selected hw solutions - uhclimenha/2015-spring-7350/hw-solutions.pdf · math 7350 selected hw...

Math 7350 Geometry of Manifolds

Dr. Vaughn Climenhaga, PGH 651A

Spring 2015

Selected HW solutions

HW 1, #1. (Lee, Problem 1-4). Locally finite coversLet M be a topological manifold, and let U be an open cover of M .

(a) Suppose each set in U intersects only finitely many others. Showthat U is locally finite – that is, every point of M has a neigh-bourhood that intersects at most finitely many of the sets in U .

Solution. Given x ∈ M , there is U ∈ U such that x ∈ U ,because U covers M . Now U is a neighbourhood of x, and bythe hypothesis, U intersects only finitely many of the sets in U .Thus U is locally finite.

(b) Give an example showing that the converse may be false.

Solution. Take any infinite but locally finite cover U (for ex-ample, U = {(n − 1, n + 1) | n ∈ Z} with M = R), and definea new cover V = U ∪ {M}. Then V is still locally finite but ithas a set (M) that intersects infinitely other cover elements.

(c) Show that the converse is true if the elements of U are precom-pact (have compact closures).

Solution. Suppose U is locally finite. Thus for every x ∈ Mthere is a neighbourhood Vx such that Vx intersects only finitelymany of the sets in U . Let U ∈ U be precompact. Then U iscompact and is covered by the open sets Vx, so it is covered byfinitely many of them, say Vx1 , . . . , Vxn . Any element of U thatintersects U must intersect one of the Vxi . Since each of theseonly intersects finitely many elements of U , the same is true ofU , and hence of U .

Math 7350 Selected HW solutions of 30

HW 1, #2. (Lee, Problem 1-6). Distinct smooth structuresLet M be a nonempty topological manifold of dimension n ≥ 1. If Mhas a smooth structure, show that it has uncountably many distinctones. [Hint: first show that for any s > 0, Fs(x) = |x|s−1x defines ahomeomorphism from the unit ball in Rn to itself, which is a diffeo-morphism if and only if s = 1.]

Solution. First we prove the hint. Given x ∈ Rn \ 0 write x = x/|x|for the normalisation of x. Note that given s > 0 and x ∈ Bn (the unitball in Rn) we have

Fs(x) = |x|sx,so in particular |Fs(x)| < 1 whenever |x| < 1, and we have Fs(Bn) ⊂ Bn.Moreover, given s, t > 0 we have

Fs ◦ Ft(x) = Fs(|x|tx) = |x|stx = Fst(x).

In particular, Fs ◦ F1/s is the identity, so Fs is a bijection from Bnto itself. Note that x 7→ |x|s = (

∑i(x

i)2)s/2 is a continuous functionBn → R (as a composition of continuous functions), and x 7→ x iscontinuous on Bn \ 0, so Fs is continuous (in fact smooth) on Bn \ 0.Moreover, Fs(x)→ 0 as x→ 0, so Fs is continuous on Bn. Its inverseF1/s is continuous as well, so Fs is a homeomorphism. If s 6= 1 then Fsis a diffeomorphism (the identity). If s < 1 then Fs is not differentiableat 0. If s > 1 then F−1

s = F1/s is not C1 at 0. Thus Fs is not adiffeomorphism for any s 6= 1.

Now we use the hint to prove the result. Let A be any smooth atlason M . Fix p ∈ M and (U,ϕ) ∈ A such that p ∈ U . Since ϕ(U) ⊂Rn is open there is r > 0 such that B(ϕ(p), r) ⊂ ϕ(U). Let V =ϕ−1(B(ϕ(p), r)) ⊂ U , and define ψ : U → Rn by

ψ(q) = (ϕ(q)− ϕ(p))/r.

Note that ψ(p) = 0 and ψ(V ) = Bn. Let A∗ = A ∪ {(V, ψ)} andnote that A∗ is smooth because every transition map involving (V, ψ)is the composition of a linear map and a transition map involving ϕ.Given (W, θ) ∈ A∗, let W ′ = W \ {p} and θ′ = θ|′W . Let A′ be theatlas obtained from A∗ by replacing every (W, θ) (except (V, ψ)) with(W ′, θ′); this is again a smooth atlas.


Given s > 0, let As be the atlas obtained from A′ by replacing (V, ψ)with (V, Fs ◦ ψ). Note that this is an atlas because Fs is a homeomor-phism from Bn = ψ(V ) to itself. It is a smooth atlas because everytransition map is the composition of Fs (away from 0) with a transitionmap from A′. Thus it defines a unique smooth structure. Finally, ifAs and At define the same smooth structure, then the transition mapbetween (V, Fs ◦ ψ) and (V, Ft ◦ ψ) must be a diffeomorphism. Thistransition map is given by

(Fs ◦ ψ) ◦ (Ft ◦ ψ)−1 = Fs ◦ F−1t = Fs/t,

so it is a diffeomorphism if and only if s = t. Thus the smooth struc-tures induced by As are distinct for s > 0, so there are uncountablymany distinct smooth structures on M .

HW 1, #4. (Lee, Problem 1-9). Complex projective n-spaceComplex projective n-space, denoted by CP n, is the set of all 1-dimensionalcomplex-linear subspaces of Cn+1, with the quotient topology inheritedfrom the natural projection π : Cn+1 \ {0} → CP n. Show that CP n isa compact 2n-dimensional topological manifold, and show how to giveit a smooth structure analogous to the one we constructed for RP n.Note that we identify Cn+1 with R2n+2 via the correspondence

(x1 + iy1, . . . , xn+1 + iyn+1)↔ (x1, y1, . . . , xn+1, yn+1).

Solution. Given 1 ≤ j ≤ n+ 1, let Uj = {(z1, . . . , zn+1) ∈ Cn+1 | zj 6=0}. Define a map ϕj : Uj → Cn by

ϕj(z1, . . . , zn+1) =

(z1

zj, . . . ,

zj−1

zj,zj+1

zj, . . . ,

zn+1

zj

).

Then ϕj is continuous and is constant on fibres of π, so there is a

unique continuous map ϕj : Uj := π(Uj)→ Cn such that the following


diagram commutes.

Cn+1 \ 0

π

��

⊃ Uj

π

��

ϕj // Cn

CP n ⊃ Uj

ϕj

??

Moreover, ϕj is a bijection since its inverse is given by (w1, . . . , wn) 7→[w1, . . . , wj−1, 1, wj+1, . . . , wn], and ϕ−1

j is continuous since it is thecomposition of continuous maps. Thus ϕj is a homeomorphism, andsince we identify Cn with R2n, we have shown that CP n is locally Eu-clidean.

To show CP n is a topological manifold it remains to show that it issecond countable and Hausdorff. The first of these follows since it isa quotient of a second countable space. For Hausdorff, note that if[z1], [z2] ∈ Uj for some j, then they can be separated by disjoint opensets since the same is true of ϕj(z1), ϕj(z2) ∈ Cn. So we consider thecase where there is no Uj containing both [z1] and [z2]. Given j 6= k letAj,k = {[z] | |zj| > |zk|} ⊂ CP n; this is open since its preimage underπ is open in Cn+1 \ 0. By the assumption on z1, z2 there are j 6= ksuch that [z1] ∈ Uj and [z2] ∈ Uk, but zj1 = zk2 = 0. Thus z1 ∈ Aj,k andz2 ∈ Ak,j. Clearly Aj,k ∩ Ak,j = ∅ so this suffices.

