'sharpens' forms to ve tors)♯ : E∗ −→ Eand assigns to ea h ω ∈ E∗ a ve tor perpendi ular to the kernel of ω.

Problem 15:a) Let e = (e1, . . . , en) be a basis of the orthogonal spa e E, and letw = (ω1, . . . , ωn) be its dual basis. Find the matri es of and ♯ inthose bases.

42 CHAPTER 1. LINEAR ALGEBRAb) Show: is an isomorphism⇔ g is nondegeneratedGive an example of a metri that when restri ted to a subspa e isdegenerated. ) Prove that whenever the metri is eu lidian

e is an orthonormal basis⇔ (e) = (e1, ..., en) is the dual basis of e.d) Is there a similar result for noneu lidian metri s?Solution:a) The i-th olumn of the matrix of has as entries the omponents of eiin the basis w:ei[ej ] = ej · ei = ei · ejThe matrix of is the metri matrix G and then the matrix of ♯ is G−1.b) is iso ⇔ det 6= 0⇔ detG 6= 0⇔ g nondegenerated.Let e = (e1, e2, e3) be a basis of a vs E of dimension 3 and onsiderthe metri

G = eT · e =

2 0 00 0 10 1 1

It is nondegenerated for detG = −2 6= 0, but its restri tion to thesubspa e F = 〈e1, e2〉 has the metri matrixG1 =

(2 00 0

)and as detG1 = 0 we see that g|V is degenerated. If we try to omputen = ae1 + be2 + ce3 ∈ F⊥ it must satisfy

n · e1 = 0 or a = 0n · e2 = 0 or c = 0and we have n = be2or F⊥ = 〈e2〉, that leaves us in the same subspa e.

1.3. FIRST CANONICAL ISOMORPHISM 43 ) As the olumns of G are the omponents of (ei) in the basis w we havee is an orthonormal basis ⇔ G = eT · e = I ⇔

⇔ [ei] = ωi ⇔ (e) is the dual basis to eIn that ase assigns to ea h ve tor the linear form with the same omponents. That's why one says that the on bases are selfdual.d) The basis e is on i the metri matrix isG = eT · e =

1(r. . .

1−1

(s. . .−1

= Irs

and we see thate is an ON basis ⇔ (e1, ..., er,−er+1, ...,−en) is the dual basis to e

E E *

e

be

e and its level sets

44 CHAPTER 1. LINEAR ALGEBRAProblem 16:Let g be a metri in E; if it is eu lidian show that if e is a basis det(eT ·e) > 0.Is noneu lidian equivalent to det(eT · e) < 0?Solution:If the spa e is eu lidian, in an on basis e the metri matrix is Ge = I anddetGe = 1 > 0. In an arbitrary basis u = eC we have Gu = CTGeC anddetGu = (detC)2 > 0.If the spa e is noneu lidian det(eT · e) an be either positive or negative;a ouple of exemples should su e:

Ge =

(1−1

), Ge =

11−1

−1

Problem 17:Let E be an orthogonal spa e, e = (e1, . . . , en) a basis, w = (ω1, . . . , ωn)its dual basis, and v = (v1, . . . ,vn) = (♯ω1, . . . , ♯ωn), the so alled re ipro albasis.a) Find the hange of basis matrix C from the basis e to the re ipro albasis v.b) Show that if u ∈ E we have u = (u · e1)v1 + · · ·+ (u · en)vn ( ompareto the ase when e is an on basis). ) Find the relation between the metri matrix Ge in the basis e and themetri matrix Gv in the re ipro al basis v.Solution:a) Let f : E → E∗ the linear map su h that f(ei) = ωi; its matrix in thebases e, w is the identity matrix I. Consider the ompositionE

f→ E∗♯→ E

1.3. FIRST CANONICAL ISOMORPHISM 45that satises (♯ f)(ei) = vi; taking the e basis in E and the w basisin E∗ the matrix of the omposition is the sought for hange of basismatrix C. So we haveC = G−1

e · I = G−1eIn the eu lidian ase if e is an on basis then Ge = I and the re ipro albasis is the basis e itself: on bases are selfre ipro al.In the noneu lidian ase if e is an on basis then Ge = Irs = G−1

e andthe re ipro al basis ise1, . . . , er,−er+1, . . . ,−er+sb) We have

u = u[e1]ω1 + · · ·+ u[en]ωn

= (u · e1)ω1 + · · ·+ (u · en)ωnAs ♯ωi = vi applying ♯ to both members we obtainu = (u · e1)v1 + · · ·+ (u · en)vnIn the eu lidian ase re all that if e is an on basis we have the expressionu = (u · e1)e1 + · · ·+ (u · en)enand in the noneu lidian ase if e is an on basis then

u = (u · e1)e1 + · · ·+ (u · er)er − (u · er+1)er+1 − · · · − (u · er+s)er+s ) From a) we know that v = eG−1e and

Gv = vT · v = (G−1e )T eT · eG−1

e = (G−1e )TGeG

−1e = (G−1

e )T = G−1e ,where we have taken into a ount that Ge is symmetri .

46 CHAPTER 1. LINEAR ALGEBRAProblem 18:Let e1 = (2, 0, 1), e2 = (−1, 1, 0), e3 = (0, 0, 1) be a basis of R3, w its dualbasis, and onsider the standard s alar produ t.a) Compute the re ipro al basis.b) Let u = (1, 2, 3) ∈ R3; ompute its omponents in the re ipro al basis. ) Compute the row of omponents of the linear form (e1 − e2) in thebasis w.Solution:a) AsGe =

5 −2 1−2 2 0

1 0 1

, G−1e =

1

2

1 1 −11 2 −1−1 −1 3

using the results in the pre eding problem we havev1 =

1

2(e1 + e2 − e3)

v2 =1

2(e1 + 2e2 − e3)

v3 =1

2(−e1 − e2 + 3e3)b) From the pre eding problem we have

u = (u · e1)v1 + (u · e2)v2 + (u · e3)v3 = 5v1 + v2 + 3v3 ) As Ge is the matrix of in the basis e, w we have(

1, −1, 0)

5 −2 1−2 2 0

1 0 1

=(

7, −4, 1)

1.3. FIRST CANONICAL ISOMORPHISM 47Problem 19: Adjoint map.Let E,F be orthogonal spa es, u ∈ E, v ∈ F and ϕ : E → F a linear map.a) Prove there is a linear mapψ : F → E alled the adjoint map of ϕ, su h that (ϕu) · v = u · (ψv). (We denoteby a dot the produ t in both vs.)b) Show that if we identify F with F ∗ and E with E∗ through the respe -tive isomorphisms, then ϕ∗ and ψ oin ide.Solution:a) Let e = (e1, . . . , en) be a basis of E, f = (f1, . . . , fm) a basis of F and

G = eT · e, Γ = fT · fits respe tive metri matri es. Let ψ be a linear map from F to E andlet A and B be the matri es of ϕ and ψ :

ϕ e = fA, ψ f = eBIf X, Y are the olumns of omponents of u ∈ E and v ∈ F ,u = eX,v = fY,then:

(ϕu) · v = ϕ(eX) · fY = fAX · fY = (AX)TΓY

u · (ψv) = eX · ψ(fY ) = eX · eBY = XTGBYSo ψ is the adjoint of ϕ i for ea h X, Y we have(AX)TΓY = XTGBY ⇔ ATΓ = GB ⇒

B = G−1ATΓNoti e we now know ψ through its matrix. In the eu lidian ase, when-ever e and f are on bases we have G = In,Γ = Im and B = AT .

48 CHAPTER 1. LINEAR ALGEBRAb) We want to see that the following diagram is ommutativeE

ϕ→ F

E∗ϕ∗

← F ∗

↑ ↑ E

ψ← FLet v ∈ F ; for ea h e ∈ E we have(ϕ∗v)[e] = v[ϕe] = v · ϕe

(ψv)[e] = ψv · eand being ψ the adjoint of ϕ, both expressions oin ide and the diagramis ommutative.

Chapter 2Alternating multilinear forms2.1 Multilinear appli ations and forms2.1.1 Basi sT Let E, V be K-vs (remind that K = R or C); an appli ation

S : E× k). . . ×E → Vis alled multilinear (or k-multilinear if needed) if it is linear in ea h entry;that is, if for ea h i = 1, . . . , k we havea)S[e1, . . . , ei + e′i, . . . , en] = S[e1, . . . , ei, . . . , en] + S[e1, . . . , e

′i, . . . , en]b)

S[e1, . . . , λei, . . . , en] = λS[e1, . . . , ei, . . . , en]Whenever S takes numeri al values we all it a multilinear form:S : E× k). . . ×E → KAs we an add two multilinear maps and multiply a multilinear map bya s alar, it is easy to see that they form a vs on K. We shall mainly dealwith multilinear forms also alled k- ovariant tensors, notated T kE; k is theorder of the tensor. For instan e T 1E = E∗, and a metri g on an R-vs is a2- ovariant tensor (symmetri and nondegenerate), g ∈ T 2E.49

50 CHAPTER 2. ALTERNATING MULTILINEAR FORMSThe tensor produ t of two multilinear forms S ∈ T kE, T ∈ T lE isS ⊗ T [v1, . . . ,vk,vk+1, . . .vk+l] := S[v1, . . . ,vk]T [vk+1, . . .vk+l]The produ t is bilinear and asso iative; it is obviously not ommutative ingeneral.

Problem 20: Tensor produ t of two 1-forms.Let E be a K-vs, e = (e1, e2) a basis and ω = (ω1, ω2) its dual basis. Ifηj = (j, j)ωT , vj = (j, j+1)eT for j = 1, 2, 3, ompute η1⊗η2⊗η3[v1,v2,v3].Solution:

η1 ⊗ η2 ⊗ η3[v1,v2,v3] = η1[v1]η2[v2]η3[v3] =

= ((

1 1)( 1

2

)) · ((

2 2)( 2

3

)) · ((

3 3)( 3

4

)) = 3 · 10 · 21 = 630

Problem 21:Let E be a vs, e = (e1, . . . , en) a basis and ω = (ω1, . . . , ωn) its dual basis.Show that the nk multilinear formsω⊗ = ωi1 ⊗ · · · ⊗ ωiki1,...,ik⊂1,...,nare a basis of T kE.Solution:We imitate the one-dimensional ase (see p.18):

• They are linearly independent, be ause if a linear ombination vanishes0 =

∑

1≤i1,...,ik≤nλi1...ikωi1 ⊗ · · · ⊗ ωik ,

2.1. MULTILINEAR APPLICATIONS AND FORMS 51applying it to ea h list of k ve tors of the basis we obtain:0 = (

∑

1≤i1,...,ik≤nλi1...ikωi1 ⊗ · · · ⊗ ωik)[ej1, . . . ejk ] =

=∑

1≤i1,...,ik≤nλi1...ikωi1[ej1 ] · · ·ωik [ejk ] = λj1...jk

• They generate T kE be ause if S ∈ T kE , letting S[ei1 , . . . , eik ] = ai1...ik(the omponents of S in the ω⊗ basis), we have the equalityS =

∑

1≤i1,...,ik≤nai1...ikωi1 ⊗ · · · ⊗ ωik .that is proved applying both terms to ea h list of k ve tors of the basis:

S[ei1 , . . .eik ] = ai1...ikand(

∑

1≤j1,...,jk≤naj1...jkωj1 ⊗ · · · ⊗ ωjk)[ei1, . . . eik ] = ai1...ik

Problem 22:If the omponents of S ∈ T kE, T ∈ T lE in a basis are respe tively ai1...ik , bj1...jl,nd the omponents of S ⊗ T .Solution:Apply S ⊗ T to the list of k + l ve tors [ei1 , . . . , eik , ej1, . . . , ejl]:ci1...ikj1...jl = (S ⊗ T )[ei1, . . . , eik , ej1, . . . , ejl] = S[ei1 , . . . , eik ]T [ej1, . . . , ejl] =

= ai1...ikbj1...jl

52 CHAPTER 2. ALTERNATING MULTILINEAR FORMSProblem 23: Identifying bilinear appli ations.De ide whether T : R2 ×R2 → R is bilinear:a) T ((x, y), (z, t)) = x+ y + z + tb) T ((x, y), (z, t)) = xt− y ) T ((x, y), (z, t)) = xt− yzd) T ((x, y), (z, t)) = xz + 2ytSolution:We may use dire tly the denition of multilinear appli ation; for instan e ina):T [(x, y) + (x′, y′), (z, t)] = T [(x+ x′, y + y′), (z, t)] =

= x+ x′ + y + y′ + z + t

T [(x, y), (z, t)] + T [(x′, y′), (z, t)] = x+ y + z + t+ x′ + y′ + z + tand we see that T is not linear in the rst omponent; we an apply the sameline of proof to the other ases.But we an as well observe that T ∈ T 2R2 must be a linear ombinationof the elements of the basis dx⊗dx, dx⊗dy, dy⊗dx, dy⊗dy where (dx, dy)is the dual basis to the anoni al one in R2. From that point of view it is lear that the only bilinear appli ations are those in ) and d) that we anwrite as dx⊗ dy − dy ⊗ dx and dx⊗ dx+ 2dy ⊗ dy.Problem 24:Let T : Rn ×Rn → R the bilinear form given by

T [(x1, . . . , xn), (y1, . . . , yn)] =n∑

i,j=1

aijxiyj, aij ∈ RExpress it in terms of the dual of the anoni al basis dx1, . . . , dxn. In par-ti ular do it for the standard s alar produ t.

2.1. MULTILINEAR APPLICATIONS AND FORMS 53Solution:T =

n∑

i,j=1

aijdxi ⊗ dxjThe standard s alar produ t is:T [(x1, . . . , xn), (y1, . . . , yn)] =

n∑

i,j=1

xiyiand its expression isT =

n∑

i=1

dxi ⊗ dxiThis last expression is also written:ds2 = dx2

1 + · · ·+ dx2n

Problem 25: Change of the omponents of a bilinear form.a) Let E be a K-vs,e = (e1, e2), u = (u1,u2) = (e1 + e2, 2e1 + 3e2),two bases, and

ω = (ω1, ω2), φ = (ϕ1, ϕ2)the respe tive dual bases. Find the expression ofT = ω1 ⊗ ω2 − 3ω2 ⊗ ω1 + ω2 ⊗ ω2in the basis φ⊗ of T 2E asso iated to the basis φ.b) Let E be a K-vs,

e = (e1, e2, e3, e4), u = (u1,u2,u3,u4) = (e1−e3, e2−e4, e1+e2+e3, e2−e3−e4),two bases, andω = (ω1, ω2, ω3, ω4), φ = (ϕ1, ϕ2, ϕ3, ϕ4)

54 CHAPTER 2. ALTERNATING MULTILINEAR FORMSthe respe tive dual bases. Find the expression ofT = ω1 ⊗ ω2 − ω2 ⊗ ω2 − 3ω2 ⊗ ω4in the basis φ⊗ of T 2E asso iated to the basis φ.

. ) Generalization: Let E be a K-vs,e = (e1, . . . , en), u = (u1, . . . ,un), u = eC,two bases, and w = (ω1, . . . , ωn), φ = (ϕ1, . . . , ϕn) the respe tive dualbases. Find the expression of T ∈ T kE in the basis φ⊗ of T kE asso i-ated to the basis φSolution:a) We may ompute in a threefold way:i) Expressing the ωj in terms of the ϕi and substituting into theexpression of T . Reminding the hange of basis in the dual (seep.23) we haveC =

(1 21 3

), φ = ω(CT )−1 ⇒ ω = φCT

(ω1, ω2) = (ϕ1, ϕ2)

(1 12 3

)= (ϕ1 + 2ϕ2, ϕ1 + 3ϕ2)Substituting we obtain:

T = −ϕ1 ⊗ ϕ1 − 4ϕ2 ⊗ ϕ1 − 3ϕ2 ⊗ ϕ2ii) Taking into a ount thatT =

∑T [ui,uj ]ϕi ⊗ ϕjand omputing T [ui,uj ] through the matrix of T in the basis e

A =

(T [e1, e1] T [e1, e2]T [e2, e1] T [e2, e2]

)

2.1. MULTILINEAR APPLICATIONS AND FORMS 55If x = eX,y = eY we haveT [x,y] = XTAYThen

T [u1,u1] = (1, 1)

(0 1−3 1

)(11

)= −1,

T [u1,u2] = (1, 1)

(0 1−3 1

)(23

)= 0,et . We obtain

T = −ϕ1 ⊗ ϕ1 − 4ϕ2 ⊗ ϕ1 − 3ϕ2 ⊗ ϕ2As the matrix is not symmetri al we should be areful with thepositions of the indexes.iii) Observing that if e = uC and x = eX = uX ′,y = eY = uY ′ thenX = CX ′, Y = CY ′ and we have

T [x,y] = XTAY = X ′TCTACY ′where we an see that the matrix of T in the basis u will beA′ = CTACIn our ase

A′ =

(1 12 3

)(0 1−3 1

)(1 21 3

)=

(−1 0−4 −3

)b) i) The hange of basis matrix isC =

1 0 1 00 1 1 1−1 0 1 −1

0 −1 0 −1

56 CHAPTER 2. ALTERNATING MULTILINEAR FORMSand that ω = φCT (see p.23), we obtain(ω1, ω2, ω3, ω4) = (ϕ1, ϕ2, ϕ3, ϕ4)

1 0 −1 00 1 0 −11 1 1 00 1 −1 −1

so

ω1 = ϕ1+ϕ3, ω2 = ϕ2+ϕ3+ϕ4, ω3 = −ϕ1+ϕ3−ϕ4, ω4 = −ϕ2−ϕ4and substituting into the expression of T we get the desired result.ii) The matrix of T in the basis ω isA =

0 1 0 00 −1 0 −30 0 0 00 0 0 0

The omponents of the ve tors of u in the basis e are

(1, 0,−1, 0)T , (0, 1, 0,−1)T , (1, 1, 1, 0)T , (0, 1,−1,−1)TThe omponent orresponding to ϕ1 ⊗ ϕ1 is T [u1,u1]

(1, 0,−1, 0)A

10−1

0

= 0,that orresponding to ϕ1 ⊗ ϕ2 is T [u1,u2]

(1, 0,−1, 0)A

010−1

= 1,et . We obtain:

T = ϕ1⊗ϕ2+ϕ1⊗ϕ3+ϕ1⊗ϕ4+2ϕ2⊗ϕ2−ϕ2⊗ϕ3+2ϕ2⊗ϕ4+3ϕ3⊗ϕ2+

+3ϕ3 ⊗ ϕ4 + 2ϕ4 ⊗ ϕ2 − ϕ4 ⊗ ϕ3 + 2ϕ4 ⊗ ϕ4

2.1. MULTILINEAR APPLICATIONS AND FORMS 57iii) We an as well use that A′ = CTAC; we leave it to the reader. ) The omponents of T in the basis φ⊗ arebj1...jk = T [uj1, . . . ,ujk ] = T [

n∑

l1=1

cl1j1el1 , . . . ,

n∑

lk=1

clkjkelk ] =

=n∑

l1,...,lk=1

cl1j1 · · · clkjkT [el1 , . . . , elk ] =

=n∑

l1,...,lk=1

cl1j1 · · · clkjk

(n∑

i1,...,ik=1

ai1...ikϕi1 ⊗ · · · ⊗ ϕik)[el1 , . . . , elk ] =

=n∑

l1,...,lk=1

cl1j1 · · · clkjkal1...lkFor instan e let T = 2ω1 ⊗ ω2 ⊗ ω2 − 3ω1 ⊗ ω2 ⊗ ω3 and let the hangeof basis matrix be

C =

1 4 72 5 83 6 9

The only nonvanishing omponents of T are a122 = 2 and a123 = −3.Let's ompute a ouple of omponents of T in the new basis:b123 = c11c

22c

23a122 + c11c

22c

33a123 = 1 · 5 · 8 · (2) + 1 · 5 · 9 · (−3) = −55

b231 = c12c23c

21a122 + c12c

23c

31a123 = 4 · 8 · 2 · (2) + 4 · 8 · 3 · (−3) = −160

T In the omputations of the omponents of a tensor the Einstein'ssummation onvention is in ommon use: the repeated indexes in positionslow-high add and one omits the sign∑. Using this onvention the pre edingresult an be written thus:bj1...jk = cl1j1 · · · c

lkjkal1...lkAll that is in luded in the so alled Ri i al ulus and is used under severalforms in the physi s texts. An interesting exposition of those questions anbe found in K.Jäni h book [Jan.

58 CHAPTER 2. ALTERNATING MULTILINEAR FORMS2.1.2 Pull-ba kT Let E,F be two K-vs and f : E → F a linear map; we an generalizethe on ept of dual appli ation (see p.27):

T kE f∗← T kFf ∗(T ) ← TDene f ∗(T ) through the a tion on a list (v1, . . . ,vk) of k ve tors in E:

f ∗(T )[v1, . . . ,vk] := T [fv1, . . . , fvk]As f ∗ a ts in a ontrary sense to that of f , it is alled a pull-ba k.Notation: We have written fv instead of f [v] to avoid weighty expressionssu h asT [f [v1], . . . , f [vk]].We shall do that whenever f is linear; but ω =

∑i λiωejωi seems ex essiveand we will write ω =

∑i λiω[ej]ωi.

Problem 26:a) Let E,F be two K-vs and f : E → F a linear map. Find a matrix off ∗ : T kF → T kE.b) Let S ∈ T kF, T ∈ T lF ; show that f ∗(S ⊗ T ) = f ∗(S)⊗ f ∗(T ).Solution:a) Let e = (e1, . . . , en) be a basis of E, u = (u1, . . . ,um) a basis of F andA the matrix of f in these bases, fe = uA.Let ω = (ω1, . . . , ωn) and φ = (ϕ1, . . . , ϕm) be the respe tive dual bases; onsider the bases of T kE, T kF

ω⊗ = ωi1 ⊗ · · · ⊗ ωik , 1 ≤ i1, . . . , ik ≤ n

φ⊗ = ϕj1 ⊗ · · · ⊗ ϕjk , 1 ≤ j1, . . . , jk ≤ m

2.1. MULTILINEAR APPLICATIONS AND FORMS 59The matrix of f ∗ has in the olumn j1, . . . , jk the omponents of f ∗(ϕj1⊗· · · ⊗ ϕjk) in the basis ω⊗; the i1, . . . , ik-th omponent isf ∗(ϕj1 ⊗ · · · ⊗ ϕjk)[ei1 , . . . , eik ] = ϕj1 ⊗ · · · ⊗ ϕjk [fei1 , . . . , feik ] =

= ϕj1[fei1 ] · · ·ϕjk [feik ] = aj1i1 · · ·ajkikIt may seem strange to design a olumn through a multiindex j1, . . . , jk,but if we onsider that the matrix has mk olumns and nk rows theidea be omes a natural one. The order of the ve tors in the basis isnot relevant but, if needed, we may use the alphabeti al order for theindexes.b) This is totally formal:

f ∗(S ⊗ T )[v1, . . . ,vk,vk+1, . . . ,vk+l] =

= S ⊗ T [f(v1), . . . , f(vk), f(vk+1), . . . , f(vk+l)] =

= S[f(v1), . . . , f(vk)] · T [f(vk+1), . . . , f(vk+l)] =

= f ∗(S)[v1, . . . ,vk]f∗(T )[vk+1, . . . ,vk+l] =

= f ∗(S)⊗ f ∗(T )[v1, . . . ,vk,vk+1, . . . ,vk+l]One says that f ∗ is natural respe t to the tensor produ t: both opera-tions ommute. It doesn't mind to do rst the tensor produ t and thenf ∗ or do f ∗ rst and then the produ t. That is, the following diagramis ommutative:

T kE × T lE f∗← T kF × T lF(P,R) (S, T )↓ ⊗E ↓ ⊗FT k+lE f∗← T k+lFP ⊗R S ⊗ T

Problem 27:Let f : C2 → C2 be f(z1, z2) = (z1 + 2z2, z1 − z2). Find a matrix of f ∗ :T 2(C2)→ T 2(C2).

60 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:C is a 1-dimensional vs over C and the ve tor z = 1 is a basis (and a2-dimensional vs over R, z1 = 1, z2 = i being a basis). Then C2 is a 2-dimensional vs over C and e1 = (1, 0), e2 = (0, 1) are a basis; let ω = (ω1, ω2)be its dual basis. The matrix of f in this basis is

M =

(1 21 −1

)A basis of T 2(C2) isω1 ⊗ ω1, ω1 ⊗ ω2, ω2 ⊗ ω1, ω2 ⊗ ω2Let's ompute the images by f ∗ :a) First olumn: omponents of f ∗(ω1 ⊗ ω1):

f ∗(ω1 ⊗ ω1)[e1, e1] = ω1 ⊗ ω1[fe1, fe1] = ω1[fe1]ω1[fe1] = 1 · 1 = 1

f ∗(ω1 ⊗ ω1)[e1, e2] = ω1 ⊗ ω1[fe1, fe2] = ω1[fe1]ω1[fe2] = 1 · 2 = 2

f ∗(ω1 ⊗ ω1)[e2, e1] = ω1 ⊗ ω1[fe2, fe1] = ω1[fe2]ω1[fe1] = 2 · 1 = 2

f ∗(ω1 ⊗ ω1)[e2, e2] = ω1 ⊗ ω1[fe2, fe2] = ω1[fe2]ω1[fe2] = 2 · 2 = 4b) Se ond olumn: omponents of f ∗(ω1 ⊗ ω2):f ∗(ω1 ⊗ ω2)[e1, e1] = ω1 ⊗ ω2[fe1, fe1] = ω1[fe1]ω2[fe1] = 1 · 1 = 1f ∗(ω1 ⊗ ω2)[e1, e2] = ω1 ⊗ ω2[fe1, fe2] = ω1[fe1]ω2[fe2] = 1 · (−1) = −1f ∗(ω1 ⊗ ω2)[e2, e1] = ω1 ⊗ ω2[fe2, fe1] = ω1[fe2]ω2[fe1] = 2 · 1 = 2f ∗(ω1 ⊗ ω2)[e2, e2] = ω1 ⊗ ω2[fe2, fe2] = ω1[fe2]ω2[fe2] = 2 · (−1) = −2,et . The matrix of f ∗ would be

M =

1 1 . . . . . .2 −1 . . . . . .2 2 . . . . . .4 −2 . . . . . .

2.2. ALTERNATING FORMS 61T Tensors in T kE are alled k- ovariant. One an as well onsidermultilinear maps

S : E∗× k). . . ×E∗ → R alled k- ontravariant tensors, and the spa e of all those tensors will be de-noted by TkE. One an even onsider mixed tensors su h asS : E× k). . . ×E× E∗× l). . . ×E∗ → Ra k- ovariant and l- ontravariant tensor; those will be designed by T kl E. Weshall fo us the attention on k- ovariant tensors.

2.2 Alternating forms2.2.1 PermutationsT Let's re all some basi fa ts about permutations.A permutation of order k is a bije tion

σ : 1, 2, . . . , k → 1, 2, . . . , k1 7→ σ(1)2 7→ σ(2). . . . . . . . .k 7→ σ(k)The notation is σ = [σ(1), σ(2), . . . , σ(k)].The produ t of permutations (non ommutative) is the omposition ofappli ations; for example

[1, 3, 5, 2, 4] · [4, 1, 5, 2, 3] = [2, 1, 4, 3, 5]Noti e the permutation on the right of the produ t tells the position of theindex we must hoose in the permutation on the left; for instan e the 4 inthe rst pla e tells us to sele t the index o upying the fourth position inthe left permutation. There we meet the index 2 that will o upy the rstposition in the produ t, et .

62 CHAPTER 2. ALTERNATING MULTILINEAR FORMSThe set of all permutations of order k is the symmetri group of order k,notated Sk.The permutations that transpose two indexes are alled transpositions; theyhave a spe ial notation:τ = [1, 2, 5, 4, 3] = (3, 5)Just to pra ti e we take a look at the produ t of a transposition and apermutation; we onsider two asesMultipli ation on the right: [4, 1, 2, 5, 3] · (3, 5) = [4, 1, 3, 5, 2]and we see that it transposes the indexes o upying the positions 3 and 5.Multipli ation on the left: (3, 5) · [4, 1, 2, 5, 3] = [4, 1, 2, 3, 5]and we see that it has transposed the indexes 3 and 5.Transpositions are important be auseEvery permutation is a produ t of transpositionsSu h a de omposition is not unique, but the parity of the number of trans-positions is unique. We an then dene the sign of a permutation throughthe parity of a de omposition: +1 if it is even and −1 if it is odd; this showsthat sgn(σ · µ) =sgn(σ)sgn(µ). There is another way to evaluate the signthat onsists in evaluating the parity of the number of inversions; we explainwith an example. The permutation

[1, 3, 2, 5, 4]has the inversions 3− 2 and 5− 4 and the sign is +.Problem 28 :Pra ti e the produ t of permutations. Find and verbalize an algorithm al-lowing to ompute the inverse of a given permutation. Compute the sign ofseveral permutations. De ompose a permutation as a produ t of transposi-tions and he k that the parity oin ides with the parity of the number ofinversions.

2.2. ALTERNATING FORMS 63Solution:Here are some examples:a) Produ t.i) [1, 2, 3, 4] · [4, 3, 1, 2] = [4, 3, 1, 2][4, 3, 1, 2] · [1, 2, 3, 4] = [4, 3, 1, 2]The permutation [1, 2, 3, 4] is the unit element of S4; we notate itId4.ii) [4, 3, 1, 2] · [2, 1, 3, 4] = [4, 3, 1, 2](1, 2) = [3, 4, 1, 2]A transposition a ting on the right transposes the indexes o u-pying the pla es told by the transposition.[1, 4, 3, 2] · [4, 3, 2, 1] = (2, 4)[4, 3, 2, 1] = [2, 3, 4, 1]A transposition a ting on the left transposes the indexes in thepermutation.iii) [4, 3, 1, 2] · [2, 1, 3, 4] = [3, 4, 1, 2][2, 1, 3, 4] · [4, 3, 1, 2] = [4, 3, 2, 1]The produ t is not ommutative.iv) p = [1, 2, 3] · [2, 3, 1] · [3, 1, 2] = ... after all [1, 2, 3] is the unit and

p = [2, 3, 1] · [3, 1, 2] = [1, 2, 3] = Id3telling us that [2, 3, 1] and [3, 1, 2] are the inverse of ea h other.b) Inverse.i) σ = [3, 1, 4, 2]−1 =? We must have [3, 1, 4, 2] · σ = [1, 2, 3, 4] andthus the rst index in σ has to sele t the pla e where the index1 is, the se ond in our ase. Then σ = [2, ·, ·, ·] and going on weobtain[3, 1, 4, 2]−1 = [2, 4, 1, 3]ii) [2, 3, 1]−1 = [3, 1, 2]

[3, 1, 2]−1 = [2, 3, 1][4, 2, 3, 1]−1 = [4, 2, 3, 1][4, 3, 5, 1, 6, 2]−1 = [4, 6, 2, 1, 3, 5] ) Sign.

64 CHAPTER 2. ALTERNATING MULTILINEAR FORMSi) [3, 1, 2] has the inversions 3-1 and 3-2; then sgn[3, 1, 2] = 1[3, 2, 1] has the inversions 3-2, 3-1 and 2-1, so sgn[3, 2, 1] = −1ii) [1, 3, 4, 2] has the inversions 3-2 and 4-2 and we have sgn[1, 3, 4, 2] =1[3, 2, 1, 4] has the inversions 3-2, 3-1, 2-1 and sgn[3, 2, 1, 4] = −1d) De omposition.We transport ea h index to the right pla e by means of transpositions:i) [3, 4, 1, 2] = [1, 4, 3, 2](1, 3) = [1, 2, 3, 4](2, 4)(1, 3) = (2, 4)(1, 3)and sgn[3, 2, 1, 4] = 1. If we look at the inversions we have 3-1,3-2, 4-1 and 4-2, an even quantity and we obtain the sign 1.ii) [5, 3, 4, 2, 1] = [1, 3, 4, 2, 5](1, 5) = [1, 2, 4, 3, 5](2, 4)(1, 5) == [1, 2, 3, 4, 5](3, 4)(2, 4)(1, 5) and sgn[5, 3, 4, 2, 1] = −1. On an-other hand the inversions are 5-3, 5-4, 5-2, 5-1, 3-2, 3-1, 4-2, 4-1 i2-1, an odd quantity leading to the sign -1.

2.2.2 A tion on lists and fun tionsT If we have an ordered list of obje ts of a set u = (u1, . . . , uk), ui ∈ U anda permutation σ ∈ Sk, the a tion of σ on u is:

uσ = (u1, . . . , uk)σ := (uσ(1), . . . , uσ(k)).Let V be a set; if we have an appli ationf : U× k). . . ×U → V,we an make a permutation σ a t on f :

(σf)(u) := f(uσ)that is:(σf)(u1, . . . , uk) = f(uσ(1), . . . , uσ(k))

2.2. ALTERNATING FORMS 65Examples:• Let u = (u1, u2, u3, u4, u5) and σ = [4, 1, 2, 3, 5]; then uσ = (u4, u1, u2, u3, u5).• If u = (u4, u2, u3, u5, u1) and σ = [4, 1, 2, 3, 5]we write u = (v1, v2, v3, v4, v5)and then

uσ = (v4, v1, v2, v3, v5) = (u5, u4, u2, u3, u1).In brief the permutation tells what positions to sele t in the list; thuswe read at on e uσ = (u5, u4, u2, u3, u1).• If f(x, y, z, t) = x2+y2+sin(zt) and σ = [2, 4, 3, 1] then (σf)(x, y, z, t) =f(y, t, z, x) = y2 + t2 + sin(zx)

Problem 29:Let u = (u1, . . . , uk), ui ∈ U be a list of obje ts in U , f : U× k). . . ×U → Van appli ation and σ, τ ∈ Sk. Show:a) (uτ)σ = u(τ · σ)b) τ(σf) = (τ · σ)fSolution:a) Using the denition:(uτ)σ = (uτ(1), . . . , uτ(k))σ = vi = uτ(i) = (v1, . . . , vk)σ = (vσ(1), . . . , vσ(k)) =

= (uτ(σ(1)), . . . , uτ(σ(k))) = (uτ ·σ(1), . . . , uτ ·σ(k)) =

= (u1, . . . , uk)(τ · σ) = u(τ · σ)b) Now we needn't write the list of obje ts:(τ(σf))(u) = (σf)(uτ) = f((uτ)σ) =

= f(u(τ · σ)) = ((τ · σ)f)(u)

66 CHAPTER 2. ALTERNATING MULTILINEAR FORMS2.2.3 Multilinear alternating formsT A multilinear form ω ∈ T kE is alternating if for ea h list of k ve tors inE we have

ω[. . . ,vj , . . . ,vi, . . . ] = −ω[. . . ,vi, . . . ,vj , . . . ], 1 ≤ i, j ≤ kWith the notation of the pre eding problem this an be written(i, j)ω = −ωThe main example of a multilinear alternating form is det ∈ T n(Rn) viewingthe olumns in a matrix A as ve tors of Rn.The set of k-multilinear alternating forms is notated ΛkE. It is obviousthat if ω, η ∈ ΛkE then ω+ η ∈ ΛkE and aω ∈ ΛkE, so ΛkE is a vs over thesame s alars as those of E.

Problem 30: Cara terization of alternating forms.Let ω ∈ T k(E) and σ ∈ Sk; show the equivalen e of the following assertionsa) ω[. . . ,vj , . . . ,vi, . . . ] = −ω[. . . ,vi, . . . ,vj , . . . ] (briey (i, j)ω = −ω).b) ω[vσ(1), . . . ,vσ(k)] = sgn (σ)ω[v1, . . . ,vk] (briey σω = sgn(σ)ω). ) ω vanishes over linearly dependent ve tors.Solution:a) ⇔ b): Assume ω alternating: τω = −ω, τ being a transposition; letσ = τ1 · · · τr be a de omposition of σ as a produ t of transpositions. Wehave:

σω = (τ1 · · · τr)ω = (τ1 · · · τr−1)(τrω) = −(τ1 · · · τr−1)ω =

= −(τ1 · · · τr−2)(τr−1ω) = (−1)2(τ1 · · · τr−2)ω = · · · == (−1)rω = sgn (σ)ω

2.2. ALTERNATING FORMS 67The re ipro al is obvious for sgn (i, j) = −1.a) ⇔ ): If ω is alternating ω[. . . ,vi, . . . ,vi, . . . ] = −ω[. . . ,vi, . . . ,vi, . . . ]and ω[. . . ,vi, . . . ,vi, . . . ] = 0. Now let, for instan e, v1 =∑k

i=2 aivi, thenω[v1, . . . ,vk] = ω[

k∑

i=2

aivi,v2, . . . ,vk] =k∑

i=2

aiω[vi,v2, . . . ,vk] = 0,for in ea h summand there are repeated ve tors.Re ipro ally if ω vanishes over linearly dependent sets we have0 = ω[. . . ,vi + vj , . . . ,vi + vj , . . . ]

= ω[. . . ,vi, . . . ,vi, . . . ] + ω[. . . ,vj, . . . ,vj, . . . ] +

+ω[. . . ,vi, . . . ,vj, . . . ] + ω[. . . ,vj, . . . ,vi, . . . ]and thenω[. . . ,vj, . . . ,vi, . . . ] = −ω[. . . ,vi, . . . ,vj, . . . ]thus ω is alternating.

T The alternator of S ∈ T kE isAlt(S) = 1k!

∑σ∈Sk

sgn(σ)σS

Problem 31: Computing alternators.Let E be a vs and ωi 1-forms. Show rst that we an ompute the alternatorof a tensor produ t of 1-forms applying the permutations with a sign to theorder of the 1-forms dire tly. Then ompute the alternator of the followingtensors:a) ω1

68 CHAPTER 2. ALTERNATING MULTILINEAR FORMSb) ω1 ⊗ ω2 ) ω1 ⊗ ω1d) ω2 ⊗ ω1e) ω1 ⊗ ω2 ⊗ ω3f) Show that Alt(ωi1 ⊗ · · · ⊗ ωik) = 0 whenever ωij = ωil for ij , il ∈(i1, . . . , ik).g) ω1 ⊗ ω2 ⊗ ω1h) ω1 ⊗ ω1 ⊗ ω2 + ω2 ⊗ ω1 ⊗ ω3 + ω2 ⊗ ω3 ⊗ ω3Solution:σ(ω1 ⊗ · · · ⊗ ωk)[v1, . . . ,vk] = ω1 ⊗ · · · ⊗ ωk[vσ(1), . . . ,vσ(k)] =

= ω1[vσ(1)] . . . ωk[vσ(k)] =

= ωσ−1(1)[v1] . . . ωσ−1(k)[vk] =

= ωσ−1(1) ⊗ · · · ⊗ ωσ−1(k)[v1, . . . ,vk]Now σ and σ−1 have the same sign (for σσ−1 = I and sgn I = 1) and sowhen σ traverses Sk so does σ−1.a) Alt (ω1) = 11!

(+ω1)b) Alt (ω1 ⊗ ω2) = 12!

(ω1 ⊗ ω2 − ω2 ⊗ ω1) ) Alt (ω1 ⊗ ω1) = 0 (apply b).d) Alt (ω2 ⊗ ω1) = −Alt (ω1 ⊗ ω2) (apply b).e) Alt (ω1 ⊗ ω2 ⊗ ω3) =1

3!(ω1 ⊗ ω2 ⊗ ω3 + ω2 ⊗ ω3 ⊗ ω1 + ω3 ⊗ ω1 ⊗ ω2 −

− ω1 ⊗ ω3 ⊗ ω2 − ω2 ⊗ ω1 ⊗ ω3 − ω3 ⊗ ω2 ⊗ ω1)

2.2. ALTERNATING FORMS 69f) For any σ ∈ Sk let τ = (ij, il)σ. Then τ(ωi1 ⊗· · ·⊗ωik) = σ(ωi1 ⊗· · ·⊗ωik); sosgn(τ)τ(ωi1⊗· · ·⊗ωik) = sgn(τ)σ(ωi1⊗· · ·⊗ωik) = −sgn(σ)σ(ωi1⊗· · ·⊗ωik)and all the permutations an be grouped in equal pairs with oppositsigns. Therefore the sum is 0.g) By the result in f) the value is 0.h) Using f) againAlt (ω1 ⊗ ω1 ⊗ ω2 + ω2 ⊗ ω1 ⊗ ω3 + ω2 ⊗ ω3 ⊗ ω3) = Alt (ω2 ⊗ ω1 ⊗ ω3) =

=1

3!(ω2 ⊗ ω1 ⊗ ω3 + ω1 ⊗ ω3 ⊗ ω2 + ω3 ⊗ ω2 ⊗ ω1 −

− ω2 ⊗ ω3 ⊗ ω1 − ω1 ⊗ ω2 ⊗ ω3 − ω3 ⊗ ω1 ⊗ ω2)

Problem 32:If T ∈ T kE show thata) Alt(T ) is alternating.b) T ∈ ΛkE i Alt(T ) = T . ) Alt(Alt(T ))= Alt(T ).Solution:a)(i, j)Alt(T ) =

1

k!

∑

σ∈Sk

sgn(σ)(i, j)(σT ) =1

k!

∑

σ∈Sk

sgn(σ)((i, j) · σ)T==

1

k!

∑

(i,j)·σ∈Sk

−sgn((i, j) · σ)((i, j) · σ)T ) = −Alt(T ),for when σ traverses Sk so does (i, j) · σ and sgn((i, j) · σ) = −sgn(σ).

70 CHAPTER 2. ALTERNATING MULTILINEAR FORMSb) If T ∈ ΛkE then σT = sgn(σ)T andAlt(T ) =1

k!

∑

σ∈Sk

sgn(σ)σT =1

k!

∑

σ∈Sk

sgn(σ)sgn(σ)T =

=1

k!k!T = TRe ípro ally if Alt(T ) = T then T is alternating by a). ) Obvious by a) and b).

Problem 33: Symmetri tensors.We all S ∈ T kE symmetri if σS = S for every σ ∈ Sk. The symmetrizedof S is Sim(S) =1

k!

∑

σ∈Sk

σSShow:a) Sim(S) is symmetri .b) S is symmetri i Sim(S) = S. ) Sim(Sim(S)) = Sim(S).d) Is it true that S is symmetri i Alt(S) = 0 ?e) Is it true that T is antisymmetri i Sim(T ) = 0 ?Solution:a)τSim(S) =

1

k!

∑

σ∈Sk

τ(σS) =1

k!

∑

σ∈Sk

(τ · σ)S =

=1

k!

∑

τ ·σ∈Sk

(τ · σ)S = Sim(S)

2.2. ALTERNATING FORMS 71b) If S is symmetri Sim(S) =1

k!

∑

σ∈Sk

σS =1

k!k!S = Sand the re iproque is a). ) Obvious by a) and b).d) If S is symmetri Alt(S) =

1

k!

∑

σ∈Sk

sgn(σ)σS =1

k!

∑

σ∈Sk

sgn(σ)S =

= (1

k!

∑

σ∈Sk

sgn(σ))S = 0but if Alt(S) = 0 we don't ne essarily have S symmetri ; for instan e lete1, e2 be a basis of E and ω1, ω2 its dual basis. Then Alt(ω1⊗ω1⊗ω2) =0 but ω1 ⊗ ω1 ⊗ ω2 is not symmetri :

ω1 ⊗ ω1 ⊗ ω2[e1, e1, e2] = 1

ω1 ⊗ ω1 ⊗ ω2[e1, e2, e1] = 0e) If T is antisymmetri Sim(T ) =1

k!

∑

σ∈Sk

σT =1

k!

∑

σ∈Sk

sgn(σ)T =

= (1

k!

∑

σ∈Sk

sgn(σ))T = 0If Sim(T ) = 0 it isn't ne essarily true that T is alternating. For instan eifT = ω1 ⊗ ω1 ⊗ ω2 − ω1 ⊗ ω2 ⊗ ω1we haveSim(T ) =

1

3!(ω1 ⊗ ω1 ⊗ ω2 + ω1 ⊗ ω2 ⊗ ω1 + ω2 ⊗ ω1 ⊗ ω1+

72 CHAPTER 2. ALTERNATING MULTILINEAR FORMS+ω1 ⊗ ω1 ⊗ ω2 + ω2 ⊗ ω1 ⊗ ω1 + ω1 ⊗ ω2 ⊗ ω1)−

− 1

k!(ω1 ⊗ ω2 ⊗ ω1 + ω2 ⊗ ω1 ⊗ ω1 + ω1 ⊗ ω1 ⊗ ω2+

+ω2 ⊗ ω1 ⊗ ω1 + ω1 ⊗ ω2 ⊗ ω1 + ω1 ⊗ ω1 ⊗ ω2) =

=2

3!(ω1 ⊗ ω1 ⊗ ω2 + ω1 ⊗ ω2 ⊗ ω1 + ω2 ⊗ ω1 ⊗ ω1)−

− 2

3!(ω1 ⊗ ω2 ⊗ ω1 + ω2 ⊗ ω1 ⊗ ω1 + ω1 ⊗ ω1 ⊗ ω2) = 0This an be seen without this al ulation be ause the permutations of

ω1 ⊗ ω1 ⊗ ω2 are the same as those of ω1 ⊗ ω2 ⊗ ω1 but with the signs hanged. In any ase T is not antisymmetri :T [e1, e1, e2] = 1that should vanish if T was antisymmetri .

Problem 34: Components of alternating and symmetri tensors.Let E be a vs, e = (e1, . . . , en) a basis and φ = (ω1, . . . , ωn) its dual basis.We look at tensors T, S ∈ T kE and their omponents in the basisφ⊗ = ωi1 ⊗ · · · ⊗ ωik : 1 ≤ i1, . . . , ik ≤ na) Remind that a tensor is alternating i σT = sgn (σ)T . Show that its omponents are alternating in the indexes.b) Remind that a tensor is symmetri i σS = ωS. Show that its ompo-nents are symmetri in the indexes.Solution:The a tion of a permutation σ ∈ Sk on k indexes i1, . . . , ik ∈ 1, . . . , n is

(i1 . . . ik)σ = (iσ(1) . . . iσ(k))

2.3. EXTERIOR PRODUCT 73a) Let T ∈ ΛkE; its omponents in the basis w⊗ are ai1...ik = T [ei1 , . . . , eik ].Then:a(i1...ik)σ = T [eiσ(1)

, . . . , eiσ(k)] =

= (σT )[ei1 , . . . , eik ] = sgn (σ)T [ei1, . . . , eik ] = sgn (σ)ai1...ikand we see that the omponents are alternating in the indexes.b) The same argument applied to a symmetri tensor S ∈ T kE givesa(i1...ik)σ = S[eiσ(1)

, . . . , eiσ(k)] =

= (σS)[ei1 , . . . , eik ] = S[ei1 , . . . , eik ] = ai1...ikshowing that the omponents are symmetri in the indexes.2.3 Exterior produ t2.3.1 Exterior produ t 1

T If ω ∈ ΛkE, η ∈ ΛlE their exterior produ t is dened as the alternating(k + l)-form:

ω ∧ η =(k + l)!

k!l!Alt(ω ⊗ η)that shows that ω ∧ η ∈ Λk+lE. We an write it also as

ω ∧ η =1

k!l!

∑

σ∈Sk+l

sgn(σ)σ(ω ⊗ η)

• The exterior produ t is bilinear and asso iative. If ω ∈ ΛkE, η ∈ ΛlEthe following ommutation rule is satisedω ∧ η = (−1)klη ∧ ωNoti e there is a hange of sign only when both degrees are odd; wemight all this nearly ommutativity. It implies, for instan e, that if

ω ∈ Λ1E then ω ∧ ω = 0, a property often used.

74 CHAPTER 2. ALTERNATING MULTILINEAR FORMS• If e = (e1, . . . , en) is a basis of E and φ = (ω1, . . . , ωn) its dual basis,the ( n

k

)k-formsφ∧ = ωi1 ∧ · · · ∧ ωik : 1 ≤ i1 < · · · < ik ≤ nare basis of ΛkE that we shall all asso iated to φ. The dimension of

ΛkE is then ( nk

).• We shall use the notation I րto refer to a stri tly in reasing multiin-dex, I = (i1, . . . , ik), 1 ≤ i1 < · · · < ik ≤ n.For a proof of those fa ts you may refer to [Spvk, p.80 and following.

Problem 35:Let e = (e1, e2, e3, e4) be a basis of E and w = (ω1, ω2, ω3, ω4) its dual basis.Write the basis w∧ of ΛkE when k = 0, 1, 2, 3, 4.Solution:• Basis of Λ0E = K: 1.• Basis of Λ1E = E∗: ω1, ω2, ω3, ω4.• Basis of Λ2E: ω1 ∧ ω2, ω1 ∧ ω3, ω1 ∧ ω4, ω2 ∧ ω3, ω2 ∧ ω4, ω3 ∧ ω4.• Basis of Λ3E: ω1 ∧ ω2 ∧ ω3, ω1 ∧ ω2 ∧ ω4, ω1 ∧ ω3 ∧ ω4, ω2 ∧ ω3 ∧ ω4.• Basis of Λ4E: ω1 ∧ ω2 ∧ ω3 ∧ ω4.

Problem 36: Exterior produ t.Let E be a vs, e = (e1, e2, e3) a basis and ω = (ω1, ω2, ω3) its dual basis. Letσ = ω1 ⊗ ω2 − ω2 ⊗ ω1 ∈ Λ2E and η = 2ω1 − ω3 ∈ Λ1E. Using the denitionof the exterior produ t ompute σ ∧ η ∈ Λ3E.

2.3. EXTERIOR PRODUCT 75Solution:Taking into a ount problem 31, f) we obtain:σ ∧ η =

(2 + 1)!

2!1!Alt ((ω1 ⊗ ω2 − ω2 ⊗ ω1)⊗ (2ω1 − ω3)) =

= 3Alt (−ω1 ⊗ ω2 ⊗ ω3 + ω2 ⊗ ω1 ⊗ ω3)== 3

1

3!(−ω1 ⊗ ω2 ⊗ ω3 − ω2 ⊗ ω3 ⊗ ω1 − ω3 ⊗ ω1 ⊗ ω2 +

+ω1 ⊗ ω3 ⊗ ω + ω2 ⊗ ω1 ⊗ ω3 + ω3 ⊗ ω2 ⊗ ω1 +

+ω2 ⊗ ω1 ⊗ ω3 + ω1 ⊗ ω3 ⊗ ω2 + ω3 ⊗ ω2 ⊗ ω1 −−ω2 ⊗ ω3 ⊗ ω1 − ω1 ⊗ ω2 ⊗ ω3 − ω3 ⊗ ω1 ⊗ ω2)and

σ ∧ η =1

22(−ω1 ⊗ ω2 ⊗ ω3 − ω2 ⊗ ω3 ⊗ ω1 − ω3 ⊗ ω1 ⊗ ω2 +

+ω1 ⊗ ω3 ⊗ ω2 + ω2 ⊗ ω1 ⊗ ω3 + ω3 ⊗ ω2 ⊗ ω1) =

= −3!Alt (ω1 ⊗ ω2 ⊗ ω3) = −ω1 ∧ ω2 ∧ ω3We an easily he k the result: observe that σ = ω1⊗ω2−ω2⊗ω1 = ω1∧ω2and thenσ ∧ η = ω1 ∧ ω2 ∧ (2ω1 − ω3) = −ω1 ∧ ω2 ∧ ω3(remind that the self produ t of a one form vanishes).

