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 be a divisor on a curve
of genus
. Then we have
.
Now, we stop a moment and say that .
Proof: Now, is the divisor associated to the sheaf
. Now, all curves are projective (in fact, all curves can be embedded in
) so we have Serre Duality, which says that the global sections of this sheaf are dual to
. So we need to show that
where
.
So we’ll proceed by induction. We start with , which means that
. So
and
, because this is Serre dual to
, 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
, so it’s good.
Now we take to be any divisor,
a point, and show that the formula holds for
if and only if it does for
. 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
the structure sheaf of the point. That is, we have a map
, and have
the direct image of the point’s structure sheaf.
So there’s a short exact sequence . We tensor with
and get a sequence
. Euler characteristic adds in short exact sequences (check this yourself, it’s pretty straightforward) and we get
, and
. So the formula holds for
iff it does for
. QED
Now, that’s a big theorem. It’s really important. Let’s use it to prove a fun theorem.
Let be an smooth curve which is not complete. Then it’s affine. Why should this be true? Well, we first embed
into
a complete curve. Now, there are finitely many points in
. Look at the divisor
. We can choose the
to all be positive and to be sufficiently large that there is a regular function
such that
and it is finite everywhere else. We get this out of Riemann-Roch, by noting that for
, we have
, and so we have the formula
. So there is some function on
which is meromorphic and has poles exactly at
of order at worst
and nowhere else.
Now, we use the linear system defined by to embed
into projective space. One thing that this accomplishes is that the divisor
becomes a hyperplane section, with the
‘s being the intersection multiplicity. So we have
for some hyperplane. Removing the hyperplane, and we turn
into
, and still have
embedded as a closed subvariety. Thus, it is affine.
That’s all for now, next time, we will think a bit about surfaces.
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