algebraic topology
De Rham Cohomology
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
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.
Theorems
Worked Examples
R^n is contractible, so by the Poincaré lemma every closed form of degree k >= 1 is exact.
A 0-form (smooth function) is closed iff df = 0 iff f is constant. So H^0_{dR}(R^n) = R (constant functions).
For k >= 1: ker(d) = im(d), so H^k_{dR}(R^n) = 0.
Answer: H^0_{dR}(R^n) = R, H^k_{dR}(R^n) = 0 for k >= 1.
Practice Problems
Compute H^*_{dR}(S^2).
Prove that omega = (x dy - y dx)/(x^2 + y^2) is a closed but not exact 1-form on R^2 \ {0}.
State the Mayer-Vietoris sequence for de Rham cohomology and use it to compute H^*_{dR}(T^2).
Common Mistakes
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).
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
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
- BookBott, R. & Tu, L. — Differential Forms in Algebraic Topology. Springer, 1982.
Mathematics