Lie Algebra Cohomology

Transcription

1 D E P A R T M E N T O F M A T H E M A T I C A L S C I E N C E S U N I V E R S I T Y O F C O P E N H A G E N Lie Algebra Cohomology Master's Project in Mathematics Mikala Ørsnes Jansen Adviser: Dustin Tate Clausen January 29 th, 2016

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3 Abstract This project deals with Lie algebra cohomology. We dene it in terms of the Chevalley-Eilenberg complex, but also consider the equivalent denition of the cohomology groups as certain Extgroups. We examine the Lie derivative as a historical motivation for the denition of the Chevalley-Eilenberg complex, and look at the relationship between De Rham cohomology and Lie algebra cohomology; to this end we prove that we can calculate the De Rham cohomology of a smooth manifold by considering the complex of dierential forms which are invariant under a smooth action of a compact connected Lie group on the manifold. Resumé Dette projekt omhandler Lie-algebra-kohomologi. Dette deneres ved hjælp af Chevalley-Eilenberg-komplekset, men vi ser også på den ækvivalente denition af kohomologigrupperne som Ext-grupper. Vi ser på Lie-derivatet som historisk motivation for denitionen af Chevalley- Eilenberg-komplekset og på forholdet mellem De Rham-kohomologi og Lie-algebra-kohomologi; undervejs viser vi, at De Rham-kohomologien af en glat mangfoldighed kan bestemmes ved at betragte komplekset af dierentialformer, der er invariante under virkningen af en kompakt sammenhængende Lie-gruppe på mangfoldigheden. Contents Preface 1 Introduction Prerequisites and Notation Lie Algebras, Universal Enveloping Algebras, and Lie Algebra Cohomology Basics Lie Algebra Cohomology Derived Functor Approach Historical Motivation I: The Lie Derivative Tensor Fields The Lie Derivative Historical Motivation II: De Rham Cohomology Lie Group Action Lie Groups and Lie Algebras Examples 27 A Appendix 30 A.1 Chevalley-Eilenberg Derivative A.2 Homological Algebra Results A.3 Integration Preserves Smoothness

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5 Preface Introduction Lie algebra cohomology was invented by Claude Chevalley and Samuel Eilenberg in the middle of the 20'th century in an attempt to compute the De Rham cohomology of compact connected Lie groups; it is dealt with in detail in their paper [3] and heavily inuenced by the work of Élie Cartan. In this project, we dene Lie algebra cohomology, consider the historical motivation for the theory, and look at some examples. In Chapter 1, we go over the denitions of Lie algebras, Lie algebra modules, and universal enveloping algebras. We go on to dene Lie algebra cohomology in terms of a chain complex and nally show that one can equivalently dene it in terms of derived functors. In Chapter 2, we dene the Lie derivative with the aim of proving the invariant formula for the exterior derivative of dierential forms; we see that we can view dierential forms as alternating multilinear maps taking vector elds as variables, which motivates the attempt to compute the De Rham cohomology of a Lie group in terms of its Lie algebra (the left-invariant vector elds on the Lie group). In Chapter 3, we explore this relationship further: We show that if a compact connected Lie group acts on a manifold, then to compute the De Rham cohomology of the manifold it suces to consider the dierential forms which are invariant under the Lie group action. As a corollary, we will see that the De Rham cohomology of a compact connected Lie group is isomorphic to the cohomology of its Lie algebra. Finally, in Chapter 4, we consider some simple examples. Prerequisites and Notation This project assumes basic knowledge of the following dierential geometry theory: Vector elds, dierential forms, De Rham cohomology, Lie groups, and Lie algebras. We refer to [11, 8, 12] for denitions and the basic theory. To make sure we agree on denitions and notation and have the relevant basics fresh in the memory, we will here briey recap the main things needed to understand this text: S k,m S k+m will denote the set of permutations σ S k+m satisfying σ(1) < < σ(k) and σ(k + 1) < < σ(k + m). We will assume that all manifolds are smooth. Let M, N be manifolds. For a smooth map of manifolds, f : M N, we denote the derivative by Df : T M T N. We may alter between the notations D p f(v) and D p,v f for the derivative of f at p M in the direction v T p M. For f C (M), we will sometimes write df instead of Df, where df Ω 1 (M), is the exterior derivative of f, and df p = D p f. In general, for a ring R and an R-module A, we denote the dual space as A = Hom R (A, R). Given a vector space V over a eld K, the exterior algebra of V, Λ(V ), is the quotient of the tensor algebra, T (V ), (see Denition Chapter 1) by the ideal I = v v v V. The exterior product is dened as v w = [v w] I. The k'th exterior power of V, denoted by Λ k (V ), is the subspace of Λ(V ) spanned by elements of the form v 1 v k, v i V. The exterior product is then a bilinear map : Λ k (V ) Λ m (V ) Λ k+m (V ). Given a map f : V W of vector spaces, we can dene a map of K-algebras Λ(f): Λ(V ) Λ(W ) by Λ(f)(v 1 v k ) = f(v 1 ) f(v k ). 1

6 2 Mikala Ørsnes Jansen A k-linear map f : V k K is said to be alternating, if f(v 1,..., v k ) = 0 whenever v i = v j for some i j. If we dene the alternating algebra of V, Alt k (V ), as the set of alternating k-linear maps V k K, and the exterior product, : Alt k (V ) Alt m (V ) Alt k+m (V ), by ω η(v 1,..., v k+m ) = ω(v σ(1),..., v σ(k) )η(v σ(k+1),..., v σ(k+m) ), σ S k,m then we have a natural isomorphism of vector spaces ϕ: Λ k (V ) Alt k (V ), ϕ(f )(v 1,..., v k ) = F (v 1 v k ). If, in addition, V is nite dimensional, then Λ k (V ) and Λ k (V ) are naturally isomorphic via the map ψ : Λ k (V ) Λ k (V ) given by ψ(f 1 f k )(v 1 v k ) = σ S k sign σ f 1 (v σ(1) ) f k (v σ(k) ). Let M be a manifold of dimension n. We opt for the denition of dierential k-forms on M as smooth sections ω : M Λ k (M) = Λ k (T p M ), where Λ k (M) is equipped with the natural smooth structure such that the projection onto M is smooth the charts are of the form p M U R (n k) Λ k (M), (x, v) Λ k ((D p θ 1 ) ) ϕ(v) Λ k (T p M ), for a chart θ : U M, and an isomorphism ϕ: R (n k) Λ k ((R n ) ). This denition is easily seen to be equivalent to the one given in [8], namely that a dierential k-form is a family, {ω p } p M, of alternating k-linear maps T p M k R, such that the pullback θ ω : U Λ k (U) = U Λ k ((R n ) ) is smooth for all charts θ : U M. The set of dierential k-forms on M is denoted by Ω k (M). Given a chart θ : R n U M, the maps x i := pr i θ 1 : U R are local coordinates on V, where pr i : R n R is projection onto the i'th coordinate. Consider the dierentials dx i : U R n R, which we shall consider as dierential 1-forms on U by evaluating in the rst coordinate, dx i : U (R n ) = Λ 1 ((R n ) ). For p = θ(x), {D x θ(e i )} n i=1 form a basis of T p M with dual basis {dx i (p)} n i=1 of T pm, where (e i ) is the standard basis of R n. Thus {dx σ(1) (p) dx σ(k) (p)} σ Sk,n k form a basis of Ω k (T p M ). Hence any map ω : M Λ k (M) can be written locally on U as ω = f σ dx σ(1) dx σ(k), σ S k,n k for some functions f σ : U R, and ω is smooth on U if and only if all the f σ are smooth. Moreover, f σ (p) = ω p (D x θ(e σ(1) ),..., D x θ(e σ(k) )), for p = θ(x). We will sometimes write dx σ := dx σ(1) dx σ(k). A smooth vector eld on M is a smooth section of the tangent bundle, X : M T M. As we are only interested in smooth vector elds, we let the smoothness assumption be implicit from now on in. We denote the set of vector elds on M by X(M). One can identify a vector eld with its action on C (M): for X X(M) and f C (M), X(f): M R given by

