# Exotic automorphisms of the fundamental group of a curve?

A while back, Jordan S. Ellenberg brought the following problem to my attention.

If $G$ is a residually finite group, let $\widehat G$ be its profinite completion. Let $S$ be a closed surface of genus $g \geq 2$, and let $\pi$ be its topological fundamental group. Let $\mathrm{Mod}(S)$ be the mapping class group of $S$.

There is homomorphism $\widehat{\mathrm{Mod}(S)} \to \mathrm{Out}(\widehat \pi)$. Is this map surjective?

In other words, does the geometric fundamental group of the moduli space surject the outer automorphisms of the geometric fundamental group of the curve?

Edit: As Jordan points out, the map is not surjective. So, the question is:

What's the closure of the image of $\mathrm{Mod}(S)$ in $\mathrm{Out}(\widehat \pi)$?

Or, more precisely, for Henry:

As Jordan explains, there is a map $\mathrm{Out}(\widehat \pi) \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) \to \widehat{\mathbb{Z}}^\star$

Is the closure of the image of $\mathrm{Mod}(S)$ in $\mathrm{Out}(\widehat \pi)$ the preimage of 1 and -1?

• I assume that we're talking about continuous automorphisms of pi hat?
– HJRW
Nov 11, 2009 at 17:59
• Yes. But I think it follows from Nikolov and Segal's theorem (that finite index subgroups of topologically finitely generated profinite groups are open) that every automorphism will be continuous. Nov 11, 2009 at 19:54

First of all, this question came to me (or someone else with my initials) via Mark Kisin, so I can't claim credit (and for all I know it came to him from elsewhere.)

Second: there's one obvious obstruction to surjectivity. Namely, the map

$$\widehat{\mathrm{Mod}(S)} \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) -det \to \widehat{\mathbb{Z}}^\star$$

has image $\mathbb{Z}^\star$, which is to say $\pm1$. On the other hand,

$$\mathrm{Out}(\widehat \pi) \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) -det \to \widehat{\mathbb{Z}}^\star$$

is surjective. So the map you ask about is definitely not surjective. The question is whether in some sense "this is the only way the map fails to be surjective." Since I don't have a precise meaning in mind for the phrase in quotes, one might just say "what is the closure of the image of the mapping class group in $\mathrm{Out}(\widehat \pi)$?"

By the way, is there a topological proof that $\mathrm{Out}(\widehat \pi) \to \widehat{\mathbb{Z}}^\star$ is surjective? The only proof I know is that if you write down an algebraic curve $X$ over $\mathbb{Q}$, the images of Frobenii in $\mathrm{Out}(\pi_1^{et}(X_\overline{\mathbb{Q}}))$ give you automorphisms of pihat with lots of different determinants. Other than this I don't know how to construct a single element of $\mathrm{Out}(\widehat \pi)$ whose determinant is not $\pm1$!

• Oops. I think you had already pointed this out to me. I readjusted the question. Nov 11, 2009 at 11:24
• Surely the precise statement of the questions should be: "Is the closure of the image of Mod(S) equal to the preimage of {+-1} under the map you define Out(pihat) - > Zhat^*?"
– HJRW
Nov 11, 2009 at 17:44
• You're right, Henry, I've added the precise question now. Nov 11, 2009 at 19:27

Just a non-thought-out thought:

For any finite simple group $S$, consider the set of maps from $\pi$ to $S$ up to conjugacy. Now, $\mathrm{Out}(\widehat{\pi})$ acts on this set. Now consider the composition of this action with the permutation character; you get a character $f(S)$ of $\mathrm{Out}(\widehat{\pi})$ valued in $\pm 1$. [If $S$ is $\mathbb{Z}/p\mathbb{Z}$, I think but didn't check that the corresponding character is the "determinant composed with the quadratic residue symbol mod $p$," if that makes sense.]

I have no idea as to the image of the map $F = \prod_{S} f(S)$, but it seems plausible to me that it is uncountable. On the other hand, since $\mathrm{Out}(\pi)$ is finitely generated, the restriction of $F$ to it must have finite image. (Slight clarification: restrict the product over $S$ to nonabelian finite simple groups, since the abelian ones provide no new information.)