Differentiable Manifold

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Differentiable Manifolds

Eckhard Meinrenken

Lecture Notes, University of Toronto, Fall 2001

The main references for these lecture notes are the first volume of Greub-Halperin-Vanstone, and the book by Bott-Tu (second edition!).


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Contents

1. Manifolds 42. Partitions of unity 143. Vector fields 174. Differential forms 235. De Rham cohomology 38

6. Mayer-Vietoris 407. Compactly supported cohomology 448. Finite-dimensionality of de Rham cohomology 469. Poincare duality 4710. Mapping degree 5311. Kuenneth formula 5412. De Rham theorem 5713. Fiber bundles 6214. The Thom class 6615. Intersection numbers 71

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1. Manifolds

1.1. Definition of manifolds. A n-dimensional manifold is a space that locallylooks like Rn. To give a precise meaning to this idea, our space first of all has to comeequipped with some topology (so that the word local makes sense). Recall that atopological space is a set M, together with a collection of subsets of M, called opensubsets, satisfying the following three axioms: (i) the empty set and the space M itselfare both open, (ii) the intersection of any finite collection of open subsets is open, (iii)the union ofanycollection of open subsets is open. The collection of open subsets ofMis also called the topology ofM. A map f : M1 M2 between topological spaces iscalledcontinuousif the pre-image of any open subset inM2 is open in M1. A continuousmap with a continuous inverse is called a homeomorphism.

One basic ingredient in the definition of a manifold is that our topological spacecomes equipped with a covering by open sets which are homeomorphic to open subsetsofRn.

Definition 1.1. LetMbe a topological space. An n-dimensional chartforM is apair (U, ) consisting of an open subsetU Rn and a continuous map : U Rn suchthat is a homeomorphism onto its image (U). Two such charts (U, ), (U, ) areC-compatibleif the transition map

1 : (U U) (U U)

is a diffeomorphism (a smooth map with smooth inverse). A coveringA = (U)A ofMby pairwise C-compatible charts is called a C-atlas.

Some people define a C

-manifold to be a topological space with a C

atlas. It ismore common, however, to restrict the class of topological spaces.

Definition 1.2 (Manifolds). A C-manifold is a Hausdorff topological space M,with countable basis, together with a C-atlas.

Remarks 1.3. (a) We recall that a topological space is called Hausdorffif anytwo points have disjoint open neighborhoods. Abasisfor a topological space Mis a collection Bof open subsets of M such that every open subset of M is aunion of open subsets in the collection B. For example, the collection of openballs B(x) in R

n define a basis. But one already has a basis if one takes onlyall ballsB(x) withx Qn and Q>0; this then defines a countable basis. A

topological space with countable basis is also called second countable.(b) The Hausdorff axiom excludes somewhat pathological examples, such as follow-

ing: Let M = R {p}, where p is a point, with the topology given by opensets in R, together with sets of the form (U\{0}) {p}, for open sets U Rcontaining 0. An open covering of M is given by the two sets U+ = R andU = R\{0} {p}. The natural projection from M to R, taking p to 0, de-scends to smooth maps + : U+ R and : U R. ThenMwith atlas


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1. MANIFOLDS 5

(U, ) satisfies all the axioms of a 1-dimensional manifold except that it is notHausdorff: Every neighborhood of 0 intersects every neighborhood ofp.

(c) It is immediate from the definitions that any open subset of a manifold is amanifold and that the direct products of two manifolds is again a manifold.

The definition of a manifold can be generalized in many ways. For instance, a man-ifold with boundary is defined in exactly the same way as a manifold, except that thecharts take values in a half space {x Rn| x1 0}. For this to make sense, one needsto define a notion of smooth maps between open subsets of half-spaces ofRn,Rm: Sucha map is called smooth if it extends to a smooth map of open subsets ofRn,Rm. Evenmore generally, one definesmanifolds with cornersmodeled on open subset of the positiveorthant{x Rn| xj 0}in R

n.A complex manifoldis a manifold where all charts take values in Cn = R2n, and all

transition maps 1

are holomorphic.

1.2. Examples of manifolds.

(a) Spheres. Let Sn Rn+1 be the unit sphere of radius 1. LetN = (1, 0, . . . , 0)be the north pole and S= (1, 0, . . . , 0) the south pole. Let U1 =Sn\{S} andU2= S

n\{N}. Define mapsj : Uj Rn by

1(x) =x (x N)N

1 x N , 2(x) =

x (x S)S

1 x S =

x (x N)N

1 + x N .

Then j : Uj Rn define the structure of an oriented manifold on Sn. Bothcharts are onto Rn, and 1(U1 U2) =2(U1 U2) = Rn\{0}. The inverse mapto 1 reads,

11 (y) =(||y||2 1)N+ 2y

1 + ||y||2 .

Thus,

2 11 (y) =

y

||y||2,

a global diffeomorphism from Rn\{0} onto itself. The 2-sphere S2 is in fact acomplex manifold: Identify R2 = C in the usual way, so that1, 2take values inC. Replace2by its complex conjugate,2(x) =2(x). In complex coordinates,2

11 (z) =z

1, which is a holomorphic function. A different complex structureis obtained by replacing 1 by it complex conjugate. Forn= 2, the spheres Sn

are not complex manifolds.(b) Projective spaces. Let RP(n) be the quotient Sn/ under the equivalence

relation x x. Let : Sn RP(n) be the quotient map. For any chart : V Rn of Sn with the property x V x V, let U = (V), and : U Rn the unique map such that = . The collection of all suchcharts defines an atlas for RP(n); the compatibility of charts follows from thatforSn.


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(c) Grassmannians. The set GrR(k, n) of allk-dimensional subspaces ofRn is called

theGrassmannian ofk-planes inRn. AC-atlas may be constructed as follows.

For any subsetI {1, . . . , n} of cardinality #I=k, let RI

Rn

be the subspaceconsisting of all x Rn with xi = 0 for i I. Thus each RI GrR(k, n).

Let UI Gr(k, n) be the set of all k-dimensional subspaces E Rn withE (RI) = {0}. There is a bijectionI : UI= L(RI, (RI))= Rk(nk) ofUIwith the space of linear maps AI : R

I (RI), where each such A correspondsto the subspace E= {x+ AI(x)| x RI}.

To check that the charts are compatible, let Idenote orthogonal projectionRn RI. We have to show that for all intersections, UI UI, the map takingAI toAIis smooth. The mapAIis determined by the equations

AI(xI) = (1 I)x, xI= Ix

forx E, andx = xI+ AIxI. Thus

AI(xI) = (I I)(AI+ 1)xI, xI= I(AI+ 1)xI.

The map I(AI+ 1) restricts to an isomorphism S(AI) : RI R I. The above

equations show,

AI= (I I)(AI+ 1)S(AI)1.

The dependence ofSon the matrix entries ofAIis smooth, by Cramers formulafor the inverse matrix. It follows that the collection of allI : UI Rk(nk)

defines on GrR(k, n) the structure of a manifold of dimension k(n k). Later,

we will give an alternative description of the manifold structure for the Grass-mannian as a homogeneous space.The discussion above can be repeated by replacing R with C everywhere.

The space GrC(k, n) is a complex manifold ofcomplexdimension k(n k), i.e.real dimension 2k(n k). In particular, GrC(1, n) = CP(n) is the complexprojective space.

(d) Flag manifolds. A (complete) flag in Rn is a sequence of subspaces {0}= V0 V1 Vn= Rn where dim Vk =k for all k . Let Fl(n) be the set of all flags.It is a manifold of dimension (n2 n)/2, as one can see from the following roughargument. Any real flag in Rn determines, and as determined by a sequenceof 1-dimensional subspaces L1, , Ln, where each Lj is orthogonal to the sum

ofLk with k = j. Indeed the flag is recovered from this by putting E1 = L1,E2 = L1 L2 and so on. There is an RP(n 1) of choices for L1. GivenL1there is an RP(n 2) of choices for L2 (since L2 is orthogonal to L1). GivenL1, . . . , Lj there is an RP(n j 1) of possibilities forLn+1. Hence we expect,

dim FlR(n) =n1j=1

(n j) =n(n 1)/2.


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1. MANIFOLDS 7

It is possible to construct an atlas for FlR(n) using this idea. Below we will givean alternative approach, by showing that the flag manifold is a homogeneous

space (see below). Similarly, one can define a complex flag manifold FlC(n), con-sisting of flags of subspaces in Cn. Also, one can define spaces FlR(k1, . . . , kl, n)ofpartialflags, consisting of subspaces {0}= E0 E1 ElEl+1 = Rn

of given dimensions k1, . . . , kl, n. Note that FlR(k, n) = Gr(k, n).(e) Klein Bottle. Let Mbe the manifold obtained as a quotient [0, 1] [0, 1]/

under the equivalence relation (x, 0) (x, 1), (0, x) (1, 1 x). Exercise: Thequotient space has natural manifold structure. Hint: WriteMas a quotient ofR2 rather than [0, 1]2. Then use charts for R2 to define charts for M.

A manifoldMis calledorientableif it admits an atlas such that the Jacobians of alltransition maps

1 have positive determinants. An manifoldMwith such an atlas

is called an orientedmanifold.

Exercise 1.4. Show that GrR(1, n+ 1) = RP(n), and GrR(k, n) = GrR(n k, n).

Exercise 1.5. Construct a manifold structure on the space M = GrorR

(k, n) ofori-entedk-planes in Rn.

Exercise 1.6. Show that RP(n) is orientable if and only ifn is odd. Any idea forwhichk, nthe Grassmannian GrR(k, n) is orientable? (Answer: If and only ifn is odd.)

Show that the Klein Bottle is non-orientable. Show that any complex manifold(viewed as a real manifold) is oriented.

1.3. Smooth maps between manifolds.

Definition 1.7. A map F : N Mbetween manifolds is called smooth (or C)if for all charts (U, ) ofNand (V, ) ofMwith F(U) V, the composite map

F 1 : (U) (V)

is smooth. The space of smooth maps fromN to M is denoted C(N, M). A smoothmapF : NMwith smooth inverse F1 : MN is called a diffeomorphism.

In the special case where the target space is the real line we write C(M) :=C(M, R). The space C(M) is an algebra under pointwise multiplication, called thealgebra of functions onM. For anyfC(M), one defines the supportoffto be the

closed set

supp(f) ={x M| f(x)= 0}.

Clearly, the composition of any two smooth maps is again smooth. In particular, anyFC(N, M) defines an algebra homomorphism

F : C(M) C(N), f Ff=f F


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called thepull-back.

R

M F

F

f

N

f

In fact, a given mapF : NMis smooth if and only if for allfC(M), the pulledback mapFf=f F is smooth. (Exercise.)

