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Differentiable Manifolds and LieGroups
Course notes for MA338, IISc Bangalore
Thomas Richard
January 24, 2014
These are the notes for the MA338 course I’ve given in 2013 on “Differentiable man-ifolds and Lie groups” at IISc Bangalore. These are far from being as polished andcomplete as I would have liked, but could still be of some use. Please feel free to reporttypos, misspellings, confusing statement at [email protected] the exercises, the *’s are a rough indication of difficulty.The prerequisites are some linear algebra, some multivariate calculus (the inverse
function theorem will be of constant use), and some point set topology (metric spaces,topological spaces, compactness, connectedness).
I don’t claim any novelty in the treatment I have made of this classical subject. I’vebeen working mainly with the textbooks [Lee02] and [Laf96] (sorry the second one isin French !). Other useful references (more advanced, and biased towards Riemanniangeometry) : [GHL04] and [Jos11]. About Lie groups, [Sti08] is really good, althoughwe won’t follow the same approach. An impressing collection of exercises can be foundin [nMM13]. The first volume of [Spi99] is also interesting, it covers all the topics we’lldiscuss here and takes great care to show how the various approaches (coordinate basedand geometric) link together. Although [Boo86] may feel slightly outdated, it is still aworthwhile reference.
I am really thankful to the students who followed this course, there questions andreactions during the lectures had a big influence on these notes.
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Contents
1 Smooth manifolds, tangent spaces and smooth maps 51.1 Quick review of differential calculus . . . . . . . . . . . . . . . . . . . . . 5
1.1.1 The differential of a map . . . . . . . . . . . . . . . . . . . . . . . 51.1.2 The inverse function theorem and its consequences . . . . . . . . 6
1.2 Smooth manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3.1 The Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.2 Submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.3.3 Product manifolds, tori . . . . . . . . . . . . . . . . . . . . . . . . 121.3.4 Quotients, tori (again) and projective spaces . . . . . . . . . . . . 13
1.4 Smooth maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4.1 Definition and first examples . . . . . . . . . . . . . . . . . . . . . 191.4.2 Building smooth maps . . . . . . . . . . . . . . . . . . . . . . . . 20
1.5 The tangent space at a point . . . . . . . . . . . . . . . . . . . . . . . . . 221.5.1 Some words about the tangent space to a submanifold in Rn . . . 221.5.2 Tangent vectors at a point to a manifold . . . . . . . . . . . . . . 231.5.3 Tangent vectors in coordinates . . . . . . . . . . . . . . . . . . . . 24
1.6 Differentiating maps between manifolds . . . . . . . . . . . . . . . . . . . 251.6.1 Definitions, first properties . . . . . . . . . . . . . . . . . . . . . . 251.6.2 Tangent vectors as derivations . . . . . . . . . . . . . . . . . . . . 27
1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2 Tangent bundle, Vector fields and Lie brackets 342.1 Vector fields in Rn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.1.1 General theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.1.2 Linear vector fields and matrix exponential . . . . . . . . . . . . . 36
2.2 Cut-off functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.3 From the tangent spaces to the tangent bundle . . . . . . . . . . . . . . . 38
2.3.1 TM as smooth manifold . . . . . . . . . . . . . . . . . . . . . . . 382.3.2 TM as a vector bundle . . . . . . . . . . . . . . . . . . . . . . . . 40
2.4 Vector fields on manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 422.5 Vector fields and global derivations . . . . . . . . . . . . . . . . . . . . . 442.6 Lie bracket of vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . 472.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
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3 Lie groups and homogenous spaces 523.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.2.1 Abelian examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.2.2 The 3-sphere as a Lie group . . . . . . . . . . . . . . . . . . . . . 523.2.3 Matrix groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.2.4 Transformation groups . . . . . . . . . . . . . . . . . . . . . . . . 53
3.3 Lie morphisms, Lie subgroups . . . . . . . . . . . . . . . . . . . . . . . . 533.4 Left invariant vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . 543.5 The Lie algebra of a Lie group . . . . . . . . . . . . . . . . . . . . . . . . 563.6 Homegeneous spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
3.6.1 Observations on topological groups . . . . . . . . . . . . . . . . . 613.6.2 Quotient of Lie groups . . . . . . . . . . . . . . . . . . . . . . . . 623.6.3 Homogeneous spaces . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4 Tensors, differential forms and Stokes Theorem 684.1 The cotangent bundle and 1-forms . . . . . . . . . . . . . . . . . . . . . . 68
4.1.1 The cotangent bundle . . . . . . . . . . . . . . . . . . . . . . . . 684.1.2 Differential 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.2 A little tensor algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.2.1 Tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.2.2 Alternating forms and wedge products . . . . . . . . . . . . . . . 70
4.3 Differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724.3.1 The differential of a p-form. . . . . . . . . . . . . . . . . . . . . . 744.3.2 Lie derivative and Cartan’s formula . . . . . . . . . . . . . . . . . 764.3.3 Closed forms, exact forms and Poicaré lemma . . . . . . . . . . . 78
4.4 Integration of differential forms . . . . . . . . . . . . . . . . . . . . . . . 804.4.1 Orientability and volume forms . . . . . . . . . . . . . . . . . . . 804.4.2 The integral of an n-form . . . . . . . . . . . . . . . . . . . . . . 83
4.5 Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.5.1 Regular domains . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.5.2 Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 884.5.3 Some vector calculus revisited . . . . . . . . . . . . . . . . . . . . 90
4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
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1 Smooth manifolds, tangentspaces and smooth maps
1.1 Quick review of differential calculus
1.1.1 The differential of a map
Definition 1.1.1. Let (E, ‖ ‖E) and (F, ‖ ‖F ) be two normed vector spaces, U ⊂ E anopen set, and f : E → F . f is said to be differentiable at a ∈ U if there exists a(continuous) linear application L : E → F such that :
f(a+ h) = f(a) + Lh+ o(h)
where ‖o(h)‖F‖h‖E
goes to 0 as h goes to zero.The linear map L is the differential of f at a, and will be denoted by Daf .
Remark 1.1.2. If E and F are finite dimensional, all linear maps are continuous and allnorms are equivalent, so the choice of the norms doesn’t matter. Although differentialcalculus in infinite dimension has a number of nice application, we won’t use it, exceptmaybe in a few optional exercises.From now on, E and F are assumed to be finite dimensional.
Remark 1.1.3. The definition implies that if f is differentiable at a then f is continuousat a.To make the link with the derivative of function of a real variable, consider a function
f : Rn → R which is differentiable at a ∈ Rn, and curve c : R→ Rn, then the derivativeof f c at 0 can be computed in the following way :
f(c(h))− f(c(0))
h=
1
h
(Daf(c(h)− c(0)) + o(c(h)− c(0))
)= Daf
(c(h)− c(0)
h
)+
1
ho(c(h)− c(0)
)t→0−−→ Daf(c′(0)) since ‖c(h)− c(0)‖ ≤ Ch.
So (f c)′(0) = Daf(c′(0)). In other words, when one evaluates the differential of v onsome vector v, one gets the variation of f along a curve that has v as a tangent vector.Example 1.1.4. Any (continuous) linear map L : E → F is differentiable, and for any
a ∈ E, Daf = L.
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Example 1.1.5. Let B : E1 × E2 → F be a bilinear map, then B is differentiable andDa1,a2B(h1, h2) = B(a1, h2) +B(h1, a2). Similar formulas hold for multilinear maps.
Example 1.1.6. det : Mn(R)→ R is differentiable, and :
DA det(H) = trace(AH)
where A is the cofactor matrix of A.Example 1.1.7. inv : GLn(R)→ GLn(R) is differentiable and :
DA inv(H) = −A−1HA−1.
Assume that E = Rn and F = Rm and write f = (f 1, · · · , fm) the components off (each of the fi is a function of (x1, · · · , xn) ∈ Rn). Then the matrix of Daf in thecanonical basis of Rn and Rm is :
∂f1
∂x1(a) · · · ∂f1
∂xn(a)
......
∂fm
∂x1(a) · · · ∂fm
∂xn(a)
.
The matrix above is called the Jacobian matrix of f at a. In order to keep notationlight, we will write ∂if j for ∂fj
∂xi.
Assume that f : U → F is differentiable at every x ∈ U . Since L(E,F ) is againa normed vector space, it makes sense to ask wether the application Df : U 3 x 7→Dxf ∈ L(E,F ) is continuous. If this holds f is said to be C1. Moreover, one can askif Df itself is differentiable. If Df is differentiable at every x ∈ U , it defines a mapD2f : U → L(E,L(E,F )), if this map is continuous, f is said to be C2. Similarly, onedefines the notion of Ck function for any integer k, and a function is C∞ (or smooth) ifit is Ck for every k.One has the following useful characterization of Ck maps :
Proposition 1.1.8. f : Rm ⊃ U → Rn is Ck is and only if it has Ck partial derivatives.
The differential of functions built by composition is easily computed using the chainrule :
Theorem 1.1.9. Let f : Rn ⊃ U → Rm differentiable at a ∈ U , g : Rm ⊃ U ′ → Rp
differentiable at f(a). Then x 7→ (g f)(x) is differentiable at a and :
Da(g f) = Df(a)g Daf.
1.1.2 The inverse function theorem and its consequences
Definition 1.1.10. Let U ⊂ Rn, V ⊂ Rm be open sets, f : U → V is a Ck-diffeomorphism(k ≥ 1) if it is a Ck bijection and it inverse f−1 : V → U is Ck on V .
Proposition 1.1.11. If f : U → V is a Ck-diffeomorphism (k ≥ 1) , then for anya ∈ U Daf : Rn → Rm is an invertible linear map. In particular n = m.
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Proof. Let g = f−1, then g f = idRn . Since g and f are differentiable, we can applythe chain rule for any a ∈ U :
idRn = Da(g f) = Df(a)g Daf,
thus Daf is invertible.
Remark 1.1.12. This also shows that we could have required in the definition of a Ck
diffeomorphism f−1 to be merely differentiable instead of Ck. Let us show this for k =1. Set g = f−1, since g is differentiable, it is continuous, hence g is a homeomorphism.Its differential is given by : Dbg = (Dg(b)f)−1. Since f is C1, a 7→ Daf is C0. Sinceg is continuous, b 7→ Dg(b)f is continuous. And using that taking the inverse of aninvertible linear map is a smooth operation, b 7→ Dbg = (Dg(b)f)−1 is continuous,hence g is C1.The inverse function theorem is a partial converse to this statement :
Theorem 1.1.13. Let f : U → Rn be a Ck function and a ∈ U such that Daf isinvertible. Then there exists an open neighborhood of a U ′ ⊂ U such that f : U ′ → f(U ′)is a Ck-diffeomorphism.
It has the drawback of being a local statement. We say that f : U → Rn is a localdiffeomorphism if any x ∈ U has an open neighborhood Ux such that f : Ux → f(Ux)is a diffeomorphism. The inverse function theorem implies the following :
Theorem 1.1.14. f : U → Rn is a local diffeomorphism if and only if for any a ∈ U ,Daf is invertible.
The following corollary is useful in the application :
Corollary 1.1.15. f : U → V is a bijection and for any a ∈ U , Daf is invertible, thenf is a diffeomorphism.
It is legitimate to ask wether something still holds when one replaces the hypothesis,Daf invertible by Daf injective or surjective. A Ck map f : Rn ⊃ U → Rm is called aCk submersion if at any a ∈ U , Daf is surjective. A Ck map f : Rn ⊃ U → Rm iscalled a Ck immersion if at any a ∈ U , Daf is injective.
Theorem 1.1.16. Let f : Rn ⊃ U → Rm be a Ck submersion, then for any a ∈ U ,one can find an neighborhood of a U ′ ⊂ U , a open set V ⊂ Rn and a diffeomorphismψ : V → U ′ such that, for any x ∈ V :
f(ψ(x1, · · · , xn)) = (x1, · · · , xm).
In other words, up to a change of coordinates in the source, a submersion is locallya linear projection. Since linear projection are open maps, and being an open map is alocal property property, the previous theorem shows that any submersion is open1.
1a map f between topological space is open if the image of any open subset of the source under f isan open subset of the target.
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Theorem 1.1.17. Let f : Rn ⊃ U → Rm be a Ck immersion, then for any a ∈ U , onecan find an neighborhood of a U ′ ⊂ U , a open set W ⊃ f(U ′) and a diffeomorphismϕ : W → ϕ(W ) ⊂ Rm such that, for all x ∈ U ′ :
ϕ(f(x1, · · · , xn)) = (x1, · · · , xn, 0, · · · , 0).
In this case, f is a linear injection up to a change of coordinates in the image space.A d-dimensional submanifold of Rn is a set that locally looks like a d-dimensional
linear subspace of Rn :
Definition 1.1.18. S ⊂ Rn is a Ck submanifold of dimension d if, for any a ∈ S,one can find a neighborhood U of a in Rn and Ck diffeomorphism Φ : U → V ⊂ Rn suchthat :
1. Φ(a) = 0,
2. ψ(S ∩ U) = Rd ∩ V .
S
U ⊂ RnΦ
Rn−d
Rd
V ⊂ Rn
Figure 1.1: The definition of a submanifold
Submanifolds can be characterized in terms of immersions or submersions in the fol-lowing way :
Theorem 1.1.19. Let S ⊂ Rn. Then the following statements are equivalent :
1. S is a submanifold of dimension d.
2. for any a ∈ S, there exists an open set U ⊂ Rn which contains a and a submersionf : U → Rn−d such that U ∩ S = f−1(0).
3. for any a ∈ S, there exists an open set U ⊂ Rn which contains a, an open setW ⊂ Rd and an immersion g : W → U such that U∩S = f(W ) and g : W → U∩Sis an homeomorphism.
The second property says that S is locally the set of solutions of a system of d equa-tions, the third one says that S admits local parametrizations.
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1.2 Smooth manifolds
Definition 1.2.1. An n-dimensional topological manifold is topological space Mwhich satisfies :
1. M is Hausdorff (for any two points x, y ∈ M , one can find disjoint open sets Uxand Uy such that x ∈ Ux and y ∈ Uy),
2. M is second countable (it is covered by a countable family of compact subsets),
3. each point m ∈ M has a neighborhood homeomorphic to a open set in Rn (M issaid to be locally euclidean).
A local chart of M is an open subset U ⊂ M together with an homeomorphismϕU : U → U ′ = φ(U) where U ′ is an open subset of Rn. One can write φU in coordinatesφU = (x1, · · · , xn), each xi is a function from U to R, the xi’s are coordinate functionson U . U is sometimes called a coordinate patch.An atlas is a collection of local charts A = (Uα, ϕUα)α∈I such that
⋃α∈I Uα = M .
Let (U,ϕU) and (V, ϕV ) be two local charts which overlap, that is U ∩V is not empty.One can then define a map :
ϕUV : ϕU(U ∩ V )→ ϕV (U ∩ V )
x 7→ ϕV (ϕ−1U (x)).
This map is called the coordinate change, or transition function from U to V .
Definition 1.2.2. Two local charts (U,ϕU) and (V, ϕV ) are said to be Ck compatibleif the transition function ϕUV is a Ck diffeomorphism.An atlas A = (Uα, ϕUα) is said to be Ck if any two charts (U,ϕU) and (V, ϕV ) in A
are Ck compatible.Two Ck atlases A and A′ are said to be compatible if there union is a Ck atlas.
M
U
V
ϕU
ϕV
ϕUV = ϕV ϕ−1U
Figure 1.2: Transition functions
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Example 1.2.3. (R, idR) is a smooth atlas of R (as soon as an atlas has only one chart,it is automatically smooth). (R, x 7→ x3) is also a smooth atlas of R. However, thesetwo atlas are C0 compatible but not Ck compatible for k ≥ 1.
Example 1.2.4. Let U ⊂ Rn be an open set, then (U, idU) is a smooth atlas of U . (U, idU)is not the only smooth atlas that one can use to define a smooth manifold structureon U , any open covering of U would do. However, note that all these atlases arecompatible.This raises the question : When should two smooth atlases A and A′ on a given
topological manifold M be considered the same ? The answer lies in the followingtheorem :
Theorem 1.2.5. Any Ck atlas A is contained in a unique maximal Ck atlas A. More-over, if A and A′ are Ck compatible, then A = A′.
Proof. Consider the set of all local charts (U, phU) which are Ck compatible with A,this is, by definition, a topological atlas A. It is then a routine check to verify that Ais indeed a Ck atlas. A is Ck compatible with A by construction, and another routinecheck shows that A is Ck compatible with A′. More details can be found in [Boo86].
This gives rise to the following definition :
Definition 1.2.6. Given a topological manifold M , a Ck structure on M is a maximalCk atlas A. A Ck manifold is a topological manifold with a Ck structure.
The previous theorem implies that given one Ck atlas on M it defines a unique Ck
structure on M , even though the particular atlas that we chose at the beginning is farfrom being unique.Remark 1.2.7. Just by changing the requirements on the transition functions, one can
define other structures on topological manifolds : there are real-analytic, holomor-phic, Lipschitz manifolds...As we will see, any construction or property which is local and well behaved under
diffeomorphism can be defined for smooth manifolds. For instance:
Definition 1.2.8. LetM be a smooth n-manifold and S ⊂M , then S is a submanifoldof dimension k in M if and only if for any chart (U,ϕU), ϕU(S ∩ U) is a submanifoldof dimension k in ϕ(U) ⊂ Rn.
1.3 Examples
1.3.1 The Sphere
The sphere Sn is defined by :
Sn =x ∈ Rn+1
∣∣ ‖x‖2 = 1.
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The north pole N and the south pole S are the points with coordinates (0, · · · , 0, 1) and(0, · · · , 0,−1), we set UN = Sn\N and US = Sn\S. Remark that Sn = UN ∪ US.For convenience we will write generic elements of Rn+1 as x = (x, xn+1) with x ∈ Rn.The (equatorial) stereographic projection with respect to N is the map : gN : UN → Rn
O
N
S
x
x
xn+1
••gN(x)
gS(x)
Figure 1.3: Stereographic projections
defined by the following relation, gN(x) is the unique (check it!) intersection of the line(Nx) with the hyperplane H = x|xn+1 = 0 ' Rn. Similarly, gS : US → RN is definedby the relation that gS(x) is the intersection of the line (Sx) with H. One can showthat :
gN(x) =x
1− xn+1, gS(x) =
x
1 + xn+1,
which imply that gN and gS are homeomorphims.Then we show (by a direct calculation or by remarking that the triangles OgS(x)S
and ONgN(x) are homothetic) :
gS g−1N (x) =
x
‖x‖2
which is a smooth diffeomorphism of Rn\0. This shows that (UN , gN) and (US, gS)form a smooth atlas of Sn.In fact g−1
N and g−1S are parametrization of Sn as a submanifold of Rn+1 (they satisfy
the third point of Theorem 1.1.19 ). This in fact a general phenomenon, any submanifoldas a canonical manifold structure, as we will show in the next section.
1.3.2 Submanifolds
Proposition 1.3.1. A d-dimensional submanifold S ⊂ Rn has a natural manifold struc-ture.
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Proof. The topology on S induced by the standard topology on Rn makes S a topologicalmanifold.
Let us show that the inverse of the parametrizations of S provide a smooth atlas. LetU, V be two open sets of Rn and US = U ∩ S, VS = V ∩ S and assume we have twoparametrizations ψU : U ′ → US ⊂ U and ψV : V ′ → VS ⊂ V with U ′ and V ′ two openssets in Rd. We need two show that the two charts (U, ψ−1
U ) and (V, ψ−1V ) are compatible.
We have that ψ−1V ψU is an homeomorphism from ψ−1
U (US ∩ VS) to ψ−1V (US ∩ VS), we
need to show it is a Ck diffeomorphism. By the inverse function theorem, it is enoughto show that ψ−1
V ψU is invertible at each point.Let m ∈ US ∩ VS, W a neighborhood of m in Rn and Φ : W → W ′ ⊂ Rn a diffeomor-
phism such thatΦ(S ∩W ) = x ∈ Rn|xd+1 = · · · = xn = 0.
Consider the map Φ ψU as map from U to Rn. It is an immersion (the differential ofΦ is bijective, the differential of ψU is injective, so their composition is injective). NowΦ ψU is indeed valued in Rd, this implies that the differential of Φ ψU is in fact alinear map from Rd to Rd. Since it is injective and the dimensions are equal, it is a linearisomorphism, and thus Φ ψU is a diffeomorphism by the inverse function theorem.Writing ψ−V 1 ψU = (Φ ψU)−1 (Φ ψU) gives that the coordinate change ψ−1
V ψUis a Ck diffeomorphism.
1.3.3 Product manifolds, tori
Proposition 1.3.2. Let Mn1 and Mm
2 be two smooth manifolds2, then the product M1×M2 is an n + m smooth manifold whose smooth structure is given by the charts (U1 ×U2, ϕU1 × ϕU2) where (Ui, ϕUi) is a chart of Mi, and
ϕU1 × ϕU2(m1,m2) = (ϕU1(m1), ϕU2(m2)) ∈ Rn × Rm.
Proof. Let’s first check that this atlas makes M1 × M2 a topological manifold. Pickany (m1,m2) ∈ M1 × M2, and charts (Ui, ϕUi) around mi on Mi. The U1 × U2 isa neighborhood of (m1,m2) which is homeomorphic to ϕU1(U1) × ϕU2(U2) which is aproduct of an open set in Rn with an open set of Rm, hence an open set in Rn+m.We now check that the atlas is smooth, let U = U1×U2 and V = V1×V2. First, since
(U1 × U2) ∩ (V1 × V2) = (U1 ∩ V1) × (U2 ∩ V2), we only have something to check whenU1 ∩ V1 6= ∅ and U2 ∩ V2 6= ∅. In such a case, the transition map ϕUV : Rn+m → Rn+m isjust given by :
ϕUV (x, y) = (ϕU1V1(x), ϕU2V2(y)).
It is easy to see that ϕUV is smooth, bijective and has a smooth inverse ϕV U , thus ϕUVis a smooth diffeomorphism.
An important example is the n-dimensional torus Tn, which is the product of n copiesof S1.
For n = 2, the smooth manifold T2 is actually homeomorphic to the surface of adoughnut in R3.
2Writing Mn is a shorthand to indicate that M has dimension n.
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1.3.4 Quotients, tori (again) and projective spaces
The quotient topology
First, let’s talk a little bit about the quotient topology.Recall that a relation R on a set X is said to be an equivalence if is :
• reflexive: xRy ⇔ xRy.
• symmetric: ∀x ∈ X xRx.
• transitive: xRy and yRz ⇒ xRz.
The equivalence class of x ∈ X is the set [x] = y ∈ X|xRy. The set of all equivalenceclasses forms a partition of X is called the quotient space XR and the quotient map isjust the map π : X → X/R which maps x to its class [x].
Definition 1.3.3. Let X be a topological space and R be an equivalence relation on X,the quotient topology on X/R is the finest topology on X/R that makes the canonicalprojection π : X → X/R continuous.
In other words, U ⊂ X/R is open if and only if π−1(U) ⊂ X is open.If we want to build manifolds using this construction, we need to ensure that X/R is
Hausdorff. This is not automatic : take R and define xRy if x and y are both positive.Then any neighborhood of the class of 1 in R/R contains 0. However we have the simplecriterion :
Proposition 1.3.4. If π : X → X/R is open and the graph of R3 is closed then X/Ris Hausdorff.
Proof. Let x = π(x) and y = π(y) be two distinct points in X/R. Since π(x) 6= π(y),(x, y) /∈ Γ(R) and, since Γ(R) is closed, we can find Ux and Uy, open sets containing xand y such that Ux×Uy ∩ Γ(R = ∅. This is shows that Ux = π(Ux) and Uy = π(Uy) aredisjoint sets in X/R which contain x and y. Since π is open, the sets are open. ThusX/R is Hausdorff.
Remark 1.3.5. Since the image of compact set under a continuous map between Haus-dorff spaces is compact, if, under the assumptions of the previous proposition, X ismoreover second countable, X/R is automatically second countable.
Back to tori
Consider the relation on Rn given by :
xRy ⇔ x− y ∈ Zn.
3defined by Γ(R) = (x, y) ∈ X|xRy
13
x1
x2
Z2
K
Figure 1.4: Rn/Zn
Then M = Rn/R is also the quotient of M = Rn/Zn of the group Rn by its subgroupZn. We will show that M is a topological manifold and can be endowed with a naturalsmooth structure. We will see in Section 1.4 that M is diffeomorphic to Tn.But first let’s try to get a more “hands on” feeling of the construction. Look at Figure
1.4. First note that π is a bijection when restricted to [0, 1)n, because quotienting by Zn islike taking the floor of each coordinate. Consider the closed cube K = [0, 1]n, we alreadyknow that that π is a bijection when restricted to the interior of K (the dashed squareon the figure). Now consider the ’left’ face x ∈ K|x1 = 0, any point (0, x2, · · · , xn) inthe left face will have the same image under π as the point (1, x2, · · · , xn) which lie onthe ’right’ face x ∈ K|x1 = 1. In other words the left and right faces of K are gluedby π. This also works along other coordinates xi. One can then say that M is obtainedfrom K by gluing opposite faces (the identification being given by translations of length1 along the coordinate axis), this is represented by the arrow tips on the side K on thefigure.
Proposition 1.3.6. M is a topological manifold.
