I was in a lecture not long ago given by C. Teleman and at some point he said "Well, since Riemann-Roch is an index problem we can do..."


Then right after that he argued in favour of such a sentence. Could anyone tell me what did he mean exactly?. That is to say, in this case what is elliptic operator like, what is the heuristic idea which such a result relies on? ...and a little bit of more details about it.


As usual references will be appreciated.


**ADD:** Thanks for the comments below, but I think they do not answer the question of title : **Why** is RR an Index problem?. Up to this point, what I can see is that two numbers happened to be the same.



Why is Riemann-Roch an Index Problem?
Here is a sketch of the argument as I learned it in a complex analysis class: For a Riemann surface $X$ and a holomorphic line bundle $L$, we want 
$$\text{dim}H^0(X,L)-\text{dim}H^0(X,L\otimes\Lambda^{0,1})=c\_1(L)+\frac{1}{2}\chi(X)$$
You have an operator $\overline{\partial}$ (differentiation with respect to $d\overline{z}$ taking $\Gamma(X,L)$ to $\Gamma(X,L\otimes\Lambda^{0,1})$. Then $H^0(X,L)$ is the kernel of $\overline{\partial}$ and $H^0(X,L\otimes\Lambda^{0,1})$ is the kernel of its adjoint, $\overline{\partial}^+$. Now define $\Delta^+=\overline{\partial}\overline{\partial}^+$ and $\Delta^-=\overline{\partial}^+\overline{\partial}$. Their spectra are the same, except for the kernels, and we get 
$$\text{Tr}(e^{-t\Delta^+})-\text{Tr}(e^{-t\Delta^-})=\text{dim}(\text{ker}\Delta^+)-\text{dim}(\text{ker}\Delta^-)$$
We also have that the kernel of $\overline{\partial}$ is the kernel of $\Delta^-$, and the kernel of $\overline{\partial}^+$ is the kernel of $\Delta^+$, so it's enough to get your hands on the left-hand side. Then you write those traces as integrals of heat kernels, take the limit as $t\rightarrow 0^+$, and show that the integrals go to $c\_1(L)+\frac{1}{2}\chi(X)$. And that's possible because we can interpret Chern classes and Euler characteristics of Riemann surfaces as integrals of curvatures of line bundles. Of course, then there's more work to turn $c\_1(L)+\frac{1}{2}\chi(X)$ into it's more familiar form.