IfMis a manifold, we say that a coordinate chart : U Rm is centered atx Mif(x) = 0.

Definition 1.8. Let F C(N, M) be a smooth map between manifolds of di-mensions n, m, and x N. The rank ofF atx, denoted rankx(F), is the rank of theJacobian

D(x)( f 1) : Rn Rm,

for any choice of charts : U Rn centered at x and : V Rm centered at F(x).The point x is called regular if rankx(F) = m, and singular (or critical) otherwise. Apointy M is called a regular valueif rankx(F) =m for allx F1(y),singular valueotherwise.

Note that the rank of the map F does not depend on the choice of coordinate charts.According to our definition, points that are not in the image ofFare regular values.

Lemma1.9. The mapM Z, xrankx(F)is lower semi-continuous: That is, foranyx0 Mthere is an open neighborhoodU aroundx0 such thatrankx(F) rankx0(F)

for x U. In particular, if r = maxxMrankx(F), the set{x M| rankx(F) = r} isopen inM.

Proof. Choose coordinate charts : U Rn centered at x0 and : V Rm

centered at F(x0). By assumption, the Jacobian D(x0) : Rn Rm has rank r =

rankx0(F). Equivalently, some r r-minor of the matrix representing D(x0) has non-zero determinant. By continuity, the same r r-minor for anyD(x), x Uhas nonzerodeterminant, provided Uis sufficiently small. This means that the rank ofF atx mustbe at leastr .

Definition 1.10. Let F C(N, M) be a smooth map between manifolds of di-mensionsn, m. The mapF is called a

submersionif rankx(F) =m for all x M. immersionif rankx(F) =n for allx M. local diffeomorphismif dim M= dim N andFis a submersion (equivalently, an

immersion).

Thus, submersions are the maximal rank maps ifm n, and immersions are themaximal rank maps ifm n.


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Theorem 1.11 (Local normal form for submersions). SupposeF C(N, M) is asubmersion,x0 N. Given any coordinate chart(V, )centered atF(x0), one can find a

coordinate chart(U, )centered atx0 such that the map F = F 1

: (U) (V)is given by

F(x1, . . . , xn) = (x1, . . . , xm).

Proof. The idea is simply to take the components ofFas the first mcomponentsj(x) of the coordinate function near x0.

Using the given coordinate chart aroundF(x0) and any coordinate chart aroundx0,we may assume that M is an open neighborhood of 0 Rm and N an open neighbor-hood of 0 Rn. We have to find a smaller neighborhood U of 0 M Rn and adiffeomorphism : U(U) Rn such that F =F 1 has the desired form.

By a linear transformation ofRn, we may assume that Fi

xj(0) =ij fori, jm. Let

: M Rn

be the map(x1, . . . , xn) =

F1(x1, . . . , xn), . . . , F m(x1, . . . , xn), xm+1, . . . , xn

.

The Jacobian of at x = 0 is just the identity matrix. Hence the inverse functiontheorem applies: There exists some smaller neighborhood U of 0 M Rn, such thatis a diffeomorphism U(U). Then (U, ) is the desired coordinate system.

Theorem 1.12 (Local normal form for immersions). SupposeF C(N, M) is animmersion, x0 N. Given any coordinate chart (U, ) centered at x0, one can find acoordinate chart(V, ) centered atF(x0) such that the map F = F 1 : (U)(V) is given by

F(x1, . . . , xn) = (x1, . . . , xn, 0, . . . , 0).

Proof. The idea is to use the given coordinates on U as coordinates on F(U),near F(x0), and supplementary m n coordinates in transversal directions to getcoordinates onM near F(x0).

Using the given coordinates chart aroundx0 and anycoordinate chart aroundF(x0),we may assume thatMis an open neighborhood of 0 Rm andNan open neighborhoodof 0 Rn. We have to find a smaller neighborhood V M, and a diffeomorphism :V (V) Rm such that F = Fhas the form F(x1, . . . , xn) = (x1, . . . , xn, 0, . . . , 0).

Let x1, . . . , xn be the coordinates on Rn and y1, . . . , ym the coordinates on Rm. Byassumption, the matrix F

i

xj(x) has maximal rank n for allx U. By a linear change of

coordinates onV, we may assume that ( Fi

xj(x0))i,jn = ij . Consider the map Consider

the map

: N Rmn Rm, (x, s)F(x) + (0, . . . , 0, sn+1, . . . , sm)

The Jacobian of at 0 is just the identity matrix. Hence, the inverse function theoremapplies, and we can find an open neighborhoodVaround 0 M Rm such that 1 is awell-defined diffeomorphism overV. The map = 1 : V (V) URmn Rm

is the desired coordinate chart.


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Exercise 1.13. Let : NMbe a surjective submersion. SupposeF : M Xis any map such that Flifts to a smooth map F : N X, i.e. F = F. Then F is

smooth.

Definition 1.14. Let M be a manifold. A subset S M is called an embeddedsubmanifold if for each x0 S there exists a coordinate system (U, ) centered at x0,such that

(U S) ={(x1, . . . , xm) (U)| xk+1 =. . .= xm = 0}

It is obvious from the definition that submanifolds inherit a manifold structure fromthe ambient space: Given a covering of S by coordinate charts (U, ) as above, onesimply takes (U S, |US) to define an atlas for S.

The following two Theorems are immediate consequences of the normal form theoremsfor submersions and immersions, respectively.

Theorem 1.15. Let F C(N, M) be a submersion. Then each level set S =F1(y) fory M is an embedded submanifold of dimensionn m.

Theorem 1.16. LetFC(N, M) be an immersion. For eachx0 N there existsa neighborhood U of x0 such that the imageS = F(U) is an embedded submanifold ofdimensionm n.

These theorems provide many new examples of manifolds. Often, manifolds areobtained as level sets for a smooth function on a Euclidean space RN. For example, wesee again that Sn Rn+1 is a manifold. Another example is the 2-torus, for 0 < r < Rthe radii of the small and big circles, given as a level setG1(r2) whereG C(R3)is the function,

G(x1, x2, x3) = (x3)2 + (

(x1)2 + (x2)2 R)2.

The 2-torus can also be described as the image of an immersion F : R2 R3, whereF(, ) = (x1, x2, x3) is given by

x1 = (R+ r cos )cos ,

x

2

= (R+ r cos )sin ,x3 = r sin

In fact, it is clear that this map descends to an embedding R2/(2Z)2 R3 as asubmanifold.

We will show below that any compact manifold can be smoothly embedded into someRN. (In fact, compactness is not necessary but we wont prove this harder result.)


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Exercise 1.17. Construct an explicit embedding of the Klein bottle into R4. Solu-tion: Given 0< r < R define F(, ) = (x1, x2, x3, x4) where

x1 = (R+ r cos )cos ,

x2 = (R+ r cos )sin ,

x3 = r sin cos /2,

x4 = r sin sin /2.

for 0 2 and 0 2.

1.4. Tangent vectors. There is a number of equivalent coordinate-free definitionsfor the tangent space TxMof a manifold x Mat some point x M. Our favoritedefinition definesTxMas the space of directional derivatives.

Definition

1.18.

LetMbe a manifold. Atangent vectoratx M is a linear mapv: C(M) Rsatisfying the Leibnitz rule (product rule)

v(f1f2) =v(f1)f2(x) + f1(x)v(f2).

The vector space of tangent vectors at x is denoted TxM, and called the tangent spaceatx.

It follows immediately from the definition that any tangent vector vanishes on con-stant functions. Indeed, if 1 denotes the constant functionf(x) = 1, the product rulegives

v(1) =v(1 1) =v(1) 1 + 1 v(1) = 2v(1)

thusv(1) = 0. Furthermore, the product rule shows that for any two functions g, hwith

g(x) =h(x) = 0, v(gh) = 0.

Lemma 1.19. If U Rn is an open subset andx0 U, the tangent spaceTx0U isisomorphic to Rn, with basis the derivatives in coordinate directions,

xi|x0 : f

f

xi(x0)

Proof. We may assume x0 = 0. Let v T0U. Given f C(U), use Taylorstheorem with remainder to write f(x) =f(0) +

mi=1

fxi (0)x

i +m

i=1 gi(x)xi, where gi

vanishes at 0. v vanishes on the constant f(0), and also on the products xigi(x). Thus

v(f) =

m

i=1

f

xi (0)v(xi

) =

m

i=1

aif

xi |x=0

whereai = v(xi).

Lemma 1.20. Let M be a manifold of dimension m. If : U M is any openneighborhood ofx, the map

: TxUTxM, v(f) =v(f|U)


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is an isomorphism. In particular, any coordinate chart(U, )aroundx gives an isomor-phismTxM= Rm.

Proof. We first show that for any v TxM, v(f) depends only on the restrictionoff to an arbitrary open neighborhood ofx. Equivalently, we have to show that iffvanishes on a neighborhood of x then v(f) = 0. Using a coordinate chart around xconstruct C(M) with = 1 on a neighborhood ofx and = 0 on a neighborhoodof the support off. Let g = 1 . Thenf g = f since g = 1 on supp(f). Since bothf, g both vanish at x, v(f) =v(f g) = 0, as required.

This result can be re-interpreted as follows: LetV Mbe an open neighborhoodofx, and FV(M) C(M) be the functions supported in V. ThenTxM can also bedefined as the space of linear maps F R satisfying the Leibnitz rule. Indeed, if issupported onV and = 1 nearx, then any functionfcoincides withf FV(M) nearx. In particular, choose V with V U. ThenFV(U) =FV(M), and it follows directlythatTxM=TxU.

Definition 1.21. LetF C(N, M) be a smooth map, and xN. The tangentmap

dxF : TxNTF(x)M

is defined as follows:(dxF(v))(f) =v(F

f), fC(M).

It is immediate from the definition that dxF is a linear map. We often writeF :TxNTF(x)Mif the base point is understood. The map F is also called push-forward.Under composition of maps, (F1 F2)= (F1) (F2).

Exercise 1.22. LetU Rm and V Rn be open subsets and F C(U, V). Forall x U, the isomorphisms TxU = Rm and TF(x)V = Rn identify the tangent mapdxF : TxUTF(x)Vwith the Jacobian DxF : R

m Rn. That is,

(dxF)(

xi) =

j

Fj

xi

yj,

wherexi, yj are the coordinates on U, V.

Thus dxFis just the coordinate-free definition of the Jacobian: any choice of charts(U, ) around x and (V, ) around F(x) identifies dxF with the Jacobian of the map

F

1

: (U) (V) at(x). In particular,rankx(F) = rank(dxF).

Fis an immersion if dxFis injective everywhere, and a submersion if dxF is surjectiveeverywhere.