Proof. We first show that M is Hausdorff and second countable. Since Rn is secondcountable, we only need to show that the hypothesis of the previous proposition aresatisfied. To see that π : Rn → M is open, consider an arbitrary open set U in Rn. Bydefinition of the quotient topology, π(U) is open if and only if π−1(π(U)) is open in Rn.We have that :
π−1(π(U)) =⋃p∈Zn
τp(U)
14
where τp is the translation by p in Rn. Since each of the τp(U) open (translations arehomeomorphism), π−1(π(U)) is open as a union of open sets. We now check that thegraph of R is closed. We have :
Γ(R) = (x, y) ∈ R2n|x− y ∈ Zn.
In other words, Γ(R) is the inverse image under the continuous application (x, y) 7→ x−yfrom R2n → Rn of Zn, which is closed. Hence Γ(R) is closed. This shows that M isHausdorff and second countable. In fact we have more than that : since M is the imageby π of K, M is compact.
We can now show that M is locally Euclidean. Consider x = π(x) ∈ M , and letB ⊂ Rn be an open ball of radius less than 1
2centered at x. Then the restriction π|B of
π to B is injective, and since π is open, π|B is also open and hence an homeomorphismon its image π(B), which is an open set containing x.
We now build an atlas of M . For all x ∈ Rn, denote by Bx of radius 12centered
at x. What we did in the previous paragraph shows that, setting ϕx = (π|Bx)−1 andUx = π(Bx), the collection (Ux, ϕx)x∈X is an atlas of M .
Proposition 1.3.7. The atlas (Ux, ϕx)x∈X is smooth.
Proof. Consider x, y ∈ Rn such that Ux ∩ Uy 6= ∅. We have the following fact : ifz ∈ Ux ∩ Uy there exists zx ∈ Bx such that π(zx) = z and zy ∈ Bx such that π(zy) = z.Since π(zx) = π(zy), p = zx − zy ∈ Zn.Consider the coordinate change from Bx to By ϕUxUy = ϕUy ϕ−1
Ux. We have that
ϕUxUy(zx) = zy. Moreover, since Bx and By are open, we can find a small ε such thatB(zx, ε) ⊂ Bx and B(zy, ε) ⊂ By. We will show that ϕUxUy is equal to τp on B(zx, ε).Let m ∈ B(zx, ε), then m′ = ϕUxUy(m) = (π|By)(π(m)). So m′ is characterized by :
m′ −m ∈ Zn and m′ ∈ By.
Since p+m satisfy these conditions, m′ = p+m. Thus ϕUxUy is equal to τp on B(zx, ε),and is therefore smooth.
The projective spaces RPn and CPn
If E is a vector space, the projectivization P(E) of E is the space of vectorial lines4 in E.Here we will see that the projectivizations RPn and CPn of Rn+1 and Cn+1 have naturalsmooth manifold structures.
To build RPn as quotient, we define, on X = Rn+1\0 an equivalence relation R by :
xRy ⇔ x and y are on the same (vectorial) line ⇔ ∃λ ∈ R∗ y = λx.
Then the equivalence class of any point x ∈ X is the vectorial line through x (minus theorigin), so that X/R is the set RPn of vectorial line. The projection π : X → RPn is just
4A caveat here : a line is a 1-dimensional vector subspace of E, in particular, if one consider complexvector spaces, a line has real dimension 2.
15
the map which sends x to the line Rx (minus its origin). It is customary in projectivegeometry to denote the the image of x = (x1, · · · , xn+1) in RPn by [x] = [x1 : · · · : xn+1],the expression on right hand side of the previous inequality is called the homogenouscoordinates representation of the point [x] ∈ RPn. Note that [x1 : · · · : xn+1] and[λx1 : · · · : λxn+1] (λ 6= 0) represent the same point in RPn.
Proposition 1.3.8. RPn is Hausdorff and compact.
Rx
xRy
y
O
Figure 1.5: The complement of Γ(R) is open.
Proof. We want to check that RPn with the quotient topology is Hausdorff and secondcountable. Like for the case of the torus, for any open set U ⊂ X :
π−1(π(U)) =⋃λ∈R∗
λU.
Thus π−1(π(U)) is open as union of open sets (homotethies are homeomorphisms), andπ is open.
We want to see that the graph of R is closed. We have :
Γ(R) =
(x, y) ∈ X2∣∣∃λ ∈ R∗ y = λx
.
We show that Γ(R) is closed by showing that its complement is open. Let (x, y) /∈ Γ(R).This means that the lines Rx and Ry intersect only at the origin. It is clear that onecan find two neighborhoods Ux and Uy of x and y such that for any x′ ∈ Ux and y′ ∈ Uy,x′ and y′ are not on the same vectorial line (look at Figure 1.5) . Then Ux × Uy doesnot intersect Γ(R). Hence Γ(R) is closed and RPn is Hausdorff.
Since any vectorial line intersects the sphere Sn+1, RPn = π(Sn+1) is compact.
16
We will now build a smooth atlas for RPn. Set, for any integer between 1 and n+ 1 :
Ui =
[x] = [x1 : · · · : xn+1] ∈ RPn∣∣xi 6= 0
and :
Hi =x = (x1, · · · , xn+1) ∈ Rn+1
∣∣xi = 1.
Ui is open as it is the image under π of the open set x ∈ Rn+1|xi 6= 0 ⊂ X under πwhich is open. Define ψi = π|Hi .
Proposition 1.3.9. For each i, ψi is a homeomorphism from Hi to Ui.
Proof. Set ϕi : Ui → Hi by :
ϕi([x1 : · · · : xn+1]) =
(x1
xi, · · · , x
i−1
xi, 1,
xi+1
xi, · · · , x
n
xi
).
ϕi is well defined because ϕi([λx]) = ϕi([x]) and has values in Hi. Moreover, it is easyto see that ϕi ψi = idHi and ψi ϕi = idUi . Hence ψi is a continuous bijection, andsince π is open, ψi is open and hence an homeomorphism.
We identify each Hi with Rn by x 7→ (x1, · · · , xi−1, xi+1, · · · , xn+1). This makes each(Ui, ϕi) a local chart. Moreover
⋃i Ui = RPn, so we have an atlas A = (Ui, ϕi)|1 ≤ i ≤
n+ 1.
Proposition 1.3.10. A is a smooth atlas.
Proof. Let (Ui, ϕi) and (Uj, ϕj) be two charts, by permuting the coordinates, we canassume that i = 1 and j = 2. We have that :
U1 ∩ U2 =
[x] = [x1 : · · · : xn+1] ∈ RPn∣∣x1 6= 0, x2 6= 0
,
hence:ϕ−1
1 (U1 ∩ U2) = x = (x2, · · · , xn+1) ∈ Rn|x2 6= 0
and:ϕ−1
2 (U1 ∩ U2) = x = (x1, x3, · · · , xn+1) ∈ Rn|x1 6= 0.
Now :
ϕ12(x2, · · · , xn+1) =ϕ2(ϕ−11 (x2, · · · , xn+1))
=ϕ2([1 : x2 : · · · : xn+1])
=
(1
x2,x3
x2, · · · , x
n+1
x2
)which is easily checked to be a smooth diffeomorphism whenever x2 6= 0.
17
Remark 1.3.11. The same construction works to define a smooth manifold structure on:
CPn = L|L ⊂ Cn+1 complex line,
seen as the quotient of Cn+1\0 by the equivalence :
xRy ⇔ ∃λ ∈ C∗ y = λx.
In fact we get more, the coordinate change will have the same form as the one inthe proof of Proposition 1.3.10, except that the real variables xi will be replaced bycomplex variables zi, and the coordinate change will therefore be a biholomorphism.CPn is the first example of a complex manifold.
R2
L
L′
z
y
L ∩ L′ ∈ RP2L ∩ L′ ∈ RP2
R2 ' z = 1 ⊂ RP2
L
L′
Figure 1.6: There are no parallels in RP2.
Remark 1.3.12. The embedding Hn+1 ' Rn → RPn allows one to see the n-dimensionalaffine space Rn as an open set in RPn. The points that are missed by the embeddingare called points at infinity of the image of Rn in RPn. In this sense, RPn is ageometric compactification of the affine space Rn.A projective line L in RP2 is the image under π of a vectorial plane P in Rn+1.
A point p in RP2 belongs to L if, as a line in Rn+1, p is included in P . Similarly,one can define a projective subspace of dimension k as the image under π of a k + 1dimensional vectorial subspace of Rn+1. It is easy to see that each k-dimensionalaffine subspace of Rn define a unique k-dimensional projective subspace of RPn.Consider the case n = 2. Then line and points in RP2 satisfy the following prop-
erties:
1. Given two distinct points, there exist a unique line which contains both.
2. Two distinct lines always intersect at a unique point.
This a geometry without parallels ! In fact, two affine lines which are parallel in R2
don’t intersect in R2 but have a unique intersection “at infinity” in RP2. See Figure1.6.
These are the first steps in projective geometry, one of the first example of non-euclidean geometry. A lot more can be found for instance in [Aud03] and [Ber09].
18
1.4 Smooth maps
1.4.1 Definition and first examples
The first advantage of smooth manifolds over topological manifolds is that they allowus to use calculus, but before really using calculus, we need to say what smooth mapsbetween manifolds are.
Definition 1.4.1. Let Mn and Np be two smooth manifolds, and f : M → N a map. fis said to be smooth if for any chart (U, φU) ofM , any chart (V, ϕV ) of N , ϕV f(ϕU)−1
is smooth as function from the open set of ϕU(f−1(V ) ∩ U) ⊂ Rn to Rk.
In other words, f is smooth if in the following commutative diagram:
M ⊃ U V ⊂ N
Rn ⊃ ϕU(U) ϕV (V ) ⊂ Rn
f
ϕV f ϕ−1U
ϕU ϕV
the bottom arrow is smooth. Note that the diagram slightly abuses notations becauseϕV f ϕ−1
U doesn’t have U as a source, but ϕU(f−1(V ) ∩ U). We will frequently makethis sort of abuse of to avoid cumbersome notations.Remark 1.4.2. It is easy to check that if f is smooth for charts in some smooth atlasesAM and AN on M and N , then it is smooth for every other choice of atlas. It isenough to check the definition for two smooth atlases AM and AN .
Remark 1.4.3. If N is Rk with its standard smooth structure, it is enough to check thatfor every chart (U,ϕU) of M , f (ϕU)−1 is smooth from U to Rk.We have the useful notion :
Definition 1.4.4. f : M → N is a diffeomorphism if f is bijective and f−1 : N →Mis smooth.
Any local property of smooth maps between open sets in Euclidean spaces can bedefined through charts, as a property of smooth maps between manifolds:
Definition 1.4.5. f : M → N is a submersion (resp. immersion) (resp. localdiffeomorphism) if for any chart (U, φU) of M , any chart (V, ϕV ) of N , ϕV f (ϕU)−1
is submersion (resp. immersion) (resp. local diffeomorphism) as a map between opensets in Euclidean spaces.
With these definitions, we have the following corollaries of Theorem 1.1.19:
Proposition 1.4.6. If f : M → N is a submersion, then, for any x ∈ N , f−1(x) is asubmanifold of M , whose codimension is the dimension of N .
19
Proposition 1.4.7. If f : M → N is an immersion and an homeomorphism from Mto f(M), then f(M) is a smooth submanifold of N , which is diffeomorphic to M .
Remark 1.4.8. In the last proposition, f is said to be a smooth embedding.We leave the proof as exercises, the guiding line should be: “ using properties of
submersion or immersion in charts, the theorem is reduced to the case where N is avectorial subspace of a vector space M , in which case it is trivial”.We now present some examples.
Example 1.4.9. The projection M1 ×M2 →M1 is a smooth submersion.Example 1.4.10. The projections Rn → Rn/Zn and Rn+1 → RPn built in Section 1.3.4
are also smooth submersions. In fact Rn → Rn/Zn is a local diffeomorphism.Example 1.4.11. The inclusion M1 →M1 ×M2 is a smooth embedding.
1.4.2 Building smooth maps
Proposition 1.4.12. if f : M → N and g : N → P are smooth, then g f : M → P issmooth.
Proof. Let a ∈M and (U,ϕU), (V, ϕV ), (W,ϕW ) be charts on M , N and P respectivelysuch that a ∈ U , f(a) ∈ V and g(f(a)) ∈ W . We want to show that ϕW (g f) ϕ−1
U
is smooth. We have the following commutative diagram :
U V W
ϕU(U) ϕV (V ) ϕV (W )
f g
ϕV f ϕ−1U ϕW f ϕ−1
V
ϕU ϕV ϕW
ϕW (g f) ϕ−1U
Thus ϕW (g f) ϕ−1U is smooth (where it is defined) as ϕV f ϕ−1
U and ϕW f ϕ−1
V are. A caveat here : what we have said only shows that ϕW (g f) ϕ−1U
is smooth on ϕU (U ∩ f−1 (g−1(W ) ∩ V )), rather than ϕU((g f)−1(W ) ∩ U). However,since smoothness is local, it is enough to check it in a neighborhood of a for every a ∈M ,and U ∩ f−1 (g−1(W ) ∩ V ) is a neighborhood of a.
Proposition 1.4.13. If N ⊂ M is a submanifold, then the inclusion N → M is asmooth immersion.
Proposition 1.4.14. If f : M → N is smooth, P ⊂ N is a submanifold of N andf(M) ⊂ P then f is smooth as a map from M to P .
20
Submersions are powerful tools to build smooth maps. If π : M → P is a submersion,the submanifold π−1(p) is called the fiber of π over p.
Theorem 1.4.15. Let π : M → P be a surjective submersion, and f : M → N be asmooth map which is constant on the fibers of π, then there exist a unique smooth mapf : P → N such that f = f π.
M
P
Nf
πf
Proof. Denote by dM , dP and dN the dimensions of M , N and P .First, pick p ∈ P and some m ∈M such that π(m) = p. Since f is constant on π−1(p),
the formula f(p) = f(m) gives a well defined map f : P → N .We now need to show it is smooth. First pick a chart (Um, ϕm) in M around m,
and a chart (Up, ϕp) around p, then ϕp π ϕ−1m is a submersion. Thus we can find a
diffeomorphism ψ : O → ϕm(Um) around ϕm(m) in RdM such that :
ϕp π ϕ−1m ψ(x1, · · · , xdM ) = (x1, · · · , xdP ).
By setting ϕm = ϕm ψ−1, we define a new chart (ϕm, Um) around m such that :
ϕp π ϕ−1m (x1, · · · , xdM ) = (x1, · · · , xdP ).
In these coordinates, the fiber in Um over some point ϕ−1p (x1, · · · , xdP ) in P is just :
ϕ−1m
((x1, · · · , xdP ) × RdM−dP
).
Pick any chart (UN , ϕN) around f(m) = f(p) in N . The fact that f is constant onthe fibers thus translates as that ϕN f ϕm doesn’t depend on xi for i > dP . Hencethere exist a smooth function f of dP variables with values in RdN such that:
ϕN f ϕ−1m (x1, · · · , xdM ) = f(x1, · · · , xdP ).
Moreover we have that:
ϕN f ϕ−1p (x1, · · · , xdP ) = ϕN f ϕ−1
m (x1, · · · , xdM ) = f(x1, · · · , xdP ).
This shows that f is smooth around p.
Example 1.4.16. The complex exponential map ϕ : R → S1 defined by ϕ(t) = eit is alocal diffeomorphism. Consider now f : R→ R a 2π periodic smooth function. f isconstant over the fibers of ϕ, and thus yields a smooth function f from S1 to R.
R
S1
Rf
ϕf
21
Example 1.4.17. We consider here the projective space RPn with its projection π :Rn+1\0 → RPn which is a submersion.Consider an invertible linear map L : Rn+1 → Rn+1, since L is invertible, Lv 6= 0
when v 6= 0. Therefore we can build the composition L′ = π L : Rn+1\0 → RPnwhich is smooth as a composition of smooth maps. Moreover, it is easy to see thatπ(Lv) = π(Lλv) for any nonzero λ. This implies that L′ is constant on the fibers ofπ, so it defines a unique smooth map : L = RPn → RPn.
Rn+1\0
RPn
Rn+1\0
RPn
L
π π
L
L′
In fact, L is a diffeomorphism, as is seen by making the same construction with L−1
and checking that the map we obtain is the reciprocal of L.These transformations are called homographies of RPn, and are the natural trans-
formations of projective geometry.
1.5 The tangent space at a point
In the preceding section, we have defined what smooth maps are, but haven’t say whattheir derivatives are. Before being able to do so, we need another construction.
The big idea of differential calculus is to approximate a map at a point by a linearmap. When f goes from Rn → Rk its differential at a point will be a linear map fromRn to Rk. However, when f is a map between two manifolds, if we want to approximatef by a linear map, we first need to understand between which vector spaces should bethe source and the target of such a linear map.
Remember that in the first section it was possible to compute the differential of amap: f : Rn → Rk at a by computing the derivative of the one variable functions f cfor every smooth curve c that pass through a. This is the idea we will follow to buildthe tangent space.
1.5.1 Some words about the tangent space to a submanifold inRn
Consider S ⊂ Rn a k-dimensional submanifold and m ∈ S. We define the tangentspace TmS to S at m is defined in the following way :
TmS =v ∈ Rn
∣∣∣∃c : R C∞−−→ S, c(0) = m, c(0) = v.
22
Lemma 1.5.1. TmS is a vectorial subspace of Rn.
Proof. To see that v ∈ TmS ⇒ λv ∈ TmS, we consider c(t) ∈ S such that c(0) = m andc(0) = v, and define c(t) = c(λt) ∈ S. Then ˙c(0) = λv.
It remains to show that v1, v2 ∈ TmS ⇒ v1 + v2 ∈ TmS. For this, we consider a localparametrization of S ψ : Rk → S such that ψ(0) = m. Let c1 and c2 be two curve in Ssuch that c1(0) = v1 and c2(0) = v2, and ci = ψ−1 ci. The chain rule implies :
D0ψ(
˙ci(0))
= ci(0) = vi.
Consider the curve c(t) = ψ(c1(t) + c2(t)). This is a curve in S that goes through m,and :
c(0) = D0ψ(
˙c1(0) + ˙c2(0))
= v1 + v2.
Thus v1 + v2 ∈ TmS.
Remark 1.5.2. This definition is not the most practical to use in applications. How-ever it is easily seen to be equivalent to the following more computation friendlycharacterizations :
• If f : Rn → Rn−k is a local equation of S around m, then TmS = ker(Dmf).
• If ψ : Rk → S is a local parametrization of S such that ψ(0) = m, then TmS =D0ψ(Rk).
Example 1.5.3. The tangent space to Sn at v is the orthogonal to v.
1.5.2 Tangent vectors at a point to a manifold
Let M be a smooth manifold and m ∈ M , we define CM(m) to be the set of smoothcurves c : R→M such that c(0) = m.Two curves c1, c2 ∈ CM(m) are said to be tangent at m if there exist a chart (U,ϕU)
around M such that :(ϕU c1)′(0) = (ϕU c2)′(0).
This doesn’t depend on the chart we used. Indeed, if (V, ϕV ) is another chart aroundm, then :
(ϕV c1)′(0) = (ϕUV ϕU c1)′(0)
= DϕU (m)ϕUV ((ϕU c1)′(0))
= DϕU (m)ϕUV ((ϕU c2)′(0))
= (ϕV c2)′(0).
Moreover, the relation “c1 is tangent to c2 at m” is an equivalence on CM(m). We willdenote it c1Rc2.
Definition 1.5.4. The set TmM of tangent vectors to M at m is the quotient ofCM(m) by R.
23
The equivalence class of a curve c ∈ CM(m) (its image under the quotient mapCM(m)→ TmM) is called the tangent vector to c at 0, and is denoted by c′(0).Example 1.5.5. The set of tangent vectors at a pointm ∈ U ⊂ Rn is naturally isomorphic
to Rn. The isomorphism is given mapping a curve c through m to c′(0) ∈ Rn andpassing to the quotient.
Proposition 1.5.6. TmM has a natural vector space structure.
Proof. Given a chart (U,ϕU) around m, we define a map θϕU : TmM → Rn by :
θϕU (ξ) = (ϕU c)′(0)
where c is a representative of ξ in TmM . This is an injection by definition. Moreover, ifv ∈ Rn and c(t) = ϕ−1
U (tv) then the image of the class of c by θϕU is v. Thus θϕU is abijection from TmM to Rn.Moreover, if (V, ϕV ) is another chart, one can check that :
θϕV θ−1ϕU
= DϕU (m)ϕUV .
Hence θϕV θ−1ϕU
is a linear isomorphism of Rn.Thus we can define a vector space structure on TmM in the following way:
ξ + η = θ−1ϕU
(θϕU (ξ) + θϕU (η)) , λξ = θ−1ϕU
(λθϕU (ξ)),
the preceding remark show that this structure is independent of the chart we used.
Definition 1.5.7. TmM with the natural vector space structure built in the precedingproposition is called the tangent space to M at m.
Example 1.5.8. Assume that S ⊂ Rn is a submanifold. Then its tangent space as asubmanifold is naturally isomorphic to its tangent space as a manifold.
1.5.3 Tangent vectors in coordinates
Let M be a manifold and (U,ϕU) a chart around m ∈M . Then the map θϕU : TmM →Rn is (by construction) a linear isomorphism. We will denote the coordinates on ϕU(U)by (x1, · · · , xn). Let BU =
(∂∂x1
∣∣m, · · · , ∂
∂xn
∣∣m
)be the image under θϕU of the standard
basis of Rn (for a justification of the notation, see section 1.6.2). BU is a basis of TmMand thus any vector v ∈ TmM can be written as :
v =n∑i=1
vi∂
∂xi
∣∣∣∣m
.
Now consider another chart (V, ϕV ) centered at m and denote by (y1, · · · , yn) thecoordinates on ϕV (V ) and by BV =
(∂∂y1
∣∣∣m, · · · , ∂
∂yn
∣∣∣m
)the corresponding basis (via
θϕV ) of TmM . Then :
v =n∑j=1
vj∂
∂yj
∣∣∣∣m
.
24
The law according to which the coordinates vi of v with respect to BU change to thecoordinates vi is given by θV θ−1
U = DϕU (m)ϕUV . If we see the change of coordinatesϕUV as a n-tuple of functions (y1(x1, · · · , xn), · · · , yn(x1, · · · , xn)), then the matrix ofDϕU (m)ϕUV with respect to the basis BU and BV has coefficients :
∂yj
∂xi.
Therefore :
vj =n∑i=1
∂yj
∂xivi.
This is the way the tangent space used to be defined in old text books : a tangentvector was an n-tuple of numbers associated to a coordinate system which changedaccording to the above law when one changed coordinates.
1.6 Differentiating maps between manifolds
1.6.1 Definitions, first properties
Definition 1.6.1. Let f : M → N be a smooth map, then the tangent map (ordifferential) of f at m ∈M is the map : Tmf : TmM → Tf(m)N defined by :
TmM 3 v 7→ Tmf(v) = (f c)′(0) ∈ Tf(m)N.
To see that this definition makes sense, remark, using a chart around m and a chartaround f(m), that if c1 and c2 are tangent at m, then f c1 and f c2 are curves whichare tangent at f(m).Another way to phrase the definition is to say that Tmf : TmM → Tf(m)N is built
from the map f : CM(m)→ CN(f(m)), c 7→ f c by quotienting by the tangency relationat the source and the target space.Example 1.6.2. By definition, the tangent map of ϕU : U → ϕU(U) ⊂ RdimM is the map
θϕU defined in the previous section.Example 1.6.3. Assume that M and N are open sets in euclidean space, then Tmf is
just the usual differential.Just like the differential, the tangent map satisfies the chain rule :
Proposition 1.6.4. If f : M → N and g : N → P are smooth, then, for any m ∈M :
Tm(g f) = Tf(m)g Tmf.
Proof. Let cM be a curve through m. Then cN = f c is a curve through f(m) andcP = g f c is a curve through g(f(m)). Denote by ξM , ξN and ξP the tangent vectorsto cM , cN and cP . We have that, by definition of the differential :
ξN = Tmf(ξM), ξP = Tf(m)g(ξN), ξP = Tm(g f)(ξM).
This ends the proof.
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Proposition 1.6.5. Tmf : TmM → Tf(m)N is linear.
Proof. Consider charts (U,ϕU) around m and (V, ϕV ) around f(m). We have the fol-lowing commutative diagram :
U
ϕU(U)
V
ϕV (V )
f
ϕU ϕV
ϕV f ϕ−1U
All the arrows of this diagram are smooth maps, using the chain rule, we then get thefollowing diagram between tangent spaces :
TmM
RdimM
Tf(m)N
RdimN
Tmf
θϕU θϕV
TϕU (m)(ϕV f ϕ−1U )
Since the arrows on the side are linear isomorphism, and the arrow on the bottom islinear, Tmf is linear.
From the proof of the previous proposition, we see that requiring some property ofDϕU (m)(ϕV f ϕ−1
U ) which which can be expressed independently of a choice of basisin RdimM and RdimN is the same as requiring this property on Tmf , thus we have :
Proposition 1.6.6. f : M → N is an immersion (resp. submersion) (resp. localdiffeomorphism) if and only if at each point m ∈ M , Tmf is injective (resp. surjective)(resp. an isomorphism).
