3
$\begingroup$

Let $X$ be a scheme over a field $k$ and $L$ be an invertible sheaf on it. Let $D$ be the scheme over dual numbers over $k$ with parameter $t$, i.e. $Spec(k[t]/(t^2)$.

Let $X':=X \times_k D$ and $i:X \rightarrow X\times D$ the natural map.

One defines a first order deformation of $L$ over $D$ as a line bundle $L'$ on $X'$, such that $i^*L'$ is isomorphic to $L$.

One knows that the iso classes of deformations of $L$ correspond to $Ext^1_{\mathcal O_X}(\mathcal O_X, \mathcal O_X)$.

One map is clear, I think: if you have a deformation $L'$, then consider the exact sequence on $D$:

$0 \rightarrow k \rightarrow \mathcal O_D \rightarrow k \rightarrow 0$, pull it back to $X'=X\times_k D$, tensor with $L'$ and push down to $X$. Please correct me if this is not the right way.

But how do I get explicitly a deformation in the sense of the above definition out of an exact sequence on $X$

$0\rightarrow L \rightarrow M \rightarrow L \rightarrow 0$?

In the books I read there are only hints I really don't understand, so I would be very glad about an answer which carefully constructs the deformation data out of this sequence.

Improve this question
$\endgroup$
2
  • $\begingroup$ A short exact sequence $0 \to L \to M \to L \to 0$ enables you to give $M$ the structure of a module over $X'$. You just need to be able to say how $\epsilon$ acts on $M$. Composing the projection $M \to L$ with the inclusion $L \to M$ tells you how $\epsilon$ acts. $\endgroup$
    – Mike Skirvin
    Commented Nov 8, 2011 at 16:28
  • $\begingroup$ Well, exactly this is what appears in the books and what I don't fully understand, even not in terms of modules. Above all, why is $i^*L'$ isomorphic to $L$? $\endgroup$
    – Veen
    Commented Nov 8, 2011 at 17:12

1 Answer 1

Reset to default
6
$\begingroup$

One important point is that $X$ and $X'$ have the same topological space, since $i:X'\rightarrow X$ is a nilpotent immersion. In particular $M$ is already a sheaf on $|X|=|X'|$ (of $\mathcal{O}_X$-modules). Since $\mathcal{O}_{X'}=\mathcal{O}_X\oplus \mathcal{O}_X[\varepsilon]$, to give $M$ the structure of a sheaf of $\mathcal{O}_{X'}$-modules you only need to define the action of $\varepsilon$, as Mike Skirvin points out in his comment.

So you define multiplication by $\varepsilon$ by the composition $M\stackrel{g}{\to} L\stackrel{f}{\to} M$ of the two maps given by the extension (notice that this is $\mathcal{O}_X$-linear, and its square is zero). This makes $M$ a sheaf of $\mathcal{O}_{X'}$-modules, and moreover using the local criterion of flatness you can check that it is flat over the base $D$.

Now since $|X|=|X'|$ and $\mathcal{O}_X=\mathcal{O}_{X'}/(\varepsilon\cdot\mathcal{O}_{X'})$, you have $i^*(M)=M\otimes_{\mathcal{O}_X'}\mathcal{O}_X=M/(\varepsilon \cdot M)=M/f(L)=L$, since by definition of the action of $\varepsilon$, $\varepsilon \cdot M=f(L)$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ @Mattia: is this $M$ with the $\mathcal O_{X'}$-structure you defined a line bundle on $X'$? It should be given the above definition of deformation. $\endgroup$
    – Veen
    Commented Nov 8, 2011 at 18:09
  • $\begingroup$ I think so, if say $X$ is loc noetherian. Take an open affine $U=Spec A \subseteq X$ (so the corresponding open of $X'$ is the spectrum of $A'=A[\varepsilon]$) and apply the local criterion of flatness for the nilpotent ideal $(\varepsilon)\subseteq A[\varepsilon]$ to the restriction of $M$ to $U$, call $N$ the corresponding $A'$-module. Namely, since $N/(\varepsilon N)$ is flat over $A$ (loc free even) and $Tor_1^{A'}(N,A)=0$ (check that $\cdot \varepsilon: A\to A'$ stays injective after tensoring with $N$), then $N$ is flat over $A'$, so it is locally free. (maybe there's an easier way..) $\endgroup$
    – Mattia Talpo
    Commented Nov 8, 2011 at 20:27
  • $\begingroup$ Well, but one needs locally free of rank one as one wants it to be a line bundle on $X'$. But I think it really is, due to local splittings of the sequence $0\rightarrow L \rightarrow M \rightarrow L \rightarrow 0$ ($L$ is locally just the structure sheaf and so you get a retraction). Does this sound reasonable? $\endgroup$
    – Veen
    Commented Nov 9, 2011 at 12:35
  • $\begingroup$ Once you know that it is locally free, it must be of rank one, since its pullback to $X$ is.. $\endgroup$
    – Mattia Talpo
    Commented Nov 9, 2011 at 13:00
  • $\begingroup$ You helped me a lot, Mattia. Thanks. $\endgroup$
    – Veen
    Commented Nov 10, 2011 at 9:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.