6.4 de Rham cohomology

De Rham cohomology bridges differential geometry and algebraic topology by studying smooth manifolds through differential forms. It provides a way to measure "holes" in manifolds and understand their topological structure using calculus-like tools.

This approach connects to the broader themes of homology and cohomology theories in the chapter. De Rham cohomology offers a concrete realization of cohomology using differential forms, illustrating how these abstract concepts apply to smooth manifolds.

Differential Forms on Manifolds

Definition and Notation

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• A differential k-form on a smooth n-manifold M is a smooth section of the kth exterior power of the cotangent bundle of M
• The set of all differential k-forms on M is denoted by $\Omega^k(M)$

Exterior Derivative and de Rham Complex

• The exterior derivative d is a linear map from $\Omega^k(M)$ to $\Omega^{k+1}(M)$ satisfying:
  • $d^2 = 0$
  • Leibniz rule: $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\deg(\alpha)} \alpha \wedge d\beta$
• The de Rham complex is the cochain complex $(\Omega^*(M), d)$ consisting of the differential forms on M with the exterior derivative as the coboundary map

Properties of Differential Forms

• Closed forms are differential forms $\alpha$ satisfying $d\alpha = 0$, while exact forms are differential forms $\alpha$ such that $\alpha = d\beta$ for some differential form $\beta$
• The wedge product $\wedge$ is a bilinear map from $\Omega^k(M) \times \Omega^l(M)$ to $\Omega^{k+l}(M)$ that is graded commutative:
  • $\alpha \wedge \beta = (-1)^{kl} \beta \wedge \alpha$ for $\alpha \in \Omega^k(M)$ and $\beta \in \Omega^l(M)$
  • Examples: $dx \wedge dy = -dy \wedge dx$, $dx \wedge dx = 0$

De Rham Cohomology Groups

Construction using the de Rham Complex

• The kth de Rham cohomology group $H^k_{dR}(M)$ is defined as the quotient of the kernel of $d: \Omega^k(M) \to \Omega^{k+1}(M)$ by the image of $d: \Omega^{k-1}(M) \to \Omega^k(M)$
  • $Z^k(M) = \{\alpha \in \Omega^k(M) | d\alpha = 0\}$ is the space of closed k-forms (kernel of d)
  • $B^k(M) = \{\alpha \in \Omega^k(M) | \alpha = d\beta \text{ for some } \beta \in \Omega^{k-1}(M)\}$ is the space of exact k-forms (image of d)
  • $H^k_{dR}(M) = Z^k(M) / B^k(M)$
• Elements of $H^k_{dR}(M)$ are equivalence classes $[\alpha]$ of closed k-forms $\alpha$, where two closed k-forms $\alpha$ and $\beta$ are equivalent if their difference $\alpha - \beta$ is exact

Induced Maps and Graded Algebra Structure

• The exterior derivative d induces a well-defined linear map from $H^k_{dR}(M)$ to $H^{k+1}_{dR}(M)$, which is the zero map due to $d^2 = 0$
• The wedge product of differential forms induces a well-defined graded commutative product on the de Rham cohomology groups, making $H^*_{dR}(M)$ a graded commutative algebra
  • Examples: $[dx] \wedge [dy] = -[dy] \wedge [dx]$, $[dx] \wedge [dx] = 0$ in $H^*_{dR}(\mathbb{R}^2)$

Cohomology Theory Axioms

Homotopy Invariance

• If $f_0, f_1: M \to N$ are homotopic smooth maps, then the induced maps $f_0^*, f_1^*: H^*_{dR}(N) \to H^*_{dR}(M)$ on de Rham cohomology are equal
  • The proof involves constructing a chain homotopy between the pullback maps $f_0^*$ and $f_1^*$ using the homotopy operator and the pullback of differential forms
  • Example: The identity map and a constant map on a manifold induce the same map on de Rham cohomology

Exactness (Mayer-Vietoris Sequence)

• For any pair of smooth manifolds M and N, the Mayer-Vietoris sequence in de Rham cohomology is exact
  • The proof involves constructing a short exact sequence of de Rham complexes using a partition of unity subordinate to an open cover of $M \cup N$ and applying the zig-zag lemma
  • Example: For $M = S^1$ and $N = (0, 2\pi)$, the Mayer-Vietoris sequence relates the cohomology of the circle to that of the interval