Example 1.6.7. Consider the canonical projection π : Rn+1\0 → RPn and f the re-striction of π to Sn. We will show that f is a local diffeomorphism. Denote byι : Sn → Rn+1 the inclusion. The differential of ι at v ∈ Sn is just the inclusionTι : v⊥ = TvSn → Rn+1.
We then have that : Tvf = TvπTvι. We want to show that Tvf is an isomorphism.We already know that Tvπ is surjective and Tvι, Thus we only need to show thatimTvι ∩ kerTvπ = 0.For dimensional reason, kerTvπ has dimension π. Moreover, if we set c(t) = (1+t)v,
we have that π(c(t)) ∈ RPn is constant. Beware here that v is viewed simultaneouslyas a point in Rn+1\0 and as a tangent vector to Rn+1 at v, this makes sense onlybecause Rn+1 is an open set on a vector space. This shows that Tvπ(v) = 0, thusKerTvπ = Rv. In particular imTvι ∩ kerTvπ = 0.
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1.6.2 Tangent vectors as derivations
This section gives another view on the tangent space at a point. The only point we willuse is the notation defined in the next paragraph. More details (including proofs !) canbe found in [Lee02].
Let p ∈M and Vp ∈ TpM . We define an action of Vp on smooth real valued functionsf : U → R on any neighborhood U of p by :
Vp · f = Tpf(Vp) ∈ R.
Remark 1.6.8. This gives a justification of the notation we introduced in section 1.5.3.With the notations of section 1.5.3, any chart (U,ϕ) around p defines vectors ∂
∂xi
∣∣p∈
TpM , one then has :
∂
∂xi
∣∣∣∣p
· f =∂(f ϕ−1)
∂xi(ϕ−1(p)).
Proposition 1.6.9. f 7→ Vp · f has the following properties :
• it is local : if f and g are equal on a neighborhood of p, Vp · f = Vp · g.
• (Vp, f) 7→ Vp · f is R-bilinear.
• it satisfies Leibnitz rule : Vp · (fg) = f(p)(Vp · f) + g(p)(Vp · f).
f 7→ Vp · f defines a map from the algebra C∞(M) of smooth functions on M to R.The third property is the defining property of a derivation at p.
Definition 1.6.10. A derivation at p ∈ M is a linear map δ : C∞(M) → R such thatfor any f, g ∈ C∞(M) :
δ(fg) = f(p)δ(g) + δ(f)g(p).
Let us denote by Dp the set of derivations on Cp. It is naturally a vector space. Then:
Proposition 1.6.11. The map L : TpM → Dp defined by Vp 7→ (f 7→ Vp · f) is a linearisomorphism.
The proof of this proposition uses a tool we haven’t introduced yet: cut-off functions,we’ll come back to it later.
This provides yet another definition of the tangent space, perhaps less intuitive thanthe one we have used. It has however the advantage that the vector space structure isobvious when one uses this definition.
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1.7 Exercises
Exercise 1.1. Let U ⊂ Rn be an open convex set. f is said to be convex if, for anyx, y ∈ U , any t ∈ [0, 1] :
f((1− t)x+ ty) ≤ (1− t)f(x) + tf(y).
Assume that f is C1. Show that f is convex if and only if, for any x, y ∈ U :
f(y)− f(x) ≥ Dxf(y − x).
Exercise 1.2. Consider f : Mn(R)→Mn(R) defined by f(A) = A2.
1. Compute, for any A ∈Mn(R), DAf .
2. Use the inverse function theorem to prove that any matrix which is close enoughto the identity has a square root.
Exercise 1.3. Using the inverse function theorem we give a proof of Theorem 1.1.16.f : Rn → Rm will be a smooth function whose differential at 0 is surjective.
1. write f = (f 1(x1, · · · , xn), · · · , fm(x1, · · · , xn)). Show that, up to a permutationof the coordinates in Rn, one can assume that the square m × m matrix whosecoefficients are ∂if j(0) for i, j ≤ m is invertible.
2. Consider the function F : Rn → Rn defined by :
F (x1, · · · , xn) = (f 1(x1, · · · , xn), · · · , fm(x1, · · · , xn), xm+1, · · · , xn).
Show that F is a local diffeomorphism at 0.
3. Show that this implies Theorem 1.1.16.
Exercise 1.4. Let f : Rn → R and g : Rn → R be two smooth function. DefineΣ = g−1(0). Assume that x ∈ Σ is a minimum of f |Σ. Show that there exist Λ ∈ R suchthat Dxf = λDxg.Hint: First prove, using the submersion theorem, that whenever Dxf 6= 0, for any
v ∈ kerDxf , there is a smooth curve γ in Σ such that γ(0) = x and γ(0) = v.
Exercise 1.5. * The goal of this exercise is to compute the differential of det : Mn(R)→R.
1. Show that det is smooth. Hint: written in coordinates, it is a polynomial.
2. Let I be the identity matrix, show that : DI detH = trace(H). Hint : Use themultilinearity of the determinant.
3. Show that whenever A is invertible, DA det(H) = det(A) trace(A−1H).
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4. Show that for any A ∈ Mn(R), DA det(H) = trace(AH), where A is the cofactormatrix of A.
Exercise 1.6. * We compute the differential of inv : GLn(R)→ GLn(R)
1. Show that GLn(R) is open.
2. Using that (I +H)−1 =∑
k≥1(−H)k for ‖H‖ ≤ 1, compute DI inv.
3. Show that DA inv(H) = −A−1HA−1.
4. Show that inv is smooth without actually computing its differential (use thatA−1 = det(A)−1A). Compute the differential of inv by differentiating the relationA× inv(A) = idRn .
Exercise 1.7. Recall that, for M ∈Mn(K) (K is R or C) :
expM =∑k=0
Mk
k!
1. Show that exp is C1.
2. Show that D0 exp = idMn(K).
Exercise 1.8. ** We will use the inverse function theorem to prove that exp : Mn(C)→GLn(C) is surjective.
1. Let G be a connected topological group5, and H be a subgroup of G which containsa neighborhood of the neutral element of G. Show that G = H Hint : H and itscomplement are open.
2. Show that (Mn(C),+) and (GLn(C),×) are topological groups.
3. Let A be a fixed element of GLn(C), and define :
C[A] = P (A) | P ∈ C[X]
Show that +, × and exp preserve C[A]. Hint: for exp, show that if M ∈ C[A],then expM is a limit of element of C[A].
4. Show that if M ∈ C[A] is invertible, then M−1 ∈ C[A]. Hint: Cayley-Hamiltoncan help.
5. Let C[A]× = C[A] ∩ GLn(C). Show that exp : (C[A],+) → C[A]× is topologicalgroup morphism.
6. Show that exp is a diffeomorphism on a neighborhood of 0.5A group G is a topological group if it is equipped with a topology which makes the group law andthe inverse continuous.
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7. Conclude that exp(C[A]) = C[A]×, and that exp : Mn(C)→ GLn(C) is surjective.
The next exercises give important example of submanifolds.
Exercise 1.9. Show that Sn is a submanifold of Rn+1.
Exercise 1.10. For which values of c is Σc = (x, y, z) ∈ R3|x2 + y2 = z2 + c asubmanifold of R3 ?
Exercise 1.11. Show that the matrix groups SL(n,R), O(n,R) and SO(n,R) are sub-manifolds of Mn(R) ' Rn×n.
Starting from now, we get to the core of the chapter, manifolds, smooth maps...
Exercise 1.12. Let E be a n dimensional vector space. For any basis B = (b1, · · · , bn)of E, we define a chart (E,ϕB) on E in the following way: for x ∈ E, ϕB(x) ∈ Rn is thevector formed by the coordinates of x in basis B.Show that A = (E,ϕB)|B is basis of E is a smooth atlas of E.
Exercise 1.13. Consider X = R × 0, 1. Define an equivalence relation on R on Xby :
(x, a)R(y, b)⇔ x 6= 0 and x = y.
X/R with the quotient topology is called the line with two origins.
1. Show that X is not Hausdorff.
2. Show that each point of X has a neighborhood homeomorphic to an open intervalin R.
Exercise 1.14. LetM1 be R with the atlas (R, idR) andM2 be R with the atlas (R, x 7→x3).
1. Show that these two atlases are not Ck compatible for any k ≥ 1.
2. Describe the smooth maps from M1 to M2 and from M2 to M1.
3. Show that M1 and M2 are diffeomorphic as smooth manifolds.
Exercise 1.15. Consider the sphere Sn ⊂ Rn+1. Set, for 1 ≤ i ≤ n + 1, U+i =
x ∈ Sn|xi > 0 and U−i = x ∈ Sn|xi > 0.Define ϕ+
i : U+i → Rn by ϕ+
i (x1, · · · , xn+1) = (x1, · · · , xi−1, xi+1, · · · , xn+1), and ϕ−i :U−i → Rn by the same formula.
1. Show that the charts (U+i , ϕ
+i ) and (U−i , ϕ
−i ) form a smooth atlas A of Sn
2. Show that this atlas define the same smooth structure on Sn as the atlas definedby the stereographic projections.
Exercise 1.16. 1. Show that RP1 is diffeomorphic to S1.
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2. Consider the Riemann sphere C = C ∪ ∞. Consider the following atlas on C:
(C, z 7→ z) (C\0, z 7→ 1/z)
with the convention that 1/∞ = 0 and 1/0 =∞. Show that this atlas is smooth.
3. Show that C is diffeomorphic to CP1 and S2.
Exercise 1.17. Consider ψ : R2 → C2 ' R4 defined by :
ψ(t, s) =1√2
(eit, eis).
1. Show that S = ψ(R2) is a submanifold of R4.
2. Show that S is in fact a submanifold of S3.
3. Show that S is diffeomorphic to T2.
Remark 1.7.1. S is called the Clifford torus. It has been recently proven by Coda-Marques and Neves that, in some precise sense, it is the least twisted torus in S3,solving a conjecture known as the Willmore conjecture.
Exercise 1.18. ** Let G be a discrete group which acts on a smooth manifold M . Weassume that:
• x 7→ g · x is smooth (the action is smooth).
• For any compact subsets K,L ⊂ M , #g ∈ G|g ·K ∩ L 6= ∅ is finite (the actionis proper).
• For any g ∈ G\e, for any x ∈M , g · x 6= x (the action is free).
We will show that the quotient6 M/G has a natural smooth manifold structure.
1. Show that the quotient map M →M/G is open. Hint: Argue as in section 1.3.4,only the smoothness of the action matters here.
2. Show thatM/G is Hausdorff. Hint: the proof is similar to section 1.3.4, propernessmust be used.
3. Show that any point x ∈M admits a neighborhood V such that, for any g ∈ G\e,g · V ∩ V = ∅. Deduce that πV is an homeomorphism from V to π(V ).
4. Since M is a manifold, one can find a chart (U,ϕU) of M around x such thatU ⊂ V . Then, setting U = π(U) and ϕU = ϕU π|−1
U , we have that (U , ϕU is achart around π(x).
Show that the charts obtained by these process form a smooth atlas of M/G.6the equivalence induced by the group action is xRy if and only if y = g · x for some g ∈ G.
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Exercise 1.19. * The goal of this exercise is to build a smooth embedding of RP2 inR4.
Consider the bilinear map B : R3 × R3 → R5 defined by :
B
xyz
,
x′y′z′
=
xx′
xy′ + x′yxz′ + yy′ + x′z
yz′ + y′zzz′
.
1. Show that B(v, v′) = 0 if and only of v = 0 or v′ = 0.
Hint: Show that, when v 6= 0, the linear map v′ 7→ B(v, v′) is injective using thatan n×m matrix has a non trivial kernel if and only if all its m×m minors vanish.
2. Consider G : R3\0 → S4 defined by:
F (v) =B(v, v)
‖B(v, v)‖.
Show that F (v) = F (v′) if and only if v′ = λv.
Hint: g(v) = g(v′)⇒ B(v, v) = λ2B(v′, v′)⇒ B(v + λv′, v − λv′) = 0.
3. Show that there exist a unique smooth f : RP2 → S4 such that f = F π whereπ : R3\0 → RP2 is the usual projection.
4. Show that f is a smooth embedding.
Hint: To see that f is an immersion, show that kerTvF = kerTvπ. The followingcommutative diagram may help:
R3\0
RP2
R5
S4
π F
f
5. Show that (0, 1, 0, 0, 0) /∈ f(RP2) and use it to define a smooth embedding of RP2
into R4.
Exercise 1.20. Let M and N be two compact manifolds of the same dimension andf : M → N a smooth map. y ∈ N is called a regular value of f if for any x ∈ f−1(y),Txf is invertible. Let y be a regular value of f .
1. Show that f−1(y) is finite.
2. Show that the cardinal of f−1(y′) is constant for y′ is some neighborhood of y.
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Exercise 1.21. * The goal of this problem is to give a proof of the fundamental theoremof algebra with the tools introduced in this chapter.
Consider P ∈ C[X] a non constant complex polynomial.
1. Show that the function P : C→ C defined by :P (z) = z if z ∈ CP (∞) =∞.
is smooth.
2. Let F = y ∈ C|y is not a regular value of P .. Show that F is finite.
3. Show that the function z 7→ #f−1(z) is constant on C\F .
4. Show that P is surjective and that this imply the fundamental theorem of algebra.
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2 Tangent bundle, Vector fields andLie brackets
2.1 Vector fields in Rn.
This part is quite fast and sketchy, a more detailed exposition can be found in [HS74].
2.1.1 General theory
Definition 2.1.1. Let U ⊂ Rn be an open set. A smooth vector field on U is asmooth map V : U → Rn.
Vector fields on U ⊂ R2 are usually plotted in the following way: at each point x, onedraws the vector V (x) with its “foot” set at x. A high level reason to do that is that oneusually sees the vector V (x) as an element of TxR2 ' R2, a tangent vector to R2 at x.We will also sometimes write the value of V at x as Vx.
x
y
Figure 2.1: The vector field V (x, y) = (−y/2, x/2) in R2
The Cauchy problem for V with initial condition x0 ∈ U is the following problem:find a function x(t) for t ∈ I ⊂ R some interval containing 0 such that:
x′(t) = V (x(t)) for all t ∈ Ix(0) = x0.
(2.1)
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A solution to (2.1) is a curve t ∈ I 7→ x(t) ∈ U which satisfies (2.1).Remark 2.1.2. If t 7→ x(t) is a solution, then x(t) = x(t + t0) is a solution with initial
condition x(0) = x(t0).Amaximal solution of (2.1) is a solution x : I → U of (2.1) which cannot be extended
further: if x : I → U is another solution of (2.1) such that I ⊃ I and x|I = x, thenI = I.We have the following important theorem:
Theorem 2.1.3. Let V be a smooth vector field on U ⊂ Rn, then for any x0 ∈ U theCauchy problem has a unique maximal solution.
The following property of maximal solutions is useful in the applications:
Proposition 2.1.4. Let x : [0, T ) → U be a solution to (2.1), with T < ∞ then eitherfor any compact K ⊂ U there exists tK ∈ [0, T ) such that x(tK) /∈ K or x can be extendedto a solution x : [0, T + ε)→ U .
Proof. Assume that there exist a compact K ⊂ U such that x(t) ∈ K for all t ∈ [0, T ),we want to show that x is not maximal. Since K is compact, V is bounded on K. Thusx1 =
∫ T0V (x(t))dt =
∫ T0x′(t)dt exists and limt→T x(t) = x1.
Let y(t)t∈[0,T ′) be the maximal solution to the Cauchy problem y(0) = x1, y′ = V (y)and set:
x(t) =
x(t) if t < T ,y(t− T ) if t ∈ [T, T + T ′).
Then x is C0 and satisfy x′ = V (x) and x(0) = x0. Thus x is not maximal.
Example 2.1.5. If one can show that any solution to (2.1) stays in a compact set, thenthe maximal solution is defined for all t ∈ R.
Definition 2.1.6. V is said to be complete if for any x0 ∈ U the maximal solutionx(t) to (2.1) with initial condition x0 is defined for all t ∈ R.
We now assume that V is a smooth vector field. We define O ⊂ U ×R by (x, t) ∈ O ifand only if the maximal solution to the Cauchy problem y′ = V (y), y(0) = x is definedat t, if V is complete, then O = U × R. For (x, t) ∈ O, we define ϕt(x) = y(t) wherey(t) is the solution to y′ = V (y), y(0) = x. (x, t) 7→ ϕt(x) is called the local flow of V .
Proposition 2.1.7. ϕt has the following property :
1. (x, t) 7→ ϕt(x) is smooth.
2. ϕt+s(x) = ϕt(ϕs(x)) whenever that makes sense. Moreover, if V is complete thenfor any t, ϕt : U → U is a smooth and satisfies ϕt+s = ϕt ϕs. In this case ϕt isa smooth diffeomorphism of U .
3. ddt |t=0
ϕt(x) = Vx.
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2.1.2 Linear vector fields and matrix exponential
Definition 2.1.8. A vector field V on Rn is linear if and only if V (x) = Ax for somelinear map A from Rn to Rn.
Definition 2.1.9. For any A ∈Mn(R), the exponential eA of A is defined by :
eA =∞∑k=0
Ak
k!.
It is easy to see that, if Mn(R) is equipped with an operator norm:∞∑k=0
∥∥∥∥Akk!
∥∥∥∥ ≤ ∞∑k=0
‖A‖k
k!≤ e‖A‖,
so this series converge absolutely and thus eA is well defined. Moreover if we interpretthe serie as a serie of functions from Mn(R) to Mn(R), then the previous inequality saysthat this serie is normally convergent on every compact set, in particular : A 7→ eA iscontinuous.
Just like functions Ceαt are all solutions to the (scalar!) differential equation x′ = αx,we have :
Proposition 2.1.10. Consider the following Cauchy problem:x′ = Ax
x(0) = x0.
The maximal solution is given by :
x(t) = etAx0 t ∈ R.
Proof. The vector field x 7→ Ax is smooth, so the previous section tells us that thereis a unique maximal solution, so we only need to check that the formula defining x(t)actually yields a solution.
To see this, set fk(t) = (tA)k
k!x0. Then x(t) =
∑k≥0 fk(t). We want to differentiate the
serie term by term, we can do it because:∑k≥0
|f ′k(t)| =∑k≥1
∣∣∣∣tk−1Akx0
(k − 1)!
∣∣∣∣ ≤∑k≥1
‖A‖∥∥∥∥ (tA)k−1
(k − 1)!
∥∥∥∥ |x0| ≤ ‖A‖et‖A‖|x0|,
and the serie of the derivatives normally converges on compacts of R.Then:
x′(t) =∑k
f ′k(t) =∑k≥1
tk−1Akx0
(k − 1)!= A
(∑k≥1
(tA)k−1
(k − 1)!
)x0 = AetAx0 = Ax(t).
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This shows that the flow of V (x) = Ax is particularly simple in this case, we have:ϕt(x) = etAx.The matrix exponential have the following properties:
Proposition 2.1.11. 1. If P is invertible, ePAP−1= PeAP−1.
2. If AB = BA, eA+B = eAeB.
3. e−A = (eA)−1.
4. det eA = etraceA.
2.2 Cut-off functions
The goal of the section it to prove the following proposition:
Proposition 2.2.1. Let M be a smooth manifold, O ⊂ M an open set and K ⊂ O acompact set. Then there exist a smooth function η : M → R such that:
• 0 ≤ η ≤ 1.
• for all x ∈ K, η(x) = 1.
• for x /∈ O, η(x) = 0.
Proof.Claim 2.2.2. The proposition is true forM = Rn, K = B(0, r) and O = B(0, R) (r < R).
First consider the function f : R→ R defined by:
f(t) =
0 if t ≤ 0,e−1/t if t > 0.
It is a classical fact that f is smooth. Setting g(t) = f(t)f(1 − t) give a nonnegativesmooth function with support [0, 1], scaling g if necessary, we can assume
∫R g(t)dt = 1.
Then G(t) =∫ t−∞ g(s)ds is nondecreasing, equal to 0 for t < 0 and equal to 1 for t > 1.
Then for any r, R > 0, one can find parameters α and β such that Gr,R(t) = G(αt+ β)is smooth, nonincrreasing, equal to 1 for t < r and equal to 0 for t > R. Finally, settingη(x) = Gr,R(|x|) does the job.Claim 2.2.3. The proposition is true if O is included in U for some chart (U,ϕU).Reducing O, we can assume the closure of O is in U . We will in fact build η on
U , we can then extend it to 0 outiside U .Since charts are smooth diffeomorphsims,we can work in ϕU(U). So we might as well assume that O ⊂ Rn. Since K ⊂ O,for each p ∈ K we can find rp such that B(p, rp) ⊂ O. Now, the B(p, rp/2) form anopen cover of K which is compact and we can find finitely many points p1, · · · , pN such
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that K ⊂⋃Ni=1B(pi, rpi/2). Let ηi : Rn → [0, 1] be a smooth function which is 1 on
B(pi, rpi/2) and 0 outiside of B(pi, rpi). Finally we set:
η = 1− (1− η1) · · · (1− ηN),
and it works.We can now prove the general result. Since K is compact, we can find a finite number
of charts (U1, ϕ1), · · · , (UN , ϕN) such that each Ui ⊂ O and K =⋃iKi where each
Ki ⊂ Ui is compact. We apply the previous claim for each pair Ki ⊂ Ui to get ηi, andset :
η = 1− (1− η1) · · · (1− ηN).
Using these cutoff functions, we can build partitions of unity:
Proposition 2.2.4. Let M be a smooth manifold and (Oi)i∈I be a locally finite1 opencover of M . Then there exists smooth functions ηi : M → R such that:
• the support of ηi is included in Oi.
•∑
i∈I ηi = 1.
Proof. First we can find an open cover Vi of M such that for each i, the closure Ki
of Vi is included in Ui. We then consider the cutoff functions ηi associated to the pairKi ⊂ Ui, and set:
ηi(x) =ηi(x)∑j∈I ηj(x)
.
2.3 From the tangent spaces to the tangent bundle
2.3.1 TM as smooth manifold
We want to study vector fields on manifolds. In order to do this we first need to definesmooth vector fields. A vector field V on a smooth n-manifold M should associate toeach point p ∈ M a vector V (p) in TpM . In particular V : M →
⋃p∈M TpM . The only
problem is that⋃p∈M TpM doesn’t have a smooth structure, so we cannot talk of the
smoothness of V . The goal of this section is to endow TM =⋃p∈M TpM with a natural
smooth structure.First we define π : TM →M by π(v) = p if v ∈ TpM , in other words π(v) is the foot
of v. Remark that a vector field V : M → TM should satisfy V π = idM (V (p) shouldbe a vector at p, not at some other point).
1any compact K ⊂M intersects a finite number of Oi’s.
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Now let’s define charts on TM . Consider a chart (U,ϕU) of M . Let TU =⋃p∈U TpM ,
and define ΦU : TU → ϕ(U)× Rn by:
ΦU(ξp) = (ϕ(p), Tpϕ(ξ)) if ξp ∈ TpM .
We saw in the previous chapter that Tpϕ : TpM → Rn is a linear isomorphism when(U,ϕU) is a smooth chart. This shows that ΦU : TU → ϕ(U)× Rn is a bijection.We haven’t defined a topology on TM yet. We will do it in the following way: a set
O ⊂ TM is said to be open if for any chart (U,ϕU) of M , Φ(TU ∩ O ⊂ ϕ(U) × Rn isopen.
Proposition 2.3.1. The statement above defines a topology on TM for which the col-lection of all (TU,ΦU) is a smooth atlas.
Proof. The fact that we have defined a topology on TM is easy : if Oi is a family of opensets in TM , then ΦU(
⋃iOi∩TU) =
⋃i ΦU(Oi∩TU), which is open in U×Rn as a union
of open sets. Thus⋃iOi is open. The same argument works for finite interesections.
We now show that ΦU : TU → U ×Rn is an homeomorphism. Since we already knowthat ΦU is a bijection, this is equivalent to the following claim:Claim 2.3.2. For any O ⊂ TU , then O is open in TM if and only if ΦU(O) is open inU × Rn.
To see this, consider another chart (V, ϕV ) of M , and the transition function ΦUV =ΦV Φ−1
U from ϕU(U ∩ V )× Rn to ϕV (U ∩ V )× Rn. We have :
ΦUV (x, v) = ΦV
((Tϕ−1
U (x)ϕU)−1(v))
= ΦV
(Tx(ϕ
−1U )(v)
)=(ϕV (ϕ−1
U (x)), Tϕ−1U (x)ϕV
(Tx(ϕ
−1U )(v)
))= (ϕUV (x), TxϕUV (v)) .
Since ϕUV is smooth, (x, v) 7→ TxϕUV (v) = DxϕUV (v) is smooth and ΦUV is smooth.Moreover, the fact that ΦUV ΦV U = id implies, since ΦV U is also smooth, that ΦUV
is a diffeomorphism. In particular ΦUV is a homeomorphism. Thus if ΦU(O) is open,ΦV (O ∩ TV ) = ΦUV (ΦU(O) ∩ (ϕU(V ∩ U)× Rn)) is also open. The claim is proved.Moreover, we have already seen that the transition functions ΦUV are diffeomorphism.
Thus TM is a smooth manifold.
Remark 2.3.3. The dimension of TM is twice the dimension of M .Remark 2.3.4. The formula for the transition functions shows that ifM is a Ck manifold,
then TM is a Ck−1 manifold.
Proposition 2.3.5. π : TM →M is a submersion.
Proof. Let (U,ΦU) be a chart if M and (TU,ΦU) be the associated chart of TM . Then:
ϕU π Φ−1U (x, v) = x
which is a submersion from R2n → Rn.