We have shown that CP n is a topological manifold of dimension 2n.To show that it is compact we let S2n+1 ⊂ Cn+1 be the unit sphere inR2n+2 (with the natural identification between R2n+2 and Cn+1), anddefine τ : Cn+1 → S2n+1 by τ(z) = z/|z|, where |z| = (

∑j |zj|2)1/2.

Writing π for the restriction of π to S2n+1, we see that π = π ◦ τ . Inparticular, CP n is a quotient space of S2n+1 via the map τ . BecauseS2n+1 is compact, it follows that CP n is as well.

Finally, we check that A = {(Uj, ϕj) | j = 1, . . . , n + 1} is a smoothatlas on CP n. If j < k, the transition map ϕk ◦ϕ−1

j : ϕj(Uj ∩Uk)→ Cn

is given by

ϕk ◦ ϕ−1j (w1, . . . , wn) = ϕk[w

1, . . . , wk−1, 1, wk+1, . . . , wn]

=

(w1

wj, . . . ,

wj−1

wj,wj+1

wj, . . . ,

wk−1

wj,

1

wj,wk+1

wj, . . . ,

wn+1

wj

).


Thus every (complex) coordinate function of ϕk ◦ ϕ−1j is of the form

w 7→ w`

wjfor some `. Writing wj = xj + iyj, we see that

w`

wj=x` + iy`

xj + iyj=

(x` + iy`)(xj − iyj)(xj)2 + (yj)2

=x`xj + y`yj + i(y`xj − yjx`)

(xj)2 + (yj)2,

so every (real) coordinate function of ϕk ◦ ϕ−1j is of the form

(x,y) 7→ x`xj + y`yj

(xj)2 + (yj)2or (x,y) 7→ y`xj − yjx`

(xj)2 + (yj)2

for some `. These are smooth as long as (xj, yj) 6= (0, 0), which holdson ϕj(Uj), so ϕk ◦ ϕ−1

j is smooth as a map on R2n.

HW 2, # 1. More on GrassmaniansLet V be a n-dimensional real vector space and recall that given aninteger 1 ≤ k ≤ n, Gk(V ) is the Grassman manifold whose elementsare all the k-dimensional subspaces of V .

(a) We have seen that Gk(V ) is a smooth manifold for each k. Provethat it is compact.

Solution. Fix a basis for V , so we work with Gk(Rn). Each S ∈Gk(Rn) is determined by a set of k orthonormal vectors in Rn.Writing these as columns of an n×k matrix gives A ∈Mn×k(R)with AtA = Ik. Let X be the set of such matrices. Then X isa closed and bounded subset of Rnk, hence X is compact. Letπ : X → Gk(Rn) be the map that assigns to each A ∈ X thesubspace spanned by its columns. Then π is surjective.

We claim that π is continuous. To this end we look at itscoordinate representation. Recall that a chart for Gk(Rn) isgiven by fixing a decomposition Rn = P ⊕Q, where dimP = k,and using the bijection between UQ = {S ∈ Gk(Rn) | S ∩ Q ={0}} and the set of linear functions from P to Q. Let RP

be projection onto P along Q, and RQ the projection onto Qalong P . Then the linear map L : P → Q associated to π(A) ∈Gk(Rn) is uniquely determined by the condition that L(RPv) =RQv for every column v of the matrix A. Now it is easy to


see that varying A continuously means varying the columnsv continuously, which in turn varies the map L continuously.Thus π is continuous.

(b) Prove that Gk(V ) and Gn−k(V ) are diffeomorphic.

Solution. Fix an inner product on V and let F : Gk(V ) →Gn−k(V ) be the map that takes a subspace S to its orthogonalcomplement S⊥. Clearly F is a bijection, we claim that it issmooth. (This is enough since its inverse is of the same form.)

To this end, let V = P⊕Q be an orthogonal decomposition ofV and let UQ be as above. Note that if S ∈ UQ then V = Q⊕S,so we have V = Q⊥ ⊕ S⊥, hence F (S) = S⊥ ∈ UQ⊥ . Thus wewrite F in local coordinates given by the charts UQ, UQ⊥ .

Let L : P → Q be linear and let S ∈ Gk(V ) be its graph.

We want to find F (L) : P⊥ → Q⊥. Given v ∈ P⊥, the image

F (L)(v) ∈ Q⊥ is determined by the condition that v+ F (L)v ∈S⊥; that is,

〈v + F (L)v, w + Lw〉 = 0 for all w ∈ P.

Expanding the inner product gives

〈v, w〉+ 〈v, Lw〉+ 〈F (L)v, w〉+ 〈F (L)v, Lw〉 = 0.

The first term always vanishes because v ∈ P⊥ and w ∈ P .Similarly for the last term (with Q⊥ and Q). Thus we have

〈F (L)v, w〉 = −〈v, Lw〉

for all v ∈ P⊥ and w ∈ P . This means that once we fix bases,F (L) is represented by the matrix −At, where A is the matrix

representing L. Thus F maps A to −At and is smooth.

HW 2, # 3. (Lee, Problem 2-9). Let p be a non-zero polynomial in onevariable with complex coefficients. Show that there is a unique contin-uous map p : CP 1 → CP 1 such that the following diagram commutes,


where G is as in the previous problem.

C G //

p

��

CP 1

p��

C G // CP 1

Show that the map p is smooth. Is it a diffeomorphism?

Solution. Recall that G(z) = [z, 1] = {(w1, w2) ∈ C2 | w1 = zw2}. Inorder for the diagram to commute, we must have

p([z, 1]) = p(G(z)) = G(p(z)) = [p(z), 1]

for all z ∈ C. Whenever w2 6= 0 we have [w1, w2] = [w1

w2 , 1] and so

p([w1, w2]) =

[p

(w1

w2

), 1

].

This defines p on CP 1 \ {[1, 0]}. Note that p is smooth on this setbecause its coordinate representation is p, which is a polynomial andhence analytic.

In order to extend p to all of CP 1, we need to use a chart containing[1, 0]. Such a chart is given by the inverse of the map F : C → CP 1

defined by F (z) = [1, z].

To find the coordinate representation of p with respect to the chartF−1, we write p(z) = a0 + a1z + · · · + anz

n for some a0, . . . , an ∈ C,and observe that given z ∈ C \ {0} for which p(1/z) 6= 0, we have

F−1 ◦ p ◦ F (z) = F−1(p([1, z])) = F−1

([p

(1

z

), 1

])= 1/p(1/z) =

1

a0 + a1z−1 + · · ·+ anz−n

=zn

a0zn + a1zn−1 + · · ·+ an.

If p(z) ≡ a0 is a constant polynomial, then we have p([0, 1]) = [a0, 1],otherwise we have p([0, 1]) = [0, 1]. In the first case p is a constantfunction and therefore smooth. In the second, there is a neighbourhood


U ⊂ C of 0 such that p(1/z) 6= 0 for all z ∈ U \ {0}, and in particular,the formula above shows that p is smooth on this neighbourhood, hencesmooth on CP 1. Note that p is a diffeomorphism if and only if p hasdegree 1 (and the linear coefficient is non-zero); for all other p, tha mapp itself is already non-injective.