Problem 37: Canoni al expression.Let E be a vs, e = (e1, . . . , en) a basis and w = (ω1, . . . , ωn) its dual basis.Let α =∑n

i=1 aiωi ∈ Λ1E and η =∑n

j,k=1 bjkωj ⊗ ωk ∈ Λ2E. Compute theexpression of α ∧ η in the basis w∧ of ΛnE.Solution:As η is alternated bii = 0 and bkj = −bjk (see problem 34 on page72) andη =

n∑

j,k=1

bjkωj ⊗ ωk =∑

1≤j<k≤nbjk(ωj ⊗ ωk − ωk ⊗ ωj) =

=∑

1≤j<k≤nbjkωj ∧ ωk

76 CHAPTER 2. ALTERNATING MULTILINEAR FORMSNow, using the nearly ommutativity, we obtainα ∧ η =

n∑

i=1

aiωi ∧ (∑

1≤j<k≤nbjkωj ∧ ωk) =

=∑

1≤i<j<k≤naibjkωi∧ωj∧ωk+

∑

1≤j<i<k≤n−aibjkωj∧ωi∧ωk+

∑

1≤j<k<i≤naibjkωj∧ωk∧ωi,the last equality be ause the terms with i = j or i = k vanish.

Problem 38: Components.Let E be a vs, e = (e1, . . . , en) a basis and w = (ω1, . . . , ωn) its dual basis.If η ∈ ΛkE ⊂ T kE, k ≤ n we an writeη =

∑

i1,...,ik

ai1...ikωi1 ⊗ · · · ⊗ ωikandη =

∑

j1<···<jk

bj1...jkωj1 ∧ · · · ∧ ωjk .Find the relation between tge omponents ai1...ik , I = (i1, . . . , ik) ⊂ 1, . . . , nand bj1...jk , J = (j1, . . . , jk) ⊂ 1, . . . , n, J ր.Solution:η =

∑

JրbJ∑

σ∈Sk

sgn(σ)ωjσ(1)⊗ · · · ⊗ ωjσ(k)If in i1, . . . , ik there are repeated indexes ai1...ik = 0 be ause η ∈ ΛkE andthe oe ients are alternating (see p.??). Assume rst that i1, . . . , ik arein reasing indexes; we have:

ai1...ik = η[ei1, . . . , eik ] =∑

JրbJ∑

σ∈Sk

sgn(σ)ωjσ(1)⊗ · · · ⊗ ωjσ(k)

[ei1 , . . . , eik ] =

=∑

Jր

∑

σ∈Sk

bJsgn(σ)δi1jσ(1)· · · δikjσ(k)

2.3. EXTERIOR PRODUCT 77and ea h summand vanishes ex ept when jσ(1) = i1, . . . , jσ(k) = ik; remindingthat J ր this happens when σ = Id and jl = il, l = 1, . . . , k. So when I րwe haveai1...ik = bi1...ikIf i1, . . . , ik are not in reasing indexes there is a permutation σ that ordersthem:

(i1, . . . , ik)σ = (j1, . . . , jk), j1 < · · · < jkThen again from the result on 72 we obtainai1...ik = sgn(σ)aj1...jk = sgn(σ)bj1...jkThere are not oe ients with repeated indexes be ause as ω ∈ ΛkE theyare alternating. For instan e if η ∈ Λ2(E) with dimE = 2 there are no termsin ω1 ⊗ ω1 nor in ω2 ⊗ ω2 and

η = a12ω1 ⊗ ω2 + a21ω2 ⊗ ω1 = b12ω1 ⊗ ω2 − b12ω2 ⊗ ω1 = b12ω1 ∧ ω2

2.3.2 De omposable formsT Those ω ∈ ΛkE, ω 6= 0 that are a produ t of 1-forms are alled de om-posable (or primitive):

ω = ϕ1 ∧ · · · ∧ ϕkThey ontain a threefold geometri al information:• They represent a k-dimensional subspa e U ⊂ E.• They give U an orientation.• They have a volume.Every n-form is de omposable and so are the k-forms in a basis w∧, andthere are others (those orresponding to another basis for instan e). Thereare nonde omposable forms as well.

78 CHAPTER 2. ALTERNATING MULTILINEAR FORMSProblem 39: De omposable forms 1.We take a look at the a tion of a de omposable form on a list of ve tors.a) Let v = (v1, . . . ,vk) be a list of ve tors in a vs E and φ = (ϕ1, . . . , ϕk)a list of 1-forms. Show thatϕ1 ∧ · · · ∧ ϕk[v1, . . . ,vk] = det

ϕ1[v1] . . . ϕ1[vk]. . . . . . . . .

ϕk[v1] . . . ϕk[vk]

b) Let e = (e1, . . . , en) be a basis of E and w = (ω1, . . . , ωn) its dualbasis. Write φ = BwT where B is a matrix (k, n) that has as rows the omponents of the forms ϕi in the basis w, and v = eA where A is amatrix (n, k) that has as olumns the omponents of the ve tors vi inthe basis e. Express the formula in a) by means of the matri es A,B. ) If k = n, A is a square matrix (n, n). Show that for every η ∈ ΛnE,η[v1, . . . ,vn] = (detA)η[e1, . . . , en].d) With the pre eding notations show:i) ϕ1, . . . , ϕk are li i ϕ1 ∧ · · · ∧ ϕk 6= 0.ii) v1, . . . ,vk are li i there is a k-form in the basis w∧ of ΛkE thatdoesn't vanish when applied to this list.e) Let E = R3 and let dx, dy, dz be the dual to the anoni al basis. Give ageometri al view of the a tion of dy∧dz, dz∧dx, dx∧dy, dx∧dy∧dz.Solution:a)

ϕ1 ∧ · · · ∧ ϕk[v1, . . . ,vk] =∑

σ∈Sk

sgn(σ)ϕσ(1)[v1] . . . ϕσ(k)[vk]That is, in the matrix

ϕ1[v1] . . . ϕ1[vk]. . . . . . . . .

ϕk[v1] . . . ϕk[vk]

2.3. EXTERIOR PRODUCT 79we sele t from the rst olumn the element in the row σ1, in the se -ond olumn we sele t the element in the row σ2, et ., assigning a signdependin on the sele tion; then we multiply these terms. Adding allprodu ts we obtain pre isely the determinantdet(ϕi[vj ])b) It su es to ompute ϕi[vj ]:

ϕi[vj ] = bi1ω1[vj] + · · ·+ binωn[vj] =

= bi1a1j + · · ·+ bina

nj = BiAjwhere Bi is the i-th row of B, Aj is the j-th olumn of A and BiAj isthe produ t 'row by olumn'. Then

ϕ1 ∧ · · · ∧ ϕk[v1, . . . ,vk] = det

B1A1 . . . B1Ak. . . . . . . . .BkA1 . . . BkAk

= det(BA) ) The n-form ω1 ∧ · · · ∧ ωn is a basis of ΛnE and η an be written η =cω1 ∧ · · · ∧ ωn. Now

η[e1, . . . , en] = cω1 ∧ · · · ∧ ωn[e1, . . . , en] = cand using b),η[v1, . . . ,vn] = η[e1, . . . , en]ω1 ∧ · · · ∧ ωn[v1, . . . ,vn] =

= η[e1, . . . , en] det(IA) = (detA)η[e1, . . . , en]d) i) Using the formula in a) we haveϕ1 ∧ · · · ∧ ϕk[ej1, . . . , ejk ] = det

ϕ1[ej1] . . . ϕ1[ejk ]. . . . . . . . .

ϕk[ej1 ] . . . ϕk[ejk ]

Taking into a ount that the bases e, ω are dualϕi[ejl] = bi1ω1[ejl] + · · ·+ binωn[ejl] = bijl

80 CHAPTER 2. ALTERNATING MULTILINEAR FORMSThenϕ1 ∧ · · · ∧ ϕk[ej1 , . . . , ejk] = det

b1j1 . . . b1jk. . . . . . . . .bkj1 . . . bkjk

= detBj1...jk

Bj1...jk being the k-minor of the matrix B obtained sele ting the olumns j1, . . . , jk. Thenϕ1, . . . , ϕk LI⇔ ∃(j1, . . . , jk) su h that detBj1...jk 6= 0⇔ ϕ1∧· · ·∧ϕk 6= 0That is to say that a ve tor in ΛkE is not null i it represents ak-dimensional spa e in E∗.ii) We use the formula in a) :

ωi1 ∧ · · · ∧ ωik [v1, . . . ,vk] = det

ωi1[v1] . . . ωi1[vk]. . . . . . . . .

ωik [v1] . . . ωik [vk]

butωil[vj ] = a1

jωil[e1] + · · ·+ anj ωil[en] = ailjand we obtainωi1 ∧ · · · ∧ ωik [v1, . . . ,vk] = det

ai11 . . . ai1k. . . . . . . . .

aik1 . . . aikk

= detAi1...ik

Ai1...ik being the k-minor of the matrix A obtained sele ting therows i1, . . . , ik. Thenv1, . . . ,vk li ⇔ ∃(i1, . . . , ik) su h that detAi1...ik 6= 0⇔

⇔ there is a basis form that doesn't vanish on the list.e) Let v1 = (a11, a

21, a

31),v2 = (a1

2, a22, a

32) ∈ R3; the matrix of these twove tors in the anoni al basis is

A =

a1

1 a12

a21 a2

2

a31 a3

2

2.3. EXTERIOR PRODUCT 81and following b) we havedy ∧ dz[v1,v2] = det

(a2

1 a22

a31 a3

2

)

dz ∧ dx[v1,v2] = det

(a3

1 a32

a11 a1

2

)

dx ∧ dy[v1,v2] = det

(a1

1 a12

a21 a2

2

)The ve tors (P1(v1),P1(v2)) = (j,k)

(a2

1 a22

a31 a3

2

) are the orthogonalproje tion of the ve tors v1,v2 onto the plane 〈j,k〉 and we see that dy∧dz[v1,v2] is the signed area of the parallelogramP = P (0;P1(v1),P1(v2)).We may say that dy∧dz sifts all what happens in its plane. Analogouslydz∧dx is the signed area of the parallelogram P = P (0;P2(v1),P2(v2))where the sign is positive if we turn from z to x in the plane xz, et .

O

area=dx^dy

plane xyf) Now we see that two ve tors v1,v2 are linearly independent i one of theareas of the proje ted parallelograms onto the oordinate planes doesn'tvanish. Clearly if the three areas vanish the ve tors are ollinear.

82 CHAPTER 2. ALTERNATING MULTILINEAR FORMSObserve that any 2-form is a linear ombination of those three proje -tions.Let v1 = (a11, a

21, a

31),v2 = (a1

2, a22, a

32),v3 = (a1

3, a23, a

33) ∈ R3 and let

A =

a1

1 a12 a1

3

a21 a2

2 a23

a31 a3

2 a33

be the matrix of the three ve tors in the anoni al basis; from the resultin ) we have:dx ∧ dy ∧ dz[v1,v2,v3] = (detA)dx ∧ dy ∧ dz[i, j,k] = detA,the signed volume of the parallelepiped generated by the three ve tors.

Problem 40: De omposable forms 2.a) If ω is de omposable show that ω ∧ ω = 0.b) If ω and η are de omposable k-forms show that(ω + η) ∧ (ω + η) =

2(ω ∧ η) if k = 20 if k = 2 + 1 ) Are the following forms de omposable? (the ϕi are independent 1-forms):i) 3ϕ1 ∧ ϕ1 + 5ϕ1 ∧ ϕ2 + 6ϕ2 ∧ ϕ1 + 10ϕ2 ∧ ϕ2ii) ϕ1 ∧ ϕ2 + ϕ3 ∧ ϕ4iii) ϕ1 ∧ ϕ2 + ϕ2 ∧ ϕ3Solution:a) Let ω = ϕ1 ∧ . . . ∧ ϕk; then:

ω ∧ ω = (ϕ1 ∧ . . . ∧ ϕk) ∧ (ϕ1 ∧ . . . ∧ ϕk) = 0,be ause the 1-form ϕ1, for instan e, appears twi e.

2.3. EXTERIOR PRODUCT 83b) Let ω and η be de omposable; then, using the nearly ommutativity:(ω + η) ∧ (ω + η) = ω ∧ η + η ∧ ω =

=

ω ∧ η + ω ∧ η = 2ω ∧ η if k = 2ω ∧ η − ω ∧ η = 0 if k = 2 + 1 ) i)

3ϕ1 ∧ ϕ1 + 5ϕ1 ∧ ϕ2 + 6ϕ2 ∧ ϕ1 + 10ϕ2 ∧ ϕ2 =

= 5ϕ1 ∧ ϕ2 + 6ϕ2 ∧ ϕ1 = −ϕ1 ∧ ϕ2and it is de omposable.ii) ω = ϕ1 ∧ ϕ2 + ϕ3 ∧ ϕ4 is not de omposable forω ∧ ω = ϕ1 ∧ ϕ2 ∧ ϕ3 ∧ ϕ4 + ϕ3 ∧ ϕ4 ∧ ϕ1 ∧ ϕ2 =

= 2ϕ1 ∧ ϕ2 ∧ ϕ3 ∧ ϕ4 6= 0iii) ω = ϕ1∧ϕ2+ϕ2∧ϕ3 satises ω∧ω = 0 and may be de omposable.It really is, forω = ϕ2 ∧ (ϕ3 − ϕ1)The guess has been easy, but we ould as well be more systemati ;put:

ω = ξ ∧ ηξ = aϕ1 + bϕ2 + cϕ3

η = a′ϕ1 + b′ϕ2 + c′ϕ3and substitutingϕ1 ∧ ϕ2 + ϕ2 ∧ ϕ3 =

= det

(a ba′ b′

)ϕ1∧ϕ2+det

(b cb′ c′

)ϕ2∧ϕ3+det

(c ac′ a′

)ϕ3∧ϕ1so we must have

det

(a ba′ b′

)= 1, det

(b cb′ c′

)= 1, det

(c ac′ a′

)= 0That is similar to nding two ve tors (a, b, c) and (a′, b′, c′) with agiven ross produ t (1, 0, 1).The ve tors should be in 〈(1, 0, 1)〉⊥ =

〈(0, 1, 0), (−1, 0, 1)〉 and we now see that

84 CHAPTER 2. ALTERNATING MULTILINEAR FORMSξ = ϕ2

η = −ϕ1 + ϕ3

Problem 41: Spa e of a de omposable form.Let E be a vs and ϕ = (ϕ1, . . . , ϕk), ψ = (ψ1, . . . , ψk), k ≤ dimE two listsof 1-forms of E su h thatλ = ϕ1 ∧ · · · ∧ ϕk 6= 0

µ = ψ1 ∧ · · · ∧ ψk 6= 0Remind that the nonvanishing of the k-forms is equivalent to li of the 1-forms. Let V = 〈ϕ1, . . . , ϕk〉 ⊂ E∗, the spa e of λ and V ′ = 〈ψ1, . . . , ψk〉 ⊂E∗ the spa e of µ. Show:a) ψ1 ∧ · · · ∧ ψk = cϕ1 ∧ · · · ∧ ϕk, c 6= 0⇔ ψ = ϕA, detA 6= 0⇔ V = V ′,

A being a matrix (k, k).b) ψ1 ∧ · · · ∧ ψk = ϕ1 ∧ · · · ∧ ϕk ⇔ ψ = ϕA, detA = 1. ) Assume E is an orthogonal spa e and let U = ♯V ; we all this k-dimensional spa e the asso iated spa e to λ. Show that U is indepen-dent of the representation of λ = ϕ1 ∧ · · · ∧ ϕk.d) Show that if U is nonsingular and vi ∈ E, thenϕ1 ∧ · · · ∧ ϕk[v1, . . . ,vk] = ϕ1 ∧ · · · ∧ ϕk[u1, . . . ,uk]the ui being the orthogonal proje tions of the vi onto U .e) Let ϕ = ϕ1∧ϕ2 6= 0 and ψ = ψ1∧ψ2 6= 0 be two de omposable 2-formsand let U,U ′ be the respe tive asso iated spa es. Show:i) If U ∩ U ′ = 0 then ϕ+ ψ is not de omposable.ii) If U∩U ′ is 1-dimensional, then ϕ+ψ is de omposable and ϕ 6= cψ.iii) If U∩U ′ is 2-dimensional, then ϕ+ψ is de omposable and ϕ = cψ.

2.3. EXTERIOR PRODUCT 85Solution:a) i) If ψ = ϕA, detA 6= 0 thenψ1 = a1

1ϕ1 + · · ·+ ak1ϕk. . . . . . . . .ψk = a1

kϕ1 + · · ·+ akkϕkandψ1 ∧ · · · ∧ ψk = (a1

1ϕ1 + · · ·+ ak1ϕk) ∧ · · · ∧ (a1kϕ1 + · · ·+ akkϕk)The nonvanishing terms in this produ t are obtained hoosing adierent ϕi in ea h fa tor:

ψ1 ∧ · · · ∧ ψk =∑

σ∈Sk

aσ(1)1 . . . a

σ(k)k ϕσ(1) ∧ · · · ∧ ϕσ(k) =

=∑

σ∈Sk

sgn(σ)aσ(1)1 . . . a

σ(k)k ϕ1 ∧ · · · ∧ ϕk =

= (detA)ϕ1 ∧ · · · ∧ ϕk = cϕ1 ∧ · · · ∧ ϕkwhere c = detA 6= 0.ii) Re ipro ally, assume that k < n and thatψ1 ∧ · · · ∧ ψk = cϕ1 ∧ · · · ∧ ϕk, c 6= 0We want to see that ea h ψi is linearly dependent on ϕ1, . . . , ϕk.Take for instan e ψ1 and assume it to be li from ϕ1, . . . , ϕk; then

ϕ1, . . . , ϕk, ψ1are a part of a basis of E∗. Let e1, . . . , ek, ek+1 be the orrespond-ing part of the dual basis. We haveϕ1 ∧ · · · ∧ ϕk[e1, . . . , ek] = det

ϕ1[e1] . . . ϕ1[ek]. . . . . . . . .ϕk[e1] . . . ϕk[ek]

=

86 CHAPTER 2. ALTERNATING MULTILINEAR FORMS= det I = 1and

ψ1 ∧ · · · ∧ ψk[e1, . . . , ek] = det

ψ1[e1] . . . ψ1[ek]. . . . . . . . .ψk[e1] . . . ψk[ek]

=

= det

0 . . . 0. . . . . . . . .ψk[e1] . . . ψk[ek]

= 0, ontradi ting that ψ1 ∧ · · · ∧ ψk = cϕ1 ∧ · · · ∧ ϕk, c 6= 0. Thisshows that ψ1 ∈ 〈ϕ1, . . . , ϕk〉; repeating the same argument withthe other forms ψi we obtain ψ = ϕA and detA 6= 0 for bothsets of ve tors are li be ause λ 6= 0, µ 6= 0. It is lear that these onditions are equivalent to V = V ′.This line of reasoning fails if k = n = dimE, but in this ase it isobvious that ϕ or ψ generate E.b) i) In a) we have seen that if ψ = ϕA thenψ1 ∧ · · · ∧ ψk = (detA)ϕ1 ∧ · · · ∧ ϕkand we know that detA = 1; so ψ1 ∧ · · · ∧ ψk = ϕ1 ∧ · · · ∧ ϕk.ii) Re ipro ally, if ϕ1 ∧ · · · ∧ ϕk = ψ1 ∧ · · · ∧ ψk the same argumentin a) ii) shows that ψ = ϕA; thenψ1 ∧ · · · ∧ ψk = (detA)ϕ1 ∧ · · · ∧ ϕkand we obtain detA = 1. This result shows that a de omposableform may have several de ompositions. ) U is independent of the representation of ϕ1∧ · · ·∧ϕk i V is indepen-dent of the representation, but that has been seen in b).d) The formula

ϕ1 ∧ · · · ∧ ϕk[v1, . . . ,vk] = det

ϕ1[v1] . . . ϕ1[vk]. . . . . . . . .

ϕk[v1] . . . ϕk[ek]

2.3. EXTERIOR PRODUCT 87tells us that it su es to show that if v = v′+v′′ with v′ ∈ U,v′′ ∈ U⊥then ϕi[v] = ϕi[v′]. So it su es to show that if e ∈ U⊥ then ϕi[e] = 0.But ϕi[e] = ♯ϕi · e = 0 be ause ♯ϕi ∈ U .e) Let u1 = ♯ϕ1,u2 = ♯ϕ2,v1 = ♯ψ1,v2 = ♯ψ2; then U = 〈u1,u2〉, U ′ =

〈v1,v2〉.i) If U ∩ U ′ = 0 we see that u1,u2,v1,v2 are li and then so arethe forms ϕ1, ϕ2, ψ1, ψ2. The form ϕ+ ψ is not de omposable for(ϕ+ ψ) ∧ (ϕ+ ψ) = (ϕ1 ∧ ϕ2 + ψ1 ∧ ψ2) ∧ (ϕ1 ∧ ϕ2 + ψ1 ∧ ψ2) =

= (ϕ1 ∧ ϕ2 ∧ ψ1 ∧ ψ2) + (ψ1 ∧ ψ2 ∧ ϕ1 ∧ ϕ2) =

= 2ϕ1 ∧ ϕ2 ∧ ψ1 ∧ ψ2 6= 0,on e taken into a ount the independen e of the forms. This guar-antees that ϕ + ψ is not de omposable due to the result in a) ofthe pre eding problem.ii) In this ase U ∩ U ′ = 〈e〉 and we let U = 〈e,v〉, U ′ = 〈e,v′〉, v,v′being li; thenV = 〈e, v〉 = 〈α, µ〉 = 〈ϕ1, ϕ2〉V ′ = 〈e, v′〉 = 〈α, ν〉 = 〈ψ1, ψ2〉By what has been seen in a) we have

ϕ1 ∧ ϕ2 = c α ∧ µψ1 ∧ ψ2 = d α ∧ νand

ϕ+ ψ = ϕ1 ∧ ϕ2 + ψ1 ∧ ψ2 = α ∧ (cµ+ dν)is a de omposable form as it is the produ t of two 1-forms. More-over we an't have ϕ = cψ be ause then we would have U = U ′.iii) If U ∩ U ′ is 2-dimensional we have U = U ′ and then ϕ = cψ andϕ+ ψ = (c+ 1)ψis a de omposable form.

88 CHAPTER 2. ALTERNATING MULTILINEAR FORMS2.3.3 Exterior produ t 2T Remind the formula for the exterior produ t

ω ∧ η =1

k!l!

∑

σ∈Sk+l

sgn(σ)σ(ω ⊗ η)or more expli itly(ω ∧ η)[v1, . . . ,vk,vk+1, . . . ,vk+l] =

=1

k!l!

∑

σ∈Sk+l

sgn(σ)ω[vσ(1), . . . ,vσ(k)]η[vσ(k+1), . . . ,vσ(k+l)]This sum an be written in a simpler form. Let's start with some fa tsabout permutations that we show by way of an example with k = 3, l = 2.Consider the permutation (noti e the bar)σ0 = [2, 3, 5|1, 4]We shall say that σ ∈ Sk+l is equivalent to σ0 if it reorders 2, 3, 5 and

1, 4 separatedly. For instan e σ1 = [5, 2, 3|4, 1] is equivalent to σ0 butσ = [2, 1, 4|3, 5] is not, be ause it has not maintained the partition. This isan equivalen e relation, ea h equivalen e lass onsisting of k!l! permutationsin Sk+l and there are ( k + l

k

) lasses. In ea h lass [σ] there is a anoni alrepresentative σ0, the permutation that in ea h part has the indexes in in- reasing order. For example σ1 ∈ [σ0] and σ0 is the anoni al representative(see [Jan, p.133 and following). The permutations σ ∈ [σ0] have the formσ = σ0σaσbwhere

σa leaves the last l indexes invariantσb leaves the rst k indexes invariantThe permutations σa, σb ommute and may be identied with permutationsof k and l indexes respe tively. For example if

σ0 = [2, 3, 5 | 1, 4], σ1 = [5, 2, 3 | 4, 1] ∈ [σ0]

2.3. EXTERIOR PRODUCT 89we have[5, 2, 3 | 4, 1] = [2, 3, 5 | 1, 4] · [3, 1, 2 | 4, 5] · [1, 2, 3 | 5, 4]

= σ0 · σa · σbAs another exampleσ0 = [1, 2, 4 | 3, 5], σ1 = [2, 4, 1 | 5, 3] ∈ [σ0]

σ1 = [2, 4, 1 | 5, 3] = [1, 2, 4 | 3, 5] · [2, 3, 1 | 4, 5] · [1, 2, 3 | 5, 4]

= σ0 · σa · σb

Problem 42: A new expression for the exterior produ tLet ω ∈ Λk(E), η ∈ Λl(E); show:ω ∧ η =

∑

σ0 an sgn(σ0)σ0(ω ⊗ η)Solution:Let's give two proofs:a) We ompute ω ∧ η =∑

σ∈Sk+lsgn(σ)σ(ω ⊗ η) applied to a list of kve tors. Let σ = σ0σaσb as above; thensgn(σ)σ(ω ⊗ η)[v1, . . . ,vk+l] =

= sgn(σ)ω[vσ(1), . . . ,vσ(k)]η[vσ(k+1), . . . ,vσ(k+l)] == sgn(σ)ω[vσ0σaσb(1), . . . ]η[vσ0σaσb(k+1), . . . ]The permutation σb is the identity on the rst k indexes and takinginto a ount that σa may be identied with a permutation of the rst

k indexes and that ω is alternating, we have:ω[vσ0σaσb(1), . . . ] = (σaω)[vσ0(1), . . . ] = sgn(σa)ω[vσ0(1), . . . ]

90 CHAPTER 2. ALTERNATING MULTILINEAR FORMSand analogouslyη[vσ0σaσb(k+1), . . . ] = (σbη)[vσ0(k+1), . . . ] = sgn(σb)η[vσ0(k+1), . . . ]Thenω ∧ η[v1, . . . ,vk] = 1

k!l!

∑σ∈Sk+l

sgn(σ)σ(ω ⊗ η)[v1, . . . ,vk] =

= 1k!l!

∑σ∈Sk+l

sgn(σ)sgn(σa)sgn(σb)ω[vσ0(1), . . . ]η[vσ0(k+1), . . . ] =

= 1k!l!

∑σ∈Sk+l

sgn(σ0)ω[vσ0(1), . . . ]η[vσ0(k+1), . . . ]as there are k!l! permutations in the lass of σ0 we obtainω ∧ η[v1, . . . ,vk] =

∑

σ0 an sgn(σ0)(ω ⊗ η)[vσ0(1), . . . ,vσ0(k+l)]b) For another style of proof putv′ = (v1, . . . ,vk),v” = (vk+1, . . . ,vk+l)Then sgn(σ)σ(ω ⊗ η)[(v′,v”)] = sgn(σ)(ω ⊗ η)[(v′,v”)σ0σaσb] =

= sgn(σ)(ω ⊗ η)[(v′σ0σa,v”σ0σb)] =

= sgn(σ)ω[v′σ0σa] · η[v”σ0σb] =

= sgn(σ)(σaω)[v′σ0] · (σbη)[v”σ0] =

= sgn(σ0σaσb)sgn(σa)ω[v′σ0]sgn(σb)η[v”σ0] =

= sgn(σ0)σ0(ω ⊗ η)[(v′,v”)]and apply the argument in a).Problem 43: Nearly ommutativity of the exterior produ t.Show that for ω ∈ ΛkE, η ∈ ΛlE we have

ω ∧ η = (−1)klη ∧ ωNoti e there is a hange of sign only when both degrees are odd.

2.3. EXTERIOR PRODUCT 91Solution:Using the new expression for the exterior produ tω ∧ η[v1, . . . ,vk,vk+1, . . . ,vk+l] =

∑

σ0 an sgn(σ0)ω[vσ0(1), . . . ]η[vσ0(k+1), . . . ]

η ∧ ω[v1, . . . ,vk+l] =∑

τ0 an sgn(τ0)η[vτ0(1), . . . ]ω[vτ0(l+1), . . . ]

σ0 and τ0 are related as in σ0 = [2, 3, 5|1, 4] and τ0 = [1, 4|2, 3, 5]; the or-responding terms only may dier in a sign if they do so. But to go fromσ0 to τ0 when we pla e the 1 in the rst position we make k transpositions,when we pla e the 4 we make k transpositions, et . The total number oftranspositions is kl and we have

(−1)klsgn(σ0) = sgn(τ0)Moreoverτ0(j) = σ0(j + k), j = 1, . . . , l

τ0(l + i) = σ0(i), i = 1, . . . , kand we shall haveη ∧ ω[v1, . . . ,vk+l] =

∑

σ0 an(−1)klsgn(σ0)η[vτ0(1), . . . ]ω[vτ0(l+1), . . . ] =

=∑

σ0 an(−1)klsgn(σ0)ω[vσ0(1), . . . ]η[vσ0(k+1), . . . ] =

= (−1)klω ∧ η[v1, . . . ,vk+l]

Problem 44: Naturality of the exterior produ t.Let f : E → F be a linear map and ω ∈ ΛkF , η ∈ ΛlF . Show thatf ∗(ω ∧ η) = f ∗(ω) ∧ f ∗(η).

92 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:f ∗(ω ∧ η)[v1, . . . ,vk+l] = ω ∧ η[fv1, . . . , fvk+l] =

=∑

σ an sgn(σ)ω[fvσ(1), . . . ] · η[fvσ(k+1), . . . ] =

=∑

σ an sgn(σ)f ∗(ω)[vσ(1), . . . ] · f ∗(η)[vσ(k+1), . . . ] =

= f ∗(ω) ∧ f ∗(η)[v1, . . . ,vk+l]

2.3.4 MedleyProblem 45: Cartan's lemma.Let E be an n-dimensional vs, ϕ1, . . . , ϕk, k ≤ n linearly independent 1-forms and ψ1, . . . , ψk 1-forms su h thatϕ1 ∧ ψ1 + · · ·+ ϕk ∧ ψk = 0.Show thatψi =

k∑

j=1

aijϕj, i = 1, . . . , kthe aij being s alars that satisfy aij = aji.Solution:Completete the forms ϕi to a basis of E∗ϕ1, . . . , ϕk, ϕk+1, . . . , ϕn,express ψi in terms of this basis

ψi = ai1ϕ1 + · · ·+ aikϕk + ai(k+1)ϕk+1 + · · ·+ ainϕnand substituting it in the given ondition0 = (a11ϕ1 ∧ ϕ1 + · · ·+ a1kϕ1 ∧ ϕk + a1(k+1)ϕ1 ∧ ϕk+1 + · · ·+ a1nϕ1 ∧ ϕn)+

2.3. EXTERIOR PRODUCT 93+(a21ϕ2 ∧ ϕ1 + · · ·+ a2kϕ2 ∧ ϕk + a2(k+1)ϕ2 ∧ ϕk+1 + · · ·+ a2nϕ2 ∧ ϕn)+

. . . . . .

+(ak1ϕk ∧ ϕ1 + · · ·+ akkϕk ∧ ϕk + ak(k+1)ϕk ∧ ϕk+1 + · · ·+ aknϕk ∧ ϕn).Writing this element of Λ2E in anoni al form0 = (a12 − a21)ϕ1 ∧ ϕ2 + · · ·+ (a(k−1)k − ak(k−1))ϕk−1 ∧ ϕk+

+a1(k+1)ϕ1 ∧ ϕk+1 + · · ·+ aknϕk ∧ ϕn,the oe ients must vanish sin e the 2-forms in this expression are a basisof Λ2E. Thenaij = 0, i = 1, . . . , k, j = k + 1, . . . , nand the ψi are linear ombinations of ϕ1, . . . , ϕk ex lusively. The remaining oe ients satisfy

aij − aji = 0 ⇒ aij = aji, 1 ≤ i < j ≤ kwhat we wanted to see.Problem 46:Let E be a K-vs, dimE = n, and ϕ ∈ E∗, ϕ 6= 0 a xed linear form. Considerthe linear maps

fr : ΛrE → Λr+1Eω 7→ ϕ ∧ ωwhere r = 0, . . . , n− 1. Show that

ΛrEfr→ Λr+1E

fr+1→ Λr+2Eis exa t, that is show that Im fr = Ker fr+1.

94 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:If θ ∈ Im fr it must be θ = ϕ ∧ ω, ω ∈ ΛrE and then fr+1(θ) = ϕ ∧ θ =ϕ ∧ (ϕ ∧ ω) = 0 and θ ∈ Ker fr+1. Re ipro ally if θ ∈ Ker fr+1 ompleteϕ 6= 0 to a basis of E∗: ϕ1 = ϕ, ϕ2, . . . , ϕn. Express θ ∈ Λr+1E in anoni form and put apart the terms ontaining ϕ:

θ =∑

1≤i1<···<ir+1≤nai1...ir+1ϕi1 ∧ · · · ∧ ϕir+1 =

=∑

2≤i2<···<ir+1≤na1i2...ir+1ϕ ∧ ϕi2 ∧ · · · ∧ ϕir+1 +

+∑

2≤i1<i2<···<ir+1≤nai1i2...ir+1ϕi1 ∧ ϕi2 ∧ · · · ∧ ϕir+1Now θ ∈ Ker fr+1 i ϕ ∧ θ = 0, that is

ϕ ∧ θ =∑

2≤i1<i2<···<ir+1≤nai1i2...ir+1ϕ ∧ ϕi1 ∧ ϕi2 ∧ · · · ∧ ϕir+1 = 0We see that the oe ients orresponding to indexes 2 ≤ i1 < i2 < · · · <

ir+1 ≤ n satisfy ai1i2...ir+1 = 0 and θ has the formθ =

∑

2≤i2<···<ir+1≤na1i2...ir+1ϕ ∧ ϕi2 ∧ · · · ∧ ϕir+1 =

= ϕ ∧ (∑

2≤i2<···<ir+1≤na1i2...ir+1ϕi2 ∧ · · · ∧ ϕir+1) =

= ϕ ∧ η, η ∈ ΛrEWe have proved thatθ ∈ Ker fr+1 ⇔ θ ∈ Im fr

Problem 47:Let E be a vs, ϕ1, . . . , ϕn a basis of E∗ and η ∈ ΛkE (k < n) a xed k-form.a) Prove that if η ∧ ϕi = 0, i = 1, . . . , n then η = 0.

2.3. EXTERIOR PRODUCT 95b) Can we relax the hypothesis? ) Consider the linear mapE∗ = Λ1E

f→ Λk+1E

ω 7→ η ∧ ωShow that if η 6= 0 then dimKerf ≤ k.d) ShowdimKerf = k ⇔ η is de omposablee) Show that η ∈ Λn−1 is des omposable.Solution:a) Express η in anoni al form:η =

∑

1≤i1<···<ik≤nai1...ikϕi1 ∧ · · · ∧ ϕikThe ondition η ∧ ϕ1 = 0 reads:

0 =∑

1≤i1<···<ik≤nai1...ikϕ1 ∧ ϕi1 ∧ · · · ∧ ϕikThe sumands with i1 = 1 vanish (ϕ1 is a repeated 1-form) and we have

0 =∑

2≤i1<···<ik≤nai1...ikϕ1 ∧ ϕi1 ∧ · · · ∧ ϕikand this means that ai1...ik = 0, 2 ≤ i1 < · · · < ik ≤ n. Equivalently,the nonvanishing terms in η must ontain ϕ1:

η =∑

2≤i2<···<ik≤na1i2...ikϕ1 ∧ ϕi2 ∧ · · · ∧ ϕikApplying su essively η ∧ ϕi = 0, i = 2, . . . , k, we arrive at

η = a1,2...kϕ1 ∧ · · · ∧ ϕkThen η ∧ ϕk+1 = 0 gives a1,2...k = 0, that is η = 0.

96 CHAPTER 2. ALTERNATING MULTILINEAR FORMSb) On one hand we have seen in a) that it su es to impose the ondi-tion for k + 1 linearly independent 1-forms. On the other hand thehypoyhesis k < n is needed be ause ifη = ϕ1 ∧ · · · ∧ ϕnthen

η ∧ ϕi = 0, i = 1, . . . , nbut η 6= 0. ) In b) we have seen that if η ∧ ϕi = 0 for k + 1 forms of E∗ then η = 0.So if η 6= 0 we have dim(Ker f) ≤ k.d) i) If η 6= 0 is de omposable, η = σ1 ∧ · · · ∧ σk 6= 0 and the 1-formsσ1, . . . , σk are linearly independent and belong to Ker f ; thendimKer f = kii) Let (σ1, . . . , σk) be a basis of Ker f ; we have seen in b) that the onditions η ∧ σi = 0, i = 1, . . . , k imply η = a1...kσ1 ∧ . . .∧ σk andη is de omposable.e) By d) it su es to show that dim Ker f = n−1. To this end we express

η in terms of the basis ϕ1, . . . , ϕn:η = a1ϕ1 ∧ ϕ2 ∧ . . . ∧ ϕn + a2ϕ1 ∧ ϕ2 ∧ . . . ∧ ϕn + . . .+ anϕ1 ∧ ϕ2 ∧ . . . ∧ ϕnand let

ω = b1ϕ1 + b2ϕ2 + . . .+ bnϕn.Putting θ = ϕ1 ∧ ϕ2 ∧ . . . ∧ ϕn we have that ω ∈ Ker f iη ∧ ω = (−1)n−1(a1b1 − a2b2 + · · ·+ anbn)θ = 0a ondition equivalent to

b1a1 − b2a2 + · · ·+ bnan = 0an equation with unknowns (b1, . . . , bn); the spa e of solutions has di-mension n− 1.

2.3. EXTERIOR PRODUCT 97

T Remind that if E is a R-vs a metri is a mapg : E × E → R

(e,v) 7→ g(e,v)bilinear, symmetri and nondegenerated: g(e,v) = 0, ∀v⇒ e = 0.If E is a 2n-dimensional R-vs, a symple ti form is a mapα : E ×E → R

(e,v) 7→ α[e,v]bilinear, antisymmetri and nondegenerated: α(e,v) = 0, ∀v ⇒ e = 0;in other words α ∈ Λ2E, α nondegenerated. A symple ti spa e is a 2n-dimensional R-vs with a symple ti form. Observe that the antisymmetryimplies α[e, e] = 0: in the language of metri s, every ve tor is isotropi .Problem 48:Let E be a 2n-dimensional R-vs, b = (e1, . . . , en; f1, . . . , fn) a basis of E and

(p1, . . . , pn; q1, . . . , qn) its dual basis.a) Show that α = p1 ∧ q1 + · · ·+ pn ∧ qn is nondegenerated.b) Compute α∧ n). . . ∧α.Solution:a) If u = (x1, . . . , xn; y1, . . . , yn)bT , v = (z1, . . . , zn; t1, . . . , tn)b

T are twove tors of E we haveα[u,v] = p1 ∧ q1[u,v] + · · ·+ pn ∧ qn[u,v] =

= det

(x1 z1y1 t1

)+ · · ·+ det

(xn znyn tn

)

98 CHAPTER 2. ALTERNATING MULTILINEAR FORMSTo see α is nondegenerate assume that u satises α[u,v] = 0, ∀v; wewant to see that u = 0. Taking v = (0, . . . , 0; 0, . . . ,i)

1, . . . , 0)bT we have0 = α[u,v] = det

(xi 0yi 1

)= xi,and taking v = (0, . . . ,−

i)

1, . . . , 0; 0, . . . , 0)bT we obtain0 = α[u,v] = det

(xi −1yi 0

)= yithat is, u = 0b) Letting αi = pi ∧ qi, i = 1, . . . , n we have α = α1 + · · · + αn; observethat the 2-forms αi ommute. Then:

α∧ n). . . ∧α = (α1 + · · ·+ αn)∧ n). . . ∧(α1 + · · ·+ αn) =

= n!α1 ∧ · · · ∧ αn = n!p1 ∧ q1 ∧ · · · ∧ pn ∧ qn

Problem 49:Let E be an n-dimensional R-vs and α ∈ Λ2E, α 6= 0; α may be degenerated.a) Show that there are subspa es V,W su h that E = V ⊕ W , α nondegenerated on W (W nonsingular), and α[v,w] = 0,v ∈ V,w ∈W .b) Show there is a basis of E∗, φ = (α1, . . . , αn) su h thatα = α1 ∧ α2 + α3 ∧ α4 + · · ·+ α2r−1 ∧ α2r

r being a positive integer su h that 2r ≤ n.

2.3. EXTERIOR PRODUCT 99Solution:a) Let V = v ∈ E : α[v, e] = 0, ∀e ∈ E, a subspa e alled the radi alof α, and let W be a suplementary spa e of V .The ondition α[v,w] = 0,v ∈ V,w ∈ W is satised by the denitionof V . W is nonsingular be ause if u ∈W is su h that α[u,w] = 0, ∀w ∈W , de ompose e ∈ E, e = v + w,v ∈ V,w ∈W to obtain

α[u, e] = α[u,v + w] = α[u,v] + α[u,w] = 0and then u ∈ V , a ontradi tion.b) i) Choose any non nul ve tor e1 ∈ W ; then there is u ∈ Wsu h that α[e1,u] 6= 0. This u is li of e1, for if u = λe1 thenα[e1,u] = λα[e1, e1] = 0. Let e2 = u

α[e1,u]; with this hoi e wehave α[e1, e2] = 1 and e1, e2 li. Noti e the matrix of α in thisbasis is (

0 1−1 0

)showing that the subspa e 〈e1, e2〉 is nonsingular.ii) Let W1 = w ∈ W : α[w, e] = 0, ∀e ∈ 〈e1, e2〉; we next showthat W = 〈e1, e2〉 ⊕W1 and that W1 is nonsingular. To this endobserve that if v ∈W we havev − α[v, e2]e1 + α[v, e1]e2 ∈W1for

α[v− α[v, e2]e1 + α[v, e1]e2, e1] = α[v, e1] + α[v, e1]α[e2, e1] = 0

α[v− α[v, e2]e1 + α[v, e1]e2, e2] = α[v, e2]− α[v, e2]α[e1, e2] = 0The sum is dire t be ause if u ∈ 〈e1, e2〉 ∩ W1 then α[u, e] =0, ∀e ∈ 〈e1, e2〉, whi h is impossible be ause 〈e1, e2〉 is nonsingular.The subspa e W1 is nonsingular be ause if there is u ∈ W1 su hthat α[u, f ] = 0, ∀f ∈ W1, then α[u,v] = 0, ∀v ∈ W but this isimpossible for W is nonsingular.

100 CHAPTER 2. ALTERNATING MULTILINEAR FORMSiii) W1 is then in the same position as W and we an iterate thepro edure that allowed the onstru tion of W1. In this way weobtain a basis of W , say e1, . . . , e2r satisfyingα[e1, e2] = α[e3, e4] = · · · = α[e2r−1, e2r] = 1We omplete this basis to a basis of E, say

e1, . . . , e2r, e2r+1, . . . , en,and let α1, . . . , αn be its dual. The expression of α in this basis isα = α[e1, e2]α1 ∧ α2 + · · ·+ α[e2r−1, e2r]α2r−1 ∧ α2r =

= α1 ∧ α2 + α3 ∧ α4 + · · ·+ α2r−1 ∧ α2r

2.4 Orientation and volume2.4.1 OrientationT In the plane we an turn lo kwise and anti lo kwise. In the spa e thereare two kinds of s rewdriver: ontinental and british. To generalize those on epts let E be an R-vs and e = (e1, . . . , en), u = (u1, . . . ,un) two basesrelated by u = eC.We say that e, u have the same orientation if detC > 0 and oppositeorientation if detC < 0. A gure:

e1

e2

e3

u1

u2

u3

2.4. ORIENTATION AND VOLUME 101The bases e and u in the gure have opposite orientations for if u = eCwe haveC =

0 1 01 0 00 0 1

, detC = −1 < 0

Problem 50:Show:a) Having the same orientation is an equivalen e relation in the set ofbases of E.b) There are two equivalen e lasses. ) A C-vs annot be oriented.Solution:a) Let e, f, h be bases of E.• e ∼ e, for e = eI and det I = 1 > 0.• e ∼ f means that f = eC, detC > 0. Then e = fC−1, detC−1 >

0 and f ∼ e.• e ∼ f , f ∼ h mean that f = eC, h = fD with detC > 0, detD >

0. Then h = eCD and det(CD) = detC ·detD > 0, that is e ∼ h.b) Let e = (e1, . . . , en) be a basis, e′ = (−e1, e2, . . . , en) and let e = e′CwhereC =

−11 . . .

1

.Let u = (u1, . . . ,un) be any other basis with u = eD.i) If detD > 0 then u ∼ e,

102 CHAPTER 2. ALTERNATING MULTILINEAR FORMSii) If detD < 0 we have u = eD = e′CD and det(CD) = (detC)(detD) =− detD > 0 and u ∼ e′. ) Now detC is a omplex number and we know that C annot be orderedas a eld of numbers.

T Let E be an R-vs; the set of all equivalent bases is an orientation andwe have shown that there are two orientations. We say that E is oriented ifwe have hoosen one of the two orientations. The hoosen lass O is alledthe positive orientation and its bases are alled positive bases. The other lass is the negative orientation and ontains the negative bases, of ourse.Problem 51: Orientation through an n-form.Let E be an n-dimensional R-vs and ω ∈ ΛnE, ω 6= 0; show that the set ofbases su h that ω[e] > 0 is an orientation.Solution:Let e and u be two bases with u = eC and ω[e] > 0; thenω[u] = (detC)ω[e]and we see that ω[u] > 0 i detC > 0.This doesn't mean that the bases e su h that ω[e] > 0 are the positivebases; we still have freedom to sele t as positive those bases su h that ω[e] <

0.Problem 52:Let E,F be two oriented n-dimensional vs. An isomorphism f : E → Fpreserves orientation if the image of a positive basis of E is a positive basisof F . Being given a matrix of f , how an we know whether f preservesorientation?

2.4. ORIENTATION AND VOLUME 103Solution:Let e = (e1, . . . , en), u = (u1, . . . ,un) be positive bases of E and F respe -tively and A the matrix of f in those bases, f(e) = uA. The matrix Ais as well the hange of basis matrix from the positive basis u to the basisuA. The latter will be a positive basis i detA > 0 and we see that underthe hypothesis that both bases are positive, f is orientation preserving idetA > 0.If e′ = eC, u′ = uD are bases and A′ is the matrix of f in those basesf(e′) = u′A′, then:

f(e) = f(e′C−1) = f(e′)C−1 = u′A′C−1 = uDA′C−1that is A = DA′C−1. We obtainf is orientation preserving ⇔ (detD)(detC)(detA′) > 0This says that if e′, u′ are both positive or both negative, f preserves ori-entation i detA′ > 0. If the bases hoosen to represent f have oppositeorientation f preserves orientation i detA′ < 0.

2.4.2 Volume and volume formVolumeT Let E be an n-dimensional eu lidian spa e and v = (v1, . . . ,vk), k ≤ n,a list of ve tors of E. The parallelepiped generated by those ve tors is

P = P (v1, . . . ,vk) = x = t1v1 + · · ·+ tkvk, 0 ≤ ti ≤ 1

104 CHAPTER 2. ALTERNATING MULTILINEAR FORMS

v1

v2 v2

v1

v3

v1 v2 v3, , )P(v1 2vP( , )

,

and its k-volume in terms of lengths and angles isVol(P ) =√

det(vT · v)(See [Xa 1, p.288 and following.)Problem 53: Volume in terms of omponents.Let v = (v1, . . . ,vk), k ≤ n be given in the basis e = (e1, . . . , en) by v = eA,

A an (n, k) matrix. Find the volume of P (v1, . . . ,vk) when:a) e is any basis.b) e is an on basis. ) e is an on basis and the parallelepiped has maximum dimension.Solution:a) Vol(P ) =√

det(vT · v) =√

det(AT eT · eA) =√

det(ATGA)where G = eT · e is the metri matrix in the basis e.

2.4. ORIENTATION AND VOLUME 105b) If e is an on basis, G = I andVol(P ) =√

det(ATA) ) If the parallelepiped has maximum dimension k = n, the matrix A issquare and: Vol(P ) =√

det 2A = | detA|

Volume formT Let E be an oriented n-dimensional vs.• Let E be eu lidian; we dene the volume form of E, a signed volumeof n-parallelepipeds:

ωE[v1, . . . ,vn] =

√det(vT · v) if v is a positive basis

−√

det(vT · v) in any other aseThen ωE ∈ ΛnE, but that is not immediate. For a very lean proof youmight want to see page 81&following in the ex ellent book [Jan.Noti e that if e is a positive on basis and v = eA, detA > 0 is anotherpositive basis, thenωE [v1, . . . ,vn] =

√det(vT · v) = | detA| = detAwhile if v is a negative basis then

ωE[v1, . . . ,vn] = −√

det(vT · v) = −| detA| = detASo in any ase ωE[v1, . . . ,vn] = detA. Obviously the result is also trueif the v are ld.• Now let E be an orhogonal spa e; the determinant of the metri matrixof a proper subspa e may vanish (see p.41) and the on ept of volumedoesn't apply. However the determinant of the metri matrix of n live tors doesn't vanish, but it may be negative. We dene the volumeform through

ωE[v1, . . . ,vn] =

−√| det(vT · v) | if v is a negative basis√| det(vT · v) | in any other ase

106 CHAPTER 2. ALTERNATING MULTILINEAR FORMSIf e is a positive on basis and v = eA, A an (n, n) matrix, we have√| det(vT · v)| =

√| det(AT IrsA)| =

√| det 2A det Irs | = | detA|and the value of the volume form applied to v is

ωE[v1, . . . ,vn] = (detA)ωE[e1, . . . , en] = (detA)√| det Irs | = detA

Problem 54: Chara terization of the volume form.a) Let E be an oriented n-dimensional orthogonal spa e. Show that thevolume form ωE ∈ ΛnE is the only n-form su h that ωE[e1, . . . , en] = 1for any positive on basis e = (e1, . . . , en).b) Let E be a real n-dimensional vs and ω ∈ ΛnE, ω 6= 0 an n-form.Show: ω is the volume form for some eu lidian s alar produ t andsome orientation.Solution:a) As dim ΛnE = 1 any form η ∈ ΛnE, η 6= ωE satises η = aωE, a 6= 1and then has a value dierent from that of ωE on a basis; this showsthe uni ity. If e is any on positive basis we have (r = n in the eu lidian ase):ωE[e1, . . . , en] =

√| det(eT · e)| =

√| det(Irs )| = 1.b) We have seen in p.102 that if we dene a basis e to be positive when

ω[e] > 0 we give E an orientation. If e = (e1, . . . , en) is a positive basisω[e1, . . . , en] > 0 and it su es to multiply e1 by a s alar to have

ω[e1, . . . , en] = 1Dene this new basis as orthonormal; that is the metri matrix in thisbasis is I. If u = eC is another basis, as the new metri matrix isGu = CT IC, the basis u will be an on basis i CTC = I.