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2.3.2 TM as a vector bundle
The charts ΦU : TU → U × Rn are more than just diffeomorphisms, they also behavewell with respect to the vector space structure of TpM . This can be summarized in thefollowing proposition, which is a direct consequence of the definition of ΦU .
Proposition 2.3.6. π : TM → M is the natural projection. We have the followingproperties :
• for any p ∈M , π−1(p) is a d-dimensional vector space.
• for any p ∈ M , there exists a neighborhood U of p such and a diffeomorphismΦU : π−1(U)→ U × Rn such that the following diagram is commutative:
π−1(U) U × Rn
U
ΦU
ππ1
where π1 is the projection (p, v) 7→ p.
• for each p ∈ U , ΦU restricted to TpM = π−1(p) is a linear isomorphism ontop × Rn.
From this situation we extract the definition of a vector bundle on a manifold, whichis a family of vector spaces smoothly attached above each point of a manifold M .
Definition 2.3.7. Let E and M be two smooth manifolds. A smooth d-dimensionalvector bundle above M is a smooth map π : E →M such that:
• for any p ∈M , Ep = π−1(p) is a d-dimensional vector space.
• for any p ∈ M , there exists a neighborhood U of p such and a diffeomorphismΦU : π−1(U)→ U × Rd such that the following diagram is commutative:
π−1(U) U × Rd
U
ΦU
ππ1
where π1 is the projection (p, v) 7→ p.
• for each p ∈ U , ΦU restricted to EpM = π−1(p) is a linear isomorphism ontop × Rd.
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Remark 2.3.8. The dimension of E is d+ dimM and π is a submersion.Example 2.3.9. TM →M is an dimM -dimensional vector bundle over M .Example 2.3.10. M × Rd → M is a vector bundle, a vector bundle is said to be trivial
if it is isomorphic to it.
Proposition 2.3.11. Consider a smooth f : M → N . Define Tf : TM → TN byTf(ξ) = Tpf(ξ) if ξ ∈ TpM . Then :
1. Tf is smooth.
2. Tf makes the following diagram commutative:
TM TN
M N
Tf
πM
f
πN
3. for each p ∈M , the restriction of Tf to TpM is linear from TpM to Tf(p)N .
These properties say that Tf is a vector bundle morphism.
Definition 2.3.12. Let π1 : E1 → M1 and π2 : E2 → M2 be two smooth vector bundle.A vector bundle morphism between E1 and E2 is a couple (f, g) of smooth mapsf : M1 →M2 and g : E1 → E2 such that :
1. f and g make the following diagram commute:
E1 E2
M1 M2
g
π1
f
π2
2. for each p ∈ M1, the restriction of g to (E1)p = π−11 (p) is linear from (E1)p to
(E2)f(p) = π−12 (f(p)).
Vector bundle morphism can be composed. A vector bundle isomorphism from E1
to E2 which has an inverse, it is equivalent to require g to be diffeomorphism. A d-dimensional vector bundle E →M is said to be trivial if it is isomorphic toM×Rd →M .
Definition 2.3.13. A smooth section of a vector bundle π : E → M is a mapσ : M → E such that π σ = idE. A section of the tangent bundle TM of M is called avector field on M .
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Remark 2.3.14. The set Γ(E) of smooth sections of E is a vector space. On can showusing bump functions on M that it has infinite dimension. In other words, vectorbundles on smooth manifolds have a lot of sections. This isn’t the case for complexmanifolds: finding holomorphic sections of complex vector bundles is in general ahard task.With this definition a smooth vector field on M is a smooth section of TM →M .The following charcterization of trivial bundles is useful:
Proposition 2.3.15. An n-dimensional vector bundle π : E →M is trivial if and onlyif there exist n sections σ1, · · · , σn such that at each p ∈M , (σ1(p), · · · , σn(p)) is a basisof Ep.
Proof. Assume first that E → M is trivial, then we have an isomorphism (f, g) fromE →M to E = M × Rn →M . Let us define smooth sections σi of E by σi(p) = (p, ei)where ei is the canonical basis of Rn, and p ∈M . We then define σi(p) = g−1 (σi(f(p))),these are smooth sections of E. Since g|Ep is a linear isomorphism, (σi(p))i=1,··· ,n is abasis of Ep.For the converse, let (σ1, · · · , σn) be n linearly independent sections E and define a
bundle morphism (f, g) from E → M to E → M by f = idM and g(p, x) =∑
i xiσi(p).
Then (f, g) is the required isomorphism.
Remark 2.3.16. In particular, if a vector bundle is trivial, it has a nowhere vanishingsection. This can be used to show that the tangent bundle to S2 is not trivial, sinceany vector field on S2 has a zero (the so-called hairy ball theorem), a theorem wewill prove later using differential forms and Stokes theorem.
2.4 Vector fields on manifolds
All the facts recalled in section 2.1 extends to manifolds. Namely:
Definition 2.4.1. Let M be a smooth manifold and V be a smooth vector field on M .A smooth curve x : I →M is said to be a solution to the Cauchy problem:
x′ = V (x)
x(0) = x0.
if it satisfies x(0) = x0 and x′(t) = V (x(t)) ∈ Tx(t)M for any t ∈ I.
Theorem 2.4.2. The Cauchy problem:x′ = V (x)
x(0) = x0.
has a unique maximal solution x : I → R.
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Definition 2.4.3. A vector field V is said to be complete if for any x0 ∈M the maximalsolution to the Cauchy problem with initial condition x0 is defined for I = R.
Proposition 2.4.4. If a vector field on a manifold has compact support, then it iscomplete.If M is compact, then any vector field on M is complete.
If V is a smooth complete vector field on M , we can define the ϕt : M → M whichwill be a one parameter group of diffeomorphisms of M . If V is not complete, then theflow (t, x) 7→ ϕt(x) is only defined on
O =
(y, t) ∈M × R
∣∣∣∣The maximal solution to x′ = V (x)with x(0) = y is defined at t.
.
Definition 2.4.5. Let f : M → N be a diffeomorphism, we define the push forwardof V by f (denoted by f∗V ) by f∗V (y) = Tf−1(y)f(Vf−1(y)), this is a smooth vector fieldon N .
Let u be a smooth function on N , then, as a derivation on C∞(N), acts by:
f∗V · u = (V · (u f)) f−1.
Proposition 2.4.6. x : I → M is an integral curve of V if and only if f x : I → Nis an integral curve of f∗V .
Proof. Set x = f x then:x′(t) = Tx(t)f (x′(t)) .
Thus x is an integral curve of f∗V if an only if
Tx(t)f (x′(t)) = f∗Vx(t) = Tf−1(x(t))f(Vf−1(x(t))
)which is equivalent to x′(t) = Vx(t).
A useful observation that follows from this is that, if we denote by ϕt the flow of Vand ψt the flow of f∗V , then ψt = f ϕt f−1. This property actually characterizes f∗V .
Proposition 2.4.7. Assume that V is a complete vector filed on M . Let ϕs be the flowof V , then (ϕs)∗V = V .
Proof. Let p ∈M , then: ((ϕt)∗V )p = D(ϕs)−1(p)V(ϕs)−1(p).. Consider the solution x to theCauchy problem x′(t) = V (t) with initial condition x(0) = (ϕs)
−1(p). Then by definitionof the flow x(s) = p and x′(s) = Vp. Moreover x(t) = ϕs(x(t)) = x(s + t) is an integralcurve of V which satisfies x(0) = p. Then
Vp = x′(0) = Dx(0)ϕs(x′(0)) = D(ϕs)−1(p)ϕsV(ϕs)−1(p) = ((ϕs)∗V )p.
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An equilibrium (or stationary) point of a vector field V is a point p ∈ M such thatVp = 0. Let us consider the constant vector field ∂1 in Rn. The following theorem tellsus that around a non-stationnary point, any vector field looks like ∂1.
Theorem 2.4.8. Let V be a vector field on M and p ∈ M such that V (p) 6= 0. Thenthere exists a neighborhood U of p and a diffeomorphism ψ : U → U ⊂ Rn such thatψ∗V = ∂1.
Proof. Since ce we are looking for a local result, we can only consider V in a coordinatechart around p. This way we can assume that V is complete Consider a chart (U, xU =(x1, · · · , xn)) around p (we assume that xU(p) = 0) and the flow Φt of V . For t small,we can assume that Φt is a diffeomorphism from V ⊂ U to Φt(V ) ⊂ U .Define a map ρ on a neighborhood of p by ρ(x1, · · · , xn) = ϕx1(0, x
2, · · · , xn).The differential of ρ at p is such that Tpρ(∂1) = V (p) and Tpρ(∂i) = ∂i for i ≥ 2. By
permuting coordinates if necessary, we can assume the the ∂1 component of V (p) is not0. This implies that Tpρ is invertible, thus ρ is a diffeomorphism on some neighborhoodof p.We now set ψ = ρ−1. Now the flow ϕt of ψ∗V is given by ψ ϕt ψ−1 and we can
show that :
ϕt(x1, · · · , xn) = ψ ϕt ρ(x1, · · · , xn)
= ψ(ϕt(ϕx1(0, x
2, · · · , xn)))
= ψ(ϕt+x1(0, x
2, · · · , xn))
= (t+ x1, x2, · · · , xn),
which is the flow of ∂1, since the flow determines the vector filed, this shows that ψ∗V =∂1.
2.5 Vector fields and global derivations
We have mentionned in the first chapter that tangent vectors at a point p ∈ M can beviewed as derivations at p. Here we will prove this result and a global version of it :vector fields on M can be viewed as derivations on M .
Let’s first recall the definition of a derivation at p ∈M
Definition 2.5.1. A linear map δ : C∞(M)→ R is a derivation at p ∈M if:
δ(fg) = δ(f)g(p) + f(p)δ(g).
The set of derivation ot p is denoted by Dp.
Proposition 2.5.2. A derivation at p satisfies the following properties:
1. If f is constant, then δ(f) = 0.
2. If f and g coincide on a neighborhood of p, then δ(f) = δ(g).
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Proof. We first prove the first statement. By linearity, we can assume that f constantequal to 1, thus f 2 = f . This implies :
δ(f) = δ(f × f) = δ(f)f(p) + f(p)δ(f) = 2δ(f),
and δ(f) = 0.For the second statement, using linearity it is enough to show that that δ(f − g) = 0.
set h = f − g and notice that h vanishes on a neighborhood U of p. Consider a compactneighborhood K of p which is included in U , and a bump function η which is equal to1 on K and equal to 0 outside of U . This implies to ηh = 0 on M . Thus:
0 = δ(ηh) = δ(η)h(p) + η(p)δ(h) = δ(h).
This proposition shows that derivations are local, if U is a neighborhood of p, we candefine δ for smooth functions f on U by setting δU(f) = δ(f) where f = ηf on U withη a bump function like in the previous proof, and extended to M by 0.
We can now prove:
Theorem 2.5.3. The map L : Xp 7→ (f 7→ Xp · f)) from Tp(M) to Dp is a linearisomorphism.
Proof. Let (U,ϕU) be a chart around p, we assume that ϕU(p) = 0 ∈ U . Consider thecoordinate functions xi : U → R.Let f : U → R be a smooth function. We make the following claim:
Claim 2.5.4. There exists smooth function hi : U → R such that f(x) = f(p) +∑i x
i(x)hi(x), moreover hi(p) = ∂if(p).For the proof of the claim, we work on ϕU(U) ⊂ Rn. Using bump function, it is
enough to prove the claim in a convex neighborhood of 0. We thus have:
f(x) = f(0) +
∫ 1
0
d
dt(f(tx))dt = f(0) +
∫ 1
0
∑i
xi∂if(tx)dt = f(0) +∑i
xi∫ 1
0
∂if(tx)dt.
Setting hi(x) =∫ 1
0xi∂if(tx)dt does the job. The claim is proved.
Now consider δ a derivation at p acting on f , we have:
δ(f) =∑i
(δ(xi)hi(p) + xi(p)δ(hi)
)=∑i
δ(xi)∂if(p).
And if we define V = δ(xi)∂i, we have that δ = L(V ). Thus L is surjective.To show that L is injective, assume that L(V ) = 0, then we have that V ·xi = 0, since
V = (V · xi)∂i, this implies that V = 0.
We are now going to globalize this construction:
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Definition 2.5.5. A (global) derivation on M is a linear map δ : C∞(M)→ C∞(M)such that:
δ(fg) = δ(f)g + fδ(g).
We denote by D(M) the space of derivations at M .
Example 2.5.6. Let V be a smooth vector field on M , then we define X · f by:
(X · f)(p) = Tpf(Xp).
f 7→ X · f is then a derivation on MAs in the case of pointwise derivations, Leibnitz implies the locality of derivations.
Proposition 2.5.7. Let δ be a derivation on M .
1. If f = g on some open set U ⊂M , then (δf)|U = (δg)|U .
2. For any open set U ⊂ M , there exist a unique derivation δ|U such that, for anyf ∈ C∞(M), δ|U(f |U) = (δf)|U .
Proof. For the first point, linearity implies that we just need to to show that if f |U = 0,then (δf)|U = 0. Let V ⊂ U be any open subset whose closure is contained in U .Consider a bump function η which is 1 on V and 0 outside U . Then fη = 0 on M and:
0 = δ(ηf) = δ(η)f + ηδ(f),
moreover δ(ηf)|V = δ(η)|V f |V + η|V δ(f)|V = δ(f)|V , thus δ(f) is 0 on any open set ofU whose closure is contained in V , hence (δf)|U = 0.For the second point, the only point is to define δ|U for smooth functions f ∈ C∞(U)
who don’t extend to smooth functions on the whole of M . To do this we consider V andη as above and set δ|U(f)|V = δ(ηf), where ηf is extended by 0 to M . The previousstatement shows that this definition doesn’t depend on the choice of η.
Theorem 2.5.8. The map L : V 7→ (f 7→ V · f) from Γ(TM) to D(M) is a linearisomorphism.
Proof. First L is injective. If L(V ) is not zero, there is a function f such that V · f isnot zero. If p ∈ M is a point where (V · f) 6= 0, then Tpf(Vp) 6= 0, in particular Vp isnot 0 and V 6= 0.
Let us now show that L is surjective. Let δ be a derivation. For any p ∈ M , f 7→(δf)(p) is a derivation at p, so there exists Vp ∈ TpM such that (δf)(p) = Vp · f . Wenow only need to show that p 7→ Vp is a smooth section. This follows from the fact that,in any chart (U,ϕU):
V (p) =∑i
δ(xi)∂i
and δ(xi) is smooth.
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2.6 Lie bracket of vector fields
Consider two vector fields V and W on M , we then have the corresponding derivationsf 7→ V · f and f 7→ W · f .
Proposition 2.6.1. The map L : f 7→ V · (W · f)−W · (V · f) from C∞(M) to C∞(M)is a derivation.
Proof. The fact that L is linear is obvious, we just need to show it satisfies Leibnitz rule:
L(fg) = V · (W · (fg))−W · (V · (fg))
= V · ((W · f)g + f(W · g))−W · ((V · f)g + f(V · g))
= (V · (W · f))g + (W · f)(V · g) + (V · f)(W · g) + f(V · (W · g))
− (W · (V · f))g − (V · f)(W · g)− (W · f)(V · g)− f(W · (V · g))
= (Lf)g − f(Lg).
Since L is a derivation, there exist a vector field X on M such that Lf = X · f forany f . X is called the Lie bracket of V and W and is usually denoted by [V,W ]. Ithas the following properties:
Proposition 2.6.2. • (V,W ) 7→ [V,W ] is bilinear.
• [V,W ](p) only depens on the value of V and W on a neighborhood of p
• [V,W ] = −[W,V ].
• [V, [W,X]] + [X, [V,W ]] + [W, [X, V ]] = 0.
• if ϕ : M → N is a diffeomorphism, [ϕ∗V, ϕ∗W ] = ϕ∗[V,W ].
Example 2.6.3. For any coordinate chart (U,ϕU), the coordinate vector fields satisfy[∂i, ∂j] = 0, since mixed partial derivatives commute.Let’s work out a formula for computing Lie brackets in coordinates: we can assume
that V = V i∂i and W = W i∂i in some coordinate charts.
[V,W ] · f = V · (W · f)−W · (V · f)
= V · (W i∂if)−W · (V i∂if)
= (V ·W i)∂if +W i(V · ∂if)− (W · V i)∂if − V i(W · ∂if)
= (V j∂jWi)∂if +W iV j∂j∂if − (W j∂jV
i)∂if − V iW j∂i∂jf
= (V j∂jWi −W j∂jV
i)∂if
thus we get[V,W ] = (V j∂jW
i −W j∂jVi)∂i.
The following proposition gives an interpretation of the Lie bracket in term of flow,informally, it shows that [W,V ] is the infinitesimal push forward of W along the flow ifV .
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Proposition 2.6.4. Let V and W be vector fields onM , and ϕt be the flow of W . Then:
[W,V ] =d
dt |t=0(ϕt)∗W.
Remark 2.6.5. There is something unclear here. If we fix p ∈ M , then t 7→ ((ϕt)∗W )pis a map from R → TM , therefore its derivatve should be an element of T (TM),the double tangent bundle. However we claim that it is [W,V ]p, an element of TpM .What happens here is that (ϕt)∗W is a time dependent vector field, and since vectorfields on M form a vector space we can define :
d
dt |t=0(ϕt)∗W = lim
t→0
(ϕt)∗W −Wt
.
Proof. We work with derivations. We have that ((ϕt)∗W ) · f = (W · (f ϕt)) ϕ−t.Consider the one parameter family of smooth functions g(x, t) = fϕt(x)−f(x). Arguingas in the proof of Claim 2.5.4, since g(x, 0) = 0, one can find a one parameter family ofsmooth functions h(x, t) such that g(x, t) = th(x, t) and h(x, 0) = d
dtf(x, 0) = V · f .
Then :
((ϕt)∗W ) · f = (W · (f + th(., t))) ϕ−t= (W · f) ϕ−t + t(W · h(., t)) ϕ−t
We now differentiate at t = 0. The first term gives −V · (W · f) and the second givesW · (h(., 0)) = W · (V · f).
The Lie bracket has the following interpretation: it measures the lack of commutativityof the flows of V and W .
Proposition 2.6.6. Let V and W be two vector fields on M and denote by ϕt and ψttheir flows. Then ψt ϕs = ϕs ψt if and only if [V,W ] = 0.
Proof. We first assume that the flows commute. We have that ψt = ϕsψt(ϕs)−1 whichimplies that (ϕs)∗W = W , taking the derivative at s = 0 we get, using the previousproposition, that [V,W ] = 0.We now assume that [V,W ] = 0. We only need to show that (ϕs)∗W = W . To see it
we compute, using that (ϕs)∗V = V :
0 = [V,W ] = [(ϕs)∗V, (ϕs)∗W ] = [V, (ϕs)∗W ].
Thus:
0 = [V, (ϕs)∗W ] =d
dt |t=0(ϕt)∗(ϕs)∗W =
d
dt |t=0(ϕt+s)∗W =
d
ds(ϕs)∗W.
And, since ϕ0 = idM , (ϕ0)∗W = 0. With the previous equality this implies that(ϕs)∗W = 0.
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2.7 Exercises
Exercise 2.1. Consider polar coordinates (r, θ) defined by x = r cos θ, y = r sin θ for(x, y) ∈ R2\0. Write the vector fields ∂r and ∂θ in term of ∂x and ∂y.
Exercise 2.2. Let ψ : (R,+)→ (GLn(R),×) be a continuous group morphism.
1. Show that there exists t0 > 0 and A0 ∈ Mn(R) such that ψ(t0) = exp(A0). Hint:use that exp : Mn(R)→ GLn(R) is a diffeomorphism on a neighborhood of 0.
2. Show that for any r ∈ Q, ψ(rt0) = erA0 .
3. Show that there exist a unique A ∈Mn(R) such that ψ(t) = etA.
Exercise 2.3. Let (x, t) 7→ ϕt(x) be a smooth map from O × R→ O such that ϕt+s =ϕt ϕs. Show that ϕt is the flow of a vector field on O. Apply this result to
ϕt(x, y) =
(cos t sin t− sin t cos t
)(xy
).
Exercise 2.4. Let V : Rn → Rn be a vector field. Assume that there is a constantC > 0 such that | 〈x, V (x)〉 | ≤ C‖x‖2. Show that V is complete.
Exercise 2.5. * Consider the vector field V (x, y) = (ax−bxy,−cy+dxy), for a, b, c, d >0 on U = (x, y)|x > 0, y > 0. The differential equation (x′, y′) = V (x, y) is calledthe Lotka-Volterra equation, it models the evolution of an two interacting populationsof preys (x(t)) and predators (y(t)).
1. Consider the function: f : U → R given by f(x, y) = dx − c log x + by − d log y.Show that f is proper.
2. Show that f(x(t), y(t)) is constant when (x(t), y(t)) is an integral curve of V .
3. Show that V is complete.
4. Show that there exists a unique (x∗, y∗) ∈ U such that V (x∗, y∗) = 0.
5. Show that (x∗, y∗) is the unique maximum of f .
6. Show that (x∗, y∗) is a stable equilibrium of V .
This means that for any neighborhood V of (x∗, y∗), there is a neighborhood V ′
of (x∗, y∗) such that for any integral curve (x(t), y(t)) satifying (x(0), y(0)) ∈ V ′,(x(t), y(t)) ∈ V for t ≥ 0.
Exercise 2.6. ** Consider f : Rn → Rn a smooth map. We will show that if f is alocal diffeomorphism and is proper, then f is a (global) diffeomorphism. We assumethat f is a proper local diffeomorphism.
1. Show that f(Rn) ⊂ Rn is open and closed. Use it to prove that f is surjective.
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2. Show that for any y ∈ Rn, f−1(y) is finite.
We want to show that f−1(y) is in fact a single point (to see that f is injective).Replacing f by f − y, we see that it is enough to show that f−1(0) is a singlepoint, the next questions give a proof of this fact.
3. Consider the vector field V (x) = (−Dxf)−1f(x), and denote by ϕt his flow. Showthat f(ϕt(x)) = e−tf(x) and use it to prove that V is complete.
4. Let f−1(0) = x1, · · · , xk denote the equilibrium points of V . Show that eachxi is an asymptotically stable equilibrium point.
This means that there exist a neighborhood V of xi such any integral curve x(t)which satisfies x(0) ∈ V satisfies limt→∞ x(t) = xi.
5. Let Oi = x0 ∈ Rn| limt→∞ ϕt(x) = xi. Show that the Oi are disjoint and that⋃iOi = Rn.
6. Show, using the smoothness of ϕt, that Oi is open. Conclude that f−1(0) isreduced to a single point.
Exercise 2.7. Use the Stone-Weierstrass theorem and bump functions to show thatthat on a compact manifold M , C∞(M) is dense in C0(M) (equipped with the supnorm).
Exercise 2.8. * Let Mn be a compact smooth manifold, (Ui, ϕi)i=1,...N be a finite atlasof M , and for each i let Vi be relatively compact open subset of Ui such that the Vi’scover M . Let ηi be be smooth function with support in Ui which is equal to 1 on Vi.
Let fi : M → Rn be defined by fi(x) = ηi(x)ϕi(x) for x ∈ Ui and fi(x) = 0 for x /∈ Ui.Define F : M → RN(n+1) by:
F (x) = (f1, . . . , fN , η1, . . . , ηN).
Show that F is a smooth embedding.Using more elaborated tools, one can show that M embeds in R2n.
Exercise 2.9. ** Let M be a smooth manifold. Using the flow of compactly supportedvector fields, show that the group of diffeomorphism of M acts transitively on M .First consider the case of balls in Rn, then use a covering of M by open sets diffeo-
morphic to open balls to globalize.
Exercise 2.10. 1. Let π1 : E1 → M1 and π2 : E2 → M2 be two vector bundles.Show that π1 × π2 : E1 × E2 →M1 ×M2 is a smooth vector bundle.
2. Let π : E → M be a smooth vector bundle and f : N → M be a smooth map.Set:
f ∗E = (p, v) ∈ N × E|f(p) = π(v)
and letπf∗E be the restriction to f ∗E of the first projection N × E → N . Showthat πf∗E : f ∗E → N is a smooth vector bundle.
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The construction in the first question is called the product of E1 and E2, the construc-tion in the second question is the pullback of E along f . An important example is whenπ1 : E1 → M and π2 : E2 → M are two vector bundles over the same base M . On canthe consider the diagonal map ∆ : M → M ×M , and the bundle ∆∗(E1 × E2) is calledthe direct sum of E1 and E2. As a set it can be desribed as:
E1 ⊕ E2 = (p, v1, v2) ∈M × E1 × E2|π1(v1) = π2(v2) = p.
Exercise 2.11. Let Sk ⊂ Rn be a smooth submanifold. The normal space at p ∈ S isdefined by:
NpS = v ∈ Rn|v ⊥ TpS.
1. Show that the normal bundle NS = (p, v)|p ∈ S, v ∈ NpS is a vector bundle ofrank n− k.
2. Show that NSn is a trivial vector bundle.
3. Show that NS ⊕ TS is a trivial vector bundle.
Exercise 2.12. Consider the skew field of quaternions H = x+iy+jz+kt|(x, y, z, t) ∈R4. Recall that ij = −ji = k, jk = −kj = l and ki = −ik = j. The modulus ofq = x + iy + jz + kt is defined by |q| = x2 + y2 + z2 + t2 and |qq′| = |q||q′|. ConsiderS3 = q ∈ H||q| = 1.
