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If I have a morphism of schemes $f:X\rightarrow Y$ and sheaves $\mathcal F,G$ on $X$, then is there a spectral sequence which relates the Ext-groups

$\mathrm{Ext}(f_* \mathcal F, f_*\mathcal G)$ on $Y$ and $\mathrm{Ext}(\mathcal F, \mathcal G)$ on $X$?

I should add, that if this is not possible in general, that my morphism $f$ is actually an affine morphism and that I want to compare the two $\mathrm{Ext}$-groups in any way. The first thing I thought of was a spectral sequence, but perhaps there are other ways you know.

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it's a shame no one gave an answer. have you made any progress with it? –  Jacob Bell Nov 23 '11 at 23:16
    
I have to confess, no. But I quite agree with you and also think that these groups are pretty incomparable... –  Cyril Dec 12 '11 at 16:41
    
have had any success in some particular cases? –  Jacob Bell Dec 13 '11 at 14:53
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2 Answers

Well, a useless thing to say is that in general you have the Grothendieck spectral sequence (cf answer by Matt Emerton Contravariant Grothendieck Spectral Sequence), (I'm thinking of A,B there as your $Rf_*F, Rf_*G$).

But as your morphism is affine then this does not help one bit.

I guess it's impossible a priori to give any comparison: essentially because $f_*$ and $\underline{Hom}$ aren't compatible (which, via $E^\vee \otimes F = \underline{Hom}(E,F)$ is another way of saying that $f_*$ and $\otimes$ aren't compatible).

But I would very like to be proven wrong, as I'm also similarly stuck!

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This probably is not an "answer" since the assumptions are going to different, but still I'd like to write down what I have worked out.

Lemma Suppose that $f: X\to Y$ is flat, then the push forward of an injective sheaf $I$ is injective.

Indeed, we need to show that $Hom_Y(-, f_*I)$ is exact. Since $f$ is flat, the pull-back functor $f^*$ is exact, so the exactness of $Hom_Y(-, f_*I)$ follows from the adjointness of $f^*$ and $f_*$.

Corollay Suppose that $f: X\to Y$ is flat, then there is an spectral sequence $$ Ext^p_Y(f_*\mathcal{F}, R^qf_*\mathcal{G}) \Rightarrow Ext_X^{p+q}(f^*f_*\mathcal{F}, \mathcal{G}).$$

Here we consider the two functors $f_*$ and $Hom_Y(f_*\mathcal{F}, -)$. Their composition is the functor $$\mathcal{G}\mapsto Hom_Y(f_*\mathcal{F}, f_*\mathcal{G})= Hom_X(f^*f_*\mathcal{F}, \mathcal{G}).$$ So the Corollary is a direct consequence of Grothendieck spectral sequence.

The edge morphism gives rise to a map $Ext^p_Y(f_*\mathcal{F}, f_*\mathcal{G})\to Ext^p_X(f^*f_*\mathcal{F}, \mathcal{G})$. On the other hand, we have the adjunction map $f^*f_*\mathcal{F}\to \mathcal{F}$. So there are maps $$Ext^p_Y(f_*\mathcal{F}, f_*\mathcal{G})\to Ext^p_X(f^*f_*\mathcal{F}, \mathcal{G}) \leftarrow Ext^p_X(\mathcal{F}, \mathcal{G}).$$ I wonder if one can relate the two "external" terms in the above formula in general.

However, if $f$ is further assumed to be affine, then $f_*$ is exact, so the spectral sequence degenerates, and we get $Ext^p_Y(f_*\mathcal{F}, f_*\mathcal{G})= Ext^p_X(f^*f_*\mathcal{F}, \mathcal{G})$, and therefore a map $Ext^p_X(\mathcal{F}, \mathcal{G}) \to Ext^p_Y(f_*\mathcal{F}, f_*\mathcal{G})$. Again, since $f_*$ is exact, the existence of this map is obvious from Yoneda's description of elements of $Ext^n$ as extensions of length $n$ (It seems flatness is not needed here?).

Anyway, say $X$ is an affine algebra over a field $k$, and let $Y=Spec(k)$. Then $Ext_Y^n(f_*\mathcal{F}, f_*\mathcal{G})=0$ for all $n>1$ since all $k$-vector spaces are injective $k$-modules. There are certainly "lose of information" by going to the $Ext$ of the push forward.

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