7 Contents 3 X(f)(p) = D p,x(p) f is smooth. X(M) is a Lie algebra when equipped with the commutator bracket with respect to the composition in End(C (M)). We may, for the sake of clarity, alter between the notations f(p) and f p for evaluation at p. As usual, a hat denotes that an element is omitted, for example (x 1,..., ˆx i,..., x n ) = (x 1,..., x i 1, x i+1,..., x n ), x 1 ˆx i x n = x 1 x i 1 x i+1 x n. N denotes the natural numbers including 0. We will write N >0 for N \ {0}. We denote De Rham cohomology by H dr ( ). Acknowledgements I would like to thank Dustin for supervision, and Matias and Kevin for proofreading.

8 1 Lie Algebras, Universal Enveloping Algebras, and Lie Algebra Cohomology In this chapter we briey recall the denitions of Lie algebras and their universal enveloping algebras. We go on to dene Lie algebra cohomology using the Chevalley-Eilenberg complex, and nally show that we can equivalently dene it as the right derived functor of the invariants functor. 1.1 Basics Let K be a eld. Definition A Lie algebra over K is a K-vector space g equipped with a bilinear map [, ]: g g g, called the Lie bracket, satisfying 1. [X, Y ] = [Y, X] for all X, Y g (anti-symmetry). 2. [[X, Y ], Z] = [X, [Y, Z]] [Y, [X, Z]] for all X, Y, Z g (Jacobi identity). A Lie algebra homomorphism is a linear map between Lie algebras, ϕ: g h, which respects the Lie bracket, i.e. ϕ([x, y] g ) = [ϕ(x), ϕ(y)] h, for all x, y g. Definition Let R be a commutative ring. An associative R-algebra is an R-module, A, equipped with an associative multiplication operation with identity which respects the R-module structure, that is the multiplication map, A A A, is R-bilinear. Remark Given an associative algebra A over a eld K, A may be turned into a Lie algebra by taking the commutator with respect to the associative product as the Lie bracket. For example, for any vector space V, the space of endomorphisms of V, End(V ), becomes a Lie algebra with the commutator bracket. Remark If G is a Lie group, then the tangent space at 1 G, T 1 G, is a Lie algebra with the Lie bracket dened as follows: Dene Ad: G GL(T 1 G), g D 1 ψ g, where ψ g : G G is conjugation by g; next, dene ad := D 1 Ad: T 1 G End(T 1 G). Finally, set [x, y] := ad(x)y for x, y T 1 G. With this g := T 1 G becomes a Lie algebra the Lie algebra of G. One can also identify g with the set of left-invariant vector elds on G, with Lie bracket the usual commutator. Definition Let g be a Lie algebra. A g-module is a K-vector space V together with a representation ρ: g End(V ) of g, i.e. ρ is a Lie algebra homomorphism. In other words, a g-module is a K-vector space V and a linear action of g on V,. : g V V, satisfying [X, Y ].v = X.(Y.v) Y.(X.v) for all X, Y g, v V. Example Examples of g-modules are 1. K with the trivial action, g End(K) the zero map. 2. g itself with the adjoint action; ad: g End(g), ad(x)(y) = [x, y]. 4

9 Chapter 1. Lie Algebras, Universal Enveloping Algebras, and Lie Algebra Cohomology 5 3. If g is the Lie algebra of a Lie group G then C (G) is a g-module: We identify g with the set of left-invariant vector elds on G, and these in turn with their action on C (G). Definition Let g be a Lie algebra over K. The universal enveloping algebra of g is an associative K-algebra U(g) and a Lie algebra homomorphism ι: g U(g) satisfying the universal property pictured in the diagram below, where A is an associative K-algebra, ϕ is a Lie algebra homomorphism and ψ is a homomorphism of associative K-algebras: ι g U(g) ϕ!ψ A Proposition Any Lie algebra has a universal enveloping algebra. To prove this, we construct the universal enveloping algebra directly. This will be done in a few steps with some intermediary results. Definition Let V be any K-vector space. For n N >0, set T n (V ) = V n, and set T 0 (V ) := K. The tensor algebra of V is T (V ) := n N T n (V ), with multiplication determined by the canonical isomorphisms T n (V ) T m (V ) T n+m (V ). There is a canonical linear map V T (V ) mapping into the second term of the direct sum, T 1 (V ) = V. For any linear map, f : V V, there is an induced map T (f): T (V ) T (V ) determined by T (f)(v 1 v n ) = f(v 1 ) f(v n ). Proposition The tensor algebra satises the following universal property, where A is an associative algebra, ϕ is linear and ψ is a homomorphism of associative K-algebras: V T (V ) ϕ!ψ A Proof. Clearly, T (V ) is an associative algebra. Given ϕ: V A, dene ψ : T (V ) A by v 1 v n ϕ(v 1 ) ϕ(v n ). Remark T is a functor from the category of K-vector spaces to the category of associative K-algebras. It is left adjoint to the forgetful functor mapping an algebra to its underlying vector space. Now, let g be a Lie algebra over K, and consider the ideal I = x y y x [x, y] x, y g T (g). Proposition U(g) := T (g)/i is the universal enveloping algebra of g. Proof. There is a canonical map ι: g U(g), namely the composition g T (g) U(g). This is a Lie algebra homomorphism by construction. Now, given an associative K-algebra A and a Lie algebra homomorphism g A, consider the diagram

10 6 Mikala Ørsnes Jansen T (g) g ι U(g) (1) ϕ A (2) The map (1) exists and is unique by the universal property of the tensor algebra, and (2) exists and is unique by the universal property of the quotient, as ϕ is a Lie algebra homomorphism, so it is trivial on the generators of I. Remark The Birkho-Witt theorem (see [7]) implies that the map ι: g U(g) is injective, so g can be seen as a subspace of U(g). Remark U is a functor from the category of Lie algebras over K to the category of associative K-algebras. It is left adjoint to the forgetful functor mapping an associative algebra to its underlying vector space equipped with the commutator bracket. Example If g is abelian, then the universal enveloping algebra of g is the symmetric algebra, Sym(g) the free commutative algebra over g. 2. If g is the Lie algebra of a Lie group G, then one can identify g with the left-invariant rst order dierential operators on G (the vector elds on G), and the universal enveloping algebra can be identied with the set of left-invariant dierential operators on G of all orders. This was how the universal enveloping algebra was originally introduced by Poincaré in 1899 (See [4] for details on this example). Remark The universal property of the universal enveloping algebra implies that a K- vector space Γ is a g-module, if and only if it is a U(g)-module in the usual sense, when viewing U(g) as a ring. If ρ: g End(Γ) is a Lie algebra homomorphism, then the ring homomorphism U(g) End(Γ) is determined by [x 1 x n ] = ρ(x 1 ) ρ(x n ). 1.2 Lie Algebra Cohomology In this section, we dene Lie algebra cohomology. To this end, we dene the Chevalley-Eilenberg chain complex. Let K be a eld, g a Lie algebra over K, and Γ a g-module. Set C n (g, Γ) := Hom K (Λ n g, Γ), n > 0, C 0 (g, Γ) := Γ. That is, C n (g, Γ) is the set of alternating n-linear maps g n Γ. These are the n-cochains of the Chevalley-Eilenberg complex. We dene the dierential d: C n (g, Γ) C n+1 (g, Γ) as follows: Given c C n (g, Γ), let dc C n+1 (g, Γ) be given by n+1 dc(x 1,..., x n+1 ) = ( 1) i+1 x i.c(x 1,..., ˆx i,..., x n+1 ) i=1 + 1 i<j n+1 ( 1) i+j c([x i, x j ], x 1,..., ˆx i,..., ˆx j,..., x n+1 ),