HW 2, # 5. Tori as quotients by latticesLet v, w ∈ R2 be independent (hence a basis), and let Λ = vZ +wZ ={av + bw | a, b ∈ Z}. We say that Λ ⊂ R2 is a lattice. The lattice Λinduces an equivalence relation on R2 by putting x ∼ y iff x − y ∈ Λ.Let MΛ = R2/Λ be the topological space obtained as the quotient ofR2 by this equivalence relation.

(a) Fix a lattice Λ and let π : R2 →MΛ be the quotient map. Given

p ∈ R2 and r ∈ (0, 12), let U r

p = Br(p) ⊂ R2, and show that π|Urpis a bijection onto its image U r

p ⊂MΛ. Let ϕrp = π|−1

Urp: U r

p → U rp .

Show that the collection A = {(U rp , ϕ

rp) | p ∈ R2, 0 < r < 1

2}

satisfies the conditions of the smooth manifold chart lemma, soMΛ is a smooth manifold with smooth structure generated bythis atlas.

Solution. As observed in class, the condition r < 1/2 needs tobe replaced by r < r0, where r0 depends on Λ.

Let r0 be such that x ∈ Λ \ {0} implies ‖x‖ > 2r0. Thengiven r ∈ (0, r0) and p ∈ R2, we observe that x, y ∈ Br(p)implies that ‖x − y‖ ≤ 2r0, and hence by the definition of r0,we have x− y /∈ Λ, so x 6∼ y. This implies that π(x) 6= π(y), soπ|Urp is 1-1.

Now we check the conditions of the smooth manifold chartlemma. We just showed that ϕrp is a bijection between U r

p and

the open set Br(p) ⊂ R2, which verifies (i). For (ii), first observethat U r

p = U rp′ for all p′ ∼ p. Given U r

p and U sq , let x ∈ Λ be such

that ‖p− (q + x)‖ is minimised. Then U rp ∩ U s

q = U rp ∩ U s

q+x =π(Br(p) ∩Bs(q + x)). We get ϕrp(U

rp ∩ U s

q ) = Br(p) ∩Bs(q + x)


and ϕsq(Urp ∩ U s

q ) = Br(p − x) ∩ Bs(q), both of which are open

in R2.

Condition (iii) follows because the transition map ϕrp ◦ (ϕsq)−1

is a translation by x, where x is as in the previous paragraph.Condition (iv) follows by choosing p to have rational coefficientsand r to be rational. Condition (v) follows since given anydistinct points [p], [q] ∈M/Λ, we can choose r such that Br(p

′)∩Br(q

′) = ∅ for all p′ ∼ p and q′ ∼ q.

(b) Let Λ1 and Λ2 be any two lattices in R2, and show that MΛ1

and MΛ2 are diffeomorphic (when equipped with the smoothstructure from the previous part). Hint: Start by finding asmooth map F : R2 → R2 such that F (Λ1) = Λ2.

Solution. Let Λ1 be generated by v, w ∈ R2 and Λ2 be gener-ated by x, y ∈ R2. Let F : R2 → R2 be the linear map takingv 7→ x and w 7→ y. Then F is smooth and F (Λ1) = Λ2. Notethat F is invertible and F−1 is also linear.

Given p ∈ R2 we have p + Λ1 ∈ R2/Λ1, and we defineF : R2/Λ1 → R2/Λ2 by

F (p+ Λ1) = F (p) + Λ2.

This is well-defined because F is linear and F (Λ1) = Λ2. It isbijective because we can define its inverse by

F−1(q + Λ2) = F−1(q) + Λ1.

It only remains to show that both F and F−1 are smooth. Withr ∈ (0, r0) as in the previous part, we can write the coordinate

representation F relative to (U rp , ϕ

rp) and (U r

F (p), ϕrF (p)), where

the first chart is in R2/Λ1, and the second in R2/Λ2. We get

F (p) = ϕrF (p) ◦ F ◦ (ϕrp)−1(p) = ϕrF (p)(F (p+ Λ1))

= ϕrF (p)(F (p) + Λ2) = F (p),

that is, the coordinate representation of F is the original linearmap F , which is smooth. The same argument shows that F−1

is smooth, so F is a diffeomorphism.


(c) Let Λ = Z2 be the lattice generated by the standard basisvectors, and show that MΛ is diffeomorphic to the torus T2 =S1 × S1 with its standard smooth structure.

Solution. Define a map p : R2 → T2 by p(x, y) = (e2πix, e2πiy).Then (x′, y′) ∈ (x, y) + Λ implies that p(x′, y′) = p(x, y), and sop passes to a map p : MΛ → T2. Moreover, p is a bijection, aswe observed in the proof that T2 is homeomorphic to the squarewith opposite edges identified.

We must show that p and its inverse are smooth. S1 is coveredby four charts, each of which is the intersection of S1 with a half-plane, and the corresponding coordinate maps are z 7→ Re(z)and z 7→ Im(z). Direct products of these charts give 16 chartsthat cover T2, all with the same form of coordinate maps. Thusthe coordinate representations of p have the form p(x, y) =(cos 2πx, cos 2πy), and similarly with one or both cos replacedby sin. All these functions are smooth, and the coordinaterepresentations of p−1 are the inverse trig functions on theirappropriate domains, which are also smooth.

HW 3, #1. Tangent space of a product manifold Let M1, . . . ,Mk

be smooth manifolds, and for each j, let πj : M1 × · · · × Mk → Mj

be the projection onto the Mj factor. Prove that for any point p =(p1, . . . , pk) ∈M1 × · · · ×Mk, the map

α : Tp(M1 × · · · ×Mk)→ Tp1M1 ⊕ · · · ⊕ TpkMk

v 7→ (d(π1)p(v), . . . , d(πk)p(v))

is an isomorphism.

Solution. The map α is a direct sum of the linear maps d(πi)p, henceit is linear. Moreover, because Tp(

∏Mi) and

⊕TpiMi have dimension∑

ni, in order to show that α is an isomorphism it suffices to showthat it is onto. (This turns out to be easier than showing 1-1.)


We prove surjectivity by exhibiting a linear map β :⊕

TpiMi → Tp(∏Mi)

such that α ◦ β is the identity on⊕

TpiMi. Given pi ∈ Mi, letτj : Mj →

∏iMi be given by

τj(x) = (p1, . . . pj−1, x, pj+1, . . . pk).

Define β by β(v1, . . . , vk) = d(τ1)p1(v1) + · · · + d(τk)pk(vk). Then β islinear since each d(τi)pi is linear, and moreover we have

α ◦ β(v1, . . . , vk) =⊕j

d(πj)p(β(v1, . . . , vk))

=⊕j

∑i

d(πj)p ◦ d(τi)pi(vi)

=⊕j

∑i

d(πj ◦ τi)pi(vi) =⊕j

vj,

where the last equality uses the fact that πj ◦ τi is the identity if i = jand a constant function otherwise. Thus α ◦ β = Id⊕

i TpiMi.