Definition 1.23. A mapFC(N, M) is called an embedding ifF is an injectiveimmersion and Fis a homeomorphism onto F(N) (with the subspace topology).


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Thus, a 1:1 immersion is an embedding if and only if the map N F(N) is openfor the subset topology. That is, one has to verify that for each open subset UofN, the

image F(U) can be written F(U) =F(N) V where V is open in M.Example 1.24. Consider the curve

: R R2, t(sin(2t), cos(t)).

Then is an immersion, with image a figure 8. LetFbe the restriction ofto theopen interval (/2, 3/2). Then F is a 1-1 immersion, but is not an embedding. Forinstance, the image of the open interval (0, ) is not open in the subspace topology. Notethat the image ofFis still the full figure 8, so it is not an embedded submanifold.

Exercise 1.25. Show that the image of an embedding is an embedded submanifold.Conversely, if N M is an embedded submanifold with the induced structure of amanifold onN, then the inclusion map NMis an embedding.

IfF is an embedding, then dxF : TxN TF(x)M is injective, so TxN is identifiedwith a subspace ofTF(x)M. In particular, ifNis an embedded submanifold ofR

m, thetangent spaces toN are canonically identified with subspaces ofRm.

Exercise 1.26. SupposeF C(N, M) hasa Mas a regular value, so F1(a) isan embedded submanifold ofN. Show that

Tx(F1(a)) = ker(dxF) TxN

for allx F1(a). (Hint: Use the normal form theorem for submersions.)

1.5. Velocity vectors for curves. If I R is an open interval, a map C(I, M) is called a parametrized curve in M. For any t I, one can define thevelocity vector

(t) T(t)M

by (t) = ()(t ). The action of the velocity vector on functions is, by definition of

push-forward,

(t)(f) = d

dtf((t)).

Exercise 1.27. Show that every v TxM is of the form v = (0) for some curve

: (, ) M with(0) =x. (Hint: Use a chart.)

Exercise 1.28. LetFC(N, M) andC(I, N) a smooth curve. Show that

F((t)) = d

dtF((t)),

the tangent vector for the curve F .


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1.6. Jet spaces. Let Cx (M) be the space of functions vanishing at x, and Cx (M)

2 Cx (M) linear combinations of products of such functions. Thus C

x (M)

2 consists of

functions on M that vanish to second orderat x. From the definition, it is clear thatany tangent vector is determined by its value onCx (M), and is zero on Cx (M)

2. Thusone has a natural linear map TxM(Cx (M)/C

x (M)

2).

Proposition 1.29. The map

TxM=(Cx (M)/C

x (M)

2)

is a vector space isomorphism.

Proof. Clearly, this map is injective. To show surjectivity, we have to show thatif v : C(M) R is any linear map vanishing on constants and on Cx (M)

2, thenv satisfies the Leibnitz rule. Given f, g write f = a+ (fa) where f(x) = a, and

g= b + (g b) where g(x) =b. Then f aandg b vanish at x. Thusv(f g) = v(ab) + av(g b) + v(f a)b + v((f a)(g b))

= av(g b) + v(f a)b

= av(g) + v(f)b.

Exercise 1.30. Show that Cx (M) is a maximal ideal in the algebra C(M), and

that every maximal ideal is of this form, for some x.

Exercise 1.31. Give the definition of the tangent map dxFfrom this point of view.

This second definition suggests a definition of higher tangent bundles: One defines,for allk = 1, 2, . . .

JkxM :=C(M)/Cx (M)

k,

the kth jet space. For any functionf Cx (M), its image Jkx (f) in J

kxM is called the

kth order jet atx. In local coordinates, Jkx (f) represents thekth order Taylor expansionatx.

Thus J0xM = R, J1xM = R T

x M. For larger k one still has projection maps

Jk+1x MJkxMbut these maps no longer split: There is no coordinate invariant meaning

of the terms of order exactly k + 1 of a Taylor expansion, unless the terms of order kvanish.

2. Partitions of unity

Before developing the theory of vector fields, differential forms etc., we need animportant technical tool known aspartitions of unity.

We recall a few notions from topology. An open cover (V)B of a topological spaceMis called a refinementof a given open cover (U)A if each V is contained in someU. A cover (U)Ais called locally finite if each point in Mhas an open neighborhoodthat intersects only finitely many Us. A topological space is called paracompact if


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every open cover has a locally finite refinement. We will show now that manifolds areparacompact. First we need:

Lemma2.1. Every manifold has an open covering(Si)Ni=1 where eachSi is compact,andSi Si+1, for all i.

Proof. Let (Ui)i=1 be a countable basis of the topology. Already the Uis with

compact closure are a basis for the topology, so passing to a subsequence we may assumethat each Ui has compact closure. LetS1 = U1. Let k(1) > 1 be an integer such thatU1, . . . , U k(1) cover S1. Put S2 := U1 . . . Uk(1). Let k(2) > k(1) be an integer such

that U1, . . . , U k(2) cover S2. Put S3 :=U1 . . . Uk(2). Proceeding in this fashion, oneproduces a sequenceSi with the required properties.

Theorem 2.2. Every manifoldM is paracompact. In fact, every open cover admitsa countable, locally finite refinement consisting of open sets with compact closures.

Proof. Let (U)A be any given cover of M. Here A is any indexing set. LetS1, S2, . . .be the sequence from Lemma 2.1. EachSiis compact, and is therefore coveredby finitely many Us. It follows that there exists a countable subsetA

A such that(U)A is a covering ofM. ReplacingA withA

, we may assume that our indexing setis A={1, 2, . . .} (possibly finite). For eachj, let k(j) be an integer such that the setsUi withi k(1) coverSj. For 1 k k(j) define

Vjk =Uk (Sj+1\Sj1)

where j is the integer such that k(j 1) < k k(j) (we put k(0) = 0 and S0 = ).Then the collection Vjks are a locally finite refinement of the given cover, and each V

jk

has compact closure.

Lemma 2.3. LetC Mbe a compact subset of some manifoldsM. For any openneighborhoodU ofCthere exists a smooth functionf C(M) with0 f 1, suchthatf= 1 onC andsupp(f) U.

Proof. For each x C, choose a function fx C(M) with supp(fx) U andfx = 1 on a neighborhood of x. (Such a function is easily constructed using a localcoordinate chart.)Let Cx = f

1x (1). Then {int(Cx)}xC are an open cover of C. By

compactness, there exists a finite subcover. Thus we can choose a finite collection ofpointsxi such that the interiors of the sets Ci = Cxi coverC. Write fi = fxi . Then

f= 1

Ni=1

(1 fi)

has all the required properties.

Lemma2.4 (Shrinking Lemma). Let(U)Abe a locally finite covering of a manifoldM by open sets with compact closure. Then there exists a covering (V)A such thatV U for all A.


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Proof. Since the covering is locally finite, it is in particular countable. (Taking anopen cover Si of M as above, where each Si is compact and contained in Si+1, only

finitely many Us meet each Si. This implies countability.) Thus we may assume thatA= N or a finite sequence, and write the given cover as (Ui)

Ni=1,N N. Inductively,

we now construct open subsets Vi with Vi Ui, such that V1, . . . , V i, Ui+1, Ui+2, . . .stillform a cover ofM. Given V1, . . . , V i, we construct Vi+1 as follows: Choose 0 f 1supported inUi+1 with f= 1 on the compact set

C=M\

ki

Vi

ki+2

Ui

Ui+1.

Then take Vi+1 to be the open set where f > 0. ThenV1, . . . , V i+1, Ui+2, . . . is an opencover. Since the original cover was locally finite, (Vi)

Ni=1 is a cover ofM.

Theorem 2.5. Let(U)A be an open covering of a manifoldM. Then there exists

a partition of unity subordinate to the cover U, that is, a collection of functions such that

(a) Each point x M has an open neighborhood U meeting the support of onlyfinitely manys.

(b) 0 1,(c) supp() U,(d)

= 1

(The sum is well-defined, since near each point only finitely many are non-zero.)

Proof. Suppose first that the cover is locally finite and that each U is compact.Choose a shrinking (V)A as in the Lemma. Choose functions 0 f 1 supported

in U with f = 1 on V. Since the covering is locally finite, the sum f = fexists, and clearly f > 0 everywhere. Put = f/f. This proves the Theorem forlocally finite covers with compact closures. In the general case, choose a locally finiterefinement (U)B consisting of open sets with compact closures. There is a function

j: B A, = j() such that UUj(), and define =

j()= .

Exercise 2.6. Show that Lemma 2.3 holds for any closed, not necessarily compact,subset ofM.

Here is a typical application of partitions of unity.

Theorem 2.7. Every manifoldMcan be embedded into Rk, fork sufficiently large.

Proof.

We will prove this only under the additional assumption that M admits afinite atlas. The existence of a finite atlas is obvious ifMis compact. It can be shownthat in fact,everymanifold admits a finite atlas but the proof is not so easy (see e.g. thebook Greub-Halperin-Vanstone). Let (Uk, k)

Nk=1 be a finite atlas. Choose a partition of

unity subordinate k, subordinate to the cover Uk. Thenkk : Uk Rm extends byzero to a function fkC(M,R

m). Let

F : M R(m+1)N, x(f1(x), . . . , f N(x), 1(x), . . . , N(x)).


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3. VECTOR FIELDS 17

We claim thatFis an embedding: That is,Fis a 1-1 immersion that is a homeomorphismonto its image. First, F is 1-1. For suppose F(x) =F(y). Thus i(x) =i(y) for all i.

Choose i with i(x) > 0. In particular,x, y Ui. Dividing the equationfi(x) = fi(y)byi(x) =i(y) we find i(x) =i(y), thus x= y . More generally, for any x N andany sequence yi N,F(yi) F(x) implies yi x. This shows that the inverse map iscontinuous, so F is a homeomorphism onto its image. Finally, let us show thatF hasmaximal rank at x: For suppose v ker(dxF). Then v(i) = 0 and v(fi) = 0 for all i.Choose isuch thati(x)>0, thus x Ui. Writing fi = ii the product rule gives

0 =v(ii) =i(x)v(i),

thusv(i) = 0. Buti is a diffeomorphism onto its image, so v = 0.

A famous theorem of Whitney (1944) says that any manifold of dimension dim M=mcan be embedded into R2m, and immersed into R2m1. See Smale, Bull.Am.Math.Soc.

69 (1963), 133-145 for a survey of results on embeddings and immersions.

3. Vector fields

3.1. Vector fields. Suppose A is an algebra over R. A derivationofA is a linearmapD : A A satisfying the Leibnitz rule

D(ab) =aD(b) + D(a)b.

