J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16
B. DIFFERENTIAL FORMS instance, if S has k elements this gives a k-dimensional vector space with S as basis. We have already seen one-forms (covector fields) on a Given vector spaces V and W, let F be the free vector space over the set V × W. (This consists of formal sums manifold. In general, a k-form is a field of alternating k- P linear forms on the tangent spaces of a manifold. Forms ai(vi, wi) but ignores all the structure we have on the set are the natural objects for integration: a k-form can be in- V × W.) Now let R ⊂ F be the linear subspace spanned by tegrated over an oriented k-submanifold. We start with ten- all elements of the form: sor products and the exterior algebra of multivectors. (v + v0, w) − (v, w) − (v0, w), (v, w + w0) − (v, w) − (v, w0), (av, w) − a(v, w), (v, aw) − a(v, w). B1. Tensor products These correspond of course to the bilinearity conditions Recall that, if V, W and X are vector spaces, then a map we started with. The quotient vector space F/R will be the b: V × W → X is called bilinear if tensor product V ⊗ W. We have started with all possible v ⊗ w as generators and thrown in just enough relations to b(v + v0, w) = b(v, w) + b(v0, w), make the map (v, w) 7→ v ⊗ w be bilinear. b(v, w + w0) = b(v, w) + b(v, w0), The tensor product is commutative: there is a natural linear isomorphism V⊗W → W⊗V such that v⊗w 7→ w⊗v. (This b(av, w) = ab(v, w) = b(v, aw). is easiest to verify using the universal property – simply factor the bilinear map (v, w) 7→ w ⊗ v through V ⊗ W to The function b is defined on the set V × W. This Cartesian give the desired isomorphism.) product of two vector spaces can be given the structure of a vector space V ⊕ W, the direct sum. But a bilinear map Similarly, the tensor product is associative: there is a natu- b: V × W → X is completely different from a linear map ral linear isomorphism V ⊗ (W ⊗ X) → (V ⊗ W) ⊗ X. Note V ⊕ W → X. that any trilinear map from V × W × X factors through this triple tensor product V ⊗ W ⊗ X. The tensor product space V ⊗ W is a vector space designed exactly so that a bilinear map b: V × W → X becomes a Of special interest are the tensor powers of a single vector ⊗k linear map V ⊗ W → X. More precisely, it can be charac- space V. We write V := V ⊗ · · · ⊗ V. If {ei} is a basis ⊗ · · · ⊗ ⊗k terized abstractly by the following “universal property”. for V, then ei1 eik is a basis for V . In particular if V has dimension m, then V⊗k has dimension mk. There Definition B1.1. The tensor product of vector spaces V is a natural k-linear map Vk → V⊗k and any k-linear map and W is a vector space V ⊗ W with a natural bilinear map Vk → W factors uniquely through V⊗k. V × W → V ⊗ W, written (v, w) 7→ v ⊗ w, with the prop- One can check that the dual of a tensor product is the tensor erty that any bilinear map b: V × W → X factors uniquely product of duals: (V ⊗ W)∗ = V∗ ⊗ W∗. In particular, through V ⊗ W. That means there exists a unique linear we have (V∗)⊗k = (V⊗k)∗. The latter is of course the set map L: V ⊗ W → X such that b(v, w) = L(v ⊗ w). of linear functionals V⊗k → R, which as we have seen is exactly the set of k-linear maps Vk → R. This does not yet show that the tensor product exists, but Definition B1.2. A graded algebra is a vector space A de- uniqueness is clear: if X and Y were both tensor products, L∞ then each defining bilinear map would factor through the composed as A = k=0 Ak together with an associative other – we get inverse linear maps between X and Y, show- bilinear multiplication operation A × A → A that respects ing they are isomorphic. the grading in the sense that the product ω · η of elements ω ∈ A and η ∈ A is an element of A . Often we consider Note that the elements of the form v ⊗ w must span V ⊗ W, k ` k+` graded algebras that are either commutative or anticommu- since otherwise L would not be unique. If {e } is a basis i tative. Here anticommutative has a special meaning: for for V and { f j} a basis for W then bilinearity gives k` ω ∈ Ak and η ∈ A` as above, we have ω · η = (−1) η · ω. X X X i j i j Example B1.3. The tensor algebra of a vector space V is v ei ⊗ w f j = v w ei ⊗ f j. i j i, j ∞ M ⊗k ⊗∗V := V . Clearly then {ei ⊗ f j} spans V ⊗ W – indeed one can check k=0 that it is a basis. This is a valid construction for the space Here of course V⊗1 V and V⊗0 R. Note that the tensor V ⊗ W – as the span of the ei ⊗ f j – but it does depend on the chosen bases. If dim V = m and dim W = n then we product is graded, but is neither commutative nor anticom- note dim V ⊗ W = mn. mutative. A much more abstract construction of V ⊗ W goes through a huge infinite dimensional space. Given any set S , the free B2. Exterior algebra vector space on S is the set of all formal finite linear com- P binations ai si with ai ∈ R and si ∈ S . (This can equally well be thought of as the set of all real-valued functions We now want to focus on antisymmetric tensors, to de- on the set S which vanish outside some finite subset.) For velop the so-called exterior algebra or Grassmann algebra
