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I want to compute $h^1(C, \mathcal{T}_C)$, where $C$ is a genus $g$ smooth projective curve.

This is where I have gotten so far: I know that, because $C$ is a curve, $\Omega_C$ is the Serre dualising sheaf. So, $h^1(\mathcal{T}_C) = h^0(\Omega^{\otimes 2}_C)$. Riemann-Roch then gives $$h^0(\Omega^{\otimes2}_C) - h^1(\Omega^{\otimes2}_C) = 2(\text{deg }\Omega_C) -g+1 = 3g-3.$$

What I do not know how to do is to compute the term $h^1(\Omega^{\otimes2}_C) =h^0(\mathcal{T}_C).$ I tried using a normal exact sequence for the map $|\omega_C| : C \to \mathbb{P}^{g-1}$, but the $\mathcal{T}_{C/\mathbb{P}^{g-1}}$ term confused me.

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In addition to Francesco's answer, can I suggest a slightly different approach?

  • If $g = 0$, then $C = \mathbb P^1$, and $T_C = \mathcal O_{\mathbb P^1}(2)$, hence $h^1(T_C) = 0$.

  • If $g = 1$, then $C$ is an elliptic curve, so $T_C = \mathcal O_C$. The dualising sheaf in this case is also $\mathcal O_C$, so $h^1(T_C) = h^1(\mathcal O_C) = h^0(\mathcal O_C) = 1$.

  • If $g \geq 2$, then ${\rm deg \ }T_C = - {\rm deg \ }\Omega_C = 2 - 2g < 0$. A line bundle with a negative degree has no global sections (Hartshorne IV.1, Lemma 1.2), so $h^0(T_C) = 0$. Therefore, by Riemann-Roch, which states that $ h^0(T_C) - h^1(T_C) = {\rm deg}(T_C) - g + 1, $ we learn that $ h^1(T_C) = 3g - 3.$

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    $\begingroup$ Silly me, I had not thought to split by cases on the genus. $\endgroup$ – user309475 Jun 27 '17 at 3:08
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The best thing to do is to compute directly $h^1(T_C)$. By Riemann-Roch we have $$h^0(T_C)-h^1(T_C)= \deg(T_C)-g+1=3-3g.$$

On the other hand, by a well-known result by Hurwitz, we know that if $g \geq 2$ then $C$ has only finitely many automorphisms. This, in turn, implies $h^0(T_C)=0$, because the dimension of $H^0(T_C)$ is precisely the dimension of $\mathrm{Aut}(C)$, so the result follows.

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