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I would like to share a (partial) computation which I made, and might be completely wrong. My goal was to compute $h^1(X,T_X)$ for a uninodal curve $X\subset \mathbb P^2$ of degree $d$. I do not even know whether to expect it to be $0$ or not. What follows is my attempt, whose conclusion is still missing.

First of all, $X$ corresponds to a point in the Hilbert scheme $H$ of plane curves of degree $d$, which is $\mathbb P^N$ for $N=d(d+3)/2$. The moduli space is smooth, so $$N=\dim_{[X]}H=h^0(X,N_{X/\mathbb P^2}).$$ From the exact sequence $$0\to T_X\to T_{\mathbb P^2}|_X\overset{d^\vee}{\to}N_{X/\mathbb P^2}\to T^1_X=\textrm{coker }d^\vee\to 0$$ we get an exact piece $$H^0(X,N_{X/\mathbb P^2})=\mathbb C^N\overset{\theta}{\to} H^0(T^1_X)=\mathbb C\overset{\phi}{\to} H^1(X,T_X)\to H^1(X,T_{\mathbb P^2}|_X)\to 0.$$ Now I briefly explored two routes:

  • $h^1(X,T_X)=0$ means $\theta$ is surjective. It might be true, but I see no reason for that. (I admit I do not know what $\theta$ does, concretely.)
  • We have, by the last displayed sequence, $$h^1(X,T_X)=h^1(X,T_{\mathbb P^2}|_X)+\dim\,(\textrm{im } \phi).$$ Of course, $\dim\,(\textrm{im }\phi)$ can be $0$ or $1$, but I do not know which. As for $h^1(X,T_{\mathbb P^2}|_X)$, from the exact sequence $$0\to \mathscr O_X\to \mathscr O_X(1)^{\oplus 3}\to T_{\mathbb P^2}|_X\to 0$$ one obtains $$H^1(X,\mathscr O_X(1)^{\oplus 3})\to H^1(X,T_{\mathbb P^2}|_X)\to 0,$$ but I do not know whether $h^1(X,\mathscr O_X(1))=0$ or not.

So my questions are:

  1. Could you please correct/comment my attempt, and thus help me to find $h^1(X,T_X)$?
  2. I assumed $X$ uninodal for the simple reason that I already computed $H^0(T^1_X)=\mathbb C$. What does change for an arbitrary $[X]\in H$?

Also, if you wish, please feel free to add your suggestions on how to think about these cohomology groups, and maps between them; I am new to this subject and I think I still miss what most of them really represent in the deformation-theoretic setting.

Thank you very much for your patience.

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Can you use Riemann Roch to compute that? – John Ma Dec 9 '13 at 1:23
RR would suggest $h^1(T_X)$ to be zero, but I am unsure on using it on singular curves. And mostly, I would like to get the result using the argument I described, just to get some practice. – Brenin Dec 9 '13 at 10:27

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