Alright, this is a follow-up to my previous question (Spectral sequences in Hypercohomology of sheaves), sorry I took so long to reply. Let $X$ be a topological space, let $F^\bullet$ be a cochain complex of sheaves, I want to compute the cohomology of the complex

$F^1(X) \rightarrow F^2(X) \rightarrow F^3(X) \rightarrow \cdots$,

specfically I need to prove that the cohomology is $0$ for $q \geq 1$ . I'm trying to use hypercohomology to obtain the cohomology. The hypercohomology of $F^\bullet$ is given by


Now let's assume that the sheaves $F^q$ are acyclic for all $q$, that means that $H^n(X,F^q) = 0$ for all $n \geq 1$ and it follows that $C^1(F^q)(X) \xrightarrow{d} C^2(F^q)(X) \xrightarrow{d} C^3(F^q)(X) \xrightarrow{d} \cdots$ is exact. This means that $''E^{p,q}_2 =$ The p-th cohomology group of the complex

$H^q_d(C^\bullet(F^0)(X)) \xrightarrow{\delta} H^q_d(C^\bullet(F^1)(X)) \xrightarrow{\delta} H^q_d(C^\bullet(F^2)(X)) \xrightarrow{\delta} \cdots$

is $0$ for $q \geq 1$,

(Here $d:C^p(F^q) \rightarrow C^{p+1}(F^q)$, and $\delta:C^p(F^q) \rightarrow C^p(F^{q+1})$).

So if I compute $''E^{p,1}_3$ I get

$''E^{p,1}_3 = \ker(''E^{p,1}_2 \rightarrow ''E^{p+2,0}_2)/Im(''E^{p-2,1+2-1}_2 \rightarrow ''E^{p,1}_2)$,

which reduces to

$''E^{p,1}_3 = \ker(0 \rightarrow ''E^{p+2,0}_2)/Im(0 \rightarrow 0)$,

but $\ker(0 \rightarrow ''E^{p+2,0}_2)$ is $0$ since $0$ is mapped to $0$ in $''E^{p+2,0}_2$ and I end up with

$''E^{p,1}_3 = 0$.

Similarly, if I solve for $''E^{p,1}_4 = 0$ I obtain

$''E^{p,1}_4 = \ker(''E^{p,1}_3 \rightarrow ''E^{p+3,1-3+1}_3)/Im(''E^{p-3,1+3-1}_3 \rightarrow ''E^{p,1}_3)$,

this reduces to

$''E^{p,1}_4 = \ker(0 \rightarrow ''E^{p+3,-1}_3)/Im(0 \rightarrow 0)$,

here again $\ker(0 \rightarrow ''E^{p+3,-1}_3)$ is $0$ so

$''E^{p,1}_4 = 0$.

If I keep going I find that $''E^{p,1}_r$ converges to $0$, which gives me

$\mathbb{H}^{p+1}(X,F^\bullet) = 0$.

Now since the sheaves $F^q$ are acyclic I have that

$\mathbb{H}^{n}(X,F^\bullet) \cong H^n(H^0(X,F^\bullet))$,

but $H^0(X,F^q) \cong F^q(X)$, and hence

$\mathbb{H}^{n}(X,F^\bullet) \cong H^n(F^\bullet(X)) = 0$

for $q \geq 1$. Which is what I'm looking for no?

Where am I wrong??


You have shown that $''E_\infty^{p,1}=0$. The same is true for $''E_\infty^{p,q}$ for all $q>0$. But it is not necessarily true for $''E_\infty^{p,0}$, which contributes to ${\mathbb H}^p(X,F^\bullet)$.

Thus you've shown that ${\mathbb H}^n(X,F^\bullet)=''E_\infty{n,0}=''E_2^{n,0}$, which in turn is the $n$th cohomology of the complex $...F^p(X)\rightarrow F^{p+1}(X)...$

And this could be anything. Which means that you cannot hope to draw any conclusions unless you make some additional assumptions.

So where you went wrong is in jumping from $''E_\infty^{p,1}=0$ to ${\mathbb H}^n=0$.

Edited to add: You seem to be interested in the cohomology of your complex $F^\bullet(X)$. This is (pretty much by definition, which your spectral sequence computation confirms) the same thing as the hypercohomology of the complex of sheaves $F^\bullet$ (given your acyclicity hypothesis). If you want to get any mileage out of this fact, you've got to use some properties of hypercohomology. For example, hypercohomology sits in exact sequences involving hypercohomology-with-supports and relative hypercohomology. That particular property might or might not be useful in your particular circumstances, but you need to use some properties if you're ever going to reach a conclusion.

  • $\begingroup$ Thank you, thank you so much for taking time to reply, I thought that I was showing that $''E_{\infty}^{p,q} = \mathbb{H}^{p+q}(X,F^\bullet) = 0$ for $q \geq 1$ and any $p$, which means that $\mathbb{H}^{0+1}(X,F^\bullet) = 0$, $\mathbb{H}^{1+1}(X,F^\bullet) = 0$, $\mathbb{H}^{2+1}(X,F^\bullet) = 0$, $\mathbb{H}^{3+1}(X,F^\bullet) = 0$..., for $p = 0,1,2,3,...$ respectively? Perhaps this assumption is where I'm wrong? can even get away with finding the cohomology for $q \geq 2$. I didn't say that $''E_{\infty}^{p,0} = \mathbb{H}^{p+0}(X,F^\bullet) = 0$, thank you again $\endgroup$ – Louis A May 1 '12 at 15:16
  • $\begingroup$ You go wrong exactly at the phrase "which means that". $\endgroup$ – Steven Landsburg May 1 '12 at 16:16

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