18 J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16 of the vector space V. is, alternating k-linear maps from Vk correspond to linear Just as we constructed V ⊗ V = V⊗2 as a quotient of a huge maps from ΛkV. (One can also phrase the universality for vector space, adding relators corresponding to the rules for all k together in terms of homomorphisms of anticommu- tative graded algebras.) bilinearity, we construct the exterior power V ∧ V = Λ2V as a further quotient. In particular, letting S ⊂ V⊗V denote So far we have developed everything abstractly and alge- span of the elements v ⊗ v for all v ∈ V, we set V ∧ V := braically. But there is a natural geometric picture of how k- (V ⊗ V)/S . We write v ∧ w for the image of v ⊗ w under the vectors in ΛkV correspond to k-planes (k-dimensional lin- quotient map. Thus v ∧ v = 0 for any v. From ear subspaces) in V. More precisely, we should talk about simple k-vectors here: those that can be written in the form 4 (v + w) ∧ (v + w) = 0 v1 ∧· · ·∧vk. We will see that, for instance, e12 +e34 ∈ Λ2R is not simple. it then follows that v ∧ w = −w ∧ v. If {e : 1 ≤ i ≤ m} is a i ∈ basis for V, then A nonzero vector v V lies in a unique oriented 1-plane (line) in V; two vectors represent the same oriented line if and only if they are positive multiples of each other. Now {ei ∧ e j : 1 ≤ i < j ≤ m} suppose we have vectors v1,..., vk ∈ V. They are linearly is a basis for V ∧ V. independent if and only if 0 , v1 ∧ · · · ∧ vk ∈ ΛkV. Two Higher exterior powers of V can be constructed in the same linearly independent k-tuples (v1,..., vk) and (w1,..., wk) way, but formally, it is easiest to construct the whole ex- represent the same oriented k-plane if and only if the L wedge products v1 ∧ · · · ∧ vk and w1 ∧ · · · ∧ wk are pos- terior algebra Λ∗V = Λ V at once, as a quotient of k itive multiples of each other, that is, if they lie in the same the tensor algebra ⊗∗V, this time by the two-sided ideal ray in ΛkV. (Indeed, the multiple here is the ratio of k- generated by the same set S = {v ⊗ v} ⊂ V ⊗ V ⊂ ⊗∗V. This means the span not just of the elements of S but also areas of the parallelepipeds spanned by the two k-tuples, of their products (on the left and right) by arbitrary other given as the determinant of the change-of-basis matrix for the k-plane.) tensors. Elements of Λ∗V are called multivectors and ele- ments of ΛkV are more specifically k-vectors. We let Gk(V) denote the set of oriented k-planes in V, called the (oriented) Grassmannian. Then the set of simple End of Lecture 30 Nov 2015 k-vectors in ΛkV can be viewed as the cone over Gk(V). (If Again we use ∧ to denote the product on the resulting (still we pick a norm on ΛkV, say induced by an inner product graded) quotient algebra. This product is called the wedge on V, then we can think of Gk(V) as the set of “unit” sim- product or more formally the exterior product. We again ple k-vectors, say those arising from an orthonormal basis get v ∧ w = −w ∧ v for v, w ∈ V. More generally, for any for some k-plane.) v1,..., vk ∈ V and any permutation σ ∈ Σk of {1,..., k}, (Often, especially in algebraic geometry, one prefers to this implies work with the unoriented Grassmannian Gk(V)/±. It is most naturally viewed as lying in the projective space vσ1 ∧ · · · ∧ vσk = (sgn σ) v1 ∧ · · · ∧ vk. P(V):= V r {0}/ R r {0}. A special case is the product of a k-vector α with an `- vector β where we use a cyclic permutation to get the anti- In algebraic geometry one typically also replaces R by C commutative law α ∧ β = (−1)k`β ∧ α. throughout.) If {ei : 1 ≤ i ≤ m} is a basis for V, then If we give V an inner product, then any k-plane has a unique orthogonal (m − k)-plane. This induces an isomor- {e := e ∧ · · · ∧ e : 1 ≤ i < ··· < i ≤ m} i1···ik i1 ik 1 k phism between GkV and Gm−kV. It extends to a linear, norm-preserving isomorphism m is a basis for ΛkV. In particular, dim ΛkV = k ; we have Λ0V = R but also ΛmV R, spanned by e12···m. For k > m ?: ΛkV → Λm−kV there are no antisymmetric tensors: Λ V = 0. The exterior k algebra has dim Λ V = Pm m = 2m. The determinant called the Hodge star operator. (Recall that both these ∗ k=0 k has a natural definition in terms of the exterior algebra: if m spaces have the same dimension k .) If v is a simple k- we have m vectors v j ∈ V given in terms of the basis {ei} vector, then ?v is a simple (m − k)-vector representing the P i as v j = i v jei then orthogonal complement. In particular, if {ei} is an oriented orthonormal basis for V, then v ∧ · · · ∧ v = det vi e . 1 m j 12···m ? e1 ∧ · · · ∧ ek = ek+1 ∧ · · · ∧ em (The components of the wedge product of k vectors vi are given by the various k×k minor determinants of the matrix and similarly each other vector in our standard basis for i v j .) ΛkV maps to a basis vector for Λm−kV, possibly with a The exterior powers of V with the natural k-linear maps minus sign. k V → ΛkV are also characterized by the following univer- Classical vector calculus in three dimensions uses the sal property. Given any alternating k-linear map Vk → X to Hodge star implicitly: instead of talking about bivectors any vector space X, it factors uniquely through ΛkV. That and trivectors, we introduce the cross product and triple
