Today we’re going after one of the most famous theorems of algebraic geometry: the Riemann-Roch Theorem. For it, we depart the world of general schemes and return to varieties. And not just any varieties, but nonsingular complete curves.

The reason we need nonsingularity is because we’re going to be working with divisors and line bundles. Specifically, we’ll be needing linear systems and the canonical divisor. We’ll go through the proof, because I feel like it, but it isn’t really all that illuminating. Here we go.

Riemann-Roch Theorem: Let $D$ be a divisor on a curve $X$ of genus $g$. Then we have $\ell(D)-\ell(K-D)=\deg D+1-g$.

Now, we stop a moment and say that $\ell(D)=\dim |D|$.

Proof: Now, $K-D$ is the divisor associated to the sheaf $\omega_X\otimes \mathscr{L}(D)^\vee$. Now, all curves are projective (in fact, all curves can be embedded in $\mathbb{P}^3$) so we have Serre Duality, which says that the global sections of this sheaf are dual to $H^1(X,\mathscr{L}(D))$. So we need to show that $\chi(\mathscr{L}(D))=\deg D+1-g$ where $\chi(D)=\dim H^0(X,\mathscr{L}(D))-\dim H^1(X,\mathscr{L}(D))$.

So we’ll proceed by induction. We start with $D=0$, which means that $\mathscr{L}(D)=\mathscr{O}_X$. So $H^0(X,\mathscr{O}_X)=k$ and $\dim H^1(X,\mathscr{O}_X)=g$, because this is Serre dual to $H^0(X,\omega_X)$, which is the number of independent 1-forms on the curve, which is a definition of the genus. It can be shown to work with our definition via the Hilbert polynomial, but it’s a bit trickier. So the formula is then $1-g=0+1-g$, so it’s good.

Now we take $D$ to be any divisor, $P$ a point, and show that the formula holds for $D$ if and only if it does for $D+P$. Then we’re done, because anything can be built from 0 in finitely many steps of adding or subtracting a point. Now, we denote by $k(P)$ the structure sheaf of the point. That is, we have a map $P\to X$, and have $k(P)$ the direct image of the point’s structure sheaf.

So there’s a short exact sequence $0\to \mathscr{L}(-P)\to \mathscr{O}_X\to k(P)\to 0$. We tensor with $\mathscr{L}(D+P)$ and get a sequence $0\to \mathscr{L}(D)\to \mathscr{L}(D+P)\to k(P)\to 0$. Euler characteristic adds in short exact sequences (check this yourself, it’s pretty straightforward) and we get $\chi(\mathscr{L}(D+P))=\chi(\mathscr{L}(D))+1$, and $\deg(D+P)=\deg D+1$. So the formula holds for $D$ iff it does for $D+P$. QED

Now, that’s a big theorem. It’s really important. Let’s use it to prove a fun theorem.

Let $X$ be an smooth curve which is not complete. Then it’s affine. Why should this be true? Well, we first embed $X$ into $\bar{X}$ a complete curve. Now, there are finitely many points in $\bar{X}\setminus X=\{P_1,\ldots,P_k\}$. Look at the divisor $D=n_1P_1+\ldots+n_kP_k$. We can choose the $n_k$ to all be positive and to be sufficiently large that there is a regular function $f:X\to \mathbb{P}^1$ such that $f^{-1}(\infty)=\{P_1,\ldots,P_k\}$ and it is finite everywhere else. We get this out of Riemann-Roch, by noting that for $\sum n_i> 2g-2$, we have $H^0(\bar{X},\mathscr{L}(K-D))=0$, and so we have the formula $H^0(\bar{X},\mathscr{L}(D))=\sum n_i+1-g$. So there is some function on $\bar{X}$ which is meromorphic and has poles exactly at $P_i$ of order at worst $n_i$ and nowhere else.

Now, we use the linear system defined by $D$ to embed $\bar{X}$ into projective space. One thing that this accomplishes is that the divisor $D$ becomes a hyperplane section, with the $n_i$‘s being the intersection multiplicity. So we have $X=\bar{X}\setminus (\bar{X}\cap H)$ for some hyperplane. Removing the hyperplane, and we turn $\mathbb{P}^n$ into $\mathbb{A}^n$, and still have $X$ embedded as a closed subvariety. Thus, it is affine.

That’s all for now, next time, we will think a bit about surfaces.