# Contravariant Grothendieck Spectral Sequence

I'm currently getting confused about indices in some spectral sequences. Assume we work in the category of modules for simplicity. Let $A^\cdot$ be a (bounded on the right) complex and let $B^\cdot$ (I don't think we have to assume anything about the boundedness of $B$). I want to compute $Ext^n(A^\cdot,B^\cdot)$, which is classically called hyperext (and sometimes denoted by $\mathbb{E}xt$.

Now, (perhaps assuming $B^\cdot$ to be bounded on the right), there exists a spectral sequence

$$E^{p,q}_2 = Ext^p(A^\cdot,H^q(B^\cdot)) \Rightarrow Ext^{p+q}(A^\cdot,B^\cdot).$$

There should be an analogous by switching A and B, but I'm unsure of the indices, so my question is

is $$E^{p,q}_2 = Ext^q(A^\cdot,H^{-p}(B^\cdot)) \Rightarrow Ext^{q-p}(A^\cdot,B^\cdot)$$ the right thing?

Thanks.

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If you write the dots as \bullet they can be seen :) –  Mariano Suárez-Alvarez Sep 8 '10 at 16:01
Yes, but I don't like the way it looks. Besides, one might as well omit the dot altogether :) –  slackenerny Sep 9 '10 at 8:30

I think that you will want $B^{\bullet}$ to be bounded below (i.e. on the left), so that after replacing either $A^{\bullet}$ by a projective resolution or $B^{\bullet}$ by an injective resolution, the complex $Hom(\text{complex},\text{complex})$ that you compute will be bounded below.

In any case, the spectral sequence you are looking for comes about by taking this $Hom$ complex (the one you get after replacing $A^{\bullet}$ or $B^{\bullet}$ by a projective or injective resolution), writing it as the total complex of a double complex, and applying the standard machine. If you do this, you will find that the second spectral sequence that you want is:

$$E_2^{p,q} = Ext^p(H^{-q}(A^{\bullet}),B^{\bullet}) \Rightarrow Ext^{p+q}(A^{\bullet},B^{\bullet}).$$

(Your first spectral sequence comes from replacing $A^{\bullet}$ by a projective resolution. This one comes from replacing $B^{\bullet}$ by an injective resolution.)

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