In this post we prove the well-known fact that for a smooth curve , the arithmetic genus agrees with the topological genus .

# Motivation

It is a common occurrence that one encounters algebraic curves before they really study any serious amount of algebraic geometry. For example, the study of elliptic curves is something that is of interest to many people well before they have heard of a sheaf, let alone a scheme.

Because of this, one often is forced to accept as fact several definitions and theorems pertaining to curves. One of the most nebulous is the genus of a geometrically integral smooth curve over . Many texts actually define elliptic curves over to be smooth projective geometrically integral curves of genus , with a distinguished -point. While projective, smooth, and geometrically integral are easy to define, the genus of a smooth projective curve is a more elusive beast.

One way people try to circumvent this issue is by declaring that attention will only be paid to smooth plane curves , and then declaring that their genus be , where is the degree of . This is somewhat unsatisfying though, because the definition seems so ad hoc, and ungeneralizable.

Another common technique to assuage ones doubts about the genus of a curve over , is to say that it is just the ‘number of holes’ the curve has. Pressed further, they say what they really mean is that when one takes the curve and moves to , the curve defines a surface, and that this surface is a torus.

In this post we prove a precise statement of this fact, to the effect of the title. Namely, that if one starts with a curve over , then the arithmetic genus is the same as the topological genus . While there are several proofs of this fact (see exercise 21.7.I in Vakil), in this post we use what seems like the most natural approach to me: Serre’s GAGA theorem.

# Proof of the theorem

In all that follows, a curve over will be a smooth projective geometrically integral curve over . We define the arithmetic genus of a curve to be .

Theorem: Let be a curve, where is any field. Then, the arithmetic genus is equal to , the topological genus of the analytification of .

We first note that since it suffices to prove the following equivalent version of the theorem:

Theorem: Let be a curve. Then, .

Now, since is algebraically closed, we know (cf. Hartshorne Corollary 7.2 in III.7) that the dualizing sheaf of is just the canonical sheaf . Then, by Serre duality we know that

and so, in particular, .

But, , where is the sheaf of holomorphic forms on . We can then apply the GAGA theorem of Serre to conclude that

and thus we’ve reduced the proof to the following lemma:

Lemma: Let be a Riemann surface. Then, , where is the topological genus of .

While this follows immediately from the general Hodge decomposition theorem for an arbitrary Kahler manifold, there is a much quicker proof of (essentially the Hodge decomposition) for Riemann surfaces.

**Proof(of Lemma):** Begin by noting, essentially by combining the -Poincare and -Poincare lemma we obtain that

is a resolution of . But, since nor are necessarily acyclic, we can’t conclude that . But, by the general theory of hypercohomology, we do know that .

The usual first spectral sequence for hypercohomology turns into the Hodge to de Rham spectral sequence

Now, while it is true in general that this spectral sequence degenerates on the first page in general, this is a very deep theorem (essentially the Hodge theorem). But, in the case of Riemann surfaces, the first page degeneration comes for free. Namely, the first page looks like the following:

But, is just differentiating constants, and so is zero, and is just Serre dual to this map, and so also zero. Thus, we do have degeneration in the first page.

So, if we denote the filtration

then, , and . So, we see that

But, from Dolbeault’s theorem, and . But, since by conjugation, we may conclude that

But, since is locally contractible, we know that

and since , the conclusion follows.

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