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 $k$?

  • $\begingroup$ The control sequence you want is \check{H}. $\endgroup$ – user64687 Feb 12 '14 at 8:58
  • 7
    $\begingroup$ On question (1): the complex of sheaves is exact, the corresponding complex of global sections usually is not. $\endgroup$ – Zhen Lin Feb 12 '14 at 9:01
  • $\begingroup$ So I am right to think that if I took the global sections of the sequence I would get the de Rham complex? $\endgroup$ – Joe Tait Feb 12 '14 at 9:03
  • $\begingroup$ You are right.. $\endgroup$ – user27126 Feb 12 '14 at 9:21
  • 1
    $\begingroup$ 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). $\endgroup$ – 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.
  • $\begingroup$ Thank you - I think I was starting to think along those lines, but it is a lot clearer now. $\endgroup$ – Joe Tait Feb 13 '14 at 10:26
  • $\begingroup$ How does this work for the degree 1 case? $\endgroup$ – Avi Steiner Apr 23 '14 at 16:33
  • $\begingroup$ Direct calculation. $\endgroup$ – Zhen Lin Apr 23 '14 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.