11 Chapter 1. Lie Algebras, Universal Enveloping Algebras, and Lie Algebra Cohomology 7 for all x 1,..., x n+1 g, where x i.c(x 1,..., ˆx i,..., x n+1 ) denotes the action of x i g on c(x 1,..., ˆx i,..., x n+1 ) Γ according to the g-module structure of Γ. A long and tedious calculation shows that d 2 = 0 (see Proposition A.1.1). Definition (C (g, Γ), d) is called the Chevalley-Eilenberg complex of g with coecients Γ, and the cohomology of g with coecients in Γ, H (g, Γ), is dened as the homology of the Chevalley-Eilenberg complex. We will give examples of Lie algebras and their cohomology in Chapter Derived Functor Approach We will now look into a more categorical approach to Lie algebra cohomology. Let K be a eld, g a Lie algebra over K. We will show that the Lie algebra cohomology is the right derived functor of the invariants functor from the category of g-modules to itself: I : Γ Γ g = {γ Γ x.γ = 0 for all x g} Note that for any two g-modules, V and W, with representations ρ: g End(V ), π : g End(W ), we have Hom U(g) (V, W ) = {T Hom K (V, W ) T ρ(x) = π(x) T for all x g}. This follows directly from the U(g)-module structure on V and W : T ([x 1 x n ].v) = T (ρ(x 1 ) ρ(x n )(v)), [x 1 x n ].T (v) = π(x 1 ) π(x n )(T (v)), for any element of the form [x 1 x n ] U(g), v V. Hence, if Γ is a g-module, then, since Hom K (K, Γ) = Γ, T T (1), we have that Γ g = Hom U(g) (K, Γ), where g acts trivially on K. This isomorphism is natural, so the two functors, I and Hom U(g) (K, ), are isomorphic. Theorem The Lie algebra cohomology of g with coecients in Γ is the right derived functor of I and can also be described as an Ext-functor: H (g, Γ) = Ext U(g) (K, Γ) = R I(Γ). Proof. By the above observations, I( ) = Hom U(g) (K, ), and thus Ext U(g) (K, Γ) = R I(Γ). Now, we calculate Ext U(g) (K, Γ) by nding a projective resolution of K. Consider the sequence Λ 3 g U(g) Λ 2 g U(g) g U(g) U(g) K where we tensor over K, and Λ n g U(g) Λ n 1 g U(g) is given by (x 1 x n ) u n ( 1) i (x 1 ˆx i x n ) x i u i=1 + ( 1) i+j+1 ([x i, x j ] x 1 ˆx i ˆx j x n ) u, 1 i<j n here we identify Λ 0 (g) U(g) = K U(g) = U(g) and Λ 1 g = g, and U(g) K is the augmentation map induced by the zero map g K.

12 8 Mikala Ørsnes Jansen This is a free and thus projective resolution of the trivial U(g)-module K: Indeed, Λ n (g) is a free K-module, being a vector space over K, and therefore Λ n (g) U(g) is a free U(g)-module. Now, for exactness of the sequence: This is a long and technical proof, so we have opted for a sketch of proof instead. For the fun of it, we prove exactness at the rst two terms of the sequence: g U(g) U(g) k 0. The augmentation map is surjective, as it restricts to the identity on K = g 0 U(g), and its kernel is the image of the subspace n N >0 g n T (g) in U(g); this is equal to ι(g)u(g) U(g), which is exactly the image of the map g U(g) U(g), x u xu, given above. So the rst part of the sequence is exact. Now, to prove exactness of the complete sequence, rst note that the composite of two succesive maps is zero by calculations similar to those used for proving that d 2 = 0 in the Chevalley- Eilenberg complex. Hence, it is a chain complex, and we can prove exactness by proving that its homology is trivial. To do this, one takes a ltration of the complex, F 0 F 1 C, where C denotes the complex dened above the Birkho-Witt theorem ([7]) is used to dene this ltration. Then one denes complexes W k := F k /F k 1 and shows that these have trivial homology by constructing a contracting chain homotopy. The short exact sequence of complexes 0 F k 1 F k W k 0 induces a long exact sequence in homology, which yields isomorphisms H (F k ) = H (F k 1 ). By denition of the ltration, F 0 is the complex 0 K K 0, which has trivial homology, implying that all the F k have trivial homology. Now, as homology respects direct limits in U(g) Mod, it follows that C = F k has trivial homology, as desired. We refer to [6] for the complete proof. This complex is called the Koszul complex. Finally, we apply Hom U(g) (, Γ) to the resolution and note that the map Hom U(g) (Λ n g U(g), Γ) Hom K (Λ n g, Γ) = C n (g, Γ), given by T T, with T (x 1 x n ) = T ((x 1 x n ) 1) is an isomorphism of chain complexes, which implies the desired result, namely that H (g, Γ) = Ext U(g) (K, Γ).

13 2 Historical Motivation I: The Lie Derivative In this chapter we dene the Lie derivative. It will not be used directly when dealing with Lie algebra cohomology, but it provides an important motivation for the denition of Lie algebra cohomology given in Chapter 1. More specically, we will prove the invariant formula for the exterior derivative of dierential forms, which is entirely dependent on the vector elds on the manifold in question (Proposition ). Here we will see what originally motivated the denition of the Chevalley-Eilenberg complex. 2.1 Tensor Fields Let M be a manifold of dimension n. Definition Set T r,s (M) := p M (T pm) r,s, where V r,s = (V Kr ) K (V Ks ) for any K-vector space V, and by convention V K0 = K. We give T r,s (M) the natural smooth structure such that the projection T r,s (M) M is smooth this is a smooth vector bundle. A tensor eld on M of type (r, s) is a smooth section S : M T r,s (M). We shall denote the set of tensor elds of type (r, s) on M by T r,s (M), and the set of all tensor elds on M by T (M) = r,s N T r,s (M). This is a C (M)-module with the obvious operations; moreover, it has a multiplication operation, namely the tensor product: If S, T T (M) of type (r, s) respectively (r, s ), then S T, given by p S(p) T (p) is a tensor eld of type (r + r, s + s ), where we use the canonical isomorphism A B = B A. Proposition The set of dierential k-forms on M, Ω k (M), is isomorphic as a C (M)- module to the set of alternating C (M)-multilinear maps X(M) k C (M), where the scalar multiplication in both cases is given by pointwise multiplication. In particular, Ω 1 (M) = X(M). Proof. Given ω Ω k (M), dene ω : X(M) k C (M) by ω(x 1,..., X k )(p) = ω p (X 1 (p),..., X k (p)), for X i X(M), p M. (2.1) ω is alternating and C (M)-multilinear, as ω p is alternating and R-multilinear for all p M. The map ω(x 1,..., X k ): M R is equal to the composite M ω X 1 X k Λ k (M) M T M M M T M ev R, where the second map is the evaluation map. Locally the evaluation map is a (k + 1)-linear map of nite-dimensional vector spaces, Alt k (R n ) (R n ) k R, and as such it is smooth. Then, since ω : M Λ k (M) is smooth, and all the X i : M T M are smooth, we conclude that ω(x 1,..., X k ) C (M). Conversely, given η : X(M) k C (M) alternating and C (M)-multilinear, dene ˆη : M Λ k (M) by ˆη p (v 1,..., v k ) = η(x 1,..., X k )(p), for some choice of X i X(M) with X i (p) = v i. Such vector elds always exist, but we must show that the denition is independent of the choice of vector elds: η(x 1,..., X k )(p) depends only on 9

