The title is pretty much the whole question. Let $S_g$ be a closed, oriented surface of genus $g$. Does there exist $n$ such that the mapping class group $\mathrm{Mod}(S_g)$ embeds as a subgroup of $\mathrm{Out}(F_n)$? (Or use $\mathrm{Aut}(F_n)$ if you prefer, up to changing $n$ this is sufficient.) For surfaces with punctures and/or boundary components, the answer is yes since the surface has free $\pi_1$, but for closed surfaces it seems unclear.
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3$\begingroup$ Aut$(F_n)$ embeds into Out$F_{n+1}$, but is it true that Out($F_n$) embeds into Aut$(F_m)$ for some $m$? $\endgroup$– YCorCommented Nov 13 at 0:51
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1$\begingroup$ @YCor Ah good point, I originally wrote "sufficient", and then absentmindedly changed it to "equivalent", assuming that direction was easy. But, right, this is unclear. I'll change it to "sufficient". $\endgroup$– Matt ZaremskyCommented Nov 13 at 1:12
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1$\begingroup$ I'm not so sure the answer is obviously "yes" if the surface has boundary components (as opposed to punctures). For instance, the mapping class group of a pair of pants is $\mathbb{Z}^3$, which does not embed in $SL_2(\mathbb{Z})$, the outer automorphism group of its fundamental group. ADDED: I guess it's OK, since you can always glue some other stuff onto the boundary components to make a punctured surface. But there is a little something to do! $\endgroup$– HJRWCommented Nov 13 at 16:06
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$\begingroup$ @HJRW Oh yes, that's true, "since the surface has free $\pi_1$,'' is not the whole story. But right, up to embedding just beef it up enough. $\endgroup$– Matt ZaremskyCommented Nov 13 at 18:10
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1 Answer
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$\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Out{Out}$It is true for $g=1,2$. $\Mod(S_1) \cong \GL_2(\mathbb{Z})\cong \Mod(S_{1,1})\cong \Out(F_2)$.
In $\Mod(S_2)$, the hyperelliptic involution is central and fixes the six Weierstrass points on any Riemann surface of genus 2. Since any mapping class group element commutes with this, it will permute the 6 Weierstrass points, and thus one may embed $\Mod(S_2)\leq \Mod(S_{6,2})$, the 6-punctured surface of genus 2 (this follows eg from Birman-Hilden theory, or see the comment on p. 2 of this paper). Hence $\Mod(S_2) < \Out(F_{9})$.
Notice that we are obtaining a splitting of the homomorphism
$\Mod(S_{6,2})\to \Mod(S_2)$. I wouldn’t expect such a splitting from $\Mod(S_{k,g})\to \Mod(S_g)$ for $g>2$, and doesn’t exist for $g>3$ by Theorem A of the above linked paper.
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$\begingroup$ That's a nice argument for $g=2$! I suppose maybe this same argument shows that for any genus, the centralizer of the hyperelliptic involution embeds in $\mathrm{Mod}(S_{2g+2,g})$ and hence in $\mathrm{Out}(F_{4g+1})$? But I don't really have a sense of how "big" these centralizers are when $g>2$, or whether that should count as "evidence" for the main question. $\endgroup$ Commented Nov 15 at 14:26
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1$\begingroup$ The centralizer of a hyperelliptic, quotient the hyperelliptic, is a mapping class group of a $2g+2$ pointed sphere. So essentially a braid group. $\endgroup$– Ian AgolCommented Nov 15 at 14:32
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$\begingroup$ Ah OK, so I guess not so helpful for higher genus, since braid groups embedding in $\mathrm{Out}(F_n)$ is "old news". $\endgroup$ Commented Nov 15 at 16:15