We remark that there is also the following alternate solution that useslocal coordinates. Let (Ui, ϕi) be charts for Mi around pi. ThenU =

∏Ui and ϕ =

∏ϕi gives a chart for M around p. Writing

x1i , . . . , x

nii : Ui → R for the coordinate functions of ϕi, we see that the

coordinate functions of ϕ are

x11, . . . , x

n11 , x

12, . . . , x

n22 , . . . , x

1k, . . . , x

nkk ,

where xji = xji ◦πi (note that we must compose with πi to get a functionU → R.) Thus a basis for Tp(

∏Mi) is given by{∂

∂xji

∣∣∣p

}i,j

.

To show that α is an isomorphism it suffices to show that

d(π`)p

(∂

∂xji

∣∣∣p

)=

∂

∂xjiif ` = i and 0 otherwise,

since then α takes the basis for Tp(∏Mi) to a basis for

⊕TpiMi. We

compute d(π`)p(∂

∂xji|p) by observing that for any f ∈ C∞(M`), we have

d(π`)p

(∂

∂xji

∣∣∣p

)(f) =

(∂

∂xji

∣∣∣p

)(f ◦ π`) =

∂g

∂xji(p),


where g = f ◦ π` ∈ C∞(M) and g = g ◦ ϕ−1 ∈ C∞(U). (Recall that

U = ϕ(U) ⊂ R∑ni .) But we also have

∂g

∂xji(p) =

∂(f ◦ π` ◦ ϕ−1)

∂xji(p) =

∂(f ◦ ϕ−1` ◦ π`)

∂xji(p)

where π` : R∑ni → Rn` is projection onto the n` coordinates corre-

sponding to xj` for j = 1, . . . , n`. The final quantity is equal to 0 if

` 6= i, and(

∂

∂xji|pi)

(f) if ` = i. This proves that α maps a basis for

Tp(∏Mi) onto a basis for

⊕TpiMi, hence it is an isomorphism.

HW 3, # 2. (Lee, Problem 3-4). A trivial tangent bundleShow that TS1 is diffeomorphic to S1 × R.

Remark: You may find it interesting to consider whether or not thesame is true for S2. Be warned that we do not yet have the machineryin this course to answer this question.

Solution. First recall that for any smooth manifold M , if (U,ϕ) is achart on M then (U , ϕ) is a chart on TM , where U =

⊔p∈U TpU and

ϕ : U → U ×Rn is given by ϕ(v) = (π(v), v1, . . . , vn), where vj = v(xj)so that v = vj ∂

∂xj|π(v). (π : TM → M is the canonical submersion.)

The maps ϕ : U → U × Rn and ϕ × Id : U × Rn → U × Rn are bothdiffeomorphisms, thus GU : U → U ×Rn given by GU = (ϕ−1× Id) ◦ ϕis a diffeomorphism.

In light of the previous paragraph, it is natural to try to define adiffeomorphism G : TM → M × Rn by fixing a smooth atlas A on Mand then for each (U,ϕ) ∈ A, putting G(v) = GU(v) for v ∈ U . Theproblem is that in general we have no reason to expect that GU |U∩V =GV |U∩V when (U,ϕ) and (V, ψ) are two overlapping charts in A.


Writing τ : U × Rn → U for projection to the first part of the orderedpair, we see that τ(GU(v)) = π(v) for every v ∈ U , which does notdepend on U . So the compatibility problems we may encounter haveto do with the coordinates that different charts induce on the tangentspaces.

The special thing about M = S1 is that we can choose charts wherethese coordinates all agree. With S1 ⊂ C as the unit circle, let U =S1 \ {−1} and let ϕ : U → (−π, π) be defined by the condition thatϕ−1(x) = eix. Similarly, let V = S1 \ {1} and let ψ : V → (0, 2π) bedefined by ψ−1(x) = eix. Let GU : U → U × R and GV : V → V × Rbe as described above. (It is an easy exercise to check that ϕ, ψ arecompatible with the usual smooth structure on S1.)

Note that given p ∈ U , every v ∈ TpS1 can be expressed as v = v1 ∂∂x

,where v1 = v(x) ∈ R. Thus GU(v) = (p, v1) ∈ U×R ⊂ S1×R. Writingx for the coordinate function relative to V , we similarly have v = v1 ∂

∂x

for v ∈ TpS1 when p ∈ V . The key is that the transition map betweenU and V is given by

x(x) =

{x x ∈ (0, π),

x+ 2π x ∈ (−π, 0).

Thus for every p ∈ U ∩ V and v ∈ TpS1, we have v(x) = v(x), hence

v1 = v1. In particular, GU = GV on U ∩ V , so this gives a well-definedbijection G : TS1 → S1 × R. Note that the coordinate representationsof G are the maps (ϕ × Id) ◦ GU ◦ ϕ−1, but each of these is just the

identity map on U × Rn, and hence is smooth. Similarly, the coor-dinate representation of G−1 is the identity, hence smooth, so G is adiffeomorphism.

HW 3, # 4. Quotient manifoldsLet G be a group and E a smooth manifold. A left action of G on Eis a map G× E → E, often written as (g, p) 7→ g.p, that satisfies

g1.(g2.p) = (g1g2).p for all g1, g2 ∈ G, p ∈ E,e.p = p for all p ∈ E.


Suppose we are given a left action of G on E such that for every g ∈ G,the map p 7→ g.p is a smooth map from E to itself. This induces anequivalence relation on E by putting x ∼ y iff there is g ∈ G such thatg.x = y.

We say that the action is “free and proper” (see p.548–549 of Lee) ifthe following are true:

(i) for every x ∈ E there is a neighbourhood U of x such thatg.U ∩ U = ∅ for every g 6= e;

(ii) for every x, y ∈ E with x 6∼ y there are neighbourhoods U of xand V of y such that g.U ∩ V = ∅ for every g ∈ G.

Write M = E/G for the quotient space of E by the relation ∼.

(a) Prove that if the action is free and proper, then M is a topo-logical manifold.

(b) Let π : E →M be the quotient map. Show that π is a coveringmap. (This means that for every x ∈ M there is a neighbour-hood U 3 x such that π−1(U) =

⊔α∈A Vα for some disjoint open

sets Vα ⊂ E such that π|Vα : Vα → U is a homeomorphism.)

(c) Prove that if E is a smooth manifold, then M has a uniquesmooth structure such that π is a smooth covering map. (“Smoothcovering map” means that ‘homeomorphism’ is replaced by ‘dif-feomorphism’ in the definition of covering map.)

Solution. To show that M is a topological manifold, we first observethat quotients of second countable spaces are themselves second count-able, so it suffices to check Hausdorff and locally Euclidean.

The Hausdorff property will follow from (ii) in a moment; first weshow that π is an open map. Indeed, given any U ⊂ E we haveπ(U) = π(G.U), where G.U =

⋃g∈U g.U . Note that for every g ∈ G,

the map p 7→ g.p is invertible with inverse q 7→ g−1.q, hence p 7→ g.p is


a diffeomorphism from E to itself; thus if U is open, then g.U is openfor every g ∈ G, so G.U is also open. Because G.U is saturated withrespect to π, it follows from the definition of the quotient topology thatπ(U) = π(G.U) is open for every open U ⊂ E, so π is an open map.

