## Kodaira-Spencer map as a “differential”

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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.

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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 at 16:44

## 1 Answer

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

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

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