# 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. $\blacksquare$