14 10 Mikala Ørsnes Jansen the values of the X i at p. By multilinearity of η, it is enough to prove the case k = 1, and it suf- ces to prove that if X(p) = 0, then η(x)(p) = 0. To see this, let θ : R n U be a chart around p, and dene vector elds on U, X i : U T M, as the composites U θ 1 R n R n {e i } Dθ T M, for i = 1,..., n. As {D x θ(e i )} n i=1 is a basis of T θ(x)m, we can uniquely write X(q) = n a q i D θ 1 (y)θ(e i ) = i=1 n a q i X i(q), for all q U, and some a q i R. i=1 Dening f i : U R by f i (q) = a q i, we see that f i C (U), as X and the X i are smooth. Assume now that X(p) = 0; then f i (p) = 0 for all i = 1,..., n. Hence, η(x)(p) = n f i (p)η(x i )(p) = 0. i=1 We conclude that ˆη is well-dened. Moreover, ˆη p is R-linear and alternating for all p M. To see that it is smooth and thus a dierential k-form, let θ : R n U be a chart on M with local coordinates x 1,..., x n, and dene vector elds, X i : U θ 1 R n R n {e i } Dθ T M, as above. Then we can write ˆη locally as ˆη U = σ S k,n k f σ dx σ(1) dx σ(k), ( ) for f σ = η Xσ(1),..., X σ(k). (Here we skip some technicalities: We can extend the X i to vector elds on M by multiplying with a suitable bump function; to do this, we may have to shrink U a little.) Then f σ C (U), so ˆη is smooth. It is clear that the maps ω ω and η ˆη are each other's inverses. Remark In view of the above proposition, we will from now on in many cases simply interpret ω Ω k (M) as an alternating k-linear map of C (M)-modules X(M) k C (M). We also need the following application of a version of the Serre-Swan Theorem: Lemma Ω 1 (M) and X(M) are nitely generated projective C (M)-modules. Proof. We refer to Theorem of [2]. Corollary The set of smooth vector elds on M, X(M), is isomorphic as a C (M)- module to the set of C (M)-linear maps Ω 1 (M) C (M). Proof. By Proposition 2.1.2, Ω 1 (M) = X(M). Being nitely generated projective, X(M) is canonically isomorphic to its double dual (cf. Proposition A.2.1). Hence, X(M) = Ω 1 (M), as desired. Tracing out the isomorphisms involved, we see that the isomorphism X(M) Ω 1 (M) is explicitly given by X X, with X(ω) = ω(x) = ω ( ) (X( )), where ω is dened in 2.1 in the proof of Proposition Proposition The set of type (r, s) tensor elds on M, T r,s (M), is isomorphic as a C (M)-module to the set of C (M)-multilinear maps Ω 1 (M) r X(M) s C (M).

15 Chapter 2. Historical Motivation I: The Lie Derivative 11 Proof. In general, we have T r,s (M) C (M) T r,s (M) = T r+r,s+s (M), by denition of the product in T (M). In particular, T r,s (M) = T 1,0 (M) r T 0,1 (M) s = X(M) r Ω 1 (M) s, where we tensor over C (M) (this will be implicit for the remainder of this proof). Hence, by Proposition and corollary T r,s (M) = X(M) r Ω 1 (M) s = ( Ω 1 (M) ) r (X(M) ) s. As X(M) and Ω 1 (M) are nitely generated projective, we have ( Ω 1 (M) ) r (X(M) ) s = ( Ω 1 (M) r X(M) s) (cf. Proposition A.2.2) = {Ω 1 (M) r X(M) s C (M), C (M)-multilinear}, where the last isomorphism comes from the universal property of the tensor product. For r = s = 0, the rst two tensor products are just C (M) which is canonically isomorphic to { C (M)}. Remark We will henceforth readily switch between the identications below and use the one most suitable to the given situation: The explicit isomorphism T r,s (M) = (X(M) r ) (Ω 1 (M) s ) = (Ω 1 (M) r ) (X(M) s ) = {Ω 1 (M) r X(M) s C (M), C (M)-multilinear}. (X(M) r ) (Ω 1 (M) s ) {Ω 1 (M) r X(M) s C (M), C (M)-multilinear} is given by X 1 X r ω 1 ω s Ψ Xi,ω i, where Ψ Xi,ω i (η 1,..., η r, Y 1,..., Y s ) = η 1 (X 1 ) η r (X r )ω 1 (Y 1 ) ω s (Y s ). Remark It follows almost directly from the denition, that we can view Ω k (M) as a submodule of T 0,k (M) this can also be seen by Propositions and From this we deduce that Ω (M) is spanned as a C (M)-algebra by Ω 1 (M): To see this, note rst that the symmetric group on k symbols, S k, acts on the set of type (0, k) tensor elds, T 0,k (M), by σ.s(x 1,..., X k ) = S(X σ(1),..., X σ(k) ), for σ S k, S T 0,k (M); on Ω k (M) this is simply acting by sign. Now, given ω Ω k (M) T 0,k (M), we can write ω as a linear combination of tensors ω 1 ω k, ω i Ω 1 (M). If ω = n i=1 f i ω1 i ωi k, ωi j Ω1 (M), f i C (M), then for any σ S k. Hence, ω(x 1,..., X k ) = sign (σ) ω(x σ(1),..., X σ(k) ) n = sign (σ) f i ω1(x i σ(1) ) ωk i (X σ(k)) = i=1 n sign (σ) f i ωσ(1) i ωi σ(k) (X 1,..., X k ). i=1 k! ω = n f i sign (σ) ωσ(1) i ωi σ(k) = σ S k i=1 n f i ω1 i ωk i i=1