Excision

• If U is an open subset of M and K is a compact subset of U, then the inclusion map $i: (M - K, U - K) \to (M, U)$ induces an isomorphism $i^*: H^*_{dR}(M, U) \to H^*_{dR}(M - K, U - K)$ on relative de Rham cohomology
  • The proof involves showing that the restriction map $\Omega^*(M, U) \to \Omega^*(M - K, U - K)$ is a quasi-isomorphism using a smooth bump function supported in U
  • Example: For $M = \mathbb{R}^2$, $U = \mathbb{R}^2 - \{0\}$, and $K = \{x \in \mathbb{R}^2 | |x| \leq 1\}$, the relative cohomology $H^*_{dR}(\mathbb{R}^2, \mathbb{R}^2 - \{0\})$ is isomorphic to $H^*_{dR}(\mathbb{R}^2 - K, \mathbb{R}^2 - \{0\})$

Topology of Smooth Manifolds

Topological Invariance and Betti Numbers

• The de Rham cohomology groups $H^k_{dR}(M)$ are topological invariants of the smooth manifold M, independent of the choice of Riemannian metric or orientation
• The dimension of $H^k_{dR}(M)$ is equal to the kth Betti number of M, which counts the number of independent k-dimensional "holes" in M
  • Examples: For the torus $T^2$, $\dim H^0_{dR}(T^2) = 1$ (connected), $\dim H^1_{dR}(T^2) = 2$ (two independent loops), and $\dim H^2_{dR}(T^2) = 1$ (one cavity)

Cup Product and Intersection Theory

• The cup product on de Rham cohomology corresponds to the intersection of submanifolds in M, providing a way to study the intersection theory of M using differential forms
  • Example: For two closed 1-forms $\alpha$ and $\beta$ on a surface, the cup product $[\alpha] \smile [\beta]$ represents the intersection of the corresponding curves

Integration and Poincaré Duality

• The integration of closed forms over cycles in M defines a non-degenerate pairing between de Rham cohomology and singular homology, which can be used to detect non-trivial topological features of M
• The de Rham cohomology of a compact oriented manifold satisfies Poincaré duality, which relates the kth de Rham cohomology group to the (n-k)th de Rham cohomology group via the wedge product and integration
  • Example: For a compact oriented surface $\Sigma$, $H^0_{dR}(\Sigma) \cong H^2_{dR}(\Sigma)$ and $H^1_{dR}(\Sigma) \cong H^1_{dR}(\Sigma)$ via the Poincaré duality pairing

De Rham vs Singular Cohomology

De Rham Theorem

• The de Rham theorem states that for any smooth manifold M, there is a natural isomorphism between the de Rham cohomology groups $H^k_{dR}(M)$ and the singular cohomology groups $H^k(M; \mathbb{R})$ with real coefficients
  • The isomorphism is given by the de Rham map, which sends a closed k-form $\alpha$ to the real cohomology class represented by the cocycle that evaluates a singular k-simplex $\sigma$ by integrating $\alpha$ over $\sigma$

Proof Steps

• The proof of the de Rham theorem involves several steps:
  • Constructing a cochain map from the de Rham complex to the singular cochain complex using the de Rham map
  • Showing that the de Rham map induces an isomorphism on cohomology by constructing a homotopy inverse using smooth singular cochains and the Whitney approximation theorem
  • Proving that the de Rham map is a ring isomorphism with respect to the wedge product on de Rham cohomology and the cup product on singular cohomology

Consequences and Interpretations

• The de Rham theorem implies that the algebraic and topological properties of the de Rham cohomology groups, such as functoriality, homotopy invariance, and Poincaré duality, also hold for the singular cohomology groups with real coefficients
• As a consequence of the de Rham theorem, the integration of closed forms over cycles in M can be interpreted as the evaluation of real cohomology classes on homology classes, providing a link between differential geometry and algebraic topology
  • Example: For a closed 1-form $\alpha$ on a circle $S^1$ and a loop $\gamma: [0, 1] \to S^1$, the integral $\int_\gamma \alpha$ computes the evaluation of the cohomology class $[\alpha] \in H^1_{dR}(S^1) \cong H^1(S^1; \mathbb{R})$ on the homology class $[\gamma] \in H_1(S^1; \mathbb{R})$