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Let $C$ be a reduced, connected, projective and purely one-dimensional scheme of finite type over a field $k$. Suppose that $C$ is rational, i.e. that the normalisation of $C$ is a disjoint union of copies of $\mathbb{P}^1_k$. Let $T_C = \mathcal{H}om(\Omega^1_C,\mathcal{O}_C)$ be the tangent sheaf of $C$.

Question: Is it true that $H^1(C,T_C)$ vanishes?

You may additionally assume that $C$ has only planar singularities, but I'm not sure this is needed. The question is true if $C$ is smooth. For context, a positive answer would imply that $C$ has no locally trivial deformations.

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Let $C$ be the union of 5 lines in general position in $\mathbb{P}^2$ (hence with 10 pairwise intersection points $P_{ij}$, $1 \le i < j \le 5$) and let $F$ be the equation of $C$. We have the standard exact sequence $$ 0 \to \mathcal{O}_C(-5) \stackrel{dF}\to \Omega_{\mathbb{P}^2}\vert_C \to \Omega_C \to 0. $$ Taking its dual we obtain an exact sequence $$ 0 \to T_C \to T_{\mathbb{P}^2}\vert_C \stackrel{dF}\to \mathcal{O}_C(5) \to \bigoplus_{1 \le i < j \le 5} \mathcal{O}_{P_{ij}} \to 0. $$ Now an easy computation of Euler characteristics gives $$ \chi(T_{\mathbb{P}^2}\vert_C) = 5, \quad \chi(\mathcal{O}_C(5)) = 20, \quad \chi(\mathcal{O}_{P_{ij}}) = 1, $$ hence $\chi(T_C) = 5 - 20 + 10 = -5$, and since $C$ is 1-dimensional, it follows that $H^1(C,T_C)$ is non-zero.

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    $\begingroup$ Doesn't the exactness on the left of the first sequence require the smoothness of $C$? $\endgroup$
    – Jef
    Commented Nov 23, 2021 at 11:50
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    $\begingroup$ @Jef: no. $dF$ is generically injective so its kernel is torsion, hence zero since $\mathscr{O}_C(-5)$ is torsion free. $\endgroup$
    – abx
    Commented Nov 23, 2021 at 17:01
  • $\begingroup$ Great, thanks for your answer! $\endgroup$
    – Jef
    Commented Nov 23, 2021 at 18:13

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