Now given [x] 6= [y] ∈ M , where x, y ∈ E and we write [x] = G.x ={g.x | g ∈ G} for the equivalence class of x, then we have x 6∼ yand so by (ii) there are neighbourhoods U 3 x and V 3 y such thatg.U ∩ V = ∅ for every g ∈ G. Since π is an open map, the sets[U ] = π(U) and [V ] = π(V ) are open in M ; since [x] ∈ [U ] and[y] ∈ [V ], to check Hausdorff it suffices to show that [U ] ∩ [V ] = ∅.Suppose [z] ∈ [U ] ∩ [V ]. Then there are g, h ∈ G such that g.z ∈ Uand h.z ∈ V . But then h−1g.z ∈ h−1g.U ∩ V , contradicting our choiceof U, V . This shows that M is Hausdorff.

To show that M is locally Euclidean, we start by proving that π isa covering map. Given any [x] ∈ M , by (i) there is a neighbourhoodU 3 x such that g.U∩U = ∅ for every g 6= e. Note that [U ] = π(G.U) =π(⋃g∈G g.U) and that G.U is saturated, so π−1[U ] =

⋃g∈G g.U . To

show that U is evenly covered it suffices to show that the union isdisjoint and that π|g.U is a homeomorphism onto its image for every g.

First we observe that if z ∈ g.U ∩ h.U for some g 6= h ∈ G, thenh−1z ∈ h−1g.U ∩ U , but h−1g 6= e, contradicting (i). Thus the unionis disjoint. Moreover, if π(y) = π(z) for some y, z ∈ g.U , then there ish ∈ G such that z = h.y, hence z ∈ hg.U ∩g.U , so g−1z ∈ g−1hg.U ∩U ,so g−1hg = e, so h = e. In particular, π|g.U is 1-1. Since π is continuousand open, this shows that πg.U is a homeomorphism onto its image, thusU is evenly covered.

Now we use the fact that π is a covering map to show that M is atopological manifold. Given [x] ∈M , let U 3 x be a neighbourhood asabove, so that π|U is a homeomorphism from U ⊂ E to [U ] = π(U) ⊂M . Because E is a topological manifold, there is a neighbourhoodV 3 x and a homeomorphism ϕ : V → V ⊂ Rn. Thus ϕ := ϕ ◦ π|−1

U isa homeomorphism from [V ]∩ [U ] to an open subset of Rn. This showsthat M is locally Euclidean, hence a topological manifold (we alreadyshowed Hausdorff and second countable).


Finally, we show that there is a unique smooth structure on M suchthat π is a smooth covering map. Observe that if there is such a smoothstructure, then because each chart ϕ on E is a diffeomorphism onto itsimage, each of the maps ϕ ◦ π|−1

U from the previous paragraph is also adiffeomorphism onto its image, so any smooth structure on M with πa smooth covering map must be compatible with the atlas defined inthe previous paragraph. It only remains to show that the charts fromthis atlas are all smoothly compatible with each other, so that we haveexistence of such a smooth structure.

Let ([U ], ϕ) and ([V ], ψ) be any two such charts on M . That is,[U ], [V ] ⊂ M are evenly covered, with U, V ⊂ E open; moreover,

ϕ = ϕ ◦ π|−1U for a smooth chart ϕ : U → Rn, and similarly for ψ.

Note that we may assume without loss of generality that π|U∩V mapsU ∩V homeomorphically to [U ]∩ [V ] (a priori we may have U ∩V = ∅even when [U ]∩ [V ] 6= ∅, but we can replace V with g.V for some g ∈ Gto get the desired property).

Now we have ψ ◦ ϕ−1 = (ψ ◦ π|−1U∩V ) ◦ (π|U∩V ◦ϕ−1) = ψ ◦ϕ−1, which is

smooth since ϕ, ψ come from a smooth atlas for E. This shows that theatlas we defined on M is smooth, thus M has a smooth structure suchthat π is a smooth covering map. We showed above that this structureis uniquely determined by the smooth structure on E.

HW 4, # 1. Let S, S be compact surfaces without boundary, andsuppose that ρ : S → S is a covering map with degree n. Prove thatχ(S) = nχ(S).

You may use without proof the following result: Given any open coverU of a surface S, there is a triangulation T of S such that every trianglein T is completely contained in some element U ∈ U .


Solution. Because ρ is a covering map of degree n, every point x ∈ Shas a neighbourhood Ux 3 x such that ρ−1(Ux) is a disjoint union ofopen sets V 1

x , . . . , Vnx ⊂ S with the property that ρ|V ix is a homeomor-

phism from V ix to Ux.

Let U = {Ux | x ∈ S}. By the result stated in the problem, there is atriangulation T of S such that every triangle is completely containedin some Ux.

Formally we defined T as a collection of triangles in R2 together withan equivalence relation on the edges, such that the quotient space isthe surface S. We may think of the triangles as being subsets of S (weidentify a triangle with its image under the quotient map), and we willdo this from now on; indeed, this is implicit in the formulation of theresult in the previous paragraph (every triangle is completely containedin some Ux).

Let T be a triangle in T , and let x be such that T ⊂ Ux. Then ρ−1(T )is a disjoint union of n triangles in S. Denote the collection of all suchtriangles in S by T . Note that the intersection of any two trianglesT1, T2 ∈ T is of the form ρ|−1

V ix(ρ(T1) ∩ ρ(T2)); since ρ(T1) and ρ(T2)

are triangles in T and ρ|V ix is a homeomorphism V ix → Ux, we see that

T1 ∩ T2 is a union of vertices and edges. Thus T is a triangulation ofS.

We showed above that the number of faces in T is n times the numberof faces in T . The same is true for vertices and edges, because eachedge and each vertex in T is contained in some Ux, which is evenlycovered by ρ. Thus we have

χ(S) = #faces in T −#edges in T + #vertices in T= n(#faces in T )− n(#edges in T ) + n(#vertices in T )

= nχ(S).


HW 4, # 2. Let Sk be the sphere with k handles. Give a necessaryand sufficient condition on k, ` ∈ N for the existence of a covering mapρ : Sk → S`.

Hint: Use the previous problem to determine a natural necessary con-dition. Then given k, ` satisfying this condition, describe a particularrealization of Sk as a surface in R3 that is symmetric under rotationby 2π/(k − 1) around the z-axis. Then for a suitable value of n, rota-tion by 2π/n around the z-axis will induce an equivalence relation onSk, whose quotient space is S`, and whose quotient map is the desiredcovering map ρ.

Solution. From the previous problem we know that if there is a cov-ering map Sk → S` then χ(Sk) = nχ(S`) for some n ∈ N. Recall fromlecture that χ(Sk) = 2 − 2k, so if a covering map exists then we have2− 2k = n(2− 2`), or in other words,

n =k − 1

`− 1∈ N.

We claim that this necessary condition is also sufficient. It suffices tocheck the case when ` 6= k, since every surface covers itself (by theidentity map). Note that if k = 0 or 1 then the condition can only besatisfied for ` = k. Thus it suffices to consider the case when k > 1.

Consider the realization of Sk in R3 sketched in the figure (for k = 5),where we start with the torus (S1) as a surface of revolution and thenadd the remaining k − 1 handles evenly spaced around the outside ofthe torus. We attach the handles symmetrically so that the surface issymmetric under rotation around the z-axis by 2π/(k − 1).


Now let n = k−1`−1

and let G be the finite cyclic group with n elements.Let G act on Sk by rotation; that is, if g is a generator for g thengm acts by rotation by 2πm/n around the z-axis. Note that this is adiffeomorphism from Sk to itself and that this gives a smooth groupaction that is free and proper; a typical orbit of the action is shownbelow. The figure also shows how a fundamental domain for the quo-tient Sk/G can be selected; cut Sk into a ‘wedge’ of angle 2π/n, andthen glue the two boundary circles together to get S`.