1. Show that z ∈ S3 7→ iz defines a smooth nowhere vanishing vector field on S3.
2. Show that TS3 is trivial.
Exercise 2.13. Let A,B ∈ Mn(R) and consider the two vector fields on Rn definedby VA(x) = Ax and VB(x) = Bx. Compute their Lie bracket and relate it to thecommutator of A and B.
Exercise 2.14. Let V (x, y) = (V x(x, y), V y(x, y)) be a vector field on R2. Show thatthe flow of V commutes with rotations around the origin if and only if:
V x + y∂xVy = 0
V y − x∂yV x = 0.
Hint: find a vector field on R2 whose flow acts by rotations around the origin.
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3 Lie groups and homogenousspaces
3.1 Definitions
Definition 3.1.1. A Lie group is a group G with a smooth manifold structure suchthat the group law (g, h) 7→ gh from G×G to G and the inverse g 7→ g−1 from G to Gare smooth.
Remark 3.1.2. One can also define topological groups, by requiring G to be equippedwith a topology such that the group law and the inverse are continuous.
3.2 Examples
3.2.1 Abelian examples
The group (R,+) is a Lie group. More generally any finite dimensional vector space Ewith the group structure given by vector addition is a Lie group.
The circle S1 is a Lie group, there are two different ways to see its group structure.The first one is to see S1 as the quotient space R/Z, since Z is a normal subgroup ofR, R/Z has a natural group structure. The smoothness of the group structure can beshown using Theorem 1.4.15.
Another way to go is to see S1 as the group of modulus 1 complex numbers. Thegroup law on S1 is then the restriction of the usual product C × C → C, and hence issmooth. The same holds for inversion.
Using products, the torus Tn of dimension n is then a Lie group. All these examplesare abelian.
3.2.2 The 3-sphere as a Lie group
Consider S3 as the unit sphere in R4 and identify R4 with the skew field of quaternionsH. Then S3 = q ∈ H| |q| = 1. Moreover, since |qq′| = |q||q′|, S3 is stable underquaternion multiplication. The inverse of a quaternion q is given by q/|q|2, so S3 is alsostable by inversion. So S3 is a subgroup of H∗, since the product and inverse are smoothon H∗, they are smooth when restricted to the smooth submanifold S3. Hence S3 is aLie group.
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3.2.3 Matrix groups
Gln(R) is at the same time a smooth manifold (as an open subset ofMn(R)) and a group.It is a classical that matrix product and inversion are smooth maps. Thus Gln(R) is aLie group. The same holds for Gln(C).Assume G ⊂ Gln(R) is at the same time a subgroup and a submanifold, then G is
automatically a Lie group (group law and inverse are smooth as restrictions of smoothmaps). We have already seen examples Sln(R), O(n,R) and SO(n,R). We can similarlyshow that Sln(C), U(n,C) and SU(n,C) are Lie groups.
3.2.4 Transformation groups
When one works with a geometric structure on some set X, it is often useful to considerthe group G of bijections of X which preserve this structure. In a lot of cases, this turnout to be a Lie group.
Consider Rn as an affine space, any affine bijection f : Rn → Rn can be uniquelywritten as f(x) = Ax + b for some A ∈ Gln(R) and some b ∈ Rn. These form a group,the affine group Aff(Rn) which can also be characterized as the group of continuousbijections of Rn which maps triplets of aligned points to triplets of aligned points. As aset Aff(Rn) is just Gln(R)×Rn, which gives it a smooth structure. On can check thatmultiplication and inversion of affine maps are smooth maps, and this gives Aff(Rn)the structure of a Lie group.We have defined homographies of RPn in chapter one. They are diffeomorphisms of
RPn given by fA([x]) = [Ax] for some A ∈ Gln+1(R). It is easy to see that A 7→ fA definea group morphism ϕ from Gln+1(R) to the group of diffeomorphisms of RPn. The kernelof this morphism is the normal subgroupH = R∗I where I is the identity. The projectivegroup PGln(R) (or group of homographies) is the image of ϕ. It is isomorphic toGln+1(R)/H. Arguing as in the construction of RPn in chapter 1, one can can show thatthat there is a unique smooth structure on PGln(R) such that ϕ : Gln+1(R)→ PGln(R)is a smooth submersion. Once this is done it is easy to show that multiplication andinversion in PGln(R) are smooth. Thus PGln(R) is a Lie group.
3.3 Lie morphisms, Lie subgroups
Definition 3.3.1. Let G and H be two Lie groups. f : G → H is a Lie groupmorphism if it is smooth and a group morphism.
Example 3.3.2. The standard submersion π : Rn → Tn is a Lie group morphism.Example 3.3.3. The determinant is a morphism from any matrix group to the Lie group
(R,+).Example 3.3.4. To each affine transformation L(x) = Ax+ b one can associate its linear
part A, this defines a Lie morphism from Aff(Rn) to Gln(R).
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Definition 3.3.5. H ⊂ G is a Lie subgroup if it is at the same time a subgroup anda submanifold of G.
Remark 3.3.6. This definition is not universal, some author define a Lie subgroup to bean injective immersion of H into G which is a group morphism. When the image ofthis immersion is moreover a submanifold, H is said to be a proper Lie subgroup.For instance matrix groups are Lie subgroups of Gln(R).The kernel of a Lie group morphism is always a Lie subgroup, this follows from the
fact that a closed subgroup of a Lie group is a Lie subgroup: this is a hard theorem thatwe won’t prove here (a proof can be found in [Lee02]). However this is not always truefor the image.Example 3.3.7. Consider the Lie group morphism f : R → R2 given by f(t) = (t, αt)
where α is some real number. Consider F = π f , where π is the projection fromR2 to the torus T2, it is a Lie group morphism from R→ T2.
If α is rational equal to p/q, then F (t) = 0 if and only if t ∈ qZ. This implies thatF pass through the quotient as a smooth group morphism F from R/qZ to T2, whichis moreover an injective immersion. So F (R) = F (R/qZ) is a submanifold of T2. SoF (R) is a Lie subgroup of T2 which is moreover isomorphic to S1 as a Lie group.
Assume now that α /∈ Q. Then it is a classical fact that the set bkαc|k ∈ Z isdense in [0, 1]. This can be used to show that F (R) is in fact dense in T2, thus ifF (R) cannot be a submanifold of dimension 1, if it was a submanifold of dimension2, it would be an open set, hence it would be the whole of T2 since it is dense. SoF (R) is ot a submanifold of T2 and hence not a Lie subgroup.
3.4 Left invariant vector fields
An important feature of Lie groups is that they have distiguished diffeomorphisms. Weset:
Lg : G→ G Rg : G→ G
h 7→ gh h 7→ hg,
these are diffeomorphisms (since Lg Lg−1 = Lg−1 Lg = idG) called left and righttranslations. The associativity of G translate into the fact that Lg Rh = Rh Lg.Another important diffeomorphism is the inversion I : G→ G.Let X be a vector field on G.
Definition 3.4.1. X is said to be left invariant (resp. right invariant) if for anyg ∈ G, (Lg)∗X = X (resp. (Rg)∗X = X).
We will develop the theory of left invariant vector fields below, the case of rightinvariant vector fields is parallel and we will omit it.
The set of left invariant vector fields is in fact linear subspace of the space of vectorfields on G. A left invariant vector field is determined by its value at the neutral elemente ∈ G.
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Proposition 3.4.2. The map X 7→ Xe from the set of left invariant vector fields on Gto TeG is a linear isomorphism.
Proof. If X is left invariant, then by definition of the push forward Xg = TeLg(Xe) forany g ∈ G. So if Xe = 0, X = 0 and X 7→ Xe is injective.Let us now fix v ∈ TeG. We set Xg = TeLg(v), then:
(Lh)∗Xg = Th−1gLh TeLh−1g(v) = Te(Lh Lh−1g)v = TeLg(v).
So we only need to show that this define a smooth vector field. In order to do this wewill show it defines a derivation on C∞(G). Let f be a smooth function, we need toshow that g 7→ Tgf(Xg) is smooth. Consider a smooth curve c(t) such that c(0) = e andc′(0) = v. Set F (t, g) = f(gc(t)) = f(Lg(c(t))), then g 7→ (t 7→ F (g, t))′(0) is smooth.To conclude remark that:
d
dt|t=0F (g, t) = Tgf(TeLg(v)) = Tgf(Xg).
Example 3.4.3. Let us consider the case of G = GLn(R). Then Lg(h) = gh is linear inh, so TeLg = Lg. And if Xe ∈ TeG 'Mn(R), then the Left invariant extension X ofXe will be given by Xg = gXe.This in particular implies that TG is a trivial bundle: pick a basis (e1, . . . , en) of TeG
and consider the left invariant vector fields Ei such that Eie = ei, these are vector fields
which form a basis of TgG at each g.
Proposition 3.4.4. 1. If X and Y are left invariant, so is [X, Y ].
2. If X is left invariant, then I∗X is right invariant.
3. If X is left invariant, so is (Rg)∗X.
Proof. The first point comes from the fact that (Lg)∗[X, Y ] = [(Lg)∗X, (Lg)∗Y ] = [X, Y ].For the second, notice that, since (gx)−1 = x−1g−1, I Lg = Rg−1 I. So, if X is left
invariant:
(Rg)∗I∗X = (Rg I)∗X = (I Lg−1)∗X = I∗(Lg−1)∗X = I∗X.
The third one comes from the fact that Lg Rh = Rh Lg.
Definition 3.4.5. A one parameter subgroup of a Lie group G is a smooth morphismh from R to G.
Example 3.4.6. We have seen in the examples that the one parameter subgroups ofGln(R) are of the form h(t) = etA.
If G is a Lie subgroup of Gln(R) and h is a one parameter subgroup of G, then h isa one parameter subgroup of Gln(R), so h(t) = etA for some A ∈Mn(R). Moreover,since h(t) ∈ G, A = h′(0) ∈ TIG. We will see that any A ∈ TIG actually gives riseto a one paramter subgroup of G.
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From a one parameter subgroup of G, we get a one parameter subgroup of diffeomor-phisms of G by setting:
ϕt(x) = xh(t) = Rh(t)(x).
Proposition 3.4.7. ϕt is the flow of a left invariant vector field X. Conversely, theflow of any left invariant vector field X is of the form ϕt(x) = xh(t) for some uniquelydetermined one parameter subgroup h.
Proof. We know that ϕt is the flow of the vector field:
Xp =d
dt |t=0ϕt(p).
We want to show that (Lg)∗X = X, but the flow of (Lg)∗X is given by
Lg ϕt L−1g = Lg Rh(t) L−1
g = Lg Lg−1 Rh(t) = ϕt.
So X and (Lg)∗X have the same flow, hence they are equal.Conversly, let h(t) be the solution to h′ = X(h), h(0) = e. We have, since (Lg)∗X = X,
that ϕt Lg = Lg ϕt, hence:
ϕt(g) = ϕt(Lg(e)) = Lg(ϕt(e)) = gh(t).
So we only need to show that h(t) is a one parameter subgroup, the fact that h(t+ s) =h(s)h(t) comes from the similar property of ϕt. We need to show that h is in fact definedon the whole of R. We know that h(t) exists for t ∈ (−ε, ε), now if h is defined up toT > 0, it is easy to see that h(t)h(eps/2) yiedls a solution defined until T + ε/2. So his defined for all t ≥ 0. The same argument works backward in time, so h(t) is definedfor any real t.
3.5 The Lie algebra of a Lie group
Definition 3.5.1. A Lie algebra on a field K is a K-vector space V endowed with anantisymmetric K-bilinear map [ , ] : V × V → V which satisfy the Jacobi identity:
[[x, y], z] + [[y, z], x] + [[z, x], y] = 0.
Example 3.5.2. The vector space of vector fields on a manifold M , with the Lie bracketof vector fields is a Lie algebra.
Example 3.5.3. If G is a Lie group, then the space of left invariant vector fields withthe Lie bracket is a Lie algebra. Since there is a canonical isomorphism between leftinvariant vector fields and TeG (given by X 7→ Xe), this gives a canonical lie algebrastructure on TeG. This Lie algebra is called the Lie algebra of G and usuallydenoted by g.
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Example 3.5.4. Let us specialise the previous discussion to G = GLn(R). ConsiderA,B ∈ TeG 'Mn(R), the associated one parameter subgroups are etA and etB whilethe flows are ϕAt : M 7→ MetA and ϕBt : M 7→ MetB. To compute the Lie bracket ofA and B, let us consider the left invariant extensions XA(g) = gA and XB(g) = gB.
We know that:[XA, XB] =
d
dt |t=0(ϕBt )∗XA
but:(ϕBt )∗XA(e) = TϕB−t(e)ϕ
Bt XA(ϕB−t(e)) = e−tABetA.
Using that etA = I + tA + . . . , we get that [XA, XB](e) = BA − AB, thus the Liealgebra structure on Mn(R) is just given by the commutator [A,B] = BA− AB.Actually, for any associative algebra A, the commutator defines a Lie bracket onA.
Definition 3.5.5. A Lie subalgebra of V is a linear subspace of V which is stableunder [ , ].
Definition 3.5.6. A Lie algebra morphism is a linear map A from (V, [ , ]V ) to(V ′, [ , ]V ′) such that:
[A(x), A(y)]V ′ = A([x, y]V ).
A Lie algebra isomorphism is a bijective Lie algbra morphism, it is then automaticthat its inverse is also a Lie algebra morphism.
Definition 3.5.7. The exponential map of a Lie group G is the map exp : g → Gwhich associates to each Xe ∈ g ϕ1(e) where ϕt is the flow of the left invariant vectorfield X such that Xe = Xe.
One can check that this generalises the matrix exponential: if A ∈ Mn(R) (which isthe Lie algebra of Gln(R), the X : g 7→ gA is the left invariant vector field such thatXI = A and c : t 7→ etA statisfies c′ = Xc and c(0) = I.
Proposition 3.5.8. For any one parameter subgroup h(t) of G, there exist a uniqueXe ∈ g such that h(t) = exp(tX)
Proof. We have seen that for any one parameter subgroup h(t) there is a left invariantvector field X such that h(t) = ϕt(e) where ϕt is the flow of X. This comes from the factthat if c(t) is a solution to c′ = X(c), then cα(t) = c(αt) is a solution to c′α = αX(cα).
In particular, the flow of a left invariant veector fieldX is given by: ϕt(x) = x exp(tXe).
Proposition 3.5.9. exp : g → G is smooth and a local diffeomorphism on a neighbor-hood of 0 ∈ g.
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Proof. The smoothness of exp comes from the fact that if Xα, where α is a parameterin an open set of a vector space, is a smooth family of vector fields then the map(x, t, α) 7→ ϕαt (x) where ϕαt is the flow of Xα is smooth. (In this α ∈ g.)We now just need to compute the differential of exp at 0. To see this consider the
curve c : t 7→ tXe in g. Then one shows that exp(tXe) is a solution to c′ = X(c) with Xthe left invariant extension of Xe, thus c′(0) = Xc(0), thus:
Te exp(Xe) = Xe
which shows that Te exp is invertible.
We have seen that the pushforward by Rg of a left invariant vector field is again leftinvariant. The following proposition shows how this translates under the identificationof the space of left invariant vector fields with TeG.
Proposition 3.5.10. Let X be a left invariant vector field, then:
(Rg)∗Xe =d
dt |t=0(g−1 exp(tXe)g).
Proof. The flow ofX is ϕt(x) = x exp(tXe), thus the flow of (Rg)∗X is ϕt = RgϕtRg−1 .Then:
(Rg)∗Xe =d
dt |t=0ϕt(e) =
d
dt |t=0(g−1 exp(tXe)g).
We define, for every g ∈ G a linear map Adg : g→ g by:
Adg(Xe) =d
dt |t=0(g exp(tXe)g
−1) = (Rg−1)∗Xe.
It follows from the properties of pushforwards of vector fields that Ad(g1g2) = Ad(g1) Ad(g2). This shows that Ad : G→ Gl(g) is a linear representation of G. Moreover:
[AdgX,Adg Y ] = Adg[X, Y ],
in other words, Adg is a Lie algebra morphism.Ad is called the adjoint representation of G.
Proposition 3.5.11.d
dt |t=0(Adexp(tX) Y ) = [X, Y ].
Proof. We have that Adexp(tX) Y = (Rexp(−tX))∗Y , but Rexp(−tX) is the flow of −X,proposition 2.6.4 then gives the answer.
We this we can show the following fundamental property of Lie group morphisms:
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Proposition 3.5.12. Let f : G → H be a Lie group morphism, then Tef : g → h is aLie algebra morphism.
Proof. Let Xe ∈ g and h(t) be the one parameter subgroup of G generated by Xe, thenk(t) = f(g(t)) is a one parameter subgroup of H, thus corresponds to a unique Ze ∈ h.To find this Ze, we consider the flow of the left invariant extension Z of Ze, we know itis given by: ϕt : x 7→ xk(t), so Ze = d
dt |t=0ϕt(e) = d
dt |t=0k(t) = Tef(Xe). In particular
f(exp(tXe)) = exp(tZe).Now for any g ∈ G, we have f(g exp(tXe)g
−1) = f(g) exp(tZe)f(g−1). Taking thederivative of the last equality with respect to t at t = 0, Proposition 3.5.10 yields:
Tef(AdgXe) = Adf(g) Ze = Adf(g) Tef(Xe).
We now take g to be exp(sYe) and differentiate with respect to s at s = 0, the lastproposition gives:
Tef([Xe, Ye]) = [Tef(Xe), Tef(Ye)].
Example 3.5.13. Consider the determinant det : Gln(R) → R∗, this is a Lie groupmorphism. So Te det = trace is a Lie algbra morphism from Mn(R) to R, sotrace([A,B]) = [trace(A), trace(B)] = 0 (since the Lie bracket on R is trivial).
Corollary 3.5.14. Let H ⊂ G be a Lie subgroup of G. Then its Lie algebra h is a Liesubalgebra of g.
Proof. The injection H → G is a Lie group morphism.
Example 3.5.15. Let us consider G ⊂ Gln(R) a Lie subgroup. Then g is a Lie subalgebraofMn(R), and the Lie bracket on g is just the restriction of the Lie bracket ofMn(R),which is the matrix commutator.
For instance if G = O(n), then g = TIO(n) is the space of antisymmetric matrices.We recover in this way that a commutator of antisymmetric matrices is antisymmet-ric.The following theorem is an example of the rigidity Lie groups enjoy:
Theorem 3.5.16. Let f : G → H be a continuous group morphism between two Liegroups, then f is smooth.
The proof of this theorem will require the following Lemmas:
Lemma 3.5.17. Let f : G→ H be a group morphism which is smooth on a neighborhoodof e ∈ G, then f is smooth on G
Proof. Pick any g ∈ G, then: Lf(g) f Lg−1 = f , the left hand side is smooth on aneighborhood of g since Lg−1(g) = e and left multiplication is smooth.
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Remark 3.5.18. We can differentiate this identity to get:
TeLf(g)Tef TgLg−1 = Tgf.
Since left multiplication is a diffeomorphism, it follows that if f is a Lie group mor-phism, then f is an immersion (resp. submersion, local diffeomorphism) if and onlyif Tef is injective (resp. surjective, bijective).
Lemma 3.5.19. Let (X1, . . . Xn) be a basis of g then the map:
ϕ : (t1, . . . , tn)→ exp(t1X1) · · · exp(tnXn)
is a diffeomorphism on a neighborhood of 0.
Proof. ϕ is smooth. Moreover we have:
T0ϕ(Xi) =d
dt |t=0exp tXi = Xi.
Thus T0ϕ : g→ g is the identity.
Lemma 3.5.20. Let h : R→ H be continuous group morphism, then there exist Y ∈ hsuch that h(t) = exp(tY ).
Proof. Consider a ball B centered at 0 in h such that exp is a diffeomorphism from Bto U = expB/2. Then f−1(U) contains the interval I = (−ε, ε) for ε > 0 small enough.Pick t0 ∈ I, then there exist Y ∈ B/2 such that f(t0) = exp(Y ). Similarly, there isY ′ ∈ B/2 such that f(t0/2) = exp(Y ′). Then we have that:
exp(2Y ′) = f(t0/2)2 = f(t0) = exp(Y ).
Since 2Y ′ and Y belong to B where exp is a diffeomrphism, Y ′ = Y /2. Repeatingthe following argument, we get that f(t0/2
k) = exp(Y /2k), and the group morphismproperty of f implies that f(nt0/2
k) = exp(nY /2k), since f is continuous we must havef(st0) = exp(sY ), which implies that f(s) = exp(sY ) for Y = Y /t0.
Proof (of Theorem 3.5.16). Consider a basis (X1, . . . , Xn) of g, then f(exp(tXi)) is acontinuous morphism from R to H, hence f(exp(tXi)) = exp(tYi). Consider ϕ as inLemma 3.5.19, then:
f ϕ(t1, . . . , tn) = f(exp(t1X1) · · · exp(tnXn)) = exp(t1Y1) · · · exp(tnYn)
is smooth from Rn to H. Since ϕ is a diffeo from a neighborhood of 0 ∈ Rn to aneighborhood of e ∈ G; f is smooth on a neighborhood of e, and hence smooth.
3.6 Homegeneous spaces
The goal of this section is to show that the quotient spaceG/H whereH is a Lie subgroupof G has a natural smooth structure. A lot of the manifolds we have encountered canactually be described as quotients G/H, this manifolds are called (smooth) homogeneousspaces.
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3.6.1 Observations on topological groups
We will need some topological properties of Lie groups which hold more generally fortopological groups. In this subsection, G will be a topological group and H will be aclosed subgroup of G.
Proposition 3.6.1. Let U be a neighborhood of e ∈ G, then there exist a neighborhoodV ⊂ U of e such that V −1 = V and a neighborhoodW of e such thatW 2 = gh|g, h ∈ W ⊂U .
Proof. Since the inversion is continuous, U−1 is open, then V = U ∩ U−1 fits.For the second point, the multiplication µ : G×G→ G is continuous, O = µ−1(U) is
open in G× G and contains (e, e), in particular O contains an open set W ×W whereW is a neighborhood of e.
For any g, h 7→ gh is an homeomorphism of G. In particular, any neighborhood of gis of the form gU for some neighborhood U of e. Using right multiplication instead, anyneighborhood of g is also of the form U ′g.In a topological group, there is a nice interplay between the group structure and
connectedness.
Proposition 3.6.2. Any open subgroup G′ of G is closed, in particular if G is connectedthen G′ = G.
Proof. We want to show that the complement of G′ is open. Consider an open neigh-borhood U of e included in G′, then if g /∈ G′, gU is an open neighborhood of g whichdoesn’t intersect G′. (If gu ∈ G′, then g ∈ G′u−1 = G′.) Thus G\G′ is open.
Proposition 3.6.3. The connected component of e ∈ G, denoted by G0 is an opennormal subgroup of G.
Proof. Consider V a connected open neighborhood of e such that V −1 = V . (To findone, just take the connected component of e in a neighborhood which is stable underinversion.) We will prove the following claim:
G0 =⋃n
V n.
Since V is connected, V n is connected and⋃n V
n is also connected, thus the inclusion⊃ is proved. Moreover
⋃n V
n is a subgroup of G, since if g ∈ V n and h ∈ V k thengh ∈ V n+k and g−1 ∈ V n. Now
⋃n V
n contains a neighborhood of the identity (V ), andis a subgroup, so
⋃n V
n is an open subgroup of G, and thus is closed. Since⋃n V
n isconnected, it is one of the connected component of G, hence
⋃n V
n = G0.To see that G0 is a normal subgroup, just remark that gG0g
−1 is connected andcontains e, and is thus included in G0.
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When H is a subgroup of G, we have the right action of H on G given by h ·x = xh =Rh(x). The relation ∼ defined by x ∼ y if and only if x = yh(= h · y) for some h ∈ His an equivalence on G. The quotient of G by this equivalence is denoted by G/H. Itselements are of the form gH. As usual π : G→ G/H will denote the quotient map.The left action of G on itself g · x = gx = Lg(x) goes down to G/H: if x = yh, then
gx = (gy)h. This action is continuous and transitive. We will again denote it by Lg, itsatisfies: Lg π = π Lg.We now show some topological properties of G/H endowed with the quotient topology.
Proposition 3.6.4. 1. π : G→ G/H is open.
2. G/H is Hausdorff.
Proof. U ⊂ G/H is open if and only if π−1(U) is open in G. Let O ⊂ G be open, thenO = π−1(π(O)) can be written as:
O =⋃g∈G
gO.
Since each gO is open, O is open.For the second point, let x = π(x) and y = π(y) be two distinct point in G/H. Then
x doesn’t belong to yH, which is closed in G. Thus there is a neighborhood U of e suchthat Ux∩yH =. This implies that UxH∩yH =: if uxh′ = yh, then ux = y(hh′−1) ∈ yH.Consider a neighborhood V ⊂ U of e such that V 2 ⊂ U and V −1 = V , then V 2xH ∩
yH =, which implies that V xH ∩V yH =. This says that π(V x) and π(V y) are disjoint,since π is open we are done.
3.6.2 Quotient of Lie groups
We now get back to the case where G is a Lie group and H is a Lie subgroup of G.This whole section is devoted to the proof of the following result:
Theorem 3.6.5. There exist a unique smooth structure on G/H such that π : G→ G/His a smooth submersion.
