Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

I first want to clarify the setting for my question.

Start with two nice schemes $S$ and $X$ over a field $k$, e.g. smooth projective varieties, together with a coherent sheaf $P$ on $S\times X$ which is flat over $S$. So one interprets it as a family of sheaves over $X$, parametrized by $S$; with $P_s$ denote the pullback of $P$ via the map

$X\simeq s\times X \rightarrow S\times X$ for a point $s\in S$. So $P_s$ is the sheaf on $X$ corresponding to the point $s$.

The Kodaira-Spencer-map is to be a k-linear map $T_s(S)\rightarrow Ext^1_{X}(P_s,P_s)$ for every closed point $s \in S$ with residue field $k$.

My problem now is the construction of this map. Therefore I recall how it is constructed in most of the books, or rather how I have understood it sofar.

A tangent vector in $s$ shall be given. This corresponds to a $k-$morphism $D \rightarrow S$, where $D$ are the dual numbers, which sends the unique point of $D$ to $s$.

Denote with $F$ the pullback of $P$ via the induced map $D\times X \rightarrow S\times X$. Note that $F$ is flat over $D$ and the pullback of $F$ to $X$ via $X \rightarrow D\times X$ is just $P_s$.

Consider the structure sequence of the unique point of $D$

$0\rightarrow k \rightarrow \mathcal O_D \rightarrow k \rightarrow 0$ on $D$.

Pulling it back to $D\times X$ and tensoring with $F$ gives a sequence of modules on $D\times X$

$0\rightarrow i_*P_s \rightarrow F \rightarrow i_*P_s \rightarrow 0$,

where $i: X \rightarrow D\times X$ is the natural map (remark: often the $i_*$ doesn't appear, but I inserted it because only this way it makes sense for me).

Then the problem: now the authors always write: "if one views this as a sequence of modules on $X$, then one gets the desired class in $Ext^1_{X}(P_s,P_s)$."

I don't understand how this works. I don't think you can just pull it back to $X$ because this doesn't preserve exactness, doesn't it? And secondly, if you pull back to $X$, then your $F$ also becomes $P_s$, which is definitely not what one wants.

I would be very glad if someone could explain to me if the above construction is right in what I said and what one has to do in the last step of the construction.

Thanks a lot!

share|improve this question
3  
You're not pulling back to $X$, but, rather, pushing forward to $X$ along the projection $D\times X\to X$; or, more concretely, you can view the sequence of $O_{D\times X}$-modules as a sequence of $O_X$-modules via the natural inclusion of sheaves $O_X\subsetO_{D\times X=O_{X[ϵ]}$ (note that both rings are sheaves over the same topological space). –  Keerthi Madapusi Pera Sep 4 '11 at 8:10
    
But does the pushforward along $D\times X \rightarrow X$ preserve the exactness of my sequence? –  Descartes Sep 4 '11 at 9:43
3  
Yes, because it is a finite morphism: as pointed out by Keerthi, here the pushforward does not change the sheaves at all since $D\times X$ and $X$ have the same underlying topological space. Alternatively, it's like saying that if we have a ring homomorphism $A \to B$ and we have an exact sequence of $B$ modules then the sequence remains exact if we view it as a sequence of $A$ modules. –  ulrich Sep 4 '11 at 12:57
    
Thank you, Keerthi and ulrich, I think I got the point. –  Descartes Sep 4 '11 at 13:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.