IfD1,D2 are derivations, then so is their commutator [D1, D2] =D1D2 D2D1. Recallthat the commutator satisfies the Jacobi identity,

[D1, [D2, D3]] + [D2, [D3, D1]] + [D3, [D1, D2]] = 0.

Thus Der(A) is a Lie subalgebra of the algebra End(A). IfA is commutative, the spaceDer(A) is an A-submodule of End(A): IfD is a derivation and x A, then xD is alsoa derivation.

Definition 3.1. A vector field on M is a derivation XDer(C(M)). That is, Xis a linear map X : C(M) C(M) satisfying the Leibnitz rule

X(f g) = (Xf)g+ f(Xg), f, g C(M).

The space of vector fields will be denoted X(M). If X, Y X(M), the vector field[X, Y] =X Y Y Xis called the Lie bracketofX andY.

Thus, the space X(M) of vector fields is a Lie algebra. They are also a C(M)-

module, that is, f X X(M) for all f C(M) and X X(M). Evaluation at anypointx Mdefines a linear map

X TxM, X Xx, Xx(f) = (Xf)(x).

Conversely Xcan be recovered from the collection of tangent vectors Xx with x M.We write

supp(X) ={x M| Xx= 0}.


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One can think of a vector field as a family of tangent vectors depending smoothly on thebase point.

Exercise 3.2 (Locality). Show that vector fields are local, in the following sense: IfUis an open subset of a manifold M, andX X(M), there exists a unique vector fieldXU X(U) (called the restriction ofX toU) such that (XU)x= Xx for all x U.

The action of X on functions f C(M) supported in U is given in terms ofthe restriction by Xf|U = XU(f|U). Using a partition of unity, this means that therestrictions ofXto coordinate charts determine the vector field X. In local coordinates,vector field have the following form:

Lemma 3.3. SupposeU Rn is an open subset. Then anyX X(U) has the form

X=n

i=1 ai

xi

whereai C(U).

Proof. From the description of tangent vectors v TxU, we know that

Xx =n

i=1

ai(x)

xi

for some function ai(x). We have to show that ai is smooth. But this is clear sinceai= X(f) where fC(U) is the coordinate function f(x) =xi.

Any diffeomorphismF C(N, M) induces a map F : X(M) X(N) of vector

fields, with F

(F(X)(f)) =X(F

f). ThusF(Xx) = (FX)F(x).

For a general map F, this equation does need not define a vector field FX, for tworeasons: 1) IfFis not surjective, the equation does not specifyFXat points away fromthe image of F, 2) IfF is not injective, the equation assigns more than one value topointsy Mhaving more than one pre-image.

Definition 3.4. LetF C(N, M). Two vector fields X X(N) and Y X(M)are called F-related if

F(Y(f)) = X(Ff)

for allfC(M). One writes XF Y.

Proposition 3.5. XF Y if and only if

F(Xx) =YF(x)

for allx N. IfX1F Y1 andX2F Y2 then

[X1, X2]F [Y1, Y2].

Proof. Exercise.


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3. VECTOR FIELDS 19

For example, ifF is the embedding of a submanifold, then X X(N) is F-relatedto a vector field Y X(M) if and only ifY is tangent to F(N) everywhere. The vector

fieldXis then just the restriction ofY toN.Definition 3.6. A Riemannian metricon a manifold M is a symmetric C(M)-

bilinear form g : X(M) X(M) C(M) which is positive definite in the sense thatfor allx M,g(X, X)(x)> 0 for Xx= 0.

Exercise 3.7. Every manifold M admits a Riemannian metric g. (Hint: Definesuch a metric in charts, and then use a partition of unity to patch the local solutionstogether.) If g is a Riemannian metric and x M, the value g(X, X)(x) dependsonly on Xx. IfU M is open, there exists a unique Riemannian metric gU such thatgU(XU, YU) =g(X, Y)|U.

3.2. The flow of a vector field. Suppose A is a finite dimensional algebra, andAut(A) its group of algebra automorphisms. A 1-parameter subgroup of automorphismsis a group homomorphism : R Aut(A), t t, i.e. a map such that t1+t2 =t1t2 for all t1, t2 R. For any 1-parameter group of automorphisms, the derivativeD= ddt |t=0t: A A is well defined. By taking the derivative of the equation

t(ab) = t(a)t(b)

one finds that D Der(A). Conversely, every derivation gives ride to a 1-parametergroup t= exp(tD) (exponential of matrices), and one check that each tis an automor-phism. Thus, if dim A< one can identify derivations of an algebra with 1-parametergroups of automorphisms.

For the infinite-dimensional algebra A = C(M), this does not directly apply. Wewill see that so-called completevector fields X X(M) define a 1-parameter group ofdiffeomorphisms t ofM, where the algebra automorphism

t ofC

(M) plays the roleof exp(tX). This flow ofX is constructed using integral curves.

Let Xbe a vector field on a manifold M. A curve : I M is called an integralcurveofX if for all t I,

(t) =X(t).

IfU Rm is an open subset andX X(U) is a vector field on U, this has the followinginterpretation. Letting xi be the local coordinates on Rm, write

X= i

ai

xi

where ai C(U). Also write (t) = (1(t), . . . , m(t)). Then the tangent vector(t) T(t)U is just

(t) =i

i

xi

(t)

.


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20 CONTENTS

To see this, apply (t) to a function fC(U):

(t)(f) =

t f((t)) = i

di

dt

f

xi (t).Thus is an integral curve if and only its components satisfy the following system offirst order ordinary differential equation:

di

dt =ai((t))

From ODE theory, it follows that for any x U, there exists a unique maximal solutionx of this system: That is, there is an open interval Ix around t = 0, and a curvex : Ix U with

x(0) =x, ix(t) =a

i(x(t))

such that any other solution of this initial value problem is obtained by restriction toa subinterval. Moreover, this solution depends smoothly on the initial value x. Byapplying this result in manifold charts, one obtains:

Theorem 3.8. Let X be a vector field on a manifold M. For each x M thereexists a unique maximal solutionx : Ix Mto the initial value problem

x(0) =x, x(t) =Xx(t).

The solution depends smoothly on the initial valuex, in the following sense:

U=xM

({x} Ix) M R.

ThenU is an open neighborhood ofM {0} inM R, and the map : UM, (x, t)x(t)

is smooth.

Here maximal solution means that any other solution is obtained by restriction tosome subinterval.

One calls the flow of the vector field X. If U = M R, that is if Ix = R forall x, the vector field is called complete, and the flow is called a global flow. A simpleexample for an incomplete vector field is X = t on M = (0, 1) R. By choosing adiffeomorphism (0, 1) = R, one obtains an incomplete vector field on R. Let us writet(x) = (x, t) fort Ix.

Exercise3.9. Show that on a compact manifold, any vector field is complete. Moregenerally, any compactly supported vector field on a manifold is complete.

Theorem3.10. SupposeX X(M)is a complete vector field, andt its flow. Theneacht is a diffeomorphism, and the map

R Diff(M), tt


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3. VECTOR FIELDS 21

is a group homomorphism. In particular,

0= Id, t1 t2 = t1+t2.

Conversely, iftt is any such group homomorphism and if the map (t, x) = t(x)is smooth, thent is the flow of a uniquely defined complete vector fieldX, called thegenerating vector field for the flow. For allfC(M) one has

X(f) =LX(f) := d

dt t=0t f.

Proof. We first prove the identity t1t2 = t1+t2 for t1, t2 R. Let x M.Givent2, both(s) = s(t2(x)) and (s) = s+t2(x) are integral curves forX. Indeed,

d

ds

(s) = d

du

u(x)|u=s+t2 =Xu(x)|u=s+t2 =X(s).

Since , have the same initial value, they coincide on their domain of definition. Inparticular, (t1) = (t1) which proves t1(t2(x)) = t1+t2(x). In particular, if X iscomplete, this equation holds for all t1, t2. Setting t1 = t, t2 = t we see that t is asmooth inverse to t. Conversely, if t is a global flow, define X byXx =

t

|t=0t(x).Using local coordinates, one checks that this defines a smooth vector field.

Exercise3.11. How much of this theorem goes though for incomplete vector fields?

Exercise 3.12. Show that the map t : Rm Rm given by multiplication by et is

a global flow, and give a formula for the generating vector field. More generally, ifAisanym mmatrix, show that the map

t(x) =etAx

is a global flow, and find its generating vector field.

Exercise 3.13. Let X X(N), Y X(M) be vector fields and F C(N, M) asmooth map. Show thatXF Y if and only if it intertwines the flows Xt ,

Yt : That is,

F Xt = Yt F.

Exercise 3.14. Show that for any vector field X X(M) and any x M withXx = 0, there exists a local chart around x in which Xis given by the constant vectorfield x1 . Hint: Show that ifSis an embedded codimension 1 submanifold, with x S

and Xx TxS, the map U (, ) Mis a diffeomorphisms onto its image, fo someopen neighborhood U ofx in S. Use the time parameter t and a chart around x U todefine a chart nearx.

For any vector field X X(M) and any diffeomorphism F C(N, M), we defineFX X(N) by

FX(Ff) =F(X(f)).


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ThusF : X(M) X(N) is just the inverse map toF: X(N) X(M). Any completevector field X X(M) with flow t gives rise to a map t : X(M) X(M). One

defines the Lie derivativeLXof a vector fieldY X(M) by

LX(Y) = d

dt

t=0

t Y X(M).

In fact, this definition also makes sense for incomplete X: (To define the restriction ofLXY to an open set U M with compact closure, let > 0 be small enough such[, ] Ix for all x U. Then t : U t(U) is defined for |t|< , and the equationabove makes sense.)

Theorem3.15. For anyX, Y X(M), the Lie derivativeLXYis just the Lie bracket[X, Y]. One has the identity

[LX, LY] =L[X,Y].

Proof. Let t = Xt be the flow ofX. ForfC

(M) we calculate,

(LXY)(f) = d

dt|t=0(

t Y)(f)

= d

dt|t=0

t (Y(

t(f)))

= d

dt|t=0(

t (Y(f)) Y(

t (f))

= X(Y(f)) Y(X(f))

= [X, Y](f).

The identity [LX, LY] =L[X,Y] just rephrases the Jacobi identity for the Lie bracket.

The definition of Lie derivative gives the formula

(1) d

dt(Xt )

Y = (Xt )(LXY)

by the calculation, ddt(Xt )

Y = ddu |u=0(Xu+t)

Y = (Xt ) ddu

|u=0(Xu)Y. From this we

obtain:

Theorem3.16. LetX, Ybe two complete vector fields, with flowsXt andYs. Then

[X, Y] = 0 if and only ifXt andYs commute for allt, s:

Xt

Ys =

Ys

Xt .