Note that because the quotient map is by a free and proper groupaction, we know from the previous assignment that it is a smoothcovering map.

HW 4, # 3. Consider the function f(x, y) = sin(4πx) cos(6πy) on thetorus T2 = R2/Z2.

(a) Prove that this is a Morse function (every critical point is non-degenerate) and calculate the number of minima, saddles, andmaxima.

(b) Describe the evolution of the sublevel sets f−1((−∞), c)) as cvaries from the lowest minimum value to the highest maximumvalue.


Solution. A straightforward computation shows that

∂f

∂x= 4π cos(4πx) cos(6πy),

∂f

∂y= −6π sin(4πx) sin(6πy),

and since cos t, sin t never vanish simultaneously, we see that (x, y) ∈ R2

is a critical point of f : R2 → R if and only if one of the following twoconditions holds:

cos(4πx) = sin(6πy) = 0,

sin(4πx) = cos(6πy) = 0.

The first condition is satisfied iff x = (2n+ 1)/8 and y = m/6 for somen,m ∈ Z; the second is satisfied iff x = n/4 and y = (2m + 1)/12 forsome n,m ∈ Z.

Note that each critical point on the torus corresponds to a critical pointin [0, 1)2 (since this is a fundamental domain for the torus), and so theset of critical points is A ∪B, where

A = {18, 3

8, 5

8, 7

8} × {0, 1

6, 1

3, 1

2, 2

3, 5

6},

B = {0, 14, 1

2, 3

4} × { 1

12, 3

12, 5

12, 7

12, 9

12, 11

12}.

Another computation gives

∂2f

∂x2= −16π2 sin(4πx) cos(6πy),

∂2f

∂x∂y= −24π2 cos(4πx) sin(6πy),

∂2f

∂y2= −36π2 sin(4πx) cos(6πy),

so the Hessian determinant is

576π2(

sin2(4πx) cos2(6πy)− cos2(4πx) sin2(6πy)).

If (x, y) ∈ A then the second term in brackets vanishes, and the first ispositive, so the determinant is positive; thus every critical point in Ais non-degenerate and is either a maximum or a minimum. Similarly,if (x, y) ∈ B then the first term vanishes and the second is positive,so the determinant is negative; thus every critical point in B is non-degenerate and is a saddle.


We have shown that f is a Morse function and that it has 24 saddles,and 24 other critical points that are either minima or maxima (thepoints in A). Note that if (x, y) ∈ A then each of sin(4πx) and cos(6πy)is either ±1, so f(x, y) is either ±1; there are 12 elements of A withf = −1, which are minima, and 12 with f = 1, which are maxima.

Now we consider the sublevel sets Sc = f−1((−∞, c)). For c < 1 this isempty; for c ∈ (−1, 0) it is the union of 12 disjoint discs (geometricallythese are ellipses). As c increases the discs grow, until at c = 0 theybecome rectangles which touch at the corners; these corners are the 24saddle points. For c ∈ (0, 1) the sublevel set is a torus with 12 discsremoved, and these discs are filled in as c increases past 1, so thatSc = T2 for c > 1. These sublevel sets are shown in the figure.

c = −23

c = −13 c = 0 c = 1

3c = 2

3

Notice that because there is more than one saddle with f = 0, passingfrom c = −ε to c = ε does not correspond to gluing on a single pair ofpants; rather, it corresponds to gluing on some kind of very complicatedset of pants with lots of legs. More precisely, a regular pair of pants isthe sphere with 3 discs removed. The “very complicated set of pantswith lots of legs” that we encounter here (that is, Sε \ S−ε) is a toruswith 24 discs removed.

HW 5, #2. Lee, Problem 4-4. Let γ : R → T2 be the curve given byγ(t) = (e2πit, e2πiαt), where α is any irrational number. Use Lemma4.21 to show that the image set γ(R) is dense in T2.


Solution. Given ε > 0, Lemma 4.21 guarantees the existence of m,n ∈Z such that |mα− n| < ε. Let β = mα− n and note that

e2πimα = e2πi(mα−n) = e2πiβ.

Given any open set V ⊂ S1 there is (a, b) ⊂ R such that

{e2πix | x ∈ (a, b)} ⊂ V.

Taking ε = b− a and letting m,n, β be as above, we have |β| < ε andit follows that there is k ∈ Z such that kβ ∈ (a, b). In particular, wehave

e2πikmα = e2πikβ ∈ V.Now we turn our attention to the curve γ : R→ T2. Recall that {U×V |U, V ⊂ S1 are open} is a basis for the topology on T2. In particular,given any open U, V ⊂ S1 we can choose s ∈ R such that e2πis ∈ Uand then use the above argument to find k ∈ Z such that

e2πikmα ∈ e−2πisαV = {e−2πisαz | z ∈ V }.Then e2πi(s+km)α ∈ V , and we note that e2πikm = 1, so

γ(s+ km) = (e2πi(s+km), e2πi(s+km)α) = (e2πis, e2πisαe2πikmα) ∈ U × V.This holds for arbitrary U, V , so γ(R) is dense in T2.

HW 6, #1. Lee, Problem 5-4. Let β : (−π, π) → R2 be the smoothcurve given by β(t) = (sin 2t, sin t). (See Example 4.19.) Show thatthe image of this curve is not an embedded submanifold of R2. Becareful: this is not the same as showing that β is not an embedding!

Solution. Let S be the image of the curve. Let U be a small nbhd of 0in R2, so S∩U is open in S (with the subspace topology). Whenever Uis small enough, (S∩U)\0 has four connected components. Thus S∩Ucannot be homeomorphic to any open ball in Rn, because an open ballwith a point removed has two connected components if n = 1, and onlyone otherwise. Thus S with the subspace topology is not a topologicalmfd.


HW 6, #4. Lee, Problem 5-18(a). Suppose M is a smooth manifoldand S ⊂M is a smooth submanifold. Show that S is embedded if andonly if every f ∈ C∞(S) has a smooth extension to a neighbourhoodof S in M .

Solution. (⇒). Suppose S is embedded; we prove every f ∈ C∞(S)can be extended to F ∈ C∞(U) for some open U ⊃ S. Since S isembedded it satisfies the local k-slice condition, so given any p ∈ Sthere is a slice chart (Up, ϕp) with p ∈ Up. Given f ∈ C∞(S) andp ∈ S, let fp ∈ C∞(Up) be the function whose coordinate representationrelative to (Up, ϕp) is

fp(x1, . . . , xn) = f(x1, . . . , xk).

Note that fp|S∩Up = f |S∩Up . Let U =⋃p Up so that U is an open

submanifold of M . Let {ψp}p∈S be a partition of unity subordinate tothe open cover (Up)p∈S of U . Then the product ψpfp gives a smoothfunction supported in Up, which we can take to be defined on U bysetting it equal to zero outside of Up. Define F : U → R by F =∑

p∈S ψpfp. Then F is smooth since the sum is finite at every point,

and for all q ∈ S we have F (q) =∑

p ψp(q)fp(q) =∑

p ψpf(q) = f(q).

