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Let $X$ and $Y$ be smooth algebraic varieties over a field $k$ of characteristic $0$. For varieties we know that $X/k$ is rigid if and only if $H^{1}(X,T_{X})=0$. But $H^{1}(X,T_{X})$ also parametrizes the first order deformations. So in fact we have that there is no infinitesimal deformations if there is no first order infinitesimal deformation. My question is that does the same hold true for deformations of morphisms? Namely, let $f_{0}:X\to Y$ be a morphism. Then we know that first order deformations are parametrized by the space $H^{0}(X,f_{0}^{*}T_{Y})$ ($T_{Y}$, tangent bundle of $Y$). Is it true that $f_{0}$ is rigid if $H^{0}(X,f_{0}^{*}T_{Y})=0$? i.e. if there is no non-trivial first order deformation? If there is a reference for this in the literature, I'd be grateful to know.

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up vote 4 down vote accepted

The space $H^{0}(X,f_{0}^{*}T_{Y})$ is the tangent space at $f_0$ to the variety $\mathrm{Hom}(X,Y)$, see J. Kollár, Rational curves on algebraic varieties. Thus if it is zero, $f_0$ is an isolated point, hence is rigid. However I don't think the converse holds, even for deformations of $X$: $X$ may be rigid with $H^1(X,T_X)\neq 0$. I don't have any example at hand but I am pretty sure that they exist.

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Thank you for your reference about deformation of $f_{0}$. I don't know of any example for which the converse does not hold, but i know that at least in the context of compact complex manifolds, the rigidity is defined as the varieties satisfying $H^{1}(X,T_{X})$. So maybe you are referring to the non-compact or singular case. – Darius Math Jan 28 '14 at 20:11
Sorry I meant as varieties satisfying $H^{1}(X,T_{X})=0$. – Darius Math Jan 28 '14 at 20:17
No, what I meant is that there are varieties with no 1-dimensional deformations but with nontrivial first order deformations. But I agree that you didn't claim that, if you take $H^1(X,T_X)=0$ as definition of rigid. – abx Jan 28 '14 at 20:55
Also, I looked it up in Kollar's book, Theorem 1.7, p.95. There it is only stated in the case $X$ is a curve. But as I went through the proof, I did not see any necessity for $X$ being a curve. – Darius Math Jan 29 '14 at 12:46

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