Proof. Suppose [X, Y] = 0. Then

d

dt(Xt )

Y = (Xt )LXY = (

Xt )

[X, Y] = 0

for all t. Hence (Xt )Y = Y for all t, which means that Y is Xt -related to itself.

It follows that Xt takes the flow Ys ofY to itself, which is just the desired equation

Xt Ys =

Ys

Xt . Conversely, by differentiating the equation

Xt

Ys =

Ys

Xt

with respect to s, t, we find that [X, Y] = 0.


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4. DIFFERENTIAL FORMS 23

Thus [X, Y] measures the extent to which Xt and Yt fail to commute. This can be

made more precise:

Exercise 3.17. Suppose X, Y are complete vector fields, and fC(M).a) Prove the formula for the kth order Taylor expansion of (Xt )

f:

(Xt )f=

kj=0

tj

j!(X)j(f) + O(tk+1).

Formally, one writes (Xt ) = exp(tX).

b) LetFt be the family of diffeomorphisms,

Ft = Xt

Yt

Xt

Yt .

Show that

F

tf=f+ t2

[X, Y](f) + O(t3

).4. Differential forms

4.1. Super-algebras. A super-vector spaceis a vector space Ewith a Z2-grading.Elements of degree 0 mod 2 are calledeven, and elements of degree 1 mod 2 are calledodd. Thus a super-vector space is simply a vector space with a decomposition E =E0 E1. Elements in E0 are called even and those in E1 are called odd. We will alsowriteE0 =Eeven and E1 =Eodd, in particularly ifEcarries other gradings that mightbe confused with the Z2-grading. The space End(E) of endomorphisms has a splitting

End(E) = End(E)0 End(E)1,

where End(E)0

consists of endomorphisms preserving the splitting E = E0

E1

, andEnd(E)1 consists of endomorphisms taking E0 to E1 and E1 to E0. A = End(E) is afirst example of a super-algebra, that is, a Z2-graded algebra

1 A = Aeven Aodd suchthat

AkAl Ak+l mod 2

fork, l {0, 1}. The sign conventions of supermathematics decrees:

Super-sign convention: A minus sign appears whenever two odd el-ements interchange their position.

Physicists thinks of elements of odd degree as fermions, and those of even degreeas bosons. Her are some examples of the super-sign convention. Thetensor productof super-algebrasA, Bis defined to be the usual tensor product A B, with Z2-grading

(A B)k = l+m=k mod 2

Al Bm,

and algebra structure

(a1 b1)(a2 b2) = (1)m1l2(a1a2) (b1b2)

1In this course, algebra always means an associative algebra over R with unit 1.


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for ai Ali , bj Bmj . Also, ifA is any super-algebra, the super-commutatorof twoelements a1 Ak1 and a2 Ak2 is defined by

[a1, a2] =a1a2 (1)k1k2a2a1 Ak1+k2.

Proposition 4.1. The super-commutator is super-skew symmetric,

[a1, a2] =(1)k1k2[a2, a1]

and satisfies the super-Jacobi identity,

[a1, [a2, a3]] + (1)k1(k2+k3)[a2, [a3, a1]] + (1)

k3(k1+k2)[a3, [a1, a2]] = 0.

Proof. Exercise.

More generally, a super-spaceEwith bracket [, ] satisfying these identities is called

asuper-Lie algebra.An derivationof a superalgebra A of degree r {0, 1} is a linear map D : A A

such that

D(ab) =D(a)b + (1)kraD(b)

for all a Ak and b Al. Note that any derivation is determined by its values ongenerators for the algebra, and that D(1) = 0 for any derivation D. We denote

Der(A) = Der0(A) Der1(A)

the space of super-derivations.

Exercise 4.2. Show that Der(A), with bracket the super-commutator of endomor-

phisms, is super-Lie algebra. IfA is super-commutative, show that Der(A) is also anA-module (by multiplication from the left).

Exercise 4.3. SupposeA is a super-algebra. Show that the map : A End(A)taking a A to the operator (a) of multiplication (from the left) by a is a homomor-phism of super-algebras, i.e. it preserves products and degrees.

Exercise 4.4. Let E, F be super-vector spaces. Define the tensor product of twoendomorphisms of finite degree, in such a way that it respects the sign convention ofsupermathematics. Show that with this definition,

End(E F) = End(E) End(F).

Many super-vector spaces arise as Z-graded algebrasE= kZ Ek, by reducing thedegree mod 2:

Eeven =kZ

E2k, Eodd =kZ

E2k+1.

In this case, we will callEa Z-graded super-vector space. Ordinary vector spaces Ecanbe viewed as super-vector spaces by putting Eeven =E, Eodd = 0.


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4.2. Exterior algebra. We now give our main example of a graded algebra. LetEbe a finite dimensional real vector space. (The example to keep in mind is E=TxM,

the tangent space to a manifold. But other examples will appear as well.)For any finite dimensional vector spaceEover R, one defines0E = R,1E =E,and more generally

kE ={antisymmetrick-linear maps : E E k times

R}.

Thus kE if it is linear in each argument and satisfies

(v(1), . . . , v(k)) = sign()(v1, . . . , vk)

for any v1, . . . , vk Eand any permutation of the index set 1, . . . , k. Each kE isa finite dimensional vector space. Ife1, . . . , en is a basis for E, any is determined by

its values on ei1 , . . . , eik for any i1 < . . . < ik. Since the number of ordered k-elementsubsets of{1, . . . , n} is n!k!(nk)! , it follows that

dim kE = n!

k!(n k)!

for k n, and kE = 0 for k > n. Note dim nkE = kE. Non-zero elements ofthe 1-dimensional vector space nE are called volume elements. The direct sum of thevector spacekE is denoted2

E =n

k=0

kE,

its dimension is dim E

= 2

n

, the number of subsets of{1, . . . , n}.The graded algebra structure on the vector space E is the so-called called wedgeproduct.

For 1 k1E and 2 k2E one defines 1 2 kE, k = k1 +k2 byanti-symmetrization

(1 2)(v1 . . . , vk) = 1

k1!k2!

Sk

sign()1(v(1), . . . , v(k1))2(v(k1+1), . . . , v(k))

whereruns over all permutations. The wedge product extends to all ofE by linearity.For example, iff1, f2 E then

(f1 f2)(v1, v2) =f1(v1)f2(v2) f1(v2)f2(v1).

More generally, iff1, . . . , f kE,

(2) (f1 fk)(v1, . . . , vk) = det(fi(vj)).

The following properties of the wedge product are left as an exercise.

2In the infinite dimensional case dim E= , one defines the exterior algebra somewhat differently.Fortunately, we will only be concerned with finite dimensions.


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In particular, any endomorphism ofEgives rise to an endomorphism ofE. For any1-parameter group At = exp(tL) of automorphisms, we obtain a 1-parameter group

A

t = exp(tL)

of automorphisms ofE

. It follows that L

: E

E

extends toa derivationDL of degree 0 ofE. One has

DL(f1 fk) =j

(1)j+1L(fj)f1 fj fk.(Dont confuse DL with the algebra homomorphism (L) : E E !).

Exercise 4.8. a) Show that the map End(E) Der0(E), L DL is a Liealgebra homomorphism.

b) For v E , L End(E), show that [DL, v] =L(v). (Hint: Since both sides arederivations, it suffices to check on generators.)

After this lengthy discussion of linear algebra, let us return to manifolds.

4.3. Differential forms.

Definition 4.9. For any manifold M and any x M, the dual space Tx M :=(TxM)

of the tangent space at x is called the cotangent spaceat x. The elements ofTx Mare calledcovectors. Elements of

kTM are called k-forms on TxM.

Any function f C(M) determines a covector dxf Tx M, called its exteriordifferential atx, by

(df)x(v) :=v(f), v TxM.

IfU Rm

is an open subset with coordinates x1

, . . . , xm

(viewed as functions on U), wecan thus define 1-forms

(dxi)xTx U.

This 1-forms are the dual basis to the basis xi

x

ofTxU: Indeed,

(dxi)x

xjx

=

xj(xi)

x

= ij .

A differential k-form on M is a family of k-forms on tangent spaces TxM dependingsmoothly on the base point, in the following sense:

Definition 4.10. A differential k-form onM is a C

(M)-linear map: X(M) . . . X(M)

k times

C(M)

that is anti-symmetric and C(M)-linear in each argument. The space of k-forms isdenoted k(M), the direct sum of all k(M) is denoted (M). In particular, 0(M) =C(M).


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For any x Mthere is a natural evaluation map k(M) kTx M, x suchthat

(X1, . . . , X k)|x= x((X1)x, . . . , (Xk)x).Similarly, ifUMis open, one can define the restriction |Uk(U) such that

(X1, . . . , X k)|U=|U((X1)|U, . . . , (Xk)|U).

IfU, Vare two open subsets, and U, Vare forms on U and V withU|UV =V|UVthen there is a unique form UV (U V) restricting to U andV, respectively.

One defines the wedge product on (M) by anti-symmetrization, similar to the wedgeproduct on E. It is uniquely defined by requiring that all the evaluation maps (M)Tx Mare algebra homomorphisms. Wedge product makes (M) into a graded super-commutative super-algebra. ForX X(M), one denotes by X=(X) Der

1((M))the operator of contraction by X, and for (M) one denotes by () the operator

of left multiplication by . As before, we have [X, Y] = 0 and [X, ()] =(X).4.4. The exterior differential. The magic fact about the algebra of differential

forms is the existence of a canonical derivation d of degree 1. For any function f C(M) one defines df1(M) by the equation,

(df)(X) =X(f).

It has the property d(f g) =fdg+gdf. In particular, ifU Rm is an open subset, wehave 1-forms dx1, . . . , dxm 1(U). Writing dxI = dxi1 dxik for any k-elementsubseti1< . . . < ik, the most general k-form on U reads

=

IIdx

I

whereI C(U) are recovered as

I =(

xi1, . . . ,

xik).

Theorem 4.11. The map d : 0(M) 1(M), f df extends uniquely to asuper-derivation of degree1 on(M), calledexterior differential, with d(df) = 0. Theexterior differential has property

d d= 0.

Proof. It suffices to prove existence and uniqueness for the restrictions to elementsof an open cover (U)A. Indeed, once we know that d|U exist and are unique, then

the forms d|U agree on overlaps U Uby uniqueness, so they define a global formd. In particular, we may take (U)A to be a covering by coordinate charts. Thisreduces the problem to open subsetsU Rm.

The derivation property of d forces us to define d : k(U) k+1(U) as follows,

d(I

IdxI) =

I

dI dxI =

j

I

I

xjdxj dxI.


