Mathematics.

algebraic topology

De Rham Cohomology

Topology120 minDifficulty9 out of 10

You should know: cohomology

Overview

De Rham cohomology computes the cohomology of a smooth manifold using differential forms rather than singular cochains. The k-th de Rham cohomology group H^k_{dR}(M) consists of closed k-forms modulo exact k-forms. The de Rham theorem establishes a canonical isomorphism between de Rham cohomology and singular cohomology with real coefficients, connecting analysis and topology.

Intuition

Integration is the bridge between analysis and topology in de Rham theory. A closed form is one whose exterior derivative is zero; an exact form is the exterior derivative of something. A closed form that is not exact detects a 'hole': integration around a loop enclosing a hole gives a nonzero result. Stokes' theorem ensures that exact forms integrate to zero on cycles, making this notion well-defined.

Formal Definition

Definition

Let M be a smooth n-manifold and Omega^k(M) the space of smooth k-forms. The exterior derivative d: Omega^k(M) -> Omega^{k+1}(M) satisfies d^2 = 0, giving the de Rham complex. The k-th de Rham cohomology group is H^k_{dR}(M) = ker(d: Omega^k -> Omega^{k+1}) / im(d: Omega^{k-1} -> Omega^k). The de Rham theorem states H^k_{dR}(M) ≅ H^k(M; R) for all k.

0Ω0(M)dΩ1(M)ddΩn(M)00 \to \Omega^0(M) \xrightarrow{d} \Omega^1(M) \xrightarrow{d} \cdots \xrightarrow{d} \Omega^n(M) \to 0
De Rham complex
HdRk(M)=ker ⁣(d:ΩkΩk+1)/im ⁣(d:Ωk1Ωk)H^k_{\mathrm{dR}}(M) = \ker\!\left(d:\Omega^k\to\Omega^{k+1}\right) \big/ \operatorname{im}\!\left(d:\Omega^{k-1}\to\Omega^k\right)
k-th de Rham cohomology
HdRk(M)Hk(M;R)H^k_{\mathrm{dR}}(M) \cong H^k(M;\mathbb{R})
De Rham's theorem
σω=σdω\int_{\partial\sigma} \omega = \int_\sigma d\omega
Stokes' theorem (key tool)

Theorems

Theorem 1: De Rham's Theorem
ForasmoothmanifoldM,thereisanaturalisomorphismHdRk(M)Hk(M;R)forallk,givenbyintegration:[ω](σσω).For a smooth manifold M, there is a natural isomorphism H^k_{\mathrm{dR}}(M) \cong H^k(M;\mathbb{R}) for all k, given by integration: [\omega] \mapsto (\sigma \mapsto \int_\sigma \omega).
Theorem 2: Poincaré Lemma
EveryclosedformonacontractibleopensubsetURnisexact:HdRk(U)=0fork1.Every closed form on a contractible open subset U \subset \mathbb{R}^n is exact: H^k_{\mathrm{dR}}(U) = 0 for k \geq 1.
Theorem 3: Hodge Decomposition
OnacompactRiemannianmanifold,everycohomologyclassinHdRk(M)hasauniqueharmonicrepresentative:Ωk=dΩk1δΩk+1Hk,whereHk=ker(Δ).On a compact Riemannian manifold, every cohomology class in H^k_{\mathrm{dR}}(M) has a unique harmonic representative: \Omega^k = d\Omega^{k-1} \oplus \delta\Omega^{k+1} \oplus \mathcal{H}^k, where \mathcal{H}^k = \ker(\Delta).

Worked Examples

  1. R^n is contractible, so by the Poincaré lemma every closed form of degree k >= 1 is exact.

  2. A 0-form (smooth function) is closed iff df = 0 iff f is constant. So H^0_{dR}(R^n) = R (constant functions).

  3. For k >= 1: ker(d) = im(d), so H^k_{dR}(R^n) = 0.

    HdRk(Rn)={Rk=00k1.H^k_{\mathrm{dR}}(\mathbb{R}^n) = \begin{cases}\mathbb{R} & k=0 \\ 0 & k\ge 1.\end{cases}

Answer: H^0_{dR}(R^n) = R, H^k_{dR}(R^n) = 0 for k >= 1.

Practice Problems

Difficulty 7/10

Compute H^*_{dR}(S^2).

Difficulty 8/10

Prove that omega = (x dy - y dx)/(x^2 + y^2) is a closed but not exact 1-form on R^2 \ {0}.

Difficulty 9/10

State the Mayer-Vietoris sequence for de Rham cohomology and use it to compute H^*_{dR}(T^2).

Common Mistakes

Common Mistake

Every closed form is exact on any manifold.

The Poincaré lemma guarantees this only on contractible sets. On a circle, dtheta is closed but not exact (its integral around S^1 is 2pi ≠ 0).

Common Mistake

De Rham cohomology captures integer topological information.

De Rham cohomology with real coefficients loses torsion information. For example, it cannot distinguish RP^2 from a point (both have H^1_{dR} = 0), while Z/2 coefficients do.

Quiz

A differential form omega is exact if:
The Poincaré lemma states that on a contractible open set:
De Rham's theorem says H^k_{dR}(M) is isomorphic to:

Summary

  • De Rham cohomology H^k_{dR}(M) is defined as closed k-forms modulo exact k-forms on a smooth manifold.
  • The de Rham theorem establishes H^k_{dR}(M) ≅ H^k(M; R), with the isomorphism given by integration via Stokes' theorem.
  • The Poincaré lemma shows H^k_{dR}(contractible) = 0 for k >= 1, which is the local triviality of the de Rham complex.
  • Hodge theory on compact Riemannian manifolds provides canonical harmonic representatives for each de Rham class.
  • De Rham cohomology is computable and provides a bridge between differential geometry (forms, curvature) and topology (cohomology, characteristic classes).

References

  1. BookBott, R. & Tu, L. — Differential Forms in Algebraic Topology. Springer, 1982.