# Why are de Rham cohomology and Cech cohomology of the constant sheaf the same

I am comfortable with de Rham cohomology, sheaves, sheaf cohomology and Cech cohomology.

I am looking to prove the following theorem:

If $M$ is a smooth manifold of dimension $m$, then we have the following isomorphism for each $k \leq m$ $$H^k_{\text{dR}}(M) \cong \check{H}^k(M; \mathbb R_M).$$ By $\mathbb R_M$ I meant the constant sheaf of $\mathbb R$ on $M$.

I'll outline the proof I am following and give my two questions as they come up.

If we let $\Omega^k$ denote the sheaf of germs of $k$ forms on $M$ then by the Poincaré lemma we have an exact sequence $$0 \rightarrow \mathbb R_M \rightarrow \Omega^0 \rightarrow \Omega^1 \rightarrow \ldots \rightarrow \Omega^m \rightarrow 0,$$ with differential maps being the exterior derivative.

Why is this sequence exact but the de Rham complex not?

In particular, I think I may not be clear on how a "sheaf of germs" is different from a sheaf? And if I took global sections would I get back the usual de Rham complex?

Anyway, assuming that the sequence is exact, we therefore get a series of short exact sequences (SES) of the form $$0 \rightarrow d\Omega^{k-1} \rightarrow \Omega^k \rightarrow d\Omega^k \rightarrow 0.$$ The sheaf $\Omega^k$ is fine, and hence its cohomology $H^i(\Omega^k)$ vanishes for $i>0$. Hence we get an isomorphism $H^i(d\Omega^{k-1}) \cong H^{i+1}(d\Omega^k)$ for each $i$. However, the last sentence of this proof is a mystery to me.

"At one end of the chain is the Čech cohomology and at the other lies the de Rham cohomology."

I can kind of see why after taking global sections we get back de Rham cohomology - i.e. the $H^0$ that come from the SES above is $H^k_{\text {dR}}(M)$. And the Cech cohomology coincides with the sheaf cohomology. But I am not sure how to relate them: should I be varying $k$ and get isomorphisms across the long exact sequences corresponding to different $k$?

• The control sequence you want is \check{H}. – user64687 Feb 12 '14 at 8:58
• On question (1): the complex of sheaves is exact, the corresponding complex of global sections usually is not. – Zhen Lin Feb 12 '14 at 9:01
• So I am right to think that if I took the global sections of the sequence I would get the de Rham complex? – Joe Tait Feb 12 '14 at 9:03
• You are right.. – user27126 Feb 12 '14 at 9:21
• Strictly speaking the de Rham complex starts at $\Omega^0$, and so it is never exact (either as a complex of sheaves or the corresponding complex of global sections). – Zhen Lin Feb 12 '14 at 10:00

1. The cochain complex of sheaves $$0 \to \mathbb{R} \to \Omega^0 \to \Omega^1 \to \cdots$$ is exact: this follows from the Poincaré lemma. (Any closed differential $(n+1)$-form on a sufficiently small open neighbourhood must be the exterior derivative of some differential $n$-form.) Thus, the cochain complex $$\Omega^0 \to \Omega^1 \to \Omega^2 \to \cdots$$ is a resolution of the constant sheaf $\mathbb{R}$. This is called the de Rham complex.
2. Let $Z^n = \ker (\Omega^n \to \Omega^{n+1})$ be the sheaf of closed differential $n$-forms. Then we have short exact sequences $$0 \to Z^n \to \Omega^n \to Z^{n+1} \to 0$$ and hence long exact sequences $$0 \to \Gamma (M, Z^n) \to \Gamma (M, \Omega^n) \to \Gamma (M, Z^{n+1}) \to H^1 (M, Z^n) \to \cdots$$ of sheaf cohomology groups. (These can be computed by Čech cohomology on a sufficiently fine open cover.) Since $M$ admits partitions of unity, $H^i (M, \Omega^n) = 0$ for $i > 0$. Thus, for $i > 0$, we have a natural isomorphism $H^i (M, Z^{n+1}) \to H^{i+1} (M, Z^n)$. In particular, $$H^{i+1} (M, Z^0) \cong H^i (M, Z^1) \cong \cdots \cong H^1 (M, Z^i)$$ but $Z^0$ is (isomorphic to) the constant sheaf $\mathbb{R}$, so we deduce $$H^{i+1} (M, \mathbb{R}) \cong \operatorname{coker} (\Gamma (M, \Omega^i) \to \Gamma (M, Z^{i+1}))$$ and since $\Gamma (M, -)$ is left exact, $$0 \to \Gamma (M, Z^{i+1}) \to \Gamma (M, \Omega^{i+1}) \to \Gamma (M, \Omega^{i+2})$$ is an exact sequence, so $\operatorname{coker} (\Gamma (M, \Omega^i) \to \Gamma (M, Z^{i+1}))$ is (isomorphic to) $H_\mathrm{dR}^{i+1} (M)$. Thus we have the de Rham isomorphism in degrees $\ge 2$.
3. In principle we still have to check that $H^0 (M, \mathbb{R}) \cong H_\mathrm{dR}^0 (M)$ and $H^1 (M, \mathbb{R}) \cong H_\mathrm{dR}^1 (M)$. But we have the exact sequence $$0 \to \Gamma (M, Z^0) \to \Gamma (M, \Omega^0) \to \Gamma (M, Z^1) \to H^1 (M, Z^0) \to 0$$ and $\Gamma (M, Z^1) \cong \ker (\Gamma (M, \Omega^1) \to \Gamma (M, \Omega^2))$, so a direct calculation completes the proof.