2.4. ORIENTATION AND VOLUME 107Then, given a metri and an orientation, there is an asso iated volumeform. And if we start from a nonvanishing n-form, it is the volume form orresponding to a ertain orientation and metri .Problem 55: Formula volume form.Let E be an oriented orthogonal spa e, e = (e1, . . . , en) a positive on basis,w = (ω1, . . . , ωn) its dual basis and ωE the volume form. Show:a) ωE = ω1 ∧ · · · ∧ ωn.b) If u = (u1, . . . ,un) is any positive basis, not ne essarily on, and α =

(α1, . . . , αn) its dual basis, thenωE =

√|gu|α1 ∧ · · · ∧ αnwhere gu = det(uT · u). ) If u = (u1, . . . ,un) is a negative basis, then

ωE = −√|gu|α1 ∧ · · · ∧ αnSolution:a) From the ara terization of the volume form in the pre eding problemit su es to show that ω1 ∧ · · · ∧ ωn takes value 1 when applied to apositive on basis; take for instan e the basis e :

ω1 ∧ · · · ∧ ωn[e1, . . . , en] = det

ω1[e1] . . . ω1[en]. . . . . . . . .ωn[e1] . . . ωn[en]

= det I = 1b) We want to see that √|gu|α1 ∧ · · ·∧αn[e] = 1. If e = uC, detC > 0 wehave√|gu|α1 ∧ · · · ∧ αn[e] = (detC)

√|gu|α1 ∧ · · · ∧ αn[u] = (detC)

√|gu|On another hand, reminding that e is on positive and u is positive,

Irs = Ge = CTGuC

108 CHAPTER 2. ALTERNATING MULTILINEAR FORMSand taking determinants, absolute values, and square roots we obtain1 = (detC)

√|gu|and then √

|gu|α1 ∧ · · · ∧ αn[e] = 1 ) If u is a negative basis, detC < 0 and1 = −(detC)

√|gu|so

−√|gu|α1 ∧ · · · ∧ αn[e] = 1

Problem 56:a) Let E be an oriented eu lidian vs and ωE its volume form; onsider aswell Rn with the standard s alar produ t and the standard orientation.Show that if an isomorphism f : Rn → E is orientation preserving andpreserves the s alar produ t, then f ∗ωE = det.b) Let f : E → E be a linear map of the n-dimensional vs E onto itself.Then ΛnE is a 1-dimensional vs and f ∗ : ΛnE → ΛnE must be theprodu t by a s alar c. Show that c = det f .Solution:a) Let u = (u1, . . . ,un) be the anoni al basis of Rn and e = f(u) =(e1, . . . , en) the image basis, an on positive basis.Then

(f ∗ωE)[u] = ωE [fu] = ωE [e] = 1But det is the volume form in Rn and sof ∗ωE = detb) det f is the determinant of any matrix representing f , fe = eA. If

ω ∈ ΛnE we have(f ∗ω)[e] = ω[fe] = ω[eA] = (detA)ω[e]⇒ f ∗ω = (detA)ω

2.4. ORIENTATION AND VOLUME 1092.4.3 Contra tionContra tionT Let E be an n-dimensional vs, T ∈ T pE a multilinear form and e =(e1, . . . , ek), k ≤ p a list of ve tors of E. The ontra tion of e with T(e1, . . . , ek)yT is

((e1, . . . , ek)yT )[u1, . . . ,up−k] := T [u1, . . . ,up−k, e1, . . . , ek]In an oriented orthogonal spa e E with metri g and volume form ω ∈ΛnE we an make the ontra tion of k ve tors (e1, . . . , ek), k ≤ n with thevolume form, thus obtaining an (n− k)-alternating form:

((e1, . . . , ek)yω)[u1, . . . ,un−k] = ω[u1, . . . ,un−k, e1, . . . , ek]With the idea of ontra tion, the rst anoni al isomorphism an be writ-ten thusE

→ Λ1E

e 7→ eyg

Problem 57: Pra ti ing ontra tion in R3.Let R3 be oriented by the anoni al basis (i, j,k) and let (π1, π2, π3) be thedual basis. The volume form is ω = π1 ∧ π2 ∧ π3 (we an write as well(dx, dy, dz) for the dual anoni al basis and ω = dx∧ dy ∧ dz for the volumeform). Dis uss the ontra tion of ω with a s alar, and with one, two or threeve tors, from the geometri al as well as from the algebrai al viewpoint.Solution:a) Dene the ontra tion with a s alar p as the three-form pyω = pω:

(pyω)[u1,u2,u3] = pω[u1,u2,u3]

110 CHAPTER 2. ALTERNATING MULTILINEAR FORMSi) Geometri ally it is p times the signed volume of the parallelepipedP (u1,u2,u3):

u3

2u

u1u1 2u u

3, , )P(

ii) Algebrai ally if e is a positive on basis and u = eA then pω[u1,u2,u3] =p detA.b) The ontra tion with a ve tor e is the 2-form

(eyω)[u1,u2] = ω[u1,u2, e]i) Geometri ally this is the signed volume of a parallelepiped withbasis the parallelogram P = P (u1,u2) and edge e.When e is unitary and perpendi ular to P , ω[u1,u2, e] = Area (P )if (u1,u2, e) is a positive basis and ω[u1,u2, e] = −Area (P ) if(u1,u2, e) is a negative basis. Noti e we are omputing signedareas using the volume form.When e is unitary but the surfa e is not perpendi ular to e then,letting h be the heigth taken from the basis of the parallelepipedP we have

ω[u1,u2, e] = Area(P )× hif (u1,u2, e) is a positive basis and we must put a negative sign if(u1,u2, e) is a negative basis.h being the heigth over the basis P .Physi ally we may view ω[u1,u2, e] as the ux of the onstantxed ve tor eld e a ross the variable oriented surfa e P (u1,u2).A ordingly eyω is alled the ux 2-form (asso iated to e).

2.4. ORIENTATION AND VOLUME 111ii) Algebrai ally the ontra tion is a 2-form of R3; it has an expressioneyω = (eyω)[i, j]π1 ∧ π2 + (eyω)[j,k]π2 ∧ π3 + (eyω)[k, i]π3 ∧ π1Let e = (a, b, c); then the oe ients are

(eyω)[i, j] = det

1 0 a0 1 b0 0 c

= cand similarly we obtain(eyω)[j,k] = a, (eyω)[k, i] = bthat is

(a, b, c)yω = aπ2 ∧ π3 + bπ3 ∧ π1 + cπ1 ∧ π2 ) The ontra tion with (e1, e2) is the 1-form(e1, e2)yω[u] = ω[u, e1, e2]i) Geometri ally this is the signed volume of a parallelepiped withxed basis P = P (e1, e2) and variable edge uii) Algebrai ally the ontra tion is a 1-form and an be expressedthus

(e1, e2)yω = ((e1, e2)yω)[i]π1+((e1, e2)yω)[j]π2+((e1, e2)yω)[k]π3Let (e1, e2) = (i, j,k)

a1

1 a12

a21 a2

2

a31 a3

2

; then(e1, e2)yω = det

1 a1

1 a12

0 a21 a2

2

0 a31 a3

2

π1 + det

0 a1

1 a12

1 a21 a2

2

0 a31 a3

2

π2 + det

0 a1

1 a12

0 a21 a2

2

1 a31 a3

2

π3 =

= det

∣∣∣∣a2

1 a22

a31 a3

2

∣∣∣∣ π1 − det

∣∣∣∣a1

1 a12

a31 a3

2

∣∣∣∣ π2 + det

∣∣∣∣a1

1 a12

a21 a2

2

∣∣∣∣ π3And this last equality shows that(e1, e2)yω = (e1 × e2) = (e1 × e2)ygor in other words ♯((e1, e2)yω) = e1 × e2.

112 CHAPTER 2. ALTERNATING MULTILINEAR FORMSd) The ontra tion with (e1, e2, e3) is the 0-form (a s alar)(e1, e2, e3)yω = ω[e1, e2, e3]i) Geometri ally this is the signed volume of the parallalepipedP (e1, e2, e3),a number.ii) Algebrai ally, if u is a positive on basis and e = uA, then ω[e1, e2, e3] =

detA.2.4.4 Volume form of subspa esProblem 58: Volume form of a (n− 1)-subspa e.Let E be an n-dimensional orthogonal oriented spa e and ω its volume form.a) Let F ⊂ E be an (n − 1)-dimensional nonsingular subspa e and n ∈

F⊥,n · n = ±1 an on basis of F⊥. Show that(nyω)|F an be seen as the volume form of F .b) Let e = (e1, . . . , en) be an on positive basis of E, w = (ω1, . . . , ωn) itsdual basis and n = n1e1+· · ·+nnen. Find the expression in oordinatesof nyω. ) Appli ation: nd the volume form of the subspa e

F = 〈(1, 0, 2), (−1, 1, 0)〉 ⊂ R3.Solution:a) On e n has been hoosen, dene a basis v1, . . . ,vn−1 of F to be positivewhenever v1, . . . ,vn−1,n is a positive basis of E.

2.4. ORIENTATION AND VOLUME 113n

F

E

Now it remains to show that nyω takes value 1 on on positive bases ofF ; let u = (u1, . . . ,un−1) be one of su h bases. We have

(nyω)[u1, . . . ,un−1] = ω[u1, . . . ,un−1,n] = 1,be ause u1, . . . ,un−1,n is a positive on basis of E and ω is the volumeform. There is an ambiguity left in the hoi e of normal ve tor thattranslates to a hange of the hoosen orientation in F and to a hangeof sign in nyω.b) We express nyω ∈ Λn−1E in the basis∧

(ω1 ∧ω2 ∧ · · · ∧ ωn), (ω1∧∧ω2 ∧ · · · ∧ ωn), . . . , (ω1 ∧ ω2 ∧ · · · ∧

∧ωn).

• The rst omponent is(nyω)[e2, . . . , en] = ω[e2, . . . , en,n] =

= det

0 . . . 0 n1

1 . . . 0 n2

. . . . . . . . . . . .0 . . . 0 nn−1

0 . . . 1 nn

= (−1)n+1n1

• The se ond omponent is(nyω)[e1, e3. . . . , en] = ω[e1, e3. . . . , en,n] =

= det

1 0 . . . 0 n1

0 0 . . . 0 n2

0 1 . . . . . . n3

. . . . . . . . . . . . . . .0 0 . . . 0 nn−1

0 0 . . . 1 nn

= (−1)n+2n2

114 CHAPTER 2. ALTERNATING MULTILINEAR FORMSet . We havenyω = (−1)n+1(n1ω2 ∧ · · · ∧ ωn − n2ω1 ∧ ω3 ∧ · · · ∧ ωn + . . . ) =

= (−1)n+1

n∑

j=1

(−1)j−1njω1 ∧ · · · ∧∧ωj ∧ · · · ∧ ωn ) A unitary normal ve tor to F is n = 1

3(2, 2,−1); we orient F as in a).Then, if π1, π2, π3 denotes the dual of the anoni al basis we obtain

ωF = nyω =2

3π2 ∧ π3 −

2

3π1 ∧ π3 −

1

3π1 ∧ π2

Problem 59: Volume form of a k-subspa e. (See [Burg, p.80.)Let E be an n-dimensional orthogonal oriented spa e and ω its volume form.a) Let F ⊂ E be a k-dimensional nonsingular subspa e and n1, . . . ,nn−kan on basis of F⊥. Show that(n1, . . . ,nn−k)yω|F an be seen as the volume form of F .b) Let e = (e1, . . . , en) be an on positive basis of E, w = (ω1, . . . , ωn) itsdual basis and (n1, . . . ,nn−k) = eA where A is the (n, n − k) matrixof the omponents of the n′s. Find the expression in oordinates of

(n1, . . . ,nn−k)yω. ) Appli ation: nd a volume form of the subspa e F = 〈v1,v2〉 ⊂ R4where v1 = (1, 0, 1, 1),v2 = (1,−1, 0, 1).Solution:a) On e (n1, . . . ,nn−k) has been hoosen, dene a basis (v1, . . . ,vk) of Fto be positive whenever (v1, . . . ,vk,n1, . . . ,nn−k) is a positive basis ofE.

2.4. ORIENTATION AND VOLUME 115Now it remains to show that (n1, . . . ,nn−k)yω takes value 1 on onpositive bases of F ; let u = (u1, . . . ,uk) be one of su h bases. We have((n1, . . . ,nn−k)yω)[u1, . . . ,uk] = ω[u1, . . . ,uk,n1, . . . ,nn−k] = 1,be ause u1, . . . ,uk,n1, . . . ,nn−k is an on positive basis of E and ω isthe volume form.There is only a hange of sign in the volume form of F when we hangethe on basis of F⊥:let (n1, . . . , nn−k) = (n1, . . . ,nn−k)C be another on basis of F⊥, Cbeing the (n−k, n−k) hange of basis matrix. We know that CT IrsC =

Irs and assume that detC > 0; then detC = 1 and we have:((n1, . . . , nn−k)yω)[v1, . . . ,vk] = ω[v1, . . . ,vk, n1, . . . , nn−k] =

= (detC)ω[v1, . . . ,vk,n1, . . . ,nn−k] =

= (n1, . . . ,nn−k)yω[v1, . . . ,vk]and as long as the new orthogonal basis is in the same orientation asthe old one, there is no ambiguity in the volume form. The volumeform hanges the sign if we hoose the new basis su h that detC < 0.b) As the volume form isω = ω1 ∧ · · · ∧ ωnwe have

(n1, . . . ,nn−k)yω =∑

Iր((n1, . . . ,nn−k)yω)[ei1, . . . , eik ]ωi1 ∧ · · · ∧ ωik =

=∑

Iրω[ei1 , . . . , eik ,n1, . . . ,nn−k]ωi1 ∧ · · · ∧ ωikWe ompute the omponents:

ω[ei1, . . . , eik ,n1, . . . ,nn−k] = det

ω1[ei1 ] . . . ω1[eik ] ω1[n1] . . . ω1[nn−k]. . . . . . . . . . . . . . . . . .

ωn[ei1 ] . . . ωn[eik ] ωn[n1] . . . ωn[n1]

=

= det

δ1i1

. . . δ1ik

a11 . . . a1

n−k. . . . . . . . . . . . . . . . . .δni1 . . . δnik an1 . . . ann−k

116 CHAPTER 2. ALTERNATING MULTILINEAR FORMS ) The on basis of F⊥, n1 = 1√3(−1,−1, 1, 0),n2 = 1√

15(2,−1, 1,−3),satises det(v1,v2,n1,n2) > 0 and v1,v2 is a positive basis of F .A ording to b) we have six oe ients of (n1,n2)yω to ompute :

((n1,n2)yω)[e1, e2] =

1 0 − 1√3

2√15

0 1 − 1√3− 1√

15

0 0 1√3

1√15

0 0 0 − 3√15

= − 3√

45= − 1√

5

((n1,n2)yω)[e1, e3] =

1 0 − 1√3

2√15

0 0 − 1√3− 1√

15

0 1 1√3

1√15

0 0 0 − 3√15

= − 3√

45= − 1√

5

((n1,n2)yω)[e1, e4] =

1 0 − 1√3

2√15

0 0 − 1√3− 1√

15

0 0 1√3

1√15

0 1 0 − 3√15

= 0

((n1,n2)yω)[e2, e3] =

0 0 − 1√3

2√15

1 0 − 1√3− 1√

15

0 1 1√3

1√15

0 0 0 − 3√15

=

3√45

=1√5

((n1,n2)yω)[e2, e4] =

0 0 − 1√3

2√15

1 0 − 1√3− 1√

15

0 0 1√3

1√15

0 1 0 − 3√15

=

3√45

=1√5

((n1,n2)yω)[e3, e4] =

0 0 − 1√3

2√15

0 0 − 1√3− 1√

15

1 0 1√3

1√15

0 1 0 − 3√15

=

3√45

=1√5and we have

(n1,n2)yω =1√5(−π1 ∧ π2 − π1 ∧ π3 + π2 ∧ π3 + π2 ∧ π4 + π3 ∧ π4)

2.4. ORIENTATION AND VOLUME 117As a he k((n1,n2)yω)[(1, 0, 1, 1), (1,−1, 0, 1)] =

1√5(1 + 1 + 1 + 1 + 1) =

√5and this is the area of the parallelogram P ((1, 0, 1, 1), (1,−1, 0, 1)) thatwe an ompute through the metri matrix

G =

(3 22 3

), g = 5and the volume is found again to be √5.

2.4.5 Cross produ tT Let us remind the basi properties of the ordinary ve tor produ t. Ife, e′ ∈ R3 the 'geometri ' denition of e× e′ uses the s alar produ t and theorientation:a) Using the metri it tells thati) e× e′ is orthogonal to e and to e′.ii) | e× e′ | is the area of the parallelogram P (e, e′).b) Using the orientation we impose (e, e′, e× e′) to be a positive basis.Noti e that:a) e× e′ lives in the one-dimensional spa e orthogonal to 〈e, e′〉.b) The measure (length) of e× e′ is the measure (area) of P (e, e′). ) e× e′ orients the orthogonal spa e.On another side we have seen in p.111 that e× e′ seen as a 1-form is

(e× e′) = (e, e′)yωand this leads to a slight generalization of the ve tor produ t in orthogonalspa es.

118 CHAPTER 2. ALTERNATING MULTILINEAR FORMSProblem 60: Cross-produ t.In a 3-dimensional oriented orthogonal spa e E with volume form ω we denea ' ross produ t'e× e′ = ♯((e, e′)yω).Show that it isa) Bilinear.b) Anti ommutative. ) Vanishing i both ve tors are linearly dependent.d) Orthogonal to e and to e′.e) If E is eu lidian and e× e′ 6= 0, then e× e′, e, e′ is a positive basis.f) If E is eu lidian then |e×e′| = |e||e′| sin θ, the area of the parallelepiped

P (e, e′).Solution:An equality between ve tors is equivalent to an equality between linear forms(through the isomorphism ). And an equality between linear forms may beestablished showing it to be true for ea h ve tor v ∈ E.a) (λe)× e′ = λ(e× e′) = e× (λe′) for((λe)× e′) · v = ω[v, λe, e′] = λω[v, e, e′] = λ(e× e′) · v,and (e + e′)× f = e× f + e′ × f for

((e + e′)× f) · v = ω[v, e + e′, f ] =

= ω[v, e, f ] + ωE[v, e′, f ] =

= (e× f) · v + (e′ × f) · vOf ourse analogous properties hold for the se ond term in the produ t.Everything omes from the multilinearity of ω.b) (e′ × e) · v = ω[v, e′, e] = −ω[v, e, e′] = −(e× e′) · v.

2.4. ORIENTATION AND VOLUME 119 ) If e′ = λe then(e× e′) · v = ω[v, e, λe] = 0, ∀vand if e, e′ are li there is a v ∈ E su h that e, e′,v is a basis and then

(e× e′) · v = ω[e, e′,v] 6= 0d) (e× e′) · e = ωE[e, e′, e] = 0, (e× e′) · e′ = ω[e, e′, e′] = 0 .e) It su es to show that ω[e× e′, e, e′] > 0. But an eu lidian metri ispositive denite andω[e× e′, e, e′] = (e× e′) · (e× e′) = |e× e′|2 > 0 .f) From the pre eding point and taking into a ount that e, e′, e× e′ is apositive basis we have (see p.105):

|e× e′|2 = ω[e× e′, e, e′] =

√√√√√det

|e× e′)|2 0 0

0 |e|2 e · e′0 e · e′ |e′|2

=

=√|e× e′|2(|e|2|e′|2 − (e · e′)2)and obtain

|e× e′| =√|e|2|e′|2 − (e · e′)2 =

√|e|2|e′|2 sin 2θ = |e||e′| sin θ .

Problem 61: Components of a ross-produ t.a) Let E be a 3-dimensional oriented orthogonal spa e and ω its volumeform. Compute the omponents of e × e′ = ♯(ω[ . , e, e′]) in a positiveog basis u = (u1,u2,u3).b) Referring to point e) in the pre eding problem, assume E is properlynoneu lidian and e×e′ 6= 0; prove that ω[e× e′, e, e′] an take positive,negative or null values.

120 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:a) Let ui · ui = λi 6= 0; and v = a1u1 + a2u2 + a3u3. Then v · ui = aiλiand we have the expression of v

v =1

λ1

(v · u1)u1 +1

λ2

(v · u2)u2 +1

λ3

(v · u3)u3Applying this to e× e′ we havee× e′ =

1

λ1((e× e′) · u1)u1 +

1

λ2((e× e′) · u2)u2 +

1

λ3((e× e′) · u3)u3Then if e = u

xyz

, e′ = u

x′

y′

z′

, and g = det(uT · u), the rst omponent of the ross produ t is1

λ1(e× e′) · u1 = 1

λ1ω[u1, e, e

′] =

=1

λ1det

1 x x′

0 y y′

0 z z′

ω[u1,u2,u3] =

= det

1λ1

0 0

x y zx′ y′ z′

√|g|and the other two are1

λ2

(e× e′) · u2 = det

0 1

λ20

x y zx′ y′ z′

√|g|

1

λ3(e× e′) · u3 = det

0 0 1

λ3

x y zx′ y′ z′

√|g|We an sum up those results in the formulae× e′ =

√|g|det

1

λ1u1

1

λ2u2

1

λ3u3

x y zx′ y′ z′

2.4. ORIENTATION AND VOLUME 121that redu es to the familiare× e′ = det

u1 u2 u3

x y zx′ y′ z′

in the ase of an on positive basis in a eu lidian spa e.b) Assume E noneu lidian with metri Gu =

1

1−1

From a) we haveω[e× e′, e, e′] = (e× e′) · (e× e′) =

= (yz′ − y′z)2 + (x′z − xz′)2 − (xy′ − x′y)2Theni) If z = z′ = 0, y = x′ = 0, x = 1, y′ = 1, that is e = u1, e′ = u2then (u1 × u2) · (u1 × u2) = −1.ii) If x = x′ = 0, y′ = z = 0, y = 1, z′ = 1, that is e = u2, e′ = u3then (u2 × u3) · (u2 × u3) = 1.iii) If x = z = 0 we obtain (e× e′) · (e× e′) = (yz′)2 − (x′y)2 = 0.

Now we go further and dene an extended ross-produ t..Problem 62: Extended ross-produ t.Let E be an n-dimensional oriented orthogonal spa e with volume form ω.Show that we an dene a ve tor produ t of n − 1 ve tors e1 × · · · × en−1and that it has properties similar to those of the ordinary ve tor produ t. Inparti ular in the eu lidian ase nd a formula for | e1 × · · · × en−1 |.

122 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:As in problem 60 we dene the produ t as a linear form and then 'sharpen'it to a ve tor:e1 × · · · × en−1 = ♯((e1, . . . , en−1)yω)and we have the formula for a mixed produ t:

(e1 × · · · × en−1) · u = ω[u, e1, . . . , en−1]Be ause of the denition through the volume form that is multilinear andalternating it is lear that this ross produ t is multilinear and alternatingin its terms; in parti ular it vanishes i the ve tors are linearly dependent.The produ t is perpendi ular to ea h of the terms:(e1 × · · · × en−1) · ei = ω[ei, e1, . . . , en−1] = 0.If E is eu lidian the basis e1 × · · · × en−1, e1, . . . , en−1 is a positive basisfor

ω[e1 × · · · × en−1, e1, . . . , en−1] =| e1 × · · · × en−1 |2> 0Moreover| e1 × · · · × en−1 |2 = ω[e1 × · · · × en−1, e1, . . . , en−1] =

=

√√√√√√√√det

| e1 × · · · × en−1 |2 0 . . . 00 e1 · e1 . . . e1 · en−1... ... ...0 en−1 · e1 . . . en−1 · en−1

=√g | e1 × · · · × en−1 |,where g = det((e1, . . . , en−1)

T · (e1, . . . , en−1)), and so| e1 × · · · × en−1 |=

√g ,the volume of the (n−1)-dimensional parallalepiped generated by the ve tors.

2.4. ORIENTATION AND VOLUME 123Problem 63: Components of a generalized ross-produ t.Let E be an n-dimensional oriented orthogonal spa e and ω its volume form.Compute the omponents of e1×· · ·×en−1 = ♯((e1, . . . , en−1)yω) in a positiveog basis u = (u1, . . . ,un).Solution:Let λi = ui · ui and ei =∑n

j=1 xjiuj . Then

e1×· · ·×en−1 = ((e1×· · ·×en−1)·u1)1

λ1

u1+· · ·+((e1×· · ·×en−1)·un)1

λnunWe have

(e1 × · · · × en−1) · u1 = ω[u1, e1, . . . , en−1] =

= det

1 x11 . . . x1

n−1

0 x21 . . . x2

n−1

. . . . . . . . . . . .0 xn1 . . . xnn−1

ω[u1, . . . ,un] =

=√| g | det

1 x11 . . . x1

n−1

0 x21 . . . x2

n−1

. . . . . . . . . . . .0 xn1 . . . xnn−1

et . where g = det(uT · u). This an be rewritten as

e1 × · · · × en−1 =√| g | det

1

λ1

u11

λ2

u2 . . .1

λnun

x11 x2

1 . . . xn1. . . . . . . . . . . .x1n−1 x2

n−1 . . . xnn−1

2.4.6 IsomorphismsT Let E be an oriented orthogonal spa e with metri g and volume formω. Consider k ve tors of E, (e1, . . . , ek); through ontra tion we an assign

124 CHAPTER 2. ALTERNATING MULTILINEAR FORMSto them a (n− k)-form:E× k). . . ×E ck→ Λn−kE(e1, . . . , ek) 7→ (e1, . . . , ek)yωOn another hand using we an assign to them a k-form:E× k). . . ×E ∧···∧→ ΛkE(e1, . . . , ek) 7→ e1 ∧ · · · ∧ eka) When k = 1 theese maps are

Eւ ց c1Λ1E Λn−1EWe know that is an isomorphism; the ontra tion c1 is linear and isalso an isomorphism. To see this just noti e that it is linear and that

dim Λn−1E =

(n

n− 1

)= n = dimE. It su es to show that it isinje tive; to this end let e 6= 0 and omplete it to a basis of E say

u1, . . . ,un−1, e. Then(eyω)[u1, . . . ,un−1] = ω[u1, . . . ,un−1, e] 6= 0,and e isn't in the kernel of c1. This isomorphism shows that everyalternating form in Λn−1E is eyω for some e ∈ E.Then we have an isomorphism

Λ1Eh→ Λn−1E

eyg 7→ eyωthat makes the following diagram ommutativeE

ւ ց c1

Λ1Eh→ Λn−1E

2.4. ORIENTATION AND VOLUME 125b) When k = n− 1 we onsider the following diagram:E× n−1. . . ×E

∧ · · · ∧ ւ ց cn−1

Λn−1Eh−1

→ Λ1E¾Is this diagram ommutative (h is the isomorphism of a)? The answeris the ontent of next problem. ) When 1 < k < n−1 the natural question is the existen e of an isomor-phism h making the following diagram ommutative.E× k. . . ×E

∧ · · · ∧ ւ ց ck

ΛkEh→ Λn−kEThe answer is in problem 83.

Problem 64:Let E be an oriented eu lidian spa e with metri g and volume form ω; lete = (e1, . . . , ek) be a list of ve tors of E. We know that any (n− 1)-form iseyω; ¾whi h is e for the (n− 1)-form e1∧ · · ·∧ en−1? Prove or disprove theguess

e1 ∧ · · · ∧ en−1 = (e1 × · · · × en−1)yωSolution:We may assume e1, . . . , en−1 li for otherwise both terms in the equalityvanish. Choose in 〈e1, . . . , en−1〉⊥ a ve tor u su h that e1, . . . , en−1,u is apositive basis. Now we prove the equality he king it on lists of n−1 ve tors hoosen among those in the basis; two ases appear:a)e1∧· · ·∧en−1[e1, . . . , en−1] = det

e1 · e1 . . . e1 · en−1

. . . . . . . . .en−1 · e1 . . . en−1 · en−1

= detGe

126 CHAPTER 2. ALTERNATING MULTILINEAR FORMSand the other member is(e1 × · · · × en−1)yω[e1, . . . , en−1] =

= ω[e1, . . . , en−1, e1 × · · · × en−1] =

= (−1)n−1ω[e1 × · · · × en−1, e1, . . . , en−1] =

= (−1)n−1 | e1 × · · · × en−1 |2= (−1)n−1 detGeb)e1 ∧ · · · ∧ en−1[e1, . . . ,

i)u, . . . , en−1] =

= det

e1 · e1 . . . e1 · u . . . e1 · en−1

. . . . . . . . . . . . . . .en−1 · e1 . . . en−1 · u . . . en−1 · en−1

= 0be ause u ∈ 〈e1, . . . , en−1〉⊥ and then the i-th olumn is nul. Also((e1 × · · · × en−1)yω)[e1, . . . ,

i)u, . . . , en−1] =

= ω[e1, . . . ,i)u, . . . , en−1, e1 × · · · × en−1] = 0be ause u is in the dire tion of e1×· · ·×en−1 and both entries are ld.Consequently the form vanishes.We have proved that

e1 ∧ · · · ∧ en−1 = (−1)n−1(e1 × · · · × en−1)yωThen the isomorphism that makes the diagram ommutative is (−1)n−1h−1:E× n−1. . . ×E

ւ ց cn−1

Λn−1E(−1)n−1h−1

→ Λ1Ee1 ∧ · · · ∧ en−1 = (e1, . . . , en−1)yω =

= (−1)n−1(e1 × · · · × en−1)yω 7→ (e1 × · · · × en−1)yg =

2.4. ORIENTATION AND VOLUME 127Problem 65:Let E be an oriented orthogonal spa e with metri g and volume form ω; lete = (e1, . . . , ek) be a list of ve tors of E. We know that any (n− 1)-form iseyω; ¾whi h is e for a given η ∈ Λn−1E? Apply the result to the (n−1)-forme1 ∧ · · · ∧ en−1 and ompare with the pre eding problem.Solution:As now omes unde onsideration the ase of E being possibly noneu lidian,the line of the pre eding problem doesn't apply be ause if

F = 〈e1, . . . , en−1〉is a singular spa e F ∩ F⊥ 6= 0. We must take another route suggestedby the present problem.Let u = (u1, . . . ,un) be a positive on basis; remind this means thatui · uj = 0 if i 6= j and ui · ui = ǫi = ±1. We are looking for a ve tor

e = λ1u1 + · · ·+ λnunsu h thatη[v1, . . . ,vn−1] = ω[v1, . . . ,vn−1, e], ∀(v1, . . . ,vn−1)This equality is satised i it is true for any sele tion of n− 1 ve tors ina basis. So (hat = omission):

η[u1,u2, . . . ,un] = ω[u1,u2, . . . ,un, λ1u1 + · · ·+ λnun]

= λ1ω[u1,u2, . . . ,un,u1] =

= (−1)n−1λ1ω[u1,u2, . . . ,un] = (−1)n−1λ1Analogouslyη[u1, u2, . . . ,un] = ω[u1, u2, . . . ,un, λ1u1 + · · ·+ λnun]

= λ2ω[u1, u2, . . . ,un,u2] = (−1)n−2λ2

= (−1)n−2λ2ω[u1,u2, . . . ,un] = (−1)n−2λ2

128 CHAPTER 2. ALTERNATING MULTILINEAR FORMSet . In this way we an ompute all of the omponents of e in the basisu. We now apply this result to η = e1∧· · ·∧ en−1; let (e1, . . . , en−1) = uA,A the (n, n − 1) matrix of the omponents of the ve tors ei in the basis u.We nd

λ1 = (−1)n−1(e1 ∧ · · · ∧ en−1)[u1,u2, . . . ,un] =

= (−1)n−1 det

e1 · u2 . . . e1 · un

. . . . . . . . .en−1 · u2 . . . en−1 · un

=

= (−1)n−1 det

ǫ2a

21 . . . ǫna

n1

. . . . . . . . .ǫ2a

2n−1 . . . ǫna

nn−1

=

= (−1)n−1(ǫ2ǫ3 . . . ǫn)ǫ11

ǫ1det

a2

1 . . . an1. . . . . . . . .a2n−1 . . . ann−1

And from the formula in problem 63 this is λ1 = (−1)n−1(−1)index (e1 ×· · · × en−1)1. Pro eeding in the same way with the other omponents weobtain:

e = (−1)n−1(−1)index e1 × · · · × en−1or, in other words,e1 ∧ · · · ∧ en−1 = (−1)n−1(−1)index e1 × · · · × en−1yω

In the present ase the linear map that makes ommutative the diagram inthe pre eding problem is (−1)n−1(−1)index h−1. The term (−1)index doesn'tappear in the eu lidian ase be ause then index = 0.

2.4. ORIENTATION AND VOLUME 1292.4.7 Se ond anoni al isomorphismT Let us spe ialize to the ase when E is a 3-dimensional oriented orthogonalspa e with volume form ωE. We have seen (p.123) that the linear map c1 isan isomorphism whi h in this 3D ontext we shall denote by 2 to remind usthat it produ es a 2-form.

E2→ Λ2E

v 7→ 2v := vyωEand we shall notate the inverse isomorphism ♯2:Λ2E

♯2→ E

η 7→ v

v being su h that vyω = η. Remind that 2v is alled the ux 2-form for ita ts on ea h pair of ve tors e, e′ measuring the ux of v through the orientedparallelogram P (e, e′).b2v

v

e

e’v

Problem 66:a) Find a matrix B of 2 in terms of a positive on basis of E.b) Find a matrix B of 2 in terms of a positive og basis of E.

130 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:a) Let e = (e1, e2, e3) be a positive on basis, w = (ω1, ω2, ω3) its dualbasis andω∧ = (ω2 ∧ ω3, ω3 ∧ ω1, ω1 ∧ ω2)the asso iated basis in Λ2E.Then if we write 2e1 = a1ω2 ∧ ω3 + b1ω3 ∧ ω1 + c1ω1 ∧ ω2, we have

a1 = 2e1[e2, e3] = ωE[e2, e3, e1] = 1

b1 = 2e1[e3, e1] = ωE[e3, e1, e1] = 0

c1 = 2e1[e1, e2] = ωE[e1, e2, e1] = 0and by similar omputations with the other ve tors of the basis weobtainB = Ib) Assume that v = (v1,v2,v3) is a positive og basis. Computing as hasbeen done in a) we get

B =

ωE[v2,v3,v1]

ωE [v3,v1,v2]ωE[v1,v2,v3]

= ωE[v1,v2,v3] IWe know (see p.107) that if gv = detGv, then ωE[v1,v2,v3] =√| gv |,that is

B =√| gv |I

Problem 67: Ve tor produ t, s alar produ t, mixed produ t, and alternat-ing forms.Let E be an oriented eu lidian 3-dimensional spa e g its metri and ωE itsvolume form. Using the anoni al isomorphisms v = vyg and 2v = vyωEshow:a) e ∧ e′ = 2(e× e′)b) e ∧ 2e′ = (e · e′)ωE ) e ∧ e′ ∧ e” = ((e× e′) · e”)ωE

2.4. ORIENTATION AND VOLUME 131Solution:a) This has been seen in problem p.125 in the n-dimensional ase; inthe present ase (−1)n−1 = (−1)2 = 1. Noti e that if the spa e isnoneu lidian the result may be false (see problem p.128). We an lookat this result in a ommutative diagram showing how the ve tor produ tis ree ted in the world of forms:E × E ×→ E(e, e′) 7→ e× e′

(, ) ↓ ↓ 2Λ1E × Λ1E

∧→ Λ2E(e, e′) 7→ e ∧ e′b) Let e = (e1, e2, e3) be a positive on basis, ω = (ω1, ω2, ω3) its dualbasis and

ω∧ = ω2 ∧ ω3, ω3 ∧ ω1, ω1 ∧ ω2the asso iated basis in Λ2E. We writee = ae1 + be2 + ce3

e′ = a′e1 + b′e2 + c′e3and ompute e, 2e′. The matrix of in the bases e, ω is the metri matrix G = I. The matrix of 2 in the bases e and ω∧ is I. Thene ∧ 2e′ = (aω1 + bω2 + cω3) ∧ (a′ω2 ∧ ω3 + b′ω3 ∧ ω1 + c′ω1 ∧ ω2) =

= (aa′ + bb′ + cc′)ω1 ∧ ω2 ∧ ω3 =

= (e · e′)ω1 ∧ ω2 ∧ ω3 = (e · e′)ωEIf we introdu e a third anoni al isomorphism3 : K → Λ3E

λ 7→ λωEwe may look at the pre eding result in a ommutative diagram showinghow the s alar produ t is ree ted in the world of forms:E ×E ·→ K

(e, e′) 7→ e · e′(, 2) ↓ ↓ 3

Λ1E × Λ2E∧→ Λ3E

(e, 2e′) e ∧ 2e′

132 CHAPTER 2. ALTERNATING MULTILINEAR FORMS )e ∧ e′ ∧ e” = 2(e× e′) ∧ e” = ((e× e′) · e”)ωE

2.5 Hodge star appli ation2.5.1 S alar produ t of k-formsS alar produ t of 1-formsT Let E be an orthogonal vs; dene the s alar produ t of two 1-formsω, ω′ ∈ Λ1E by:

(ω, ω′) := (♯ω, ♯ω′)Let e be a basis of E, α its dual basis and G = eT · e the metri matrix. Letus nd the metri matrix for the produ t of 1-forms. To this end let Y, Y ′the olumns of omponents of the forms ω, ω′ that is ω = αY, ω′ = αY ′. Itss alar produ t is:♯ω = eG−1Y, ♯ω′ = eG−1Y ′,

(ω, ω′) = (♯ω, ♯ω′) = (G−1Y )TG(G−1Y ′) = Y T (G−1)TGG−1Y ′ = Y TG−1Y ′that tells us that the metri matrix for the produ t of 1-forms in the basis αis G−1. Whenever e is an og basis and ǫi = ei · ei then αi · αi = 1/ǫi and allmixed produ ts vanish; in parti ular if e is on ǫi = ±1 and α is on.Problem 68:Consider R3 with the standard s alar produ t; let (π1, π2, π3) be the dual ofthe anoni al basis.a) Compute the s alar produ t of ω = π1 + 2π3 and ω′ = π1 − π2 + π3.b) Let e be the basis e1 = (1, 0, 1), e2 = (0,−1, 0), e3 = (0, 0, 1) and

ϕ = (ϕ1, ϕ2, ϕ3) its dual basis. Compute the s alar produ t of ω =−ϕ1 + 2ϕ3 and ω′ = ϕ1 + ϕ2 + ϕ3.

2.5. HODGE STAR APPLICATION 133Solution:a) The metri matrix of the eu lidian produ t in the anoni al basis isG = I = G−1. That is, we do the produ t as we do it with ve tors:

(π1 + 2π3, π1 − π2 + π3) = (1, 0, 2)I

1−1

1

= 3b) The metri matrix of the eu lidian produ t in the basis e and its inverseareG =

2 0 10 1 01 0 1

, G−1 =

1 0 −10 1 0−1 0 2

and the produ t is(ω, ω′) = (−1, 0, 2)

1 0 −10 1 0−1 0 2

111

= (−1, 0, 2)

011

= 2

S alar produ t of k-formsT Let E be an orthogonal vs; the s alar produ t of two k-forms λ, µ ∈ ΛkEis dened rst on de omposable forms: let λ = ω1∧· · ·∧ωk, µ = σ1∧· · ·∧σkbe two k-forms with onstituents the 1-forms ωi, σi, i = 1, . . . , k; then dene

(λ, µ) := det

(ω1, σ1) . . . (ω1, σk). . . . . . . . .

(ωk, σ1) . . . (ωk, σk)

This denition is independent of the representation of λ, µ for it is a by-produ t of a anoni al denition, as an be seen in [Jan, p.220 and [Xa 2,p.196.Now for two general k-forms su h as λ =∑

Iր ai1...ikσi1 ∧ · · · ∧ σik andµ =

∑Jր bj1...jkσj1 ∧ · · · ∧ σjk we dene

(λ, µ) :=∑

Iր

∑

Jրai1...ikbj1...jk(σi1 ∧ · · · ∧ σik , σj1 ∧ · · · ∧ σjk)

134 CHAPTER 2. ALTERNATING MULTILINEAR FORMSProblem 69: Just an example.Consider in R3 the standard s alar produ t; denote by (π1, π2, π3) the dualof the anoni al basis.a) Compute the s alar produ t of λ = 2π1 ∧ π2 + π2 ∧ π3 and µ = π1 ∧π2 + 2π3 ∧ π1.b) Let e be the basis e1 = (1, 0, 1), e2 = (0,−1, 0), e3 = (0, 0, 1) andϕ = (ϕ1, ϕ2, ϕ3) its dual basis. Compute the s alar produ t of ω =−ϕ1 ∧ ϕ2 + 2ϕ3 ∧ ϕ1 and ω′ = ϕ2 ∧ ϕ1.Solution:a) We ompute rst the produ t of the de omposable forms

(π1 ∧ π2, π1 ∧ π2) = det

((π1, π1) (π1, π2)(π2, π1) (π2, π2)

)= det

(1 00 1

)= 1

(π1 ∧ π2, π3 ∧ π1) = det

((π1, π3) (π1, π1)(π2, π3) (π2, π1)

)= det

(0 10 0

)= 0and similarly one sees that

(π2 ∧ π3, π1 ∧ π2) = (π2 ∧ π3, π3 ∧ π1) = 0,in fa t the basis π1 ∧ π2, π2 ∧ π3, π3 ∧ π1 is an on basis of Λ2R3. Thenthe produ t is as we were multiplying ve tors:(λ, µ) = 2b) We know from the pre eding problem that the metri matrix of theprodu t of 1-forms is

G−1 =

1 0 −10 1 0−1 0 2

Then(ϕ1 ∧ ϕ2, ϕ1 ∧ ϕ2) = det

((ϕ1, ϕ1) (ϕ1, ϕ2)(ϕ2, ϕ1) (ϕ2, ϕ2)

)= det

(1 00 1

)= 1

2.5. HODGE STAR APPLICATION 135(ϕ3 ∧ ϕ1, ϕ2 ∧ ϕ1) = det

((ϕ3, ϕ2) (ϕ3, ϕ1)(ϕ1, ϕ2) (ϕ1, ϕ1)

)= det

(0 −10 1

)= 0and

(λ, µ) = (ϕ1 ∧ ϕ2, ϕ1 ∧ ϕ2) + 2(ϕ3 ∧ ϕ1, ϕ2 ∧ ϕ1) = 1

on bases of ΛkE

T Let E be an orthogonal vs; we look for orthonormal bases in ΛkE. Weshall use the notations:a) A stri tly in reasing k-multiindexI = (i1, . . . , ik), 1 ≤ i1 < · · · < ik ≤ nwill be written as I ր.b) Let ω = (ω1, . . . , ωn) be a basis of Λ1E; for ea h stri tly in reasing

k-multiindex I ր deneωI = ωi1 ∧ · · · ∧ ωik ) If I, J are stri tly in reasing k-multiindexs deneδJI =

1 if I = J0 otherwised) If ω = (ω1, . . . , ωn) is an on basis of Λ1E and I, J as above then

(ωI , ωJ) = ±δJIfor if J 6= I in the determinant giving the produ t there is, at least, a olumn (and a row) of zeros, and if J = I, the only nonvanishing termsare in the diagonal and its value is ±1. We have seen:(ω1, . . . , ωn) ON basis of E∗ ⇒ ω∧ = ωI , I ր ON basis of ΛkE

136 CHAPTER 2. ALTERNATING MULTILINEAR FORMSi) A usual produ t in ΛnE is that of the volume form ωE :(ωE, ωE) = (ω1∧· · ·∧ωn, ω1∧· · ·∧ωn) = (ω1, ω1) · · · (ωn, ωn) = (−1)swhere s =index of the metri .ii) Also if λ, µ ∈ ΛkE their expressions in the basis ω∧ are

λ =∑

Iրλ(eI)ωI

µ =∑

Iրµ(eI)ωIand their s alar produ t is

(λ, µ) =∑

Iրλ(eI)µ(eI)

2.5.2 Geometri al denitionProblem 70: A geometri al view of Hodge's star appli ation. (See [do Car.)Let E be an oriented eu lidian spa e, and η = ϕ1 ∧ · · · ∧ ϕk 6= 0 a k-form. Remind that V = 〈ϕ1, . . . , ϕk〉 is the spa e of η (independent of therepresentation of η, see p.84); let φ′ = (ϕ1, . . . , ϕk) and hoose in V ⊥ an onbasis φ′′ = (ϕk+1, . . . , ϕn) su h thatφ = (ϕ1, . . . , ϕk, ϕk+1, . . . , ϕn)is a positive basis of E∗ (:= ♯φ is a positive basis of E)a) Dene the volume of η asVol(ϕ1 ∧ · · · ∧ ϕk) =

√det(φ′T · φ′)Show that it is independent of the representation of η.

2.5. HODGE STAR APPLICATION 137b) Dene the Hodge star appli ation ∗ : ΛkE → Λn−kE rst on a basis φ∧∗η = Vol(η)ϕk+1 ∧ · · · ∧ ϕnand extend it linearly to ΛkE. Noti e that the spa e of ∗η is orthogonalto the spa e of η and the volume of ∗η is the same as that of η (be auseas φ′′ is on we have Vol (ϕk+1 ∧ · · · ∧ ϕn) = 1). Show that η ∧ ∗η is a

n-form that takes positive values on positive bases. ) Show that ∗η is well dened.d) ¾Can the pre eding denitions and results be extended to an orientednoneu lidian spa e E?Solution:Next gure in E∗ shows three basis forms ωi, the parallelepiped P (ϕ1, ϕ2)whose volume is Vol (η), V the spa e of η and V ⊥ with an on basis ϕ3.

ω1

ω2

ω3

ϕ3

ϕ1

ϕ2

V

Va) If η = β1 ∧ · · · ∧ βk is another de omposition we have (see p.82):ψ = (β1, . . . , βk) = (ϕ1, . . . , ϕk)A, detA = 1

138 CHAPTER 2. ALTERNATING MULTILINEAR FORMSThendet(ψT · ψ) = det(ATφ′T · φ′A) = (detA)2 det(φ′T · φ′),that isVol(β1∧· · ·∧βk) =

√det(ψT · ψ) =

√det(φ′T · φ′) = Vol(ϕ1∧· · ·∧ϕk).b) Let vi = ♯ϕi, i = 1, . . . , n; it is a positive basis v of E; noti e that

ϕi[vj] = vi[vj ] = vi · vj = ϕi · ϕjThen(η ∧ ∗η)[v1, . . . ,vk,vk+1, . . . ,vn] = Vol(η)ϕ1 ∧ · · · ∧ ϕn[v1, . . . ,vn] =

= Vol(η) det

ϕ1[v1] . . . ϕ1[vk] ϕ1[vk+1] . . . ϕ1[vn]. . . . . . . . . . . . . . . . . .

ϕn[v1] . . . ϕn[vk] ϕn[vk+1] . . . ϕn[vn]

=

= Vol(η) det

ϕ1 · ϕ1 . . . ϕ1 · ϕk. . . . . . . . . 0ϕk · ϕ1 ϕk · ϕk

1 0. . . . . . . . .0 1

=

= Vol (η)Vol2(η) > 0If u = vC is another positive basis we have(η ∧ ∗η)[u] = (detC)(η ∧ ∗η)[v] > 0,for detC > 0 as the basis u is positive. ) i) First of all we show that it is independent of the on basis hoosenin V ⊥. If ψ′′ is another basis of V ⊥, it is related to φ′′ by

ψ′′ = (ψk+1, . . . , ψn) = (ϕk+1, . . . , ϕn)C, detC = 1

2.5. HODGE STAR APPLICATION 139and from the result in problem p.84 we haveϕk+1 ∧ · · · ∧ ϕn = ψk+1 ∧ · · · ∧ ψnBoth onstru tions of ∗η oin ide:Vol (η)ϕk+1 ∧ · · · ∧ ϕn = Vol (η)ψk+1 ∧ · · · ∧ ψnii) Assume ψ1∧· · ·∧ψk is another representation of η. We know that

ψ′ = (ψ1, . . . , ψk) = (ϕ1, . . . , ϕk)C, detC = 1and the spa e of ψ1∧· · ·∧ψk is V . Due to the independen e of the hoi e of a basis in V ⊥ shown in i) we obtain the independen e ofthe representation.d) We annot do the pre eding onstru tion in a noneu lidian spa e, be- ause the volume of a form may vanish; the asso iated spa e wouldthen be singular and we don't have a spa e orthogonal to V . But seein the next two problems a way to ontour this di ulty.Note the ressemblan es with the ve tor produ t (see p.117):• The spa e of ∗η o uppies the orthogonal spa e of η.• ∗η has the same volume as η.• The hoosen bases form a positive basis of E∗.

Problem 71: Fundamental property of Hodge's appli ation.Let E be an oriented eu lidian spa e with volume form ωE and let η ∈ΛkE,µ ∈ Λn−kE. Show that Hodge's appli ation satises the fundamentalproperty

η ∧ µ = (∗η, µ)ωE(Compare with F ∧ 2G = (F ·G)ωR3.)

140 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:Let φ = (ϕ1, . . . , ϕn) be an on positive basis of E∗, I ր a stri tly in reasingk-multiindex, I ր the stri tly in reasing (n−k)- omplementary multiindex.If η = ϕI = ϕi1 ∧ · · · ∧ ϕik 6= 0 the forms orresponding to the in-dexs in I, ϕik+1

, . . . , ϕin, are an on basis of 〈ϕi1 , . . . , ϕik〉⊥. The basis(ϕi1, . . . , ϕik , ϕik+1

, . . . , ϕin) is positive or negative depending on sign(I, I).Then, noting that Vol(η) = 1, we have ∗η = sign(I, I)ϕik+1∧ · · · ∧ ϕin =sign(I, I)ϕI .Express µ as µ =

∑Jր aJϕJ , J an (n−k)-multiindex; due to the linearityin µ of the fundamental formula, it su es to he k it for µ = ϕJ .Let J = (jk+1, . . . , jn) and express µ as µ =

∑Jր aJϕJ ; due to thelinearity in µ of the fundamental formula, it su es to he k it for µ = ϕJ .Moreover if J 6= I the left term vanishes for there will be repeated 1-formsand the s alar produ t is (∗η, µ) = (ϕI , ϕJ) = 0 as well. We prove theformula for η = ϕI and µ = ϕI :

η ∧ µ = ϕI ∧ ϕI = sign(I, I)ωEand the right term is(∗η, µ)ωE = (sign(I, I)ϕI , ϕI)ωE = sign(I, I)ωEThe formula has been proven for µ arbitrary and η = ϕI ; for η =

∑I aIϕIwe have

η ∧ µ = (∑

I

aIϕI) ∧ µ =∑

I

aI(ϕI ∧ µ) =

=∑

I

aI(∗ϕI , µ)ωE = (∗∑

I

(aIϕI), µ)ωE =

= (∗η, µ)ωEWe have prooved that the star appli ation dened geometri ally satises thefundamental formula.

2.5. HODGE STAR APPLICATION 1412.5.3 General denitionT The fundamental formula allows the denition of Hodge star appli ationin the noneu lidian ase; in the onstru tion we follow [Fla. For anotherinteresting way to pro eed, see [Jän p.222.Let E be an oriented orthogonal spa e (we no more assume it eu lidian),with volume form ωE . The Hodge star operator is a linear map

ΛkE∗→ Λn−kE

η 7→ ∗ηsatisfying for any µ ∈ Λn−kE the fundamental propertyη ∧ µ = (∗η, µ)ωETo see the existen e of su h operator x η ∈ ΛkE; then the appli ationΛn−kE → ΛnEµ 7→ η ∧ µis linear and η ∧ µ is a multiple of ωE; then η ∧ µ = fη(µ)ωE for a ertainlinear fun tion

fη : Λn−kE → RSo fη ∈ (Λn−kE)∗ and the rst anoni al isomorphism (see p.40) applied tothe orthogonal spa e Λn−kE tells us that fη(.) = (∗η, . ) for a ertain ∗η ∈Λn−kE. This onstru tive argument shows that Hodge's operator satisesthe fundamental property. Moreover it is unique (see next problem).We an look at it as onverting the exterior produ t into a s alar produ tthat a ts as a weight of the volume form. Or we an think that we haveseparated η ∧ µ into a spatial part, ωE , and a numeri al value (∗η, µ) .

Problem 72:Using only the fundamental property η ∧ µ = (∗η, µ)ωE, show that Hodge'soperatorΛkE

∗→ Λn−kE

η 7→ ∗ηis the only linear appli ation satisfying it.

142 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:a) It is linear:i) If t ∈ R th fundamental property is:(tη) ∧ µ = (∗(tη), µ)ωE

(tη) ∧ µ = t(η ∧ µ) = t(∗η, µ)ωE = (t(∗η), µ)ωEThen(∗(tη), µ) = (t(∗η), µ)⇒ (∗(tη)− t(∗η), µ) = 0, ∀µ ∈ Λn−kEand as the s alar produ t of forms is nondegenerated we have

∗(tη)− t ∗ η = 0⇒ ∗(tη) = t ∗ ηii) If η, η′ ∈ ΛkE the fundamental property is:(η + η′) ∧ µ = (∗(η + η′), µ)ωE

(η+η′)∧µ = η∧µ+η′∧µ = (∗v, µ)ωE+(∗η′, µ)ωE = (∗η+∗ηω′, µ)ωEThen, as above∗(η + η′) = ∗η + ∗η′b) It is the only one:Let ⋆ : ΛkE

⋆→ Λn−kE be another appli ation satisfying η ∧ µ =(⋆η, µ)ωE; then it is linear and (∗η, µ) = (⋆η, µ). This implies that

∗η = ⋆ηN.B.: The geometri onstru tion of ∗ in eu lidian spa es satises the fun-damental property. The geometri al and algebrai al onstru tions give thesame result. Moreover now we have a ∗ operator in the noneu lidian ase.

