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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: \begin{equation} h^0(D) - h^0(K-D) = \mathrm{deg}(D) + 1 - g \end{equation} 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.

Maxim
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    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? – KReiser Apr 17 '20 at 23:51
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    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). – KReiser Apr 18 '20 at 04:55

2 Answers2

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Not sure if that helps you, but I remember it as follows:

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.

For your other concerns, I would take a look at the comment of KReiser. I think the classical proof is what at least gives me the intuition.

Con
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If you want geometric intuition on the growth of the left hand side, recall that for $D$ large, $K-D$ is negative and thus has zero global sections, so the question of growth becomes why we have $\dim|D| ≥ d-g$, where $|D|$ is the linear series parametrized by the projective space defined by the space of global sections.

This comes from Abel's theorem, that linear series of degree d are fibers of the map by abelian integrals: $Sym(d)(C)-->Jac(C)$, from the $d$-fold symmetric product of the curve, to the $g$-dimensional Jacobian variety. Hence all fibers have dimension $≥ d-g$. The map also has maximal possible image dimension, so equality holds generically when $d ≥ g$, and holds always once $K-D$ is negative, i.e. $d > 2g-2$. That's the easy part, but more is true.

In fact the fibers are smooth, so the codimension $d-dim|D|$ of the fiber $|D|$ in $Sym(d)(C)$ equals the codimension of its tangent space at $D$, i.e. the dimension of the image $I$ of the tangent space at $D$ of $Sym(d)(C)$, in the tangent space of $Jac(C)$. But differentials on $C$ define linear forms on the tangent space to $Jac(C)$, and that image $I$ is defined by the differentials that vanish on $D$, so we have the precise RRT: $d-dim|D| = g-h^0(K-D)$.

This precise statement is due to Mattuck and Mayer http://www.numdam.org/item/ASNSP_1963_3_17_3_223_0/.

To flesh this out a little more, the Abel map is defined by integrating differential forms on C, so its derivative is given by evaluating them. Hence the projective derivative of the degree $1$ Abel map $C-->Jac(C)$ is just the canonical map of the curve into the projective space defined by the tangent space to $Jac(C)$ at the origin. I.e. it takes the tangent line to $C$ at $p$ to the canonical image of the point $p$. Similarly, the derivative of the Abel map on $Sym(d)(C)$, takes the projective tangent space at $D$ to the span of the points of the divisor $D$ in canonical space. Since linear forms on this space are just differential forms on $C$, that span is cut out by sections of $K-D$, i.e. by hyperplanes in canonical space that contain $D$. Then RRT says the points of $D$ in canonical space, fail to be in general position by exactly $\dim|D|$, i.e. their span has dimension $(g-1)-h^0(K-D) = (d-1)-\dim|D|$.

roy smith
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