# $Ext$ functor, filtered complexes and spectral sequences

Let $\mathcal{A}$ an abelian category. Take $M$ an object of $\mathcal{A}$, and $K_*$ a bounded complex in $\mathcal{A}$ equipped with a bounded increasing filtration $F$. By using homological and functorial properties of the $Ext$ functor, one can construct, for all $k \in \mathbb{Z}$, a bigraded complex

$(Ext^{k-\alpha}(M,\frac{F_{\alpha}K_{\beta}}{F_{\alpha-1}K_{\beta}}))_{(\alpha,\beta) \in \mathbb{Z} \times \mathbb{Z}}$

with decreasing indices $\alpha$ and $\beta$.

This bigraded complex (like any bigraded complex) induces two spectral sequences and, since it is bounded (the complex $K_*$ and its filtration $F$ are bounded), they both converge to the homology of the associated total complex, but a priori this last one has not really a nice expression, and remains a kind of "theoritical" abutment.

My question is then the following : is there a "simple" or "nice" expression for the abutment of the spectral sequences induced by the bigraded complex $(Ext^{k-\alpha}(M,\frac{F_{\alpha}K_{\beta}}{F_{\alpha-1}K_{\beta}}))_{(\alpha,\beta) \in \mathbb{Z} \times \mathbb{Z}}$ ?

By advance, thank you very much.

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How $F_\alpha A_\beta$ is related to the complex $K$? –  Sasha Jun 6 '11 at 17:29
Sorry, it was a mistake, I edited my question : it was a confusion between notations... Thank you for your comment. –  Fabien Priziac Jun 7 '11 at 5:22
I'm curious what exactly you are hoping for? Do you want a "nice" expression in any special cases, or do you want to work in the stated generality. –  Karl Schwede Jun 8 '11 at 14:38
I would like to obtain an abutment "depending" somewhat on the homology of the complexes $\frac{F_{alpha}K_*}{F_{\alpha-1}K_*}$. Actually, the abelian category I am working in is the category of $k$-vector spaces equipped with a linear action of a group $G$ (and equivariant morphisms) and I take $M=k$ so that the double complex above is now $(H^{k-\alpha}(G,\frac{F_{alpha}K_{\beta}}{F_{\alpha-1}K_{\beta}}))$. But I chose to make my question in a more general context because there can be people who know about some results on $Ext$ functor (maybe in litterature) without knowing group cohomology. –  Fabien Priziac Jun 9 '11 at 7:16