2.5. HODGE STAR APPLICATION 143Problem 73:Compute Hodge's operator in the extreme ases:a) Λ0E∗→ ΛnEb) ΛnE∗→ Λ0ESolution:a) We have

Λ0E∗→ ΛnE

1 7→ ∗1 = cωEThe fundamental property applied to 1 and µ = ωE givesωE = 1 ∧ ωE = (∗1, ωE)ωE = c(ωE, ωE)ωE ⇒ c = (ωE, ωE) = (−1)indexand

∗1 = (−1)indexωEb) Now we ompute ∗ωE:ΛnE

∗→ Λ0E

ωE 7→ ∗ωE = cThe fundamental property applied to 1 givesωE = ωE ∧ 1 = (∗ωE, 1)ωE = cωE ⇒ c = 1and

∗ωE = 1

T We now ompute the a tion of Hodge's star operator on the k-forms ofa basis of ΛkE when E is an orthogonal spa e, using the fundamentalformula alone.

144 CHAPTER 2. ALTERNATING MULTILINEAR FORMSProblem 74: A tion of the ∗ operator on basis.Find a formula for the omputation of ∗ :a) When u = (u1, . . . ,un) is an on positive basis of E with ǫi = ui · ui =±1 and w = (ω1, . . . , ωn) its dual basis.b) When u = (u1, . . . ,un) is an og positive basis of E with λi = ui ·ui 6= 0and w = (ω1, . . . , ωn) its dual basis.Solution:It su es to know the a tion of ∗ on the forms of a basis of ΛkE.a) Remind that w∧ = ωi1 ∧ · · · ∧ωik , I = (i1, . . . , ik), I ր is an on basisof ΛkE, and ωjk+1

∧ · · · ∧ ωjn, J = (jk+1, . . . , jn), J ր an on basisof Λn−kE. Then for the form ωI = ωi1 ∧ · · · ∧ ωik we an write theexpression∗ωI =

∑

JրaJωJAs the basis is on we have aJ = ǫJ(∗ωI , ωJ) where ǫJ = ǫjk+1

· · · ǫjn .Then the fundamental property gives(∗ωI , ωJ)ωE = ωI ∧ ωJ =

0 if J 6= I

ωI ∧ ωI = sgn (I, I)ωE if J = Ibe ause the volume form is ωE = ω1 ∧ · · · ∧ ωn and for if J 6= I therewill be repeated indi es; then∗ωI = sgn (I, I)ǫIωIb) Let u = (u1, . . . ,un) be an og positive basis of E, w = (ω1, . . . , ωn) itsdual basis, λi = ui · ui 6= 0, λI = λi1 · · ·λik and ωI = ωi1 ∧ · · · ∧ ωik . Itsu es to adapt the on ase; the volume form is ωE =

√| g |ω1∧· · ·∧ωn(p.107) and aJ = 1

λJ(∗ωI , ωJ). Then

(∗ωI , ωJ)ωE = ωI ∧ ωJ =

0 if J 6= I

ωI ∧ ωI = sgn (I, I) 1√|g|ωE if J = I

2.5. HODGE STAR APPLICATION 145(∗ωI , ωI) = sgn (I, I)

1√| g |So

∗ωI =1√| g |

sgn (I, I)1

λIωI

Problem 75:Let E be an oriented orthogonal spa e, G the metri matrix in a sele tedbasis, g = detG and ωE the volume form. If ω, η ∈ Λk(E) show:a) ∗ ∗ ω = (−1)k(n−k)(−1)indexω.b) ∗ω ∧ η = ∗η ∧ ω = (−1)k(n−k)(−1)index(ω, η)ωE. ) ω ∧ ∗η = η ∧ ∗ω = (−1)index(ω, η)ωE.Solution:Let w = (ω1, . . . , ωn) be an on basis of E∗.a) Due to the linearity of ∗ it su es to he k the equality when ω = ωI ,a basis k-form:∗ ∗ ωI = ∗(sgn (I, I)ǫIωI) =

= sgn (I, I)ǫI ∗ ωINow observe that to go from the permutation (I, I) to (I, I) we mustdo k(n− k) transpositions; then(I, I) = (I, I)τ1 · · · τk(n−k)sgn (I, I) = sgn (I, I)(−1)k(n−k)sgn (I, I)sgn (I, I) = (sgn (I, I))2(−1)k(n−k) = (−1)k(n−k)and

∗ ∗ ωI = sgn (I, I)ǫIsgn (I, I)ǫIωI =

= (−1)k(n−k)ǫIǫIωI =

= (−1)k(n−k)(−1)indexωI

146 CHAPTER 2. ALTERNATING MULTILINEAR FORMSb) We apply a):∗ω ∧ η = (∗ ∗ ω, η)ωE = (−1)k(n−k)(−1)index(ω, η)ωEand the right hand term is symmetri . ) We use the nearly ommutativity of the exterior produ t and b):ω ∧ ∗η = (−1)k(n−k) ∗ η ∧ ω =

= (−1)k(n−k)(−1)k(n−k)(−1)index(η, ω)ωE =

= (−1)index(η, ω)ωEand the symmetry of the right hand term shows the result.Problem 76: Pythagoras theorem revisited (on e more, true; but don't missthis one!)a) Related to the next gure (in R3) and as an introdu tion to point b)show this Pythagoras theorem in spa e:/home/dalmau/BOOK 2/Pitagoras gral/pitagoras gralb.fig not found!

Area2(ABC) = Area2(OAB) + Area2(OBC) + Area2(OCA)This result has been found again and again along time. For instan eJ.P. de Gua de Malves, A adémie de Cien es de Paris, 1783. It seemssurprising not to nd it presented more often.b) Let e = (e1, . . . , en) be the anoni al basis of Rn and (π1, . . . , πn) itsdual basis. Take k ≤ n ve tors in Rn, v = (v1, . . . ,vk) = eA, A being

2.5. HODGE STAR APPLICATION 147the (n, k) matrix of the omponents of the vi. The k-parallelepipedgenerated by those points, with a vertex at 0 isP = P (v1, . . . ,vk) = x =

k∑

i=1

tivi, 0 ≤ ti ≤ 1Let I be an in reasing k-multiindex,I = (i1, . . . , ik), 1 ≤ i1 < · · · < ik ≤ nand letΠI be the orthogonal proje tion of Rn onto the subspa e 〈ei1 , . . . , eik〉(it leaves the i1, . . . , ik omponents of a point invariant and sends theothers to 0). Let PI = ΠI(P ); show thatVol2k(P ) =

∑

IրVol2k(PI) ) Compute Volk(PI) in terms of Volk(P ) and ertain dire tion osinus ofthe ve tors orthogonal to P .Solution:a) Pythagoras theorem in spa e.Let A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c) and u = B − A =

(−a, b, 0),v = C − A = (−a, 0, c); the area of ABC isArea2(ABC) =1

4|u ∧ v |2= 1

4| (bc, ac, ab) |2=

=1

4(b2c2 + c2a2 + a2b2)and the result is proved. A more original (:= loser to the origins) proofuses the Heron formula, giving the area of a triangle with sides a, b, cand semiperimeter p = 1

2(a+ b+ c):Area(ABC) =

√p(p− a)(p− b)(p− c)b) If u1, . . . ,un is an on basis of a eu lidian spa e, for a ve tor

u = λ1u1 + · · ·+ λnun

148 CHAPTER 2. ALTERNATING MULTILINEAR FORMSwe have Pythagoras theorem:u · u = λ2

1 + · · ·+ λ2nConsider now the eu lidian spa e ΛkRn and the on basis

π∧ = πI = πi1 ∧ · · · ∧ πikI=(i1,...,ik)րWe apply Pythagoras theorem to the k-formω = v1 ∧ · · · ∧ vk =

∑

IրaIπIWe have

ω · ω = (v1 ∧ · · · ∧ vk) · (v1 ∧ · · · ∧ vk) =

= det

v1 · v1 . . . v1 · vk. . . . . . . . .

vk · v1 . . . vk · vk

= det(vT · v) =

= Vol2k(P )The omponents of ω are:aI = ω[eI ] = det

v1[ei1 ] . . . v1[eik ]. . . . . . . . .

vk[ei1 ] . . . vk[eik ]

= det

vi11 . . . vik1. . . . . . . . .

vi1k . . . vikk

= ±Volk(PI)and Pythagoras theorem shows thatVol2k(P ) =∑

IրVol2k(PI)The lue, as in R3 and in the ase k = 1 in Rn, lies in the orthonormalityof the basis. The only pri e we pay for k > 1 is that the algebrai representatives of P are forms instead of ve tors; the rest of the settingis the same.Noti e how mu h the proof rests upon the orthonormality of the basisof forms. That suggests that one should view Pythagoras theorem asin the next gure (the segment c is the same all around):

2.5. HODGE STAR APPLICATION 149/home/dalmau/BOOK 2/Pitagoras gral/pitagoras new ter.fig not found!and no so mu h as in the lassi al/home/dalmau/BOOK 2/Pitagoras gral/pitagoras new.fig not found!.Another point is that the fa t that we took parallelepipeds with vertexat 0 is not relevant, for a translation doesn't modify the volume of Pnor that of its proje tions.Bonus 1: Let u1, . . . ,uk, 1 ≤ k ≤ n be an orthonormal set of ve torsof Rn whose matrix of omponents in the anoni al basis is

u11 . . . u1

k

u21 . . . u2

k

. . . . . . . . .un1 . . . unk

and let J = (j1, . . . , jk) ր be a k-multiindex. Then using Pythagorastheorem we obtain

∑

Jրdet2 uj11 . . . uj1k

. . . . . . . . .

ujk1 . . . ujkk

= 1,be ause Volk(P (u1, . . . ,uk)) = 1.To give an example let (u1,u2,u3) be an orthonormal system in R3.Theni) for k = 1 the result readsve tor u1:(u1

1)2 + (u2

1)2 + (u3

1)2 = 1

150 CHAPTER 2. ALTERNATING MULTILINEAR FORMSve tor u2:(u1

2)2 + (u2

2)2 + (u3

2)2+ = 1ve tor u3:

(u13)

2 + (u23)

2 + (u33)

2 = 1ii) for k = 2 we haveve tors u1,u2:det2( u11 u1

2

u21 u2

2

)+ det2( u1

1 u12

u31 u3

2

)+ det2( u2

1 u22

u31 u3

2

)= 1ve tors u1,u3:det2( u1

1 u13

u21 u2

3

)+ det2( u1

1 u13

u31 u3

3

)+ det2( u2

1 u23

u31 u3

3

)= 1ve tors u2,u3:det2( u1

2 u13

u22 u2

3

)+ det2( u1

2 u13

u32 u3

3

)+ det2( u2

2 u23

u32 u3

3

)= 1iii) and taking k = 3ve tors u1,u2,u3: det2 u1

1 u12 u1

3

u21 u2

2 u23

u31 u3

2 u33

= 1 ) It's a natural question to nd Volk(PI) in terms of Vol k(P ) and theposition of P in spa e. Let V = 〈v1, . . . ,vk〉 be the subspa e of P ,and let u = (vk+1, . . . ,vn) be an orthonormal basis of V ⊥ su h that(v, u) = (v1, . . . ,vk,vk+1, . . . ,vn) is a positive basis of Rn. Let η =vk+1∧· · ·∧vn; from the geometri al denition of Hodge's appli ation∗ω = Volk(P )η and

ω ∧ η =1Volk(P )

ω ∧ (∗ω) =1Volk(P )

(ω, ω)ωRn = Volk(P )ωRn

2.5. HODGE STAR APPLICATION 151Write eI = (ei1 , . . . , eik) and apply ω ∧ η to (eI , u) in two ways; we dothe ase I = (1, . . . , k):ω ∧ η[eI , u] = Volk(P )det 1 . . . 0 v1

k+1 . . . v1n

. . . . . . . . . . . . . . . . . .0 . . . 1 vkk+1 . . . vkn0 . . . 0 vk+1

k+1 . . . vk+1n

. . . . . . . . . . . . . . . . . .0 . . . 0 vnk+1 . . . vnn

=

= Volk(P ) det

vk+1k+1 . . . vk+1

n

. . . . . . . . .vnk+1 . . . vnn

= Volk(P ) detT(k+1,...,n)andω ∧ η[eI , u] = ( v1 ∧ · · · ∧ vk ∧ vk+1 ∧ · · · ∧ vn)[eI , v] =

= det

v1 · e1 . . . v1 · ek 0 . . . 0. . . . . . . . . . . . . . . . . .

vk · e1 . . . vk · ek 0 . . . 0− . . . − 1 . . . 0. . . . . . . . . . . . . . . . . .− . . . − 0 . . . 1

=

= det

v11 . . . v1

k 0 . . . 0. . . . . . . . . . . . . . . . . .vk1 . . . vkk 0 . . . 0− . . . − 1 . . . 0. . . . . . . . . . . . . . . . . .− . . . − 0 . . . 1

= ±Volk(P(1...k))The matri es TI ontain the information about the position of P inspa e be ause in the ase at hand for instan e, we proje ted on the thesubspa e 〈e1, . . . , ek〉 whose perpendi ular spa e is 〈ek+1, . . . , en〉 andthen we take into a ountvlj = el · vj = cos(el,vj), l, j = k + 1, . . . , nthat is the dire tion osinus of the normal ve tors to P with respe t tothe normal dire tion of the spa e where we proje t.If we all Ithe (n− k)- omplementary in reasing multiindex to I the

152 CHAPTER 2. ALTERNATING MULTILINEAR FORMSformula will be: Volk(PI) = | detTI|Vol(P )Noti e that it may happen that Volk(PI) = 0 whenever detTI = 0 andthis means that the proje tion of P is degenerated.Bonus 2: Let A be the matrixA =

v11 . . . v1

n

. . . . . . . . .vn1 . . . vnn

Think about the olumn i as the omponents of a ve tor vi in the anoni albasis e. Let I be a stri tly in reasing k-multindex , J a stri tly in reasing(n− k)-multiindex and onsider the determinants

det( vI) = det

vi11 . . . vi1k. . . . . . . . .

vik1 . . . vikk

det( uJ) = det

vj1k+1 . . . vj1n. . . . . . . . .

vjn−k

k+1 . . . vjn−kn

Put πI = πi1 ∧ · · · ∧ πik , ω = v1 ∧ · · · ∧ vk =∑

Iր det(vI)πI , and η =

vk+1 ∧ · · · ∧ vn =∑

Jր det(uJ)πJ . Thenω ∧ η = (

∑

Iրdet(vI)πI) ∧ (

∑

Jրdet(uJ)πJ)

=∑

I,Jրdet(vI) det(uJ)πI ∧ πJ =

= (∑

Iրsgn(I, I) det(vI) det(uI

))ωRnwhere the jump to the last line is be ause if J 6= I then in the produ tπI ∧ πJ there are repeated 1-forms and it vanishes.Applying ω ∧ η to the anoni al basis we have

(ω ∧ η)[e1, . . . , en] =

2.5. HODGE STAR APPLICATION 153= det

v1[e1] . . . v1[ek] v1[ek+1] . . . v1[en]. . . . . . . . . . . . . . . . . .

vk[e1] . . . vk[ek] vk[ek+1] . . . vk[en]vk+1[e1] . . . vk+1[ek] vk+1[vk+1] . . . vk+1[vn]. . . . . . . . . . . . . . . . . .

vn[e1] . . . vn[ek] vn[vk+1] . . . vn[vn]

= det

v1 · e1 . . . v1 · ek v1 · ek+1 . . . v1 · en. . . . . . . . . . . . . . . . . .

vk · e1 . . . vk · ek vk · ek+1 . . . vk · ek+1

vk+1 · e1 . . . vk+1 · ek vk+1 · ek+1 . . . vk+1 · ek+1

. . . . . . . . . . . . . . . . . .vn · e1 . . . vn · ek vn · ek+1 . . . vn · en

= detAWe have obtaineddetA =

∑

Iրsgn(I, I

o) det(vI) det(uI

o) ,Lapla e's rule for the omputation of a determinant.An example:

A =

v11 v1

2 v13 v1

4

v21 v2

2 v23 v2

4

v31 v3

2 v33 v3

4

v41 v4

2 v43 v4

4

If k = 1 we have:

I = 1, I = (2, 3, 4), sign(I, I) = 1I = 2, I = (1, 3, 4), sign(I, I) = −1I = 3, I = (1, 2, 4), sign(I, I) = 1I = 4, I = (1, 2, 3), sign(I, I) = −1and

detA = v11 det u(234) − v2

1 det u(134) + v31 det u(124) − v4

1 det u123the usual rule of the development of the determinant trough the rst olumn.If k = 2 we have:I = (1, 2), I(3, 4), sign(I, I) = 1

154 CHAPTER 2. ALTERNATING MULTILINEAR FORMSI = (1, 3), I(2, 4), sign(I, I) = −1I = (1, 4), I(2, 3), sign(I, I) = 1I = (2, 3), I(1, 4), sign(I, I) = 1I = (2, 4), I(1, 3), sign(I, I) = −1I = (3, 4), I(1, 2), sign(I, I) = 1anddetA = det v(12) det u(34)−det v(13) det u(24)+det v(14) det u(23)+det v(23) det u(14)−

det v(24) det u(13) + det v(34) det u(12)

T In the next problems the dual of the anoni al basis of Rn will benotated (dx1, . . . , dxn) or (dx, dy, dz) in the 3D ase.Problem 77:In R3 with the eu lidian metri , the usual orientation and volume form

ωE = dx ∧ dy ∧ dz, ompute Hodge's operator applied to on bases ofΛ0(R3),Λ1(R3),Λ2(R3),Λ3(R3).Solution:Remind the metri matrix is G = I ; thena) ∗1 = (−1)indexdx ∧ dy ∧ dz = dx ∧ dy ∧ dzb) ∗dx = sgn (1, 2, 3)ǫ23dy ∧ dz = dy ∧ dz

∗dy = sgn (2, 1, 3)ǫ13dx ∧ dz = −dx ∧ dz = dz ∧ dx∗dz = sgn (3, 1, 2)ǫ12dx ∧ dy = dx ∧ dy ) ∗dx ∧ dy = sgn (1, 2, 3)ǫ3dz = dz∗dy ∧ dz = sgn (2, 3, 1)ǫ1dx = dx∗dz ∧ dx = sgn (3, 1, 2)ǫ2dy = dyd) ∗dx ∧ dy ∧ dz = 1

2.5. HODGE STAR APPLICATION 155Problem 78:In R4 with the eu lidian metri , the usual orientation and volume formωE = dx ∧ dy ∧ dz ∧ dt, ompute Hodge's operator applied to on basesof Λ0(R4),Λ1(R4),Λ2(R4),Λ3(R4),Λ4(R4).Solution:Again the metri matrix is G = I and we havea) ∗1 = (−1)indexdx ∧ dy ∧ dz ∧ dt = dx ∧ dy ∧ dz ∧ dtb) ∗dx = sgn (1, 2, 3, 4)ǫ234dy ∧ dz ∧ dt = dy ∧ dz ∧ dt

∗dy = sgn (2, 1, 3, 4)ǫ134dx ∧ dz ∧ dt = −dx ∧ dz ∧ dt∗dz = sgn (3, 1, 2, 4)ǫ124dx ∧ dy ∧ dt = dx ∧ dy ∧ dt∗dt = sgn (4, 1, 2, 3)ǫ123dx ∧ dy ∧ dz = −dx ∧ dy ∧ dz ) ∗dx ∧ dy = sgn (1, 2, 3, 4)ǫ34dz ∧ dt = dz ∧ dt∗dx ∧ dz = sgn (1, 3, 2, 4)ǫ24dy ∧ dt = −dy ∧ dt∗dx ∧ dt = sgn (1, 4, 2, 3)ǫ23dy ∧ dz = dy ∧ dz∗dy ∧ dz = sgn (2, 3, 1, 4)ǫ14dx ∧ dt = dx ∧ dt∗dy ∧ dt = sgn (2, 4, 1, 3)ǫ13dx ∧ dz = −dx ∧ dz∗dz ∧ dt = sgn (3, 4, 1, 2)ǫ12dx ∧ dy = dx ∧ dyd) ∗dx ∧ dy ∧ dz = sgn (1, 2, 3, 4)ǫ4dt = dt∗dx ∧ dy ∧ dt = sgn (1, 2, 4, 3)ǫ3dz = −dz∗dx ∧ dz ∧ dt = sgn (1, 3, 4, 2)ǫ2dy = dy∗dy ∧ dz ∧ dt = sgn (2, 3, 4, 1)ǫ1dx = −dxe) ∗dx ∧ dy ∧ dz ∧ dt = 1

Problem 79:In R4 with the Lorentz metri , the usual orientation and volume form ωE =dx∧dy∧dz∧dt ompute Hodge's operator applied to on bases of Λi(R4), i =1, . . . , 4.

156 CHAPTER 2. ALTERNATING MULTILINEAR FORMSSolution:The metri matrix isG =

11

1−1

and ǫ1 = ǫ2 = ǫ3 = 1, ǫ4 = −1; we also have ǫ1234 = −1 and

ǫij =

1 if none of the indi es is 4−1 if any of the indi es is 4

ǫijk =

1 if none of the indi es is 4−1 if any of the indi es is 4Thena) ∗1 = (−1)indexωE = −dx ∧ dy ∧ dz ∧ dtb)

∗dx = sgn (1, 2, 3, 4)ǫ234dy ∧ dz ∧ dt =

= −dy ∧ dz ∧ dt

∗dy = sgn (2, 1, 3, 4)ǫ134dx ∧ dz ∧ dt =

= −(−dx ∧ dz ∧ dt)

∗dz = sgn (3, 1, 2, 4)ǫ124dx ∧ dy ∧ dt =

= −dx ∧ dy ∧ dt

∗dt = sgn (4, 1, 2, 3)ǫ123dx ∧ dy ∧ dz =

= −dx ∧ dy ∧ dz ) ∗dx ∧ dy = sgn (1, 2, 3, 4)ǫ34dz ∧ dt = −dz ∧ dt∗dz ∧ dt = sgn (3, 4, 1, 2)ǫ12dx ∧ dy = dx ∧ dy et .d) ∗dx ∧ dy ∧ dz = sgn (1, 2, 3, 4)ǫ4dt = −dt∗dx ∧ dy ∧ dt = sgn (1, 2, 4, 3)ǫ3dz = −dz∗dy ∧ dz ∧ dt = sgn (2, 3, 4, 1)ǫ1dx = −dxet .e) ∗dx ∧ dy ∧ dz ∧ dt = 1

2.5. HODGE STAR APPLICATION 157Problem 80:Consider in R4 the metri that in the anoni al basis i, j,k, l has the matrixG =

11−1

−1

,the orientation given by the anoni al basis and volume form ωE = dx∧dy∧

dz∧dt . Compute Hodge's operator applied to on bases ofΛ0(R4),Λ1(R4),Λ2(R4),Λ3(R4),Λ4(R4).Solution:As in the pre eding problem we havea) ∗1 = (−1)indexωE = dx ∧ dy ∧ dz ∧ dtb)∗dx = sgn (1, 2, 3, 4)ǫ234dy ∧ dz ∧ dt =

= dy ∧ dz ∧ dt

∗dy = sgn (2, 1, 3, 4)ǫ134dx ∧ dz ∧ dt =

= −dx ∧ dz ∧ dt

∗dz = sgn (3, 1, 2, 4)ǫ124dx ∧ dy ∧ dt =

= −dx ∧ dy ∧ dt

∗dt = sgn (4, 1, 2, 3)ǫ123dx ∧ dy ∧ dz =

= dx ∧ dy ∧ dz ) ∗dx ∧ dy = sgn (1, 2, 3, 4)ǫ34dz ∧ dt = dz ∧ dt∗dx ∧ dt = sgn (1, 4, 2, 3)ǫ23dy ∧ dz = −dy ∧ dz∗dz ∧ dt = sgn (3, 4, 1, 2)ǫ12dx ∧ dy = dx ∧ dyet .

158 CHAPTER 2. ALTERNATING MULTILINEAR FORMSd) ∗dx ∧ dy ∧ dz = sgn (1, 2, 3, 4)ǫ4dt = −dt∗dx ∧ dy ∧ dt = sgn (1, 2, 4, 3)ǫ3dz = −(−dz)∗dy ∧ dz ∧ dt = sgn (2, 3, 4, 1)ǫ1dx = −dxet .e) ∗dx ∧ dy ∧ dz ∧ dt = 1

2.5.4 Hodge and anoni al isomorphismsProblem 81:The geometri al onstru tion of Hodge's appli ation suggests that if e,v ∈R3, ∗(e ∧ v) might well be the ve tor produ t in the world of forms, thatis: ∗(e∧ v) = (e×v). (We are of ourse onsidering the usual orientationand metri .) Show that this is true.Solution:We know that e ∧ v = 2(e× v); applying ∗ we have

∗(e ∧ v) = ∗2(e× v)Let us ompute the right hand term; if f = f1u1+f2u2+f3u3 (u the anoni albasis) then2f = f1dy ∧ dz + f2dz ∧ dx+ f3dx ∧ dyand as ∗(dy∧dz) = dx, ∗(dz∧dx) = dy, ∗(dx∧dy) = dz, we obtain ∗(2f) = f .Then∗(e ∧ v) = ∗2(e× v) = (e× v)

In next problem we generalize the pre eding result to an arbitrary orientedorthogonal spa e.

2.5. HODGE STAR APPLICATION 159Problem 82:Let E be an oriented orthogonal spa e and ωE its volume form. Let u =(u1, . . . ,un) be a positively oriented on basis with ui · ui = ǫi, and letw = (ω1, . . . , ωn) be its dual basis. Consider a ve tore = a1u1 + · · ·+ anun.Compute:a) eyωE.b) ∗(eyωE). ) Let e1, . . . , en−1 be ve tors in E ; apply the pre eding result to

∗(e1 × · · · × en−1yωE).d) Compute ∗(e1 ∧ · · · ∧ en−1).Solution:a)eyωE[u1,u2, . . . ,un] = ωE [u1,u2, . . . ,un, a1u1 + · · ·+ anun]

= ωE [u1,u2, . . . ,un, a1u1] =

= (−1)n−1a1

eyω[u1, u2, . . . ,un] = ωE [u1, u2, . . . ,un, a1u1 + · · ·+ anun]

= ωE [u1, u2, . . . ,un, a2u2] =

= (−1)n−2a2et . TheneyωE = (−1)n−1(a1ω2 ∧ · · · ∧ ωn − a2ω1 ∧ ω2 ∧ · · · ∧ ωn + . . . )b)

∗(ω2 ∧ · · · ∧ ωn) = sgn (2, 3, . . . , n, 1)ǫ1ω1 =

= (−1)n−1ǫ1ω1

160 CHAPTER 2. ALTERNATING MULTILINEAR FORMS∗(ω1 ∧ ω2 ∧ · · · ∧ ωn) = sgn (1, 2, . . . , n, 2)ǫ2ω2 =

= (−1)n−2ǫ2ω2et . We have obtained∗(eyωE) = (−1)n−1(−1)n−1(ǫ1a1ω1 + ǫ2a2ω2 + . . . ) = ebe ause the matrix of in the bases u and w is

G =

ǫ1ǫ2 . . .

ǫn

. )

∗(e1 × · · · × en−1yωE) = (e1 × · · · × en−1)d) We know from problem p.127 thate1 ∧ · · · ∧ en−1 = (−1)n−1(−1)indexe1 × · · · × en−1yωEand then

∗(e1 ∧ · · · ∧ en−1) = (−1)n−1(−1)index ∗ (e1 × · · · × en−1yω) =

= (−1)n−1(−1)index(e1 × · · · × en−1)

In p.125 we asked about the existen e of a linear map f making thefollowing diagram ommutative:E× k. . . ×E

∧ · · · ∧ ւ ց ck

ΛkEf→ Λn−kE

2.5. HODGE STAR APPLICATION 161Problem 83:Let E be an oriented eu lidian spa e with volume form ωE and letv1, . . . ,vk, 1 < k < n− 1ve tors in E. Is it true that

∗(v1 ∧ · · · ∧ vk) = (v1, . . . ,vk)yωE?What an we say of ∗((v1, . . . ,vk)yωE)?Solution:Let uk+1, . . . ,un be an on basis of 〈v1, . . . ,vk〉⊥ su h that v = (v1, . . . ,vk,uk+1, . . . ,un)is a positive basis of E. Then to show the equality it su es to he k it whenapplied to e1, . . . , en−k hoosen among the ve tors in the basis v.a) If (e1, . . . , en−k) = (uk+1, . . . ,un) and η = v1 ∧ · · · ∧ vk we have∗(v1 ∧ · · · ∧ vk)[uk+1, . . . ,un] = Vol (η)(uk+1 ∧ · · · ∧ un)[uk+1, . . . ,un] =

= Vol (η) det(uT · u) = Vol (η)and((v1, . . . ,vk)yωE)[uk+1, . . . ,un] = ωE[uk+1, . . . ,un,v1, . . . ,vk] = (−1)k(n−k)Vol (η)b) Assume now that among the hoosen ve tors there are one or morefrom the vi. Then∗(v1∧· · ·∧vk)[uk+1, . . . ,vi, . . . ,un] = Vol (η)(uk+1∧· · ·∧un)[uk+1, . . . ,vi, . . . ,un] = 0be ause there will be a olumn of zeros in the determinant. On anotherhand(v1, . . . ,vk)yωE[uk+1, . . . ,vi, . . . ,un] = ωE[uk+1, . . . ,vi, . . . ,un,v1, . . . ,vk] = 0for at least one of the ve tors is repeated.

162 CHAPTER 2. ALTERNATING MULTILINEAR FORMSThen we see that the right formula is∗(v1 ∧ · · · ∧ vk) = (−1)k(n−k)(v1, . . . ,vk)yωETo answer the other question we apply property a) in p.145:

∗((v1, . . . ,vk)yωE) = (−1)k(n−k) ∗ ∗(v1 ∧ · · · ∧ vk) =

= (−1)k(n−k)(−1)k(n−k)(v1 ∧ · · · ∧ vk)that is∗((v1, . . . ,vk)yωE) = v1 ∧ · · · ∧ vk

Problem 84:Let E be an orthogonal spa e with volume form ωE and let v = (v1, . . . ,vk), 1 <k < n− 1 be ve tors in E. ¾Is it still true that

∗(v1 ∧ · · · ∧ vk) = (−1)k(n−k)(v1, . . . ,vk)yω?Solution:Remind that now the spa e 〈v1, . . . ,vk〉may be singular and we annot followthe line of the pre eding problem.Let u = (u1, . . . ,un) be an on positive basis of E , ui · ui = ǫi = ±1,w = (ω1, . . . , ωn) its dual basis, and A the matrix of the omponents of theve tors in the basis u:

(v1, . . . ,vk) = u

c11 . . . c1k. . . . . . . . .cn1 . . . cnk

= uA

2.5. HODGE STAR APPLICATION 163Then we have the expression v1 ∧ · · · ∧ vk =∑

Jր aJωJ whereaJ = (v1 ∧ · · · ∧ vk)[uj1 , . . . ,ujk ] =

= det

v1 · uj1 . . . v1 · ujk. . . . . . . . .

vk · ujk . . . vk · ujk

=

= det

ǫj1c

j11 . . . ǫj1c

jk1

. . . . . . . . .

ǫjkcj1k . . . ǫjkc

jkk

=

= ǫj1 . . . ǫjk det

cj11 . . . cj1k. . . . . . . . .

cjk1 . . . cjkk

= ǫJ detAJsov1 ∧ · · · ∧ vk =

∑

JրǫJ(detAJ)ωJ

To ompute the star operator let as usual J = (jk+1, . . . , jn) be thein reasing omplementary (n− k)-multiindex to J ; we have∗ωJ = sgn(J, J)ǫJωJand

∗(v1 ∧ · · · ∧ vk) =∑

JրǫJǫJ(detAJ)sgn(J, J)ωJ =

= (−1)index∑Jր

sgn(J, J)(detAJ)ωJ

Now let's ompute (v1, . . . ,vk)yω; its omponent orresponding to thebasis elementωJ = ωjk+1

∧ · · · ∧ ωjn

164 CHAPTER 2. ALTERNATING MULTILINEAR FORMSis(v1, . . . ,vk)yωE[ujk+1

, . . . ,ujn] = ωE[ujk+1, . . . ,ujn,v1, . . . ,vk] =

= (−1)k(n−k)ωE[v1, . . . ,vk,ujk+1, . . . ,ujn] =

= (−1)k(n−k) det

a11 . . . a1

k 0 . . . 0. . . . . . . . . . . . . . . . . .. . . . . . . . . 1 . . . 0. . . . . . . . . . . . . . . . . .. . . . . . . . . 0 . . . 0. . . . . . . . . . . . . . . . . .. . . . . . . . . 0 . . . 1. . . . . . . . . . . . . . . . . .an1 . . . ank 0 . . . 0

jk+1

jn

=

= (−1)k(n−k)(−1)k+1+jk+1 . . . (−1)n+jn detAJLet s = j1 + · · ·+ jk and assume proved that surprising property:sgn(J, J) =

(−1)s if 1 + · · ·+ k = 2(−1)s−1 if 1 + · · ·+ k = 2 + 1

(∗)Then as jk+1 + · · ·+ jn = (1+n)n2− (j1 + · · ·+ jk) = (1+n)n

2− s we have

(−1)k+1+jk+1 . . . (−1)n+jn = (−1)(k+1)+···+n(−1)jk+1+···+jn =

= (−1)(k+1+n)(n−k)

2 (−1)(1+n)n

2 (−1)s =

= (−1)(k+1)k

2 (−1)(k+1)n

2 (−1)n(n−k)

2 (−1)(1+n)n

2 (−1)s =

= (−1)(k+1)k

2 (−1)(k+1)n

2 (−1)n(n−k−n−1)

2 (−1)s =

= (−1)(k+1)k

2 (−1)s = sgn(J, J)and we have proved that(v1, . . . ,vk)yω[ujk+1

, . . . ,ujn] =∑

Jր(−1)k(n−k)sgn(J, J)(detAJ)ωJFinally

∗(v1 ∧ · · · ∧ vk) = (−1)index(−1)k(n−k)∑

Jրsgn(J, J)(detAJ)ωJ =

= (−1)k(n−k)(−1)index(v1, . . . ,vk)yω

2.5. HODGE STAR APPLICATION 165∗(v1 ∧ · · · ∧ vk) = (−1)k(n−k)(−1)index(v1, . . . ,vk)yωand

∗(v1, . . . ,vk)yω = (−1)indexv1 ∧ · · · ∧ vkLet us do a ouple of examples of formula (*):a) Let k = 2, n = 4. Then 1 + 2 = 3 and sgn(I, I) = (−1)s−1:sgn(24|13) = (−1)2+4−1 = −1sgn(23|14) = (−1)2+3−1 = 1b) Let k = 3, n = 5. Then 1 + 2 + 3 = 6 and sgn(I, I) = (−1)ssgn(245|13) = (−1)2+4+5 = −1sgn(134|25) = (−1)1+3+4 = 1The author doesn't know how to prove that fa t that strongly seems to betrue.

166 CHAPTER 2. ALTERNATING MULTILINEAR FORMS

Chapter 3Dierential forms3.1 Dierential forms3.1.1 Basi sT Following [Spvk p.86 we dene the tangent spa e to Rn at the point pas:

Rnp = (v,p) : v ∈ Rn,p ∈ RnWe shall write vp instead of (v,p) with the impli it idea that it is a ve tor'at the point p' (despite being simply a ve tor in Rn).In the same way as a ve tor eld in an open set U ⊂ Rn is a map thatassigns to ea h point in U a ve tor (properly speaking we assign to ea hpoint p a ve tor in Rn

p), a dierential k-form in U assigns to ea h p ∈ U analternating k-form in Rnp:ω : U → ∪p∈UΛk(Rn

p)

p 7→ ω(p)Writing dxi for the dierential form that assigns to ea h point p ∈ U , thei-th proje tion of Rn

p

πi : Rnp → R

(v,p) 7→ vi,any dierential k-form in U ⊂ Rn has an expression

ω(x) =∑

1≤i1<···<in≤nai1...ik(x)dxi1 ∧ · · · ∧ dxik167

168 CHAPTER 3. DIFFERENTIAL FORMSω is of lass Cr(U) if the fun tions ai1...ik are in Cr(U). We notate Ωk(U)the set of dierential k-forms in U and shall assume them dierentiable asneeded. As we an add (point by point) two dierential k-forms and multiplya dierential k-form by a s alar, we see that Ωk(U) is a vs.

Problem 85:Letϕ : D → R3be a regular parametrization of a surfa e S in R3 and let n be the asso iatedeld of unit normal ve tors. If θ1, θ2, θ3 are the angles that n has with theaxes, assuming that n is nowhere orthogonal to none of the axes, show thatthe dierential forms

ω1 =1

cos θ1dy ∧ dz, ω2 =

1

cos θ2dz ∧ dx, ω3 =

1

cos θ3dx ∧ dy oin ide on every plane tangent to S.Solution:The unit normal eld n is

n =ϕu ∧ ϕv|ϕu ∧ ϕv|

= (cos θ1, cos θ2, cos θ3)and letting ϕ = (ϕ1, ϕ2, ϕ3), we may express the pre eding equality in om-ponents:ϕ2uϕ

3v − ϕ3

uϕ2v = cos θ1|ϕu ∧ ϕv|

ϕ3uϕ

1v − ϕ1

uϕ3v = cos θ2|ϕu ∧ ϕv|

ϕ1uϕ

2v − ϕ2

uϕ1v = cos θ3|ϕu ∧ ϕv|We want to show that the three forms oin ide in ea h tangent spa e and tothis end it su es to he k that they oin ide when applied to a basis. Wehave:

ω1[ϕu, ϕv] =1

cos θ1dy ∧ dz[ϕu, ϕv] =

1

cos θ1det

(ϕ2u ϕ2

v

ϕ3u ϕ3

v

)= |ϕu ∧ ϕv|and the same is true for ω2 and ω3.

3.1. DIFFERENTIAL FORMS 1693.1.2 Exterior derivativeT The exterior derivative, or E. Cartan's derivative, is the only (see [Janp.138) linear appli ation

d : Ωk(U)→ Ωk+1(U)su h that:a) If f ∈ Ω0(U) then df is the usual dierential.b) d d = 0, whi h we may write d2 = 0. ) d(ω ∧ η) = dω ∧ η + (−1)degree(ω)ω ∧ dη.Using those properties we obtain the formulad(∑

Iրai1...ikdxi1 ∧ · · · ∧ dxik) =

∑

Iրd(ai1...ik) ∧ dxi1 ∧ · · · ∧ dxikA dierential form ω ∈ Ωk(U) is losed whenever dw = 0.Problem 86: Exterior derivative.Compute the exterior derivative of the following dierential forms:a) ω = xdxb) ω = ydy ) ω = ydxd) ω = xydxe) ω = f(x)dxf) ω = f(y)dxg) ω = (arctan y)dx+ x2dz

170 CHAPTER 3. DIFFERENTIAL FORMSSolution:a) dω = dx ∧ dx = 0b) dω = dy ∧ dy = 0 ) dω = dy ∧ dxd) dω = xdy ∧ dxe) dω = f ′(x)dx ∧ dx = 0f) dω = f ′(y)dy ∧ dxg) dω = 11+y2

dy ∧ dx+ 2xdx ∧ dz

Problem 87:Compute the exterior derivative of:a) ω = xydx ∧ dyb) ω = xzdx ∧ dy ) ω = xydx ∧ dzd) ω = exydx ∧ dye) ω = f(x, y)dx ∧ dyf) ω = f(x, z)dx ∧ dyg) ω = f(x, y)dx ∧ dz

3.1. DIFFERENTIAL FORMS 171Solution:a) dω = ydx ∧ dx ∧ dy + xdy ∧ dx ∧ dy = 0b) dω = xdz ∧ dx ∧ dy ) dω = xdy ∧ dx ∧ dzd) dω = 0e) dω = 0f) dω = ∂zf dz ∧ dx ∧ dyg) dω = ∂yf dy ∧ dx ∧ dz

Problem 88:Compute the exterior derivative.a) θ = dω ∧ η − ω ∧ dηb) θ = dω ∧ ξ ∧ η + ω ∧ dξ ∧ η + ω ∧ ξ ∧ dη, ω, ξ being of even degree.Solution:a) We use the formula d(ω∧η) = dω∧η+(−1)deg (ω)ω∧dη = dω∧η+ω∧dη;then:d(dω ∧ η − ω ∧ dη) = d2ω ∧ η − dω ∧ dη − dω ∧ dη − ω ∧ d2η =

= −2dω ∧ dηb) We do it by pie es:α = d(dω ∧ ξ ∧ η) = d(dω ∧ ξ) ∧ η + (−1)deg (dω∧ξ)dω ∧ ξ ∧ dηand as deg dω is odd,

α = d(dω ∧ ξ) ∧ η − dω ∧ ξ ∧ dη =

172 CHAPTER 3. DIFFERENTIAL FORMS= −dω ∧ dξ ∧ η − dω ∧ ξ ∧ dηOn another hand

β = d((ω ∧ dξ) ∧ η) = d(ω ∧ dξ) ∧ η − ω ∧ dξ ∧ dη =

= dω ∧ dξ ∧ η − ω ∧ dξ ∧ dηandγ = d((ω ∧ ξ) ∧ dη) = d(ω ∧ ξ) ∧ dη + ω ∧ ξ ∧ d2η =

= dω ∧ ξ ∧ dη + ω ∧ dξ ∧ dηFinallydθ = α + β + γ = 0

Problem 89: Flux 2-form of the gravitational eld .A dierential form ω is losed if dω = 0. Show that the 2-form in U =R3 − 0

ω =1

r3(xdy ∧ dz + ydz ∧ dx+ zdx ∧ dy), r = |(x, y, z)|is losed.Solution:Remind that

∂xr =x

r, ∂yr =

y

r, ∂zr =

z

r.Then

∂x(x

r3) =

r3 − 3r2 xrx

r6=r2 − 3x2

r5, ∂y(

y

r3) =

r2 − 3y2

r5, ∂z(

z

r3) =

r2 − 3z2

r5anddω =

1

r5((r2−3x2)dx∧dy∧dz+(r2−3y2)dy∧dz∧dx+(r2−3z2)dz∧dx∧dy) =

=1

r5(3r2 − 3(x2 + y2 + z2))dx ∧ dy ∧ dz = 0

3.1. DIFFERENTIAL FORMS 173Problem 90: Ele tromagnetism and dierential forms.When there are harges the spa e a quires a 'state of ele tromagneti tension'that Mi hael Faraday (1791-1867) des ribed by means of the idea of a eld.James Clerk Maxwell (1831-1875) was the rst to establish the ompleteequations of the ele tromagnetism (see [Feyn vol. II):Maxwell equations∇ ·E =

ρ

ǫ0, ∇× E = −∂tB

∇ ·B = 0, c2∇×B = ∂tE +j

ǫ0Two ve tor elds, the ele tri eld E(x, y, z, t) and the magneti eldB(x, y, z, t), des ribe that 'state of ele tromagneti tension' at point (x, y, z)at the instant t. The sour es of that spa e tension are the ele tri harges, de-s ribed by means of a s alar fun tion, the ele tri harge density ρ(x, y, z, t),and the ele tri urrents, des ribed by means of a ve tor fun tion, the ele tri urrent density j(x, y, z, t).Given the sour es Maxwell equations allow (at least in prin iple) the omputation of the ele tri eld E = (Ex, Ey, Ez) and the magneti eldB = (Bx, By, Bz). On e the elds are known, we an ompute the for e theyexert on a harge q that moves with a velo ity v through Lorentz's formula:

F = q(E + v ×B)Then Newton equation allows the omputation of the movement of the harge.Notation: In this problem and the next one we shall notate the els bymeans of an arrow, thus: −→F = (F1, F2, F3). We shall use boldfa e lettersto notate dierential forms. Using ad ho units we an rewrite Maxwellequations in the form (noti e the symmetries):∇ · −→B = 0 , ∇×−→E + ∂t

−→B = 0

∇ · −→E = ρ , ∇×−→B − ∂t−→E =

−→j

174 CHAPTER 3. DIFFERENTIAL FORMSa) The rst two equations ∇·−→B = 0,∇×−→E +∂t−→B = 0 don't ontain anytra e of the sour es. We dene the ele tri eld 1-form, the magneti eld 2-form, and the ele tromagneti tensor or Faraday's 2-form by:

E = Exdx+ Eydy + Ezdz

B = Bxdy ∧ dz +Bydz ∧ dx+Bzdx ∧ dyF = E ∧ dt+ BShow the equivalen e∇ · −→B = 0 , ∇×−→E + ∂t

−→B = 0 ⇔ dF = 0b) The se ond group of equations ∇ · −→E = ρ,∇ × −→B − ∂t

−→E =

−→j on-tains the ontribution of the sour es. Here we introdu e the followingdierential forms:

E = Exdy ∧ dz + Eydz ∧ dx+ Ezdx ∧ dyB = Bxdx+Bydy +Bzdz

F = −B ∧ dt+ Eas well as the urrent 2-form, the density 3-form and a 3-form alledthe quadridimensional urrent ve tor:j = jxdy ∧ dz + jydz ∧ dx+ jzdx ∧ dyr = ρdx ∧ dy ∧ dzJ = j ∧ dt− rShow the equivalen e

∇ · −→E = ρ,∇×−→B − ∂t−→E =

−→j ⇔ dF = −JNoti e that we have redu ed the four Maxwell equations to only two equa-tions.Solution:a) For the time being we do an exer ise about the exterior derivative; lateron (see p.195) we have a better approa h. We omit momentarily the

3.1. DIFFERENTIAL FORMS 175sign ∧ in the exterior produ ts but without forgetting that the order isimportant; for instan e dxdydz will mean dx ∧ dy ∧ dz et .We have:dF = dE ∧ dt+ dBLet's ompute rst ea h term:

dE ∧ dt = ∂yExdydxdt+ ∂zExdzdxdt+ ∂xEydxdydt+

+ ∂zEydzdydt+ ∂xEzdxdzdt+ ∂yEzdydzdtanddB = ∂xBxdxdydz + ∂yBydydzdx+ ∂zBzdzdxdy +

+ ∂tBxdtdydz + ∂tBydtdzdx+ ∂tBzdtdxdyThen:dF = dE ∧ dt+ dB =

= (∂yEz − ∂zEy + ∂tBx)dydzdt+ (∂zEx − ∂xEz + ∂tBy)dzdxdt+

+ (∂xEy − ∂yEx + ∂tBz)dxdydt+ (∂xBx + ∂yBy + ∂zBz)dxdydzand we an read in this expression for dF thatdF = 0⇔ div −→B = 0, rot −→E + ∂t

−→B = 0b) We have

dF = −dB ∧ dt+ dECompute rst ea h term:−dB ∧ dt = −(∂yBxdydx+ ∂zBxdzdx+ ∂xBydxdy +

+ ∂zBydzdy + ∂xBzdxdz + ∂yBzdydz)dt

dE = ∂xExdxdydz + ∂yEydydzdx+ ∂zEzdzdxdy +

+ ∂tExdtdydz + ∂tEydtdzdx+ ∂tEzdtdxdyThendF = (∂zBy − ∂yBz + ∂tE1)dydzdt+ (∂xBz − ∂zB1 + ∂tEy)dzdxdt+

+ (∂yBx − ∂xBy + ∂tEz)dxdydt+

+ (∂xEx + ∂yEy + ∂zEz)dxdydzand we an read in this expression for dF thatdF = −J⇔ div −→E = ρ,−rot −→B + ∂t

−→E = −−→j

176 CHAPTER 3. DIFFERENTIAL FORMSProblem 91:Let ωi ∈ Ω1(Rn), i = 1, . . . , r < n−1 be dierential 1-forms linearly indepen-dent at 0 ∈ Rn. Show that in a neigborhood of 0 the following statementsare equivalent:a) There are 1-forms αij ∈ Ω1(Rn), i, j = 1, . . . , r su h thatdωi =

r∑

j=1

αij ∧ ωj, i = 1, . . . , rWe may write these relations in a more pi turesque form:

dω1

· · ·dωr

=

α11 · · · α1r

· · · · · · · · ·αr1 · · · αrr

∧

ω1

· · ·ωr

b) There is a 1-form λ su h thatd(ω1 ∧ · · · ∧ ωr) = λ ∧ ω1 ∧ · · · ∧ ωr ) dωi ∧ ω1 ∧ · · · ∧ ωr = 0, i = 1, . . . , rSolution:a)⇒b):

d(ω1∧· · ·∧ωr) = (dω1∧ω2∧· · ·∧ωr)−(ω1∧dω2∧· · ·∧ωr)+· · ·+(−1)r−1(ω1∧ω2∧· · ·∧dωr) =

= ((r∑

j1=1

α1j1 ∧ ωj1)∧ ω2 ∧ · · · ∧ ωr)− (ω1 ∧ (r∑

j2=1

α2j2 ∧ ωj2)∧ · · · ∧ ωr) + · · ·+

+(−1)r−1(ω1 ∧ ω2 ∧ · · · ∧ (r∑

jr=1

αrjr ∧ ωjr)) =

= (α11∧ω1∧ω2∧· · ·∧ωr)−(ω1∧α22∧ω2∧· · ·∧ωr)+· · ·+(−1)r−1(ω1∧ω2∧· · ·∧αrr∧ωr) =

= (α11 + · · ·+ αrr) ∧ ω1 ∧ ω2 ∧ · · · ∧ ωrand then λ = α11 + · · ·+ αrr (the tra e of the matrix of the 1-forms (αij)).