We already know that G/H is Hausdorff, the second countable property then follows.We set M = G/H.We now need to build charts onM , for this we will use the exponential map. Consider
a suppelment m to h in g. We have g = m⊕ h.
Lemma 3.6.6. The map ψ : m⊕ h→ G defined by:
(X, Y ) 7→ exp(X) exp(Y )
is a local diffeomorphism on a neighborhood of 0.
Proof. One sees easily that T0ψ((X, 0)) = (X, 0) and T0ψ((0, Y )) = (0, Y ). Thus T0ψ isan isomorphism from m⊕ h to g, the inverse function theorem applies.
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Let V ⊂ h and U ⊂ m be neighborhoods of 0 such that ψ is a diffeomorphism fromU × V to its image.
Lemma 3.6.7. There exists a neighborhood U ′ ⊂ U of 0 ∈ m such that for any g ∈ G,the map ϕg : U ′ →M defined by:
ϕg(X) = π(g exp(X))
is an homeomorphism onto its image.
Proof. Since ϕg = Lg ϕe and Lg : G/H → G/H is an homeomorphism, we only needto work with ϕe.
We will first restrict U in the following way: first choose W a neighborhood of e ∈ Gsuch that W 3 ⊂ ψ(U × V ) and W−1 = W , and choose U ′ and V ′ neighborhoods of 0 inh and m such that U ′ × V ′ ⊂ ψ−1(W ).We now show that ϕe is injective on U ′. Pick X and X ′ in U ′ such that ϕe(X) =
ϕe(X′). This means that there exists h ∈ H such that exp(X) = exp(X ′)h, moreover
h = (expX)−1 exp(X) ∈ W 2, thus exp(X ′)h ∈ W 3 ⊂ ψ(U × V ). We can then apply theprevious lemma to conclude that X = X ′.To prove that ϕe is an homeomorphism on its image, we only need to show that ϕe is
open. To see this, just remark that if O ⊂ U ′ is open, then O×V ′ is open and ψ(O×V ′)is also open since ψ is a diffeomorphism. But ϕe(O) = π(ψ(O × V ′)) and π is open, soϕe(O) is open.
Proof (of Theorem 3.6.5). The (π(g exp(U ′)), (ϕg)−1) from the previous lemma form a
topological atlas M , to show that it is smooth we need to show that ϕ−1g2ϕg2 is smooth.
Asssume that O = π(g1 exp(U ′)) ∩ π(g2 exp(U ′)) is not empty, then for any x ∈ Othere exist unique X1 and X2 such that:
x = π(g1 exp(X1)) = π(g2 exp(X2)),
and we need to show X1 7→ X2 is a diffeomorphism.Fix some x ∈ O and let X1 and X2 such that ϕg1(X1) = ϕg2(X2), then there exist
h ∈ H such that g1 exp(X1)h = g2 exp(X2).Consider the neighborhood W of e from the proof of the previous lemma. Then
g2 exp(X2) = g1 exp(X1)h ∈ g2W , so g−12 g1 exp(X1)h ∈ W . Hence for X1 in some neigh-
borhood U of X1, g−12 g1 exp(X1)h ∈ W . Hence there exist X ∈ m and Y ∈ h, depending
smoothly on X1 ∈ U , such that: g−12 g1 exp(X1)h = ψ(X, Y ) = exp(X) exp(Y ). So:
g2 exp(X) exp(Y ) = g1 exp(X1)h.
Moreover, we have:
ϕg2(X) = π(g2 exp(X)) = π(g2 exp X exp Y )
= π(g1 exp(X1)h) = π(g1 exp(X1))
= ϕg1(X1) = x
= ϕg2(X2).
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So X = X2 since ϕg2 is injective. Since X depends smoothly on X1, we have defined asmooth structure on M .
We now prove that π : G→ G/H is a submersion. Consider the charts of G given bythe inverse of the maps ψg : (X, Y ) 7→ gψ(X, Y ), in these charts arguing as before wehave ϕ−1
g π ψg(X, Y ) = X, which is a submersion.For the uniqueness part, the structure of topological manifold on M has to be given
by the atlas we considered (since the toplogy on M is the quotient topology), and therequirement that π is smooth forces the maps ϕg to be smooth.
3.6.3 Homogeneous spaces
On M = G/H, since π is a submersion, the right action Rg of G on itself goes throughthe quotient to a smooth action of G on M which is transitive. A manifold on whicha Lie group acts smoothly transitively is called a smooth G-homeneous space. Thefollowing theorem says that all such manifolds are of the form G/H.
Theorem 3.6.8. Let M be a smooth G-homogeneous space and x ∈ M , set Gx = g ∈G|g ·x = x (the stabilizer of x), then Gx is a Lie subgroup of G and M is diffeomorphicto G/Gx.The diffeomorphism is built by quotienting the map F (g) = g · x.
The proof uses some tools that we haven’t introduced and won’t be given here. How-ever a lot of the examples of manifolds we have seen so far can be seen in this way.Example 3.6.9. Consider the sphere Sn in Rn+1. The Lie group G = SO(n + 1) acts
transitively on Sn. The stabilizer of (1, 0, . . . , 0) is given by the subgroupH consistingof matrices of the form: (
1 00 h
)where h ∈ SO(n). We then have a diffeomorphsim from G/H to Sn.
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3.7 Exercises
Exercise 3.1. Consider Cn with its usual hermitian structure (〈z1, z2〉 =∑
i zi1zi2).
Recall the adjoint M∗ of M ∈Mn(R) is defined by M∗ = MT .The unitary group U(n) is the group of matrices M ∈ GLn(C) such that MM∗ = I,
the special unitary group SU(n) is the intersection of U(n) with SLn(C).Show that U(n) and SU(n) are Lie subgroups of GLn(C), and describe their Lie
algebras.
Exercise 3.2. Let E be a finite dimensional (real or complex) vector space and q be anon degenerate quadratic form on E, set:
O(q) = A ∈ Gl(E)|∀x ∈ E, q(Ax) = q(x)
We will denote by b : E × E → R the associated symmetric bilinear form.The goal of this exercise is to prove that O(q) is a Lie group.
1. Show that for anyM ∈ L(E), there exist a uniqueM∗ ∈ L(E) such that b(Mx, y) =b(x,M∗y) for any x, y ∈ E. Show that M 7→M∗ is linear (and hence smooth).
2. Show that M ∈ O(q) if and only if MM∗ = I.
3. Let ϕ : M 7→ MM∗ and S = M ∈ L(E)|M = M∗. Show that ϕ(M) ∈ S andshow that for any M ∈ O(q), TMϕ : E → S is surjective.
4. Show that O(q) is a Lie subgroup of GL(E) and compute its tangent space at I.
Nondegenerate quadratic forms on Rn are classified by their signature, they are allcongruent to one of the forms q(x) =
∑pi=1(xi)2 −
∑ni=p+1(xi)2. The orthogonal group
of this particular quadratic form is denoted by O(p, n− p).
Exercise 3.3. Let E be the space of 2 symmetric matrices.
1. Show that det is a quadratic form on E of signature (2, 1).
2. For any A ∈ SL2(R), define ϕA : E → E by ϕA(S) = ASAT . Show that A 7→ ϕAis a Lie group morphism from SL2(R) to O(det).
3. Show that O0(2, 1) is isomorphic to SL2(R)/I,−I
Exercise 3.4. Show that a vector field on (Rn,+) is left invariant if and only if it isconstant. Describe left invariant vector fields of GLn(R).
Exercise 3.5. Show that the vector fields on S3 ⊂ H given by X1 : x 7→ ix, X2 : x 7→ jxand X3 : x 7→ kx and form a basis of the space of left invariant vector fields. Compute[X1, X2].
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Exercise 3.6. * Consider the affine group G = Aff(Rn). We define a map ϕ : G →GLn+1(R) by sending the affine map f(x) = Ax+ b to the matrix:(
1 0b A
).
Show that ϕ is an injective Lie group morphism and that ϕ(G) is a Lie subgroup ofGLn(R). Use it to describe the Lie bracket on TeG in terms of A and b.
Exercise 3.7. Show that on G is a matrix subgroup, then for g ∈ G and X ∈g,Adg(X) = gXg−1. Prove directly that, for [X, Y ] ∈ g:
d
dtAdexp(tY )X = [Y,X].
Exercise 3.8. Consider the set of matrices in A = M2(C) of the form:(α β−β α
)1. Show that A is a complex subalgebra of M2(C).
2. Consider the map f : A → H: (α β−β α
)7→ α + βj
Show that f is a C-linear isomorphism. (H is viewed as a C vector space with theleft multiplication.)
3. Show that f is moreover an algebra isomorphism.
4. Use this isomorphism to show that S3 is isomorphic to SU(2).
Exercise 3.9. For any unit quaternion q ∈ S3, consider the the transformation ϕq of Hgiven by: ϕq(x) = qxq−1.
1. Show that ϕq is an isometry for the scalar product 〈q1, q2〉 = <(q1q2).
2. Show that ϕq preserves the space of purely imaginary quaternions.
3. Use this fact to build a Lie morphism S3 → SO(3,R), what is its kernel ?
4. Show that SO(3) is diffeomorphic to RP3.
Exercise 3.10. For q1, q2 ∈ S3 × S3, consider the transformation of H defined byϕq1,q2(x) = q1xq
−12 .
1. Use it to define a surjective Lie group morphism S3 × S3 → SO(4), compute itskernel.
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2. Show that SO(4) is not simple. (It has non trivial normal subgroups.)
Exercise 3.11. * Let H ⊂ G be a normal Lie subgroup. Show that h is an ideal of g,that is for any X ∈ g, Y ∈ h, [X, Y ] ∈ h. Hint: "differentiate" the relation ghg−1 ∈ H.
Exercise 3.12. ** Let G be an Abelian connected Lie group.
1. Show that for any X, Y ∈ g, [X, Y ] = 0.
2. Show that exp : (g,+)→ (G,×) is a Lie group morphism.
3. Use the fact that exp is a local diffeomrphism and that G is connected to provethat exp is surjective.
4. Show that the kernel K of exp is a discrete subgroup of (g,+).
Discrete subgroup of vector spaces are of the form Ze1 ⊕ · · · ⊕ Zek where the ei’s arelinearly independant, from this one can deduce that G ' g/K is actually isomorphic to(S1)k × Rn−k.
Exercise 3.13. * For each of the sets below, find a transitive Lie group action and useit to describe it as a homegeneous space:
1. The projective space RPn.
2. The set of affine lines in Rn.
3. The set of k dimensional vector spaces in Rn (this set is called the Grassmannianof k-planes).
4. The of flags in a real vector space E of dimension n. (A flag in E is a sequenceE0, E1, . . . , En of linear subspaces of E such that Ei ⊂ Ei+1 (stricly), in particular,dim(Ei) = i.)
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4 Tensors, differential forms andStokes Theorem
4.1 The cotangent bundle and 1-forms
One of the most important construction in linear algebra is duality, it also has its im-portance in differential geometry. Recall that if E is a real vector space, its dual E∗ isjust the vector space of linear maps from E to R.Also recall that if f : E → F is a linear map, it induces a linear map f ∗ : F ∗ → E∗,
called the transpose of f by setting, for α ∈ F ∗, f ∗α : x ∈ E 7→ α(f(x)) ∈ R, or moreconcisely f ∗α = α f . f ∗ is an isomorphism if and only if f is, and (f g)∗ = g∗ f ∗.
4.1.1 The cotangent bundle
The cotangent bundle of a smooth manifold M , denoted by T ∗M , is, as a set theunion of the duals to the tangent spaces:
T ∗M =⋃p∈M
(TpM)∗.
An element of T ∗M is thus a linear form on some TpM .In the same way as we did for the tangent bundle, we can define a smooth structure on
T ∗M such that the natural projection π : T ∗M → M defines a vector bundle structureon T ∗M .Recall that the charts on TM are given ΦU : TU → U → Rn where (U,ϕU) is a chart
of M and:ΦU : v ∈ TpU → (ϕU(p), TpϕU(v)).
Let us define T ∗U =⋃p∈U(TpM)∗ and:
ΨU : α ∈ TpU → (ϕU(p), (Tpϕ−1U )∗(α)) ∈ U × (Rn)∗.
Since TpϕU is an isomorphism, ((TpϕU)−1)∗ is an isomorphism and ΨU is a bijection. thetopology on T ∗M is built in the same way the topology on TM was. The ΨU form atopological atlas for this topology.We then compute the transition function given by ΨUV = ΨV Ψ−1
U , if ΨV comes froma smooth chart (V, ϕV ):
ΨUV (x, α) =(ϕUV (x), (TϕU (x)ϕ
−1V )∗
((TϕU (x)ϕ
−1U )∗
)−1α)
=(ϕUV (x), ((TxϕUV )∗)−1 α
).
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These are smooth diffeomorphisms. Moreover, ΨU restricted to (TpM)∗ is a linear iso-morphism from (TpM)∗ to (Rn)∗.Thus we have shown:
Proposition 4.1.1. T ∗M with the projection π : T ∗M →M is a smooth vector bundleover M .
4.1.2 Differential 1-forms
Definition 4.1.2. A (smooth) differential 1-form onM is a smooth section of T ∗M .
The vector space of smooth 1-forms on M is usually denoted by Ω1(M).Example 4.1.3. Let f be a smooth function on M , then for each p ∈M , Tpf is a linear
form on TpM . Thus Tf : p 7→ Tpf can be seen as a section of T ∗M , which is usuallydenoted by df .To see that it is smooth we compute:
ΨU df ϕ−1U (x) = ψU(Tϕ−1
U (x)f)
=(x, ((Tϕ−1
U (x)ϕU)−1)∗Tϕ−1U (x)f
)=(x, Tϕ−1
U (x)f Txϕ−1U
)=(x, Tx(f ϕ−1
U ))
which is smooth since f ϕ−1U is.
To express differential forms in coordinates ϕU(p) = (x1(p), . . . , xn(p)), we have anatural basis for each TpM :
(∂∂x1 |p, . . . ,
∂∂x1 |p
). This gives us a dual basis of T ∗pM , that
we denote by (dx1|p, . . . , dx
n|p). p 7→ dxi|p is a actually the differential of the coordinate
function p 7→ xi(p).From this we deduce that any smooth one form α can be written as α =
∑i αidx
i,where the αi’s are smooth functions on U . Moreover if α = df for some smooth functionf : M → R, then df =
∑i(∂i · f)dxi, just as in the case of functions on an open set of
Rn.Let f : M → N be a smooth map and α ∈ Ω1(N).
Definition 4.1.4. The pullback of α by f is the smooth 1-form on M defined by:
f ∗α : v ∈ TpM 7→ α(Tpf(v)).
4.2 A little tensor algebra
4.2.1 Tensor product
Let E and F be two finite dimensional real vector spaces. We will define tensor prod-ucts only for dual spaces, which is enough for our purpose. However this restriction isunnecessary.
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Definition 4.2.1. The tensor product E∗⊗F ∗ is the vector space of bilinear maps fromE × F to R.
There is a natural bilinear map: E∗ × F ∗ → E∗ ⊗ F ∗ defined by:
(f, g) ∈ E∗ × F ∗ 7→ (f ⊗ g : (u, v) 7→ f(u)g(v)).
Assume that E has a basis (ei)i and F has a basis (fj)j, and denote by (ei)i and (f j)jthe corresponding dual basis. Let b ∈ E∗ ⊗ F ∗, u =
∑i u
iei ∈ E and v =∑
j vjfj ∈ F ,
then: b(u, v) =∑
i,j uivjb(ei, fj). Recall that by definition of dual basis: ei ⊗ f j(u, v) =
ei(u)f j(v) = uivj. This show that: b =∑
i,j bijei ⊗ f j with bij = b(ei, fj). In particular
the (ei ⊗ f j) form a basis of E∗ ⊗ F ∗, so dim(E∗ ⊗ F ∗) = dim(E∗)× dim(F ∗).Similarly one defines
⊗ni=1E
∗i as the vector space of n-linear maps from
⊕ni=1Ei to
R. It has dimension dim(E∗1) × · · · × dim(E∗n), and a basis is given by elements of theei11 ⊗ · · · ⊗ einn if (eikk )ik is a basis of (Ek)
∗.Of particular interest to us is the tensor product
⊗k E∗ of E∗ with itself k times, it isthe vector space of k-linear maps from Ek to R (also called k-linear forms). From nowon, n will denote the dimension of EOne can define the tensor product of α ∈
⊗k E∗ and β ∈⊗lE∗ by:
α⊗ β(v1, . . . , vk+l) = α(v1, . . . , vk)β(vk+1, . . . , vk+l).
α ⊗ β is thus an element of⊗k+lE∗. Note that this tensor product is associative but
not commutative.The tensor product turns T (E∗) =
⊕k
⊗k into an associative algebra called thetensor algebra of E∗.
4.2.2 Alternating forms and wedge products
Definition 4.2.2. An n-linear form α ∈⊗k E∗ is said to be alternating if for any
permutation σ ∈ Sk:
α(vσ(1), . . . , vσ(k)) = sign(σ)α(v1, . . . , vk),
where sign(σ) is the signature of the permutation σ.
Proposition 4.2.3. Let α ∈⊗k E∗,
1. α is alternating if and only if α(v1, . . . , vk) = 0 as soon as two of the vi’s are equal.
2. If α is alternating then α(v1, . . . , vk) = 0 as soon as the vk’s are linearly dependant.
Proof. For the first part, since transpositions generate Sk as a group, it is enough to showthat exchanging two of the vk’s in α(v1, . . . , vk) changes the sign, since the signature of atransposition is −1. So we only need to show the condition in (1) is equivalent to the factthat β : (vi, vj) 7→ α(v1, . . . , vk) is antisymmetric, which is equivalent to β(vi, vj) = 0 ifvi = vj.
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For the second part, we might as well assume that v1 =∑
i≥2 λivi. Then:
α(v1, . . . , vk) =∑i≥2
λiα(vi, v2, . . . , vk),
and each term of the sum vanishes since vi appears two times in it.
We will denote by ΛkE∗ the space of alternating k-forms on E. Any 1-form is alter-nating, so Λ1E∗ = E∗, we will set (it is purely a convention) Λ0E∗ = R. We will alsoset ΛE∗ =
⊕k ΛkE∗.
Proposition 4.2.4. Let (ei) be a basis of E and α ∈ ΛkE∗, then:
α(v1, . . . , vk) =∑
1≤i1<···<ik≤n
det
vi11 · · · vi1k...
...vik1 · · · vikk
α(ei1 , . . . , eik)
where vl =∑
i vilei.
Proof. The k-linearity gives:
α(v1, . . . , vk) =∑
1≤i1,...,ik≤n
vi11 vi22 · · · v
ikk α(ei1 , . . . , eik).
The alternating property gives that we can only consider k-tuples (i1, . . . , ik) whereall the il are different. We then get the required formula by regrouping the k-tuplesthat corresponds to the same subsets of 1, . . . , n and applying the formula for thedeterminant of a matrix in term of its coefficients.
Corollary 4.2.5. dim(ΛkE∗) =
(nk
).
In particular ΛkE∗ is non trivial only for k between 0 and n, also we get thatdim(ΛE∗) = 2n.
Definition 4.2.6. Alt :⊗k E∗ → ΛkE∗ is defined by:
Alt(α)(v1, . . . , vk) =1
k!
∑σ∈Sk
sign(σ)α(vσ(1), . . . , vσ(k)).
We have that α is alternating if and only if Alt(α) = α.Example 4.2.7. If β =
⊗2E∗, then:
Alt(β)(v, w) =1
2(β(v, w)− β(w, v)).
We want to turn ΛE∗ into an algebra. Unfortunately the tensor product of twoalternating forms is not alternating in general, so we need to define a new product,called the wedge product:
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Definition 4.2.8. Let α ∈ ΛkE∗ and β ∈ ΛlE∗. We define α ∧ β ∈ Λk+lE∗ by:
α ∧ β =(k + l)!
k!l!Alt(α⊗ β).
Example 4.2.9. If α, β ∈ Λ1E∗, then:
α ∧ β(v, w) = α(v)β(w)− α(w)β(v).
The wedge product has the following properties:
Proposition 4.2.10. 1. (α, β) 7→ α ∧ β is bilinear.
2. (α ∧ β) ∧ γ = α ∧ (β ∧ γ).
3. If α ∈ ΛkE∗ and β ∈ ΛlE∗, then α ∧ β = (−1)klβ ∧ α.
Example 4.2.11. One can show that if (ei) is a basis of E and ei is the dual basis of E∗,then:
ei1 ∧ · · · ∧ eik(v1, . . . vk) = det
vi11 · · · vi1k...
...vik1 · · · vikk
where vl = vilei.The first and second properties show that the wedge product can be extended on the
whole ΛE∗, and turns it into an associative algebra.Let f : F → E be a linear map. And α ∈ ΛkE∗ then f ∗α is the element of ΛkF ∗
defined by:f ∗α(u1, . . . , uk) = α(f(u1), . . . , f(uk)).
It is easy to check that f ∗(α ∧ β) = f ∗α ∧ f ∗β and that (f g)∗ = g∗ f ∗.Example 4.2.12. Consider the one dimensional vector space ΛnE∗, then for any f ∈
L(E), f ∗ defines a linear map from ΛnE∗ to itself. Thus there exists λ ∈ R suchthat f ∗ω = λω for any ω ∈ ΛnE∗. One can actually show that λ = det(f). This is atotally coordinate free definition of the determinant. The fact that (f g)∗ = g∗ f ∗gives that det(f g) = det(g) det(f).
4.3 Differential forms
One defines the bundle of p-form ΛpT ∗M as a set by the formula:
ΛpT ∗M =⋃m∈M
Λp(TmM∗),
it comes with the natural projection π : ΛpT ∗M → M which sends each Λp(TmM∗) to
m.
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Then for any chart of (U,ϕU) of M , one defines a chart on ΛpT ∗U =⋃m∈U Λp(TmM
∗)by setting:
ΨU(α) = (π(α), (Tpϕ−1U )∗α).
As in the case of the cotangent bundle, one easily checks that it defines a smoothstructure on ΛpT ∗M such that π : ΛpT ∗M →M is a smooth vector bundle over M .
Definition 4.3.1. A smooth (differential) p-form on M is a smooth section ΛpT ∗M .The vector space (in fact it is also a C∞(M) module) of sections of ΛpT ∗M is denotedby ΩpM , and Ω(M) denotes
⊕p ΩpM .
Note that Ω0M is just the vector space of smooth functions. If α ∈ ΩpM , we willdenote by αm its value on m ∈M (which is an alternating linear form on TmM).Example 4.3.2. Assume M is an open set U in Rn, then the bundle ΛpT ∗U is trivial,
and the p-forms:m ∈ U 7→ dxi1 ∧ · · · ∧ dxip
for p-tuples such that i1 < · · · < ip gives a basis of ΩpU as a C∞(U)-module. Inparticular any p-form α on U can be written as:
α =∑
i1<···<ip
αi1...ipdxi1 ∧ · · · ∧ dxip ,
where the αi1...ip are smooth functions on U . This shows that ΩU is generated bysmooth functions and their differential as en algebra.If f : M1 →M2 is a smooth map and α ∈ ΩpM2, we define the pull back of α by f by
the formula:(f ∗α)m1 = (Tm1f)∗αf(m1).
f ∗α is a smooth p-form on M2.Moreover, f ∗(α ∧ β) = f ∗α ∧ f ∗β. If η ∈ Ω0M = C∞(M), f ∗η = η f .
Remark 4.3.3. When computing in coordinates, the pull back is actually just a “changeof variables”. Assume that f : Rn → Rm and maps x = (x1, . . . , xn) to y(x) =(y1(x1, . . . , xn), . . . , ym(x1, . . . , xn)). Then a p-form α on Rm can be written as:
αy =∑
0≤i1<···<ip≤m
αi1...ip(y1, . . . , ym)dyi1 ∧ · · · ∧ dyip .
Then f ∗α is easily computed by changing yi to yi(x1, . . . , xn) and computing thedifferentials dyi as
∑i ∂jy
idxj.For instance consider f(r, θ, φ) = (r sin θ cosφ, r sin θ sinφ, r cos θ) (spherical coor-
dinates in R3), and α = dx ∧ dy ∧ dz, then:
f ∗α =d(r sin θ cosφ) ∧ d(r sin θ sinφ) ∧ d(r cos θ)
=((sin θ cosϕ)dr + (r cos θ cosϕ)dθ − (r sin θ sinϕ)dϕ)
∧ ((sin θ sinϕ)dr + (r cos θ sinϕ)dθ + (r sin θ cosϕ)dϕ)
∧ (cos θdr − r sin θdθ)
=r2 sin θdr ∧ dθ ∧ dϕ.
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Proposition 4.3.4. Let M be a compact manifold, then ΩM is generated by smoothfunctions on M and their differential as an algebra.
Proof. We only need to show that any p-form α can be written as a sum of forms of thetype gdf1 ∧ · · · ∧ dfp. Consider a finite cover of M by charts (Ui, ϕi), and ηi a partitionof unity subordinated to the cover Ui. Then if we set αi = ηiα,
∑i αi = α. So we only
need to check the property for each αi. We know that the property is true on open setsof Rn, so, in ϕi(Ui) :
αi = (ϕ−1i )∗αi =
∑k1,...,kp
(αi)k1...kpdxk1 ∧ · · · ∧ dxkp .
