# Kodaira Spencer map and versal deformation

First I want to clarify what I mean by the Kodaira-Spencer map.

Let's have a family of deformations $\pi:\mathcal{X}\rightarrow B$ of a complex manifold $X=\mathcal{X}_0:=f^{-1}(0)$ (by that I mean that $f$ is an holomorphic proper submersion between the complex manifolds $\mathcal{X}$ and $B$). Then the Kodaira-Spencer map $\rho:T_{B,b}\rightarrow H^1(\mathcal{X}_b,T_{\mathcal{X}_b})$ is the map induced in cohomology by the exact sequence of sheaves $$0\rightarrow T_{\mathcal{X}_b}\rightarrow T_{\mathcal{X}|\mathcal{X}_b}\rightarrow f^*T_B|\mathcal{X}_b\rightarrow 0$$ By "versal deformation" $Def(X)$ I mean that any other deformation $\mathcal{X}\rightarrow B$ of $X$ can be obtained as the pullback deformation for a certain map $g:B\rightarrow Def(X)$ and the differential of $g$ is unique.

My question is: why is the Kodaira-Spencer map for the versal deformation bijective?

This is not clear to me. Also, some authors define the versal deformation as the deformation for which the Kodaira-Spencer map is bijective.

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You must be missing an hypothesis: perhaps the vanishing of $h^2(\mathcal{X}_b,T_{\mathcal{X}_b})$? Without a hypothesis such as this, there may be examples of a complex manifold $\mathcal{X}_0$ that admits nontrivial first-order deformations, i.e., elements in $H^1(\mathcal{X}_b,T_{\mathcal{X}_b})$, none of which deform to higher order. In that case, taking $B$ to be a point, the morphism satisfies your "versal property". Yet $T_B$ is zero-dimensional. –  Jason Starr Jul 13 '13 at 22:40

This is basically a tautology. A tangent vector $v$ to $Def(X)$ at $0$ is a map $D:=Spec(k[t]/(t^2))\to Def(X)$ mapping the closed point to 0, i.e. a family $X_v\to D$, i.e. a first order deformation of $X_0$. These are classified by the vector space $H^1(X, T_X)$, so we get a bijection $T_0 Def(X) \to H^1(X, T_X)$ which turns out to be the Kodaira-Spencer map.

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What you say would be perfectly correct if the OP did not insist that $B$ be a complex manifold, instead of merely a complex analytic space. However, with the hypothesis that $B$ be a complex manifold, the map need not be bijective. –  Jason Starr Jul 14 '13 at 1:44
@Jason you're completely right! I didn't read the question carefully and thought of schemes the whole time... –  Piotr Achinger Jul 14 '13 at 8:04