19 J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16 product: Λ1V = V∗ we set v × w := ?(v ∧ w), [u, v, w]:= hu, v × wi = ?(u ∧ v ∧ w). ω ∧ η := ω ⊗ η − η ⊗ ω. k ` But even physicists noticed that such vectors and scalars More generally, for ω ∈ Λ V and η ∈ Λ V we use an transform differently (say under reflection) than ordinary alternating sum over all permutations σ ∈ Σk+`: vectors and scalars, and thus refer to them as pseudovec- tors and pseudoscalars. (ω ∧ η)(v1,..., vk+`):= 1 X For dim V = m, we can use these terms as follows: (sgn σ) ω(v ,..., v ) η(v ,..., v ). k!`! σ1 σk σ(k+1) σ(k+`) σ • scalars are elements of R = Λ0V, i • vectors are elements of V = Λ1V, The factor is chosen so that if {ei} is a basis for V and {ω } is the dual basis for Λ1V = V∗ then • pseudovectors are elements of ?V = Λm−1V, and i ···i i i • pseudoscalars are elements of ?R = ΛmV. ω 1 k := ω 1 ∧ · · · ∧ ω k
k Of course, these are in a sense the easy cases. For these k, is the basis of Λ V dual to the basis {ei1···ik } for ΛkV. any k-vector is simple. We can identify both G1V and Putting these spaces together, we get an anticommutative Gm−1V as the unit sphere in V = Λ1V Λm−1V. For graded algebra 2 ≤ k ≤ m − 2 on the other hand, not all k-vectors are m simple, and GkV has lower dimension than the unit sphere M Λ∗V := ΛkV. in ΛkV. Indeed, it can be shown that the set of simple k- k=0 vectors (the cone over GkV) is given as the solutions to a certain set of quadratic equations called the Grassmann– Again the dimension of each summand is m so the whole P i j ∈ 4 k Plücker relations. For instance a ei j Λ2R is a simple algebra has dimension 2m. 2-vector if and only if If L: V → W is a linear map, then for each k we get an ∗ a12a34 − a13a24 + a14a23 = 0. induced map L : ΛkW → ΛkV defined naturally by ∗ 4 L ω(v ,..., v ) = ω(Lv ,..., Lv ). This shows that G2R is a smooth 4-submanifold in the 1 k 1 k 5 4 unit sphere S ⊂ Λ2R . Of course, we have introduced these ideas in order to apply If we choose an inner product on V, then thinking about them to the tangent spaces T M to a manifold Mm. We get how oriented orthonormal bases for a k-plane and its or- p dual bundles Λ TM and ΛkTM of rank m . thogonal complement fit together, we see that we can iden- k k tify GkV = SO(m)/ SO(k) × SO(m − k) . In particular, it Definition B3.1. A (differential) k-form on a manifold Mm is a smooth manifold of dimension k(m − k). is a (smooth) section of the bundle ΛkTM. We write Ωk M = Γ(ΛkTM) for the space of all k-forms, which is a module over C∞ M = Ω0 M. Similarly we write L B3. Differential forms Ω∗ M = Γ(Λ∗TM) = Ωk M for the exterior algebra of M. Many textbooks omit discussion of multivectors and con- If ω ∈ Ωk M is a k-form, then at each point p ∈ M the sider only the dual spaces. (This is presumably because the value ω ∈ ΛkT M is an alternating k-linear form on T M abstract definition of tensor powers and then exterior pow- p p p or equivalently a linear functional on Λ T M. That is, for ers as quotient spaces seems difficult.) Recall that vector k p any k vectors X ,..., X ∈ T M we can evaluate subspaces and quotient spaces are dual operations, in the 1 k p ∗ sense that if Y ⊂ X is a subspace, then the dual (X/Y) ωp(X1,..., Xk) = ω(X1 ∧ · · · ∧ Xk) ∈ R. of the quotient can be naturally identified with a subspace ∗ o ∗ of X , namely with the annihilator Y of X , consisting of In particular, ωp naturally takes values on (weighted) k- those linear functionals on X that vanish on Y: planes in T p M; as we have mentioned, k-forms are the nat- ural objects to integrate over k-dimensional submanifolds (X/Y)∗ Yo ⊂ X∗. in M. If f : Mm → Nn is a smooth map and ω ∈ ΩkN is a k-form, Using this, we find that then we can pull back ω to get a k-form f ∗ω on M defined k ∗ ⊗k∗ by Λ V := (ΛkV) ⊂ V ( f ∗ω) (X ,..., X ) = ω ((D f )X ,..., (D f )X ). is the subspace of those k-linear maps Vk → R that are p 1 k f (p) p 1 p k alternating. (Of course this vanishes if k > m.) As a special case, if While it is easy to construct the wedge product on multi- f : M → N is the embedding of a submanifold, then f ∗ω = vectors as the image of the tensor product under the quo- ω|M is the restriction of ω to the submanifold M, in the ∗ tient map, the dual wedge product on Λ V requires con- sense that we consider only the values of ωp(X1,..., Xk) structing a map to the alternating subspace. For ω, η ∈ for p ∈ M ⊂ N and Xi ∈ T p M ⊂ T pN.