3.1. DIFFERENTIAL FORMS 177b)⇒ ):We show, for instan e, that dω1 ∧ ω1 ∧ ω2 ∧ · · · ∧ ωr = 0:0 = ω1 ∧ λ ∧ ω1 ∧ ω2 ∧ · · · ∧ ωr = ω1 ∧ d(ω1 ∧ · · · ∧ ωr) =

= ω1 ∧ (dω1 ∧ ω2 ∧ · · · ∧ ωr)− ω1 ∧ (ω1 ∧ dω2 ∧ · · · ∧ ωr) + · · ·++(−1)r−1ω1 ∧ (ω1 ∧ ω2 ∧ · · · ∧ dωr) =

= ω1 ∧ dω1 ∧ ω2 ∧ · · · ∧ ωr = dω1 ∧ ω1 ∧ ω2 ∧ · · · ∧ ωr. )⇒a): As the r linear forms ωi(0), i = 1, . . . r are li in (Rn0)∗, if its ompo-nents are

ω1(0) = a11(0)dx1 + · · ·+ a1n(0)dxn. . . . . .ωr(0) = ar1(0)dx1 + · · ·+ arn(0)dxnthere will be an r-minor that doesn't vanish at 0, let's say ∆ = det(aij(0)) 6=

0, i, j = 1, . . . , r. Be ause of the ontinuity this minor won't vanish in U ,a ertain neighbourhood of 0. So we may solve dierentiably the systemand express dx1, . . . , dxr in terms of ω1, . . . , ωr, dxr+1, . . . dxn. Writing ωi =dxi, i = r+1, . . . , n every dierentiable form an be written in U in terms ofthe basis ω1, . . . , ωn. For instan e we an write

dωk =∑

1≤i<j≤nbkijωi ∧ ωj, k = 1, . . . , rand substituting it into the ondition dωk ∧ω1 ∧ · · · ∧ωr = 0, k = 1, . . . , r weobtain

(∑

1≤i<j≤nbkijωi ∧ ωj) ∧ ω1 ∧ · · · ∧ ωr = 0Be ause of repetitions in the produ t this is equivalent to

(∑

r+1≤i<j≤nbkijωi ∧ ωj) ∧ ω1 ∧ · · · ∧ ωr = 0and as the

ω1 ∧ · · · ∧ ωr ∧ ωr+1 ∧ ωr+2, . . . , ω1 ∧ · · · ∧ ωr ∧ ωn−1 ∧ ωn

178 CHAPTER 3. DIFFERENTIAL FORMSare a system of generators of Ωr+2(U) we see that bkij = 0, r + 1 ≤ i < j ≤ nand thendωk =

∑

1≤i<j≤rbkijωi ∧ ωj =

∑

1≤i≤r−1

ωi ∧ (∑

i+1≤j≤rbkijωj)We see that we must hoose

αkl = −n∑

j=l+1

bkljωj, l = 1, . . . , r − 1

Problem 92:Let p ∈ R3 and ω = ady ∧ dz + bdz ∧ dx+ cdx ∧ dy ∈ Ω2(R3) a dierential2-form su h that ω(p) 6= 0. Show that if in a neighborhood of p there are afun tion f and a 1-form η su h that ω = η∧df then df(p) 6= 0 and f satisesan equation relating its partial derivatives. Is there a re iproque?Solution:From ω = η ∧ df we must have df(p) 6= 0 be ause otherwise ω(p) = 0. Letη = Pdx+Qdy +Rdz; then the equality an be writtenady∧dz+bdz∧dx+cdx∧dy = (Pdx+Qdy+Rdz)∧ (

∂f

∂xdx+

∂f

∂ydy+

∂f

∂zdz)that is seen to be equivalent to:

(∂zf)Q− (∂yf)R = a(∂xf)R− (∂zf)P = b(∂yf)P − (∂xf)Q = c

The determinant of the matrix of the system vanishes:detA = det

0 ∂zf −∂yf−∂zf 0 ∂xf∂yf −∂xf 0

= 0Three minors of the matrix are (∂xf)2, (∂yf)2, (∂zf)2 whose sum is (∂xf)2 +(∂yf)2 +(∂zf)2 6= 0 at the point p be ause df(p) 6= 0, and by ontinuity this

3.1. DIFFERENTIAL FORMS 179doesn't vanish in a neighborhood U of p. The matrix of the system has rank2 in U ; assume for instan e that ∂zf 6= 0. As we know that η(p) 6= 0 thesystem has a solution and we haverank 0 ∂zf a−∂zf 0 b∂yf −∂xf c

= 2 ≡ afxfz + bfyfz + c(fz)2 = 0the sought for equation that f must satisfy.Re ipro ally if f is a fun tion satisfying the pre eding equation and if df(p) 6=

0, the system is ompatible and there is η su h that ω = η ∧ df .3.1.3 Pull-ba k for dierential forms

T Whenever we have a dierentiable mapf : U ⊂ Rn → Rmits dierential at a point is a linear map between tangent spa es

(Df)(p) : Rnp → Rm

f(p)that we shall notate (f∗)p. Then for ea h p we may onsider the mapΛk(Rn

p)← Λk(Rmf(p)) : (f ∗)f(p)that we introdu ed in problem p.58. Joining all those f ∗ we have

Ωk(U)f∗← Ωk(Rm)

f ∗ω ← ωdened by (f ∗ω)(p) = f ∗(ω(f(p)) that we may write with the notation(f ∗ω)p = f ∗(ωf(p)), pla ing the points where the forms are evaluated assubs ripts.

180 CHAPTER 3. DIFFERENTIAL FORMSProblem 93: Computations with f ∗.Let f : U ⊂ Rn → Rm, g : Rm → R, h : Rm → Rp be dierentiable fun tionsand ω, ω1, ω2, η dierential forms of Rm, ω1 and ω2 of the same degree. Showthat:a) f ∗(ω1 + ω2) = f ∗(ω1) + f ∗(ω2)b) f ∗(gω) = (g f)(f ∗ω) ) f ∗(ω ∧ η) = f ∗ω ∧ f ∗ηd) (h f)∗ = f ∗ h∗e) If (x1, . . . , xn) are oordinates in Rn and (y1, . . . , ym) are oordinatesin Rm show thatf ∗(dyi) = d(f ∗yi)Solution:a) Let p ∈ U and v1, . . . ,vk ve tors in Rn

p; thenf ∗((ω1 + ω2))p[v1, . . . ,vk] = (ω1 + ω2)f(p)[(f∗)pv1, . . . (f∗)pvk] =

= (ω1)f(p)[(f∗)pv1, . . . (f∗)pvk] +

+ (ω2)f(p)[(f∗)pv1, . . . (f∗)pvk] =

= ( f ∗(ω1)p + f ∗(ω2)p)[v1, . . . ,vk]b) The omposition is:U ⊂ Rn f→ Rm g→ RLet ω ∈ Ωk(Rm), p ∈ U and v1, . . . ,vk ∈ Rn

pand f∗ = DfpWe have(f ∗(gω))p[v1, . . . ,vk] = (gω)f(p)[(f∗)pv1, . . . (f∗)pvk] =

= g(f(p))ωf(p)[(f∗)pv1, . . . (f∗)pvk] =

= (g f)(p)(f ∗(ω))p[v1, . . . ,vk]This result shows that to ' hange variables' in gω we must ' hangevariables' in g (obtaining a new fun tion) and multiply by the result of

3.1. DIFFERENTIAL FORMS 181' hanging variables' in ω.Here we an see the usefulness of the notation that distinguishes thelinear a tion on the tangent spa e at a point (bra kets) from the pointwhere the form is evaluated (parentheses). ) This is an algebrai property satised in ea h Λk(Rmq ) (see p.91).d) The omposition is:

U ⊂ Rn f−→ Rm h

−→ RpIf ω ∈ Ωk(Rp), p ∈ U , v1, . . . ,vk ∈ Rnp, we have

((h f)∗ω)p[v1, . . . ,vk] = ωh(f(p))[(h f)∗v1, . . . , (h f)∗vk] =

= ωh(f(p))[h∗(f∗v1), . . . , h∗(f∗vk)] =

= h∗(ω)f(p)[f∗v1, . . . , f∗vk] =

= f ∗(h∗(ω))p[v1, . . . ,vk] =

= ((f ∗ h∗)ω)p[v1, . . . ,vk]e) Let (dx1, . . . , dxn) be the dual basis of the anoni al basis in Rn, (dy1, . . . , dym)the dual basis of the anoni al basis in Rm, and v ∈ Rnp. Then

(f ∗(dyi))p[v] = dyi[(f∗)pv]

= dyi[Dpf1v, . . . , Dpf

mv] =

= DpfivWe nd the omponents of f ∗(dyi) in the basis (dx1, . . . , dxn) applyingthe 1-form to the ve tors (u1, . . . ,um) of the anoni al basis of Rm.Using the pre eding equality we have

f ∗(dyi)[uj ] = Dpfiuj =

∂f i

∂xj|pThen

f ∗(dyi) =∂f i

∂x1dx1 + · · ·+ ∂f i

∂xndxn = df i = d(yi f) = d(f ∗yi)This property is alled the naturality of f ∗ respe t to d.

182 CHAPTER 3. DIFFERENTIAL FORMSProblem 94:Call r, θ the oordinates in a rst R2 and x, y the oordinates in a se ond R2.Consider the fun tionf : R2 → R2

(r, θ) 7→ (r cos θ, r sin θ)and the dierential 1-forms ω = (x2 + y2)dx, η = (x − y)dy. Computef ∗(ω), f ∗(η).Solution:Using the results in the pre eding problem we have

f ∗(ω) = r2f ∗(dx) =

= r2d(r cos θ) =

= r2 cos θdr − r3 sin θdθ

f ∗(η) = (r cos θ − r sin θ)f ∗(dy) =

= (r cos θ − r sin θ)d(r sin θ) =

= (r cos θ − r sin θ)(sin θdr + r cos θdθ)

Problem 95:Let f : R2 → R2 be f(x, y) = (xy, 1). Compute f ∗(dx), f ∗(dy), f ∗(ydx), f ∗(xdx).Solution:f ∗(dx) = d(f ∗x) = d(x f) = d(xy) = ydx+ xdy

f ∗(dy) = d(f ∗y) = d(y f) = d(1) = 0

f ∗(ydx) = (y f)f ∗(dx) = 1(ydx+ xdy)

f ∗(xdx) = (x f)f ∗(dx) = xy(ydx+ xdy)

3.1. DIFFERENTIAL FORMS 183Problem 96:Letf : Rn → Rnbe f(X) = AXT +BT where X = (x1, . . . , xn), B = (b1, . . . , bn), A an (n, n)matrix. Compute f ∗(dx1 ∧ · · · ∧ dxn).Solution:Let (e1, . . . , en) be the anoni al basis of Rn; using the denition of f ∗ wehave:

(f ∗(dx1∧· · ·∧dxn))p[e1, . . . , en]p = (dx1∧· · ·∧dxn)f(p)[(f∗)p[e1], . . . , (f∗)p[en]]but (f∗)p = Dpf and the matrix of this linear map in the anoni al basis isA. Then we have at ea h p

((f∗)p[e1], . . . , (f∗)p[en]) = (e1, . . . , en)A

(f ∗(dx1∧· · ·∧dxn))p[e1, . . . , en]p = (detA)(dx1∧· · ·∧dxn)f(p)[e1, . . . , en]f(p) = detAor, brieyf ∗(dx1 ∧ · · · ∧ dxn) = (detA)(dx1 ∧ · · · ∧ dxn)

T Let's ompute f ∗(ω) in three frequent ases:a) One variablef : R → R3

u 7→

f 1(u)f 2(u)f 3(u)

The matrix of the dierential of f in the anoni al basis isf ′ =

(f 1)′

(f 2)′

(f 3)′

← f ∗(dx)← f ∗(dy)← f ∗(dz)

184 CHAPTER 3. DIFFERENTIAL FORMSand we see that the rows are the respe tive omponents of f ∗(dx), f ∗(dy), f ∗(dz)in the basis du of R∗. If we have a 1-form ω = Pdx+Qdy+Rdz, thenf ∗(ω) = f ∗(Pdx+Qdy +Rdz) =

= (P f)f ∗(dx) + (Q f)f ∗(dy) + (R f)f ∗(dz)and in the matrix of f ′ we see that f ∗(dx) = (f 1)′du, et .Observation: If η is a 2-form f ∗(η) = 0; why?b) Two variables:f : R2 → R3

(u, v) 7→

f 1(u, v)f 2(u, v)f 3(u, v)

The matrix of the dierential of f in the anoni al basis isf ′ =

∂uf

1 ∂vf1

∂uf2 ∂vf

2

∂uf3 ∂vf

3

← f ∗(dx)← f ∗(dy)← f ∗(dz)and we see that the rows are the respe tive omponents of f ∗(dx),

f ∗(dy), f ∗(dz), now in the basis du, dv .• If we have a 1-form α = Pdx + Qdy + Rdz, let's look at thetransform of a term, say Qdy:

f ∗(Qdy) = (Q f)f ∗(dy) =

= (Q f)(∂uf2du+ ∂vf

2dv)

• If we have a 2-form ω = Pdy ∧ dz + Qdz ∧ dx + Rdx ∧ dy, let'slook at the transform of a term, say Pdy ∧ dz:f ∗(Pdy ∧ dz) = (P f)f ∗(dy ∧ dz) =

= (P f) f ∗(dy) ∧ f ∗(dz)

3.1. DIFFERENTIAL FORMS 185Nowf ∗(dy)∧ f ∗(dz) = det

(∂uf

2 ∂vf2

∂uf3 ∂vf

3

)du∧dv =

∂(f 2, f 3)

∂(u, v)du∧dvAnd analogously

f ∗(dz)∧ f ∗(dx) = det

(∂uf

3 ∂vf3

∂uf1 ∂vf

1

)du∧dv =

∂(f 3, f 1)

∂(u, v)du∧dv

f ∗(dx)∧ f ∗(dy) = det

(∂uf

1 ∂vf1

∂uf2 ∂vf

2

)du∧dv =

∂(f 1, f 2)

∂(u, v)du∧dv ) Three variables:

f : R3 → R3

(u, v, t) 7→

f 1(u, v, t)f 2(u, v, t)f 3(u, v, t)

f ′ =

∂uf

1 ∂vf1 ∂tf

1

∂uf2 ∂vf

2 ∂tf2

∂uf3 ∂vf

3 ∂tf3

← f ∗(dx)← f ∗(dy)← f ∗(dz)and we see that the rows are the respe tive omponents of f ∗(dx), f ∗(dy), f ∗(dz)in the basis du, dv, dt.

• If we have a 1-formω = Pdx+Qdy +Rdz

f ∗(Pdx+Qdy+Rdz) = (P f)f ∗(dx)+(Qf)f ∗(dy)+(Rf)f ∗(dz)that we an read in the matrix of f ′.• If we have a 2-formω = Pdy∧dz+Qdz∧dx+Rdx∧dy, let's lookat the transform of a term, say Pdy ∧ dz

f ∗(Pdy ∧ dz) = (P f) f ∗(dy) ∧ f ∗(dz) =

= (P f)(∂(f 2, f 3)

∂(u, v)du∧ dv+

∂(f 2, f 3)

∂(v, t)dv ∧ dt+ ∂(f 2, f 3)

∂(u, t)du∧ dt)and analogous formulae for f ∗(dz ∧ dx), f ∗(dx ∧ dy).

186 CHAPTER 3. DIFFERENTIAL FORMS• If we have a 3-formω = Pdx ∧ dy ∧ dz,

f ∗(ω) = f ∗(Pdx ∧ dy ∧ dz) = (P f)f ∗(dx ∧ dy ∧ dz) =

= (P f)(det f ′)du ∧ dv ∧ dtThe two following problems are examples of the hange of oordinates to ylindri al oordinates and spheri al oordinates (see those oordinate sys-tems in p.197) of a form given in artesian oordinates.Problem 97:Let f(r, θ, z) = (r cos θ, r sin θ, z) and h : R3 → R a fun tion. Compute:a) f ∗(hdx ∧ dy ∧ dz)b) f ∗(xdy ∧ dz)Solution:The ja obian matrix of f is:f ′(r, θ, z) =

cos θ −r sin θ 0sin θ r cos θ 0

0 0 1

, det f ′ = ra) As we know from the pre eding problemf ∗(hdx ∧ dy ∧ dz) = (h f)rdr ∧ dθ ∧ dzb) Following the same method

f ∗(xdy ∧ dz) = r cos θ(det

(sin θ r cos θ

0 0

)dr ∧ dθ+

+ det

(sin θ 0

0 1

)dr ∧ dz + det

(r cos θ 0

0 1

)dθ ∧ dz) =

= r cos θ sin θdr ∧ dz + r2 cos 2θdθ ∧ dz

3.1. DIFFERENTIAL FORMS 187Problem 98:Let f(r, ϕ, θ) = (r sinϕ cos θ, r sinϕ sin θ, r cosϕ). Compute:a) f ∗(dx), f ∗(dy), f ∗(dz)b) f ∗(dx ∧ dy), f ∗(dy ∧ dz), f ∗(dz ∧ dx) ) f ∗(dx ∧ dy ∧ dz)Solution:The ja obian matrix of f is:f ′(r, ϕ, θ) =

sinϕ cos θ r cosϕ cos θ −r sinϕ sin θsinϕ sin θ r cosϕ sin θ r sinϕ cos θ

cosϕ −r sinϕ 0

, det f ′ = r2 sinϕa) We an read dire tly in f ′ thati)f ∗(dx) = sinϕ cos θdr + r cosϕ cos θdϕ− r sinϕ sin θdθii)f ∗(dy) = sinϕ sin θdr + r cosϕ sin θdϕ+ r sinϕ cos θdθiii)

f ∗(dz) = cosϕdr − r sinϕdϕb) As in pre eding problems we have:i)f ∗(dx∧ dy) =

∂(x, y)

∂(r, ϕ)dr ∧ dϕ+

∂(x, y)

∂(r, θ)dr ∧ dθ+

∂(x, y)

∂(ϕ, θ)dϕ∧ dθ =

= r sin 2ϕdr ∧ dθ + r2 sinϕ cosϕdϕ ∧ dθii)f ∗(dy∧dz) = −r sin θdr∧dϕ−r sinϕ cosϕ cos θdr∧dθ+r2 sin 2ϕ cos θdϕ∧dθ

188 CHAPTER 3. DIFFERENTIAL FORMSiii)f ∗(dx∧dz) = −r cos θdr∧dϕ+r sinϕ cosϕ sin θdr∧dθ−r2 sin 2ϕ sin θdϕ∧dθ ) Andf ∗(dx ∧ dy ∧ dz) = det f ′dr ∧ dϕ ∧ dθ = r2 sinϕdr ∧ dϕ ∧ dθ

T In many omputations with f ∗ the naturality property respe t to theexterior derivative is useful (we have seen an example in p.180): wheneverwe have a dierentiable fun tion f : U ⊂ Rn → Rm, the following diagramis ommutativeΩk(U)

d→ Ωk+1(U)f ∗ ↑ ↑ f ∗

Ωk(Rm)d→ Ωk+1(Rm)that is, if ω ∈ Ωk(Rm) then d(f ∗ω) = f ∗(dω).

Problem 99: Che king.Letω =

(xdy − ydx) ∧ dzx2 + y2 + z2

,a 2-form of U = R3 − 0 and let f be the same fun tion of the pre edingproblem. Che k that f ∗(dω) = d(f ∗(ω)).Solution:Rewrite ωω =

x

r2dy ∧ dz − y

r2dx ∧ dza)

dω =r2 − 2x2

r4dx ∧ dy ∧ dz − r2 − 2y2

r4dy ∧ dx ∧ dz =

=2z2

r4dx ∧ dy ∧ dz

3.2. TRANSLATION ISOMORPHISMS 189and thenf ∗(dω) =

2r2 cos 2ϕ

r4f ∗(dx ∧ dy ∧ dz) =

2 cos 2ϕ

r2r2 sinϕdr ∧ dϕ ∧ dθ =

= 2 cos 2ϕ sinϕdr ∧ dϕ ∧ dθb)f ∗(ω) =

r sinϕ cos θ

r2f ∗(dy ∧ dz)− r sinϕ sin θ

r2f ∗(dx ∧ dz) =

=r sinϕ cos θ

r2(−r sin θ dr∧dϕ−r sinϕ cosϕ cos θ dr∧dθ+r2 sin 2ϕ cos θ dϕ∧dθ)−

−r sinϕ sin θ

r2(−r cos θ dr∧dϕ+r sinϕ cosϕ sin θ dr∧dθ−r2 sin 2ϕ sin θ dϕ∧dθ) =

= − sin 2ϕ cosϕdr ∧ dθ + r sin 3ϕdϕ ∧ dθandd(f ∗(ω)) = (−2 sinϕ cos 2ϕ+sin 3ϕ) dϕ∧dr∧dθ+sin 3ϕdr∧dϕ∧dθ =

= 2 sinϕ cos 2ϕdr ∧ dϕ ∧ dθ

3.2 Translation isomorphisms3.2.1 Metri s in an open setT Remind (see p.28) that a metri in a vs E is a g ∈ T 2E symmetri andnondegenerated. If U ⊂ Rn is an open set, a metri in U is to give a metri in ea h tangent spa e Rn

p, p ∈ U and this in a dierentiable way (:= its oe ients in a oordinate expression must be dierentiable):g : U → ∪pT 2(Rn

p)p 7→ gpwhere for ea h p ∈ U , gp ∈ T 2(Rn

p) is a bilinear symmetri and nondegener-ated tensor.

190 CHAPTER 3. DIFFERENTIAL FORMSExample: Eu lidian standard metri .It is dened on U = Rn bygp(x,y) = x1y1 + · · ·+ xnynWe may write it as well without any mention to the ve tors

gp = dx1 ⊗ dx1 + · · ·+ dxn ⊗ dxnOne sees this written as well in the formds2 = dx2

1 + · · ·+ dx2nExample: Lorentz metri .It is dened on U = R4 by

gp(x,y) = x1y1 + x2y2 + x3y3 − x4y4that we may write asgp = dx1 ⊗ dx1 + dx2 ⊗ dx2 + dx3 ⊗ dx3 − dx4 ⊗ dx4or

ds2 = dx21 + dx2

2 + dx23 − dx2

4

3.2.2 Canoni al isomorphismsT Consider a metri in U ⊂ Rn; for ea h p ∈ U we have a metri in Rn

pand we have there a rst anoni al isomorphism p : Rnp → (Rn

p)∗. Joining althese isomorphisms we obtain a linear map from V(U), the ve tor elds inU , to Ω1(U), the 1-forms in U

: V(U) → Ω1(U)X 7→ X

,where (X)p = p(Xp). This map is also alled rst anoni al isomorphism.

3.2. TRANSLATION ISOMORPHISMS 191If we have a metri in U ⊂ R3 and at ea h R3p we have an orientation hoosed in a ontinuous manner, then at ea h R3

p we have a volume form ωpand for ea h p ∈ U we have a se ond anoni al isomorphism (2)p. Joiningall these we have a linear map2 : V(U) → Ω2(U)

X 7→ 2X,where (2X)p = (2)p(Xp). It is alled se ond anoni al isomorphism.Problem 100:Take at ea h R3

p the standard eu lidian metri and the standard orientation.Compute and 2 in anoni al bases.Solution:The metri matrix in the anoni al bases is I and we obtainV(U)

−→ Ω1(U)

X = X1e1 +X2e2 +X3e3 7→ αX = X1dx+X2dy +X3dzandV(U)

2−→ Ω2(U)

X = X1e1 +X2e2 +X3e3 7→ βX = X1dy ∧ dz +X2dz ∧ dx+X3dx ∧ dy

Problem 101:Compute the exterior derivative of X and 2X.Solution:a)d(X) = (∂yX3 − ∂zX2)dy ∧ dz + (∂zX1 − ∂xX3)dz ∧ dx+ (∂xX2 − ∂yX1)dx ∧ dy =

= 2(rotX)

192 CHAPTER 3. DIFFERENTIAL FORMSb)d(2X) = (∂xX1 + ∂yX2 + ∂zX3)dx ∧ dy ∧ dz = (divX)dx ∧ dy ∧ dzWe see that the dieretial operators rot and div are subsumed in the exteriorderivative.

T Let U be an open set in R3 andF(U) = f : U → R, f dierentiable.Completing the isomorphisms , 2 with the following ones

F(U)Id→ Ω0(U)

f 7→ fandF(U)

3→ Ω3(U)f 7→ fdVwhere dV = dx ∧ dy ∧ dz is the anoni al volume form in R3, we an sumup the information in the pre eding problem in the following fundamentaldiagram :

0 → F(U)grad→ V(U)

rot→ V(U)div→ F(U) → 0Id ↓ ↓ 2 ↓ 3 ↓

0 → Ω0(U)d→ Ω1(U)

d→ Ω2(U)d→ Ω3(U) → 0The sequen e of Ωk(U) and d is alled the de Rham omplex of the open set

U .Problem 102: Ve tor identities (Leibniz rule).Let f, g : U ⊂ R3 → R be dierentiable fun tions and F,G ∈ V(U) dif-ferentiable ve tor elds. Using the pre eding diagram and reminding thatF ∧ G = 2(F×G) and that F ∧ 2G = F ·GdV , (see p.130) show:

3.2. TRANSLATION ISOMORPHISMS 193a) ∇(fg) = (∇f)g + f(∇g)b) ∇ · (fF) = ∇f · F + f∇ · F ) ∇× (fF) = (∇f)× F + f(∇× F)d) ∇ · (F×G) = (∇× F) ·G− F · (∇×G)Solution:a)∇(fg) = ♯d(fg) = ♯(gdf + fdg) =

= g♯df + f♯dg = g(∇f) + f(∇g)b)∇ · (fF)dV = d(2(fF)) = d(f2F)

= df ∧ 2F + f ∧ d(2F) =

= (gradf) ∧ 2F + f(divF)dV =

= (gradf · F + f(divF))dV )∇× (fF) = ♯2d((fF)) = ♯2d(fF) =

= ♯2(df ∧ F + f ∧ 2rotF) =

= ♯2(gradf ∧ F) + ♯2(f ∧ 2rotF) =

= ♯2(2(gradf × F)) + f♯2(2rotF) =

= gradf × F + frotF

194 CHAPTER 3. DIFFERENTIAL FORMSd)∇ · (F×G)dV = d(2(F×G)) = d(F ∧ bG) =

= d(F) ∧ G− F ∧ d(G) =

= (2rotF) ∧ G− F ∧ (2rotG) =

= G ∧ (2rotF)− F ∧ (2rotG) =

= ((rotF) ·G− F · (rotG))dV

Problem 103:Let e = (e1, e2, e3) be the anoni al basis of R3, ω = (ω1, ω2, ω3) li dieren-tial 1-forms, and F = Xe1 + Y e2 + Ze3 a dierentiable ve tor eld. Showthatd(F) =

1

∆(ω1[rotF]ω2∧ω3+ω2[rotF]ω3∧ω1+ω3[rotF]ω1∧ω2),∆ = det(ωi[ej ])Hint: ∆ = ω1 ∧ ω2 ∧ ω3[e1, e2, e3].Solution:We want the omponents (a, b, c) of d(F) in the basis ω∧:

d(F) = aω2 ∧ ω3 + bω3 ∧ ω1 + cω1 ∧ ω2Using the hint we haveω1 ∧ d(F) = aω1 ∧ ω2 ∧ ω3,and apply it to (e1, e2, e3):

ω1 ∧ d(F)[e1, e2, e3] = aω1 ∧ ω2 ∧ ω3[e1, e2, e3] = a∆

a =1

∆ω1 ∧ d(F)[e1, e2, e3]So it su es to show that ω1 ∧ d(F)[e1, e2, e3] = ω1[rotF] (and analogousresults for the other omponents); reminding that d(F) = 2(rot F) andwriting G = ♯ω1, and dV for the volume form, we obtain

ω1 ∧ d(F) = G ∧ 2(rot F) = (G · rotF)dV

3.2. TRANSLATION ISOMORPHISMS 195ω1 ∧ d(F)[e1, e2, e3] = (G · rotF)dV [e1, e2, e3] =

= G · rotF = ♯ω1 · rotF = ω1[rotF]

Problem 104: Maxwell equations and dierential forms.Remind (see p.173) that, from the harge density, the urrent density, theele tri eld and the magneti eld we onstru ted the dierential formsF = E ∧ dt +B

F = −B ∧ dt +E

J = j ∧ dt −rwhereE =

−→E , B = 2

−→B

E = 2−→E , B =

−→B

j = 2−→j , r = 3ρand , 2, 3 are the anoni al isomorphisms with t xed.Using the anoni al isomorphisms show that Maxwell equations are equiva-lent to

dF = 0, dF = −JSolution:Note that E,B,J depend on the four variables x, y, z, t. But, for ea h t xed,we may use the anoni al isomorphisms for the ase of three variables:F = (

−→E ) ∧ dt+ 2

−→B

F = −−→B ∧ dt+ 2−→E

J = 2−→j ∧ dt− ρdVWe ompute the exterior derivatives of F,F:

196 CHAPTER 3. DIFFERENTIAL FORMSa)dF = d(

−→E ) ∧ dt+ d(2

−→B ) =

= 2(rot −→E ) ∧ dt+ div −→B dx ∧ dy ∧ dz+

+∂tBxdt ∧ dy ∧ dz + ∂tBydt ∧ dz ∧ dx+ ∂tBzdt ∧ dx ∧ dy =

= 2(rot −→E ) ∧ dt+ div −→B dx ∧ dy ∧ dz + 2(∂t−→B ) ∧ dtand the ondition dF = 0 readsrot −→E + ∂t

−→B = 0div −→B = 0b)

dF = −d(−→B ) ∧ dt+ d(2−→E ) =

= −2(rot−→B ) ∧ dt+ div−→E dx ∧ dy ∧ dz + 2(∂t−→E ) ∧ dt

= (−2(rot−→B ) + 2(∂t−→E )) ∧ dt+ div−→E dx ∧ dy ∧ dzand the ondition dF = −J reads

(−2(rot−→B ) + 2(∂t−→E )) ∧ dt+ div−→E dV = −(2

−→j ∧ dt− ρdV )or

(−2(rot−→B ) + 2(∂t−→E ) + 2

−→j ) ∧ dt+ (div−→E − ρ) dV = 0i) The oe ient of dV must vanish:div−→E − ρ = 0⇒ div−→E = ρii) On another hand

2(−rot−→B + ∂t−→E +

−→j ) ∧ dt = 0The 2-form 2(−rot−→B +∂t

−→E +−→j ) in the variables dx, dy, dz mustvanish but as 2 is an isomorphism it is equivalent to:

−rot−→B + ∂t−→E +

−→j = 0⇔ rot−→B = ∂t

−→E +

−→j

3.3. COORDINATES 1973.3 Coordinates3.3.1 Coordinates and oordinate urvesT The set of all ordered lists of n real numbers is Rn; the addition of twosu h lists and the produ t of a number by a list gives the set a vs stru ture.Then one may think ea h list as a ve tor, with the usual adhered geometri al on epts.This vs has a spe ial basis in orporated, ei = (0, . . . , 0,

i)

1, 0, . . . , 0), i =1, . . . , n and the singularity of this basis is that the omponents of a ve tor(the omponents of a list) in this basis, are pre isely the numbers in the list.This is handy to single out a ve tor: simply exhibit the list the ve tor is.Polar oordinates systemWe treat in detail polar oordinates as a model for other systems of oordi-nates.In the following gure we see how to assign the polar oordinates (r, θ)to a point P = (x, y) (here we are using the anoni al basis of R2):

x

y

θ

P

r

The oordinate r is learly r(x, y) =√x2 + y2, but the oordinate θgenerates some problems. For instan e the point (0, 0) has no angular oor-dinate; to ontour this di ulty we may delete (0, 0) from our onsiderationsand try to assign an angular oordinate only to the points of R2 \ 0. Weare still in di ulty be ause if we assign θ = 0 to the points on the semiaxis

L = x ≥ 0, y = 0, when we arrive at points of L from below, after giving a omplete winding, the points of L would have an angular oordinate θ = 2πand they would have two angular oordinates:

198 CHAPTER 3. DIFFERENTIAL FORMSθ 2π

θ 0

y

x .

This we annot allow and we may de ide to assign those points the valueθ = 0 and not θ = 2π; the angular oordinate of the points in R2 \ (0, 0)would then be in the interval 0 ≤ θ < 2π.But a problem remains be ause the angular oordinate fun tion wouldnot be ontinuous at the points of L, for we have assigned to them the valueθ = 0 while points near L with y < 0 have an angular oordinate near 2π.The way out of this impasse is to assign an angular oordinate only to thepoints in the open set U = R2 \ L. Of ourse a similar onstru tion an bemade ex luding any other ray emanating from 0 instead of L.Whi h is the fun tion θ(x, y)? We ould think at rst sight that θ(x, y) =arccos(x

r) but the image of U by this fun tion is (0, π) and we need anglesgreater than π for points su h that y < 0. We dene

θ(x, y) =

arccos(xr) if y > 0

π if x < 0, y = 02π − arccos(x

r) if y < 0and one an see that θ is a ontinuous fun tion on U and even a dierentiableone (the deli ate points are those near x < 0, y = 0 of ourse).Briey the assignation of polar oordinates to the points of U = R2 \ Lis a bije tion from U onto the open set U ′ = (0,+∞)× (0, 2π):

k : U → U ′

(x, y) 7→ (r, θ)The inverse map is the expression of a point in terms of its polar oordinates:h : U ′ → U

(r, θ) 7→ (x, y)

3.3. COORDINATES 199given by h(r, θ) = (r cos θ, r sin θ). This fun tion is dierentiable and theja obian h′(r,θ) has a determinanth′(r,θ) =

(cos θ −r sin θsin θ r cos θ

), det h′ = r > 0The inverse fun tion theorem asserts that h has a dierentiable lo al inverseat ea h point, but the global inverse k must oin ide with ea h lo al inverseand so k is a dierentiable fun tion. Moreover we now know that det k′ = 1

r.Coordinate urves and lo al basis in polar oordinatesIn artesian oordinates (x, y) we fo us the attention on the points (x, b),straight lines parallel to the x axis, as well as on the points (a, y), straightlines parallel to the y axis. It is then a sound question to ask what is thegure drawed by the points with polar oordinates (r, b) (with a onstantangular oordinate) and what the gure drawed by the points (a, θ) (with a onstant radial oordinate). This is easy to answer as they are the points onthe urves:

γ(r) = h(r, b) = (r cos b, r sin b) = r(cos b, sin b)

ϕ(θ) = h(a, θ) = (a cos θ, a sin θ) alled oordinate urves. The rst is a ray emanating from the origin withslope sin b / cos b and the se ond is a ir umferen e of radius a entered atthe origin (with the ex eption of the point (a, 0)).γ(r)

h

k

U

.x

y

r

(r,θ)

U’θ2π ϕ(θ)

The tangent ve tors to these urves are respe tivelyγ′(r) =

∂h

∂r(r, b) = (cos b, sin b) = er

ϕ′(θ) =∂h

∂θ(a, θ) = (−a sin θ, a cos θ) = eθ

200 CHAPTER 3. DIFFERENTIAL FORMSThose ve tors are li for they are the olumns of h′ that has a nonvanishingdeterminant; we all them the lo al basis asso iated to the polar oordinates.They are losely similar to the ve tors i, j of R2 in the fa t that they allowto express any ve tor in R2p but they vary from point to point:er

er

eθ

eθ

y

x

pq

Notation: It is reasonable to write r = r(x, y), θ = θ(x, y) where wethink that r, θ are fun tions and x, y are numbers; in this ontext we maywrite k = (r, θ) as an expression of the omponent fun tions of k.In the expressions x = x(r, θ), y = y(r, θ) we think that x, y are fun tionsand r, θ are numbers; here we shall write h = (x, y).From a pra ti al point of view the artesian oordinates are interestingto the lo al geographer or the ar hite t while polar oordinates are useful forthe airport ontrollers.Problem 105:Express the angular oordinate of the polar oordinates by means of the

arctan fun tion.

3.3. COORDINATES 201Solution:The fun tion arctan has its values in (−π/2, π/2) and we must make somekind of adjustment to over the whole interval (0, 2π). Deneθ(x, y) =

arctan yx

if x > 0, y > 0π/2 if x = 0, y > 0π + arctan y

xif x < 0, y > 0

3π/2 if x = 0, y < 0π + arctan y

xif x < 0, y < 0

2π + arctan yx

if x > 0, y < 0or briey:θ(x, y) =

arctan yx

if x > 0, y > 0π + arctan y

xif x < 0

2π + arctan yx

if x > 0, y < 0π/2 if x = 0, y > 03π/2 if x = 0, y < 0We shall write θ(x, y) =′ arctan( y

x)′ to denote this fun tion.

T The general denitions mimi k what has been seen in the ase of thepolar oordinates. We hoose to dene the oordinate systems through theexpression of a point, that is through a fun tion h as that in polar oordinates.• A system of oordinates in an open set U ⊂ Rn is a bije tion h froman open set U ′ ⊂ Rn onto U

h : U ′ → Uthat is dierentiable and has a dierentiable inverse k = h−1 (briey:h is a dieomorphism).We denote points in U ′ by y = (y1, . . . yn) and points in U by x =(x1, . . . , xn); a ordingly we notate the omponent fun tions of h, kthus

h = (x1, . . . , xn), xi : U ′ → R, i = 1, . . . , n

k = (y1, . . . , yn), yi : U → R, i = 1, . . . , n

202 CHAPTER 3. DIFFERENTIAL FORMS• Choose a point p ∈ U and let k(p) = (y0

1, . . . , y0n) ∈ U ′. The i-th oordinate urve through p is

γi(t) = h(y01, . . . , y

0i + t, . . . , y0

n), γi(0) = p

• The tangent ve tors to the oordinate urves are the olumns of h′k(p)

(e1)p = (∂h

∂y1)k(p), . . . , (en)p = (

∂h

∂yn)k(p)and they are li for h′k(p) has a nonvanishing determinant; we all themthe lo al basis at point p ∈ U .

• The exterior derivatives of the fun tions yi : U → R are dierential1-forms in U ; at p ∈ U the 1-forms(dy1)p, . . . , (dyn)pare the dual basis of the lo al basis at p for

(dyi)p[(ej)p] = (n∑

k=1

∂yi∂xk

dxk)p[(ej)p] =

=n∑

k=1

∂yi∂xk|p (πk[(ej)p]) =

=n∑

k=1

∂yi∂xk|p (πk[

∂h

∂yj|k(p)]) =

=

n∑

k=1

∂yi∂xk|p∂xk∂yj|k(p)= δijwhere the last equality omes from kh = I ⇒ (Dk)h(k(p)) (Dh)k(p) =

(Dk)p (Dh)k(p) = Id or in matrix form

∂y1∂x1

. . . ∂y1∂xn

. . . . . .∂yn

∂x1. . . ∂yn

∂xn

p

∂x1

∂yj

. . .∂xn

∂yj

k(p)

=

0. . .1. . .0

3.3. COORDINATES 203Problem 106: Cylindri al and spheri al oordinates.Write the dieomorphisms h and k for the ylindri al and the spheri al o-ordinates, giving expli itly the open sets U and U ′. Des ribe geometri allythe oordinate urves of both systems.Solution:a) Cylindri al oordinates.i) The elementary gure for ylindri al oordinates isθ

x

r

P(x,y,z)

z

y

z

.

Choose U ′ = r > 0 × (0, 2π)× R, U = R3 \ (x, 0, z) : x ≥ 0;the ylindri al system ish(r, θ, z) = (r cos θ, r sin θ, z)

x(r, θ, z) = r cos θy(r, θ, z) = r sin θz(r, θ, z) = z

k(x, y, z) = (√x2 + y2 + z2,′ arctan

y

x

′, z)

r(x, y, z) =√x2 + y2 + z2

θ(x, y, z) = ′ arctan yx′

z(x, y, z) = z

204 CHAPTER 3. DIFFERENTIAL FORMSii) The oordinate urves are:θ = θ0, z = z0 : γ1(r) = (r cos θ0, r sin θ0, z0)r = r0, z = z0 : γ2(θ) = (r0 cos θ, r0 sin θ, z0)r = r0, θ = θ0 : γ3(z) = (r0 cos θ0, r0 sin θ0, z),that is half lines in the plane z = z0 emanating from the point

(0, 0, z0), ir umferen es of radius r0 in the plane z = z0 , andstraight lines perpendi ular to z = 0 through the point (r0 cos θ0, r0 sin θ0)respe tively.b) Spheri al oordinates.i) The elementary gure for spheri al oordinates isr

y

x

z P(x,y,z)

θ

ϕ

Choose U ′ = r > 0×(0, π)×(0, 2π), U = R3−(x, 0, z) : x ≥ 0the spheri al system ish(r, ϕ, θ) = (r sinϕ cos θ, r sinϕ sin θ, r cosϕ)

x(r, ϕ, θ) = r sinϕ cos θy(r, ϕ, θ) = r sinϕ sin θz(r, ϕ, θ) = r cosϕ

k(x, y, z) = (√x2 + y2 + z2, arccos

z√x2 + y2 + z2

,′ arctany

x

′)

r(x, y, z) =√x2 + y2 + z2

ϕ(x, y, z) = arccos z√x2+y2+z2

θ(x, y, z) = ′ arctan yx′

3.3. COORDINATES 205ii) The oordinate urves are:θ = θ0, ϕ = ϕ0 : γ1(r) = (r sinϕ0 cos θ0, r sinϕ0 sin θ0, r cosϕ0)

r = r0, θ = θ0 : γ2(ϕ) = (r0 sinϕ cos θ0, r0 sinϕ sin θ0, r0 cosϕ)

r = r0, ϕ = ϕ0 : γ3(θ) = (r0 sinϕ0 cos θ, r0 sinϕ0 sin θ, r0 cosϕ0).that is half lines in the plane z = z0 emanating from the originwhose dire tion ve tor is(sinϕ0 cos θ0, sinϕ0 sin θ0, cosϕ0),half meridian ir les (remind that 0 < ϕ < π), and the parallel ata olatitude ϕ0 of radius r0 sinϕ0 in the plane z = r0 cosϕ0.

xy

z

xy

z

y

z

x

Problem 107: Spheri al oordinates in Rn.We dene indu tively the spheri al oordinates (r, θ1, . . . , θn−1) of a pointx = (x1, . . . , xn) ∈ Rn:If n = 2 they are the polar oordinates. If n > 2 let r =| x | and letθ1 be the angle that x makes with the Ox1 axis so that x1 = r cos θ1; theremaining oordinates are the spheri al oordinates of (x2, . . . , xn) in Rn−1.Write out the expression of a point in spheri al oordinates for n = 3, 4, 5.Solution:A gure may help:

206 CHAPTER 3. DIFFERENTIAL FORMSθ

1

1x

, ... ,nx

, ... , nx

θ1

.r

x( )2

sin

1x( )

r

a) n = 3. Noti e that | (x2, . . . , xn) |= r sin θ1; then in U ′ = R+× (0, π)×(0, 2π) we have

x1 = r cos θ1x2 = r sin θ1(cos θ2)x3 = r sin θ1(sin θ2)b) n = 4. Now U ′ = R+ × (0, π)× (0, π)× (0, 2π)

x1 = r cos θ1x2 = r sin θ1(cos θ2)x3 = r sin θ1(sin θ2 cos θ3)x4 = r sin θ1(sin θ2 sin θ3) ) n = 5. U ′ = R+ × (0, π)× (0, π)× (0, π)× (0, 2π)

x1 = r cos θ1x2 = r sin θ1(cos θ2)x3 = r sin θ1(sin θ2 cos θ3)x4 = r sin θ1(sin θ2 sin θ3 cos θ4)x5 = r sin θ1(sin θ2 sin θ3 sin θ4)

3.3. COORDINATES 207Problem 108:a) Compute the lo al basis of the polar oordinates and he k that thedual basis is the one that has been said above.b) Compute the lo al basis of the ylindri al oordinates. ) Compute the lo al basis of the spheri al oordinates.In ea h ase ompute the metri matrix of the eu lidian produ t in R2 andR3 in terms of the dual of the lo al basis.Solution:a) Polar oordinates.Lo al basis

er = (cos θ, sin θ)eθ = (−r sin θ, r cos θ)Dual basis

dr, dθThe dierential of the fun tion r =√x2 + y2 is the dierential 1-form

dr = 1r(xdx+ ydy) and

dr[er] =1

r(xdx+ydy)[er] =

1

r(x cos θ+y sin θ) =

1

r(xx

r+y

y

r) =

x2 + y2

r2= 1

dr[eθ] =1

r(xdx+ydy)[eθ] =

1

r(x(−r sin θ)+y(r cos θ)) =

1

r(−xy+yx) = 0The dierential of the fun tion θ =′ arctan y

x′ dened in the open set

U = R2 \ L is the dierential 1-form dθ = 1r2

(−ydx+ xdy) anddθ[er] =

1

r2(−ydx+xdy)[er] =

1

r2(−y cos θ+x sin θ) =

1

r2(−yx

r+x

y

r) = 0

dθ[eθ] =1

r2(−ydx+xdy)[eθ] =

1

r2(−y(−r sin θ)+x(r cos θ)) =

1

r2(y2+x2) = 1The metri matrix of the eu lidian metri is

Gh(r,θ) =

(1 00 r2

)and its expression in the basis dr, dθ isGh(r,θ) = dr ⊗ dr + r2dθ ⊗ dθ

208 CHAPTER 3. DIFFERENTIAL FORMSb) Cylindri al oordinates.Lo al basiser = (cos θ, sin θ, 0)eθ = (−r sin θ, r cos θ, 0)ez = (0, 0, 1)Dual basis

dr, dθ, dzThe metri matrix of the eu lidian metri isGh(r,θ,z) =

1 0 00 r2 00 0 1

and its expression in the basis dr, dθ, dz isGh(r,θ,z) = dr ⊗ dr + r2dθ ⊗ dθ + dz ⊗ dz ) Spheri al oordinates.Lo al basis

er = (sinϕ cos θ, sinϕ sin θ, cosϕ)eϕ = (r cosϕ cos θ, r cosϕ sin θ,−r sinϕ)eθ = (−r sinϕ sin θ, r sinϕ cos θ, 0)Dual basis

dr, dϕ, dθThe metri matrix of the eu lidian metri isGh(r,ϕ,θ) =

1 0 00 r2 00 0 r2 sin 2ϕ

and its expression in the basis dr, dϕ, dθ isGh(r,ϕ,θ) = dr ⊗ dr + r2dϕ⊗ dϕ+ r2 sin 2ϕdθ ⊗ dθ

3.3. COORDINATES 209Problem 109: Ellipti al oordinates.For yet another oordinate system, the formulaex = cosh u cos θy = sinh u sin θ

, 0 < u, 0 < θ < 2πdene the ellipti al oordinates.a) Find the oordinate urves and show how to assign the oordinates u, θto a point in the plane. Whi h points of the xy plane are omitted?b) Show that h(u, θ) = (cosh u cos θ, sinh u sin θ) is a dieomorphism from

u > 0 × (0, 2π) onto an open set U ⊂ R2 to be found. ) Find the lo al basis and show it is og; write the metri matrix of theeu lidian produ t in the dual basis of the lo al basis.Solution:a) Remind that the ellipse x2

a2+ y2

b2= 1 an be parametrized by α(θ) =

(a cos θ, b sin θ), 0 < θ < 2π. The right bran h of the hyperbola x2

a2−

y2

b2= 1, a > 0 an be parametrized by β(u) = (a cosh u, b sinh u) andthe left bran h an be parametrized by β(u) = (a cosh u, b sinh u)i) The u = u0 urves are α(θ) = (cosh u0 cos θ, sinh u0 sin θ), 0 < θ <

2π. Those are ellipsesx2

cosh2 u0

+y2

sinh2 u0

= 1with semiaxes cosh u, sinh u. The fo al distan e of these ellipses isc2 = cosh2 u0 − sinh2 u0 = 1showing that all the members of the family have the same fo us,at (−1, 0) and (1, 0). The ex entri ity is

e =1

cosh u0

210 CHAPTER 3. DIFFERENTIAL FORMSand limu0→+∞ e = 0 shows that the ellipses get loser to ir umfer-en es as u0 grows. Similarly limu0→0 e = 1 shows that the ellipsesare more oblate as u0 approa hes 0. As γ(θ) = (cosh u0 cos θ, sinh u0 sin θ)is a parametrization of these ellipses we see that they degenerateinto the segment [−1, 1] for u0 = 0 (and then this segment in thexy plane is omitted, be ause we assume u > 0):

x.

1−1

y

ii) The θ = θ0 urves are β(u) = (cos θ0 cosh u, sin θ0 sinh u), u > 0.With the ex eption of the values θ = π/2, π, 3π/2 those are thehyperbolasx2

cos2 θ0− y2

sin2 θ0= 1whose assymptotes have a slope tan θ. For the ex eption valueswe have respe tively:

θ0 = π/2 : β(u) = (0, sinh u), the positive y axisθ0 = π : β(u) = (− cosh u, 0) the negative x axis

θ0 = 3π/2 : β(u) = (0,− sinh u) the negative y axis

3.3. COORDINATES 211y

1−1

x

.

iii) From the pre eding we see that through ea h point in U = R2 \(x, y) : −1 ≤ x, y = 0 passes a θ- urve and a u- urve and onlyone su h, thus assigning the point the oordinates u, θ. The pre- eding ondiferations show that the ellipti al oordinates establisha bije tion

h : U ′ = u > 0 × (0, 2π)→ R2 \ (x, y) : −1 ≤ x, y = 0b) The ja obian of h(θ, u) = (cosh u cos θ, sinh u sin θ) is:h′(u, θ) =

(sinh u cos θ − cosh u sin θcosh u sin θ sinh u cos θ

), deth′ = sinh2 u cos2 θ+cosh2 u sin2 θ > 0The inverse fun tion theorem asserts that h has a lo al inverse dif-ferentiable, but we know that h has a global inverse whi h is thendierentiable. ) The lo al basis is

eu = (sinh u cos θ, cosh u sin θ)

eθ = (− cosh u sin θ, sinh u cos θ)and the dual basis isdu, dθ

212 CHAPTER 3. DIFFERENTIAL FORMSThe metri matrix isGh(u,θ) =

(sinh2 u cos2 θ + cosh2 u sin2 θ 0

0 cosh2 u sin2 θ + sinh2 u cos2 θ

)

=

(cosh2 u− cos2 θ 0

0 cosh2 u− cos2 θ

)where we see that the lo al basis is og.Gh(u,θ) = (cosh2 u− cos2 θ)(du⊗ du+ dθ ⊗ dθ)

Outline of oordinate systems (for referen e)Polar r, θh(r, θ) = (r cos θ, r sin θ)

h′(r, θ) =

(cos θ −r sin θsin θ r cos θ

), det h′ = r > 0

er = (cos θ, sin θ)eθ = (−r sin θ, r cos θ)

Gh(r,θ) =

(1 00 r2

)Ellipti al u, θh(u, θ) = (cosh u cos θ, sinh u sin θ)

h′(u, θ) =

(sinh u cos θ − cosh u sin θcosh u sin θ sinh u cos θ

), det h′ = cosh2 u− cos2 θ > 0

eu = (sinh u cos θ, cosh u sin θ)eθ = (− cosh u sin θ, sinh u cos θ)

Gh(u,θ) = (cosh2 u−cos2 θ)

(1 00 1

)

Cylindri al r, θ, zh(r, θ, z) = (r cos θ, r sin θ, z)

3.3. COORDINATES 213h′(r, θ, z) =

cos θ −r sin θ 0sin θ r cos θ 0

0 0 1

, det h′ = r > 0

er = (cos θ, sin θ, 0)eθ = (−r sin θ, r cos θ, 0)ez = (0, 0, 1)

Gh(r,θ,z) =

1 0 00 r2 00 0 1

Spheri al r, ϕ, θh(r, ϕ, θ) = (r sinϕ cos θ, r sinϕ sin θ, r cosϕ)

h′(r, ϕ, θ) =

sinϕ cos θ r cosϕ cos θ −r sinϕ sin θsinϕ sin θ r cosϕ sin θ r sinϕ cos θ

cosϕ −r sinϕ 0

, det h′ = r2 sinϕ > 0

er = (sinϕ cos θ, sinϕ sin θ, cosϕ)eϕ = (r cosϕ cos θ, r cosϕ sin θ,−r sinϕ)eθ = (−r sinϕ sin θ, r sinϕ cos θ, 0)

Gh(r,ϕ,θ) =

1 0 00 r2 00 0 r2 sin 2ϕ

All lo al bases are orthogonal and positively oriented in the standardorientation of R2 or R3.Change of oordinates

T The need to make a hange of variable may arise in two ways:a) Consider the equationsx2 − x− 1 = 0

x6 − x3 − 1 = 0Making the hange of variables y = x3 in the se ond it be omesy2 − y − 1 = 0The new variable y is a fun tion of the old one.

214 CHAPTER 3. DIFFERENTIAL FORMSb) If we have a fun tion f(x, y) = x+ y2 we an express f in terms of thepolar oordinates of x, y:F (r, θ) = r cos θ + r2 sin2 θHere the old variables x, y are known fun tions of the new ones r, θ.Assume we have two oordinate systems in an open set U ⊂ Rn, h : U ′ →

U, h = (x1, . . . , xn) and l : U ′′ → U, l = (X1, . . . , Xn):U ′

ϕ−→ U ′′

h

ցl

ւUThe dieomorphism ϕ = l−1 h is alled the transition map (or the hangeof variables) from the variables yi to the variables Yi.

Problem 110:Let h be the ylindri al system and l the spheri al system in R3. Computethe transition map from the ylindri al to the spheri al oordinates.Solution:In fa t what we want is to nd the spheri al oordinates (ρ, ϕ, θ) of a pointhaving the ylindri al oordinates (r, θ, z). That is, we must solve the systemρ sinϕ cos θ = r cos θρ sinϕ sin θ = r sin θρ cosϕ = z

In view of the following gure

3.3. COORDINATES 215(r,θ ,z)P

θ

ϕ

θ

z

ρ

P(ρ,ϕ,θ )z

x

y

r

x

z

y

we haveϕ(r, θ, z) = (

√r2 + z2, arccos

z√r2 + z2

, θ)The ylindri al oordinates aredened in U ′ = r > 0 × (0, 2π)×R, thespheri al in U ′′ = ρ > 0 × (0, π)× (0, 2π), and the transition map isϕ : U ′ → U ′′.