16 12 Mikala Ørsnes Jansen Interpreting Ω k (M) as a submodule of T 0,k (M) does not, however, relate the multiplication operations in the two modules, respectively. We shall instead interpret Ω k (M) as a quotient of T 0,k (M). We claim that Ω k (M) is the largest quotient of T 0,k (M) on which S k acts by sign, i.e. it satises the universal property pictured in the diagram below, where A is a C (M)- module on which S k acts by sign and all maps are equivariant: T 0,k (M) p A q! Ω k (M) The map q : T 0,k (M) Ω k (M) is given by ω 1 ω k ω 1 ω k, ω i Ω 1 (M). This is surjective, as we observed above that Ω k (M) is spanned by elements of the form ω 1 ω k, ω i Ω 1 (M). Note that q has a section, namely the map s: Ω k (M) T 0,k (M), given by s(ω 1 ω k ) = 1 sign (σ) ω k! σ(1) ω σ(k). σ S k Now, given p as in the diagram above, set ϕ := ps. Clearly, this is the unique map which makes the diagram commute, so all we need to show, is that it is surjective. By linearity, it suces to show that ϕ hits p(ω 1 ω k ) for any ω 1 ω k T 0,k (M). Given ω 1 ω k T 0,k (M), we see that ϕ(ω 1 ω k ) = 1 sign (σ) p(ω k! σ(1) ω σ(k) ) = 1 sign (σ) σ.p(ω 1 ω k ) k! σ S k σ S k = 1 (sign (σ)) 2 p(ω 1 ω k ) = p(ω 1 ω k ). k! σ S k As q respects the multiplication operations, we get the desired relationship between them. In view of this observation, we will often interpret a dierential k-form as a vector eld of type (0, k), even though we are in fact looking at an equivalence class of vector elds. 2.2 The Lie Derivative Let M be a manifold. Definition Let X X(M). The Lie derivative with respect to X is the type-preserving map L X : T (M) T (M) satisfying the following axioms: 1. L X is R-linear. 2. L X (f) = X(f) for any f C (M) = T 0,0 (M). 3. L X (S T ) = (L X S) T + S (L X T ) for any S, T T (M). (Leibniz' rule) 4. L X (η(y )) = (L X η)(y ) + η(l X Y ) for any η Ω 1 (M), Y X(M). (Leibniz' rule with respect to contractions) 5. L X commutes with the exterior derivative on C (M) = Ω 0 (M). Before proving existence of such a map, we will show that given it exists, it must be unique.

17 Chapter 2. Historical Motivation I: The Lie Derivative 13 Proposition If the Lie derivative exists, it is unique. Let X X(M). Assume that L X : T (M) T (M) is type-preserving and satises Proof. axioms 1-5 of Denition Being type-preserving and R-linear, it suces to show that it is unique on T r,s (M) for any choice of r, s. Now, the fact that T r,s (M) = (X(M) r ) (Ω 1 (M) s ) and axiom 3 reduces the problem to showing that L X is unique on X(M) and Ω 1 (M). If L X is unique on X(M), then axioms 2 and 4 imply uniqueness on Ω 1 (M). So we just need to show that L X : X(M) X(M) is unique. We shall identify X(M) with its action on C (M): Z X(M) denes a map C (M) C (M), g Z(g) given by Z(g): p D p,z(p) g. Note that for any g C (M) and Z X(M), we have Z(g) = dg(z), where dg Ω 1 (M) = X(M). Using this and axioms 1, 4 and 5, we see that for Y X(M) and f C (M), XY (f) = X(Y (f)) = L X (Y (f)) = L X (df(y )) = (L X df)(y ) + df(l X Y ) = (d(l X f))(y ) + (L X Y )(f) = d(x(f))(y ) + (L X Y )(f) = Y (X(f)) + (L X Y )(f) = Y X(f) + (L X Y )(f). Hence, L X Y = [X, Y ]. In particular, L X is uniquely given on X(M). We will need the following denition in the existence proof: Definition Given a vector eld X X(M) and a dieomorphism f : M N, the pushforward of X along f is the vector eld f X := df X f 1 X(N). Definition Given a dieomorphism f : N M, dene a map f : T (M) T (N), such that f is R-linear andf (S T ) = f S f T for any S, T T (M), and such that f agrees with the usual pullback on C (M) and Ω 1 (M), and f X = f 1 X for X X(M). This uniquely determines f, and we dene the pullback of a tensor eld S T (M) along f as f S T (N). Proposition Given X X(M), there exists a type-preserving map L X : T (M) T (M) satisfying the axioms of Denition Let S T r,s (M). We will dene L X S locally. Let q M, and let U M be a Proof. chart with q U. Consider a bump function λ: M R such that suppλ U is compact and there is an open V U such that q V and λ V = 1; then λx is a compactly supported vector eld on U, and therefore complete. Let Φ: U R U denote the ow of λx, and dene ϕ t := Φ U {t} : U U for all t R; {ϕ t } t R is a one-parameter group of dieomorphisms on U, and ϕ 0 = id U. The map (t, p) (ϕ t S) p denes a smooth map R U T r,s (M). Dene L X S T r,s (M) by (L X S) p := d dt (ϕ t S) p t=0, for p V. Uniqueness of ows implies that Φ V R coincides with the ow of X on V, which in turn implies that L X S is well-dened. It is enough to show the that the axioms hold locally, so we x a U, V and {ϕ t } t R as above. Clearly, L X is type-preserving and R-linear, as the pullback is R-linear. Note that (L X f) p = d t=0 dt (ϕ t f) p = d dt f ϕ t(p) = D p,x(p) f = X(f)(p), t=0 for all p V. Hence, L X f = X(f), so axiom 2 holds.

18 14 Mikala Ørsnes Jansen Let S, T T (M) of type (r, s) respectively (r, s ). Then, using the chain rule and the fact that : T r,s (T p M) T r,s (T p M) T r+r,s+s (T p M) is R-bilinear, we have that L X (S T ) p = d dt (ϕ t (S T )) p t=0 = d dt (ϕ t S) p (ϕ t T ) p t=0 = d dt (ϕ t S) p t=0 (ϕ t T ) p t=0 + (ϕ t S) p t=0 d dt (ϕ t T ) p t=0 = (L X S) p T p + S p (L X T ) p, for all p V. So axiom 3, Leibniz' rule, holds. Let c: T 1,1 (M) = X(M) Ω 1 (M) T 0,0 (M) = C (M) denote the contraction given by c(y ω) = ω(y ). We claim that c and L X commute: Indeed, c commutes with the pullback (this is easily checked), and thus, as c is linear L X c(y ω) = d dt ϕ t (c(y ω)) = d t=0 dt c(ϕ t (Y ω)) = c d t=0 dt ϕ t (Y ω) = cl X (Y ω). t=0 Using that we have already proved axiom 3, it follows that L X (ω(y )) = cl X (Y ω) = c(y L X ω + L X Y ω) = (L X ω)(y ) + ω(l X Y ), and thus axiom 4 holds. To prove axiom 5, we need an intermediary result, namely that L X Y = [X, Y ] for any Y X(M). Recall that Y (f) = df(y ) for f C (M), Y X(M). Given Y X(M), note that dϕ t (ϕ t ) Y = Y ϕ t, as ϕ 1 t = ϕ t, and thus for any f C (M) ((ϕ t ) Y )(f ϕ t ) = df dϕ t (ϕ t ) Y = df Y ϕ t = Y (f) ϕ t. Given p V, note that the map ɛ: T p M C (M) C (M), (v, f) D p,v f = Y (f) p, Y X(M) with Y (p) = v, is an R-bilinear map of nite dimensional vector spaces. Dierentiating both sides of the equation (Y (f) ϕ t ) p = (((ϕ t ) Y )(f ϕ t )) p, we get X(Y (f)) p = d dt (Y (f) ϕ t=0 t) p = d dt (((ϕ t=0 t) Y )(f ϕ t )) p = d ( dt ɛ((ϕ d ) ( t) Y (p), f ϕ t ) = ɛ t=0 dt (ϕ t) Y (p), f + ɛ Y (p), d ) t=0 dt f ϕ t t=0 = ɛ((l X Y ) p, f) + ɛ(y (p), X(f)) = ((L X Y )(f)) p + Y (X(f)) p. This holds for all p V ; thus L X Y = [X, Y ], as desired. Using this and axioms 1 and 4, we see that for f C (M), L X (df)(y ) = L X (df(y )) df(l X Y ) = X(Y (f)) [X, Y ](f) = Y (X(f)) = Y (L X f) = (dl X f)(y ). Thus, as desired, L X commutes with d on C (M), axiom 5. Definition Given X X(M), L X : Ω (M) Ω (M) is given by L X (ω 1 ω k ) := q(l X (ω 1 ω k )). This is well-dened, as we have a section of q, and it is the unique type-preserving map satisfying: 1. L X is R-linear.