Thus on Ui:αi =
∑k1,...,kp
((αi)k1...kp ϕi
)ϕ∗i dx
k1 ∧ · · · ∧ ϕ∗i dxkp .
But if we write ϕi = (ϕ1i , . . . , ϕ
ni ), then ϕ∗i dxk = dϕki . So:
αi =∑
k1,...,kp
((αi)k1...kp ϕi
)dϕk1i ∧ · · · ∧ dϕ
kpi .
We are almost done. The only problem here is that that the ϕki are only defined on Ui.To overcome this, pick a smooth ψi which is 0 outside of Ui and 1 on the support ofηi, then since αi vanishes outside of the support of ηi, we can safely replace the dϕki byd(ψiϕ
ki ) and ψiϕki extended by 0 outside of Ui is a smooth function on the whole M .
This result is useful in the following situation: assume we want to prove a linear for-mula about differential forms, and that this formula behaves well under wedge products,then it is enough to check it for smooth functions f and their differentials df .
4.3.1 The differential of a p-form.
Theorem 4.3.5. There exists a unique R-linear map: d : ΩM → ΩM such that:
1. d(ΩpM) ⊂ Ωp+1M ,
2. if f ∈ Ω0M is a smooth function, then df coincides with the differential of f ,
3. if α ∈ ΩpM , d(α ∧ β) = dα ∧ β + (−1)pα ∧ dβ,
4. d d = 0.
We will prove the theorem in several steps.
Lemma 4.3.6. The theorem is true if M is an open set U ⊂ Rn.
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Proof. We need to have that d(dxi) = 0 by the last property. Then property 3 and anelementary induction imply that d(dxi1 ∧ · · · ∧ dxip) = 0. Write any α ∈ ΩpU as:
α =∑
i1<···<ip
αi1...ipdxi1 ∧ · · · ∧ dxip ,
Then property 3 and linearity imply:
dα =∑
i1<···<ip
dαi1...ip ∧ dxi1 ∧ · · · ∧ dxip ,
but αi1...ip is a smooth function, so by property 2 dαi1...ip is just the usual differential.So the formula above is the only possible definition for d.It is then a routine check to see that it satisfies the 4 required properties. Let us see
that d d = 0 (assumeing the other ones). We have that:
d2α =∑
i1<···<ip
d2αi1...ip ∧ dxi1 ∧ · · · ∧ dxip ,
so it is enough to check that d2f = 0 for any smooth function. We have then that:
d2f = d(∑i
∂ifdxi) =
∑i,j
∂jifdxi ∧ dxj.
This is zero because dxi ∧ dxi = 0 and dxi ∧ dxj = −dxj ∧ dxi.
The following property is central, and shows that d is "natural".
Proposition 4.3.7. Let dM and dN be linear maps on ΩM and ΩN which satisfy theassumptions f the theorem, and f : M → N be a smooth map. Then:
f ∗ dN = dM f ∗.
Proof. First the formula is true if α = η ∈ Ω0N = C∞(N) because:
f ∗dNη = f ∗Tη = Tη Tf = T (η f) = dM(η f) = dMf∗η.
Assume now that α = dη = dNη ∈ Ω1N (dN coincides with the usual differential d onsmooth functions). Then dM(f ∗dη) = dM(d(f ∗η)) because η is a smooth function, hencedM(f ∗dη) = d2
M(f ∗η) = 0 since d2M = 0. Moreover f ∗(dNdη) =∗ (d2
Nη) = 0. So:
f ∗dN(dη) = dMf∗(dη).
Now using the fact that ΩN is generated as an algebra by smooth functions and theirdifferential and the formulas for f ∗(α ∧ β) and d(α ∧ β), one can prove the formula forevery p-form.
Considering the special case where U ⊂ M is open and f : U → M is the inclusion,we get:
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Lemma 4.3.8. Let U ⊂ M be an open set, and dU , dM be operators on ΩU and ΩMsatisfying the hypothesis of the theorem. Then for any α ∈ ΩM :
(dMα)|U = dU(α|U).
We can now prove the theorem.
Proof. The uniqueness of dM comes from the uniqueness on Rn. For any α ∈ ΩM andany chart (U,ϕ), we must have, since ϕ is a diffeomorphism:
(dMα)|U = ϕ∗d((ϕ−1)∗α),
since d is unique on ϕ(U) ⊂ Rn, the right hand side is the unique possible definition ofdM on U .To get existence, we need to show that the definition we gave does not depend on
charts, but if (V, ψ) is another chart:
ψ∗d((ψ−1)∗α) = (ψ ϕ−1 ϕ)∗d((ϕ−1 ϕ ψ−1)∗α)
= ϕ∗(ψ ϕ−1)∗d(ϕ ψ−1)∗(ϕ−1)∗α
= ϕ∗d((ϕ−1)∗α)
since (ψ ϕ−1)∗d = d(ψ ϕ−1)∗.
Definition 4.3.9. The operator d whose existence and uniqueness is given by the pre-vious theorem is called the exterior differential.
4.3.2 Lie derivative and Cartan’s formula
Definition 4.3.10. Let X be a vector field on M and α be a p-form on M , the Liederivative of α along X is the p-form defined by:
LXα =d
dt |t=0ϕ∗tα,
where ϕt is the flow of X.
Proposition 4.3.11. LX is charaterised by the following properties:
1. for η ∈ Ω0M = C∞(M), LXη = dη(X),
2. LX d = d LX ,
3. LX(α ∧ β) = LXα ∧ β + α ∧ LXβ.
Proof. Properties 1,2 and 3 follow from taking the time derivative of the similar prop-erties of the pullback: ϕ∗tη = η ϕt, ϕ∗td = dϕ∗t and ϕ∗t (α ∧ β) = ϕ∗tα ∧ ϕ∗tβ.To see that these properties charcterize LX , we just observe that properties 1 and 2
define LX and functions and their differentials and property 3 tells us how to computethe Lie derivative of a wedge propduct.
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We can actually compute the Lie derivative in terms of the Lie Bracket.
Proposition 4.3.12. Let X be a vector field on M , and α be a p-form on M . Then forany smooth vector fields X1, . . . , Xp, we have:
LXα(X1, . . . , Xp) = X · (α(X1, . . . , Xn)) +∑i
α(X1, . . . , Xi−1, [Xi, X], Xi+1, . . . , Xp).
Proof. Consider the smooth function: f : m ∈ M 7→ αm((X1)m, . . . , (Xp)m). Let ϕt bethe flow of X. Then:
f ϕt(m) =αϕt(m)((X1)ϕt(m), . . . , (Xp)ϕt(m))
=ϕ∗tαm(Tϕt(m)ϕ−t(X1ϕt(m)), . . . , Tϕt(m)ϕ−t(Xpϕt(m)))
=ϕ∗tαm((ϕ−t∗(X1))m, . . . , (ϕ−t∗(Xp))m).
Thus:
X · f(m) =LXα(X1, . . . , Xp) +∑i
α(X1, . . . , Xi−1,d
dt |t=0ϕ−t∗Xi, Xi+1, . . . , Xp)
=LXα(X1, . . . , Xp) +∑i
α(X1, . . . , Xi−1, [Xi,−X], Xi+1, . . . , Xp)
Remark 4.3.13. If we set LXY = [X, Y ], which is justified by the fact that [X, Y ] =ddt |t=0
ϕ−t∗Y = ddt |t=0
ϕ∗tY , we get that:
X · (α(Y, Z)) = LXα(Y, Z) + α(LXY, Z) + α(Y, LXZ).
which should be seen as another occurence of Leibnitz rule.We will need a third operation on differential forms, the interior product:
Definition 4.3.14. Let X be a vector field on M and α be a p-form, the interiorproduct of α with X is the p− 1-form ιXα defined by:
(ιXα)m(v1, . . . , vp−1) = αm(Xm, v1, . . . , vp−1)
for any m ∈M and v1, . . . , vp−1 ∈ TmM . If η is a smooth function, we set ιXη = 0.
The interior product and the wedge product satisfy are related by the following for-mula:
ιX(α ∧ β) = ιXα ∧ β + (−1)kα ∧ ιXβ.
The Lie derivative, interior product and exterior differential are linked by Cartan’sformula.
Theorem 4.3.15. LX = d ιX + ιX d.
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Proof. Set LX = d ιX + ιX d.We will show that LX satisfy the charcteristic propertiesof LX from the previous proposition. For smooth functions η:
LXη = ιX(dη) = dη(X) = LXη.
Now, since d2 = 0: d LX = d ιX d = LX d. To end the proof we use the formulafor ιX(α ∧ β) together with the similar identity for d, this implies that LX(α ∧ β) =LXα ∧ β + α ∧ LXβ.
This allows us to define recursively d in without any reference to coordinates. We willwork this out only for 1-forms, this works in full generality but the formula gets messier.Consider a 1-form α and two vector fields X and Y , we have:
dα(X, Y ) =ιXdα(Y )
=LXα(Y )− d(ιXα)(Y )
=X · (α(Y ))− Y · (α(X))− α([X, Y ]).
4.3.3 Closed forms, exact forms and Poicaré lemma
Definition 4.3.16. A p-form α is said to be closed if dα = 0.
Definition 4.3.17. A p-form α is said to be exact if there exist a (p− 1)- form β suchthat dβ = α.
Since d2 = 0, any exact form is closed. However, the converse is not true.Example 4.3.18. Consider the 1-form α on R2\0 given by:
α =−ydx+ xdy
x2 + y2.
We have that:
dα =−(2dx ∧ dy)(x2 + y2)− (−ydx+ xdy) ∧ (2xdx+ 2ydy)
(x2 + y2)2= 0.
So α is closed. Assume that there is some smooth function η on R2\0 such thatdη = α, and consider the standard inclusion: i : S1 = x2 + y2 = 1 → R2\0.Then i∗α = i∗dη = d(i∗η) = d(η i). In particular, since S1 is compact, i∗α has tovanish at some point. But for any p = (x, y) ∈ S1, vp = (−y, x) ∈ TpS1 and:
i∗α(vp) = α(vp) = 1,
so that i∗α never vanishes. Hence α is not closed.The relation between closed and exact forms can actually be read on the topology of
M , through what is called de Rahm cohomology. We won’t develop this theory here,however let us mention Poincaré lemma, which says that if the topology of M is simple,any closed differential form is exact.
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Theorem 4.3.19. Let O ⊂ Rn be a starshaped1 open set, then any closed differentialform on O is exact.
For the proof we will use the following property:
Lemma 4.3.20. Let ϕt be the (local) flow of some vector field X on U and α ∈ Ωp(U)be a closed p-form, then for any t1, t0 :
ϕ∗t1α− ϕ∗t0α
is an exact form.
Proof. We first write:
ϕ∗t1α− ϕ∗t0α =
∫ t1
t0
d
dtϕ∗tαdt.
Then:d
ds s=tϕ∗sα =
d
ds s=0ϕ∗t+sα = LXϕ
∗tα.
We now use Cartan’s formula: LXϕ∗tα = d(ιXϕ∗tα) + ιXd(ϕ∗tα). Thus, since d(ϕ∗tα) =
ϕ∗tdα = 0, we get:
ϕ∗t1α− ϕ∗t0α =
∫ t1
t0
d(ιXϕ∗tα)dt = d
(∫ t1
t0
(ιXϕ∗tα)dt
).
We can now prove the Poincaré Lemma:
Proof. We assume that O is starshaped with respect to the origin in Rn, and let α bea closed k-form on U . Consider the radial vector field Xp = p, and let ϕt be its flow.Then for any t < 0, ϕt maps U inside U . Moreover, by the previous lemma:
α− ϕ∗tα = dωt
where:
ωt =
∫ 0
t
(ιXϕ∗tα)dt.
It is now just a matter of easy computations to check that as t goes to −∞, ϕ∗tα goesto 0 and ωt converges to some smooth form ω−∞ such that:
α = dω−∞.
Remark 4.3.21. The topologically minded reader will notice that ϕt is actually an ho-motopy between the constant map (at t = −∞) and the identity (at t = 0).
1A domain O is starshaped if there exist o ∈ O such that for any x ∈ O, [o, x] ⊂ O.
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4.4 Integration of differential forms
4.4.1 Orientability and volume forms
Let us first consider the case of finite dimensional vector spaces. When E is a vector spaceover R, the group Gl(E) has two connected components, and the connected componentof the identity is given by:
Gl+(E) = g ∈ Gl(E)| det g > 0.
The group Gl(E) acts transitively on the set of (ordered) basis of E by g · (v1, . . . , vn) =(gv1, . . . , gvn). Two basis v = (v1, . . . , vn) and v = (v1, . . . , vk) of E are said to definethe same orientation if v = g · v for some g ∈ Gl+(E). Defining the same orientationis an equivalence relation on the set of basis of E, which has exactly two equivalenceclasses. For this reason, linear maps of Gl+(E) are said to be orientation preserving.
Definition 4.4.1. An orientation of E is the choice of one equivalence class of basisunder the "defining the same orientation" relation. A vector space on which an orien-tation has been given is said to be oriented.
Given an orientation of E, a basis is said to be positive if it belongs to the equivalenceclass of basis that defines the orientation.Recall that the space of n-forms on E has dimension 1. Let ω be a non zero element
in ΛnE∗, then the set of basis v = (v1, . . . , vn) such that ω(v1, . . . , vn) > 0 defines anorientation on E. This comes from the fact that if v and v satisfy ω(v1, . . . , vn) > 0 andω(v1, . . . , vn) > 0, and we define f : E → E by setting f(vi) = vi, then:
ω(v1, . . . , vn) = ω(f(v1), . . . , f(vn))
= f ∗ω(v1, . . . , vn)
= det(f)ω(v1, . . . , vn).
To summarize, orienting a vector space is equivalent to choosing a nonzero n-form upto scaling by a positive real. Now whenever we encounter the vector space Rn, we willendow it with its orientation coming from the standard basis, or equivalently from then-form given by the determinant.
We now move to manifolds. Let U be an open set in a vector space E, and f : U → Ebe a diffeomorphism. f is said to be orientation preserving if for any x ∈ E, Txf ∈Gl+(E). For instance if f : Rn → Rn, it is equivalent to the positivity of the determinantof the Jacobian matrix of f . The connectedness of Gl+(E) implies that it is enough tocheck that Txf ∈ Gl+(E) at only one point x.
Definition 4.4.2. An orientation atlas of a smooth manifold is a smooth atlas suchthat all transition functions are orientation preserving diffeomorphisms. A manifoldwhich admits an orientation atlas is said to be orientable.
Two orientation atlases (U,ϕ) and (V, ψ) are said to be compatible if their unionis again an orientation atlas. Compatibility is an equivalence relation on orientationatlases, on a given connected manifold their are exactly two classes for this relation.
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Example 4.4.3. On Sn consider the atlas given by the stereographic projections aroundthe north and south pole. The transition function is: x 7→ x
‖x‖2 , which is orientationreversing. In particular this is not an orientation atlas. However, if we compose thestereographic projection around the south pole with an orientation reversing map ofRn, we get an orientation atlas. Thus Sn is orientable.
Definition 4.4.4. An orientation on an orientable manifold M is an equivalence classof compatible orientation atlases.An orientation atlas of an oriented manifold is an atlas which belongs to the equiv-
alence class which defines the orientation of M .
Given an orientation of M , one defines an orientation on each TpM in the followingway. One first pick some orientation atlas in the given equivalence class of orientation at-lases ofM , and for each p ∈M , one defines an orientation on TpM by picking a chart fromthe orientation atlas (U,ϕ) around p and by declaring that the basis
(∂∂x1 |p, . . . ,
∂∂xn |p
)is positive. The fact that the chosen atlas is an orientation atlas implies that this ori-entation doesn’t depend on the chosen chart. Similarly, the orientation won’t change ifwe replace (U,ϕ) by a compatible atlas.A volume form ω on an n-manifold Mn is a smooth n-form such that ωp 6= 0 for
any p ∈ M . It is clear that the pulback of a volume form by a local diffeomorphismis a volume form. Since the bundle ΛnT ∗M has rank 1, the existence of volume formis equivalent to the triviality of ΛnT ∗M . In particular, once a volume form is chosen,every n-form can be written as fω for some smooth function f .
Theorem 4.4.5. A compact manifold M is orientable if and only if it admits a volumeform.
Remark 4.4.6. The theorem is also valid for second countable manifolds, however theproof requires some topological subtleties.
Proof. First assume that M is orientable and choose a finite orientation atlas (ϕi, Ui),consider on each Ui the n-form ωi = ϕ∗i (dx
1 ∧ · · · ∧ dxn). If Uj intersects Ui, then atp ∈ Ui ∩ Uj:
(ωj)p = ϕ∗j(dx1 ∧ · · · ∧ dxn)
= (ϕi ϕ−1i ϕj)∗(dx1 ∧ · · · ∧ dxn)
= (ϕ−1i ϕj)∗(ωi)p
= fij(p)(ωi)p,
where fij is smooth and positive on Ui∩Uj. The fact that the atlas we are working withis an orientation atlas imply that f(p) > 0. We now choose a partition of unity ηi and
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set ω =∑
i ηiωi. Then for any p ∈ Uj:
ωp =∑i
ηiωi
= ηjωj +∑
i 6=j|p∈Ui
ηiωi
= ηjωj +∑
i 6=j|p∈Ui
ηifji(p)ωj
=
ηj(p) +∑
i 6=j|p∈Ui
ηi(p)fji(p)
ωi.
The coefficient in front of ωi is positive, so ω is not 0 and is thus a volume form.We now assume that we are given a volume form ω on M . Let (Ui, ϕi) be a finite
atlas of M , for any i, we have on Ui that:
ω = fiϕ∗i (dx
1 ∧ · · · ∧ dxn).
Where the smooth function fi doesn’t change sign. Exchanging two of the coordinatesxi if necessary, we can assume that fi > 0. Now we write:
dx1 ∧ · · · ∧ dxn =1
fj(ϕ−1
j )∗fiϕ∗i (dx
1 ∧ · · · ∧ dxn) =fifj
(ϕi ϕ−1j )∗(dx1 ∧ · · · ∧ dxn).
Since on n-forms, (ϕiϕ−1j )∗ is just multiplication by the jacobian determinant of ϕiϕ−1
j ,the jacobian determinant has to be fj/fi > 0.
Example 4.4.7. Let ν = dx1 ∧ · · · ∧ dxn+1 be the standard volume form on Rn+1, andconsider α = ιXν, the interior product of ν with the vector field X given by Xp = p.Consider then the usual inclusion i : Sn → Rn+1, and let ω0 = i∗α. One can checkthat ω0 is never zero, and is thus a volume form on Sn.ω0 has a special property (which actually charcterise it up to scaling): ω0 is invari-
ant under the action of the special orthogonal group SO(n + 1,R), that is for anyg ∈ SO(n+ 1,R), g∗ω0 = ω0. To prove that we first remak that g∗ν = det(g)ν = ν.Now pick p ∈ Sn, and v1, . . . , vn ∈ TpSn:
(g∗ω0)p(v1, . . . , vn) = (ω0)g(p)(Tpg(v1), . . . , Tpg(vn))
= (i∗ιXν)g(p)(gv1, . . . , gvn)
= νg(p)(Xg(p), gv1, . . . , gvn)
= νg(p)(gXp, v1 . . . , vn)
= (g∗ν)p(Xp, v1, . . . , vn)
= νp(Xp, v1, . . . , vn)
= (i∗ιXν)p(v1, . . . , vn) = ω0(v1, . . . , vn).
Similarly if g ∈ O(n+ 1)\SO(n+ 1), g∗ω0 = −ω0.
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Proposition 4.4.8. RPn is orientable if and only if n is odd.
Proof. For this proof we will see RPn as the quotient Sn/I, σ, where σ : x 7→ −x, anddenote by π : Sn → RPn the projection map, which is a local diffeomorphism.Let us first assume that n is even and that ω is a volume form on RPn. Then π∗ω is a
volume form on Sn, therefore there is a nowhere vanishing smooth function f : Sn → Rsuch that:
π∗ω = fω0,
where ω0 is the volume form built in the previous example.But π σ = π, so σ∗(π∗ω0) = π∗ω0, hence:
fω0 = σ∗(fω0) = (f σ)σ∗ω0 = −(f σ)ω0,
because σ ∈ O(n+ 1)\SO(n+ 1) since n+ 1 is odd.In particular, f(−x) = −f(x) for all x ∈ Sn, so f take both positive and negative
values, and must vanish at some point, which contradicts the fact that f doesn’t vanish.We now assume that n is odd. Consider an open set U ⊂ Sn where π is a diffeomor-
phism. On π(U), we can define a volume form by setting ωU = (π−1|U )∗ω0. Now assume
that V is another open set such that: π(U)∩ π(V ) is not empty, and pick some x in theintersection, now either x = π(x) for some x ∈ U ∩ V , or x ∈ U and −x ∈ V . In theformer case, we trivially have that (ωU)x = (ωV )x. In the latter case, since σ∗ω0 = ω0
when the dimension is odd, we also get that: (ωU)x = (ωV )x. This shows that the n-formω defined by setting ω|U = ωU is actually a well defined volume form on RPn.
4.4.2 The integral of an n-form
When one inspects the change of variable formula for integrals of several variables func-tions, it is clear that this notion is not invariant by diffeomorphisms (because of theJacobian determinant term). Therefore one cannot integrate functions on a smoothmanifold in a meaningful way without some extra structure.
However consider an n-form α on Rn with compact support (p|αp 6= 0 has compactclosure) in some U ⊂ Rn. Then there is a unique smooth function f with compactsupport in U such that:
α = fdx1 ∧ · · · ∧ dxn.We define: ∫
U
α =
∫U
fdx1 . . . dxn,
where the right hand side is just the usual Riemann or Lebesgue integral of f .This definition may seem purely conventional and highly unimpressive, but it turns
out to have the following important property:
Proposition 4.4.9. Let ϕ : U ⊂ Rn → ϕ(U) ⊂ Rn be an orientation preservingdiffeomorphism and α be differential form with support in ϕ(U), then:∫
U
ϕ∗α =
∫ϕ(U)
α.
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Proof. Let us write α as fdx1 ∧ dxn, then:
ϕ∗α = (f ϕ)ϕ∗dx1 ∧ · · · ∧ dxn = (f ϕ) det(Dϕ)dx1 ∧ · · · dxn.
In particular: ∫U
ϕ∗α =
∫U
(f ϕ) det(Dϕ)dx1 . . . dxn.
On the other hand, the usual change of variable formula for integrals of functions on Rd
tells us that: ∫ϕ(U)
α =
∫ϕ(U)
fdx1 . . . dxn =
∫U
(f ϕ)| det(Dϕ)|dx1 . . . dxn.
Now since ϕ is orientation preserving, det(Dϕ) > 0 and thus det(Dϕ) = | det(Dϕ)|which gives the required equality.
Now consider an oriented manifold M . We will assume for simplicity that M iscompact. Take a finite orientation atlas (Ui, ϕi) of M , and a partition of unitity ηisuborditated to the cover Ui.
We set, for any n-form α on M :∫M
α =∑i
∫ϕi(Ui)
(ϕ−1i )∗(ηiα).
Proposition 4.4.10. The expression above doesn’t depend on the chosen orientationatlas and partition of unity.
Proof. Let us first notice that α =∑
i(ηiα), and that each αi = ηiα is compactlysupported in Ui.Now let (Vj, ψj) be another orientation atlas and νi and associated partition of unity.
We want to check that:∑i
∫ϕi(Ui)
(ϕ−1i )∗(ηiα) =
∑j
∫ψj(Vj)
(ψ−1j )∗(νjα).
We have: ∑i
∫ϕi(Ui)
(ϕ−1i )∗(ηiα) =
∑i
∫ϕi(Ui)
(ϕ−1i )∗((
∑j
νj)ηiα) (4.1)
=∑i,j
∫ϕi(Ui∩Vj)
(ϕ−1i )∗(νjηiα). (4.2)
Now ϕi ψ−1j is an orientation preserving diffeomorphism, so:∫
ϕi(Ui∩Vj)(ϕ−1
i )∗(νjηiα) =
∫ψj(Ui∩Vj)
(ϕi ψ−1j )∗(ϕ−1
i )∗(ηiνjα)
=
∫ψj(Ui∩Vj)
(ψ−1j )∗(ηiνjα).
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Plugging this back in equation (4.1), we get:∑i
∫ϕi(Ui)
(ϕ−1i )∗(ηiα) =
∑i,j
∫ψj(Ui∩Vj)
(ψ−1j )∗(ηiνjα).
=∑j
∫ψj(Vj)
(ψ−1j )∗((
∑i
ηi)νjα)
=∑j
∫ψj(Vj)
(ψ−1j )∗(νjα),
which is the required equality.
Definition 4.4.11.∫Mα is the integral of the n-form α.
We will also need to integrate differential forms on compact subsets K ⊂M . We willdefine: ∫
K
α =∑i
∫ϕi(Ui∩K)
(ϕ−1i )∗(ηiα).
One shows as before that this depends neither on the orientation atlas, nor on thepartition of unity that we have used.Volume forms play the role of non vanishing functions:
Proposition 4.4.12. If ω is a volume form compatible with the orientation of M then∫Mω > 0.
