Question

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.

Answer 1

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.

Answer 2

The Riemann-Roch theorem (and HRR, GRR, GRR for stacks, etc), over $\mathbb{C}$, associates topological information on one side $\chi(X,F)$, the Euler characteristic, to analytic information on the other side. If you want to relate analysis to topology, the biggest, baddest tool in your arsenal is The Atiyah-Singer Index Theorem. Then it's just a matter of finding an operator that shows up naturally that should apply. In my answer on that one, I linked to the original paper, where Atiyah and Singer explicitly do HRR as Theorem 3.

Answer 3

Hirzebruch-Riemann-Roch is a special case of Atiyah-Singer. On the RHS you have the Euler characteristic of a holomorphic vector bundle, and on the LHS you have an integral of some characteristic classes, which are topological invariants. The RHS is an analytic index, the LHS is a topological index. The reason why the Euler characteristic is equal to the analytic index is basically because of Dolbeault cohomology, i.e. you can compute sheaf cohomology using $\overline{\partial}$.

Edit: I'm not sure what kind of answer you're looking for. RR says that two numbers, which are defined purely algebraically, are equal. We are interested in computing these numbers, especially the LHS number, the Euler characteristic. On the other hand, these numbers also happen to be equal to analytic and topological indexes of an elliptic operator. So we can equivalently say that we are interested in computing these indexes, especially the analytic index.

Answer 4

I know nothing about the Atiyah-Singer index theorem, but wikipedia seems to have a pretty decent introduction. It even tells you the elliptic operator for the Riemann-Roch case. (So as not to leave you hanging, I'll just tell you: it's $D = \bar{\partial} + \bar{\partial}^*$)

Answer 5

Riemann-Roch in the version I know it:

Let $(E,\bar{\partial})$ be a holomorphic bundle over a compact Riemann surface $M$. Then $$index(\bar{\partial}) = deg E -(g-1)rank E$$

Here $index(\bar{\partial}) = dim H^0 (E) - dim H^0(KE^*)$.
This version arises, if you proof the fundamental theorem for elliptic operators ($\bar{\partial}$ is an elliptic operator, and Serre duality states, that two $\bar{\partial}$ operators on a complex vector bundle have the same index). (By the way: This is just a reformulation of RR, stated with divisiors (use Kodaira and Chow (?) )