19 Chapter 2. Historical Motivation I: The Lie Derivative L X (f) = X(f) for any f Ω 0 (M) = C (M). 3. L X (ω η) = (L X ω) η + ω (L X η) for any ω, η Ω (M). 4. L X (η(y )) = (L X η)(y ) + η([x, Y ]) for any η Ω 1 (M), Y X(M). 5. L X commutes with the exterior derivative on Ω 0 (M) = C (M). It is this Lie derivative, which we will need. Remark The denition given in the above existence proof could also be used to directly dene the Lie derivative and is important in understanding the geometric interpretation: It measures the change of a tensor eld along the ow of the given vector eld X. Another interpretation is the following: In general, an action of a Lie algebra g on a manifold M is a Lie algebra homomorphism g X(M) such that the corresponding map g M M is smooth. Such an action induces an action on Ω k (M): Let X g, ω Ω (M), and let V X with ow {ϕ t } denote the vector eld corresponding to X under the given action. Then t (ϕ t ω) p is a smooth curve in Λ(T p M ), and we dene (X.ω) p = d dt (ϕ t ω) p t=0. There are of course some details left to prove to see that X.ω Ω (M), but we will not go into this here. The point is that the Lie derivative is simply the action of X(M) on Ω (M) induced by the trivial action of X(M) on M, id: X(M) X(M), L X (ω) = X.ω. We have opted for the algebraic denition for the sake of clarity and less technicalities. We go on to prove some properties of the Lie derivative. Definition Given X X(M), dene a map i X : Ω (M) Ω 1 (M), called the interior product (with respect to X) by i X ω(x 1,..., X k ) = ω(x, X 1,..., X k ), for any ω Ω k+1 (M), X i X(M), k 1, and i X f = 0, f Ω 0 (M) = C (M). Lemma For X X(M), ω Ω k (M), η Ω m (M), we have i X (ω η) = (i X ω) η + ( 1) k ω (i X η). Proof. For X 1,..., X k+m 1 X(M), we have i X (ω η)(x 1,..., X k+m 1 ) = (ω η)(x, X 1,..., X k+m 1 ) = σ S k 1,m sign (σ) ω(x, X σ(1),..., X σ(k 1) )η(x σ(k),..., X σ(k+m 1) ) + σ S k,m 1 ( 1) k sign (σ) ω(x σ(1),..., X σ(k) )η(x, X σ(k+1),..., X σ(k+m 1) ) = ((i X ω) η)(x 1,..., X k+m 1 ) + ( 1) k (ω (i X η))(x 1,..., X k+m 1 ). Proposition (Cartan's magic formula). We have the following identity for all X X(M), ω Ω (M) L X ω = i X dω + di X ω. Proof. To prove the claim, we shall prove that i X d + di X : Ω (M) Ω (M) satises axioms 1-5 of Denition That i X d + di X is R-linear is clear.

20 16 Mikala Ørsnes Jansen 2. Let f C (M). As i X f = 0, we have (i X d + di X )(f) = i X df = df(x) = X(f). 3. Let ω Ω k (M), η Ω m (M). Then using Lemma and the fact that the exterior derivative also has this relationship with the exterior product, we get: (i X d + di X )(ω η) = i X (dω η + ( 1) k ω dη) + d(i X ω η + ( 1) k ω i X η) = i X dω η + ( 1) k+1 dω i X η + ( 1) k i X ω dη + ( 1) 2k ω i X dη + di X ω η + ( 1) k 1 i X ω dη + ( 1) k dω i X η + ( 1) 2k ω di X η = (i X dω + di X ω) η + ω (i X dη + di X η). 4. Now it suces to prove the identity locally, and as we have already proved R-linearity, we can consider η Ω 1 (M), which satises η = fdx on some chart U of M and some local coordinate x on U. Let Y X(M). Then dη = df dx, and (i X d + di X )(η)(y ) + η([x, Y ]) = dη(x, Y ) + d(η(x))(y ) + η([x, Y ]) which at p U gives = (df dx)(x, Y ) + Y (fdx(x)) + fdx([x, Y ]) = df(x)dx(y ) df(y )dx(x) + Y (fx(x)) + f([x, Y ](x)) = X(f)Y (x) Y (f)x(x) + Y (fx(x)) + fx(y (x)) fy (X(x)), D p,x(p) fd p,y (p) x D p,y (p) fd p,x(p) x + D p,y (p) (fx(x)) + f(p)d p,x(p) Y (x) f(p)d p,y (p) X(x) = D p,x(p) fd p,y (p) x + f(p)d p,x(p) Y (x) = D p,x(p) (fy (x)), which is the value at p of X(fY (x)) = X(η(Y )) = (i X d + di X )(η(y )) by axiom For any ω Ω (M), (i X d + di X )(dω) = i X d 2 ω + di X dω = di X dω + d 2 i X ω = d(i X d + di X )(ω). Remark Cartan's magic fomula implies that L X : Ω (M) Ω (M) is a chain map, and that it is null-homotopic. Lemma For any X, Y X(M), i X i Y = i Y i X and i X = i X. Proof. This is obvious. Lemma Given X, Y X(M), we have the identity L X i Y i Y L X = i [X,Y ]. Proof. Direct calculations using Lemma and axiom 3 of Denition 2.2.6, show that for ω Ω k (M), η Ω m (M), we have (L X i Y i Y L X )(ω η) = ((L X i Y i Y L X )ω) η + ( 1) k ω ((L X i Y i Y L X )η).

