Recently, I encountered the version of the Riemann-Roch theorem for line bundles $\mathscr L$ on a compact Riemann surface $X$: $$\dim H^0(X,\mathscr L) - \dim H^1(X,\mathscr L) = 1 - g + c_1(\mathscr L),$$

with $c_1$ the first Chern-number.
I wonder if we can use this formula to derrive properties of the tangent bundle and the vector fields on $X$?

I first wanted to use that the top Chern class is the Euler-class, i.e. the first Chern number of the tangent bundle is the Euler characteristics. But then I realized, that the tangent bundle is only complex, but not holomorphic, so in fact, we need to set $\mathscr L = T^{(1,0)}X$, the holomorphic part of $TX$ in order to apply Riemann-Roch.

Can we still make some general statements about $c_1(\mathscr L)$ and if yes, what does the resulting formula tell us?

Edit: I made some progress on this questionm however I am not sure about the meaning of the results. Maybe someone can provide some background information.

As the holomorphic tangent bundle $\mathcal{T}_X$ is the dual of the canonical bundle $\Omega_X$, it follows that $c_1(\mathcal T_X)=-c_1(\Omega_X) = 2-2g=\chi(X)$.

Thus we get $c_1(\mathcal T_X)=c_1(TX) = c_1(T^{1,0}X \oplus T^{0,1}X) = c_1(T^{1,0}X)+c_1(T^{0,1}X)$, so $c_1(T^{0,1}X)=0$. What does this mean (maybe in regards to curvature)?

This also implies, that for $g\ge 2$, $\dim H^0(X,\mathcal T_X)=0$, so there is no global holomorphic section of the tangent bundle in this case, i.e. no global holomorphic vector field.

Using Riemann-Roch, we get: $$\dim H^0(X,\mathcal T_X)- \dim H^1(X,\mathcal T_X) = 1-g+c_1(\mathcal T_X)=3-3g. $$ As for $g=1$, $\mathcal T_X$ is trivial, it follows that $\dim H^0(X,\mathcal T_X)=1.$
For $g=0$, $X=\mathbb P^1$ and $\Omega_{\mathbb P^1}=O(-2)$, so $\mathcal T_{\mathbb P^1} = O(2)$, which is given by the homogeneous polynomials in $z_0,z_1$ of degree $2$. Thus $\dim H^0(X,\mathcal T_X) = 3.$

In conclusion: $$\dim H^1(X,\mathcal T_X) = \begin{cases} 0 &&g=0 ,\\1 && g=1, \\3g-3 && g\ge 2. \end{cases}$$ Is this information any useful or can this be motivated by some geometric insights? Maybe there is a short exact sequence of vector bundles/sheaves starting with $\mathcal T_X$, so atleast in the case of $g=0$ one can conclude surjectivity of map.?

  • $\begingroup$ Riemann-Roch theorem can be found using Atiyah-Singer's theorem. Maybe use AS thm to get the generalization you want ? $\endgroup$ – NAC Nov 17 '17 at 12:41
  • $\begingroup$ Tangent bundle is holomorphic. It is also isomorphic to $T^{1,0}X$. $\endgroup$ – Moishe Kohan Nov 28 '17 at 1:13
  • 1
    $\begingroup$ If you weren't aware, $H^1$ of the holomorphic tangent bundle is the space of first-order holomorphic deformations of $X$, which (at least when $Aut(X) = \emptyset$) is the tangent space at $[X]$ to the moduli space $\mathcal M_g$ (this is only a stack when $g < 2$ so assume otherwise as needed). So the cohomology calculations you have done are telling you the dimension of the moduli space. Unfortunately, I cannot relate this to anything about the smooth tangent bundle. $\endgroup$ – Tabes Bridges Dec 1 '17 at 6:21
  • $\begingroup$ Thx, this is interesting. Would you mind to write down the argument or give me a reference? $\endgroup$ – klirk Dec 1 '17 at 11:06
  • $\begingroup$ Harris and Morrison's Moduli of Curves discusses this sort of thing. Chapter 3, section B discusses deformation theory. $\endgroup$ – Tabes Bridges Dec 1 '17 at 18:10

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