Proof. In each chart (Ui, ϕi) of an orientation atlas, we know that (ϕ−1i )∗ω = fidx
1 ∧· · · ∧ dxn for some smooth positive function fi on ϕi(Ui), so:∑
i
∫ϕi(Ui)
(ϕ−1i )∗(ηiα) =
∑i
∫ϕi(Ui)
ηifidx1 . . . dxn.
Each of the terms in the sum are nonnegative, and except if ηi is vanishing, they areactually positive, since at least one of the ηi doesn’t vanish, the ntegral is positive.
This framework gives a really geometric interpretation of the divergence of a vectorfield. Consider a manifoldM with a volume form ω, andX a vector field onM . Considerthe flow ϕt of X. The question we want to answer is the following: how does the volumeof Kt = ϕt(K) evolves with respect to t ? Here by volume we mean
∫Ktω? To see this
we write: ∫Kt=ϕt(K)
ω =
∫K
ϕ∗tω.
Now we want to differentiate the right hand side with respect to t at t = 0, differentiatingunder the integral we get
∫KLXω. Now LXω is an n form, so LXω = fω for some smooth
f , f is called the divergence of X (with respect to the volume form ω), in other wordsLXω = div(X)ω is the definition of the divergence in this context. From a more physicalpoint of view, div(X) measures the change of volume when one moves along X.
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Let us now assume that M = Rn with the standard volume form ω = dx1 ∧ · · · ∧ dxn,and that X = X i∂i is a vector field. We want to compute the divergence of X in terms ofits components X i. We use Cartan’s formula, which, since dω = 0 (dω is an n+1-form!)writes as:
LXω = d(ιXω) =∑i
d(X iι∂iω).
Now:ι∂iω = (−1)idx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxn,
and d(X i) =∑
j ∂jXidxj. Using that dι∂iω = 0, we get:
LXω = d(ιXω) =
(∑i
∂iXi
)ω.
Hence div(X) =∑
i ∂iXi, the usual formula.
4.5 Stokes Theorem
4.5.1 Regular domains
Stokes theorem reads as “∫Ddω =
∫∂Dω”, before getting to this, we need to understand
who D and ∂D are.WhenD is a subset ofM , we will denote by D it closure, D its interior, and ∂D = D\D
its boundary.
Definition 4.5.1. A connected compact D in a manifold Mn is said to be a regulardomain if it is equal to the closure of its interior, and its boundary ∂D is a smoothsubmanifold of dimension n− 1 in M .
Example 4.5.2. If M is a compact smooth manifold, then M itself is a regular domain(with empty boundary).
The unit ball in Rn is a regular domain with boundary the unit sphere Sn−1.The annulus Ar1,r2 = x ∈ Rn|r1 ≤ ‖x‖ ≤ r2 is a regular domain, its boundary is
the union of the spheres of radius r1 and r2.The square in the plane is not a regular domain: its boundary is not a smooth
submanifold.
Proposition 4.5.3. For each p ∈ ∂D there is a chart (U,ϕ) ofM such that ϕ(U∩∂D) =x ∈ ϕ(U)|x1 = 0 and ϕ(U ∩D) = x ∈ ϕ(U)|x1 ≤ 0.
Proof. The first point is just the fact that ∂D is submanifold of M . We may assumethat ϕ(U) is the unit ball B in Rn. Consider the open sets O = ϕ(U ∩ D) and O′ =B\x1 = 0. We have that O ⊂ O′ and, since D is a regular domain, the frontier of O′as a subset of O′ is empty. Thus O is closed and open in O′. Hence O has to be a unionof connected components of O′, which are B ∩ x1 < 0 and B ∩ x1 > 0.
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However, we know that the frontier of O in B is not empty (and equal to x1 = 0),so it has to be one of the connected components. Therefore, composing ϕ with thesymmetry with respect to x1 = 0 if necessary, we get that: ϕ(U ∩ D) = B ∩x1 < 0.Taking the closure we have ϕ(U ∩D) = B ∩ x1 ≤ 0.
We want to integrate n-forms on D ⊂ M , so we need M to be orientable. We alsowant to integrate n − 1 forms on ∂D, so we also need ∂D to be orientable. This isactually automatic.
Proposition 4.5.4. Let D be a regular domain in an orientable manifold, then ∂D isorientable.
Proof. We consider a chart (U,ϕ) as in the previous lemma. Changing the x2 coordinateto −x2 if necessary, we can ensure that (U,ϕ) is compatible with orientation of M .We denote by ϕ = (x1, . . . , xn) the coordinates of ϕ. We consider the smooth map:ϕ : U = ∂D ∩ U → Rn−1 given by ϕ(p) = (x2(p), . . . , xn(p)). We will show that these ϕform an orientation atlas of ∂D.
Consider another orientation chart of M from the lemma (V, ψ), and denote byψ = (y1, . . . , yn) the coordinates of ψ. The change of chart ψ ϕ−1 is then the map(x1, . . . , xn) 7→ (y1, . . . , yn). Its Jacobian matrix is thus given by
(∂yj
∂xi
)ij.
Moreover, since ϕ(∂D∩U) = x1 = 0 and ψ(∂D∩V ) = y1 = 0, ψϕ−1(0, x2, . . . , xn) =
(0, y2, . . . , yn), so ∂y1
∂xi(0, x2, . . . , xn) = 0 for i ≥ 2. Also, using that ϕ(D∩U) = x1 ≤ 0
and ψ(D ∩ U) = y1 ≤ 0, we get that y1(x1, . . . , xn) ≥ 0 when x1 ≥ 0, in particular(using that ψ ϕ−1 is a diffeomorphism) we get that ∂y1
∂x1(0, x2, . . . , xn) > 0.
In particular tthe Jacobian matrix of ψ ϕ−1 is of the form:∂y1
∂x10 · · · 0
∂y2
∂x1∂y2
∂x2· · · ∂y2
∂xn......
...∂yn
∂x1∂yn
∂x2· · · ∂yn
∂xn
and at points (0, x2, . . . , xn), the determinant of the submatrix
(∂yj
∂xi
)i,j≥2
is positive.
To end the proof, we just remark that ψ ϕ−1 maps (x2, . . . , xn) to (y2, . . . , yn), andwe just observed that the Jacobain of this map is positive.
The proof not only shows that ∂D is orientable, but also gives it an orientation (theone defined by the atlas of all the (U , ϕ). This orientation is the one coming from an"outward normal". To see this we work in one of the chart of M (U,ϕ) considered inthe proof. A basis of the tangent space at each p ∈ U , and we consider the vector fieldX on U given by (ϕ−1
∗∂∂x1
, notice that on ∂D, X is pointing towards the outside of D.The orientation on ∂D can then be defined in the following way: A basis (v2, . . . , vn) ofTp∂D is positive if and only (Xp, v2, . . . , vn) is a positive basis of TpM .
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D
Figure 4.1: Orienting ∂D.
For example if D is the unit disk, the oriention on the boundary S1 is the "anticlockwise" one. Furthermore, if we remove a disk from the interior of D, the orientationof the inner component of the boundary will be reversed.
An important particular case is when D = [a, b] ⊂ R, then ∂[a, b] = a ∪ b, aunion of two 0-dimensional submanifolds. A 0-form on a point is just a constant, soa orientation is just the choice of a sign (+ or −). On [a, b], consider the orientationgiven by the 1-form dx. The vector X+ = ∂x is outward pointing at b, the proof aboveshows that the orientation on the boundary is given by the volume form ιXω, where Xis an outward pointing vector and ω is a volume form in M . So the orientation on bis given by ιXdx = 1, so is +. Similarly, since −∂x is an outard pointing vector at a, theorientation on a is −. This explains the notation ∂[a, b] = b − a.
4.5.2 Stokes Theorem
Theorem 4.5.5. Let D be a regular domain in a smooth oriented n-manifold M and αbe an n− 1-form on M . We denote by i : ∂D →M the inclusion. Then:∫
D
dα =
∫∂D
i∗α.
This is a generalization of the fundamental theorem of calculus. Let f be a smoothfunction on R and α = f(x) seen as a 0 form. Then:∫
[a,b]
f ′(x)dx =
∫[a,b]
dα =
∫∂[a,b]
α = f(b)− f(a).
Another important case is the case when D = M where M is a compact manifold,thus the boundary of D is empty and:∫
M
dα = 0.
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In other words the integral of an exact n-form on a compact manifold is zero. As acorollary, we get that a volume form is never exact.
Proof. Using a partition of unity, it is enough to show the result for compactly supportedforms whose support is included in some chart.
We have three cases to deal with. First if the support of α doesn’t meet D, the resultis trivial.
Second, assume that the support of α is included in the interior D. Then∫∂Dα = 0,
and we only need to show that∫Ddα = 0. Let (U,ϕ) be a chart such that the support
of α is included in U , then: ∫D
dα =
∫U
dα =
∫Rnd(ϕ−1)∗α
Since (ϕ−1)∗α has support in ϕ(U). We set β = (ϕ−1)∗α.Let ηk = dx1 ∧ · · · ∧ dxk−1 ∧ dxk+1 ∧ · · · dxn. The ηk form a basis of Ωn−1Rn as a
C∞(Rn) module. So there are smooth functions fk such that β =∑
k fkηk. Moreover
each fk is compactly supported and:
dβ = (∑k
(−1)k+1∂kfk)dx1 ∧ · · · ∧ dxn.
∫x1≤0
dβ =∑k
∫x1≤0
(−1)k+1∂kfkdx1 . . . dxn.
Now for any k:∫Rn∂kfkdx
1 . . . dxn =
∫Rn−1
(
∫R∂kfkdx
k)dx1 . . . dxk−1dxk+1 . . . dxn = 0,
since fk is compactly supported. This comes form the fundamental theorem of calculus:∫R∂kfk(x
1, . . . , xn)dxk =
∫[−A,A]
∂kfk(x1, . . . , xn)dxk
=fk(x1, . . . , xk−1, A, xk+1, . . . , xn)
− fk(x1, . . . , xk−1,−A, xk+1, . . . , xn)
=0.
for A big enough. Thus∫Rn dβ = 0.
The last case is the case when the support of α meets ∂D. Using another partition ofunity, we can assume that the support of α is contained in a chart given by Proposition4.5.3. Let (U,ϕ) be such a chart. We can assume withouth loss of generality that ϕ(U)is included the unit ball B in Rn and that ϕ(D) = x1 ≤ 0. Then:∫
D
dα =
∫ϕ(D)
(ϕ−1)∗dα =
∫x1≤0
d(ϕ−1)∗α.
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Similarly, using the orientation charts for ∂D built in the previous section:∫∂D
i∗α =
∫x1=0
(ϕ−1)∗i∗α =
∫x1=0
i∗(ϕ−1)∗α
where i is the inclusion of x1 = 0 in Rn. Setting β = (ϕ−1)∗α, we only need to showthat
∫x1≤0 dβ =
∫x1=0 i
∗β.With the same notation as the previous case, we write β =
∑k fkη
k. We get:∫x1≤0
dβ =∑i
∫x1≤0
(−1)k+1∂kfkdx1 . . . dxn.
If k 6= 1:∫x1≤0
∂kfkdx1 . . . dxn =
∫Rn−1
(
∫R∂kfkdx
k)dx1 . . . dxk−1dxk+1 . . . dxn = 0,
since fk is compactly supported.Now for k = 1, we have:∫x1≤0
∂1f1dx1 . . . dxn =
∫Rn−1
(
∫x1≤0
∂1f1dx1)dx2 . . . dxn
=
∫Rn−1
(
∫[−A,0]
∂1f1dx1)dx2 . . . dxn
=
∫Rn−1
(f1(0, x2, . . . , xn)− f1(−A, x2, . . . , xn))dx2 . . . dxn
=
∫Rn−1
f1(0, x2, . . . , xn)dx2 . . . dxn
=
∫x1=0
i∗(f1η1).
Moreover∫x1=0 fk i
∗ηk = 0 for k 6= 1, because i∗ηk = 0 in this case. Putting all thesetogether we finally get: ∫
x1≤0dβ =
∫x1=0
i∗β.
4.5.3 Some vector calculus revisited
Let us derive a couple of consequences. First we have the Green Riemann formula:
Corollary 4.5.6. Let D ⊂ R2 be a regular domain and P,Q be smooth function on R2.Then: ∫
D
∂Q
∂x− ∂P
∂ydxdy =
∫∂D
Pdx+Qdy.
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Proof. We just compute:
d(Pdx+Qdy) = dP ∧ dx+ dQ ∧ dy =∂Q
∂x− ∂P
∂ydx ∧ dy
Let us look at the divergence theorem.
Corollary 4.5.7. Let X be a vector field on a compact oriented manifold M and ω bea volume form, then: ∫
M
div(X)ω = 0.
Proof. By definition, div(X)ω = LXω. Now, using Cartan’s formula:
LXω = d(ιXω) + ιX(dω) = d(ιXω).
In particular div(X)ω is an exact n-form, so its integral onM is zero by Stokes Theorem.
What happens when we integrate on a regular domain D ? Stokes theorem gives:∫D
div(X)ω =
∫∂D
ιXω.
Let us assume that D is a regular domain in Rn and that ω is the usual volume formdx1 ∧ · · · ∧ dxn. The left hand side of the integral is just the integral of the real valuedfunction div(X) on D. To understand the right hand side, we need a definition:
Definition 4.5.8. Let D ⊂ Rn be a regular domain. The canonical volume form of∂D is the the n− 1-form on ∂D given by σ = i∗ιNω, where N is a unit outward normalvector field on ∂D and i : ∂D → Rn is the inclusion.
Remark 4.5.9. We have used in a crucial way the euclidean structure here, so this defini-tion doesn’t make sense for smooth manifolds, however it makes sense for Riemannianmanifolds.Using this canonical volume form, we can integrate function on ∂D. In R2, ∂D is an
oriented curve and we get the concept of line integral, in R3, ∂D is an oriented surfaceand we get the concept of surface integral.
Now, let us consider p ∈ ∂M and v1, . . . , vn−1 be vectors in Tp∂D, then:
(ιXω)p(v1, . . . , vn−1) = ωp(Xp, v1, . . . , vn−1)
= ω(〈Xp, Np〉Np + (Xp − 〈Xp, Np〉Np), v1, . . . , vn−1)
= 〈Xp, Np〉ω(Np, v1, . . . , vn−1)
= 〈Xp, Np〉σ(v1, . . . , vn−1)
since (Xp−〈Xp, Np〉Np) belong to Tp∂D, the space of the vi’s. So that i∗ιXω = 〈X,N〉σ.To summarize we get the equality:∫
D
div(X)ω =
∫∂D
〈X,N〉σ.
The right hand is called the flux of X through ∂D.
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4.6 Exercises
Exercise 4.1. Let M be a smooth manifold. Recall that the space of vector fieldsΓ(TM) is a C∞(M) module. Consider a map µ : Γ(TM) → C∞(M) which is C∞Mlinear (µ(fX + gY ) = fµ(X) + gµ(Y ) for f, g ∈ C∞(M) and X, Y ∈ Γ(TM)). Showthat there exist a unique 1-form α such that for any vector field X:
µ(X)(p) = αp(Xp).
First use a partition of unity to show that it is enough to consider the case where M issome open set in Rn. Second decompose any X as a sum
∑iX
i∂i to see what α shouldbe.
Exercise 4.2. Let α = xdy + ydx, show that α is exact. Show that β = xdy − ydx isnot exact.
Exercise 4.3. * Consider a one form α on Rn. Write α =∑
i gidxi and assume that
for i 6= j, ∂igj = ∂jgi. Set:
f(x) =∑i
xi∫ 1
0
gi(tx)dt.
Show that df = α.Show that ∂jf(x) =
∫ 1
0ddt
(tgj(tx))dt.
Exercise 4.4. Consider a smooth curve c : [a, b]→M , and α a 1-form on M , we definethe integral of α along c by: ∫
c
α =
∫ b
a
αc(t)(c′(t))dt.
1. Let ϕ : [c, d]→ [a, b] be a smooth increasing diffeomorphism and c = c ϕ. Showthat: ∫
c
α =
∫c
α.
2. Assume that α = df for some smooth function f on M , show that:∫c
α = f(c(b))− f(c(a)).
3. Consider the smooth 1-form on R2\0 given by:
α =ydx− xdyx2 + y2
.
Show that α is not exact on R2\0.
4. Show that however α is exact on U = (x, y)|y > 0. (Hint:α is sometimes denotedby dθ.)
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Exercise 4.5. An alternating p-form α ∈ ΛpE∗ is said to be simple if it can be writtenas the wedge product of p linear forms.
1. Show that any n or n− 1 alternating for is simple.
2. Show that α = θ ∧ α′ for some linear form θ if and only if θ ∧ α = 0.
3. Let α, β, γ, δ be linearly independent linear forms. Show that ω = α∧ β + γ ∧ δ isnot simple.
4. Show that ω ∧ ω 6= 0.
Exercise 4.6. ** Let E = Λ2R4∗. And let ω be a non-zero element in Λ4R4∗.
1. For any u, v ∈ E, u∧ v ∈ Λ4R4∗, hence there exist B(u, v) ∈ R such that: u∧ v =B(u, v)ω. Show that (u, v) 7→ B(u, v) is a symmetric bilinear form. What is itsignature ?
2. For any g ∈ Sl4(R), show that B(u, v) = B(g∗u, g∗v).
3. Show that O0(3, 3) is isomorphic to a quotient of Sl4(R).
Exercise 4.7. Any 2-form ω ∈ Λ2E∗ defines a map ϕ : E → E∗ by ϕ(x) = (y 7→ω(x, y)). ω is said to be non degenerate if ϕ is an isomorphism.
1. Show that if E has odd dimension, then any 2 form is degenerate.
2. On E = R2n, set:α = dx1 ∧ dxn+1 + · · ·+ dxn ∧ dx2n.
Show that α is nondegenerate.
3. Set Sp(n,R) = u ∈ Gl2n(R)|∀x, y ω(x, y) = ω(u(x), u(y)). Show that u ∈Sp(n,R) if and only if its matrix A satisfies J = ATJA, where:
J =
(0 In−In 0
).
4. Show that Sp(n,R) is a Lie group, and determine its Lie algebra.
A non degenerate linear 2-form is called a (linear) symplectic form, the one we havestudied here happens to be the only one (up to a change of basis). Sp(n,R) is called thesymplectic group.
Exercise 4.8. Let F (r, θ) = (r cos θ, r sin θ), compute F ∗(dx ∧ dy), F ∗(xdx+ ydy) andF ∗(ydx− xdy).
Exercise 4.9. Let α be a 1-form on a smooth manifold M , X be a smooth vector fieldand ϕ : M →M be a diffeomorphism. Show that:
ϕ∗(α(X)) = ϕ∗α(ϕ∗X),
where we have set ϕ∗X = (ϕ−1)∗X.
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Exercise 4.10. We work onM = R2. A differential form α on R2 is said to be SO(2,R)invariant if g∗α = α for any g ∈ SO(2,R)
1. Let α = xdx+ydy and β = ydx−xdy. Show that α and β are SO(2,R)) invariant.
2. Let γ be an SO(2) invariant 1-form on R2\0. Show that there exist two smoothfunctions f, g : R∗+ → R such that:
γ = f(r)α + g(r)β,
where r =√x2 + y2.
3. What are the SO(2) invariant 1-forms on R2.
Exercise 4.11. For E a vector space, we denote by S2E∗ the space of symmetric bilinearforms on E. If f ∈ L(E,F ) and B ∈ S2F ∗ we set:
f ∗B(u, v) = B(f(u), f(v)),
so that f ∗B ∈ S2E∗.As in the case of say 2-forms, the set formed by the union of the S2(TpM)∗ is a smooth
vector bundle whose charts are given by:
B ∈ S2(TpM)∗ 7→ (ϕ(p), ((Tpϕ)−1)∗B).
A Riemannian metric on M is a smooth section g of S2T ∗M such that gp is aninner product on TpM for every p, a manifold with a Riemannian metric is called aRiemannian manifold. Let (M, g) be a Riemannian manifold.
1. Show that if f : N →M is an immersion, then f ∗g is a Riemannian metric on N .
2. Show that every compact manifold admits a Riemannian metric.
3. Show that for any smooth vector field X on M , αp : vp ∈ TM 7→ gp(Xp, vp) is asmooth 1-form on M (usually denoted by X[)
4. Show that for any 1-form α on M , there exist a smooth vector field α# on M suchthat for any vp ∈ TpM :
gp(α#, vp) = α(X).
When α = df , df# is called the gradient of f and usually denoted by ∇f .
Exercise 4.12. We work on R3, with the volume form ω = dx ∧ dy ∧ dy, and innerproduct 〈 , 〉, which actually defines a Riemannian metric on R3.
1. Let X =∑
iXi∂i be smooth vector field on R3. Compute X[ in terms of X i and
dxi.
94
2. Let X be a vector field on R3. Show that rot(X) is charcterized by:
ιrot(X)ω = d(X[),
and that div(X) is characterized by:
div(X)ω = d(ιXω).
3. Show, using the properties of d, that rot(∇f) = 0 and div(rotX) = 0.
Exercise 4.13. Show that on p-forms:
LX(LY α)− LY (LXα) = L[X,Y ]α
and:LX(ιY α)− ιY (LXα) = ι[X,Y ]α.
Exercise 4.14. Let S be an hypersurface in U ⊂ Rn which is given by S = f−1(0)where f : U → R is a submersion. Show that S is orientable.Show that at any point p ∈ S, ∇f is orthogonal to TpS. Use it to build a volume form
on S from a volume form on U .
Exercise 4.15. * Let G be a connected Lie group. We denote by (Ωp(G))` the vec-tor space of p-forms α on G such that )Lg)∗α = α for every g ∈ G (or equivalently(Rg−1)∗α = α). Those p-forms are called left invariant.
1. Show that α ∈ (Ωp(G))` 7→ αe ∈ Λpg is a vector space isomorphism.
2. Show that any Lie group is orientable.
3. Let α ∈ Ω1(G)`. Show that for any two left invariant vector fields:
dα(X, Y ) = α([X, Y ]).
Use that p 7→ αp(Xp) is constant.
4. Derive a similar formula for left invariant 2-forms.
What we see here is that on Left invariant forms, the d operator can be recovered fromthe Lie bracket, and thus from the data of the Lie algebra of G only.
Exercise 4.16. * We use the same notations as in the previous exercise.
1. Let ω be a volume form in (Ωn(G))`. Show that, for any compactly supported fand any g ∈ G: ∫
G
(f Lg)ω =
∫G
fω.
2. Show that for any g ∈ G, (Rg)∗ω ∈ (Ωn(G))`, and that there exists a real mod(g)such (Rg)∗ω = mod(g)ω.
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3. Show that mod is morphism from G to (R∗,×).
4. Show that if G is compact and connected, for any g ∈ G, Rg∗ω = ω.
ω is thus left and right invariant, it is called bi-invariant.
5. By considering the subgroup of Gl2(R) of matrices of the form(a b0 1
), show that
this property can fail when G is not compact.
6. Consider the morphism G→ Gl(G) given by g 7→ Adg, where Adg(X) = (Rg)∗X.Show that ω is right invariant if and only if det(Adg) = 1 for every g.
7. ** Show that G admits a bi-invariant volume form if and only the linear mapadX : Y ∈ g 7→ [X, Y ] ∈ g has zero trace for any X ∈ g.
The left invariant volume form on G is unique up to scaling, it is called the Haarmeasure on G. The last question show that the property that G admits a bi-invariantvolume form can actually be read from the Lie algebra of G.
Exercise 4.17. Let G be a compact Lie subgroup of Gln(R). Let ω be a Haar measureon G.
1. Show that, for v, w ∈ Rn:
〈v, w〉G =
∫G
〈gv, gw〉ω
defines an inner product on Rn.
2. Show that for any g ∈ G:〈gv, gw〉G = 〈v, w〉G .
3. Show that there exist h ∈ Gln(R) such that:
hGh−1 ⊂ O(n)
Exercise 4.18. Consider a surface S in R3 and Σ be a regular domain in S. We denoteby C the boundary of Σ.
1. Let c : I → R3 be a smooth curve. Show that:∫c
X[ =
∫I
⟨Xc(t), c
′(t)⟩dt.
The right hand side is called the circulation of X around c, denoted by∮cX.
2. Show the following corollary of Stokes theorem:∫Σ
〈rot(X), N〉σ =
∮c
X,
Namely, the flux of rot(X) along Σ is equal to the circulation of X along ∂Σ
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Exercise 4.19. We want to compute the volumes of spheres and balls. We considerSn ⊂ Rn+1. Let ω = dx1 ∧ · · · ∧ dxn+1 and σ = ιXω|Snbe the canonical volume form onSn, where Xp = p is the radial vector field.
We set Voln(Sn) =∫Sn σ.
1. Let F : R∗+ × Sn → Rn+1\0 be defined by: F (ru) = ru. Show that F ∗ω =rndr ∧ σ.
2. By computing∫Rn+1 e
−|x|2dx1 . . . dxn in two different ways, show that:(∫Re−t
2
dt
)n= Voln Sn
∫R+
e−r2
rndr.
3. Show that:
Voln Sn = 2πn+12
Γ(n+1
2
) ,where Γ(t)
∫R+e−sts−1ds.
4. Show that (n+ 1)ω = d(ιXω).
5. Use Stokes theorem to show that:∫Bn+1
ω = VolBn+1 =1
n+ 1Vol Sn,
where Bn+1 is the unit ball in Rn+1.
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