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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