Problem 111:Let (r, θ) be the polar oordinates and (u,Θ) the ellipti al oordinates. Com-pute the transition map from ellipti al to polar oordinates and the ja obianof this transition map.Solution:As r =√x2 + y2 , θ = arctan y

xwe have

r =√

cosh2 u cos2 Θ + sinh2 u sin2 Θ =√

cosh2 u− sin2 Θ

θ = arctansinh u sin Θ

cosh u cos Θ= arctan(tanhu tanΘ)or

ϕ(u,Θ) = (√

cosh2 u− sin2 Θ , arctan(tanhu tanΘ))

216 CHAPTER 3. DIFFERENTIAL FORMSThenϕ′(u,Θ) =

(coshu sinhu√cosh2 u−sin2 Θ

− cos Θ sinΘ√cosh2 u−sin2 Θ

cos Θ sinΘcosh2 u−sin2 Θ

coshu sinhucosh2 u−sin2 Θ

)

Problem 112:The basis e1 = (1, 1), e2 = (−1, 1) of R2 determines a oordinate systemh(u, v) = (x(u, v), y(u, v)). Compute the transition map from the oordinates(u, v) to the polar variables oordinates (r, θ).Solution:We have

h(u, v) = ue1 + ve2 = (u− v, u+ v)and the polar oordinates of this point arer =

√(u− v)2 + (u+ v)2, θ = arctan(

u+ v

u− v )The transition map is thenϕ(u, v) = (

√2√u2 + v2, arctan(

u+ v

u− v ))As the polar oordinates system gives oordinates only to R2 − (x, 0) :x ≥ 0, we must shrink the domain of ϕ to U ′ = R2\(u, v) : u = −v, u > v.

Expression of fun tionsT Consider a fun tion f : U → R dened in an open set U ⊂ Rn andlet h : U ′ → U , be a oordinate system. The representation of f in the oordinates (y1, . . . , yn) is the fun tion f : U ′ → R dened by f = f h. Ina diagram:

U ′ fh ↓ ց

Uf→ R

3.3. COORDINATES 217For instan e if f(x, y) = x2 + y2 its representation in polar oordinates isf(r, θ) = (r cos θ)2 + (r sin θ)2 = r2Components of ve tor eldsLet h : U ′ → U be a oordinate system in an open set U ⊂ Rn and F ave tor eld in U whose expression in the lo al basis isFp = F1(p)(e1)p + · · ·+ Fn(p)(en)pHow do the omponents vary when we make a hange of oordinates?We reason in the ase of two variables; the general ase is similar. Let

h = (x1, x2), l = (X1, X2) be two oordinate systems in U and ϕ the transitionmap:U ′

ϕ−→ U ′′

(y1, y2) (Y1, Y2)h

ցl

ւUFrom h = l ϕ we have the matrix equality h′ = l′ ϕ′:

(∂x1

∂y1∂x1

∂y2∂x2

∂y1

∂x2

∂y2

)=

( ∂X1

∂Y1

∂X1

∂Y2∂X2

∂Y1

∂X2

∂Y2

)· ϕ′or

(e1, e2) = (E1,E2) · ϕ′Then for a ve tor eld:F = (e1, e2)

(f1

f2

)= (E1,E2) · ϕ′

(f1

f2

)= (E1,E2)

(F1

F2

)or (F1

F2

)= ϕ′

(f1

f2

)If it's easier to write the transition map the other way around, ψ : U ′′ → U ′,the relation of omponents would be(f1

f2

)= ψ′

(F1

F2

)

218 CHAPTER 3. DIFFERENTIAL FORMSProblem 113:Compute the ve tor eld F(x, y) = (x, y) in polar oordinates and in ellipti al oordinates.Solution:a) Polar oordinates.i) We have F(x, y) = xi+yj and we sear h an expression of the formF(r, θ) = F1(r, θ)er + F2(r, θ)eθThe transition map from artesian to polar oordinates is ϕ(x, y) =

(√x2 + y2, arctan y

x) whose ja obian matrix is

ϕ′ =

(xr

yr

− yr2

xr2

)We have(F1

F2

)=

(xr

yr

− yr2

xr2

)(xy

)=

(r0

)andF(r, θ) = rerii) We may as well reason from ψ(r, θ) = (r cos θ, r sin θ), with ja o-bian

C =

(cos θ −r sin θsin θ r cos θ

), C−1 =

1

r

(r cos θ r sin θ− sin θ cos θ

)and then (xy

)=

(cos θ −r sin θsin θ r cos θ

)(F1

F2

)or (F1

F2

)=

(cos θ sin θ

−1rsin θ 1

rcos θ

)(xy

)=

=

(x cos θ + y sin θ

1r(−x sin θ + y cos θ)

)=

(r cos 2θ + r sin 2θ

1r(−r cos θ sin θ + r sin θ cos θ)

)=

(r0

)

3.3. COORDINATES 219b) Ellipti al oordinates.We have F(x, y) = xi + yj and we sear h an expression of the formF(u, θ) = Fu(u, θ)eu + Fθ(u, θ)eθThe transition map from artesian to ellipti al oordinates is not easyto obtain. We use instead the transition map from ellipti al to artesian oordinates:ψ(u, θ) = (cosh u cos θ, sinh u sin θ)with ja obian

ψ′ =

(sinh u cos θ − cosh u sin θcosh u sin θ sinh u cos θ

), detψ′ = cosh2 u− cos2 θand

(FuFθ

)=

1

cosh2 u− cos2 θ

(sinh u cos θ cosh u sin θ− cosh u sin θ sinh u cos θ

)(cosh u cos θsinh u sin θ

)

F(u, θ) =sinh u coshu

cosh2 u− cos2 θeu +

sin θ cos θ

cosh2 u− cos2 θeθ

Problem 114:Compute the gravitational eld g(r) = − rr3

in ylindri al oordinates. Re-mind that r = (x, y, z) and r = |r| =√x2 + y2 + z2; we shall use ρ =√

x2 + y2 in ylindri al oordinates to avoid onfusions.Solution:We have g(x, y, z) = − 1

(√x2+y2+z2)3

(xi+yj+zk) and we sear h an expressionsu h asg(ρ, θ, z) = F1(ρ, θ, z)eρ + F2(ρ, θ, z)eθ + F3(ρ, θ, z)ezWe ompute as in a) of the pre eding problem:

ϕ(x, y, z) = (√x2 + y2, arctan

y

x, z)

220 CHAPTER 3. DIFFERENTIAL FORMSwith ja obian matrixϕ′ =

xρ

yρ

0

− yρ2

xρ2

0

0 0 1

and obtain

F1

F2

F3

=

xρ

yρ

0

− yρ2

xρ2

0

0 0 1

−x(√x2+y2+z2)3

−y(√x2+y2+z2)3

−z(√x2+y2+z2)3

=

−(x2+y2)

ρ(√x2+y2+z2)3

0−z

(√x2+y2+z2)3

=

−ρ(√

(ρ2+z2)3

0−z

(√

(ρ2+z2)3

g(ρ, θ, z) = − ρ

(√ρ2 + z2)3

eρ −z

(√ρ2 + z2)3

ez =

Problem 115:Compute the gravitational eld g(r) = − rr3

in spheri al oordinates.Solution:It is obvious that the result will be g(r, ϕ, θ) = − 1r2

er but let's do the om-putation. The transition map isϕ(x, y, z) = (

√x2 + y2 + z2, arccos

z√x2 + y2 + z2

, arctany

x)with ja obian matrix

ϕ′ =

xr

yr

zr

xzr2√r2−z2

yz

r2√r2−z2

z2−r2r2√r2−z2

− yx2+y2

xx2+y2

0

and the omponents in the lo al basis are:

xr

yr

zr

xzr2√r2−z2

yz

r2√r2−z2

z2−r2r2√r2−z2

− yx2+y2

xx2+y2

0

− xr3

− yr3

− zr3

=

− 1r2

00

3.3. COORDINATES 221and the eld is, as expe ted,g(r, ϕ, θ) = − 1

r2er

Components of formsTo do a hange of variables in a dierential form one uses the pull-ba k map(see p.186).Problem 116:Compute the expression in polar and ellipti al oordinates of the dierential1-form of R2, ω = xdx+ ydy .Solution:a) Remind the map h : U ′ = r > 0 × (0, 2π) → R2 \ L, h(r, θ) =(r cos θ, r sin θ) whose ja obian is

h′ =

(cos θ −r sin θsin θ r cos θ

)Thenh∗(ω) = r cos θ(cos θdr − r sin θdθ) + r sin θ(sin θdr + r cos θdθ)

= rdrb) For ellipti al oordinates h : U ′ = r > 0 × (0, 2π) → R2 \ [−1, 1],h(θ, u) = (cosh u cos θ, sinh u sin θ) and

h′(u, θ) =

(sinh u cos θ − cosh u sin θcosh u sin θ sinh u cos θ

)Thenh∗(ω) = cosh u cos θ(sinh u cos θdu− cosh u sin θdθ)

+ sinh u sin θ(cosh u sin θdu+ sinh u cos θdθ) =

= cosh u sinh udu− sin θ cos θdθ

222 CHAPTER 3. DIFFERENTIAL FORMSProblem 117:Let ω = rdr ∧ dθ, v1 = (1, 0),v2 = (1, 1); ompute ω[v1,v2].Solution:a) We may express both ve tors in the lo al basis; the hange of basismatrix from the anoni al basis (i, j) to the lo al basis (er, eθ) isC =

(cos θ −r sin θsin θ r cos θ

)The omponents of v1,v2 in the basis (er, eθ) are(

cos θ sin θ−1rsin θ 1

rcos θ

)(10

)=

(cos θ−1rsin θ

)

(cos θ sin θ−1rsin θ 1

rcos θ

)(11

)=

(cos θ + sin θ

−1rsin θ + 1

rcos θ

)Thenω[v1,v2] = rdr∧dθ[v1,v2] = r det

(cos θ cos θ + sin θ−1rsin θ −1

rsin θ + 1

rcos θ

)= 1b) We express ω in artesian oordinates; we maintain r as an intermediatevariable:

ω = rd(r) ∧ d(arctany

x) = r(

x

rdx+

y

rdy) ∧ (

−ydx+ xdy

r2) =

=1

r2(xdx+ ydy) ∧ (−ydx+ xdy) =

=1

r2(x2 + y2)dx ∧ dy = dx ∧ dyThen

ω[(1, 0), (1, 1)] = det

(1 10 1

)= 1

3.3. COORDINATES 223Volume form in oordinatesProblem 118:Take in R2 and R3 the standard eu lidian produ t and the standard orienta-tion. Compute the volume form of R2 in polar and ellipti al oordinates andthe volume form of R3 in ylindri al and spheri al oordinates,a) Using the formula in p.107.b) Doing a hange of variables in ωR2 = dx ∧ dy and ωR3 = dx ∧ dy ∧ dz.Solution:a) Remind the formula ωE =√|gu|α1 ∧ · · · ∧ αn that gives the volumeform in terms of the dual basis of an og basisi) Polar oordinates.The eu lidian metri matrix is

G =

(1

r2

), detG = r2and the volume form (a tually the area form) is

ωR2 = rdr ∧ dθii) Ellipti al oordinates.The eu lidian metri matrix isG =

(cosh2 u− cos2 θ 0

0 cosh2 u− cos2 θ

)and the volume form isωR2 = (cosh2 u− cos2 θ)du ∧ dθiii) Cylindri al oordinates.The eu lidian metri matrix isG =

1

r2

1

, detG = r2and the volume form isωR3 = rdr ∧ dθ ∧ dz

224 CHAPTER 3. DIFFERENTIAL FORMSiv) Spheri al oordinates.The eu lidian metri matrix isG =

1

r2

r2 sin 2ϕ

, detG = r4 sin2 ϕand the volume form isωR3 = r2 sinϕdr ∧ dϕ ∧ dθb) Let's do it making the hange of oordinatesi) Polar oordinates.

ωR2 = dx ∧ dy = (cos θdr − r sin θdθ) ∧ (sin θdr + r cos θdθ) =

= r(cos 2θ + sin 2θ)dr ∧ dθ = rdr ∧ dθii) Ellipti al oordinates.h′(u, θ) =

(sinh u cos θ − cosh u sin θcosh u sin θ sinh u cos θ

)

ωR2 = dx ∧ dy = (sinh u cos θdu− cosh u sin θdθ) ∧∧ (cosh u sin θdu+ sinh u cos θdθ) =

= (sinh2 u cos2 θ+cosh2 u sin2 θ)du∧dθ = (cosh2 u−cos2 θ)du∧dθthe standard eu lidian produ te and the standard orientation.iii) Cylindri al oordinates.ωR3 = dx∧dy∧dz = (cos θdr−r sin θdθ)∧(sin θdr+r cos θdθ)∧dz =

= r(cos 2θ + sin 2θ)dr ∧ dθ ∧ dz = rdr ∧ dθ ∧ dziv) Spheri al oordinates.(dx, dy, dz) = (dr, dϕ, dθ)CT =

= (dr, dϕ, dθ)

sinϕ cos θ sinϕ sin θ cosϕr cosϕ cos θ r cosϕ sin θ −r sinϕ−r sinϕ sin θ r sinϕ cos θ 0

3.3. COORDINATES 225We havedx = sinϕ cos θdr + r cosϕ cos θdϕ− r sinϕ sin θdθ

dy = sinϕ sin θdr + r cosϕ sin θdϕ+ r sinϕ cos θdθ

dz = −r sinϕ sin θd.r + r sinϕ cos θdϕand the volume form isωR3 = dx ∧ dy ∧ dz = r2 sinϕdr ∧ dϕ ∧ dθ

Problem 119:Take in R3,R4 and R5 the standard eu lidian produ t and the standardorientation. Compute the metri matrix and the volume form of those spa esin spheri al oordinates in Rn (see p.205).Solution:As the lo al bases are og and positive we use the formulaω =√gdα1 ∧ · · · ∧ dαna) In R3 the lo al basis ve tors are

er =

cos θ1sin θ1 cos θ2sin θ1 sin θ2

, eθ1 =

−r sin θ1r cos θ1 cos θ2r cos θ1 sin θ2

, eθ2 =

0−r sin θ1 sin θ2r sin θ1 cos θ2

G =

1

r2

r2 sin2 θ1

, g = detG = r4 sin2 θ1

ωR3 = r2 sin θ1dr ∧ dθ1 ∧ dθ2b) In R4 the lo al basis ve tors areer =

cos θ1sin θ1 cos θ2sin θ1 sin θ2 cos θ3sin θ1 sin θ2 sin θ3

, eθ1 =

−r sin θ1r cos θ1 cos θ2r cos θ1 sin θ2 cos θ3r cos θ1 sin θ2 sin θ3

226 CHAPTER 3. DIFFERENTIAL FORMSeθ2 =

0−r sin θ1 sin θ2r sin θ1 cos θ2 cos θ3r sin θ1 cos θ2 sin θ3

, eθ3 =

00−r sin θ1 sin θ2 sin θ3r sin θ1 sin θ2 cos θ3

G =

1r2

r2 sin2 θ1r2 sin2 θ1 sin2 θ2

, g = detG = r6 sin4 θ1 sin2 θ2

ωR4 = r3 sin2 θ1 sin θ2dr ∧ dθ1 ∧ dθ2 ∧ dθ3 ) In R5 we pro eed by analogy:ωR5 = r4 sin3 θ1 sin2 θ2 sin θ3dr ∧ dθ1 ∧ dθ2 ∧ dθ3 ∧ dθ4

3.3.2 Canoni al isomorphisms in oordinatesT Consider in R3 the eu lidian metri and the standard orientation. Leth = (x1, x2, x3) a oordinate system in an open set U , k = (y1, y2, y3) theinverse dieomorphism, e = (e1, e2, e3) the lo al basis of h that we shallassume og positive, and dy = (dy1, dy2, dy3) the dual basis of e. Remind(see p.129) that the matri es of , 2, ♯ and ♯2 in those bases are:

R3p

→ (R3p)∗

e dyM = G, M♯ = G−1

R3p

2→ Λ2(R3p)

e dy∧M2 =

√gI, M♯2 =

1√gIwhere g = detG = λ1λ2λ3, λi = ei · ei and dy∧ = (dy2 ∧ dy3, dy3 ∧ dy1, dy1 ∧

dy2).

Problem 120:Write expli itly the isomorphisms , 2 in polar and ellipti al oordinates inthe plane and in ylindri al and spheri al oordinates in the spa e. Thisexer ise will be useful in the omputations of the dierential operators in oordinates.

3.3. COORDINATES 227Solution:a) Polar oordinates.F = F1er + F2eθ

7→ F = F1dr + r2F2dθThe lo al basis is not on; the orresponding on basis isur = er,uθ =

1

reθwhose dual basis is

dr, rdθIn the basis u = (ur,uθ) and its dual we haveF = Frur + Fθuθ

7→ F = Frdr + Fθrdθ

2 doesn't exist in this ase.b) Ellipti al oordinates.F = F1eu + F2eθ

7→ F = (cosh2 u− cos2 θ)F1du+ (cosh2 u− cos2 θ)F2dθThe lo al basis isn't on; the orresponding orthonormal basis isuu =

1√cosh2 u− cos2 θ

eu,uθ =1√

cosh2 u− cos2 θeθand the orresponding dual basis is

√cosh2 u− cos2 θdu,

√cosh2 u− cos2 θdθThen

F = Fuuu + Fθuθ7→ F =

√cosh2 u− cos2 θ(Fudu+ Fθdθ) ) Cartesian oordinates.i)

F = Xi + Y j + Zk7→ F = Xdx+ Y dy + Zdz

228 CHAPTER 3. DIFFERENTIAL FORMSii)F = Xi + Y j + Zk

27→ 2F = Xdy ∧ dz + Y dz ∧ dx+ Zdx ∧ dyd) Cylindri al oordinates.i)F = F1er + F2eθ + Fzez

7→ F = F1dr + r2F2dθ + F3dzThe lo al basis isn't on; the orresponding orthonormal basis isur = er,uθ =

1

reθ,uz = ezwhose dual basis is

dr, rdθ, dz

F = Frur + Fθuθ + Fzuz7→ F = Frdr + rFθdθ + Fzdzii) As the matrix of 2 in the bases (er, eθ, ez) (non on) and (dθ ∧

dz, dz ∧ dr, dr ∧ dθ) is M =√

detGI = rI, we haveF = F1er+F2eθ+Fzez

27→ 2F = rF1dθ∧dz+rF2dz∧dr+rF3dr∧dθIn the on basis (ur,uθ,uz) with dual (dr, rdθ, dz), and the basis(rdθ ∧ dz, dz ∧ dr, rdr ∧ dθ) the matrix of b is M = I an thenF = Frur+Fθuθ+Fzuz

27→ 2F = Frrdθ∧dz+Fθdz∧dr+Fzrdr∧dθe) Spheri al oordinates.i)F = F1er + F2eϕ + Fzeθ

7→ F = F1dr + r2F2dϕ+ r2 sin 2ϕF3dθThe lo al basis isn't on; the orresponding orthonormal basis isur = er,uϕ =

1

reϕ,uθ =

1

r sinϕeθand its dual basis is

dr, rdϕ, r sinϕdθ

F = Frur + Fϕuϕ + Fθuθ7→ F = Frdr + rFϕdϕ+ r sinϕFθdθ

3.3. COORDINATES 229ii) As the matrix of 2 in the basis (er, eϕ, eθ) (non on) and (dϕ ∧dθ, dθ ∧ dr, dr ∧ dϕ) is M =

√detGI = r2 sinϕI, we have

F = F1er + F2eϕ + F3eθ

2F = r2 sinϕF1dϕ ∧ dθ + r2 sinϕF2dθ ∧ dr + r2 sinϕF3dr ∧ dϕIn the on basis (ur,uϕ,uθ) with dual basis (dr, rdϕ, r sinϕdθ),and the basis (r2 sinϕdϕ∧ dθ, r sinϕdθ∧ dr, rdr∧ dϕ), the matrixof 2 is M = I andF = Frur + Fϕuϕ + Fθuθ

2F = Frr2 sinϕdϕ ∧ dθ + Fϕr sinϕdθ ∧ dr + Fθrdr ∧ dϕ

3.3.3 Dierential operators in oordinatesT For referen e we rewrite here the fundamental diagram involvingthree of the dierential operators:

0 → Ω0(U)d→ Ω1(U)

d→ Ω2(U)d→ Ω3(U) → 0Id ↑ ↑ 2 ↑ 3 ↑

0 → F(U)grad→ V(U)

rot→ V(U)div→ F(U) → 0Noti e that the left half of the diagram, that involving Id and , uses themetri alone, while the right half uses the orientation as well as the metri .Now let h : U ′ → U a oordinate system in U su h that the lo al basis

e = (e1, e2, e3) is positive and og (with eu lidian produ t and usual orienta-tion, say), with metri matrixG =

λ1

λ2

λ3

, λi = ei · ei, g = detG

230 CHAPTER 3. DIFFERENTIAL FORMSand dual basis (dy1, dy2, dy3).The expression of f : U → R in the oordinate system h is f = f h orf(yi) = f(h(yi)); for simpli ity we shall write f(yi) for that form of f . Theexpression for a ve tor eld X in U is

X = X1e1 +X2e2 +X3e3GradientT For f ∈ F(U), we read in the diagram that gradf = ♯df , and as

df =∂f

∂y1

dy1 +∂f

∂y2

dy2 +∂f

∂y3

dy3we havegradf = eTG−1

∂f∂y1∂f∂y2∂f∂y3

= eT

1λ1

1λ2

1λ3

∂f∂y1∂f∂y2∂f∂y3

= eT

1λ1

∂f∂y1

1λ2

∂f∂y2

1λ3

∂f∂y3

and the expression of gradf in the lo al basis is:gradf =

1

λ1

∂f

∂y1e1 +

1

λ2

∂f

∂y2e2 +

1

λ3

∂f

∂y3e3We shall write the result in the asso iated on basis, ui = 1√

λiei, for ompar-ison: gradf =

1√λ1

∂f

∂y1u1 +

1√λ2

∂f

∂y2u2 +

1√λ3

∂f

∂y3u3

Problem 121:Compute the gradient of a fun tion in the system of polar oordinates, ellip-ti al oordinates, artesian oordinates, ylindri al oordinates and spheri al oordinates.

3.3. COORDINATES 231Solution:In the ase of polar and ellipti al oordinates we may easily adapt the pre- eding, thus:a) Polar oordinates.The eu lidian metri matrix isG =

(1

r2

)and we have gradf =∂f

∂rer +

1

r2

∂f

∂θeθIn the asso iated on basisgradf =

∂f

∂rur +

1

r

∂f

∂θuθb) Ellipti al oordinates.The eu lidian metri matrix is

G =

(cosh2 u− cos2 θ 0

0 cosh2 u− cos2 θ

)and we have gradf =1

cosh2 u− cos2 θ(∂f

∂ueu +

∂f

∂θeθ)In the asso iated on basisgradf =

1√cosh2 u− cos2 θ

(∂f

∂uuu +

∂f

∂θuθ) ) Cartesian oordinates.The eu lidian metri matrix is

G =

1

11

and we have gradf =∂f

∂y1e1 +

∂f

∂y2e2 +

∂f

∂y3e3The lo al basis is on.

232 CHAPTER 3. DIFFERENTIAL FORMSd) Cylindri al oordinates.The eu lidian metri matrix isG =

1

r2

1

and we have gradf =∂f

∂rer +

1

r2

∂f

∂θeθ +

∂f

∂zezIn the asso iated on basisgradf =

∂f

∂rur +

1

r

∂f

∂θuθ +

∂f

∂zuze) Spheri al oordinates.The eu lidian metri matrix is

G =

1

r2

r2 sin 2ϕ

and we have gradf =∂f

∂rer +

1

r2

∂f

∂ϕeϕ +

1

r2 sin 2ϕ

∂f

∂θeθIn the asso iated on basisgradf =

∂f

∂rur +

1

r

∂f

∂ϕuϕ +

1

r sinϕ

∂f

∂θuθ

Rota ionalT For X ∈ V(U) we read in the diagram that rotX = ♯2(d(X)) and, fromthe known matri es of the anoni al isomorphisms we have:

X = X1e1 +X2e2 +X3e3

X = λ1X1dy1 + λ2X2dy2 + λ3X3dy3

3.3. COORDINATES 233d(X) = (

∂(λ3X3)

∂y2− ∂(λ2X2)

∂y3)dy2 ∧ dy3 + . . .rotX = ♯2d(X) =

1√g((∂(λ3X3)

∂y2− ∂(λ2X2)

∂y3)e1 + . . . ) =

=1√g

det

e1 e2 e3∂∂y1

∂∂y2

∂∂y3

λ1X1 λ2X2 λ3X3

If the lo al basis is an on basis the formula is the well knownrotX = det

e1 e2 e3∂∂y1

∂∂y2

∂∂y3

X1 X2 X3

If the eld is given in the on basis asso iated to the lo al basis, X = X ′1u1 +X ′2u2 +X ′3u3, applying the pre eding formula (g = detGe) we have

X =X ′1√λ1

e1 +X ′2√λ2

e2 +X ′3√λ3

e3rotX =1√g

det

√λ1u1 . . . . . .∂∂y1

. . . . . .√λ1X

′1 . . . . . .

Problem 122:Compute the rotational in artesian, ylindri al and spheri al oordinates(for the metri matri es see p.212)Solution:We use the same pre eding notations.a) Cartesian oordinates.λ1 = λ2 = λ3 = detG = 1 and we obtain the usual expression.

234 CHAPTER 3. DIFFERENTIAL FORMSb) Cylindri al oordinates.λ1 = λ3 = 1, λ2 = r2, detG = r2 and we obtain

X = Xrer +Xθeθ +XzezrotX =1

rdet

er eθ ez∂∂r

∂∂θ

∂∂z

Xr r2Xθ Xz

and in the asso iated on basisrotX =1

rdet

ur ruθ uz∂∂r

∂∂θ

∂∂z

X ′r rX ′θ X ′z

) Spheri al oordinates.λ1 = 1, λ2 = r2, λ3 = r2 sin2 ϕ, detG = r4 sin2 ϕ and

X = Xrer +Xϕeϕ +XθeθrotX =1

r2 sinϕdet

er eϕ eθ∂∂r

∂∂ϕ

∂∂θ

Xr r2Xϕ r2 sin2 ϕXθ

and in the asso iated on basisrotX =1

r2 sinϕdet

ur ruϕ r sinϕuθ∂∂r

∂∂ϕ

∂∂θ

X ′r rX ′ϕ r sinϕX ′θ

Divergen eT For X ∈ V(U) we read in the diagram that (divX)dV = d(2X); as weknow that the volume element is dV =

√gdy1 ∧ dy2 ∧ dy3 we have

X = X1e1 +X2e2 +X3e3

2X =√g(X1dy2 ∧ dy3 +X2dy3 ∧ dy1 +X3dy1 ∧ dy2)

3.3. COORDINATES 235d(2X) =

∂(√gX1)

∂y1dy1 ∧ dy2 ∧ dy3 +

∂(√gX2)

∂y2dy2 ∧ dy3 ∧ dy1 + · · · =

= (∂(√gX1)

∂y1+ . . . )dy1 ∧ dy2 ∧ dy3 =

=1√g(∂(√gX1)

∂y1+ . . . )dVand we see that divX =

1√g(∂(√gX1)

∂y1+ . . . )Again if the system is on we have the usual formula. If the eld is given inthe asso iated on basis

X = X ′1u1 +X ′2u2 +X ′3u3thenX =

X ′1√λ1

e1 +X ′2√λ2

e2 +X ′3√λ3

e3and, applying the pre eding formula, we obtaindivX =1√g(∂(√gX′

1√λ1

)

∂y1+ . . . ) =

1√g(∂(√λ2λ3X

′1)

∂y1+ . . . )

Problem 123:Compute the divergen e in artesian, ylindri al and spheri al oordinates.What about the divergen e in the plane?Solution:With the pre eding notations we have:a) Cartesian oordinates.divX =∂X1

∂y1+∂X2

∂y2+∂X3

∂y3

236 CHAPTER 3. DIFFERENTIAL FORMSb) Cylindri al oordinates.divX =1

r(∂(rXr)

∂r+∂(rXθ)

∂θ+∂(rXz)

∂z) =

=1

r

∂(rXr)

∂r+∂Xθ

∂θ+∂Xz

∂zIf the eld is given in the asso iated on basis thendivX =1

r(∂(rX ′r)

∂r+∂(X ′θ)

∂θ+∂(rX ′z)

∂z) ) Spheri al oordinates.divX =

1

r2 sinϕ(∂(r2 sinϕXr)

∂r+∂(r2 sinϕXϕ)

∂ϕ+∂(r2 sinϕXθ)

∂θ) =

=1

r2

∂(r2Xr)

∂r+

1

sinϕ

∂(sinϕXϕ)

∂ϕ+∂Xθ

∂θIf the eld is given in the asso iated on basis thendivX =1

r2 sinϕ(∂(r2 sinϕX ′r)

∂r+∂(r sinϕX ′ϕ)

∂ϕ+∂(rX ′θ)

∂θ) =

=1

r2

∂(r2X ′r)

∂r+

1

r sinϕ

∂(sinϕX ′ϕ)

∂ϕ+

1

r sinϕ

∂X ′θ∂θd) For the divergen e in the plane, observe the following linear operationson a ve tor eld X = (X, Y ):

V(U)α→ V(U)

→ Ω1(U)d→ Ω2(U)

β→ F(U)(X, Y ) 7→ (−Y,X) 7→ −Y dx+Xdy 7→ (∂X

∂x+ ∂Y

∂y)dV 7→ ∂X

∂x+ ∂Y

∂y= divXThe matrix of α in the anoni al basis is

M =

(0 −11 0

)and, as the hange of basis matrix from the anoni al to the polar lo albasis isC =

(cos θ −r sin θsin θ r cos θ

),

3.3. COORDINATES 237we haveM ′ = C−1MC =

1

r

(r cos θ r sin θ− sin θ cos θ

)(0 −11 0

)(cos θ −r sin θsin θ r cos θ

)=

=1

r

(r sin θ −r cos θcos θ sin θ

)(cos θ −r sin θsin θ r cos θ

)=

1

r

(0 −r2

1 0

)Then, if X = Xrer +Xθeθ in the lo al basis of the polar oordinates,α(X) = −rXθer +

Xr

reθ

(α(X)) = −rXθdr + r2Xr

rdθ

d((α(X))) = −∂(rXθ)

∂θdθ ∧ dr +

∂(rXr)

∂rdr ∧ dθ =

= (∂(rXr)

∂r+∂(rXθ)

∂θ)dr ∧ dθ =

= (∂(rXr)

∂r+∂(rXθ)

∂θ)1

rdVand we read that divX =

1

r

∂(rXr)

∂r+∂Xθ

∂θe) We try the divergen e in ellipti oordinates. Now the hange of basismatrix from the anoni al to the ellipti al lo al basis isC =

(sinh u cos θ − cosh u sin θcosh u sin θ sinh u cos θ

), detC = cosh2 u− cos2 θ = κ

M ′ =1

κ

(sinh u cos θ cosh u sin θ− cosh u sin θ sinh u cos θ

)(0 −11 0

)(sinh u cos θ − cosh u sin θcosh u sin θ sinh u cos θ

)=

=1

κ

(cosh u sin θ − sinh u cos θsinh u cos θ cosh u sin θ

)(sinh u cos θ − cosh u sin θcosh u sin θ sinh u cos θ

)=

=1

κ

(0 −κκ 0

)=

(0 −11 0

)

238 CHAPTER 3. DIFFERENTIAL FORMSThen, if X = Xueu+Xθeθ in the lo al basis of the ellipti al oordinates,reminding that in that basisG = κ

(1 00 1

), dV = κdu ∧ dθ,we have

α(X) = −Xθeu +Xueθ

(α(X)) = κ(−Xθdu+Xrdθ)

d((α(X))) = −∂(κXθ)

∂θdθ ∧ du+

∂(κXr)

∂udu ∧ dθ =

= (∂(κXr)

∂u+∂(κXθ)

∂θ)du ∧ dθ =

=1

κ(∂(κXr)

∂u+∂(κXθ)

∂θ)dVand we read that divX =

1

κ(∂(κXr)

∂u+∂(κXθ)

∂θ)

Lapla ianT Let f : U ⊂ R3 → R a s alar eld in U and dV the volume form of R3.Then, with the usual notations

(∇2f)dV = div(gradf)dV = div(♯df)dV = d(2(♯df))gradf =1

λ1

∂f

∂y1e1 +

1

λ2

∂f

∂y2e2 +

1

λ3

∂f

∂y3e3

2(gradf) =√g(

1

λ1

∂f

∂y1

dy2 ∧ dy3 + . . . )

d(2(gradf)) = (∂(√g

λ1

∂f∂y1

)

∂y1+ . . . )dy1 ∧ dy2 ∧ dy3 =

=1√g(∂(√g

λ1

∂f∂y1

)

∂y1

+ . . . )dV

3.3. COORDINATES 239and nally∇2f =

1√g(∂(√g

λ1

∂f∂y1

)

∂y1

+ . . . )

Problem 124:Compute the lapla ian in artesian, ylindri al, spheri al, polar and ellipti al oordinates.Solution:a) Cartesian oordinates.∇2f =

∂2f

∂y21

+∂2f

∂y22

+∂2f

∂y23b) Cylindri al oordinates.

∇2f =1

r(∂

∂r(r∂f

∂r) +

∂

∂θ(1

r

∂f

∂θ) +

∂

∂z(r∂f

∂z)) =

=1

r

∂

∂r(r∂f

∂r) +

1

r2

∂2f

∂θ2+∂2f

∂z2 ) Spheri al oordinates.∇2f =

1

r2 sinϕ(∂

∂r(r2 sinϕ

∂f

∂r) +

∂

∂ϕ(sinϕ

∂f

∂ϕ) +

∂

∂θ(

1

sinϕ

∂f

∂θ)) =

=1

r2

∂

∂r(r2∂f

∂r) +

1

r2 sinϕ

∂

∂ϕ(sinϕ

∂f

∂ϕ) +

1

r2 sin 2ϕ

∂2f

∂θ2d) Polar oordinates.As we know that the divergen e of a eld X = Xrer +Xθeθ is divX =1r∂(rXr)∂r

+ ∂Xθ

∂θ(see p.235), we apply it to

X = ♯df =∂f

∂rer +

1

r2

∂f

∂θeθand obtain

∇2f =1

r

∂

∂r(r∂f

∂r) +

1

r2

∂2f

∂θ2

240 CHAPTER 3. DIFFERENTIAL FORMSe) Ellipti al oordinates.Now divX = 1κ(∂(κXr)

∂u+ ∂(κXθ)

∂θ) and we apply it to

X = ♯df =1

κ(∂f

∂uer +

∂f

∂θeθ)and obtain

∇2f =1

κ(∂2f

∂u2+∂2f

∂θ2)

3.4 Hodge's operatorProdu t of dierential 1-formsT Let be given a metri in the open set U ⊂ Rn, and let η, µ ∈ Ω1(U) betwo dierential 1-forms; dene its produ t as the fun tion(η, µ)(p) := (ηp, µp):

U → K

p 7→ (ηp, µp)Noti e that the appli ation : V(U) → Ω1(U)preserves the s alar produ t g(X,Y) = (X) · (Y), the dot being here (fora short moment) the produ t of dierential 1-forms.Problem 125:With the s alar produ t in R2, ompute the produ t of the dierential 1-forms given in polar oordinates ω = rdr + sin θdθ and ω′ = sin θdr + rdθ.Solution:Remind that the matrix of the eu lidian metri in the lo al basis isG =

(1 00 r2

)

3.4. HODGE'S OPERATOR 241The metri matrix of the produ t of 1-forms in the dual basis dr, dθ is theinverse:G−1 =

(1 00 1

r2

)Then(ω, ω′) = (r, sin θ)

(1 00 1

r2

)(sin θr

)= (r,

sin θ

r2)

(sin θr

)= r sin θ+

sin θ

r

Produ t of dierential k-formsT Let U ⊂ Rn be an open set and η, µ ∈ Ωk(U); dene its produ t as thefun tion (η, µ) := (ηp, µp), as for 1-forms (see p.132).Let e = (e1, . . . , en) a system of generators of V(U) (with fun tions as oe ients) and G the metri matrix. If dx = (dx1, . . . , dxn) is the systemof generators of Ω1(U) dual of e then its metri matrix is is G−1. Considerthe system of generators of Ωk(U)

dxi1 ∧ · · · ∧ dxikIրThe metri matrix of the produ t of k-forms is(dxi1 ∧ · · · ∧ dxik , dxj1 ∧ · · · ∧ dxjk) = det(G−1)j1...jki1...ik

Problem 126:Consider R2 with the standard s alar produ t. Let ω = r sin θdr ∧ dθ, η =rdr ∧ dθ; omputea) (ω, ω)b) (η, η) ) (ω, η)

242 CHAPTER 3. DIFFERENTIAL FORMSSolution:a)(ω, ω) = r2 sin2 θ det

((dr, dr) (dr, dθ)(dθ, dr) (dθ, dθ)

)=

= r2 sin2 θ det

(1 00 1

r2

)= sin2 θb)

(η, η) = r2 det

((dr, dr) (dr, dθ)(dθ, dr) (dθ, dθ)

)= r2 det

(1 00 1

r2

)= 1 )

(ω, η) = det

(sin θ(dr, dr) sin θ(dr, dθ)r2 sin θ(dθ, dr) r2 sin θ(dθ, dθ)

)=

= r2 sin θ det

(1 00 1

r2

)= sin θ

Problem 127:In R3 with the standard s alar produ t omputea) (dy ∧ dz, dy ∧ dz)b) (dx ∧ dy, dx ∧ dy)Solution:a) (dy ∧ dz, dy ∧ dz) = det

((dy, dy) (dy, dz)(dz, dy) (dz, dz)

)= det

(1 00 1

)= 1b) (dx ∧ dy, dx ∧ dy) = det

((dx, dx) (dx, dy)(dy, dx) (dy, dy)

)= det

(1 00 1

)= 1 ) (dx ∧ dy, dx ∧ dz) = det

((dx, dx) (dx, dz)(dy, dx) (dy, dz)

)= det

(1 00 0

)= 0

3.4. HODGE'S OPERATOR 243Problem 128:In polar oordinates ompute the produ t (rdr ∧ dθ, rdr ∧ dθ).Solution:(rdr ∧ dθ, rdr ∧ dθ) = r2(dr ∧ dθ, dr ∧ dθ) = r2 det

((dr, dr) (dr, dθ)(dθ, dr) (dθ, dθ)

)=

= r2 det

(1

1r2

)= 1

on bases of Ωk(U)

T

• U = Rn, artesian oordinates: dx1, . . . , dxn is an on basis at ea hpoint and it is an on system of generators of Ω1(Rn). The volume formωRn = dx1 ∧ · · · ∧ dxn satises (ωRn, ωRn) = 1.• U = R2 − L, polar oordinates: dr, rdθ is an on basis at ea h pointand it is an on system of generators of Ω1(U). The volume form ωR2 =rdr ∧ dθ satises (ωR2 , ωR2) = 1.• U ⊂ R3, ylindri al oordinates: dr, rdθ, dz is an on basis at ea hpoint and it is an on system of generators of Ω1(U). And rdr∧dθ, dr∧dz, rdθ ∧ dz is an on system of generators of Ω2(U). The volume formωR3 = rdr ∧ dθ ∧ dz satises (ωR3 , ωR3) = 1.• U ⊂ R3, spheri al oordinates: dr, rdϕ, r sinϕdθ is an on basis atea h point and it is an on system of generators of Ω1(U). Moreoverrdr∧dϕ, r sinϕdr∧dθ, r2 sinϕdϕ∧dθ is an on system of generators ofΩ2(U). The volume form ωR3 = r2 sinϕdr∧dϕ∧dθ satises (ωR3, ωR3) =1.

244 CHAPTER 3. DIFFERENTIAL FORMS3.4.1 Hodge's operatorProblem 129:Consider in R2 the eu lidian metri in polar oordinates with on lo al basisur = er,uθ = 1

reθ, dual basis dr, rdθ, and volume form ωE = rdr ∧ dθsatisfying (ωE , ωE) = (rdr ∧ dθ, rdr ∧ dθ) = 1 . Compute Hodge's operatorapplied to the on systems of Ωi(R2), i = 0, 1, 2 asso iated to the dual basis.Solution:a) ∗1 = (−1)indexωE = rdr ∧ dθb) ∗dr = sgn (1, 2)rdθ = rdθ

∗rdθ = sgn (2, 1)dr = −dr ) ∗rdr ∧ dθ = 1

Problem 130:Consider in R3 the eu lidian metri in ylindri al oordinates with on lo albasis ur = er,uθ = 1reθ,uz = ez, dual basis dr, rdθ, dz, and volume form

ωE = rdr ∧ dθ ∧ dz satisfying (ωE, ωE) = (rdr ∧ dθ ∧ dz, rdr ∧ dθ ∧ dz) = 1.Compute Hodge's operator applied to on bases of Ωi(R3), i = 0, 1, 2, 3.Solution:a) ∗1 = (−1)indexωE = rdr ∧ dθ ∧ dzb) ∗dr = sgn (1, 2, 3)rdθ ∧ dz = rdθ ∧ dz∗rdθ = sgn (2, 1, 3)dr ∧ dz = −dr ∧ dz∗dz = sgn (3, 1, 2)rdr ∧ dθ = rdr ∧ dθ ) ∗rdr ∧ dθ = sgn (1, 2, 3)ǫ3dz = dz∗dr ∧ dz = sgn (1, 3, 2)ǫ2rdθ = −rdθ∗rdθ ∧ dz = sgn (2, 3, 1)ǫ1dr = drd) ∗rdr ∧ dθ ∧ dz = 1

3.4. HODGE'S OPERATOR 245Problem 131:Consider in R3 the eu lidian metri in spheri al oordinates, on lo al basisur = er,uϕ = 1

reϕ,uθ = 1

r sinϕeθ, dual basis dr, rdϕ, r sinϕdθ, and volumeform ωE = r2 sinϕdr ∧ dϕ ∧ dθ satisfying

(ωE , ωE) = (r2 sinϕdr ∧ dϕ ∧ dθ, r2 sinϕdr ∧ dϕ ∧ dθ) = 1Compute Hodge's operator applied to on bases of Ωi(R3), i = 0, 1, 2, 3.Solution:a) ∗1 = (−1)indexωE = r2 sinϕdr ∧ dϕ ∧ dθb) ∗dr = sgn (1, 2, 3)r2 sinϕdϕ ∧ dθ = r2 sinϕdϕ ∧ dθ∗rdϕ = sgn (2, 1, 3)r sinϕdr ∧ dθ = −r sinϕdr ∧ dθ∗r sinϕdθ = sgn (3, 1, 2)rdr ∧ dϕ = rdr ∧ dϕ ) ∗rdr ∧ dϕ = sgn (1, 2, 3)r sinϕdθ = r sinϕdθ∗r sinϕdr ∧ dθ = sgn (1, 3, 2)rdϕ = −rdϕ∗r2 sinϕdϕ ∧ dθ = sgn (2, 3, 1)dr = drd) ∗r2 sinϕdr ∧ dϕ ∧ dθ = 1

3.4.2 Hodge and dierential operatorsRotational and divergen e of formsT In V(U), the world of ve tor elds in an open set U ⊂ R3, we have thedierential operator alled rotational

V(U)rot→ V(U)

F 7→ rot FWe ask what operation does that in the world of dierential 1-forms, that is:Ω1(U)

?→ Ω1(U)

F 7→ (rot F)

246 CHAPTER 3. DIFFERENTIAL FORMSWe know that d(F) = 2(rot F), an equality between 2-forms; as we lookfor a mapping between 1-forms it seems reasonable to use Hodge's operator:∗d(F) = ∗2(rot F)We ompute the a tion of ∗2 on a eld G in the anoni al system:

∗2G = ∗2(Xi + Y j + Zk) = ∗(Xdy ∧ dz + Y dz ∧ dx+ Zdx ∧ dy) =

= Xdx+ Y dy + Zdz = Gand obtain∗d(F) = (rot F)Now we see that ∗d is the sought for operation. We generalize to dene therotational of a k-form α as the (n− k − 1)-formrot α = ∗(dα)In V(U) we have as well another dierential operator alled the divergen eV(U)

div→ F(U)

F 7→ div Fand now we ask what operation does that in the world of dierential1-forms, that is:Ω1(U)

?→ Ω0(U) = F(U)

F 7→ div FNow as d(2F) = (div F)ωE and 2 = ∗

d(2F) = d(∗F) = (div F)ωE ⇒ ∗d(∗F) = (div F) ∗ ωE = div FThis shows that ∗d∗ is the sought for operation. We generalize to dene thedivergen e of a k-form α as the (k − 1)-formdiv α = (∗d∗)α

3.4. HODGE'S OPERATOR 247Problem 132:In R3 with the eu lidian metri , the usual orientation and volume form ωE =dx ∧ dy ∧ dz, ompute the rotational and the divergen e of the forms:a) α = adx+ bdy + cdzb) β = Ady ∧ dz +Bdz ∧ dx+ Cdx ∧ dySolution:a) Rotational

dα = (∂c

∂y− ∂b

∂z)dy ∧ dz + (

∂a

∂z− ∂c

∂x)dz ∧ dx+ (

∂b

∂x− ∂a

∂y)dx ∧ dyrot α = ∗dα = (

∂c

∂y− ∂b

∂z)dx+ (

∂a

∂z− ∂c

∂x)dy + (

∂b

∂x− ∂a

∂y)dzDivergen e

∗α = ady ∧ dz + bdz ∧ dx+ cdx ∧ dy

d ∗ α = (∂a

∂x+∂b

∂y+∂b

∂z)dx ∧ dy ∧ dzdiv α = ∗d ∗ α =

∂a

∂x+∂b

∂y+∂b

∂zNoti e that if we take into a ount the motivation for the denition ofrot and div those omputations are in fa t not ne essary.b) i) The rotational isdβ = (

∂A

∂x+∂B

∂y+∂C

∂z)dx ∧ dy ∧ dzrot β = ∗dβ =

∂A

∂x+∂B

∂y+∂C

∂zand the divergen e∗β = Adx+Bdy + Cdz

248 CHAPTER 3. DIFFERENTIAL FORMSd∗β = (

∂C

∂y− ∂B∂z

)dy∧dz+(∂A

∂z− ∂C∂x

)dz∧dx+(∂B

∂x− ∂A∂y

)dx∧dydiv β = ∗d ∗ β = (∂C

∂y− ∂B

∂z)dx+ (

∂A

∂z− ∂C

∂x)dy + (

∂B

∂x− ∂A

∂y)dzii) We an as well follow another line of reasoning; let F = Ai+Bj+

Ck and thenβ = 2F⇒ dβ = d(2F) = (div F)ωE ⇒ ∗dβ = div FThat is rot β =

∂A

∂x+∂B

∂y+∂C

∂zThe divergen e is∗β = ∗2F = F⇒ d ∗ β = d(F) = 2(rot F)

∗d ∗ β = ∗2(rot F) = (rot F)and we obtaindiv β = (∂C

∂y− ∂B

∂z)dx+ (

∂A

∂z− ∂C

∂x)dy + (

∂B

∂x− ∂A

∂y)dz

Problem 133:In R3 with the eu lidian metri and the usual orientation, remind that theon positive lo al basis in ylindri al oordinates is ur = er,uθ = 1reθ,uz = ezand the dual basis is dr, rdθ, dz. The volume form ωE = rdr∧dθ∧dz satises

(ωE, ωE) = 1. Let α = Frdr+Fθrdθ+Fzdz; ompute the rotational and thedivergen e of α.Solution:We show two ways; rst the rotational:

3.4. HODGE'S OPERATOR 249a)dα = (

∂Fz∂θ−∂(Fθr)

∂z)dθ∧dz+(

∂Fr∂z−∂Fz∂r

)dz∧dr+(∂(rFθ)

∂r−∂Fz∂θ

)dr∧dθrot α = ∗dα = (∂Fz∂θ−∂(Fθr)

∂z)1

rdr+(

∂Fr∂z−∂Fz∂r

)rdθ+(∂(Fθr)

∂r−∂Fz∂θ

)1

rdzb) Taking into a ount how we arrived at the denition we know thatrot α = (rot F) where F = Frur + Fθuθ + Fzuz. Thenrot F = (

∂Fz∂θ− ∂(Fθr)

∂z)1

rur + (

∂Fr∂z− ∂Fz

∂r)uθ + (

∂(Fθr)

∂r− ∂Fz

∂θ)1

ruz

(rot F) = (∂Fz∂θ− ∂(Fθr)

∂z)1

rdr+(

∂Fr∂z− ∂Fz

∂r)rdθ+(

∂(Fθr)

∂r− ∂Fz∂θ

)1

rdzThe divergen e an also be omputed in two ways:a)

∗α = Frrdθ ∧ dz + Fθr−1

rdr ∧ dz + Fzrdr ∧ dθ

d(∗α) = (∂(rFr)

∂r+∂Fθ∂θ

+∂(rFz)

∂z)dr ∧ dθ ∧ dzdiv α = ∗d ∗ α =

1

r(∂(rFr)

∂r+∂Fθ∂θ

+∂(rFz)

∂z)b) From the denition we know that divα = div ♯α:

♯α = Frur + Fθuθ + Fzuzbut we have omputed the divergen e of a ve tor eld given in an onbasis (see p.235); thendiv α = div ♯α =1

r(∂(rFr)

∂r+∂Fθ∂θ

+∂(rFz)

∂z)

250 CHAPTER 3. DIFFERENTIAL FORMSProblem 134:In R3 with the eu lidian metri and the usual orientation, the on lo al basisin spheri al oordinates is ur = er,uϕ = 1reϕ,uθ = 1

r sinϕeθ, with dual basis

dr, rdϕ, r sinϕdθ, and volume form ωE = r2 sinϕdr ∧ dϕ ∧ dθ that satises(ωE, ωE) = 1. Compute the rotational and the divergen e of the 1-formα = Frdr + Fϕrdϕ+ Fθr sinϕdθ.Solution:As in the pre eding problem we an do the omputations in two ways; rstthe rotational:a)

dα = (∂(Fθr sinϕ)

∂ϕ− ∂(Fϕr)

∂θ)dϕ ∧ dθ + (

∂Fr∂θ− ∂(Fθr sinϕ)

∂r)dθ ∧ dr +

+ (∂(Fϕr)

∂r− ∂Fr

∂ϕ)dr ∧ dϕrot α = ∗dα = (

∂(Fθr sinϕ)

∂ϕ− ∂(Fϕr)

∂θ)

1

r2 sinϕdr + (

∂Fr∂θ− ∂(Fθr sinϕ)

∂r)

1

sinϕdϕ+

+ (∂(Fϕr)

∂r− ∂Fr

∂ϕ) sinϕdθb) As α = F where F = Frur + Fϕuϕ + Fθuθ, we have rot α = (rot F):rot F = (

∂(Fθr sinϕ)

∂ϕ−∂(Fϕr)

∂θ)

1

r2 sinϕur+(

∂Fr∂θ−∂(Fθr sinϕ)

∂r)

1

r sinϕuϕ+

+(∂(Fϕr)

∂r− ∂Fr

∂ϕ)1

ruθ

(rot F) = (∂(Fθr sinϕ)

∂ϕ− ∂(Fϕr)

∂θ)

1

r2 sinϕdr + (

∂Fr∂θ− ∂(Fθr sinϕ)

∂r)

1

sinϕdϕ+

+ (∂(Fϕr)

∂r− ∂Fr

∂ϕ) sinϕdθFor the divergen e we have

3.4. HODGE'S OPERATOR 251a)∗α = Frr

2 sinϕdϕ ∧ dθ − Fϕr sinϕdθ ∧ dr + Fθrdr ∧ dϕ

d ∗ α = (∂(Frr

2 sinϕ)

∂r+∂(Fϕr sinϕ)

∂ϕ+∂(Fθr)

∂θ)dr ∧ dϕ ∧ dθdiv α = ∗d ∗ α =

1

r2 sinϕ(∂(Frr

2 sinϕ)

∂r+∂(Fϕr sinϕ)

∂ϕ+∂(Fθr)

∂θ)b) As (div F)ωE = d(2F) we have

2F = Frr2 sinϕdϕ ∧ dθ + Fϕr sinϕdθ ∧ dr + Fθrdr ∧ dϕ

d(2F) = (∂(Frr

2 sinϕ)

∂r+∂(Fϕr sinϕ)

∂ϕ+∂(Fθr)

∂θ)dr ∧ dϕ ∧ dθdiv F =

1

r2 sinϕ(∂(Frr

2 sinϕ)

∂r+∂(Fϕr sinϕ)

∂ϕ+∂(Fθr)

∂θ)

Problem 135:Consider in R4 the basis t = (1, 0, 0, 0), i = (0, 1, 0, 0), j = (0, 0, 1, 0),k =(0, 0, 0, 1) and let it orient the spa e. Call the asso iated oordinates (t, x, y, z)and take the Lorentz metri

G =

−11

11

Compute the rotational and the divergen e of the forms:a) α = adx+ bdy + cdz.b) β = Ady ∧ dz +Bdz ∧ dx+ Cdx ∧ dy.Remind that the oe ients a, b, c, A,B, C depend on t.

