Exotic automorphisms of the fundamental group of a curve?

Question

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?

Answer 1

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 $\pm1$!

Answer 2

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.)