20 J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16
Exercise B3.2. Pullback commutes with wedge product in Theorem B4.3. For any manifold Mm, the differential map the sense that d : Ω0 M → Ω1 M has a unique R-linear extension to an antiderivation d : Ω∗ M → Ω∗ M satisfying d2 = d ◦ d = 0. f ∗(ω ∧ η) = ( f ∗ω) ∧ ( f ∗η) This antiderivation has degree 1 in the sense that it sends Ωk M to Ωk+1 M; it is called the exterior derivative. for f : M → N and ω, η ∈ Ω∗N. Proof. First suppose g, f i ∈ C∞ M so that g d f 1∧· · ·∧d f k ∈ In a coordinate chart (U, ϕ) we have discussed the coordi- Ωk M. The two conditions on d together automatically im- nate bases {∂ } and {dxi} for T M and T ∗ M, respectively, i p p ply that the pullbacks under ϕ of the standard bases on Rm. Simi- larly, d g d f 1 ∧ · · · ∧ d f k = dg ∧ d f 1 ∧ · · · ∧ d f k ∈ Ωk+1 M. n o i1 ik dx ∧ · · · ∧ dx : 1 ≤ i1 < ··· < ik ≤ m In a coordinate chart (U, ϕ) of course every k-form ω can be expressed as a sum of terms of this form. The propo- forms the standard coordinate basis for k-forms; any ω ∈ sition above shows we can work locally in such a chart. k Ω (M) (or more properly its restriction to U) can be ex- Thus we know the exterior derivative (if it exists) must be pressed uniquely as given in coordinates by X X X X X | i1 ∧ · · · ∧ ik I I i I ω U = ωi1···ik dx dx d ωIdx = dωI ∧ dx = ∂iωI dx ∧ dx i1<···
End of Lecture 7 Dec 2015 f ∗ω = ( f ∗a) d f i1 ∧ · · · ∧ d f ik
21 J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16 and so An equivalent definition of orientation (analogous to that of smooth structures) is through a coherently oriented atlas d( f ∗ω) = d( f ∗a) ∧ d f i1 ∧ · · · ∧ d f ik for M. Here two charts (U, ϕ) and (V, ψ) are coherently −1 = f ∗(da) ∧ d f i1 ∧ · · · ∧ d f ik oriented if the transition function ϕ ◦ ψ is an orientation- preserving diffeomorphism of Rm. ∗ i i ∗ = f da ∧ dy 1 ∧ · · · ∧ dy k = f (dω). Suppose now Mm is an oriented Riemannian manifold. At m any p ∈ M there is a unique Ωp ∈ Λ T p M such that The contraction of a form with a vector field (also known Ωp(e1,..., em) = +1 for any oriented orthonormal basis as interior multiplication) has a seemingly trivial defini- { } k k−1 e1,..., em for T p M. These fit together to give the Rie- tion: if ω ∈ Ω M and X ∈ X(M) then ιXω ∈ Ω is given mannian volume form Ω ∈ Ωm M. In terms of the Hodge by star, we have Ω = ?1. Given an oriented coordinate chart (U, ϕ) then at any p ∈ ιXω(X2,..., Xk):= ω(X, X2,..., Xk). U we have the coordinate basis {∂i} for T p M but can also choose an oriented orthonormal basis {e }. Then of course First note that this is a purely pointwise operation, so we k for some matrix A = ak we have ∂ = P ake . Since could define it on ΛkV for a single vector space – even i i k i k he , e i = δ , we get proving the next proposition at that level – but we won’t k ` k` bother. (It is the adjoint of the operator on Λ V given by DX X E X k g = ∂ , ∂ = ake , a`, e = akak. left multiplication by X.) i j i j i k j ` i j k 0 Next note that for a 1-form, ιX(ω) = ω(X) ∈ Ω M. For a 0 ∞ T 0-form f ∈ Ω M = C M we set ιX f = 0. As a matrix equation, we can write gi j = A A, which implies det(g ) = (det A)2. Since both bases are positively i j p Proposition B4.5. For any X, the operation ιX is an an- oriented, we know det A > 0, so det A = + det g. (Note tiderivation on Ω∗ M of degree −1 whose square is zero. that while abbreviating det(gi j) as det g is common, it un- fortunately hides the fact that this is an expression in par- Proof. It is clear that ιX ◦ ιX = 0 since ticular coordinates.) Now we compute ιXιXω(...) = ω(X, X,...) = 0. p The antiderivation property is Ωp(∂1, . . . , ∂m) = (det A) Ωp(e1,..., em) = det A = det g.
k Equivalently, we have the coordinate expression ιX(ω ∧ η) = (ιXω) ∧ η + (−1) ω ∧ (ιXη) p Ω = det g dx1 ∧ · · · ∧ dxm. for ω ∈ Ωk M; we leave the proof as an exercise. On an oriented Riemannian manifold (Mm, g), any m-form We will later discuss Cartan’s Magic Formula, relating this ω is a multiple ω = f Ω = ? f of the volume form Ω, with contraction to exterior and Lie derivatives. f ∈ C∞ M.
