H⁰(M) is the space of smooth functions f on M satisfying df = 0. For a connected manifold, what is H⁰(M)?
AThe zero vector space
Bℝ (the constant functions)
CC∞(M) (all smooth functions)
Dℝⁿ where n = dim(M)
A function f with df = 0 has zero derivative everywhere, hence is locally constant. On a connected manifold, locally constant functions are globally constant. So H⁰(M) = {constant functions} ≅ ℝ. For a manifold with k connected components, H⁰(M) ≅ ℝᵏ — one constant for each component. This is the simplest example of how de Rham cohomology encodes topology.
Question 2 True / False
The Poincaré lemma states that on a contractible open set U, every closed form is exact: Hᵏ(U) = 0 for k ≥ 1. This means de Rham cohomology is a global invariant, not a local one.
TTrue
FFalse
Answer: True
Locally (on any sufficiently small open ball), there are no nontrivial cohomology classes — every closed form can be written as dα. Nontrivial cohomology arises only from the global topology of M. For instance, the 1-form dθ on S¹ is closed but not exact (its integral around S¹ is 2π ≠ 0), and this reflects the nontrivial topology of the circle (it has a hole). The Poincaré lemma means de Rham cohomology is entirely a global phenomenon.
Question 3 Short Answer
The de Rham theorem establishes that the de Rham cohomology H*_dR(M) is isomorphic to which other cohomology theory?
Think about your answer, then reveal below.
Model answer: The singular cohomology H*_sing(M; ℝ) with real coefficients. The isomorphism is given by the integration pairing: a closed k-form ω and a singular k-cycle σ pair to give the number ∫_σ ω. Stokes' theorem ensures this pairing is well-defined on cohomology and homology classes. The de Rham theorem says this pairing is a perfect pairing, establishing an isomorphism H^k_dR(M) ≅ H^k_sing(M; ℝ). This means topological invariants (Betti numbers, cup products) can be computed using differential forms.
The de Rham theorem is remarkable because it equates a construction from analysis (differential forms, exterior derivative) with a construction from algebraic topology (singular chains, boundary operator). It implies that the Betti numbers bₖ = dim Hᵏ(M) are finite for compact manifolds and satisfy Poincaré duality. The theorem justifies using differential forms as a computational tool for topology.
Question 4 True / False
For the n-torus Tⁿ = (S¹)ⁿ, the de Rham cohomology is Hᵏ(Tⁿ) ≅ ℝ^{C(n,k)}, where C(n,k) is the binomial coefficient.
TTrue
FFalse
Answer: True
The n-torus has coordinates (θ₁,...,θₙ), and the closed forms dθ_{i₁} ∧ ... ∧ dθ_{iₖ} for i₁ < ... < iₖ represent a basis for Hᵏ(Tⁿ). There are C(n,k) such forms, so dim Hᵏ(Tⁿ) = C(n,k). The total dimension Σ_k dim Hᵏ = Σ_k C(n,k) = 2ⁿ, and the Euler characteristic χ = Σ(-1)ᵏ C(n,k) = 0 for n ≥ 1. The Künneth formula (cohomology of a product is the tensor product of cohomologies) makes this computation systematic.