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

Using the laguage of derived category, the Kodaira-Spencer map

$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$

can be described as a Fourier-Mukai transformation $\Phi_{\mathcal F}$.

$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $

Here, $x \in X$ is a point of a smooth projective variety over an algebraically closed field.

$\mathcal F$ is a coherent sheaf on $X \times X$ which is flat over the first factor, and $\mathcal F_x$ means a $\mathcal {i}^*_{x\times X} \mathcal F$.

(which is equal to $\Phi_{\mathcal F} (k(x))$)

Of course, KS map is not defined as a differential of some morphism, but please consider the following argument ;

Suppose $\mathcal F_x$ is concentrated at $x$ for all $x \in X$. It means the map $f:x \mapsto \mathcal F_x$ is injective. Hence $\kappa(x) := df(x)$ is injective at generic point.

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)

Here, the definition of $f$ is a nonsense. It is not even clear where the image of $f$ lives. But the argument is quite persuasive. And it is even geometrically intuitive, because it describes a KS map as a differential of some "function".

I think this must be a shadow of rigorous mathematical contents(probably a deformation theory), but I failed to make it so. Could someone explain to me what's going on? Thanks in advance.

share|improve this question
1  
This can be made totally rigorous. There are a ton of details to check, but I think it is fairly straightforward when (as you point out) you know what the target of $f$ is supposed to be. Let $P$ be the Hilbert polynomial of $\mathcal{F}_x$. The image is $Hilb^P(X)$ (the scheme representing the functor $S\to X \mapsto $ flat quotients of $\mathcal{O}_{S\times X}$ with Hilbert polynomial $P$). Commutativity of the appropriate diagram and identification of tangent spaces with the Ext groups using deformation theory shows that $df$ is exactly $\kappa (x)$. –  Matt Oct 16 '12 at 16:44

1 Answer 1

You can view the sheaf $\mathcal{F}$ on $X \times X$ instead as a map from $X$ (the first factor) to the moduli stack $M$ of sheaves on $X$ (the second factor). This induces a map on the tangent spaces. The tangent space to $M$ at a sheaf $F$ is the space of first order deformations of $F$, which is $\mathrm{Ext}^1(F, F)$.

Using the exact sequence

$0 \rightarrow I \rightarrow \mathcal{O}_X \rightarrow k(x) \rightarrow 0$

you get an identification between $T_x X = \mathrm{Hom}_{\mathcal{O}_X}(I, k(x)) = \mathrm{Ext}^1(k(x),k(x))$. Therefore the differential gives a map

$T_x X = \mathrm{Ext}^1(k(x),k(x)) \rightarrow T_{\mathcal{F}_x} M = \mathrm{Ext}^1(\mathcal{F}_x, \mathcal{F}_x)$.

share|improve this answer

Your Answer

 
discard

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

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