B5. Volume forms and orientation B6. Integration
An orientation on an m-dimensional vector space V is a We will base our integration theory on the Riemann inte- choice of component of ΛmV r {0} R r {0}, that is a gral. Recall that given an arbitrary real-valued function f choice of a nonzero m-form in ω ∈ ΛmV (up to positive on a box B = [a , b ] × · · · × [a , b ] ⊂ Rm we define up- real multiples). If V is oriented by ω, then an ordered ba- 1 1 m m per and lower Riemann sums over arbitrary partitions into sis {e ,..., e } for V is said to be positively oriented if 1 m small boxes – the function f is Riemann integrable if these ω(e ,..., e ) > 0. Often an orientation on V is defined 1 m have the same limiting value, which we call through such a basis (to avoid the machinery of the exte- rior algebra). Z f dx1 ··· dxm. A volume form on a is a nowhere vanishing m-form ω ∈ B Ωm M. We say M is orientable if it admits a volume form. (The Möbius strip and the Klein bottle are exam- Recall also that A ⊂ Rm has (Lebesgue) measure zero if ples of nonorientable 2-manifolds.) An orientation of M for each ε > 0 there is a covering of A by countably many is a choice of volume form, up to pointwise multiplication boxes of total volume less than ε. The image of a set of by positive smooth functions λ > 0 ∈ C∞ M. This is the measure zero under a diffeomorphism (or indeed under any same as a continuous choice of orientations of the tangent locally Lipschitz map) again has measure zero. Thus we spaces T p M. A connected orientable manifold has exactly can also speak of subsets of measure zero in a manifold M. two orientations. Given a function f : D → R with D ⊂ Rm, we define its The standard orientation on Rm is given by dx1 ∧· · ·∧dxm, extension by zero f¯: Rm → R by setting f¯ = f on D so that {e1,..., em} is an oriented basis for each T p M. and f¯ = 0 elsewhere. Lebesgue proved the following: A
22 J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16 bounded function f : D → R defined on a bounded do- First consider a single (oriented) chart (U, ϕ) and assume m ¯ m main D ⊂ R is Riemann integrable if and only if f is ω ∈ Ωc U. Then we define continuous almost everywhere, meaning that its set of dis- Z Z continuities has measure zero. ω := ϕ−1∗ω. U ϕ(U) For instance, the characteristic function χD is Riemann in- tegrable if D is bounded and its boundary ∂D has measure We claim this is independent of ϕ: if (U, ψ) is another ori- zero. Then we call D a domain of integration. ented chart, then using the diffeomorphism ϕ ◦ ψ−1 we find Because a continuous function f on a compact set is Z Z Z −1∗ −1∗ −1∗ −1∗ bounded, we find: If U ⊂ Rm is open and f : U → R ϕ ω = ϕ ◦ ψ ϕ ω = ψ ω. has compact support in U, then f is Riemann integrable. ϕ(U) ψ(U) ψ(U) k k We write Ωc M ⊂ Ω M for the subspace of k-forms with In general, we choose a partition of unity { fα} subordinate compact support. (If M is compact, then of course Ωk = m c to an oriented atlas (Uα, ϕα) . For any ω ∈ Ωc M, note k P Ω .) that ω = α fαω is a finite sum and each summand has compact support in the respective U . We define Definition B6.1. If ω ∈ ΩmU is an m-form with compact α c Z Z support in U ⊂ Rm then of course we can write uniquely X ω := f ω. ω = f dx1 ∧ · · · dxm. We define α M α Uα Z Z Z We just need to check this is independent of the choice of ω = f dx1 ∧ · · · ∧ dxm := f dx1 ··· dxm. U U U atlas and partition of unity. So suppose {gβ} is a partition of unity subordinate to an- m m Note that we use the standard basis element for Λ R other oriented atlas (Vβ, ϕβ) . Then we have R 2 1 here. Otherwise we have for instance f dx ∧ dx = Z Z R 1 2 X X X − f dx dx . fαω = fα gβω α Uα α Uα β Lemma B6.2. If ϕ: U → V is a diffeomorphism of con- Z Z m X X X X nected open sets in R and ω an m-form with compact = fαgβω = fαgβω. support in V, then α β Uα α β Uα∩Vβ Z Z But by symmetry, we see that the last expression also ϕ∗ω = ± ω, P R equals β gβω, as desired. U V Vβ − Note: If M denotesR the manifoldR M with opposite orien- where the sign depends on whether ϕ is orientation- tation, then we have ω = − ω preserving or not. −M M Note: for m = 0, a compact oriented 0-manifold is a finite collection of points with signs ±1: we write M = P p − Proof. Use xi for the standard coordinates on U and y j for P i 1 m q j. (Here we cannot use charts to test orientation.) The those on V. Then ω = f dy ∧ · · · ∧ dy for some func- integral of a zero-form (function) f : M → R is defined to i i R tion f . Writing ϕ = y ◦ ϕ, the Jacobian matrix of ϕ is be f = P f (p ) − P f (q ). J := (∂ϕi/∂x j). We have dϕi = ϕ∗dyi and so M i i j j We have developed this theory for smooth