Let $X$ denote an algebraic scheme over $\operatorname{Spec} k$ such that its deformation functor $\operatorname{Def}_X$ has a semi-universal couple $(R,u)$, where $R$ is an Artinian $k$-algebra and $u \in \operatorname{Def}_X(R)$.

On pg. 91 of the book "Deformations of algebraic schemes" by E. Sernesi the following automorphism functor is introduced

$$ \operatorname{Aut}_u: \mathcal{A}_R \to \operatorname{Set} $$

$$ \operatorname{Aut}_u(A) = \text{the group of automorphisms of the deformations} \ \mathcal{X}_A$$

My question concerns the following proposition:

$\textbf{Proposition 2.6.2} $ If $X$ is projective, then $\operatorname{Aut}_u$ has $H^0(X,T_X)$ as tangent space.

The proof of the proposition concludes (Eqn 2.29, pg. 92) with the claim that $$Aut_u(k[\epsilon]) \cong H^0(X, T_X).$$

However, $k[\epsilon]$ does not have a natural $R$-algebra structure so writing $Aut_u(k[\epsilon])$ doesn't really make sense.

What did Sernesi mean when he wrote this?

  • 1
    $\begingroup$ It seems to me that a $k$-algebra homomorphism $R \longrightarrow k[\varepsilon]$ exist by semi-universality, or am I missing something? $\endgroup$ Jul 1, 2020 at 8:13
  • $\begingroup$ It appear that the same problem would occur at the beginning of the proof, when Sernesi shows that $\mathrm{Aut}_u$ satisfies condition $H_0$ (as in pag45), that is $\mathrm{Aut}_u(k)$ is the identity. $\endgroup$
    – Aurelio
    Jul 1, 2020 at 8:16
  • $\begingroup$ In my mind, geometrically, in the case of complex projective varieties this corresponds to a morphism of schemes $\mathrm{Spec}(k[\varepsilon]) \longrightarrow \mathrm{Def}(X)$, where $\mathrm{Def}(X)$ is (the germ of) the Kuranishi family of $X$. $\endgroup$ Jul 1, 2020 at 8:21

1 Answer 1


In fact, there is a natural such structure, namely $$ R \to k \to k[\varepsilon], $$ corresponding to the constant deformation $X\otimes k[\varepsilon]$, and indeed the automorphisms of $X\otimes k[\varepsilon]$ over $k[\varepsilon]$ restricting to the identity on $X$ correspond to derivations $\mathcal{O}_X \to \mathcal{O}_X$.


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