Elementary Proof of Riemann-Roch for Compact Riemann Surfaces I am supposed to give a talk about the Riemann-Roch theorem to a seminar of first and second year graduate students. I want to do Riemann-Roch for compact Riemann surfaces, but I am open to perhaps doing the version for projective curves.
I assume the crowd knows little algebraic geometry. I assume however, knowledge of elementary differential geometry, sheaves, cohomology and complex analysis. 
I am looking for an elementary proof (especially because I want time to be able to work out some examples). Any references?
 A: The proof given in Otto Forster, Lectures on Riemann Surfaces (Graduate Texts in Mathematics 81), chapter 16, seems very much suited to your list of prerequisites. 
A: There are (at least) 2 kinds of proofs: analytic ones (which use the existence of Abelian differentials with certain properties) and algebraic ones.
The proofs of the first kind use powerful analytic machinery, so it is a question what are you willing to accept as prerequisites.
The simplest analytic proof that I know is in Hurwitz-Courant (exists in German and Russian, but unfortunately not in English). Of course all prerequisites are in the same book, starting from definition of complex numbers. The proof of the Riemann-Roch itself is about 1 page, once existence
of Abelian differentials is established.
The simplest algebraic proof that I know is in Lang's book Algebraic and Abelian functions, which is almost self-contained (refers only to Algebra book of the same author).
A: There is a U of Chicago REU by a Valeriya Talovikova which does everything from the beginning in 10 pages.
A: I wrote up notes for the 4 lectures I did going through a completely algebraic proof at the end of a Shavarevich based algebraic geometry course. I think this is a nice approach in that it introduces in an elementary way the cohomological tools which will come up sophisticated ways in the following term. I didn't allocate nearly enough time to explain why we cared, though.
A: RRT
There is a big difference in difficulty between the compact Riemann surface case and the projective curve case, for reasons already mentioned.  Namely a projective curve comes equipped with a large supply of meromorphic functions, but the proof that they exist for a compact Riemann surface is a major step.
I suggest you decide what your goal is.  E.g., do you want to make clear why the theorem is true, without giving all steps of verification, or do you want to show how some of the trivial steps are derived easily using modern sheaf techniques, or perhaps give complete derivations of some significant parts of the statement?
if you use sheaf theory, it is trivial to show that chi(D)-chi(O) = deg(D) for any divisor D, where chi is the holomorphic Euler characteristic: chi(D) = h^0(D)-h^1(D).  To get full RRT from this one needs to compute chi(O) = 1-g, where g is the topological genus, and then to prove duality, that h^1(D) = h^0(K-D), where K is the canonical divisor.
In 4 pages of the notes on my web page, I take an argument from Bill Fulton to do one of these for plane curves, namely that chi(O) = 1-g, with some hand waving over the computation of topological Euler characteristics by deformation.
see pages 38-42:
http://alpha.math.uga.edu/%7Eroy/rrt.pdf
If you just want to show why the result is true with some arguments omitted, I feel nothing beats Riemann’s own exposition.  Riemann himself proved the theorem in clear natural stages:  1st the theorem in the special case of the canonical divisor, i.e. he proved that there are exactly g independent holomorphic differential forms on a compact Riemann surface of genus g.  Then he proved that for each point p, there is one meromorphic differential with a double pole at p and zero residue, equivalently he proved the RRT for divisors of form K+2p, i.e. that h^0(K+2p) = g+1, so that in addition to the g holomorphic forms there is one with a double pole at p, (and necessarily zero residue).  (Riemann also allows himself the simplification of assuming all points of the divisor considered are distinct.)  Then, from the existence of these basic types of differentials, one deduces the converse of the residue theorem.
Even if you barely read German, you can get the idea of what is most important just from deciphering the headings of the first two paragraphs in Riemann's treatment of the theorem in his great paper "On Abelian Integrals"; (since differentials the things you integrate, he speaks of their integrals): "Integrale erster Gattung" (integrals of first kind, i.e. of holomorphic differentials); "Integrale zweiter Gattung" (integrals of second kind, i.e. of differentials with double pole at one point)....
With this information, one can deduce the RRT in two steps, as clearly explained in Griffiths and Harris’s book, (see in particular pages 233, 244-5).  Namely for each effective divisor D, to compute the number of independent meromorphic functions with pole divisor dominated by D, is equivalent to computing the number of meromorphic differential forms that would occur as their differentials.  The assumed facts give the number of meromorphic differentials with the right singularities, and then one has only to compute how many of those differentials are exact, which is a period computation.
As in Riemann’s original paper they give this calculation in terms of the periods of integrals, and as in Roch’s follow-up these periods are computed in terms of residues using Green’s theorem.  All this is done nicely in Griffiths Harris, where they depend on the Kodaira vanishing theorem for the requisite existence of differential forms of first and second kinds.  (They also discuss this deep theorem earlier in their book.)