forms, partly dϕ1 ∧ · · · ∧ dϕm = det J dx1 ∧ · · · ∧ dxm. just because we have no notation for possibly discontin- uous sections of ΛmTM. As long as ω is bounded, van- Thus ishes outside some compact set and is continuous almost everywhere, we can repeat the calculations above with no Z Z R ∗ changes to define ω. ϕ ω = ( f ◦ ϕ) dϕ1 ∧ · · · ∧ dϕm (1) M U U Z End of Lecture 14 Dec 2015 = ( f ◦ ϕ) det J dx1 ∧ · · · ∧ dxm. (2) U On an oriented Riemannian manifold M (or any manifold with a specified volume form Ω), we define the volume On the other hand, the standard change-of-variabes for- integral of a function f ∈ C∞ M as mula says Z Z Z Z Z Z f d vol := f Ω = ? f. ω = f dy1 ··· dym = ( f ◦ϕ) | det J| dx1 ∧· · ·∧dxm. M M M V V U Note that if we switch orientation, the volume form on −M is −Ω, so the volume integral is independent of orientation: Since U is connected, det J has a constant sign, depending R R f d vol = f d vol. on whether ϕ is orientation-preserving. −M M For a domain D ⊂ M (compact with boundary of measure zero) we define its volume to be Now suppose Mm is an oriented manifold, and ω ∈ Ωm M c Z Z Z isR a compactly supported m-form. Then we will define ω ∈ R vol D := 1 d vol = Ω := χDΩ > 0. M . D D M
23 J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16
R The volume of the manifold is vol(M):= M 1 d vol = These special cases are of course normally formulated not R with differential forms and the exterior derivative, but with M Ω. This works directly if M is compact; for a non- compact manifold we can take a limit over an appropriate gradients of functions, and divergence and curl of vector compact exhaustion and reach either a finite value or +∞. fields. More precisely, on any Riemannian manifold, we ∗ use the inner product to identify T p M and T p M and thus vector fields with one-forms. The gradient ∇ f of a function ∞ B7. Manifolds with boundary f ∈ C M is the vector field corresponding in this way to d f . In particular, for any vector field X, we have
Suppose Mm is an manifold with boundary; its boundary g(∇ f, X) = h∇ f, Xi = d f (X) = X f. ∂M is an (m − 1)-manifold. At p ∈ ∂M ⊂ M we see On R3 we further use the Hodge star to identify vectors that T p∂M ⊂ T p M is a hyperplane, cutting T p M into two parts, consisting of the inward- and outward-pointing vec- with pseudovectors and thus one-forms with two-forms, tors at p. and to identify scalars with pseudoscalars and thus zero- forms with three-forms. Then div, grad and curl are all An orientation on M induces an orientation on ∂M as fol- just the exterior derivative. Explicitly, we identify both the lows. Suppose (v, v1,..., vm−1) is an oriented basis for one-form p dx + q dy + r dz and the two-form p dy ∧ dz + T p M, where v is outward-pointing and vi ∈ T p∂M. Then q dz ∧ dx + r dx ∧ dy with the vector field p∂x + q∂y + r∂z, (v1,..., vm−1) is by definition an oriented basis for T p∂M. and the three-form f dx∧dy∧dz with the function f . Then (There are four obvious possible conventions here – either d : Ω0 → Ω1 is the gradient as above, d : Ω1 → Ω2 is the and inward- or outward-pointing vector could be put either curl, and d : Ω2 → Ω3 is the divergence. before or after the basis for T p∂M. Our convention works best for Stokes’ Theorem. Our version of Stokes’ theorem is (as mentioned above) certainly not the most general. For instance, we could eas- Equivalently, suppose the orientation of M is given by a ily allow “manifolds with corners”, like compact domains ∈ X volume form Ω, and we pick a vector field X (M) with piecewise smooth boundaries. (It should be clear that which is outward-pointing along ∂M. Then the contraction the divergence theorem in R3 is valid for a cube as well as ιX(Ω) restricted to ∂M is a volume form on the boundary a sphere.) which defines its orientation. Theorem B8.2 (Stokes). Suppose Mm is an oriented man- ifold with boundary and ω is an (m − 1)-form on M with B8. Stokes’ Theorem compact support. Then Z Z m dω = ω. Suppose M is an oriented manifold with boundary and ω M ∂M − is an (m 1)-formR with compactR support on M. Stokes’ ω ω 2 Proof. Both sides are linear and integrals are defined via Theorem then says M d = ∂M . We see d = 0 is dual to the condition that ∂(∂M) = ∅: partitions of unity. In particular Z Z Z X X X X 0 = d2η = dη = η = 0. dω = d( fαω) = d fα ω + fα dω = fα dω, M ∂M ∂∂M we see that it suffices to consider the case when ω is Stokes’ Theorem is quite fundamental, and can be used compactly supported inside one oriented coordinate chart for instance to define dω for nonsmooth forms, or ∂M for (U, ϕ). We may also assume that ϕ(U) = Rm or ϕ(U) = generalized surfaces M. m R H , depending on