21 Chapter 2. Historical Motivation I: The Lie Derivative 17 We prove the claim by induction on the degree of the dierential form. On Ω 0 (M) the left- and right-hand side are both zero. For ω Ω 1 (M), axiom 4 of Denition gives us L X i Y ω i Y L X ω = L X (ω(y )) (L X ω)(y ) = ω([x, Y ]) = i [X,Y ] ω. Now, assume the equality holds on Ω k 1 (M) for some k > 1. To show equality on Ω k (M), it suces to show it on an element of the form ω η Ω k (M), for ω Ω 1 (M), η Ω k 1 (M): (L X i Y i Y L X )(ω η) = ((L X i Y i Y L X )ω) η + ( 1) k ω ((L X i Y i Y L X )η) = (i [X,Y ] ω) η + ( 1) k ω (i [X,Y ] η) = i [X,Y ] (ω η). Proposition (Invariant formula). For any ω Ω k (M), X 0,..., X k X(M), we have k dω(x 0,..., X k ) = ( 1) i X i (ω(x 0,..., X i,..., X k )) i=0 + i<j ( 1) i+j ω([x i, X j ], X 0,..., X i,..., X j,..., X k ). Proof. First we prove that L X0 ω(x 1,..., X k ) = X 0 (ω(x 1,..., X k )) k ω(x 1,..., X i 1, [X 0, X i ], X i+1,..., X k ). (2.2) i=1 We do this by induction on k. For k = 0 and k = 1, this is axiom 2, respectively, axiom 4 of Denition Now, assume that it holds for k 1 for some k > 1, and let ω Ω k (M). The induction hypothesis implies that X 0 (ω(x 1,..., X k )) = X 0 ((i X1 ω)(x 2,..., X k )) = (L X0 i X1 ω)(x 2,..., X k ) + Now, by Cartan's magic formula and Lemmas and , k (i X1 ω)(x 2,..., X i 1, [X 0, X i ], X i+1,..., X k ). i=2 L X0 i X1 = i X0 di X1 + di X0 i X1 = i X0 di X1 di X1 i X0 = i X0 L X1 i X0 i X1 d L X1 i X0 + i X1 di X0 = (L X1 i X0 i X0 L X1 ) + i X1 (i X0 d + di X0 ) = i [X1,X 0 ] + i X1 L X0 = i [X0,X 1 ] + i X1 L X0. Hence, as desired X 0 (ω(x 1,..., X k )) = i [X0,X 1 ]ω(x 1,..., X k ) + (i X1 L X0 ω)(x 2,..., X k ) k + (i X1 ω)(x 2,..., X i 1, [X 0, X i ], X i+1,..., X k ) i=2 = L X0 ω(x 1,..., X k ) + k ω(x 1,..., X i 1, [X 0, X i ], X i+1,..., X k ). i=1

22 18 Mikala Ørsnes Jansen Finally, we prove the invariant formula, also by induction on k. For k = 0, the equation reads df(x) = X(f), which is true. Now assume that it holds for k 1 for some k 1, and let ω Ω k (M). Then Cartan's magic formula, the induction hypothesis and equation 2.2 yield dω(x 0,..., X k ) = i X0 dω(x 1,..., X k ) = L X0 ω(x 1,..., X k ) di X0 ω(x 1,..., X k ) k = X 0 (ω(x 1,..., X k )) ( 1) i 1 ω([x 0, X i ], X 1,..., X i,..., X k ) i=1 k i=1( 1) i+1 X i (i X0 ω(x 1,.., X i,.., X k )) ( 1) i+j i X0 ω([x i, X j ], X 1,.., X i,.., X j,.., X k ) i<j k = i=0( 1) i X i (ω(x 0,..., X i,..., X k )) + ( 1) i+j ω([x i, X j ], X 0,..., X i,..., X j,..., X k ). i<j We immediately see the motivation for the dierential in the Chevalley-Eilenberg complex in this formula. In the case of a Lie group G, its Lie algebra g can be interpreted as the set of left-invariant vector elds on G. Then C (G) is a g-module, and a dierential k-form, ω Ω k (G), restricts to an alternating k-linear map c := ω g k : g k C (G), i.e. an element of C k (g, C (G)). The derivative of c in the Chevalley-Eilenberg complex is, in view of the invariant formula, exactly the exterior derivative of ω restricted to g k. This suggests an intricate relationship between the Lie algebra cohomology of g and the de Rham cohomology of G, which we will see more explicitly in the following chapter.

23 3 Historical Motivation II: De Rham Cohomology In this chapter, we consider the relationship between Lie algebra cohomology and De Rham cohomology. As mentioned in the preface, the theory of Lie algebra cohomology was developed in an attempt to calculate the De Rham Cohomology of a compact Lie group. Corollary below shows that the De Rham Cohomology of a compact connected Lie group can be computed purely in terms of its Lie algebra. To this end, we will prove an in itself important theorem about De Rham cohomology: If a compact connected Lie group acts on a manifold, then any element of the De Rham Cohomology of the manifold can be represented by an invariant dierential form. 3.1 Lie Group Action Let G be a Lie group acting smoothly on M, G M M, where we will also denote by g the smooth map M M corresponding to g G by the above action. Definition Let Ω k (M) G denote the set of invariant forms on M, i.e. the forms ω Ω k (M) satisfying g ω = ω. Note that dω is invariant, if ω is invariant; hence, Ω (M) G is a chain complex with dierential the restriction of the exterior derivative. Lemma Let G be a compact Lie group with some xed volume form ω vol and associated measure µ, let M be a manifold, and f : G M R be a smooth map. Then the map F : M R dened as F (p) = G f(g, p) dµ is smooth and D pf (v) = G D p(f i g )(v) dµ, where i g : M G M is the inclusion p (g, p). Proof. Let σ : R n U be an arbitrary chart on M, and let {σ α : R m V α } be a nite open cover of G by charts such that σαω vol = dx 1 dx m is the standard Euclidean volume form, and let {ϕ α } a smooth partition of unity subordinate to {V α }. Then F σ(x) = f(g, σ(x)) dµ = ϕ α (g)f(g, σ(x)) dµ G α V α = ϕ α σ α (y)f(σ α (y), σ(x)) dλ α R m = ϕ α σ α (y)f(σ α (y), σ(x)) dλ, α suppϕ α σ α where λ denotes the usual Lebesgue measure on R m. Now, (x, y) ϕ α σ α (y)f(σ α (y), σ(x)) is a smooth map R n R m R and suppϕ α σ α is compact. An application of Lebesgue's Dominated Convergence Theorem (see Corollary A.3.2) implies that F σ is smooth with x j (F σ)(x) = G x j f(g, σ(x)) dµ. It follows that F is smooth with D p F (v) = G D p(f i g )(v) dµ. 19

24 20 Mikala Ørsnes Jansen Proposition If M is a manifold and G is a connected compact Lie group acting on M by α: G M M, then the inclusion ι: Ω (M) G Ω (M) is a quasi-isomorphism. Proof. Clearly, ι is a chain map. Suppose G is of dimension m, and let ω vol Ω m (G) be a right-invariant volume form on G satisfying G ω vol = 1 with associated Haar measure µ. Let τ : G M G denote the projection onto the rst coordinate, and π : G M M the projection onto the second coordinate. We will for notational reasons denote tangent vectors on G M by v w T (g,p) (G M) = T g G T p M for v T g G, w T p M. Dene a map π : Ω (G M) Ω m (M) as follows: On Ω k (G M), k < m, π is identically zero. For k m, let ω Ω k (G M), p M, v 1,..., v k m T p M and dene (π ω) p (v 1,..., v k m ) = ω p,(vi), where ω p,(v i) Ω m (G) is given by ω p,(v i) g (w 1,..., w m ) = ω (g,p) (0 v 1,..., 0 v k m, w 1 0,..., w m 0). To see that p (π ω) p is smooth requires some technical work: Let U be a chart on M with local coordinates x i, {V α } a nite open cover of G by charts with local coordinates yi α, and let {ϕ α} be a smooth partition of unity subordinate to this cover. We can extend x 1,..., x n, y1 α,..., yα m to U V α in an obvious way such that they become local coordinates on U V α, and therefore write ω = fσ,τ α dx σ dyτ α, on U V α, for fσ,τ α C (U V α ). r+s=k σ S r,n r τ S s,m s Given p U, v 1,..., v k m T p (M), g V α, and w 1,..., w m T g G, we have ω p,(v i) g (w 1,.., w m ) = r+s=k σ S r,n r τ S s,m s = G fσ,τ α (g, p) dx σ (p) dyτ α (g)(0 v 1,.., 0 v k m, w 1 0,.., w m 0) σ S k m,n+m k f α σ,id (g, p) dx σ(p)(v 1,..., v k m ) dy α id (g)(w 1,..., w m ). Then (π ω) p (v 1,..., v k m ) = α ϕ α ω p,(vi) = ( ) ϕ α (g)f α (g, p) dyα dx σ,id id σ (p)(v 1,..., v k m ). V α α,σ V α Now, for all choices of σ and α, (g, p) ϕ α (g)f α (g, p) extends smoothly to G U R by σ,id setting it to be zero outside of V α. Then Lemma implies that f σ := α V α ϕ α (g)f α σ,id (g, p) dyα id is a smooth map on U, using that dy α is a volume form on G. We conclude that π id ω is smooth. In addition, Lemma tells us that D p f σ (v) = ϕ α (g)d p f α (g, p)(v) dyα. σ,id id α V α

