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Suppose $C$ is an abelian category and I am trying to compute $Ext^i(M,N)$ for some objects $M,N$. Suppose there is an exact sequence

$0 \rightarrow A_1 \rightarrow A_2 ... \rightarrow A_n \rightarrow M \rightarrow 0$.

Is it possible to get a double complex where the rows are

$0 \rightarrow Ext^i(A_n,N) \rightarrow ... \rightarrow Ext^i(A_1,N) \rightarrow 0$

and where the vertical rows are $\{Ext^*(A_i,N)\}$, such that the spectral sequence of the double sequence converges to $Ext^*(M,N)$?

I was reading a paper and the author seems to have done this.

Not a grad student
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1 Answers1

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The spectral sequence is not exactly what you said (there would be no vertical rows). In fact, it is one of the hypercohomology spectral sequence, and you described exactly the $E_1$-page.

Let me change the order of your exact sequence so that I get the indices right. So you have a exact sequence $$0\rightarrow A_n\rightarrow\dots\rightarrow A_0\rightarrow M\rightarrow 0.$$ This can be seen as an quasi-isomorphism of complexes $A_*\rightarrow M[0]$ where $M[0]$ is the complex with $M$ concentrated in degree 0.

Because hypercohomology is invariant by quasi-isomorphism, there is an isomorphism $$\operatorname{Ext}^i(M,N)\overset{\sim}\rightarrow\mathbb{Ext}^i(A_*,N)$$ and now the hypercohomology spectral sequence takes the form $$E_1^{pq}=\operatorname{Ext}^q(A_p,N)\Rightarrow \mathbb{Ext}^{p+q}(A_*,N)$$

Roland
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