whether U is disjoint from ∂M or not. Remark B8.1. Of course in ω, the integrand is really Since the statement of the theorem is invariant under pull- ∂M ∗ the restriction or pullback ω|∂M = i ω of ω to ∂M. This is back by a diffeomorphism, we have shown it suffices to now a top-dimensional form on the (m − 1)-manifold ∂M. consider the cases (a) M = Rm and (b) M = Hm. ∅ After scaling, we can assume that ω is compactly sup- When M is a manifold without boundaryR (∂M = ) of ω ported within the cube (a) Q := (−1, 0)m or (b) Q := course Stokes’ Theorem reduces to M d = 0. It turns out m (−1, 0] × (−1, 0)m−1. In either case, we write that on a connected orientable closed manifold M , anR m- form η can be written as dω for some ω if and only if η M Xm vanishes; we will return to such questions after proving the ω = (−1) j−1ω j dx1 ∧ · · · ∧ dxcj ∧ · · · ∧ dxm theorem. j=1 Stokes himself would probably not recognize this gener- alized version of his theorem. The modern formulation in with supp ω j ⊂ Q, so that terms of differential forms is due mainly to Élie Cartan. X ∂ω j The classical cases are those in low dimensions. For M an 1 ∧ · · · ∧ m dω = j dx dx , interval (m = 1), we just have the fundamental theorem ∂x of calculus; for a domain in R2, we have Green’s theorem; meaning that for a domain in R3, we have Gauss’s divergence theorem; Z Z R3 X ∂ω j and for a surface with boundary in we have the theorem 1 ··· m dω = j dx dx . attributed to Stokes. M Q ∂x
24 J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16
Now for each j we have Theorem B9.2. If Mm is an orientable closed manifold with n components, then H0(M) Rn. Z ∂ω j Z 0 Z 0Z 0 ∂ω j ··· j 1 ··· cj ··· m j = j dx dx dx dx . 0 0 0 Q ∂x −1 −1 −1 ∂x Proof. Note that B = 0 so H = Z , which is the space of functions with vanishing differential. But these are just By the fundamental theorem of calculus, the inner inte- the locally constant functions, so it is clear this space is gral in parentheses equals ω j(..., 0,...) − ω j(..., −1,...). n-dimensional. Since ω has compact support in Q, this vanishes for j > 1. In caseR (a) it vanishes even for j = 1, completing the proof For orientable closed manifolds Mm, Poincaré duality (re- that dω = 0. M lated to the Hodge star operation) gives a connection be- In case (b) we have obtained tween co/homology in complementary dimensions. As Z Z 0 Z 0 an example, if such a manifold has n components, then dω = ··· ω1(0, x2,..., xm) dx2 ··· dxm. Hm(M) Rn. We prove the dimension is at least this big. Hm −1 −1 m m Theorem B9.3. If M is an orientable closed manifold Now consider the restriction of ω to ∂H , the pullback with n components, then Hm(M) has dimension at least n. under the inclusion map i. Since i∗dx1 = 0 we immediately get Proof. Denote theR components by Mi. By Stokes, inte- ∗ 1 2 m gration ω 7→ ω over each component gives a map i ω = ω dx ∧ · · · ∧ dx . Mi m m m m R Ω = Z → R vanishing on B , and thus a map H → R; ω n Comparing this to the formula for M d shows we are together these give a map to R . Choosing a Riemannian done. metric on Mi, its volume form has positive integral; these n forms show that our map Hm → Rn is surjective.
B9. De Rham cohomology B10. Lie derivatives Definition B9.1. We say a k-form ω on Mm is closed if dω = 0; we say ω is exact if there is a (k − 1)-form η such Earlier we defined the Lie derivative of a vector field Y k k+1 that dη = ω. For clarity, write dk := d|Ωk : Ω → Ω . with respect to a vector field X. This is a derivative along We write Bk(M) for the space of exact forms and Zk(M) the integral curves of X, where we use (pushforwards un- k for the space of closed forms. That is, Z = ker dk and der) the flow ϕt of X to move vectors of Y between different k B = Im dk−1. points along these curves. Since by definition d2 = 0, it is clear that exact forms are The Lie derivative of a differential k-form ω is defined in closed. (Algebraically, we have Bk ⊂ Zk ⊂ Ωk.) An in- the same way, except that the pushforward under ϕ−t is teresting question is to what extent the converse fails to be replaced by a pullback under ϕt. That is, we define: true. The answer is measured by the de Rham cohomology d ∗ d ∗ k k k H (M):= Z /B , the quotient vector space (over R). A LXω p := ϕt ωϕt(p) = (ϕt ω)p. dt t=0 dt t=0 typical element is the equivalence class [ω] = {ω + dη} of a closed k-form ω. Note that this is again a k-form. In the particular case of ∞ If we consider all degrees k together, we set k = 0 where ω = f ∈ C M we can ignore the pullback – LX f is simply the derivative of f along the integral curve, 0 m 0 m Z := Z ⊕ · · · ⊕ Z = ker d, B := B ⊕ · · · ⊕ B = Im d. that is, LX f = X f .