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We have

d2(I IdxI) = jk I

2I

xkxjdxk dxj dxI = 0,

by equality of mixed partials. It remains to check d is a derivation. For = fdxI k

and=gdxJ l we have

d( ) = d(fdxI gdxJ)

= d(f g) dxI dxJ

= (fdg+ gdf) dxI dxJ

= df dxI gdxJ + (1)kf dxI dg dxJ

= (d) + (1)k d.

Let us write out the exterior differential on M = R3, with coordinates x, y,z. Onfunctions,

df=f

xdx+

f

ydy+

f

zdz,

so d is essentially the gradient. On 1-forms = fdx + gdy+ hdz,

d=g

x

f

y

dx dy+

hy

g

z

dy dz+

fz

h

x

dz dx,

so d is essentially the curl, and on 2-forms = fdy dz+ gdz dx + hdx dy,

d= fx

+ gy

+ hzdx dy dz,

so d is essentially thedivergence. In general, one should think of d as the proper abstractsetting for grad, curl, div.

4.5. Functoriality. Any smooth mapFC(N, M) between manifolds gives riseto an algebra homomorphism

F : (M) (N)

such that (F)x= F(x).

Theorem 4.12. F

commutes with exterior differential:F d= d F.

Proof. The algebra (M) is generated by all functions fC(M) together withall differentials df. That is, any differential form can be written as a finite linear com-bination of expressions

f0df1 dfk,


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30 CONTENTS

and finally by linearity it holds everywhere. This follows by choosing a finite atlas ofMand a subordinate partition of unity. 3 Hence, it suffices to check the the identity on

functionsfand differentials df.ForfC(M), and anyx M, v TxM, we have

(F(df))x(v) = (df)F(x)(dxF(v)) = (dxF(v))(f) =v(Ff) = d(Ff)x(v).

ThusF d = d F on functions. On differentials df:

F(d(df)) = 0, d(F(df) = d(d(Ff)) = 0.

For instance, ifU Rm and V Rn are open subsets with coordinates xi, yj, andFC(U, V), with components Fj =Fyj, we have

Fdyj = dFyj = dFj = i

F

j

xidxi.

If dim U= dim V =m, we obtain in particular

F(dy1 dym) = det(Fj

xi) (dx1 dxm).

Suppose X X(M) is a complete vector field, with flow t. Then we have a 1-parameter group of automorphisms t of (M). We define the Lie derivative to be itsgenerator:

LX= d

dt|t=0

t .

The definition extends to incomplete vector fields: IfUis any open subset with compactclosure, the map t : U Mexists for|t|< , and the equation defines LX|U.

We have now defined three kinds of derivations of (M): The contraction X, theLie derivativeLX, and the exterior differential d. Recall that the graded commutator oftwo derivations is again a derivation.

Theorem 4.13. One has the following identities:

[d, d] = 0,

[LX, d] = 0,

[X, d] = LX,

[X, Y] = 0,[LX, LY] = L[X,Y],

[LX, Y] = [X,Y].

3We mentioned earlier that any manifold admits a finite atlas, although we never proved this. Thefollowing proof can be easily modified to close this gap: it is enough to check the identity on functionssupported in coordinate charts, where the fi for i >0 can be taken as coordinate functions.


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Proof. The first equation is just [d, d] = 2d2 = 0. Since (tX) and d commute,

[LX, d] = 0 by definition ofLX. The other identities can be checked on generators f, df

of (M) (using that two derivations are equal if and only if they agree on generators).For instance, the third equation is verified by the calculations,

[X, d](f) =Xdf= (df)(X) =X(f) =LXf

and[X, d](df) =Xddf+ dXdf= dLXf=LXdf.

The fourth equation is obvious since both sides vanish on generators. The fifth equation,on functions f, is just the definition of the Lie bracket, and on df follows since Liederivatives and d commute. In the last equation, both sides vanish on functions, and ondfwe have

[LX

, Y

](df) = [[LX

, Y

], d](f) = [LX

, [Y

, d]](f) = [LX

, LY

](f) =L[X,Y]

(f) =[X,Y]

df.

(Alternatively, the last equation follows by direct application of the definition of the Liederivative.)

These identities are of fundamental importance in the Cartans calculus of differentialforms. It is somewhat remarkable that contractions, Lie derivatives and d form a gradedLie subalgebra of the graded Lie algebra of derivations!

Proposition 4.14. For any k(M) and anyX0, . . . , X k X(M),

(d)(X0, . . . , X k) =j

(1)jXj((X0, . . . ,Xj, . . . , X k))+

i


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32 CONTENTS

Exercise 4.15. Let Ube an open subset ofRm, and X =

Xi

xi a vector field.Let = dx1 dxm be the standard volume form. Show that

LX = (i

Xixi

) .

Conclude that the flow ofXis volume preserving if and only if the divergence div (X) =(

iXixi) vanishes everywhere.

4.6. Orientation of manifolds. A diffeomorphism F : U V between opensubsets ofRm is called orientation preserving if for all x U, the Jacobian DxF haspositive determinant.

An atlas (U, ) for a manifold M is said to be orientedif all transition functions

1 are orientation preserving. If such an atlas exists, M is called orientable.

An orientable manifold M with an oriented atlas is called an oriented manifold. A

MapF C(N, M) between oriented manifolds of the same dimension are orientationpreserving if its expressions in oriented charts are orientation preserving.

Orientability of manifolds M is closely related to the existence of volume forms. Avolume form is an m-form m(M), with x = 0 for all x M.

Theorem 4.16. A manifoldM is orientable if and only if it admits a volume form.Any volume form on a manifold M defines a unique orientation, with the propertythat for all oriented charts(U, ),

(dx1 dxn) =f|U

where f > 0 everywhere. Two volume forms , define the same orientation if and

only if

=f withf >0.Proof. SupposeMis orientable, and let (U, )Abe an oriented atlas. Choose a

partition of unitysubordinate to the cover (U)A. The form = (dx

1 dxm)is a volume form on U. Set

=

.

Then is a volume form on M. Indeed, if v1 . . . , vm is an oriented basis of TmM,each of the forms ()x with xU takes positive values on v1 . . . , vm. Hence so does

(x) ()x.

Exercise 4.17 (Oriented double cover). LetMbe a connected manifold. The tan-

gent space at any pointx Mhas exactly two orientations. The choice of an orientationon x also induces orientations on nearby points. Let : M Mbe the map withfibers1(x) the two possible orientations onx. Show that Mhas a natural structure ofan oriented manifold, with a local diffeomorphism. Show thatM is orientable if andonly ifMis disconnected, and the choice of an orientation is equivalent to the choice ofa component ofM. One calls M the oriented double coverofM. What is the orienteddouble cover for RP(2)? For the Klein bottle?


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If M, N are two oriented manifolds, with orientations defined by volume formsM, N, the direct product MN carries an orientation defined by the volume form

MM

NN, where M, Nare the projections from M N to the two factors.Exercise 4.18. LetF : NMbe a smooth map between oriented manifolds, and

suppose a M is a regular value. Then any choice of volume forms N, M defines avolume form Son the level set S= F

1(a). In particular, S is oriented.

Suppose M is an oriented manifold with boundary M. We define an orientationon Mas follows: Given x M, let v1 TxM\TxMbe an outwardpointing vector.That is,v1(f) 0 for any functionfC(M) withf0 onM. Call a basisv2, . . . , vmofTxMoriented ifv1, . . . , vm is an oriented basis ofTxM.

4.7. Integration on manifolds. LetU Rm be open. The integral of a compactlysupported differential form k(U) is defined to be 0 unless k = m. For

= fdx1 dxm m(U),

with fa compactly supported smooth function on U, one definesU

=

f(x1, . . . , xm) dx1dx2 dxm

as a Riemann integral.Recall the change of variables formula for the Riemann integral: IfF : V U is a

diffeomorphism, thenU

=

(Ff)(y1, . . . , ym) det(

Fi

yj)dy1dy2 dym

Since det(Fiyj)dy1dy2 dym is just the pull-back under F of the form dx1dx2 dxm,

this can be written in more compact form,U

=

F1(U)

F.

This formula shows that integration is independent of the choice of coordinates, and isused to extend integration to manifolds:

Theorem 4.19. LetMbe an oriented manifold of dimensionm. There is a uniquelinear map

M

: comp(M) R such that for all oriented charts(U, ) ofM, and anyform with compact support inU,

(4) M

= (U)

(1).

Proof. Choose an atlas (U, ) and a partition of unity subordinate to thecover {U}. For any compactly supported form on M, define

M

:=

(U)

(1 )().


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34 CONTENTS

This has the required property (4), since if is supported in a chart (U, ), the changeof variables formula shows

(U)

(1 )() =

(U)

(1)() = (U)

(1).

Conversely, this equation also shows uniqueness of an integration map satisfying (4).

We will often use the following notation: IfM is any manifold, and S M is anoriented embedded submanifold, and : S Mthe inclusion map, one defines

S

: (M) R,

S

:=

S

.

4.8. Integration over the fiber. Let M, B be manifolds of dimension m, b, andlet

: MB

be a submersion. We have proved that each fiber1(y) Mof is a smooth embeddedsubmanifold of dimensionf=m b. Moreover, for anyx Mthere exists a local chart(U, ) around x such, in the coordinates x1, . . . , xm defined by , the map is justprojection onto the last bcoordinates.

To define the operation integration over the fibers we need to assume that eachfiber 1(y) is oriented, and that the orientation depends smoothly on y.

Definition 4.20. A fiberwise orientation for the submersion : M B is anequivalence class of forms mb(M) such that for each y B, the pull-back of to the fiber

1(y) is a volume form for 1(y). Two such forms ,

are calledequivalent if their pull-backs to each fiber 1(y) differ by a positive function on1(y).

For example, ifM, B are oriented by volume forms M, B, one obtains a fiberwiseorientation by letting anymb-form with M = B. The form is not uniquelydefined by this property, but its pull-back to the fibers is.)

Theorem 4.21. Let M, B be manifolds of dimension m, b, and : M B be asubmersion with smoothly oriented fibers. There exists a unique linear map

: comp(M) comp(B)

of degreeb m with the following properties:

a) For ally B the diagramcomp(M)

comp(1(y))

R

comp(B) ({y}) = R

commutes. Here the horizontal maps are pull-back under the inclusion.


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b) For allcomp(M) and all (B),

= ( ).

The map is called integration over the fiber,

Proof. Since has degree f = (m b), it vanishes on forms of degree lessthenf. On forms of degree fit is determined by property (a). In general, any form onMcan be written as a locally finite sum off-forms on M, wedged with pull-backs offorms on B . (This is easily seen locally, and the global statement follows by choosing apartition of unity.) Hence, uniqueness is clear and it suffices to prove existence. Again,by partition of unity it suffices to prove existence in charts. By the normal form theoremfor submersions, we can cover Mby chartsU, with image a direct product of open sets(U) = V W, such that becomes projection to the first f m coordinates. This

reduces the theorem to the product situation.

