# Remembering Riemann-Roch

Embarrassingly, I've always struggled to remember the form of the Riemann-Roch theorem for curves. Does anyone have any intuition to share about how to remember the some of the terms in the formula?

Recall that for $$C$$ a Riemann surface and $$D$$ a divisor on $$C$$, the Riemann-Roch theorem says that: $$$$h^0(D) - h^0(K-D) = \mathrm{deg}(D) + 1 - g$$$$ where $$K$$ is the canonical divisor on $$C$$. I'm happy with the interpretation of the terms on the left hand side (it's some kind of Euler characteristic), but is anyone able to give an informal explanation for the quantity on the right hand side? Why $$\mathrm{deg}(D) + 1 - g$$? In particular, why should I expect the left hand side to grow like $$\mathrm{deg}(D)$$, with a correction of $$1-g$$? I understand that there's some very classical way to think about this, but I've never seen it explained anywhere.

• Are you familiar with the proof? The strategy in many/most versions of the proof is to show that the claim is true for a divisor $D$ iff it's true for a divisor $D+P$, so this means it's enough to prove the claim for $D=0$ and then everything is very straightforwards via Serre Duality identifying $H^0(C,K)$ and $H^1(C,\mathcal{O}_C)$. Is this "intuition" enough for you, or is it not so helpful? Apr 17, 2020 at 23:51
• I mean, this is the proof that the statement for $D$ is equivalent to the statement for $D+P$: you write the short exact sequence $$0\to \mathcal{L}(D)\to\mathcal{L}(D+P)\to k(P) \to 0$$ and observe that the Euler characteristic adds over short exact sequences, and the Euler characteristic of the $\mathcal{L}$s is the LHS while the Euler characteristic of the skyscraper sheaf is one. I guess this is less intuition than proof, but the proof is somewhat nice (or at least I think so). Apr 18, 2020 at 4:55

Riemann-Roch for a curve $$C$$ (or Riemann surface if you prefer) says, that given some line bundle $$\mathscr{L} \in \text{Pic}(C)$$ (or some divisor), we have the equality $$\chi(\mathscr{L}) = \text{deg}(\mathscr{L}) + \chi(\mathscr{O}_C).$$ Actually I prefer to state it as $$\text{deg}(\mathscr{L}) = \chi(\mathscr{L}) - \chi(\mathscr{O}_C)$$ since I can more easily remember the sign and to sort of use it as the definition of the degree as well (it somehow also seems more aesthetic to have all the Euler characteristics on one side).
Now under the usual mild assumptions, we have that $$h^0(\mathscr{O}_C) = 1$$ and we have that $$h^1(\mathscr{O}_C) = g$$ is the genus of the curve. Plugging that in the first equation yields the Riemann-Roch that you stated. So I guess what I am trying to say is that rephrasing both the left hand side and the $$1 - g$$ by some Euler characteristic makes it quite easy to remember.