252 CHAPTER 3. DIFFERENTIAL FORMSSolution:a) i)dα = (

∂c

∂y− ∂b

∂z)dy ∧ dz + (

∂a

∂z− ∂c

∂x)dz ∧ dx+ (

∂b

∂x− ∂a

∂y)dx ∧ dy +

+∂a

∂tdt ∧ dx+

∂b

∂tdt ∧ dy +

∂c

∂tdt ∧ dzTo ompute ∗ we assign the index 0 to the oordinate t :

∗dy ∧ dz = sgn (2, 3, 0, 1)ǫ01dt ∧ dx = −dt ∧ dx

∗dz ∧ dx = sgn (3, 1, 0, 2)ǫ02dt ∧ dy = −dt ∧ dy

∗dx ∧ dy = sgn (1, 2, 0, 3)ǫ03dt ∧ dz = −dt ∧ dz

∗dt ∧ dx = sgn (0, 1, 2, 3)ǫ23dy ∧ dz = dy ∧ dz

∗dt ∧ dy = sgn (0, 2, 1, 3)ǫ13dx ∧ dz = dz ∧ dx

∗dt ∧ dz = sgn (0, 3, 1, 2)ǫ12dx ∧ dy = dx ∧ dyand the rotational of α is:∗dα = −(

∂c

∂y− ∂b

∂z)dt ∧ dx− (

∂a

∂z− ∂c

∂x)dt ∧ dy − (

∂b

∂x− ∂a

∂y)dt ∧ dz +

+∂a

∂tdy ∧ dz +

∂b

∂tdz ∧ dx+

∂c

∂tdx ∧ dy =

= −dt ∧ ((∂c

∂y− ∂b

∂z)dx+ (

∂a

∂z− ∂c

∂x)dy + (

∂b

∂x− ∂a

∂y)dz) +

+∂a

∂tdy ∧ dz +

∂b

∂tdz ∧ dx+

∂c

∂tdx ∧ dythat we may express thusrot α = (rotSα) ∧ dt+ ∗S∂tαwhere rotSα, ∗S are omputed with respe t to the 'spa e' variables

(x, y, z) and that's why they have the subindex S.

3.4. HODGE'S OPERATOR 253ii) To ompute the divergen e we rst nd ∗:∗dx = sgn(1, 0, 2, 3)ǫ023dt ∧ dy ∧ dz = dt ∧ dy ∧ dz

∗dy = sgn(2, 0, 1, 3)ǫ013dt ∧ dx ∧ dz = −dt ∧ dx ∧ dz

∗dz = sgn(3, 0, 1, 2)ǫ012dt ∧ dx ∧ dy = dt ∧ dx ∧ dyand then∗α = a dt ∧ dy ∧ dz − b dt ∧ dx ∧ dz + c dt ∧ dx ∧ dy =

= dt ∧ (ady ∧ dz + bdz ∧ dx+ cdx ∧ dy) = dt ∧ ∗SαAs dt is of degree 1 we haved(∗α) = −dt ∧ d(∗Sα) = −dt ∧ (dS(∗Sα) + dt ∧ ∂t(∗Sα)) =

= −dt ∧ dS(∗Sα)anddS(∗Sα) = (

∂a

∂x+∂b

∂y+∂b

∂z)dx ∧ dy ∧ dz

−dt ∧ dS(∗Sα) = −(∂a

∂x+∂b

∂y+∂b

∂z)dt ∧ dx ∧ dy ∧ dzApplying ∗ we obtain

∗d ∗ α = −(∂a

∂x+∂b

∂y+∂b

∂z)We express the result thusdiv α = −divSαwhere the rst div is in four dimensions and divS is omputedwith respe t to the variables (x, y, z).b)

254 CHAPTER 3. DIFFERENTIAL FORMSi)dβ = (

∂A

∂x+∂B

∂y+∂C

∂z)dx∧dy∧dz+dt∧(

∂A

∂tdy∧dz+∂B

∂tdz∧dx+∂C

∂tdx∧dy)We must ompute ∗:

∗dx ∧ dy ∧ dz = sgn (1, 2, 3, 0)ǫ0dt = dt

∗dt ∧ dx ∧ dy = sgn (0, 1, 2, 3)ǫ3dz = dz

∗dt ∧ dy ∧ dz = sgn (0, 2, 3, 1)ǫ1dx = dx

∗dt ∧ dz ∧ dx = sgn (0, 3, 1, 2)ǫ2dy = dyThenrot β = ∗dβ = (∂A

∂x+∂B

∂y+∂C

∂z)dt+ (

∂A

∂tdx+

∂B

∂tdy +

∂C

∂tdz)that we an rewrite as (see p.247)rot β = rotSβ + ∂t ∗S βii) To ompute the divergen e, from the values of ∗ obtained in a),we have :

∗β = −(Adt ∧ dx+Bdt ∧ dy + Cdt ∧ dz)= −dt ∧ (Adx+Bdy + Cdz)

d(∗β) = dt∧((∂C

∂y−∂B∂z

)dy∧dz+(∂A

∂z−∂C∂x

)dz∧dx+(∂B

∂x−∂A∂y

)dx∧dy+

+∂A

∂tdt ∧ dx+

∂B

∂tdt ∧ dy +

∂C

∂tdt ∧ dz) =

= dt∧((∂C

∂y− ∂B∂z

)dy∧dz+(∂A

∂z− ∂C∂x

)dz∧dx+(∂B

∂x− ∂A∂y

)dx∧dyBy b) i)∗d ∗ β = (

∂C

∂y− ∂B

∂z)dx+ (

∂A

∂z− ∂C

∂x)dy + (

∂B

∂x− ∂A

∂y)dzthat we an write as div β = divSβ

3.4. HODGE'S OPERATOR 255Lapla ianT The odierential of a k-form α is the (k− 1)-form (see [Nav p.31, [Jänp.225 and [Flem p.275 to appre iate the diversity of signs in the denition)

δα = (−1)n(k+1)+1div α = (−1)n(k+1)+1 ∗ d ∗ αand the lapla ian a ting on k-forms is∆ = δd+ dδNote:The origin of the name o-dierential lies in the following diagram

Ωn−k d→ Ωn−k+1

∗ ↑∼= ∼=↓ ∗Ωk ∗d∗→ Ωk−1that shows how ∗d∗ lowers one unity the degree of the form.

Problem 136:Taking in R3 the usual s alar produ t and the usual orientation apply thepre eding denition to a fun tion f : R3 → R to see whether we obtain the lassi al expression or not. Investigate the ase f : Rn → R.Solution:As f an be seen as a 0-form we have(∗d∗)f = ∗d(fdx ∧ dy ∧ dz) = 0⇒ δf = 0On the other hand

df =∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz

∗df =∂f

∂xdy ∧ dz +

∂f

∂ydz ∧ dx+

∂f

∂zdx ∧ dy

256 CHAPTER 3. DIFFERENTIAL FORMSd(∗df) = (

∂2f

∂x2+∂2f

∂y2+∂2f

∂z2)dx ∧ dy ∧ dz

(∗d∗)df =∂2f

∂x2+∂2f

∂y2+∂2f

∂z2Finally∆f = (δd+ dδ)f = δdf = (−1)3(1+1)+1div (df) =

= −(∂2f

∂x2+∂2f

∂y2+∂2f

∂z2)The ase of Rn produ es δf = 0 and the omputations are similar.

Problem 137:Compute in ylindri al oordinates the lapla ian of a fun tion f dened inR3.Solution:The operator ∗ in ylindri al oordinates has been worked out in p.244. Againδf = 0, for

(∗d∗)f = ∗d(f rdr ∧ dθ ∧ dz) = 0On another handdf =

∂f

∂rdr +

∂f

∂θdθ +

∂f

∂zdz

∗df =∂f

∂rrdθ ∧ dz +

∂f

∂θ

−1

rdr ∧ dz +

∂f

∂zrdr ∧ dθ

d(∗df) = (∂

∂r(∂f

∂rr) +

∂2f

∂θ2

1

r+∂2f

∂z2r)dr ∧ dθ ∧ dz

= (1

r

∂

∂r(∂f

∂rr) +

∂2f

∂θ2

1

r2+∂2f

∂z2)rdr ∧ dθ ∧ dz

(∗d∗)df =1

r

∂

∂r(∂f

∂rr) +

1

r2

∂2f

∂θ2+∂2f

∂z2Finally∆f = (δd+ dδ)f = δ(df) = (−1)3(1+1)+1div (df) = −div (df) =

= −(1

r

∂

∂r(∂f

∂rr) +

1

r2

∂2f

∂θ2+∂2f

∂z2)

3.4. HODGE'S OPERATOR 257Problem 138:Compute in spheri al oordinates the lapla ian of a fun tion f dened in R3.Solution:The operator ∗ in spheri al oordinates is worked out in p.245. As in thetwo pre eding problems δf = 0, for div f = 0 (as a 0-form). Then

df =∂f

∂rdr +

∂f

∂ϕdϕ+

∂f

∂θdθ

∗df =∂f

∂rr2 sinϕdϕ ∧ dθ +

∂f

∂ϕ(− sinϕ)dr ∧ dθ +

∂f

∂θ

1

sinϕdr ∧ dϕ

d(∗df) = (∂

∂r(r2∂f

∂r) sinϕ+

∂

∂ϕ(sinϕ

∂f

∂ϕ) +

1

sinϕ

∂2f

∂θ2)dr ∧ dϕ ∧ dθdiv (df) = (∗d∗)df =

1

r2 sinϕ((∂

∂r(r2∂f

∂r) sinϕ+

∂

∂ϕ(sinϕ

∂f

∂ϕ) +

1

sinϕ

∂2f

∂θ2)Finally:

∆f = (δd+ dδ)f = δ(df) = (−1)3(1+1)+1div (df) = −div (df) =

= − 1

r2 sinϕ((∂

∂r(r2∂f

∂r) sinϕ+

∂

∂ϕ(sinϕ

∂f

∂ϕ) +

1

sinϕ

∂2f

∂θ2)

Problem 139:Let the basis t = (1, 0, 0, 0), i = (0, 1, 0, 0), j = (0, 0, 1, 0),k = (0, 0, 01) orientR4, all the asso iated oordinates (t, x, y, z), (dt, dx, dy, dz) the dual basisand take the Lorentz metri

G =

−11

11

The volume form ωR4 = dt ∧ dx ∧ dy ∧ dz satises (ωR4, ωR4) = −1.Compute in artesian oordinates the lapla ian of a fun tion f .

258 CHAPTER 3. DIFFERENTIAL FORMSSolution:Remind that ∆α = (dδ + δd)α and that δα = (−1)4(k+1)+1 ∗ d ∗ α whenα ∈ Ωk(R4). First δf = (−1)4(0+1)+1 ∗ d ∗ f = − ∗ d(−fωR4) = 0 ; let's ompute δdf :

df =∂f

∂tdt+

∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz

∗df =∂f

∂tdx ∧ dy ∧ dz +

∂f

∂xdt ∧ dy ∧ dz +

∂f

∂ydt ∧ dz ∧ dx+

∂f

∂zdt ∧ dx ∧ dy

d ∗ (df) = (∂2f

∂t2− ∂2f

∂x2− ∂2f

∂y2− ∂2f

∂z2)dt ∧ dx ∧ dy ∧ dz

(∗d∗)df = −∂2f

∂t2+∂f

∂x2+∂2f

∂y2+∂2f

∂z2Finally∆f = (δd+ dδ)f = δ(df) = (−1)4(1+1)+1 ∗ d ∗ (df) = − ∗ d ∗ (df) =

=∂2f

∂t2− ∂f

∂x2− ∂2f

∂y2− ∂2f

∂z2The equation ∆f = 0, that is∂2f

∂x2+∂2f

∂y2+∂2f

∂z2− ∂2f

∂t2= 0is alled the wave equation and the operator f = ∂2f

∂x2 + ∂2f∂y2

+ ∂2f∂z2− ∂2f

∂t2is alled the d'alembertian. The solutions of f = 0 are the waves; the usualwaves satify this equation.Problem 140:a) Show that ∆α = (−1)n+1rot rot α − grad div α for a 1-form α ∈

Ω1(Rn).b) To dene the lapla ian of a ve tor eld F ∈ V(U), U ⊂ R3 we ompute∆F; if F = (a, b, c) de ide wether ∆F = (∆a)dx + (∆b)dy + (∆c)dzor not.

3.4. HODGE'S OPERATOR 259Solution:a) rot α = ∗dα, rot (rot α) = ∗d ∗ dαdiv α = ∗d ∗ α, grad (div α) = d ∗ d ∗ αOn another handδα = (−1)n(1+1)+1 ∗ d ∗ α = − ∗ d ∗ αdδα = −d ∗ d ∗ α = −grad (div α)

δ(dα) = (−1)n(2+1)+1 ∗ d ∗ dα = (−1)n+1 ∗ d ∗ dα =

= (−1)n+1rot (rot α)and∆α = δ(dα) + dδα = (−1)n+1rot (rot α)− grad (div α)b) It su es to he k the ase F = (a, 0, 0) ; we have

δ(adx) = (−1)3(1+1)+1 ∗ d ∗ (adx) =

= − ∗ d(ady ∧ dz) =

= − ∗ ∂a∂xdx ∧ dy ∧ dz =

= −∂a∂xand

dδ(adx) = −(∂2a

∂x2dx+

∂2a

∂x∂ydy +

∂2a

∂x∂zdz)On another hand

δd(adx) = (−1)3(2+1)+1 ∗ d ∗ (∂a

∂ydy ∧ dx+

∂a

∂zdz ∧ dx) =

= ∗d(−∂a∂ydz +

∂a

∂zdy) =

= ∗( ∂2a

∂z∂xdx ∧ dy − (

∂2a

∂y2+∂2a

∂z2)dy ∧ dz +

∂2a

∂y∂xdz ∧ dx) =

= −(∂2a

∂y2+∂2a

∂z2)dx+

∂2a

∂y∂xdy +

∂2a

∂z∂xdz

260 CHAPTER 3. DIFFERENTIAL FORMSFinally∆(adx) = dδ(adx) + δd(adx) =

= −(∂2a

∂x2+∂2a

∂y2+∂2a

∂z2)dx =

= (∆a)dxSumming up we have∆F = (∆a)dx+ (∆b)dy + (∆c)dz

Problem 141:If ω ∈ Ωk(Rn), show thata) δ ∗ ω = (sgn g)(−1)n−k+1 ∗ dω.b) ∗δω = (−1)kd ∗ ω.Solution:Let w = (ω1, . . . , ωn) be an on basis and let ω = ωi1 ∧ · · · ∧ ωik ; we havea)δ ∗ ω = (−1)n(n−(n−k)+1)+1 ∗ d ∗ ∗ω =

= (−1)n(k+1)+1 ∗ d((−1)k(n−k)(sgn g)ω) =

= (sgn g)(−1)nk+n+1+nk−k2 ∗ dω =

= (sgn g)(−1)n−k+1 ∗ dωb)∗δω = (−1)n(k+1)+1 ∗ ∗d ∗ ω =

= (−1)n(k+1)+1(−1)(n−k+1)(k−1)d ∗ ω =

= (−1)n(k+1)+1+(n−k+1)(k−1)d ∗ ω =

= (−1)kd ∗ ω

Chapter 4Stokes theorem4.1 Closed and exa t forms4.1.1 Closed and exa t formsT A dierential form ω ∈ Ωk(U) in an open set U ⊂ Rn is losed wheneverdω = 0 and is exa t if there exists η ∈ Ωk−1(U) ( alled a primitive of ω) su hthat dη = ω. As d2 = 0, every exa t form is losed.

Problem 142: Closed and exa t forms.a) Show that the exterior produ t of two losed forms is losed.b) Show that the exterior produ t of a losed form and an exa t form isexa t. ) A fun tion f is an integrating fa tor of a 1-form ω in U ⊂ Rn if f(x) ∈C1(U), f(x) 6= 0 for every x ∈ U and fω is losed. Show that if ω hasan integrating fa tor then ω ∧ dω = 0.d) Let F and G be ve tor elds in R3, F = ∇V and ∇×G = 0. Find ave tor potential ofF×G (:= a ve tor eldM su h that∇×M = F×G).261

262 CHAPTER 4. STOKES THEOREMSolution:a) If dω = 0, dη = 0 thend(ω ∧ η) = dω ∧ η + (−1)deg ωω ∧ dη = 0and ω ∧ η is losed.b) If dω = 0 and η = dθ thend(ω ∧ θ) = dω ∧ θ + (−1)deg ωω ∧ dθ =

= (−1)deg ωω ∧ dθ = (−1)deg ωω ∧ ηand we see that (−1)deg ωω ∧ θ is a primitive of ω ∧ η. ) Let f be an integrating fa tor of ω. Then fω is losed:0 = d(fω) = df ∧ ω + fdωSo

dω = −1

fdf ∧ ωand

ω ∧ dω = −ω ∧ (1

fdf ∧ ω) = 0,for as ω is a 1-form we have ω ∧ ω = 0.d) Let ω1 = F, ω2 = G; the hypotheses in the language of forms are

ω1 = dV, dω2 = 0,that is ω1 is exa t and ω2 is losed. By the result in b) ω1 ∧ω2 is exa tand V ω2 is a primitive:d(V ω2) = d(V G) = d(VG) = 2(rot VG)

d(V ω2) = ω1 ∧ ω2 = F ∧ G = 2(F×G)that is rot VG = F×G.

4.1. CLOSED AND EXACT FORMS 263Problem 143:Let f : R→ R be a ontinuous fun tion and ω = f(|x|)(x1dx1+· · ·+xndxn) ∈Ω1(Rn \0) a 1-form (that is to say that the ve tor eld ♯ω is entral). Finda primitive of ω. (Hint: use a primitive of uf(u).)Solution:As f is ontinuous xf(x) has a primitive, all it G(x); then φ(x) = G(|x|) isa primitive of ω for

dφ = G′(|x|)( 1

|x|x1dx1 + · · ·+ 1

|x|xndxn) =

= |x|f(|x|)( 1

|x|x1dx1 + · · ·+ 1

|x|xndxn) =

= f(|x|)(x1dx1 + · · ·+ xndxn)

Problem 144:Let p : R → R be a ontinuous fun tion in (a, b); onsider the 1-form ω =p(x)ydx + dy ∈ Ω1(U) where U = (x, y) : a < x < b ⊂ R2. If P (x) is aprimitive of p, show that f(x) = eP (x) is an integrating fa tor of ω. Find aprimitive of fω.Solution:f doesn't vanish and we must see that fω = f(x)p(x)ydx+ f(x)dy is losed;noti ing that f ′(x) = eP (x)p(x) = f(x)p(x) we have

d(fω) = f(x)p(x)dy ∧ dx+ f(x)p(x)dx ∧ dy = 0To nd a primitive we must nd a fun tion V (x, y)su h that∂xV = f(x)p(x)y∂yV = f(x)Integrating respe t to x the rst equation we have

V (x, y) = f(x)y + C(y)

264 CHAPTER 4. STOKES THEOREMand substituting into the se ond we ndf(x) + C ′(y) = f(x)that is C ′(y) = 0, y ∈ R and then C = onst. A primitive isV (x, y) = f(x)y

Problem 145:Let f : U ⊂ Rn → f(U) ⊂ Rn be a dieomorphism and assume that every losed form in f(U) is exa t. Show that the same is true in U .Solution:We use the naturality of d respe t to f ∗; let us remind the intervening appli- ations:U

f−→ f(U)

Ωk(U)f

∗

←− Ωk(f(U))

Ωk(U)(f−1)∗

−→ Ωk(f(U))

ω 7−→ (f−1)∗ωLet ω be a losed k-dierential form in U . Then (f−1)∗ω is losed in f(U)ford((f−1)∗ω) = (f−1)∗dω = (f−1)∗0 = 0The hypothese says that this losed form in f(U) is exa t, that is there is aform η1 ∈ Ωk−1(f(U)) su h that dη1 = (f−1)∗ω. Then η = f ∗η1 is a primitiveof ω :

dη = d(f ∗η1) = f ∗(dη1) = f ∗((f−1)∗ω) = (f−1 f)∗ω = Id∗ω = ω

4.1. CLOSED AND EXACT FORMS 265Problem 146:Let ϕ1, . . . , ϕn ∈ Ω1(Rn) be linearly independent 1-forms (:=at ea h pointp ∈ Rn) and let ω ∈ Ω1(Rn) be su h that

dϕi = ω ∧ ϕi, i = 1, . . . , nShow that ω is losed.Solution:0 = d2ϕi = d(dϕi) = dω ∧ ϕi − ω ∧ dϕi =

= dω ∧ ϕi − ω ∧ ω ∧ ϕi = dω ∧ ϕiand the result in the problem on p.94 asserts that dω = 0.4.1.2 Poin aré formula

T An open set U ⊂ Rn is ontra tile to a point p ∈ U if there is a dieren-tiable mapφ : [0, 1] ×U → U

t x 7→ φ(t,x)satisfyingφ(0,x) = p

φ(1,x) = xThe appli ation φ is alled a ontra tion to p.

266 CHAPTER 4. STOKES THEOREMU [0,1]

φU

pt

U tφ(U t )

A set U ⊂ Rn is star-shaped respe t to p ∈ U if ∀x ∈ U we have[p,x] ⊂ U . For a star shaped set U respe t to p we have the ontra tion

φ(t,x) = (1− t)p + txIn parti ular if U is star-shaped with respe t to 0 (for instan e Rn obvi-ously is) we have the retra tion φ(t,x) = tx. In the gure below we see astar-shaped set in the plane

x

y

pU

In an open set U ⊂ Rn ontra tile to a point we dene a mapI : Ωk([0, 1]× U)→ Ωk−1(U)

4.1. CLOSED AND EXACT FORMS 267through its value on generators:I(a(t,x)dxi1 ∧ · · · ∧ dxik) = 0

I(b(t,x)dt ∧ dxi1 ∧ · · · ∧ dxik−1) = (

∫ 1

0

b(t,x)dt)dxi1 ∧ · · · ∧ dxik−1One proves (see[Jan p.202, [Fla p.27) that whenever U is ontra tile toa point, ω ∈ Ωk(U) is losed, and φ is a ontra tion, the form I(φ∗ω) is aprimitive of ω. Then if U is ontra tile to a point, every losed form in U isexa t. In parti ular we have:Poin aré lemma:U ⊂ Rn open, star-shaped set ⇒ every losed form in U is exa t.

Problem 147:a) Find all the dierential forms ω of Rn su h that dω = ω ∧ ω.b) Whi h forms ω of Rn satisfy dω ∧ dω = 0?Solution:a) If ω ∈ Ωk(Rn) and dω = ω ∧ ω, equating the degrees we see thatk + 1 = 2k, k = 1; then dω = ω ∧ ω = 0 and ω is losed. As Rn isstar shaped (respe t any point) applying Poin aré lemma we see thatω = df .b) If deg (ω) is even then deg (dω) is odd and dω ∧ dω = −dω ∧ dω thatis dω ∧ dω = 0 and any form of even degree is a solution. On the otherhand if deg (ω) is odd, there are forms not satisfying the ondition; forinstan e in R4

ω = xdy + zdt

dω = dx ∧ dy + dz ∧ dtanddω ∧ dω = dx ∧ dy ∧ dz ∧ dt+ dz ∧ dt ∧ dx ∧ dy =

= 2dx ∧ dy ∧ dz ∧ dt 6= 0

268 CHAPTER 4. STOKES THEOREMAgain if n = 2 every 1-form is a solution and in general if deg (ω) = kthen deg (dω ∧ dω) = 2k + 2 and the produ t vanishes if 2k + 2 > n.Problem 148:Find a formula to ompute primitives of losed dierential forms in a starshaped open set U .Solution:We want to ompute I(φ∗ω); assume without loss of generality that U isstar shaped respe t to 0 and take the ontra tion φ(t,x) = tx. If ω =

a(x)dxi1 ∧ · · · ∧ dxik , using the naturality of d we haveφ∗(a(x)dxi1 ∧ · · · ∧ dxik) = a(tx)((xi1dt+ tdxi1) ∧ · · · ∧ (xikdt+ tdxik))Be ause of the denition of I it su es to look at terms ontaining dt

φ∗ω = a(tx)(xi1tk−1dt∧

∧dxi1 ∧dxi2 ∧ · · · ∧ dxik − xi2tk−1dt ∧ dxi1∧

∧dxi2 ∧ · · · ∧ dxik + . . . )Finally we obtain Poin aré formula

I(φ∗(a(x)dxi1∧· · ·∧dxik)) = (

∫ 1

0

a(tx)tk−1dt)k∑

j=1

(−1)j−1xijdxi1∧· · · ∧∧dxij ∧ · · ·∧dxikthat we write thus

(Iω)(x) = (

∫ 1

0

a(tx)tk−1dt)

k∑

j=1

(−1)j−1xijdxi1 ∧ · · · ∧∧dxij ∧ · · · ∧ dxikAs any k-form is the addition of forms like ω, we an extend I to Ωk(U).

4.1. CLOSED AND EXACT FORMS 269Problem 149:Compute a primitive for those of the following forms that are losed:a) ω = xdx+ ydyb) ω = xydy + xdz ) ω = sin xdy + dzd) ω = eydx+ ydye) ω = exydx ∧ dyf) ω = sin(xyz) dx ∧ dy ∧ dzSolution:a) dω = 0 and a primitive is:I(xdx) = (

∫ 1

0

txdt)x =x2

2

I(ydy) = (

∫ 1

0

tydt)y =y2

2and nallyI(xdx+ ydy) =

x2

2+y2

2b) It is not losed. ) It is not losed.d) It is not losed.e) i) If x 6= 0, y 6= 0 we haveI(exydx∧dy)(x,y) = (

∫ 1

0

tet2xydt)(xdy−ydx) =

et2xy

2xy|t=1t=0(xdy−ydx) =

=exy − 1

2xy(xdy − ydx)

270 CHAPTER 4. STOKES THEOREMii) If x = 0 or y = 0 we haveI(exydx ∧ dy)(x,y) = (

∫ 1

0

te0dt)(xdy − ydx) =1

2(xdy − ydx)As the formula of Poin aré produ es dierentiable forms we haveseen that the fun tion

f(x, y) =

exy−12xy

if x 6= 0, y 6= 012

if x = 0 o y = 0is dierentiable.f) i) As in the pre eding point, if x 6= 0, y 6= 0, z 6= 0 we haveI(sin(xyz) dx∧dy∧dz)(x,y) = (

∫ 1

0

t2 sin(t3xyz)dt)(xdy∧dz−ydx∧dz+zdx∧dy) =

=1− cosxyz

3xyz(xdy ∧ dz − ydx ∧ dz + zdx ∧ dy)ii) If x = 0 or y = 0 or z = 0 then

I(sin(xyz) dx∧dy∧dz)(x,y) = (

∫ 1

0

t2·0dt)(xdy∧dz−ydx∧dz+zdx∧dy) = 0and we an make the same omment as in the pre eding point.Problem 150:A fun tion g : R3 → R is homogeneous of degree k if g(tx) = tkg(x), t > 0.Let a, b, c be homogeneous fun tions of degree k. Compute primitives of thoseof the following forms that are losed:a) ω = adx+ bdy + cdz.b) ω = ady ∧ dz + bdz ∧ dx+ cdx ∧ dy.

4.1. CLOSED AND EXACT FORMS 271Solution:a) ω is losed idω = ∂ya dy∧dx+∂za dz∧dx+∂xb dx∧dy+∂zb dz∧dy+∂xc dx∧dz+∂yc dy∧dz =

= (∂yc− ∂zb)dy ∧ dz + (∂za− ∂xc)dz ∧ dx+ (∂xb− ∂ya)dx ∧ dy = 0In that ase we an ompute a primitive:Iω = (

∫ 1

0

a(tx)dt)x+ (

∫ 1

0

b(tx)dt)y + (

∫ 1

0

c(tx)dt)z =

= (

∫ 1

0

tka(x)dt)x+ (

∫ 1

0

tkb(x)dt)y + (

∫ 1

0

tkc(x)dt)z =

=1

k + 1(a(x)x+ b(x)y + c(x)z)b) ω is losed i

dω = ∂xa dx ∧ dy ∧ dz + ∂yb dy ∧ dz ∧ dx+ ∂zc dz ∧ dx ∧ dy =

= (∂xa+ ∂yb+ ∂zc)dx ∧ dy ∧ dz = 0In that ase we an ompute a primitive:Iω = (

∫ 1

0

ta(tx)dt)(ydz − zdy) + (

∫ 1

0

tb(tx)dt)(zdx − xdz) +

+ (

∫ 1

0

tc(tx)dt)(xdy − ydx) =

= (

∫ 1

0

tk+1a(x)dt)(ydz − zdy) + (

∫ 1

0

tk+1b(x)dt)(zdx− xdz) +

+ (

∫ 1

0

tk+1c(x)dt)(xdy − ydx) =

=1

k + 2((b(x)z − c(x)y)dx+ (c(x)x− a(x)z)dy + (a(x)y − b(x)x)dz)The ondition for the forms to be losed an be expressed in terms of theve tor eld F = (a, b, c); in the rst ase the ondition is rotF = 0 and inthe se ond ase it is divF = 0. In a), if the ve tor eld F has rotF = 0 wehave found a poten ial and in b) if divF = 0 we have found a ve tor eld Gsu h that rotG = F, a ve tor potential of F.

272 CHAPTER 4. STOKES THEOREMProblem 151: Classi operators vs Cartan derivative.a) Using exterior derivatives show thati) rot (grad f) = 0ii) div (rot F) = 0b) What about the questions suggested by point a)i) If rot F = 0, is F = grad f for some f ?ii) If div F = 0, is F = rot G for some ve tor eld G ?Solution:a) Reminding the fundamental diagram0 → F(U)

grad→ V(U)rot→ V(U)

div→ F(U) → 0Id ↓ ↓ 2 ↓ 3 ↓0 → Ω0(U)

d→ Ω1(U)d→ Ω2(U)

d→ Ω3(U) → 0both results are an immediate onsequen e of the fa t that d2 = 0; itsu es to go down to the baseline through the orresponding isomor-phisms.b) i) rot F = 0 is equivalent to d(F) = 0, that is to say that F is a losed 1-form. If we are in a set ontra tile to a point we knowthat F is exa t and there is a 0-form V ( omputable throughPoin aré formula if we are in a star shaped open set) su h thatdV = F. In the language of ve tor elds grad V = F: the eldhas a potential.ii) div F = 0 is equivalent to d(2F) = 0, that is to say that 2F isa losed 2-form. If we are in a ontra tile to a point set we knowthat 2F is exa t and there is a 1-form η su h that dη = 2F. Inthe language of ve tor elds if G = ♯η we have rot G = F: G isa ve tor potential of F.

4.1. CLOSED AND EXACT FORMS 273Problem 152:a) Let c(s, t) = (x(s, t), y(s, t), z(s, t)) be a parametri surfa e and F(x, y, z) =(X(x, y, z), Y (x, y, z), Z(x, y, z)) a ve tor eld in R3. Show that c∗(F)is losed i rotF is tangent to the surfa e.b) If c(s, t) = (s, t, (1 − s)(1 − t)) and F(x, y, z) = (yz, zx, x + y), showthat c∗(bF) is losed and ompute a primitive.Solution:a) Let ω = F and let's ompute d(c∗ω) = c∗(dω):

ω = Xdx+ Y dy + Zdz

dω = (rotF)1dy ∧ dz + (rotF)2dz ∧ dx+ (rotF)3dx ∧ dy

c∗(dω) = ((rotF)1∂(y, z)

∂(s, t)+ (rotF)2

∂(z, x)

∂(s, t)+ (rotF)3

∂(x, y)

∂(s, t))ds ∧ dtThen

c∗(dω) = 0 ⇔ (rotF)1∂(y, z)

∂(s, t)+ (rotF)2

∂(z, x)

∂(s, t)+ (rotF)3

∂(x, y)

∂(s, t)=

= rotF · n = 0

n being the normal ve tor asso iated to c.b) We omputerotF = det

i j k

∂x ∂y ∂zyz zx x+ y

= (1− x, y − 1, 0)and as the tangent plane is 〈(1, 0, t− 1), (0, 1, s− 1)〉 to de ide if rot Fis tangent to the surfa e it su es to evaluate the determinantdet

1 0 1− s0 1 t− 1

t− 1 s− 1 0

= −(1− s)(t− 1)− (s− 1)(t− 1) = 0Then from a) we know that c∗(bF) is losed; we ompute a primitivebF = yzdx+ zxdy + (x+ y)dz

274 CHAPTER 4. STOKES THEOREMc∗(bF) = t(1−s)(1−t)ds+s(1−s)(1−t)dt+(s+t)((t−1)ds+(s−1)dt) =

= s(t2 − 1)ds+ t(s2 − 1)dtand by Poin aré formulaI(bF) = (

∫ 1

0

λs((λt)2 − 1)dλ)s+ (

∫ 1

0

λt((λs)2 − 1)dλ)t

= (

∫ 1

0

(λ3st2 − λs)dλ)s+ (

∫ 1

0

(λ3ts2 − λt)dλ)t =

= (st2

4− s

2)s+ (

ts2

4− t

2)t =

s2t2 − s2 − t24

Problem 153:Let ω =∑3

i,j=1 aijxidxj ∈ Ω1(R3) and A = (aij), i, j = 1, 2, 3 a real number(3, 3) matrix.a) Show that ω is losed i A is symmetri al.b) Assuming that ω is losed, ompute a primitive that has value 1 at theorigin .Solution:a)

dω = d(3∑

i,j=1

aijxidxj) =

= (a12 − a21)dx1 ∧ dx2 + (a23 − a32)dx2 ∧ dx3 + (a31 − a13)dx3 ∧ dx3and we see that ω is losed i A is symmetri al.b) We use Poin aré formula to obtain(Iω)(x) =

3∑

i,j=1

I(aijxidxj)(x) =

3∑

i,j=1

aij(

∫ 1

0

txidt)xj =

=

3∑

i,j=1

1

2aijxixj =

∑

1≤i≤j≤3

aijxixj ,

4.1. CLOSED AND EXACT FORMS 275a primitive having value 0 at the origin; thenV =

∑

1≤i≤j≤3

aijxixj + 1has the required value 1.Problem 154:If U ⊂ Rn is a star shaped open set then any losed form has a primitive,but this su ient ondition is not ne essary as the following example shows.Let A = [−1, 1]×(−∞,−1]∪ [1,+∞) ⊂ R2 and U = R2 \A; prove that

U is not star shaped respe t to any of its points but show that any losedform in U is exa t.Solution:The gure shows that U is not star shaped with respe t to any of its points:if p is any point in U its easily seen that there are rays emanating from pthat don't a eed to all of the points in U . In a more pi turesque form we ansay that, viewing the rays as light rays, the set A always produ es shadows.x

y

U

A

A

.

a) Assume ω = Pdx + Qdy is a losed 1-form; we onstru t a primitive,that is a fun tion su h that its dierential is ω. We may apply Poin aré

276 CHAPTER 4. STOKES THEOREMformula to onstru t a primitive ϕ0 of ω in the star shaped set (respe tto any of its points)U0 = (x, y) : 1 < x

U0

y

A

A

x

b) Analogously we an onstru t a primitive ϕ1 of ω in the star shapedset respe t to 0

U1 = (−a, a)× (−1, 1), a > 1

x

y

A

A

.U1

In U0∩U1 we have d(ϕ1−ϕ0) = ω−ω = 0 and as U0∩U1 is onne tedϕ1 − ϕ0 = c. Redening if ne essary ϕ1 as ϕ1 − c both primitives oin ide in U0 ∩U1 and in this way we have a primitive V01 in U0 ∪U1. ) Using again Poin aré formula we an onstru t a primitive ϕ2 of ω inthe star shaped set (respe t to any of its points)

U2 = (x, y) : x < −1

4.2. STOKES THEOREM 277x

y

A

A

.U2

In U2∩ (U0∪U1) we have d(ϕ2−V01) = ω−ω = 0 and as we have donein b) we an redene V01 if ne essary to make it oin ide with ϕ2. Inthis way we have a primitive of ω.Assume now that ω = Rdx ∧ dy (a losed form for any R); pro eeding as ina) we an onstru t a primitive η0 of ω in U0 and a primitive η1 in U1. Thenin U0 ∩ U1we have d(η1 − η0) = ω − ω = 0; the form η1 − η0 is losed andthe result in a) asserts the existen e of a fun tion ϕ su h that η1 − η0 = dϕ.Redening if ne essary η1 as η1 − dϕ we obtain a primitive of ω in U0 ∪ U1.Carrying out the same tri k on U2 we arrive at a primitive η of ω in the wholeof U .4.2 Stokes theorem4.2.1 Chains

T We follow here [Burg p.99 ; the ube Qk = [0, 1]k ⊂ Rk is alled thek-dimensional basi ube or the basi k- ube. We may look at it as adierentiable fun tion

I : Qk → Rk

(s1, . . . , sk) 7→ (s1, . . . , sk)

278 CHAPTER 4. STOKES THEOREMThe i-th fa es of Qk are the (k − 1)- ubesI(i,0) : Qk−1 → Rk

(s1, . . . , si, . . . sk) 7→ (s1, . . . ,i)

0, . . . , sk)

I(i,1) : Qk−1 → Rk

(s1, . . . , si, . . . sk) 7→ (s1, . . . ,i)

1, . . . , sk)and the boundary of the basi k- ube is the formal sum of (k − 1)- ubes∂I =

k∑

i=1

∑

α=0,1

(−1)i+αI(i,α)In the following gure we see the boundary of Q2:c(1,1)

c(2,0)

c(1,0)

−

c(2,1)

2

−

Q

A k- ube in the open set U ⊂ Rn is a dierentiable mapc : Qk → U

(s1, . . . , sk) 7→ c(s1, . . . , sk)

Q2

c

and its fa es are the (k − 1)- ubes c(i,0) = c I(i,0) and c(i,1) = c I(i,1),expli itly:

4.2. STOKES THEOREM 279c(i,0) : Qk−1 → U

(s1, . . . , sk−1) 7→ c(s1, . . . ,i)

0, . . . , sk)

c(i,1) : Qk−1 → U

(s1, . . . , sk−1) 7→ c(s1, . . . ,i)

1, . . . , sk)and the boundary of c is the formal sum of (k − 1)- ubes∂c =

k∑

i=1

∑

α=0,1

(−1)i+αc I(i,α)A k- hain is a formal sum of k- ubesc = n1c1 + · · ·+ niciwhere nj ∈ Z. One an addition, subtra t and multiply by integers the

k- hains. The boundary of a hain is the formal sum of boundaries ∂c =n1∂c1 + · · ·+ ni∂ci and a fundamental property of ∂ is that ∂2 = 0.

Problem 155:a) Write expli itly the fa es and the boundary of a 3- ube c : [0, 1]3 → Rnb) Compute the fa es of the 2- ube c : [0, 1]× [0, 1]→ R2 wherec(r, θ) = (r cos 2πθ, r sin 2πθ)This 2- ube parametrizes the unit dis . ) Compute the fa es and the boundary of the 3- ube c : [0, 1]× [0, 1] ×

[0, 1]→ R3 wherec(r, ϕ, θ) = ((a+rb cos 2πϕ) cos 2πθ, (a+rb cos 2πϕ) sin 2πθ, rb sin 2πϕ)This 3- ube parametrizes the solid torus.d) Compute the boundary of the boundary of c and he k that it vanishes.

280 CHAPTER 4. STOKES THEOREMSolution:a) Let (s, t, u) be the variables; the fa es are:c(1,0)(t, u) = c(0, t, u)

c(1,1)(t, u) = c(1, t, u)

c(2,0)(s, u) = c(s, 0, u)

c(2,1)(s, u) = c(s, 1, u)

c(3,0)(s, t) = c(s, t, 0)

c(3,1)(s, t) = c(s, t, 1)and the boundary is∂c = −c(1,0) + c(1,1) + c(2,0) − c(2,1) − c(3,0) + c(3,1)b) The fa es are

c(1,0)(θ) = (0, 0)

c(1,1)(θ) = (cos 2πθ, sin 2πθ)

c(2,0)(r) = (r, 0)

c(2,1)(r) = (r, 0)and the boundary is∂c = c(1,1) − c(1,0)for the other two fa es being equal and appearing with ontrary signin ∂c are eliminated. Observe there an be some ollapsing, as in c(1,0). )

c(1,0)(ϕ, θ) = (a cos 2πθ, a sin 2πθ, 0)

c(1,1)(ϕ, θ) = ((a+ b cos 2πϕ) cos 2πθ, (a+ b cos 2πϕ) sin 2πθ, b sin 2πϕ)

c(2,0)(r, θ) = ((a+ rb) cos 2πθ, (a+ rb) sin 2πθ, 0)

c(2,1)(r, θ) = ((a+ rb) cos 2πθ, (a+ rb) sin 2πθ, 0)

c(3,0)(r, ϕ) = (a + rb cos 2πϕ, 0, rb sin 2πϕ)

c(3,1)(r, ϕ) = (a + rb cos 2πϕ, 0, rb sin 2πϕ)

4.2. STOKES THEOREM 281The boundary is∂c = −c(1,0) + c(1,1) + c(2,0) − c(2,1) − c(3,0) + c(3,1)

= c(1,1) − c(1,0),for c(2,0) and c(2,1) are the same 2- ube and they appear in the boundarywith opposite signs and the same happens with c(3,0) and c(3,1).d)(c(1,1))(1,0)(θ) = ((a+ b) cos 2πθ, (a+ b) sin 2πθ, 0)

(c(1,1))(1,1)(θ) = ((a+ b) cos 2πθ, (a+ b) sin 2πθ, 0)

(c(1,1))(2,0)(ϕ) = (a + b cos 2πϕ, 0, b sin 2πϕ)

(c(1,1))(2,1)(ϕ) = (a + b cos 2πϕ, 0, b sin 2πϕ)and we obtain∂(c(1,1)) = −(c(1,1))(1,0) + (c(1,1))(1,1) + (c(1,1))(2,0) − (c(1,1))(2,1) = 0Analogously one an see that ∂(c(1,0)) = 0. Finally ∂(∂c) = 0.

4.2.2 Integration on hainsT Let ω ∈ Ωk(Rk) and let (dt1, . . . , dtk) be the dual of the anoni al basis;then ω has an expression ω = f(t)dt1 ∧ · · · ∧ dtk where f : Rk → R. Dene

∫

Qk

ω :=

∫

[0,1]kf(t)dt1 . . . dtkIf ω ∈ Ωk(U) and c is a k- ube in U dene

∫

c

ω :=

∫

[0,1]kc∗(ω)and if c = n1c1 + · · ·+ nici is a hain in U then

∫

c

ω :=i∑

j=1

∫

cj

ω

282 CHAPTER 4. STOKES THEOREMProblem 156:Let ω = xydx + (x2 − y2)dy ∈ Ω1(R2) and c a 1- ube with the indi atedimage. In ea h ase ompute ∫cω.a) The unit ir umferen e.b) The ar of the parabola y2 = x between the points (1,−1) and (1, 1). ) The ar of the urve y2 = x3 between the points (0, 0) and (1, 1).d) The ellipse x2 + 2y2 = 4.Solution:Observe we are allowed to hoose the 1- ube that has the given image; forinstan e we an traverse the ir umferen e as many times as we want. Whatfollows is then one among many possible solutions.a) c(θ) = (cos θ, sin θ), 0 ≤ θ ≤ 2π is a 1- ube with image the unit ir um-feren e. Now:

c∗(ω) = cos θ sin θ(− sin θ)dθ + (cos 2θ − sin 2θ) cos θdθ =

= (−2 cos θ sin 2θ + cos 3θ)dθ∫

c

ω =

∫

[0,2π]

c∗(ω) =

∫

[0,2π]

(−2 cos θ sin 2θ + cos 3θ)dθ =

= 2

∫ 2π

0

− cos θ sin 2θdθ +

∫ 2π

0

(1− sin 2θ) cos θdθ =

= −2sin 3θ

3|2π0 + 2− sin 3θ

3|2π0 = 2b) c(t) = (t2, t), −1 ≤ t ≤ 1 is a 1- ube with image the stated ar of theparabola. Then:

c∗(ω) = t2t2tdt+ (t4 − t2)dt = (3t4 − t2)dt∫

c

ω =

∫

[−1,1]

c∗(ω) =

∫

[−1,1]

(3t4 − t2)dt =4

15

4.2. STOKES THEOREM 283 ) Observe rst that none point of the urve has x < 0. The equation de-nes impli itly two dierentiable fun tions y = x3/2 and y = −x3/2;to go from the origin to (1, 1) we hoose y = x3/2. Then c(t) =(t, t3/2), 0 ≤ t ≤ 1 is a 1- ube with image the stated ar and we have:

c∗(ω) = t5/2dt+ (t2 − t3)32t1/2dt = (

5

2t5/2 − 3

2t7/2)dt

∫

c

ω =

∫

[0,1]

c∗(ω) =

∫

[0,1]

(5

2t5/2 − 3

2t7/2)dt =

5

7− 1

3d) We know that c(θ) = (2 cos θ,√

2 sin θ), 0 ≤ θ ≤ 2π is a 1- ube withimage the ellipse traversed on e. Thenc∗(ω) = (−4

√2 cos θ sin 2θ + (4 cos 2θ − 2 sin 2θ)

√2 cos θ)dθ =

= (−6√

2 cos θ sin 2θ + 4√

2 cos 3θ)dθ∫

c

ω =

∫

[0,2π]

c∗(ω) =

∫

[0,2π]

(−6√

2 cos θ sin 2θ + 4√

2 cos 3θ)dθ = 8√

2

Problem 157:Compute ∫cω whena) ω = x2ydx+xdy and c is a 1- hain traversing the triangle with verti es

(0, 0), (1, 0), (0, 1).b) ω = yzdx+ zxdy + xydz and c(t) = (cos 2πt, sin 2πt, 2πat), 0 ≤ t ≤ 1. ) ω = ydx− xdy and c(t) = (t− sin t, 1− cos t), t ∈ [0, 2π].Solution:a) Letting c1(t) = (t, 0), c2(t) = (t, 1− t), c3(t) = (0, 1− t), 0 ≤ t ≤ 1 wehave three 1- ubes and the hainc = c1(t) + c2(t) + c3(t)

284 CHAPTER 4. STOKES THEOREMtraverses the whole triangle. Thenc∗1(ω) = 0, c∗2(ω) = (t2(1− t)− t)dt, c∗3(ω) = 0

∫

c

ω =

∫

c1

ω +

∫

c2

ω +

∫

c3

ω =

∫

[0,1]

c∗2(ω) =

=

∫

[0,1]

(t2(1− t)− t)dt =1

3− 1

4− 1

2= − 5

12b) c(t) = (cos 2πt, sin 2πt, 2πat), 0 ≤ t ≤ 1

c∗(ω) = (2πat sin 2πt(−2π sin 2πt)+2πat cos 2πt(2π cos 2πt)+cos 2πt sin 2πt·2πa)dt =

= (4π2at(− sin 22πt+ cos 22πat) + 2πa sin 4πt)dt =

= (4π2at(cos 4πt− 1

2+

cos 4πt+ 1

2) + 2πa sin 4πt)dt =

= (4π2at cos 4πt+ 2πa sin 4πt)dt∫

c

ω =

∫

[0,1]

c∗(ω) =

∫

[0,1]

(4π2at cos 4πt+ 2πa sin 4πt)dt = 0Alternatively we may observe that f(x, y, z) = xyz is a primitive of ωand use Stokes theorem (see p.289):∫

c

ω =

∫

c

df =

∫

∂c

f = f(c(1))− f(c(0)) =

= f(1, 0, 2πa)− f(1, 0, 0) = 0 ) c(t) = (t− sin t, 1− cos t), t ∈ [0, 2π]

c∗(ω) = ((1− cos t)2 − (t− sin t) sin t)dt =

= (1 + cos 2t− 2 cos t− (t− sin t) sin t)dt = (2− 2 cos t− t sin t)dt∫

c

ω =

∫

[0,2π]

c∗(ω) =

∫

[0,2π]

(2− 2 cos t− t sin t)dt = 6π

4.2. STOKES THEOREM 285Problem 158:Let ω = x2dy ∧ dz + y2dz ∧ dx+ z2dx ∧ dy; integrate ω on the 2- ubec(ϕ, θ) = (sinϕ cos θ, sinϕ sin θ, cosϕ), 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2πSolution:

c∗(ω) = (sin 2ϕ cos 2θ∂(y, z)

∂(ϕ, θ)+ sin 2ϕ sin 2θ

∂(z, x)

∂(ϕ, θ)+ cos 2ϕ

∂(x, y)

∂(ϕ, θ))dϕ ∧ dθ =

= (sin4 ϕ cos3 θ + sin4 ϕ sin3 θ + cos3 ϕ sinϕ)dϕ ∧ dθ∫

c

ω =

∫

[0,π]×[0,2π]

c∗(ω) =

=

∫ π

0

∫ 2π

0

(sin4 ϕ cos3 θ + sin4 ϕ sin3 θ + cos3 ϕ sinϕ)dϕdθNow ∫ 2π

0

cos3 θdθ =

∫ 2π

0

sin3 θdθ = 0and ∫ π

0

cos3 ϕ sinϕdϕ = −cos4 ϕ

4|π0= 0Finally ∫

c

ω = 0

Problem 159:Let D = (x, y, z, w) : x2 + y2 = 1, z2 + w2 = 1.a) View D as a 2- ube c of R4.b) Compute I =∫cdz ∧ dw + xzdy ∧ dw.