25 Chapter 3. Historical Motivation II: De Rham Cohomology 21 Using this, we see that π commutes with d, as (dπ ω) p (v 1,..., v k m+1 ) = σ df σ (p) dx σ (p)(v 1,..., v k m+1 ) on U, and = σ k m+1 i=1 ( 1) i 1 D p f σ (v i )dx σ (p)(v 1,..., v i,..., v k m+1 ) ( dω p,(v i) )g (w 1,.., w m ) = σ,τ df α σ,τ (g, p) dx σ (p) dy α τ (g)(0 v 1,.., 0 v k m+1, w 1 0,.., w m 0) = σ on V α, and thus k m+1 i=1 (π dω) p (v 1,.., v k m+1 )= σ,i ( 1) i 1 D p f α σ,id (v i) dx σ (p)(v 1,..., v i,..., v k m+1 ) dy α id (g)(w 1,..., w m ) ( 1) i 1 ( α V α ϕ α (g)d p f α σ,id (g, p)(v i)dy α id ) dx σ (p)(v 1,.., v i,.., v k m+1 ). Now, we dene a map I : Ω k (G M) Ω k (M) by I(ω) = π (ω τ ω vol ). This is a chain map, as ω vol is a top form on G, so dω vol = 0, and thus I(dω) = π (dω τ ω vol ) = π (d(ω τ ω vol ) ( 1) k ω dτ ω vol ) = π (d(ω τ ω vol ) ( 1) k ω τ dω vol ) = π (d(ω τ ω vol )) = dπ (ω τ ω vol ) = di(ω). To ease the calculations to come, let p M, v 1,..., v k T p M, g G, w 1,..., w m T g G, and set β := (ω τ ω vol ) p,(v i) Ω m (G), t i := 0 v i for i = 1,... k, and t i := w i k 0 for i = k + 1,..., k + m. Note that D (g,p) τ : T g G T p M T g G is the projection onto the rst coordinate. Then β g (w 1,..., w m ) = (ω τ ω vol ) (g,p) (t 1,..., t k+m ) = σ S k,m sign σ ω (g,p) (t σ(1),..., t σ(k) )τ ω vol(t σ(k+1),..., t σ(k+m) ) = ω (g,p) (0 v 1,..., 0 v k )(ω vol ) g (w 1,..., w m ) so β g = ω (g,p) (0 v 1,..., 0 v k )(ω vol ) g, and thus I(ω) p (v 1,..., v k ) = ω g,p (0 v 1,..., 0 v k ) dµ. G Aside: We have not dened I like this directly, as we will use the actual denition in some of the calculations below. Moreover, the construction above can be generalised: the map π is a bre integral, which can be dened for any smooth vector bundle E M with compact oriented bres. However, as we are only interested in the trivial bundle π : G M M, it seems unnecessary to spend much time on it. Dene a chain map ρ: Ω (M) Ω (M) G as ρ := I α. Note that ρ is explicitly given as "averaging the dierential forms over G": For ω Ω (M), p M, v 1,..., v k T p M, (α ω) (g,p) (0 v 1,..., 0 v k ) = ω g(p) (D p g(v 1 ),..., D p g(v k )) = (g ω) p (v 1,..., v k ),

26 22 Mikala Ørsnes Jansen where we use that D (g,p) α: T g G T p M T g M is given by D (g,p) α(v w) = D g ev p (v)+d p g(w), with ev p : G M evaluation at p. Hence, ρ(ω) p (v 1,..., v k ) = I(α ω) p (v 1,..., v k ) = (α ω) g,p (0 v 1,..., 0 v k ) dµ = G G (g ω) p (v 1,..., v k ) dµ. We also write ρ(ω) = G g ω dµ. To show that ρ is well-dened, we must show that ρ(ω) is G-invariant for any ω Ω (M): For any h G, the above explicit expression of ρ and rightinvariance of µ gives h ρ(ω) = h ( G ) g ω dµ = h g ω dµ = (gh) ω dµ = g ω dµ = ρ(ω). G G G We will show that on cohomology, H(ρ) = H(ι) 1. It is seen directly that ρ ι = id. To show that H(ι) H(ρ) = id requires some more work. We claim that in our denition of π, we may change the domain of integration to a neighbourhood U of 1 G and obtain a map I U, which is homotopic to I. To see this, let U G be a neighbourhood of 1 and let λ: G R be a bump function with support contained in U such that G λω vol = 1. Let I U : Ω k (G M) M denote the map arising from the above construction, replacing ω vol by λω vol. As ω vol and λω vol integrate to the same, they dier by an exact form: Indeed, G : Hm (G) R is an isomorphism as G is compact and connected (cf. Theorem [8]). Let η Ω m 1 (G) such that ω vol λω vol = dη, and consider the map h: Ω (G M) Ω 1 (M), h(ω) = ( 1) k π (ω τ η), for ω Ω k (G M). h is a chain homotopy from I to I U : For any ω Ω k (G M), h(dω) + dh(ω) = ( 1) k+1 π (dω τ η) + ( 1) k dπ (ω τ ( η) ) = ( 1) k+1 π (d(ω τ η) ( 1) k ω τ dη) + ( 1) k dπ (ω τ η) = π (ω τ dη) = π (ω τ (ω vol λω vol )) = I(ω) I U (ω). The advantage of this is that we can restrict I U to Ω k (U M), so to speak: Let i: U M G M denote the inclusion, and dene Ĩ U : Ω k (U M) Ω k (M) as ĨU(ω) p (v 1,..., v k ) = λω (g,p) (0 v 1,..., 0 v k ) dµ, where λ is the bump function from above, and we extend ω (not necessarily continuously) to G M by dening it to be zero outside of U M. Then I U = ĨU i. Moreover, ĨU π U = id, where π U : U M M is the restriction of π: Indeed, for any ω Ω k (U M), p M, v 1,..., v k T p M, Ĩ U π U(ω) p (v 1,..., v k ) = λ(π Uω) (g,p) (0 v 1,..., 0 v k ) dµ G = λω p (v 1,..., v k ) dµ = ω p (v 1,..., v k ) λdµ = ω p (v 1,..., v k ). G If we take U to be a contractible neighbourhood of 1, and let j : M U M denote the inclusion p (1, p), then the composite j π U is homotopic to the identity on U M. Combining the above results, we see that H(ι) H(ρ) = H(I) H(α ) = H(I U ) H(α ) = H(ĨU) H(i ) H(α ) = H(ĨU) H(π U) H(j ) H(i ) H(α ) = H((α i j) ) = H(id) = id. G G