(⇐). Suppose every f ∈ C∞(S) has a smooth extension to a neighbour-hood of S in M . We show that S is embedded by verifying the localk-slice condition. Given p ∈ S there is a nbhd V 3 p in S (with respectto the topology making S an immersed submfd) such that ι : V → Mis an embedding. (This is a general property of immersions.) In par-ticular, there is a chart U around p in which V ∩ U is a k-slice.

Let f ∈ C∞(S) be a bump function supported in V , with f(p) > 0. Bythe hypothesis there is an open (in M) set W ⊃ S and F ∈ C∞(W )such that F |S = f . Because F is continuous, W ′ := {q ∈ W | F (q) >0} is an open set. Moreover, W ′ ∩ S ⊂ supp(f) ⊂ V . Restricting thechart U to U ∩W ′, we see that U ∩W ′ ∩ S ⊂ U ∩W ′ ∩ V is a k-slice.


HW 6, #5. Lee, Problem 5-20. Show by giving a counterexample thatthe conclusion of Proposition 5.37 may be false if S is merely immersed.That is, give an example of a smooth immersed submanifold S ⊂ Mand p ∈ S for which TpS does not coincide with {v ∈ TpM | vf =0 whenever f ∈ C∞(M) and f |S = 0}.

Solution. Here is one example: let S ⊂ R2 be the ‘figure eight’from #1 of this homework assignment. Let S be the same set with thetopology and smooth structure corresponding to β(t) = (− sin 2t, sin t).Then T0S ⊂ R2 is the line spanned by (2, 1), while T0S is the linespanned by (−2, 1). Let v, v be the derivations on C∞(R2) correspond-ing to directional derivatives in these two directions. Note that bothv, v annihilate any f ∈ C∞(R2) with f |S = 0, but only one of themcan be contained in T0S.

Another example is the line with irrational slope on the torus. Let Sbe this submanifold and note that TpS is one-dimensional. But anysmooth function vanishing on S must vanish on T2 since S is dense,and so vf = 0 for every v ∈ TpT2.

HW 7, #1. Lee, Problem 5-6. Suppose M ⊂ Rn is an embedded m-dimensional submanifold, and let UM ⊂ TRn be the set of all unittangent vectors to M :

UM := {(x, v) ∈ TRn | x ∈M, v ∈ TxM, |v| = 1}.It is called the unit tangent bundle of M . Prove that UM is an em-bedded (2m− 1)-dimensional submanifold of TRn ≈ Rn × Rn.


Solution. Because M is embedded it satisfies the local m-slice condi-tion: for every x ∈M there is a chart (U,ϕ) around x with ϕ(U∩M) ={x ∈ ϕ(U) | xm+1 = · · · = xn = 0}. Let (U , ϕ) be the associated chartfor TRn ≈ Rn × Rn, so U = π−1(U), and note that ϕ(U ∩ TM) ={(x, v) ∈ Rn × Rn | xm+1 = · · · = xn = 0, vm+1 = · · · = vn = 0}, soTM satisfies the local 2m-slice condition. This shows that TM is a2m-dimensional embedded submanifold of R2n.

Now consider the function f : R2n → R defined by f(x, v) = |v|2 =∑ni=1(vi)2. Note that f is smooth, hence f restricts to a smooth func-

tion TM → R. Moreover, UM = f−1(1), so to show that UM is anembedded submanifold of TM (and hence also of R2n), it suffices toshow that f |TM has rank 1 at every (x, v) ∈ UM .

To this end, it is enough to produce a curve γ : (−ε, ε) → TM withγ(0) = (x, v) such that df(x,v)γ

′(0) 6= 0. Let γ(t) = (x, (1 + t)v), then

df(x,v)γ′(0)(Id) = (f ◦ γ)′(0) = d

dt|(1 + t)v|2|t=0 = 2. This shows that

f |TM has rank 1 at every (x, v) ∈ UM , hence UM is a codimension 1embedded submanifold of TM , which completes the proof.

HW 7, #3. Lee, Problem 7-1. Show that for any Lie group G, themultiplication map m : G×G→ G is a smooth submersion. Hint: uselocal sections

Solution. Given g ∈ G, define a map σg : G → G × G by σg(h) =(hg−1, g), so that m ◦ σg(h) = hg−1g = h. Thus σg is a smooth sectionof m. Given any (h, g) ∈ G×G, we have σg(hg) = (hgg−1, g) = (h, g),so (h, g) is in the image of the smooth section σg. This shows that mis a smooth submersion.


HW 7, #5. Lee, Problem 7-11. Considering S2n+1 as the unit spherein Cn+1, define an action of S1 on S2n+1, called the Hopf action, by

z · (w1, . . . , wn+1) = (zw1, . . . , zwn+1).

Show that this action is smooth and its orbits are disjoint unit circlesin Cn+1 whose union is S2n+1.

Solution. Given any z, the map (w1, . . . , wn+1) 7→ (zw1, . . . , zwn+1)is smooth because multiplication is smooth, and the orbits form a par-tition of S2n+1 by basic properties of group actions. The only real taskis to show that every orbit is a unit circle. To this end we observe thatgiven (w1, . . . , wn+1) ∈ S2n+1 ⊂ Cn+1, the map F : C→ Cn+1 given byF (z) = (zw1, . . . , zwn+1) is linear and 1-1, so its image is a subspaceof Cn+1 with real dimension 2. The orbit of (w1, . . . , wn+1) is the in-tersection of that image with S2n+1; that is, it is the intersection of aplane with S2n+1, which is a unit circle.

HW 8, #2. Lee, Problem 8-21. Prove that up to isomorphism, thereare exactly one 1-dimensional Lie algebra and two 2-dimensional Liealgebras. Show that all three of these algebras are isomorphic to Liesubalgebras of gl(2,R).

Solution. Let g be a one-dimensional Lie algebra and let q ∈ g benonzero. Define a linear map f : g → gl(2,R) by f(cq) = ( c 0

0 0 ). Notethat [q, q] = −[q, q] = 0 by anti-symmetry and hence [cq, c′q] = 0 forall c, c′ ∈ R by bi-linearity. Thus g is abelian and f is a Lie algebrahomomorphism. The image of f is a 1-dimensional subspace of gl(2,R)in which all matrices commute, hence f is a Lie algebra isomorphismonto its image ( R 0

0 0 ), so all 1-dimensional Lie algebras are abelian andare isomorphic to this one.


Now suppose g is a two-dimensional abelian Lie algebra. Let v, w ∈ gbe linearly independent and define f : g → gl(2,R) by f(v) = ( 1 0

0 0 )and f(w) = ( 0 0

0 1 ). Let h be the image of f , then h is a two-dimensionalabelian Lie subalgebra of gl(2,R) and f : g→ h is a Lie algebra isomor-phism. Thus every two-dimensional abelian Lie algebra is isomorphicto h, the Lie algebra of diagonal matrices in gl(2,R).

Finally, suppose g is a two-dimensional non-abelian Lie algebra. Thenthere are v, w ∈ g such that [v, w] 6= 0. As above, this implies that v, ware linearly independent. Thus they form a basis for g. Let u = [v, w]and note that one of the sets {u, v} and {u,w} is linearly independent(otherwise v, w are scalar multiples of each other). Without loss of gen-erality assume u, v are linearly independent, hence a basis. Moreover,writing u = av + bw we have [u, v] = [av + bw, v] = b[w, v] = −bu.