286 CHAPTER 4. STOKES THEOREMSolution:a) For instan e we may hoosec(s, t) = (cos s, sin s, cos t, sin t), (s, t) ∈ [0, 2π]2b) We have to omputec∗(dz ∧ dw) = − sin t dt ∧ cos t dt = 0

c∗(xzdy ∧ dw) = cos s cos t cos s ds ∧ cos t dt =

= cos2 s cos2 t ds ∧ dtandI =

∫ 2π

0

∫ 2π

0

cos2 s cos2 t dsdt =

= (

∫ 2π

0

cos2 u du)2 = π2

Problem 160:Let f : [0, 1] → Rn the fun tion f(t) = (t, t2, . . . , tn) and ω = x1dx1 + · · ·+xndxn; ompute

I =

∫

[0,1]

f ∗(ω)Solution:f ∗(ω) = f ∗(x1dx1 + . . . ) = (x1 f)d(x1 f) + · · · =

= td(t) + t2d(t2) + · · · = tdt+ 2t3dt+ · · · == (t+ 2t3 + 3t5 + . . . )dtand∫ 1

0

f ∗(ω) =

∫ 1

0

(t+ 2t3 + 3t5 + . . . )dt

=1

2+

2

4+

3

6+ · · ·+ n

2n=n

2

4.2. STOKES THEOREM 287Problem 161:Let OMP a triangle, O being the origin and P the orthogonal proje tion ofM on the plane z = 0. Compute the volume of the region V swept by thetriangle when M traverses the ar of the urve

γ(t) = (2 cos2 t, 2 sin t cos t, 2 sin t cos2 t), t ∈ [0,π

2]Solution:A gure:

x

y

z

P

M

O

We rst parametrize the segment OM for ea h xed M :Γ(s) = (2s cos2 t, 2s sin t cos t, 2s sin t cos2 t), s ∈ [0, 1]and for ea h point of Γ(s) we parametrize the verti al segment between Γ(s)and its proje tion on z = 0. Letting then M vary along the urve we haveas a parametrization of the volume

c(t, s, u) = (2s cos2 t, 2s sin t cos t, 2su sin t cos2 t), (t, s, u) ∈ [0,π

2]× [0, 1]2This parametrization is obviously inje tive and the volume will beVol (V ) =

∫ ∫ ∫

c

dx ∧ dy ∧ dz =

∫ π/2

0

∫ 1

0

∫ 1

0

det c′(t, s, u)dtdsdu

288 CHAPTER 4. STOKES THEOREMWe havedet c′ = det

−4s sin t cos t 2 cos2 t 0

2s(cos2 t− sin2 t) 2 sin t cos t 0. . . . . . 2s sin t cos2 t

=

= −8s2 sin t cos4 tand Vol (V ) =

∫ π/2

0

∫ 1

0

∫ 1

0

−8s2 sin t cos4 tdtdsdu =

= −8

3

∫ π/2

0

sin t cos4 tdt =8

3

1

5cos5 t |t=π/2t=0 = − 8

15The minus sign is due to an inversion of the orientation when doing theparametrization; it is lear that dening c(s, t, u) as before we shall obtainVol (V ) = 815

Problem 162:We have already used intervals of parameters other than the [0, 1]k; is thatright?a) For 1- ubes we an revert to the interval [0, 1] by means of a linearmap:h : [0, 1] → [a, b]

τ 7→ t = a+ τ(b− a)If c1 = c h the point is whether ∫cω =

∫c1ω.b) For 2- ubes we have

h : [0, 1]2 → [a, b]× [a′, b′](σ, τ) 7→ (s, t) = (a+ σ(b− a), a′ + τ(b′ − a′))and if c1 = c h we ask again whether ∫

cω =

∫c1ω. ) Generaly speaking if c : [a1, b1]× · · · × [ak, bk]→ Rn is a map (that weview as a k- ube) and h a dieomorphism:

h : [0, 1]k → [a1, b1]× · · · × [ak, bk]then if c1 = c h is it true that ∫cω =

∫c1ω?

4.2. STOKES THEOREM 289Solution:a) If c∗(ω) = f(t)dt we have (c1)∗(ω) = (h∗ c∗)(ω) = h∗(f(t)dt) =

f(h(τ))h′(τ)dτ .∫

c

ω =

∫

[a,b]

c∗(ω) =

∫

[a,b]

f(t)dt =

∫ b

a

f(t)dt

∫

c1

ω =

∫

[0,1]

c∗1(ω) =

∫

[0,1]

f(h(τ))h′(τ)dτ =

∫ 1

0

f(h(τ))h′(τ)dτIf h(0) = a, h(1) = b, as it happens with the given h, the hange ofvariables theorem for one variable integrals tells us that both integrals oin ide and then ∫cω =

∫c1ω .b) Let c∗(ω) = f(s, t)ds ∧ dt ; then (c1)

∗(ω) = h∗(f(s, t)ds ∧ dt) =

(c1)∗(ω) = h∗(f(s, t)ds ∧ dt) = f(s(σ, τ), t(σ, τ)) det h′(σ, τ)The hange of variables theorem gives

∫

c

ω =

∫ b

a

∫ b′

a′f(s, t)dsdt =

∫ 1

0

∫ 1

0

f(s(σ, τ), t(σ, τ)) | det h′(σ, τ) |and in our ase det h′(σ, τ) = (b− a)(b′ − a′) > 0 and the last integralis ∫c1ω. ) In fa t in a) and b) we have used only that h([a, b]) = [0, 1] or that

h([0, 1]× [0, 1]) = [a, b]× [a′, b′] together with the fa t that det h′ > 0.Now the same arguments show that if det h′(ξ) > 0 both integrals oin ide.4.2.3 Stokes theorem

T Stokes theoremIf ω ∈ Ωk−1(U) and c is a k- hain in U then∫

c

dω =

∫

∂c

ω

290 CHAPTER 4. STOKES THEOREMProblem 163:Che k Stokes theorem when ω = −yexdx+ xeydy and c is the basi 2- ube.Solution:• c(s, t) = (s, t), (s, t) ∈ [0, 1]× [0, 1] is the basi 2- ube and its boundaryis

c(1,0)(t) = (0, t), c(1,1)(t) = (1, t), c(2,0)(s) = (s, 0), c(2,1)(s) = (s, 1)

∂c = −c(1,0) + c(1,1) + c(2,0) − c(2,1)

• We ompute the integral of ω on ∂c :c∗(1,0)(ω) = 0, c∗(1,1)(ω) = etdt, c∗(2,0)(ω) = 0, c∗(2,1)(ω) = −esds

∫

∂c

ω =

∫ 1

0

etdt−∫ 1

0

−esds = 2(e− 1)

• On another handdω = −exdy ∧ dx+ eydx ∧ dy = (ey + ex)dx ∧ dy

c∗(dω) = (et + es)ds ∧ dt∫

c

dω =

∫

[0,1]2(et + es)ds ∧ dt =

∫ 1

0

∫ 1

0

(et + es)dsdt = 2(e− 1)and we have he ked Stokes theorem.Problem 164:Che k Stokes theorem if ω = xdx+ ydy + zdz and c is the 2- ube

c(s, t) = (s cos 2πt, s sin 2πt, 2πt), (s, t) ∈ [0, 1]2

4.2. STOKES THEOREM 291Solution:• The fa es of c are

c(1,0)(t) = (0, 0, 2πt)c(1,1)(t) = (cos 2πt, sin 2πt, 2πt)c(2,0)(s) = (s, 0, 0)c(2,1)(s) = (s, 0, 2π)The boundary is∂c = −c(1,0) + c(1,1) + c(2,0) − c(2,1)

• We ompute the integral of ω on ∂c:a)c∗(1,0)(ω) = 4π2tdt

∫c∗(1,0)ω =

∫ 1

0

4π2tdt = 2π2b)c∗(1,1)(ω) = (cos 2πt(−2π sin 2πt+sin 2πt 2π cos 2πt+2πt2π)dt = 4π2tdt

∫c∗(1,1)ω = 2π2 )

c∗(2,0)(ω) = sds

∫c∗(2,0)ω =

∫ 1

0

sds =1

2d)c∗(2,1)(ω) = sds∫c∗(2,1)ω =

1

2and we obtain ∫∂cω = 0.

292 CHAPTER 4. STOKES THEOREM• On another hand

dω = 0∫

c

dω = 0We have he ked Stokes theorem.Problem 165:Constru t a 2- hain of R3 with image the pie e of the surfa e of the paraboloid

2z = x2 + y2 limited by the plane z = 2. Let F(x, y, z) = (3y,−xz, yz2); he k Stokes theorem for F.Solution:On one hand we have the 1-form F = 3ydx − xzdy + yz2dz; on the otherhand we must onstru t the 2- ube stated. The interse tion of the paraboloidwith the plane z = 2 is the ir umferen e x2 + y2 = 4, z = 2. To parametrizethe surfa e we do it for the dis x2 + y2 = 4, z = 0 and then limb to thesurfa e:c(u, v) = (2u cos 2πv, 2u sin 2πv, 2u2), (u, v) ∈ [0, 1]2

• The fa es of the boundary arec(1,0)(v) = (0, 0, 0)

c(1,1)(v) = (2 cos 2πv, 2 sin 2πv, 2)

c(2,0)(u) = (2u, 0, 2u2)

c(2,1)(u) = (2u, 0, 2u2)The boundary is∂c = c(1,1) − c(1,0)and it su es to integrate on c(1,1).

4.2. STOKES THEOREM 293• The integral of F is:

∫

∂c

F =

∫

c(1,1)

F =

∫

[0,1]

c∗(1,1)(F)But c∗(1,1)(F) = (−24π sin2 2πv − 16π cos2 2πv)dv and it is easily seenthat ∫∂cF = −20π.

• To know ∫cd(F) we have

d(F) = 3dy∧dx−zdx∧dy−xdz∧dy+z2dy∧dz = (z2+x)dy∧dz−(z+3)dx∧dy

c∗(F) = (4u4 + 2u cos 2πv) c∗(dy ∧ dz)− (2u2 + 3)c∗(dx ∧ dy)Now we ompute the c∗:c∗(dy∧dz) = det

(2 sin 2πv 4πu cos 2πv

4u 0

)du∧dv = (−16πu2 cos 2πv)du∧dv

c∗(dx ∧ dy) = det

(2 cos 2πv −4πu sin 2πv2 sin 2πv 4πu cos 2πv

)du ∧ dv = 8πu du ∧ dvand

∫

c

d(F) =

∫ 1

0

∫ 1

0

((4u4+2u cos 2πv)(−16πu2 cos 2πv)−(2u2+3)8πu)dudvThis gives −20π and we have he ked Stokes theorem.Problem 166:Consider the 3- ube c(r, ϕ, θ) = (r sinϕ cos θ, r sinϕ sin θ, r cosϕ), 0 ≤ r ≤

1, 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π and the 2-form ω = x2dy∧dz+y2dz∧dx+z2dx∧dy.Che k Stokes theorem.

294 CHAPTER 4. STOKES THEOREMSolution:• We ompute the boundary of c; rst the fa es:

c(1,0)(ϕ, θ) = c(0, ϕ, θ) = (0, 0, 0)c(1,1)(ϕ, θ) = (sinϕ cos θ, sinϕ sin θ, cosϕ)c(2,0)(r, θ) = c(r, 0, θ) = (0, 0, r)c(2,1)(r, θ) = c(r, π, θ) = (0, 0,−r)c(3,0)(r, ϕ) = c(r, ϕ, 0) = (r sinϕ, 0, r cosϕ)c(3,1)(r, ϕ) = c(r, ϕ, 2π) = (r sinϕ, 0, r cosϕ)The boundary is:

∂c = −c(1,0) + c(1,1) + c(2,0) − c(2,1) − c(3,0) + c(3,1)and as c(3,0) = c(3,1) the last two terms an el:∂c = −c(1,0) + c(1,1) + c(2,0) − c(2,1)Those remaining are: the enter of the sphere, the surfa e of the sphere,the radius to the north pole and the radius to the south pole. We write

∂c = −(origen)+(S2) + (radi N)−(radi S)• To integrate ω on ∂c, ompute (∂c)∗(ω):

c∗(1,0)(ω) = ω(0, 0, 0) = 0dy ∧ dz + 0dz ∧ dx+ 0dx ∧ dy = 0

c∗(1,1)(ω) = (sinϕ cos θ)2c∗(1,1)(dy ∧ dz) + (sinϕ sin θ)2c∗(1,1)(dz ∧ dx) +

+ cos 2ϕc∗(1,1)(dx ∧ dy)Now we ompute following the method; write the matrix (c(1,1))′:

(c(1,1))′ =

cosϕ cos θ − sinϕ sin θcosϕ sin θ sinϕ cos θ− sinϕ 0

c∗(1,1)(dy ∧ dz) = sin 2ϕ cos θdϕ ∧ dθc∗(1,1)(dz ∧ dx) = sin 2ϕ sin θdϕ ∧ dθ

4.2. STOKES THEOREM 295c∗(1,1)(dx ∧ dy) = sinϕ cosϕdϕ ∧ dθSo:

c∗(1,1)(ω) = ((sinϕ cos θ)2 sin 2ϕ cos θ + (sinϕ sin θ)2 sin 2ϕ sin θ +

+ cos 2ϕ sinϕ cosϕ)dϕ ∧ dθ =

= (sin4 ϕ cos3 θ + sin4 ϕ sin3 θ + cos3 ϕ sinϕ)dϕ ∧ dθ =

= (sin4 ϕ(cos3 θ + sin3 θ) + cos3 ϕ sinϕ)dϕ ∧ dθHappily the other fa es are easier; we suspe t that when integratinga 2-form on a degenerate 2- ube we shall obtain 0, whi h we showanaliti ally:c∗(2,0)(ω) = r2c∗(2,0)(dx) ∧ c∗(2,0)(dy) = 0andc∗(2,1)(ω) = r2c∗(2,1)(dx) ∧ c∗(2,1)(dy) = 0

• We integrate∫

∂c

ω =

∫

c(1,1)

ω =

∫

[0,π]×[0,2π]

(sin4 ϕ(cos3 θ+sin3 θ)+cos3 ϕ sinϕ)dϕ∧dθ =

=

∫ π

0

∫ 2π

0

(sin4 ϕ(cos3 θ + sin3 θ) + cos3 ϕ sinϕ)dϕdθ =

=

∫ 2π

0

∫ π

0

(sin4 ϕ(cos3 θ + sin3 θ))dϕdθ =

=3

8π

∫ 2π

0

(cos3 θ + sin3 θ)dϕdθ = 0

• Its easy to see that dω = 2(x+ y + z)dx ∧ dy ∧ dz and we obtainc∗(dω) = 2(r sinϕ cos θ + r sinϕ sin θ + r cosϕ)(det c′)dr ∧ dϕ ∧ dθ =

= (2r(sinϕ(cos θ + sin θ) + cosϕ)r2 sinϕ)dr ∧ dϕ ∧ dθ∫

c

dω =

∫

[0,1]×[0,π]×[0,2π]

c∗(dω) =

=

∫ 1

0

∫ π

0

∫ 2π

0

(2r3 sinϕ(sinϕ(cos θ + sin θ) + cosϕ))drdϕdθ =

296 CHAPTER 4. STOKES THEOREM=

1

2

∫ 2π

0

∫ π

0

(sin 2ϕ(cos θ + sin θ) + sinϕ cosϕ))dϕdθ =

=1

2

∫ 2π

0

π

2(cos θ + sin θ)dθ = 0

167:The 3- ubec(r, θ, ϕ) = ((a+ rb cosϕ) cos θ, (a+ rb cosϕ) sin θ, rb sinϕ),

(r, θ, ϕ) ∈ D = [0, 1]× [0, 2π]2parametrizes a solid torus. Reminding the volume of a torus he k Stokestheorem for the 2-form ω = zdx ∧ dy.Solution:We ompute the boundary of cc(1,0)(θ, ϕ) = (a cos θ, a sin θ, 0)

c(1,1)(θ, ϕ) = ((a+ b cosϕ) cos θ, (a+ b cosϕ) sin θ, b sinϕ))

c(2,0)(r, ϕ) = (a+ rb cosϕ, 0, rb sinϕ)

c(2,1)(r, ϕ) = (a+ rb cosϕ, 0, rb sinϕ)

c(3,0)(r, θ) = ((a + rb) cos θ, (a+ rb) sin θ, 0)

c(3,1)(r, θ) = ((a + rb) cos θ, (a+ rb) sin θ, 0)and the boundary is∂c = c(1,1) − c(1,0).We al ulate ∫

∂cω:

c∗(1,1)(ω) = b sinϕ c∗(1,1)(dx ∧ dy) = (a+ b cosϕ)b2 sin2 ϕdθ ∧ dϕ

c∗(1,1)(ω) = 0

4.2. STOKES THEOREM 297∫

∂c

ω =

∫ 2π

0

∫ 2π

0

(a + b cosϕ)b2 sin2 ϕdθdϕ = πab22πOn another hand∫

c

dω =

∫

c

dx ∧ dy ∧ dz =

∫

[0,1]×[0,2π]2(det c′)drdθdϕand due to the hange of variables theorem, taking into a ount that det c′ >

0, the value of the last integral is∫

c(D)

1dxdydz = Vol (c(D)) = 2π2ab2Stokes theorem is he ked..Problem 168:Let

ω =n∑

i=1

(−1)i−1 xirαdx1 ∧ · · · ∧

∧dxi ∧ · · · ∧ dxn ∈ Ωn−1(Rn − 0)a) Cal ulate α so as ω is losed.b) For n = 3 and for the α in a), al ulate ∫

SRω. ) Dedu e from b) that ω is not exa t.Solution:a)

dω =

n∑

i=1

(−1)i−1 rα − αrα−1 x

2i

r

r2αdxi ∧ dx1 ∧ · · · ∧

∧dxi ∧ · · · ∧ dxn =

= (

n∑

i=1

1− αr−2x2i

rα)dx1 ∧ · · · ∧ dxn =

n− αrα

dx1 ∧ · · · ∧ dxniand dω = 0 i α = n.

298 CHAPTER 4. STOKES THEOREMb) Consider ω = 1r3

(xdy ∧ dz + ydz ∧ dx+ zdx ∧ dy) and the 2- ubec(ϕ, θ) = (R sinϕ cos θ, R sinϕ sin θ, R cosϕ), ϕ ∈ [0, π], θ ∈ [0, 2π]

c∗(dy ∧ dz) = det

(R cosϕ sin θ R sinϕ cos θ−R sinϕ 0

)dϕ ∧ dθ =

= R2 sin2 ϕ cos θ dϕ ∧ dθ

c∗(dz ∧ dx) = det

(−R sinϕ 0

R cosϕ cos θ −R sinϕ sin θ

)dϕ ∧ dθ =

= R2 sin2 ϕ sin θ dϕ ∧ dθ

c∗(dx ∧ dy) = det

(R cosϕ cos θ −R sinϕ sin θR cosϕ sin θ R sinϕ cos θ

)dϕ ∧ dθ =

= R2 sinϕ cosϕdϕ ∧ dθNowc∗(ω) =

1

R3(R sinϕ cos θ R2 sin2 ϕ cos θ +R sinϕ sin θ R2 sin2 ϕ sin θ +

+ R cosϕR2 sinϕ cosϕ)dϕ ∧ dθ =

= (sin3 ϕ(cos2 θ + sin2 ϕ) + sinϕ cos2 ϕ)dϕ ∧ dθ =

= (sin3 ϕ+ sinϕ cos2 ϕ)dϕ ∧ dθ = sinϕdϕ ∧ dθFinally ∫

c

ω =

∫ π

0

∫ 2π

0

sinϕdϕdθ = 4πThe result was predi table if we remind that we are omputing the uxof the eld rr3. ) If ω was exa t there would be a 1-form η dened in U = Rn−0 su hthat dη = ω and by Stokes theorem

∫

c

ω =

∫

c

dη =

∫

∂c

ηIt's easy to see that ∂c = 0 and the we have ∫cω = 0, a ontradi tion.

4.2. STOKES THEOREM 299Problem 169 :Let D = (x, y, z, w) : x2 + y2 + z2 ≤ w2, 0 ≤ w ≤ 1.a) Parametrize D thus viewing it as a 3- ube c.b) Compute ∫∂c

(y + w)dx ∧ dy ∧ dz. ) Compute∫∂c

(x2 + y2 + z2 + w2)dx ∧ dy ∧ dz.Solution:R2

R3

4R

a) To see how to parametrize D we an start lowering the dimension:C = (x, y, z) : x2 + y2 ≤ z2, 0 ≤ z ≤ 1,a solid one of height 1. The points of C at a height z0 are in the dis of radius z0

(x, y, z0) : x2 + y2 ≤ z02,Parametrizing ea h dis we obtain a parametrization of C:

c(r, θ, z) = (zr cos θ, zr sin θ, z) 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 1.By analogy we see that D is a (hiper) one that has as se tion at a(hiper)height w a ball of radius w. Parametrizing ea h ball we obtaina parametrization of D:c(r, ϕ, θ, w) = (wr sinϕ cos θ, wr sinϕ sin θ, wr cosϕ,w)

(r, ϕ, θ, w) ∈ R = [0, 1]× [0, π]× [0, 2π]× [0, 1]

300 CHAPTER 4. STOKES THEOREMb) Using Stokes theorem we have:∫

∂c

(y+w)dx∧dy∧dz =

∫

c

dw∧dx∧dy∧dz = −∫

c

dx∧dy∧dz∧dw = −Vol(D)Let's rst do a onje ture starting from low dimensions:• R2 : Area= 1

2Lenght(basis)×height.

• R3 : Volume= 13Area(basis)×height.

• R4 : (hiper)Volume= 14Volume(basis)×height= 1

443π = π

3?To ompute the volume we need the ja obian matrix of c:

c′ =

w sinϕ cos θ wr cosϕ cos θ −wr sinϕ sin θ r sinϕ cos θw sinϕ sin θ wr cosϕ sin θ wr sinϕ cos θ r sinϕ sin θw cosϕ −wr sinϕ 0 r cosϕ

0 0 0 1

det c′ = w cosϕ(w2r2 sinϕ cosϕ) + wr sinϕ(w2r sin 2ϕ) =

= w3r2 sinϕ∫

c

dx ∧ dy ∧ dz ∧ dw =

∫

R

w3r2 sinϕdr ∧ dϕ ∧ dθ ∧ dw =

=

∫ 1

0

dr

∫ π

0

dϕ

∫ 2π

0

dθ

∫ 1

0

w3r2 sinϕdw =1

32(2π)

1

4=π

3and ∫

∂c

(y + w)dx ∧ dy ∧ dz = −π3 ) Using Stokes theorem

∫

∂c

(x2 + y2 + z2 + w2)dx ∧ dy ∧ dz =

∫

c

2wdw ∧ dx ∧ dy ∧ dz =

=

∫

R

2ww3r2 sinϕdrdϕdθdw = 21

32(2π)

1

5=

8

15π

4.2. STOKES THEOREM 301Problem 170:Compute the integral of ω = zdx∧ dy on the surfa e of the torus T obtainedwinding the ir umferen e of equations(y − b)2 + z2 = a2, x = 0 (b > a > 0)around the Oz axis.Solution:To use Stokes theorem we view T as the boundary of the 3- ube

c(r, θ, ϕ) = ((a+ rb cosϕ) cos θ, (a + rb cosϕ) sin θ, rb sinϕ)

(r, ϕ, θ) ∈ D = [0, 1]× [0, 2π]× [0, 2π]The boundary of c is (see p.296)∂c = c(1,1) − c(1,0)

c(1,1) being the surfa e of the torus and c(1,0) being a degenerate 2- ube thatdoesn't ontribute to the integral. We obtain∫

∂c

ω =

∫

c

dz ∧ dx ∧ dz =

∫

c

dx ∧ dy ∧ dz = Vol (V ) = 2π2ab2

Problem 171:Letω = x(z2 − y2)dy ∧ dz + y(x2 − z2)dz ∧ dx+ z(y2 − x2)dx ∧ dy.a) If c is a 2- ube of R3, express ∫

cω as an integral over a 1- hain.b) Let c be a parametrization of the solid torus obtained winding the dis

(0, y, z) : (y − 1)2 + z2 ≤ 1/4 around the Oz axis; ompute ∫∂cω. ) Let c be a 2- ube that parametrizes S = (x, y, z) : x2 + y2 + z2 =

4, z > 1, using Stokes theorem ompute ∫cω.

302 CHAPTER 4. STOKES THEOREMSolution:a) To use Stokes theorem we ompute a primitive of ω. We rst he kthat ω is losed:dω = (z2−y2)dx∧dy∧dz+(x2−z2)dy∧dz∧dx+(y2−x2)dz∧dx∧dy =

= (z2 − y2 + x2 − z2 + y2 − x2)dx ∧ dy ∧ dz = 0As R3 is star shaped respe t the origin we an use Poin aré formula:I(x(z2 − y2)dy ∧ dz) = (

∫ 1

0

t(tx)(t2z2 − t2y2)dt)(ydz − zdy) =

=1

5x(z2 − y2)(ydz − zdy)and

I(y(x2 − z2)dz ∧ dx) =1

5y(x2 − z2)(zdx− xdz)

I(z(y2 − x2)dx ∧ dy) =1

5z(y2 − x2)(xdy − ydx)We obtain:

η =1

5(x(z2 − y2)(ydz − zdy) + y(x2 − z2)(zdx− xdz) + z(y2 − x2)(xdy − ydx)) =

=1

5(yz(2x2 − y2 − z2)dx+ xz(2y2 − x2 − z2)dy + xy(2z2 − x2 − y2)dz)To transform ∫

cω into an integral over a 1- hain we use Stokes theorem:

∫

c

ω =

∫

c

dη =

∫

∂c

ηb) As dω = 0 we need no omputations:∫

∂c

ω =

∫

c

dω = 0. ) A parametrization of the region of the sphere S isc(ϕ, θ) = (2 sinϕ cos θ, 2 sinϕ sin θ, 2 cosϕ), (ϕ, θ) ∈ [0, π/3]× [0, 2π]

4.2. STOKES THEOREM 303The fa es of the boundary and the boundary arec(1,0)(θ) = (0, 0, 2)

c(1,1)(θ) = (√

3 cos θ,√

3 sin θ, 1)

c(2,0)(ϕ) = (2 sinϕ, 0, 2 cosϕ)

c(2,1)(ϕ) = (2 sinϕ, 0, 2 cosϕ)

∂c = c(1,1) − c(1,0)and Stokes theorem gives∫

c

ω =

∫

c

dη =

∫

∂c

η =

∫

c(1,1)

ηNow, introdu ing the η omputed in a), we have:∫

c

ω =

∫

c(1,1)

η =1

5

∫ 2π

0

(√

3 sin θ(6 cos2 θ − 3 sin2 θ − 1)(−√

3 sin θ) +

+√

3 cos θ(6 sin2 θ − 3 cos2 θ − 1)(√

3 cos θ))dθ =

=3

5

∫ 2π

0

(3 sin4 θ − 3 cos4 θ + sin2 θ − cos2 θ)dθ =

= 0

Problem 172:Let F = (f, g, h) : U ⊂ R3 → S2 ⊂ R3, F ∈ C2(U), U an open neighbourhoodof the unit ball B = x :| x |≤ 1. Compute ∫S2 fdg ∧ dh and dedu e thatthere is no F su h that its restri tion to the unit sphere S2 is the identity.Solution:We view B as a 3- ube, simply parametrizing it by means of spheri al o-ordinates. It has been seen on p.293 that as far as the integration of formsis on erned, the boundary of B is the orresponding parmetrization of S2.Then by Stokes theorem we have

∫

S2

fdg ∧ dh =

∫

B

df ∧ fdg ∧ dh

304 CHAPTER 4. STOKES THEOREMThat F(x) ∈ S2 means that f 2 + g2 + h2 = 1 and dierentiating su es-sively respe t to x, y, z gives2f ∂f

∂x+ 2g ∂g

∂x+ 2h∂h

∂x= 0

2f ∂f∂y

+ 2g ∂g∂y

+ 2h∂h∂y

= 0

2f ∂f∂z

+ 2g ∂g∂z

+ 2h∂h∂z

= 0

,an homogeneous system in the unknowns f, g, h; this system has nonvanishingsolutions at ea h point. The forms df, dg, dh are linearly dependent or, whatis the same, df ∧ dg ∧ dh = 0 and then∫

S2

fdg ∧ dh = 0If F(x, y, z) = (x, y, z) on S2 then∫

S2

fdg ∧ dh =

∫

S2

xdy ∧ dz =

∫

B

dx ∧ dy ∧ dz = Vol (B) =4

3πand we have a ontradi tion.

Problem 173:Let cR(t) = (R cos t, R sin t), t ∈ [0, 2π] and ω = −ydx+xdyx2+y2

. Compute ∫cRωand use Stokes theorem to prove that c is not the boundary of a 2- hain in

U = Rn − 0.Solution:c∗R(ω) =

1

R2(−R sin t(−R sin t) +R cos t(R cos t))dt = dt

∫

cR

ω =

∫ 2π

0

dt = 2πIf cR = ∂c for some 2- hain of U = Rn − 0 we would have∫

cR

ω =

∫

∂c

ω =

∫

c

dω =

∫

c

0 = 0a ontradi tion.

4.2. STOKES THEOREM 305Problem 174:Let f, g ∈ C1(R3) and assume that the ve tor eld F = (X, Y, Z) = ∇f ×∇gdoesn't vanish at every point. Show thata) df ∧ dg = Xdy ∧ dz + Y dz ∧ dx+ Zdx ∧ dy.b) df ∧ dg[rot F,F] = 0. ) Compute the integral ∫∂SXdy ∧ dz + Y dz ∧ dx + Zdx ∧ dy (∂S is aboundary).Solution:a) Using the anoni al isomorphisms

df ∧ dg = 2(∇f ×∇g) = 2F = Xdy ∧ dz + Y dz ∧ dx+ Zdx ∧ dyb) df ∧ dg[rot F,F] = 2F[rot F,F] = ωR3[rot F,F,F] = 0. ) ∫∂SXdy ∧ dz + Y dz ∧ dx + Zdx ∧ dy =

∫∂Sdf ∧ dg =

∫Sd(df ∧ dg) =∫

S0 = 0

Problem 175:Let ω = −ydx+xdyx2+y2

, γ(t) = (cos 2πt, sin 2πt), t ∈ [0, 1] a 1- ube and Γ : [0, 1]→R2, a losed simple urve su h that the segment [γ(t),Γ(t)] doesn't passthrough 0 for every t. Show that ∫

Γω = 2π.Solution:The 2- ube c(s, t) = sΓ(t) + (1− s)γ(t) is ontained in R2−0; we an useStokes theorem

c(1,0)(t) = γ(t)

c(1,1)(t) = Γ(t)

c(2,0)(s) = sΓ(0) + (1− s)γ(0)

c(2,1)(s) = sΓ(1) + (1− s)γ(1)

306 CHAPTER 4. STOKES THEOREMAs the urves are losed the two last 1- ubes are the same and the boundaryof c is∂c = Γ− γStokes theorem gives ∫

∂c

ω =

∫

c

dω =

∫

c

0 = 0

∫

Γ−γω = 0⇒

∫

Γ

ω =

∫

γ

ω = 2π

Volume of balls and area of spheresT To ompute the volume of the balls

Bn = (x1, . . . , xn) : x21 + · · ·+ x2

n ≤ R ⊂ Rnwe remind the fun tions Γ and β :Γ(u) =

∫∞0xu−1e−xdx, u > 0

β(u, v) = 2∫ π/20

cos2u−1 θ sin2v−1 θdθ, u, v > 0and some of its properties:a) It is lear that Γ(1) = 1.b) Integrating by parts we see that Γ(u + 1) = uΓ(u); noti e that thisimplies Γ(n+ 1) = n!. ) Substiting x = t2

2we obtain another expression for Γ:

Γ(u) =1

2u−1

∫ ∞

0

t2u−1e−t2

2 dt.d) The produ t Γ(u)Γ(v) may be evaluated onverting it into an integralon the rst quadrant by means of a hange to polar oordinates; weobtain Γ(u)Γ(v) = Γ(u + v)β(u, v), that we an write in the moresymmetri formβ(u, v) =

Γ(u)Γ(v)

Γ(u+ v)As β(12, 1

2) = π we obtain Γ(1

2) =√π.

4.2. STOKES THEOREM 307Problem 176:Using spheri al oordinates in Rn express in terms of the Γ fun tion thevolume of the ballsBn = (x1, . . . , xn) : x2

1 + · · ·+ x2n ≤ R2 ⊂ Rn, n = 1, 2, . . .How an we ompute the area of the spheres

Sn−1 = (x1, . . . , xn) : x21 + · · ·+ x2

n = R2 ⊂ Rn, n = 1, 2, . . .?Solution:We view Bn as the n- ubec : D = (0, R)× (0, π)× n−2). . . ×(0, π)× (0, 2π)→ Rngiven by the spheri al oordinates in Rn (see p.205). Let ωn be the volumeform of Rn; the volume of the ball is:Vol (Bn) =

∫

c

ωn =

∫

D

c∗(ωn)Here c∗(ωn) is the expression in spheri al oordinates of the volume form ofRn that we have omputed in p.225.Let's start with low dimensions, as it is easier:a) n = 1 no spheri al oordinates be ause not enough dimension, butobviously in this ase Vol (B1) = 2R.b) n = 2 , c∗(ω2) = rdr ∧ dθVol (B2) =

∫

D

rdr ∧ dθ =

∫ R

0

∫ 2π

0

rdrdθ =

=1

2R22π =

1

2R22π · 1

308 CHAPTER 4. STOKES THEOREM ) n = 3 , c∗(ω3) = r2 sin θ1dr ∧ dθ1 ∧ dθ2Vol (B3) =

∫

D

r2 sin θ1dr ∧ dθ1 ∧ dθ2 =

∫ R

0

∫ π

0

∫ 2π

0

r2 sin θ1drdθ1dθ2 =

=1

3R32π · 2At this point a pattern seems to emerge. We try one more dimensionto he k:d) n = 4, c∗(ω4) = r3 sin2 θ1 sin θ2dr ∧ dθ1 ∧ dθ2 ∧ dθ3Vol (B4) =

∫

D

r3 sin2 θ1 sin θ2dr ∧ dθ1 ∧ dθ2 ∧ dθ3 =

=

∫ R

0

∫ π

0

∫ π

0

∫ 2π

0

r3 sin2 θ1 sin θ2drdθ1dθ2dθ3 =1

4R42π

∫ π

0

∫ π

0

sin2 θ1 sin θ2dθ1dθ2 =

=1

4R42π · π

2· 2And we must dismiss the guess; what we an see isVol (Bn) =

1

nRn2π(

∫ π

0

sinn−2 θdθ)(

∫ π

0

sinn−3 θdθ) . . . (

∫ π

0

sin θdθ))Vol (Bn−1) =1

n− 1Rn−12π(

∫ π

0

sinn−3 θdθ)(

∫ π

0

sinn−4 θdθ) . . . (

∫ π

0

sin θdθ))and then Vol (Bn)Vol (Bn−1)=n− 1

nR

∫ π

0

sinn−2 θdθThe integral is∫ π

0

sinn−2 θdθ = 2

∫ π/2

0

sinn−2 θdθ =

= β(1

2,n− 1

2) =

Γ(12)Γ(n−1

2)

Γ(n2)

=

√πΓ(n−1

2)

Γ(n2)

4.2. STOKES THEOREM 309And then Vol (Bn)Vol (Bn−1)=n− 1

nR

√πΓ(n−1

2)

Γ(n2)Vol (Bn)Vol (B1)

=Vol (Bn)Vol (Bn−1)

· Vol (Bn−1)Vol (Bn−2)· · · Vol (B2)Vol (B1)

=

=n− 1

n· n− 2

n− 1· · · 1

2Rn−1

√πΓ(n−1

2)

Γ(n2)·√πΓ(n−2

2)

Γ(n−12

)· · ·√πΓ(1

2)

Γ(1)=

=1

nRn−1 (

√π)n−1Γ(1

2)

Γ(n2)

=1

nRn−1 (π)n/2

Γ(n2)As Vol (B1) = 2R we obtainVol (Bn) = Rn πn/2

(n/2)Γ(n2)We an do some he king in low dimensions (and known results):

• n = 1 The volume is 2R, the length of the segment (−R,R), and theobtained formula givesVol (B1) = 2R

√π

Γ(12)

= 2R

• n = 2 The volume is πR2, the area of the dis of radius R, and againthe formula agrees Vol (B2) = R2 (√π)2

Γ(1)= πR2

• n = 3 The volume of the ball of radius R is 43πR3, and the result of theformula is Vol (B3) =

2

3R3 (√π)3

Γ(32)but Γ(3

2) = Γ(1

2+ 1) = 1

2Γ(1

2) = 1

2

√π and thenVol (B3) =

2

3R3π√π

12

√π

=4

3πR3

310 CHAPTER 4. STOKES THEOREMTo ompute the area of the spheres we must integrate the volume form ofSn−1.At a point x = (x1, . . . , xn) ∈ Sn−1 the tangent hiperplane Tx has a unitexterior normal

n =1

R(x1, . . . , xn)The volume form ωRn = dx1 ∧ · · · ∧ dxn indu es in ea h Tx the form (seep.114)

ωx =1

R

n∑

i=1

(−1)i−1xidx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxnand we obtain the area of Sn−1 integrating ωx on an (n−1)- hain parametriz-ing Sn−1. We an apply Stokes theorem noti ing that, essentially, Sn−1 is theboundary of c, the n- ube parametrizing Bn:Area (Sn−1) =

∫

∂c

ωx =

∫

c

dωx =n

R

∫

c

ωE =n

RVol Bn = 2Rn−1 π

n/2

Γ(n2)Area (Sn−1) = 2Rn−1 π

n/2

Γ(n2)We an do some he king in low dimensions:

• n = 2 the length of S1is 2πR and the formula gives Long (S1) = 2Rπ.• n = 3 the area of S2 is 4πR2 and the formula gives Area (S2) =

2R2 π3/2

Γ(3/2)= 2R2 π3/2

(1/2)Γ(1/2)= 4R2π

4.3 Maxwell equations and dierential formsT We write the elds with an arrow and the dierential forms with boldfa e.We write the equations in the following form

∇ · −→B = 0 , ∇×−→E + ∂t−→B = 0

∇ · −→E = ρ , ∇×−→B − ∂t−→E =

−→jThe elds in Maxwell equations live in R4 and we all (t, x, y, z) the oordi-nates.

4.3. MAXWELL EQUATIONS AND DIFFERENTIAL FORMS 3114.3.1 The rst two equationsT As the divergen e is related to the derivative of a 2-form, we assign themagneti eld the ux 2-form:

B = Bxdy ∧ dz +Bydz ∧ dx+Bzdx ∧ dyAnalogously the rotational is related to the derivative of a 1-form and weassign to it the work 1-form:E = Exdx+ Eydy + EzdzWhenever ω is a form not ontaining dt we an write it as ω =

∑I ωIdx

I(here the ωI are fun tions of t, x, y, z) and thendω =

∑

I

(∂tωIdt ∧ dxI +3∑

i=1

∂iωIdxi ∧ dxI) = dt ∧ ∂tω + dSωwhere we have separated the Spa e derivative a ting on the variables (x, y, z)from the t ime derivative:

∂tω =∑

I

(∂tωI)dxI

dSω =∑

I

(3∑

i=1

∂iωIdxi ∧ dxI)

Problem 177:In the stati ase all the time derivatives vanish and the rst two equationsare∇ · −→B = 0,∇×−→E = 0.Show they are equivalent to dB = 0 and dE = 0.

312 CHAPTER 4. STOKES THEOREMSolution:We use the anoni al isomorphisms respe t to the variables (x, y, z) and treatt as a parameter. As the elds are stati we have dE = dSE+dt∧∂tE = dSEand dB = dSB; now

E = −→E ,B = 2

−→B

∇×−→E = 0 ⇔ dS(−→E ) = 0 ⇔ dE = dS(

−→E ) = 0

∇ · −→B = 0 ⇔ dS(2−→B ) = 0 ⇔ dB = dS(2

−→B ) = 0

T We revert to the general ase (non stati ); the rst two equations are:∇ · −→B = 0,∇×−→E + ∂t

−→B = 0We dene the ele tromagneti eld 2-form (F for Mi hael Faraday I guess)

F = B + E ∧ dtIn oordinatesF = Bxdy ∧ dz +Bydz ∧ dx+Bzdx ∧ dy +

+ Exdx ∧ dt+ Eydy ∧ dt+ Ezdz ∧ dtThe variables in F are (t, x, y, z) and they establish an order in thevariables; so we write:F = −Exdt ∧ dx− Eydt ∧ dy − Ezdt ∧ dz

+Bzdx ∧ dy +Bxdy ∧ dz − Bydx ∧ dzExpressing this 2-form as F = 12Fµνdx

µ ∧ dxν( we use here the so alledEinstein's onvention that implies summation over repeated indexes inopposite positions; see p.57) it has a matrix:Fµν =

0 −Ex −Ey −EzEx 0 Bz −By

Ey −Bz 0 Bx

Ez By −Bx 0

4.3. MAXWELL EQUATIONS AND DIFFERENTIAL FORMS 313Problem 178:Show that the rst two Maxwell equations are equivalent to dF = 0.Solution:We havedF = dB + dE ∧ dt =

= dSB + dt ∧ ∂tB + (dSE + dt ∧ ∂tE) ∧ dt =

= dSB + (∂tB + dSE) ∧ dtand as dSB does not ontain dt we see thatdF = 0⇔ dSB = 0 and ∂tB + dSE = 0butdSB = dS(2

−→B ) = (div −→B )dx ∧ dy ∧ dz

dSE = 2(rot −→E ) = 2(rot −→E )

∂tB + dSE = ∂tBxdy ∧ dz + ∂tBydz ∧ dx+ ∂tBzdx ∧ dy +

+ (rot −→E )xdy ∧ dz + (rot −→E )ydz ∧ dx+ (rot −→E )zdx ∧ dyand then that dF = 0 is equivalent todiv −→B = 0, rot −→E + ∂tB = 0

4.3.2 The last two equationsT We study now the se ond pair of Maxwell equations

∇ · −→E = ρ , ∇×−→B − ∂t−→E =

−→jLet ∗S be Hodge's operator respe t to the variables x, y, z. We onsider nowthe 1-form j = jxdx+ jydy + jzdz.

314 CHAPTER 4. STOKES THEOREMProblem 179:Show that the se ond pair of Maxwell equations is equivalent to∗SdS ∗S E = ρ

−∂tE + ∗SdS ∗S B = jSolution:We have seen in p.245 that for a 1-form as E is we have(∗SdS∗S)E = div E = div ♯E = div −→EIn this formula the rst 'div' is the orresponding operator a ting on the1-form E and the others 'div' are the usual ones for ve tor elds. We haveseen that

∇ · −→E = ρ⇔ ∗SdS ∗S E = ρAlso, in p.247 we have the omputation of div B:∗SdS ∗S B = div B = (∂yBz − ∂zBy)dx+ (∂zBx − ∂xBz)dy + (∂xBy − ∂yBx)dzand we see that ∗SdS ∗S B− ∂tE = j is equivalent to ∇×−→B − ∂t−→E =

−→j .

Problem 180:Consider in R4 the anoni al basis t, i, j,k with variables (t, x, y, z) the dualbasis dt, dx, dy, dz and the Lorentz metri . We dene the 1-form alled 'the urrent'J = jxdx+ jydy + jzdz − ρdtShow that the se ond pair of equatios are equivalent to

∗d ∗ F = Jthat we may write as div F = J

4.3. MAXWELL EQUATIONS AND DIFFERENTIAL FORMS 315Solution:∗d ∗ F = ∗d ∗B + ∗d ∗ (E ∧ dt)On one hand, reminding the problem in p.251 we have

∗d ∗B = (∂Bz

∂y− ∂By

∂z)dx+ (

∂Bx

∂z− ∂Bz

∂x)dy + (

∂By

∂x− ∂Bx

∂y)dzCompute the other term:

∗(E ∧ dt) = ∗(Exdx ∧ dt+ Eydy ∧ dt+ Ezdz ∧ dt)= −(Exdy ∧ dz + Eydz ∧ dx+ Ezdx ∧ dy)

d ∗ (E ∧ dt) = −(∂Ex∂x

+∂Ey∂y

+∂Ez∂z

)dx ∧ dy ∧ dz −

− dt ∧ (∂Ex∂t

dy ∧ dz +∂Ey∂t

dz ∧ dx+∂Ez∂t

dx ∧ dy)

∗d ∗ (E ∧ dt) = −(∂Ex∂x

+∂Ey∂y

+∂Ez∂z

)dt−

− (∂Ex∂t

dx+∂Ey∂t

dy +∂Ez∂t

dz)Finally∗d ∗ F = (

∂Bz

∂y− ∂By

∂z− ∂Ex

∂t)dx+ (

∂Bx

∂z− ∂Bz

∂x− ∂Ey

∂t)dy + (

∂By

∂x− ∂Bx

∂y− ∂Ez

∂t)dz −

− (∂Ex∂x

+∂Ey∂y

+∂Ez∂z

)dt− (∂Ex∂t

dx+∂Ey∂t

dy +∂Ez∂t

dz)and the ondition ∗d ∗ F = J is∂Bz

∂y− ∂By

∂z− ∂Ex

∂t= jx

∂Bx

∂z− ∂Bz

∂x− ∂Ey

∂t= jy

∂By

∂x− ∂Bx

∂y− ∂Ez

∂t= jz

≡ ∇×−→B − ∂t

−→E =

−→j

−(∂Ex∂x

+∂Ey∂y

+∂Ez∂z

) = −ρ ≡ ∇ · −→E = ρ

316 CHAPTER 4. STOKES THEOREMProblem 181: Continuity equation.Show that the ontinuity equationdiv −→j +∂ρ

∂t= 0is equivalent to ∗d ∗ J = 0 or div J = 0.Solution:

J = j− ρdt

∗d ∗ J = ∗d ∗ j− ∗d ∗ ρdtUsing the problem in p.251 we have∗d ∗ j = −(

∂jx∂x

+∂jy∂y

+∂jz∂z

)Compute the other term:∗ρdt = ρdx ∧ dy ∧ dz

d ∗ ρdt =∂ρ

∂tdt ∧ dx ∧ dy ∧ dz

∗d ∗ ρdt = −∂ρ∂tFinally

∗d ∗ J = −(∂jx∂x

+∂jy∂y

+∂jz∂z

)− ∂ρ

∂twhere we read that div J = 0⇔ div −→j +∂ρ

∂t= 0

Chapter 5Graphi al bibliographyReminding times where the types and the binding oered the author a spa efor his artisti side and as a kind antidote to the TEX terseness, we oer agraphi al bibliography that is at your disposal starting at next page.

317

318 CHAPTER 5. GRAPHICAL BIBLIOGRAPHY

319

320 CHAPTER 5. GRAPHICAL BIBLIOGRAPHY

321

322 CHAPTER 5. GRAPHICAL BIBLIOGRAPHY

323

324 CHAPTER 5. GRAPHICAL BIBLIOGRAPHY

Bibliography[Arn Arnold V., Méthodes Mathématiques de la mé anique lassique. Éditions Mir. Mos ou 1976.[Artin Artin E., Algèbre géométrique. Gauthier-Villars. Paris1967.[Ba-Mu Baez J., Muniain J.P., Gauge fields, knots and Gravity.World S ienti Publishing Co. Singapore 1994.[Bam-Stern Bamberg P., Sternberg S., A ourse in Mathemati s forstudents of Physi s. Cambridge University Press. Cam-bridge 1988.[Burg Burguès J.M., Integra ió i àl ul ve torial. Servei dePubli a ions UAB. Col. Materials 94. Bellaterra 2002.[Car Cartan H., Formes différentielles. Hermann. Colle tionMéthodes. Paris 1967.[Corw Corwin L.J., Sz zarba R.H., multivariable al ulus. Mar- el Dekker. 1982.[do Car do Carmo M.P., Differential forms and appli ations.Springer-Verlag. Universitext. Berlin 1994.[Feyn Feynman R., Leighton M., Sands G., The Feynman le -tures on Physi s. Addison-Wesley. Reading, Massa husetts1965.[Fla Flanders H., Differential forms with appli ations tothe physi al s ien es. Dover. NewYork 1989.325

326 BIBLIOGRAPHY[Flem Fleming W.H., Fun iones de varias variables. Ed.CECSA. Méxi o 1976.[Jan Jäni h K., Ve tor analysis. Springer-Verlag. Undergraduatetexts in Mathemati s. 2000.[Kell Kellog O.D., Potential theory. Dover. NewYork 1954.(Original J.Springer 1929.)[Marsd Marsden J.E., Tromba A., Ve tor al ulus. W.H.Freemanand o. NewYork 1988.[Mxw Maxwell J.C., A treatise on ele tri ity and mag-netism. Dover. NewYork 1954. (Original Clarendon Press1891.)[Nav Navarro V., Geometría y teoría de ampos. CPDA.Bar elona.[Pol Polya G., Matemáti as y razonamiento Plausible, Te -nos S.A. Madrid 1966.[Spvk Spivak M., Cal ulus on manifolds. W.A. Benjamin, in .NewYork 1965.[Xa 1 Xambó S., Algebra lineal y geometrías lineales, VolI. Eunibar. Bar elona 1984.[Xa 2 Xambó S., Algebra lineal y geometrías lineales, VolII. Eunibar. Bar elona 1984.

Index adjoint map, 47alpha and beta fun tions, 306alternating multilinear form, 66alternator, 67area n-sphere, 307asso iated spa e to a k-form, 84basi k- ube, 277basis, 11bidual spa e, 26boundary of a ube, 278 anoni al isomorphisms in oords, 226 hange of basis matrix, 12 hange of oordinates, 213 hange of variable in a form, 186 hange of variable in elds, 217 losed form, 172, 261 odierential, 255 omponents of a k-form, 51 ontra tile to point, 265 ontra tion, 109 oordinate urves, 199, 202 oordinate system, 17, 201 oordinates, 197 oordinates of a basis, 16 ross produ t, 117 ross-produ t, 118 ylindri al oordinates, 203de Rham omplex, 192

de omposable form, 77degenerated metri , 29dierential form, 167divergen e in oords, 234divergen e of a form, 246dual basis, 18dual map, 27dual spa e, 18Einstein's onvention, 57ele tri eld, 173ellipti al oordinates, 209eu lidian spa e, 29exa t form, 261extended ross-produ t, 121exterior derivative, 169exterior produ t, 73, 89rst anoni al iso in a set, 190rst anoni al isomorphism, 40ux 2-form, 110grad,rot,div in oordinates, 229gradient, 230gravitational eld, 172Hodge's appli ation, 136Hodge's operator, 240, 244index of a metri , 36integrating fa tor, 261isomorphism at, 41327

328 INDEXisomorphism sharp, 41isomorphism sharp-2, 129isotropi ve tor, 28k- hain, 279k- ontravariant tensor, 61k- ovariant tensor, 49k- ube, 278Lapla e's rule, 153lapla ian, 238lapla ian of a form, 255Leibniz rule, 192linear form, 16linear map, 13lo al basis, 199, 202Lorentz metri , 30, 155, 190, 251, 257Lorentz's formula, 173magneti eld, 173matrix of a linear map, 14Maxwell equations, 173, 195, 310metri , 28metri in a set, 189metri matrix, 29mixed tensor, 61multilinear appli ation, 49multilinear form, 49n-spheri al oordinates, 205naturality, 59, 91, 188nonsingular spa e, 29operators and exterior derivative, 272orientation, 100orientation preserving, 102oriented spa e, 102orthogonal basis, 35orthogonal spa e, 29

orthonormal basis, 29, 35permutation, 61Poin are formula, 265, 268Poin are lemma, 267polar oordinates, 197potential, 271primitive of a form, 261produ t of dierential forms, 240pull-ba k, 58, 179Pythagoras theorem, 146radi al, 99re ipro al basis, 44retra tion, 266rotational in oords, 232rotational of a form, 246s alar produ t, 28s alar produ t of 1-forms, 132s alar produ t of k-forms, 133se ond anoni al iso in an set, 191se ond anoni al isomorphism, 129sign of a permutation, 62signature of a metri , 36spa e of a k-form, 84spheri al oordinates, 203standard metri , 30star-shaped, 266Stokes theorem, 289symmetrizer, 70symple ti form, 97tangent spa e, 167tensor produ t, 50transition map, 214transpositions, 62ve tor eld, 167

INDEX 329ve tor identities, 192ve tor potential, 261, 271volume, 103volume form, 105volume form formula, 107volume form n-spheri al oords, 225volume n-ball, 307volume of a form, 136wave equation, 258