How would a topologist explain "every Riemann surface of genus $g$ is hyperelliptic if and only if $g=2$"?

Question

Recall that a compact Riemann surface/algebraic curve $C$ is hyperelliptic if it admits a branched double cover $C \to \mathbb P^1$, where $\mathbb P^1$ is the complex projective line/Riemann sphere. Among those curves of hyperbolic type ($g \ge 2$), the only genus that admits such a double cover in general is $g = 2$. While this fact is essentially trivial from the perspective of Hartshorne's fourth chapter, I have always found this proof to feel like a nuclear flyswatter (at least over $\mathbb C$).

Intuitively, a hyperelliptic Riemann surface of genus $g > 2$ is one that can be "drawn conformally" (whatever exactly that may mean) in such a way that all $g$ of its holes are "lined up." As such, my intuition suggests that the proposition in the title is somehow analogous to the trivial statement from Euclidean plane geometry that every configuration of $n > 1$ points is collinear if and only if $n=2$, where the holes in the surface are treated as analogous to points in the plane. So my question is this: is there a "holomorphic geometric topological" proof of said proposition that proceeds roughly as described in this paragraph rather than using any big theorems like Riemann-Roch?

Answer 1

Here is a small variant on Eremenko's answer.

The "Fenchel-Nielsen" coordinates on the space of hyperbolic metrics on a surface $\Sigma_g$ can be described via a pants decomposition. This is a decomposition of the surface along a collection of curves, that split the surface into a union of disjoint $3$-punctured spheres.

Hyperbolic metrics on 3-punctured spheres making the boundary into totally geodesic curves are specified by the cuff lengths, i.e. three real parameters.

So if you have a surface $\Sigma_g$, take a pants decomposition. Notice that you can choose pants decomposition equivariant with respect to a hyperelliptic involution. So if you think about the constraints on Fenchel-Nielsen coordinates coming from the surface being hyperelliptic, these only occur for $g > 2$. With $g=2$ all the curves are preserved and pants exchanged via the hyperelliptic involution, i.e. there are no additional constraints to be hyperelliptic.

But when $g>2$, there are curves off the hyperelliptic axis in your pants decomposition, so there are cuff lengths that are constrained by the hyperelliptic involution.

This is a long version of my comment.

Answer 2

A 19th century topologist would explain this by dimension count. By Riemann-Hurwitz, a surface of genus $g$ covering the sphere with $2$ sheets has $2g+2$ ramification points which gives $2g-1$ free complex parameters, since $3$ ramification points can be fixed. On the other hand, according to Riemann, when $g>1$ the space of Riemann surfaces of genus $g$ depends on $3g-3$ parameters which is strictly more than $2g-1$ when $g>2$. I suppose this is how this fact was discovered.

Answer 3

Among closed, oriented, connected surfaces $S_g$ of genus two or higher, the genus two surface $S_2$ is the only one whose mapping class group has non-trivial centre. This centre is a copy of $\mathbb{Z}_2$, generated by a (and thus the) hyperelliptic involution $\tau$.

The proof is topological. We follow Humphries to find, in $S_g$, a collection of $2g + 1$ Dehn twists that generate. In genus two all of these twists commute with $\tau$; thus $\tau$ is central. In higher genus, $2g$ of the twists form a chain, and thus commute with some hyperelliptic element. However the final twist does not so commute.

[This business with chains also produces the centre of the mapping class group of the genus one surface, which is again the unique (hyper)elliptic element. That is, the negative of the identity in $\mathrm{SL}(2, \mathbb{Z})$.]

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ADDED: As virkkunen points out in the comments below, there is something missing. Recall that an involution $\tau$ of a (closed, connected, oriented) topological surface $S$ is hyperelliptic if it fixes exactly $2g + 2$ points. These are all conjugate; thus their centralisers (the symmetric mapping class group) are all conjugate. The discussion above says that, when $S$ has genus $g$ at least three, there are (infinitely) many hyperelliptic involutions in the mapping class group for $S$.

But I've not yet proven that there is some Riemann surface which is not hyperelliptic. (The various other answers do this by dimension counting, which was topological in the 1800's, but today would be called Teichmüller theory.) However, David points out a way to save my proof.

Recall that a marked Riemann surface is a pair $(X, f)$ where $X$ is a Riemann surface and $f : X \to S$ is an orientation-preserving homeomorphism.

Lemma: Suppose that $(X, f)$ is a marked Riemann surface homeomorphic. Then $(X, f)$ admits at most one (biholomorphic) hyperelliptic involution.

Given the lemma, suppose that $X$ and $Y$ are marked Riemann surfaces. Suppose that $X$ and $Y$ admit distinct hyperelliptic involutions. Then connect $X$ to $Y$ by a path $(X_t)$ of marked Riemann surfaces with $X = X_0$ and $Y = X_1$. By the lemma there is some $t$ so that the marked Riemann surface $X_t$ has no hyperelliptic involution.

All that remains is to give a topologist's proof of David's lemma. Here is a nice tool (I think due to Nielsen, in the very early 1900's).

Proposition: Non-trivial periodic mapping classes act non-trivially on $H_1(S, \mathbb{Z})$.

Now suppose that $\tau$ is a hyperelliptic involutions. As an easy exercise, the image of $\tau$ in $\mathrm{Sp}(2g, \mathbb{Z})$ is the negative of the identity matrix.

Suppose that $\tau'$ is also a hyperelliptic element. Thus $\tau \circ \tau'$ is either the identity, or is non-torsion, in the mapping class group. Since the group of biholomorphic transformations of a Riemann surface is finite, we deduce that if $\tau$ and $\tau'$ are simultaneously biholomorphic for a marked surface $X$, then $\tau = \tau'$. This proves David's lemma.

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ADDED MORE: Well, dang. The argument above beginning

Then connect $X$ to $Y$ by a path $(X_t)$ of marked Riemann surfaces with $X = X_0$ and $Y = X_1$. By the lemma there is some $t$ so that the marked Riemann surface $X_t$ has no hyperelliptic involution.

is incomplete. We need each hyperelliptic locus to be a submanifold. We proved above that there are only countably many such, and they do not intersect. If these loci are submanifolds (they are) then they either are codimension zero or positive codimension. We've proved the former leads to contradiction, so the latter is holds, and we win. This is not the same as dimension counting (we just bound the dimension above) but it is getting uncomfortably close... so I think I will give up now.

Answer 4

The fact that every curve of genus $2$ is hyperelliptic comes from the fact that the canonical $g_2^1$ induces a hyperelliptic involution.

If $g \geq 3$, the general curve is not hyperelliptic. In fact, something stronger is true: the general curve $C$ of genus $\geq 3$ has trivial automorphism group (in particular, $\operatorname{Aut}(C)$ contains no involution at all). A proof can be given by using a variation of the Riemann-Hurwitz argument used in Alexandre Eremenko's answer, see this MO question.