Defining Proposition B10.1. The Lie derivative LX on forms satis- fies the following properties: H := Z/B = H0 ⊕ · · · ⊕ Hm 1. it is a derivation on Ω∗ M, that is, an R-linear map we find this cohomology ring is not just a vector space but satisfying indeed an algebra under the wedge product. To check the 0 details, start by noting that if ω is closed, then LX(ω ∧ η) = (LXω) ∧ η + ω ∧ (LXη); 0 0 0 (ω + dη) ∧ ω = ω ∧ ω + d(η ∧ ω ). 2. it commutes with the exterior derivative, that is,
An important theorem in the topology of manifolds says LX(dω) = d(LXω); that this cohomology agrees with other standard defini- tions, in particular that it is dual to singular homology. 3. it satisfies the “product” formula – for a k-form ω (This is defined via cycles of simplices modulo boundaries, applied to k vector fields Yi ∈ X(M) we have and can be thought of as counting loops or handles in di- mension k.) The key here is Stokes’ Theorem: a closed LX ω(Y1,..., Yk) form integrates to zero over any boundary, so closed forms k X can be integrated over homology classes. Furthermore an = (LXω)(Y1,..., Yk) + ω Y1,..., LXYi,..., Yk . exact form integrates to zero over any cycle. i=1
25 J.M. Sullivan, TU Berlin B: Differential Forms Diff Geom II, WS 2015/16
Proof. 1. This follows directly from the fact that pull- Proof. We know that LX is a derivation commuting with d. back commutes with wedge product and from the Since d2 = 0, it is easy to check the right-hand side also product rule for d/dt: commutes with d. Furthermore it is a derivation: for ω ∈ Ωk M we get
d ∗ LX(ω ∧ η) p = ϕt (ω ∧ η) p dt t=0 dιX(ω ∧ η) + ιXd(ω ∧ η) d k ∗ ∗ = d((ιXω) ∧ η) + (−1) d(ω ∧ ιXη) = (ϕt ω)p ∧ (ϕt η)p dt t=0 k ! ! + ιX((dω) ∧ η) + (−1) ιX(ω ∧ dη) d d ∗ ∗ k = (ϕt ω)p ∧ ηp + ωp ∧ (ϕt η)p = (dιXω) ∧ η + (−1) (ιXω) ∧ (dη) + ··· dt t=0 dt t=0 = (dιXω) ∧ η + (ιXdω) ∧ η + ω ∧ (dιXη) + ω ∧ (ιXdη). = (Lxω)p ∧ ηp + ωp ∧ (LXη)p. Thus if the formula holds for ω and η, it also holds for 2. This follows from the fact that d is linear and com- ω ∧ η and for dω. By linearity and locality, this means it is mutes with pullback: enough to check it for 0-forms:
d ∗ d ∗ (dι + ι d) f = ι d f = (d f )(X) = X f = L f. LXdω = ϕt dω = dϕt ω X X X X dt t=0 dt t=0
d ∗ Proposition B10.3. Suppose X and Y are vector fields on = d ϕt ω = dLXω. m dt t=0 M and ω is a 1-form. Then 3. The proof follows (as for the product rule for d/dt) dω(X, Y) = Xω(Y) − Yω(X) − ω([X, Y]). from a clever splitting of one difference quotient into two or more. We will write out the proof only for Proof. We use Cartan’s Magic Formula and the product k = 1, considering ω(Y). We find rule for LXω: 1 dω(X, Y) = (ιXdω)(Y) LX ω(Y) = lim ωϕ p Yϕ p − ωp(Yp) p t→0 t t t = (LXω)(Y) − (dιXω)(Y) 1 = lim ωϕ p Yϕ p − ωp ϕ−t∗Yϕ p t→0 t t t t = X ω(Y) − ω [X, Y] − d ω(X) (Y) 1 = X ω(Y) − ω [X, Y] − Y ω(X) . + lim ωp ϕ−t∗Yϕ p − ωp(Yp) . t→0 t t Note that by linearity and locality it suffices to con- Here the second limit clearly gives sider ω = f dg. So an alternate proof simply computes each term for this case, getting for instance Xω(Y) = d ωp ϕ−t∗Yϕt p = ωp LXY . X( f dg(Y)) = X( f Yg) = (X f )(Yg) + f XYg. dt t=0 Theorem B10.4. Suppose ω ∈ Ωk(Mm) is a k-form and For the first limit, we can rewrite the first term as ∗ X0,..., Xk ∈ X(M) are k + 1 vector fields. Then ϕt ωϕt p ϕ−t∗Yϕt p , so that both terms are applied to the same vector. The limit becomes (dω)(X0,..., Xk) ∗ X ϕ (ωϕ p) − ωp i t t = (−1) Xi ω(X0,..., Xbi,..., Xk) lim ϕ−t∗Yϕt p , t→0 t 0≤i≤k X − i+ jω , , ,..., ,..., ,..., . where the form clearly limits to (LXω)p and the vec- + ( 1) [Xi X j] X0 Xbi Xbj Xk) 0≤i< j≤k tor to Yp. Note that the case k = 0 is simply d f (X) = X f , and the case k = 1 is the last proposition. The general proof by End of Lecture 4 Jan 2016 induction on k is left as an exercise; the hint is to use Car- tan’s magic formula as in the proof of the proposition to Since the Lie derivatives of functions and vector fields are write known, we can rewrite the product formula as a formula for LXω as follows: (dω)(X0,..., Xk)
= (LX ω)(X1,..., Xk) − (dιX ω)(X1,..., Xk). (LXω)(Y1,..., Yk) 0 0 k X = X ω(Y1,..., Yk) − ω Y1,..., [X, Yi],..., Yk . i=1
Proposition B10.2 (Cartan’s magic formula). For any vector field X we have LX = dιX + ιXd.
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