4.9. Stokes theorem. Integration can be extended to manifoldsMwith boundary.Recall that manifolds with boundary are defined similar to manifolds, but taking the halfspace x1 0 as the target space for the coordinate charts. The boundaryM ofM isa manifold of dimension m 1. It inherits an orientation from a given orientation onM, as follows: (...) Letx M, and v1 TXMan outward-pointing tangent vector atx. A basis v2, . . . , vn a basis ofTx(M) is oriented if and only ifv1, . . . , vn is oriented.That is, in local coordinates identifyingUwith an open subset of{x1 0}and Uwithan open subset of the subspace {x1 = 0}, the orientation is given by the volume formdx2 . . . dxm.

Theorem 4.22 (Stokes). If M is an oriented manifold with boundary, and is acompactly supported form, then

M

d=

M

.

Proof. We may assume that the support ofis contained in some coordinate chart(U, ), where (U) is an open subset of the half space {x1 0}. Write

=mi=1

fidx1 dxi dxm

where each fi has compact support. Then

d=mi=1

(1)i+1fixi

dx1 dxm.

Using Fubinis theorem, we can change the order of integration, and in the ith termintegrate over the xi-variable first. Thus all terms, except the term for i = 1, vanish.


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36 CONTENTS

We find

U d =

0

f1

x1 dx

1 dx2 dxm=

U

f1dx2 dxm =

U

.

One can view Stokes theorem as a geometric version of Greens formula from calculus.Two mapsF0, F1 C(S, M) are called (smoothly) homotopic if there exists a map

FC([0, 1] S, M) such thatF(0, ) =F0 andF(1, ).

Corollary 4.23. Suppose S is an oriented manifold, and F0, F1 : S M twosmooth maps which are smoothly homotopic. For anyclosed form onM,

S

F1 =S

F0 .

Proof. Using Stokes theorem,S

F1 F

0 =

[0,1]S

d(Ft) =

[0,1]S

Ftd= 0.

The following generalizes Stokes theorem to submersions.

Theorem 4.24. If: MB is a submersion between oriented manifolds (without

boundary),(1)mbd = d.

More generally, ifMhas a boundary, and := |Mis also a submersion, then

(1)mbd() =(d) ().

4.10. Homotopy operators. Suppose F0, F1 C(N, M) are called (smoothly)homotopic. Let

h:= F : (M) (N)

be the operator of degree1, defined as a composition of pull-back byFand integrationover fibers of : [0, 1] NN.

Theorem 4.25. The map h is ahomotopy operator betweenF0 , F

1 . That is,

h d+d h= F1 F

0 : (M) (N).

Proof. Let0, 1denote the inclusion of the two boundary componentsN{0}, N{1}. Since the fibers of are 1-dimensional, we have

d + d = ()= 1

0.


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Therefore

hd + dh = Fd + dF

= (d + d)F

= () F

= (F 1) (F 0)

= F1 F

0 .

Corollary 4.26 (Poincare Lemma). SupposeU Rm is an open ball of radiusRaround0, possiblyR= . If k(U), withk >0, is closed (i.e. d = 0) then isexact, i.e. = d for some.

Proof. Let F : [0, 1] U U be the map F(t, x) = tx. ThenF1

is the identitymap, and F0 is the constant map taking everything to the orgin. Thus F

0 = 0 since

we assume the degree of is positive, and of course F1 = . Let h be the homotopyoperator forF. Then

= F1 F

0 = dh+ hd= dh= d

where = h.

Note that the homotopy operator provides an explicit primitive for the closed form.

Examples 4.27. Let h be the homotopy operator for Rm, corresponding to theretractionF(t, x) =tx. Consider the volume form = dx1 dxm on Rm. Then

F = j

(1)j+1tm1dt j

(1)jxjdx1 dxj dxm + . . .where. . . denotes terms not involving dt. Thus

h = 1

m

j

(1)j+1xjdx1 dxj dxmThe right hand sides is proportional to the volume form on Sm1. More precisely, onehas

h =c rm

wherer = ||x|| and cis a certain constant.

Next, consider a 1-form = i fidxi. ThenF=

i

xifi(tx)dt + . . .

so that

h=i

xi 1

0

fi(tx)dt.


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38 CONTENTS

If is closed, one checks that dh= .

Exercise 4.28. LetE =jxj

xj be the Euler vector fieldon R

m. Show that if

is a closed k-form =IIdxI where the coefficients are polynomials of degree l, wehave

h= 1

k+ l(E).

5. De Rham cohomology

5.1. Definition. Let M be a manifold. The exterior differential d on (M) givesus a complex

00(M) d1(M)

d

dm(M)0.

called thede Rham complex. The subspace of closed forms

Zk(M) ={ k(M)| d= 0}

is called the space ofcocycles, and the space of exact forms

Bk(M) = d(k1(M)) k(M)

is the space ofcoboundaries. One writes B(M) =m

k=0 Bk(M) and Z(M) =

mk=0 Z

k(M).Since d squares to zero, every k-coboundary is a k-cocycle. The converse is not true ingeneral, and the obstruction is measured by the cohomology

Hk(M) =Zk(M)/Bk(M)

We write Z(M) =

mk=0 Z

k(M) andB(M) =

mk=0 B

k(M) and call

H(M) =m

k=0

Hk(M)

the the de Rham cohomologyofM. The numbers

bk = dim Hk(M)

are called the Betti numbersof the manifold M. (We will see later that bk 0 is also exact. Thus

Hk(Rm) =

R fork= 0,

0 otherwise .


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5. DE RHAM COHOMOLOGY 39

Examples 5.1. (a) Since B0(M) = 0, the zeroth cohomology space is justH0(M) = Z0(M). A function f 0(M) is a cocycle if and only is f is

constant on connected components ofM. ThusZ0

(M) is the space of locallyconstant functions, and b0 is simply the number of connected components.(b) SupposeMis a manifold of dimension m, d vanishes onm-forms since there are

no m + 1-forms. Thus

Zm(M) = m(M).

When is such a form exact? Suppose Mis connected, orientable and compact,and let Mbe a volume form on M defining the orientation. Then

M

> 0. If = dfor somem1-form, then we would have

M =

Md= 0 by Stokes

theorem. This contradiction shows that represents a non-trivial cohomologyclass in Hm(M). We will show later that indeed, ifM is compact, orientable

and connectedHm(M) is 1-dimensional, and is spanned by the class of . IfMis non-compact or non-orientable this is usually false.

(c) Let us try to get a feeling for H1(M). If 1(M) is exact, = df thenthe integral

S1

along any closed path : S1 M vanishes, by Stokestheorem:

S1=

S1

df=

S1

df= 0.

The converse is true as well: If 1(M) is such thatS1

is always 0, thenthe integral of along a path : [0, 1] M depends only on the end points.One can then define a function fby fixing x0 M and setting f(x) =

[0,1]

for any path from x0 tox. It is not hard to see that the function fobtained inthis way is smooth and satisfies df =. (Indeed, in a coordinate chart aroundx0, the functionfis just the image under the homotopy operator from PoincaresLemma.)

Similarly, is closed if and only if the integral along closed paths does notchange under smooth homotopies of the path. We has already seen one directionof this statement. The opposite direction can be seen as follows: Suppose isnot closed. then there existsx0 Mwith d= 0 atx0. ChooseX X(M) suchthat(X)d= 0 nearx0, and a loop0never tangent toX. Using the flow ofX(with >0 sufficiently small), we can construct a homotopy : [0, 1] S1 Mof this loop. Then

[0,1]S1

d = 0, showing by Stokes theorem that the

integral ofalong t changes.Exercise 5.2. Show more generally that a k-form is closed if and only if for all

mapsF : Sk M, the integral ofF depends only on the homotopy class ofF.

5.2. Homotopy invariance. For any F C(N, M), the pull-back map F :(M) (N) is a homomorphism of graded algebras commuting with d. Hence itdefines an algebra homomorphism F : H(M) H(N).


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40 CONTENTS

Theorem 5.3 (Homotopy invariance). If F0, F1 C(N, M) are smoothly homo-topic, then the induced mapsF0 , F

1 : H(M) H(N) are equal.

Proof. We have to show that the map F1 F

0 : (M) (N) induces the zeromap in cohomology. Thus, we have to show that if Z(M) thenF1 F

0 B(N).

Let h : (N) (M) be the homotopy operator for a smooth homotopy betweenF0, F1:

F1 F

0 =h d + d h: (M) (N).

Apply this equation to , using d= 0:

F1 F

0 = d(h).

Two maps F : N M and G : M Nare called homotopy inverses ifF G ishomotopic to IdM and G Fhomotopic to IdN. In this case, the theorem shows:

Corollary 5.4. Suppose F : N M and G : M N are homotopy inverses.ThenF : H(M) H(N) is an isomorphism with inverseG : H(N) H(M).

Proof. According to the theorem, the maps GF : H(M) H(M) andFG :H(N) H(N) are the identity maps.

Definition 5.5. Let N be a submanifold of a manifold M. A smooth map F :[0, 1] M M such that F(0, ) = IdM and F(1, x) N for all x M is called adeformation retraction from M onto N. It is called a strong deformation retraction ifF(t, x) =x for all t [0, 1] and x N.

As a special case, one has:

Proposition 5.6. If M admits a strong deformation retraction onto an embeddedsubmanifoldN, then the inclusion map : NM gives an isomorphism : H(M)H(N).

Proof. LetF : [0, 1]MMbe a strong deformation retraction. Let : MNbe the map thus obtained, i.e. = F(1, ) : M N. ThenF gives a homotopybetween IdMand. But = IdN. Thusandare homotopy inverses; in particular is an isomorphism in cohomology.

This Proposition once again explains Poincares Lemma: Since Rm

admits a strongdeformation retraction onto a point, its cohomology is trivial.

6. Mayer-Vietoris

6.1. Exact sequences. A (finite or infinite) sequence of vector spaces and maps

EE E


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6. MAYER-VIETORIS 41

is called exact atE if the kernel of the map E E is equal to the image of the mapE E. The sequence is called exact if it is exact everywhere. An exact sequence of

the form0F

E

G0

is called short exact. This has the following interpretation: Exactness at Fmeans thatthe map: F Eis injective. That is, we can think ofFas a subspace ofE. ExactnessatG means that the map : EG is surjective. Thus we may think ofG as a quotientspaceG= E/ ker(). Exactness atEtells us that ker() = im()=F. Putting all thistogether, we conclude that F is a subspace ofEand G = E/F.

