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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?

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    $\begingroup$ How would you prove that not all configurations of $n>2$ points are collinear? By explicit example. I think you could prove that not all curves of genus $g>2$ are hyperelliptic by explicit example. Does the analogy go deeper than that? Maybe you have some kind of moduli dimension counting in mind? $\endgroup$
    – user178279
    Aug 28, 2021 at 6:38
  • $\begingroup$ Or maybe you mean there is some surjective map from configurations of points to curves such that non-collinear configurations get sent to non-hyperelliptic curves? I can't think of a natural map like that. $\endgroup$
    – user178279
    Aug 28, 2021 at 6:41
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    $\begingroup$ A topologist would not answer your question. Topologically, every surface of genus $\geq 2$ is hyperelliptic — i.e. can be realized as a double covering of the sphere branched at $2g+2$ points. $\endgroup$
    – abx
    Aug 28, 2021 at 6:46
  • $\begingroup$ @abx - yes, but in genus greater than two the hyperelliptic elements (there are infinitely many) are not central in the mapping class group. $\endgroup$
    – Sam Nead
    Aug 28, 2021 at 8:53
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    $\begingroup$ @TabesBridges : I agree that your question is probably best interpreted using hyperbolic geometry. But perhaps it’s worth pointing out that the equivalence with hyperbolic geometry uses Uniformisation, which of course is another “big theorem”! $\endgroup$
    – HJRW
    Aug 29, 2021 at 7:18

4 Answers 4

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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.

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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.

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    $\begingroup$ I can't think of someone I would call "a 19th century topologist". $\endgroup$
    – Ben McKay
    Aug 28, 2021 at 12:08
  • $\begingroup$ Strictly speaking you would need to prove that infinitesimal deformations "integrate" to actual deformations. I don't know if 19th century topologists were concerned with such things. $\endgroup$
    – user178279
    Aug 28, 2021 at 12:43
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    $\begingroup$ @BenMcKay: Poincare's work started appearing towards the end of the 19th century. I thought he was generally-acknowledged to be the founder of the subject. There were certainly earlier contributors: Riemann, Gauss, Schoenflies, etc. $\endgroup$ Aug 28, 2021 at 17:03
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    $\begingroup$ @Ben McKay: "19 century topologist" was joke, of course. But they did use topological arguments. $\endgroup$ Aug 28, 2021 at 21:47
  • $\begingroup$ @vikkunen: they were concerned. And this concern lead to the development of a) dimension theory, and b) moduli spaces in the beginning of 20th century. $\endgroup$ Aug 29, 2021 at 12:16
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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 $\mathrm{SL}(2, \mathbb{Z})$.]


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.


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.

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    $\begingroup$ Can you expand on how this is an explanation for OP? What I see is two phenomena specific to $g=2$, one in OP and one in your post. I'm not sure why one of them should be an explanation for the other. If anything the impression I get from your remark is that there are so many different hyperelliptic maps that we can't select a natural one. $\endgroup$
    – user178279
    Aug 28, 2021 at 9:07
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    $\begingroup$ You ask: In genus two, any chain $C = (c_i)$ of five curves determines a hyperelliptic element $\tau$. If $C'$ is another chain, then it gives $\tau'$. Why should $\tau'$ be equal to $\tau$ (up to isotopy)? To which I reply: the Dehn twists $\gamma_i$ about the $c_i$ generate the mapping class group (in genus two). So there is some product $G$ that sends $C$ to $C'$. Thus $G$ conjugates $\tau$ to $\tau'$. But the $\gamma_i$ commute with $\tau$, thus $\tau' = \tau$. $\endgroup$
    – Sam Nead
    Aug 28, 2021 at 14:52
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    $\begingroup$ This proof exactly does not work in genus three and higher. There a chain of $g+1$ curves only generates the hyperelliptic (also called symmetric) subgroup of the mapping class group. To avoid circularity, we prove this using Humphries original proof, showing that the hyperelliptic subgroup does not surject $\textrm{Sp}(2g, \mathbb{Z})$ (again, when $g \geq 3$). $\endgroup$
    – Sam Nead
    Aug 28, 2021 at 14:56
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    $\begingroup$ Gahh. In the immediately previous comment the “$g+1$ chain” should be a $2g+1$ chain. $\endgroup$
    – Sam Nead
    Aug 28, 2021 at 19:15
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    $\begingroup$ @virkkunen It sounds like the step you are missing is that a curve $X$ of genus $\geq 2$ can have at most one hyperelliptic involution $\sigma$ (meaning an involution such that $X/\sigma \cong \mathbb{P}^1$). Indeed, if $\sigma$ is a hyperelliptic involution, then $\sigma^{\ast} \mathcal{O}(g-1) \cong \omega_X$, which means that $X \to X/\sigma \cong \mathbb{P}^1$ can be recovered intrinsically from $X$ as the canonical map $X \to H^0(X, \omega_X)^{\vee}$. $\endgroup$ Aug 29, 2021 at 17:53
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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.

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