cohomology of a tangent bundle

Question

Suppose that $C$ is a complex riemann surface of positive genus lying in a complex algebraic surface of general type. Let $T_C$ the tangent bundle to the curve $C$. Is there a way to compute the cohomology of this bundle?
i.e. the spaces $H^i(C,T_C)$ for $i=0,1$?

Answer 1

As John Ma mentioned, one can do this using Riemann-Roch theorem. The answer depends only on the genus of the curve.

For the case of a curve Riemann-Roch implies the following: for any linear bundle $L$ on $C$ $$ \chi(C, L) = h^0(C, L) - h^1(C, L) = \operatorname{deg}(L) + \chi(C, \mathcal{O}_C). $$

For any curve of genus $g$ one always has $\chi(C, \mathcal{O}_C) = 1-g$ and $\operatorname{deg}(TC) = 2-2g$.

Therefore we know that $h^0(TC) - h^1(TC) = 3-3g$ and it is sufficient to find at least one of these numbers.

If g = 1, then $C$ is an elliptic curve, $TC$ is trivial and $h^0(TC) = h^0(\mathcal{O}_C) = 1$. Thus $h^1(TC)$ also equals $1$.

If $g>1$ then $\deg(TC) < 0$. However, a divisor of negative degree can never has global sections. This means that $h^0(TC) = 0$ and $h^1(TC) = -(3-3g) = 3g-3$.

Perhaps you have heard that the dimension of the moduli space of curves of genus $g$ also equals $3g-3$. This is not a coincidence: first cohomologies of the tangent bundle describe all infinitisemal deformations of $C$, that is a tangent space to the moduli space.