Let x = −v/b (note that b 6= 0 since g is non-abelian), then [u, x] =−b−1[u, v] = u. Define f : g → gl(2,R) by f(u) = ( 0 1

0 0 ) and f(v) =( 0 0

0 1 ), and note that [f(u), f(v)] = f(u) = f([u, v]) so f is a Lie algebrahomomorphism. In particular it is a Lie algebra isomorphism betweeng and h := f(g) since f is 1-1. Thus every two-dimensional non-abelianLie algebra is isomorphic to h = {( 0 a

0 b ) | a, b ∈ R}.

HW 8, #3. Lee, Problem 8-28. Considering det : GL(n,R) → R∗ asa Lie group homomorphism, show that its induced Lie algebra homo-morphism is Tr: gl(n,R) → R. Hint: see Problem 7-4, which was #4from the last assignment.

Solution. Recall that gl(n,R) = M(n,R) with the commutator bracket.This is identified with TInGL(n,R) by identifying A ∈ gl(n,R) with thetangent vector at t = 0 to the curve γA(t) = In + tA, which is con-tained in GL(n,R) for sufficiently small values of t. Then TInGL(n,R)is identified with the space of left-invariant vector fields on GL(n,R) by


associating to v ∈ TInGL(n,R) the left-invariant vector field (vL)X =d(LX)In(v) it generates (here X ∈ GL(n,R)).

Similarly, the space of left-invariant vector fields on R∗ is identifiedwith T1R∗ by associating v ∈ T1R∗ with (vL)x = d(Lx)In(v). In thecanonical chart on R∗, where TxR∗ is identified with R via the basis∂∂x

, this gives (vL)x = xv for x ∈ R∗ and v ∈ R.

Up to the identification of left-invariant vector fields with their rep-resentatives at the identity, the Lie algebra homomorphism inducedby det is, by definition, the linear map TInGL(n,R) → T1R∗ givenby d(det)In . It was shown on the last assignment that d(det)In(A) =Tr(A). Thus trace is the Lie algebra homomorphism induced by deter-minant.

HW 9, #1. Let V be a finite-dimensional vector space. Recall thatgiven two tensors ω ∈ T kV ∗ and η ∈ T `V ∗, the tensor product ω⊗ η ∈T k+`V ∗ is defined by

ω ⊗ η(v1, . . . , vk, vk+1, . . . , vk+`) = ω(v1, . . . , vk)η(vk+1, . . . , vk+`).

Let det ∈ T 2(R2)∗ be the 2-tensor defined by det(v, w) = det(v1 w1

v2 w2

)=

v1w2−v2w1, where v = viEi and w = wiEi. Recall that if (e1, e2) is thestandard basis for (R2)∗, then det = e1⊗ e2− e2⊗ e1. Determine (withproof) whether or not there are 1-tensors (covectors) ω, η ∈ T 1(R2)∗ =(R2)∗ such that det = ω ⊗ η.

Solution. If det = ω ⊗ η then we would have

0 = det(E1, E1) = ω(E1)η(E1),

0 = det(E2, E2) = ω(E2)η(E2).

Since det 6= 0 we have ω 6= 0 and η 6= 0, hence either ω(E1) = η(E2) = 0or ω(E2) = η(E1) = 0. But then we either have det(E1, E2) = 0 ordet(E2, E1) = 0, a contradiction.


HW 9, #2. Let (e1, e2, e3) be the standard dual basis for (R3)∗. Showthat e1 ⊗ e2 ⊗ e3 is not equal to a sum of an alternating tensor and asymmetric tensor.

Solution. Suppose ω, η ∈ T 3(R3)∗ are alternating and symmetric,respectively. Then since (1, 2, 3) 7→ (2, 3, 1) is an even permutation wehave

ω(E1, E2, E3) = ω(E2, E3, E1),

η(E1, E2, E3) = η(E2, E3, E1),

and hence the same symmetry holds for ω+η. The result follows observ-ing that (e1⊗e2⊗e3)(E1, E2, E3) = 1 but (e1⊗e2⊗e3)(E2, E3, E1) = 0.

HW 9, #4. Let ω1, . . . , ωk be covectors on a finite-dimensional vectorspace V .

(a) Lee, Problem 14-1. Show that ω1, . . . , ωk are linearly dependentif and only if ω1 ∧ · · · ∧ ωk = 0.

Solution. (⇒). Without loss of generality suppose ω1 =∑ki=2 aiω

i for some ai ∈ R. Then

ω1 ∧ · · · ∧ ωk =k∑i=2

aiωi ∧ ω2 ∧ · · · ∧ ωk = 0

because each term has a repeated ωi.

(⇐). Suppose ω1, . . . , ωk are linearly independent, then theyextend to a basis ω1, . . . , ωn for V ∗. Let v1, . . . , vn be the ba-sis for V that is dual to this. (Formally, take the dual basis


for (V ∗)∗ corresponding to {ωi}, then use the canonical isomor-phism between V and (V ∗)∗.) Then (ω1∧· · ·∧ωn)(v1, . . . , vn) =1, hence ω1 ∧ · · · ∧ ωn 6= 0, hence ω1 ∧ · · · ∧ ωk 6= 0.

(b) Suppose ω1, . . . , ωk are linearly independent, and so is the col-lection of covectors η1, . . . , ηk ∈ V ∗. Prove that span(ω1, . . . , ωk) =span(η1, . . . , ηk) if and only if there is some nonzero real numberc such that ω1 ∧ · · · ∧ ωk = c η1 ∧ · · · ∧ ηk.

Solution. (⇒). Since the spans are the same there are aij ∈ Rsuch that ωi = aijη

j for every i. Then we get

ω1 ∧ · · · ∧ ωk = (a1j1ηj1) ∧ · · · ∧ (akjkη

jk) = AJηJ

where for a multiindex J = (j1, . . . , jk) we write AJ = a1j1· · · akjk

and ηJ = ηj1 ∧ · · · ejk , and the sum is over all multi-indices (notjust increasing ones). Note that the wedge product vanisheswhenever J contains a repeated index, so this is actually a sumover all permutations, and if J is a permutation of {1, . . . , k}then the wedge product is a scalar multiple of η1 ∧ · · · ∧ ηk. Itfollows that ω1 ∧ · · · ∧ ωk = cη1 ∧ · · · ∧ ηk for some c ∈ R, andwe observe that c 6= 0 since ω1, . . . , ωk are linearly independenthence the left hand side is not 0, by (a).

(⇐). Suppose c ∈ R\0 is such that ω1∧· · ·∧ωk = cη1∧· · ·∧ηk.Then for every i we have

ω1 ∧ · · · ∧ ωk ∧ ηi = cη1 ∧ · · · ∧ ηk ∧ ηi = 0

where the last equality is because ηi is repeated. By part (a) weconclude that {ω1, . . . , ωk, ηi} is linearly dependent. Since theωj are linearly independent this implies that ηi ∈ span{ω1, . . . , ωk}.This holds for all i, hence span{η1, . . . , ηk} ⊂ span{ω1, . . . , ωk}.The reverse inclusion is similar.

selected hw solutions - uhclimenha/2015-spring-7350/hw-solutions.pdf · math 7350 selected hw...

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