Exercise 6.1. LetA : FEbe any linear map. Show that the sequence

0ker(A)F AEE/ im(A)0

is exact.Suppose the sequence

0E E

kE

is exact. Then E can be thought of as a subspace ofE, and the map k : E E

vanishes exactly on E E. That is, k descends to an injective map E/EE. Thuswe obtain a new exact sequence

0E/E kE

Exercise 6.2. Show that if the sequence 0 E0 Ek 0 is exact, thenthe alternating sum of dimensions vanishes:

i

(1)i dim Ei = 0.

6.2. Differential complexes. The de Rham complex ((M), d) is an example ofadifferential complex. Since we will encounter many more examples of differential com-plexes later on, it is useful to treat them in generality.

LetE=

kZ Ek be a graded vector space, and d : EEa linear map of degree

1. That is, d is a collection of maps

(5) Ek dEk+1

dEk+2

One calls (E, d) a differential complex if d squares to 0. The kernel of the map d :

Ek Ek+1 is called the space ofk-cocycles, and is denoted Zk(E). The image of themap d : Ek1 Ek is called the space ofk-coboundariesand is denoted Bk(E). Sinced d = 0,Bk(E) is a subspace ofZk(E). The quotient space

Hk(E) =Zk(E)/Bk(E)

is called the kth cohomology of the complex (E, d). We denote H(E) =

kZ Hk(E).

Note thatH(E) = 0 if and only if the sequence (5) is exact.


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42 CONTENTS

(In the special case E= (M), we will continue to writeB(M), Z(M), H(M) ratherthanB((M)), Z((M), H((M)).

Definition 6.3. Let (E, d) and (E, d) be two differential complexes. A linear map : E E of degree 0 is called a homomorphism of differential complexesor cochainmap, if d = d .

A cochain map : E E takes cocycles to cocycles, hence it gives a map incohomology, which we denote by the same letter:

: H(E) H(E).

For example, ifF : M N is a smooth map of manifolds, it defines a cochain map = F : (N) (M), and therefore H(N) H(M).

Definition 6.4. A homotopy operator(or chain homotopy) between two cochain

maps 0, 1 : E E is a linear maph : E Eof degree1, with property

h d + d h= 1 0.

If such a maphexists, 0, 1are calledchain homotopic. Two cochain maps : EE

and : E Eare called homotopy inverseif : F Fand : EEareboth homotopic to the identity, if such maps exist, E, E are calledhomotopy equivalent.

For example, if F0, F1 : N M are smoothly homotopic, we had constructed ahomotopy operator between 1= F

1 and 0 = F

0 . As in this special case, we see:

Proposition6.5. Chain homotopic cochain maps induce equal maps in cohomology.If two differential complexesE, E are homotopy equivalent, their cohomologies are equal.

Suppose now that E,F,G are differential complexes, and let

0E jF

kG0

be a short exact sequence where all maps are homomorphisms of differential complexes.We will construct a map : H(G) H(E) of degree +1 called the connecting ho-momorphism. This is done as follows: Let [] Hk(G) be a given cohomology class,represented by Zk(G). Sincek : F G is surjective, we can choose, Fk withk() =. Then

k(d) = d(k()) = d= 0.

Thus, d ker(k) = im(j). Since j is injective, there is a unique Ek+1 with

j() = d. Then j (d) = dj() = dd= 0, so d= 0. Set

([]) = [].

One has to check that this definition does not depend on the choices of and . Forexample, if Fk is another choice with k() = , then k( ) = 0, so ker(k) = im(j). Sincej is injective, there is a unique element Ek withj() = .Thus if Ek+1 is the unique element such that j() = d, we have = d,


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6. MAYER-VIETORIS 43

so [] = []. Similarly, one checks that the definition does not depend on the choice ofrepresentative for [].

Theorem 6.6. Let0E

jF

kG0

be a short exact sequence of homomorphisms of differential complexes with connectinghomomorphism. Then there is a long exact sequence in cohomology,

Hk(E) jHk(F)

kHk(G)

Hk+1(E)

where the connecting homomorphism : Hk(G) Hk+1(E) is defined as follows: ForZk(G), ([]) = [] where Zk+1(E) is an element withj() =d fork() =.

Proof. Let us check exactness at Hk(G). We have to check two inclusions, (i)im(k)ker() and (ii) ker() im(k).

(i) Suppose Zk(F). We have to show that[k()] = 0. By construction of theconnecting homomorphism,[k()] = [] where j () = d= 0. Thus = 0.

(ii) Suppose Zk(G) with[] = 0. We will show that there exists a closedelement Zk(F) such that = k(). Start by choosingany Fk with k() = , and let Ek+1 be the unique element with j () = d. By definition of, 0 =[] = [], thus= d for some Ek. Define =j(). Then is closed:

d = dj(d) = dj() = 0,

and k() = k() = since kj = 0. Exactness at Hk(F) and Hk(E) is checkedsimilarly.

Theorem 6.7 (Mayer-Vietoris). SupposeM = U V whereU, V are open, and letU, Vbe a partition of unity subordinate to the cover ofU, V. The sequence

0p(U V) jp(U) p(V)

kp(U V)0

wherej() = (|U, |V) andk(, ) =|UV |UV is exact. It induces a long exactsequence

0 Hp(UV)Hp(U)Hp(V)Hp(UV) Hp+1(UV) 0

The connecting homomorphism takes the cohomology class ofZp(U V)to the classof() Zp+1(U V), where

()|U=d(V), ()|V =d(U).

The formula for() makes sense, since on the overlap U V,

d(V) (d(U)) = d((U+ V)) = d= 0.

Proof. Exactness at p(M) is obvious since the map p(M) p(U) p(V) isclearly injective. The kernel of the map p(U) p(V)p(U V) consists of forms, with|UV =|UV. These are exactly the forms which patch together to a globalform on M, with =|U and =|V. This shows exactness at p(U) p(V). It


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remains to check surjectivity of the last map : p(U) p(V) p(U V). Let p(U V). Choose a partition of unityU, Vsubordinate to the cover U, V. Then

V is zero in a neighborhood ofU\UV, hence it extends by 0 to a form on U.SimilarlyU is zero in a neighborhood ofV\U V, hence it extends by 0 to a form onV.

OnU V we have |UV |UV =. This shows thatk is surjective. Any shortexact sequence of chain complexes induces a long exact sequence in cohomology.

As a first application of the Mayer-Vietoris sequence, we can now compute the coho-mology of the sphere.

Example 6.8. Let M = Sm, and U, V the covering by the complements of thesouth pole and north pole, respectively. The intersection U V smoothly retracts ontothe equatorial Sm1. Since Hk(U) = Hk(V) = 0 f or k > 0, Mayer-Vietoris gives

isomorphismsHk(Sn)=Hk1(U V) =Hk1(Sn1)

for k > 1. By induction, we get Hk(Sn) = H1(Snk+1). In low degrees, if n > 1,Mayer-Vietoris is an exact sequence

0 R R R R H1(Sn)0.

Here the map R R Ris easily seen to be surjective, so R H1(Sn) must be thezero map, so H1(Sn) = 0. Forn = 1 we have H1(S1) = 0. We conclude,

Hk(Sn) =

R fork = 0, n,

0 otherwise .

Thusp(t) = 1 + tn

is the Poincare polynomial.7. Compactly supported cohomology

IfMis non-compact, one can also study the compactly supportedde Rham cohomol-ogy: Working with the complex of compactly supported forms

00comp(M) d1comp(M)

d

dmcomp(M)0

one defines Hpcomp(M) as the cohomology of this complex,

Hpcomp(M) =Hp(mcomp(M)).

If M is compact, Hcomp(M) agrees with the usual cohomology, but for non-compact

manifolds it is quite different. For instance, ifMhas no compact component,H0comp(M) = 0.

This follows since Z0comp(M) consists of locally constant functions of compact support,and there are no such if M has no compact component. It is also different in higherdegree: For instance,

(6) H1comp(R) = R


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Similarly, ifUMis an open subset, one has a natural restriction map

(M) (U),

but this map does not take comp(M) into comp(U) (unlessUis a connected componentofM, the inclusion map is not proper.) On the other hand, one has a natural extensionby 0 map

comp(U) comp(M),

but there is no such map from (U) to (M). As a consequence, the Mayer-Vietorissequence for comp() looks different from that for ().

Theorem7.3 (Mayer-Vietoris for cohomology with compact support). SupposeM=UV whereU, Vare open, and letU, Vbe a partition of unity subordinate to the cover{U, V}. The sequence

0pcomp(U V) jpcomp(U)

pcomp(V)

kpcomp(U V)0,

where j() = (, ) and k(, ) = +, is exact for all p. Hence there is a longexact sequence in compactly supported cohomology,

Hpcomp(U V) Hpcomp(U) H

pcomp(V) H

pcomp(U V)

Hp+1comp(U V) .

The connecting homomorphism : Hpcomp(UV) Hp+1comp(UV) takes the class of

Zpcomp(U V) to the class of

() :=dU = dV .

Note that the formula for() is well-defined since dU=dV 1comp(U V).

Proof. This is very similar to the proof of Mayer-Vietoris for (), and is left as anexercise.

Again, the direct sumHcomp(M) is an algebra under wedge product. Since the wedgeproduct of a compactly supported form with any form is compactly supported, we seethatHcomp(M) is a module for the algebra H(M).

8. Finite-dimensionality of de Rham cohomology

Using the Mayer-Vietoris sequence, we will now show that H(M) is for any compact

manifoldM, or more generally for all manifolds of so-called finite type.Definition 8.1. A cover {U}A is called a good cover if all finite non-empty in-

tersections of the sets U are diffeomorphic to Rm. A manifold admitting a finite good

cover is called offinite type.

Theorem 8.2. Every given cover of a manifoldMhas a refinement which is a goodcover.


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9. POINCARE DUALITY 47

Sketch of proof. We wont give the full proof here, which requires some elementsfrom Riemannian geometry. The idea is as follows. Choose a Riemannian metricg on

M. That is, g is a symmetric C

(M)-bilinear form X(M) X(M) C

(M), withthe property that g(X, X)x >0 ifXx = 0. Using g one can define the length of curves

: [a, b] Mas length() =ba

g(, )1/2 dt. An open set UMis called geodesicallyconvex, that is any two points in Uare joined by a unique (up to re-parametrization)curve : IU of shortest length.

Geodesically convex open sets are diffeomorphic to Rm, and the intersection of twogeodesically convex sets is again geodesically convex. Thus any cover consisting ofgeodesically convex open sets is a good cover. It can be shown that for any pointx M, there exists >0 such that the set Uof

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