Topological genus agrees with arithmetic genus

In this post we prove the well-known fact that for a smooth curve X/\mathbb{C}, the arithmetic genus g_a(X) agrees with the topological genus g(X^\text{an}).

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 E/\mathbb{Q} 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 \mathbb{Q}. Many texts actually define elliptic curves over k to be smooth projective geometrically integral curves of genus 1, with a distinguished k-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 C\subseteq\mathbb{P}^2, and then declaring that their genus be \displaystyle \frac{(d-1)(d-2)}{2}, where d is the degree of C. 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 \mathbb{Q}, 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 C and moves to \mathbb{C}, 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 C over \mathbb{Q}, then the arithmetic genus g_a(C) is the same as the topological genus g\left(C_\mathbb{C}^\text{an}\right). 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 k will be a smooth projective geometrically integral curve over k. We define the arithmetic genus of a curve X/k to be g_a(X):=\dim_k H^1(X,\mathcal{O}_X).

Theorem: Let X/k be a curve, where k\subseteq\mathbb{C} is any field. Then, the arithmetic genus g_a(X) is equal to g(X_\mathbb{C}^\text{an}), the topological genus of the analytification of X_\mathbb{C}.

We first note that since \dim_k H^1(X,\mathcal{O}_X)=\dim_\mathbb{C} H^1(X_\mathbb{C},\mathcal{O}_{X_\mathbb{C}}) it suffices to prove the following equivalent version of the theorem:

Theorem: Let X/\mathbb{C} be a curve. Then, g_a(X)=g(X^\text{an}).

Now, since \mathbb{C} is algebraically closed, we know (cf. Hartshorne Corollary 7.2 in III.7) that the dualizing sheaf of X is just the canonical sheaf \Omega^1_{X/\mathbb{C}}. Then, by Serre duality we know that

H^1(X,\mathcal{O}_X)\cong H^0(X,\Omega^1_{X/\mathbb{C}})^\vee

and so, in particular, g_a(X)=\dim_\mathbb{C} H^0(X,\Omega^1_{X/\mathbb{C}}).

But, (\Omega^1_{X/\mathbb{C}})^\text{an}=\Omega^1_{\text{hol}}, where \Omega^1_\text{hol} is the sheaf of holomorphic forms on X^\text{an}. We can then apply the GAGA theorem of Serre to conclude that

H^0(X,\Omega^1_{X/\mathbb{C}})\cong H^0(X^\text{an},\Omega^1_\text{hol})

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

Lemma: Let X be a Riemann surface. Then, \dim_\mathbb{C} H^0(X,\Omega^1_\text{hol})=g, where g=g(X) is the topological genus of X.

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 d-Poincare and \overline{\partial}-Poincare lemma we obtain that

0\to \underline{\mathbb{C}}\to \mathcal{O}_X\to\Omega^1_\text{hol}\to 0

is a resolution of \underline{\mathbb{C}}. But, since \Omega^0_\text{hol}:=\mathcal{O}_X nor \Omega^1_\text{hol} are necessarily acyclic, we can’t conclude that H^1(X,\underline{\mathbb{C}})=H^1(\Omega^\bullet_\text{hol}(X)). But, by the general theory of hypercohomology, we do know that H^1(X,\underline{\mathbb{C}})\cong \mathbb{H}^1(X,\Omega^\bullet_\text{hol}).

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

E^{p,q}_1=H^q(X,\Omega^p_\text{hol})\implies \mathbb{H}^{p+q}(X,\Omega^\bullet_\text{hol})\cong H^{p+q}(X,\underline{\mathbb{C}})

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:

\begin{matrix}\vdots & \vdots & \vdots & \vdots & \\ 0 & 0 & 0 & 0 & \cdots\\ H^1(X,\mathcal{O}_X) & \to & H^1(X,\Omega^1_\text{hol}) & 0 & \cdots\\ H^0(X,\mathcal{O}_X) & \to & H^0(X,\Omega^1_\text{hol}) & 0 & \cdots\end{matrix}

But, H^0(X,\mathcal{O}_X)\to H^0(X,\Omega^1_\text{hol}) is just differentiating constants, and so is zero, and H^1(X,\mathcal{O}_X)\to H^1(X,\Omega^1_\text{hol}) 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

0=F_0\subseteq F_1\subseteq F_2=H^1(X,\underline{\mathbb{C}})

then, F_2/F_1\cong H^0(X,\Omega^1_\text{hol}), and F_1=H^1(X,\mathcal{O}_X). So, we see that

\dim_\mathbb{C} H^1(X,\underline{\mathbb{C}})=\dim_\mathbb{C} H^0(X,\Omega^1_\text{hol})+\dim_\mathbb{C} H^1(X,\mathcal{O}_X)

But, from Dolbeault’s theorem, H^0(X,\Omega^1_\text{hol})=H^{1,0}(X) and H^1(X,\mathcal{O}_X)\cong H^{0,1}(X). But, since H^{1,0}(X)\cong H^{0,1}(X) by conjugation, we may conclude that

\dim_\mathbb{C} H^1(X,\underline{\mathbb{C}})=2\dim_\mathbb{C} H^0(X,\Omega^1_\text{hol})

But, since X is locally contractible, we know that

H^1(X,\underline{\mathbb{C}})\cong H^1_\text{sing}(X,\mathbb{C})

and since H^1_\text{sing}(X,\mathbb{C})\cong \mathbb{C}^{2g}, the conclusion follows. \blacksquare

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