The point is that the deduction of RRT from the known existence of the right number of differential forms of first and second kinds is very clear and elementary complex calculus.  Hence it only remains to provide an elementary proof of these prerequisite facts about forms.  Now if one restricts attention to plane curves, at least the holomorphic forms can be immediately written down in coordinates, as Riemann himself points out. He also says one can write them down in the meromorphic case as well, but does not do so.  I believe this can indeed be done in an elementary way, as indicated in the book of Brieskorn and Knorrer, but I have not done it except in special cases.
In fact one can finesse the existence of the forms of second kind by the trick of Brill and Noether; i.e. one can use duality to rely exclusively on the existence of holomorphic forms for the theorem.  This argument is explained in the books of Arbarello, Cornalba, Grifiths and Harris (appendix A, chapter 1), and in the book on algebraic curves by Griffiths.  I have also written out this argument in the notes for my course in pdf form, at the link below: 
http://alpha.math.uga.edu/%7Eroy/8320.pdf
If you want a complete argument for compact Riemann surfaces, a nice treatment using sheaves is in the book Lectures on Riemann surfaces by Robert Gunning, including some nice Hilbert space arguments to deduce the finiteness of cohomology groups.  I like the old book, but a more current version of his notes are posted on his web site at:
https://web.math.princeton.edu/~gunning/
Finally David Mumford recommended reading the argument by George Kempf, in Crelle’s Journal, and reproduced in his book Algebraic Varieties, using sheaf theory heavily for the algebraic case.  I found this rather terse, but aspire to understanding it.  Here is a link to it:
http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002194015&physid=PHYS_0046
Please forgive the somewhat rushed answer, I hope some of the references are useful.  I apologize if my correction of an error causes a bump, but I hope it does not.
A: Here is one that is not elementary.  Joe Polchinski mentions a derivation of Riemann-Roch via Feynman Path integrals.  Unfortunately most people don't know what "ghosts" are and try to do String Theory without them. 
https://physics.stackexchange.com/questions/110322/number-of-zero-modes-on-the-sphere
It is worked out in Chapter 5 of Polchinski's String Theory Vol 1 - he is trying to talk about the Moduli space of Riemann Surfaces
A: Joe Harris, as recorded in his course notes here, gives the following slick proof when both $D$ and $K-D$ are effective; it has the advantage of never mentioning $H^1$. See lecture 1 for this argument, and 6-7 for removing the effectiveness hypothesis.
Let $X$ be a complete connected curve of genus $g$. Let $K$ be the canonical class, and let $D = x_1 + x_2 + \cdots + x_n$ be an effective divisor. For notational simplicity, assume the $x_i$ are distinct. Assume also that we already know $\dim H^0(X, \Omega^1)=g$ and $\deg K = 2g-2$. For each $x_i$, choose a uniformizer $z_i$ near $x_i$, so that we have a residue map $\mathrm{Res}_i$ from meromorphic functions on $X$ to $\mathbb{C}$. An element of $\mathcal{O}(D)$ which has no residue at any $x_i$ is a global holomorphic function, hence a constant, so we have an exact sequence:
$$0 \to \mathbb{C} \to H^0(\mathcal{O}(D)) \to \mathbb{C}^n$$
where the last map is the direct sum of the residue maps. 
Similarly, we have a short exact sequence
$$0 \to H^0(\Omega^1(-D)) \to H^0(\Omega^1) \to \mathbb{C}^n$$
where the $i$-th component of the last map sends a differential form $\omega$ to $\left. \tfrac{\omega}{d z_i} \right|_{x_i}$. Here $\tfrac{\omega}{d z_i}$ is a ratio of $1$-forms, hence a meromorphic function, which is defined and hence evaluable at $x_i$.
I claim that the images of these maps are perpindicular under the standard inner product on $\mathbb{C}^n$. This is because, for $f \in H^0(\mathcal{O}(D))$ and $\omega \in H^0(\Omega^1)$, the pairing $\sum \mathrm{Res}_i (f dz_i) \cdot \left(\left. \tfrac{\omega}{d z_i} \right|_{x_i}\right)$ is simply the sum of the residues of $f \omega$, and hence $0$ by Stokes theorem. 
If we knew that the images were orthogonal complements, we would deduce
$$(\ell(D)-1) + (g-\ell(K-D)) = n = \deg D.$$
Rearranging, we would get Riemmann-Roch and, in fact, Riemann-Roch is true and they are orthogonal complements.
Without knowing this, simply the fact that they are orthogonal gives the inequality
$$(\ell(D)-1) + (g-\ell(K-D)) \leq \deg D.$$
But, now consider the same argument for $K-D$ in place of $D$ (here is where we assume $K-D$ effective). We deduce
$$(\ell(K-D)-1) + (g-\ell(D)) \leq \deg K-D.$$
Adding both sides, $2g-2 \leq \deg D+\deg K-D = \deg K = 2g-2$. So every step must have been equality, and we have proved the result.
