13
$\begingroup$

Let $\mathrm{Out}(F_g)$ denote the automorphism group of a free group, and $\mathrm{Mod}_g$ the mapping class group of a closed oriented genus $g$ surface. Is there a map, as indicated with the dashed arrow below, making the following diagram commute?

$$ \begin{array}{ccc} \mathrm{Out}(F_g) & \dashrightarrow & \mathrm{Mod}_g \\ \downarrow & & \downarrow \\ \mathrm{GL}(g,\mathbf Z) & \to & \mathrm{Sp}(2g,\mathbf Z) \\ \end{array} $$ The lower horizontal map is induced by the functor $V \mapsto V \oplus V^\ast$ taking a vector space of dimension $g$ to a $2g$-dimensional symplectic vector space.

I do have a reason for asking but it's too long of a story to write here.

$\endgroup$
8
  • 2
    $\begingroup$ Don't we even expect that $\mathrm{Out}(F_4)$ doesn't embed into any mapping class group? More precisely, if we consider the HNN-extension of $F_2\times F_2$ by the partial isomorphism $(x,x)\mapsto (x,1)$, does it embed into any MCG? If I remember correctly, it embeds into $\mathrm{Out}(F_4)$ and this is used to prove the nonlinearity of the latter. $\endgroup$
    – YCor
    Jun 1, 2022 at 20:11
  • 2
    $\begingroup$ @YCor That's a nice comment, thanks. So we shouldn't expect to have a "cheap" construction of an injection from $\mathrm{Out}(F_g)$ to $\mathrm{Mod}_g$, since for $g\geq 4$ it would resolve a big open problem. On the other hand I did not require the map to be injective. $\endgroup$ Jun 1, 2022 at 20:41
  • 4
    $\begingroup$ If you have any homomorphism with infinite image, you get an infinite subgroup of MCG with Property T, and this is unknown too. $\endgroup$
    – YCor
    Jun 1, 2022 at 20:51
  • 3
    $\begingroup$ Very curious about the long story if you can say a bit! $\endgroup$ Jun 2, 2022 at 0:34
  • 1
    $\begingroup$ Just a thought on how one might try to rule this out—such a diagram would induce a map on relative completions, in the sense of Hain. I don’t know if anyone has worked out the lower central series of the unipotent radical of the relative completion of Out(F_n), but the Mod_g side is relatively well-understood, by Hain… $\endgroup$ Jun 2, 2022 at 13:06

2 Answers 2

9
$\begingroup$

As pointed out by YCor, $\mathrm{Out}(F_g)$ is not linear (for $g \geq 3$). Also, the linearity of $\mathrm{Mod}(S_g)$ is unknown. So the existence of such an embedding would solve a long-standing open question. I suspect that there is no such embedding.

However, perhaps you would be interested in a substitute. Let $V_g$ be the three-dimensional handlebody of genus $g$. Note that $\partial V_g = S_g$. Let $\mathrm{Mod}(V_g)$ be the resulting mapping class group. Restricting to the boundary gives a monomorphism $r \colon \mathrm{Mod}(V_g) \to \mathrm{Mod}(S_g)$. On the other hand, mapping classes act (via outer automorphism) on the space's fundamental group. So we have a epimorphism (as it turns out) $f \colon \mathrm{Mod}(V_g) \to \mathrm{Out}(\pi_1(V_g)) \cong \mathrm{Out}(F_g)$. Thus instead of a commuting square there is a pentagon.

$\endgroup$
4
  • 3
    $\begingroup$ The restriction map $Mod(V_g) \rightarrow Mod(S_g)$ is not surjective (but it is injective). $\endgroup$
    – Adrien
    Jun 2, 2022 at 6:54
  • $\begingroup$ Whoops, absolutely. Fixed. Thank you. $\endgroup$
    – Sam Nead
    Jun 2, 2022 at 10:27
  • 2
    $\begingroup$ This was very helpful, thanks. The map from Mod(V_g) is what I should have been thinking about all along. $\endgroup$ Jun 2, 2022 at 19:28
  • 1
    $\begingroup$ Sebastian Hensel has claimed that there is no section from $Out{F_g)$ to $Mod(V_g)$, but it appears that the paper was taken down. See reference 22 in this paper: mathematik.uni-muenchen.de/~hensel/papers/hno4.pdf $\endgroup$
    – Ian Agol
    Jun 2, 2022 at 21:41
4
$\begingroup$

This works for $g=2$, but it’s a very special case. $Out(F_2)\cong GL_2( \mathbb{Z}) \cong Mod_1 \cong Mod_{1,1}$, the mapping class group of a pointed torus. This is realized by the linear action of $GL_2(\mathbb{Z})$ on $T^2=\mathbb{R}^2/\mathbb{Z}^2$ fixing the origin. Taking the oriented blowup of the action at the origin (blowup by rays), one obtains an action of $GL_2(\mathbb{Z})$ on the surface $\Sigma_{1,1}$, a genus 1 surface with one boundary component. Then $GL_2(\mathbb{Z})$ acts on $\Sigma_{1,1}\times [-1,1]$, which is homeomorphic to a genus 2 handlebody and boundary homeomorphic to the double of $\Sigma_{1,1}$ along its boundary, ie $\Sigma_2$ the closed connected orientable surface of genus 2. The first homology splits as a direct sum into $H_1(\Sigma_{1,1} \times \{1\})$ and $H_1(\Sigma_{1,1}\times \{-1\})$, in such a way that the action of $GL_2(\mathbb{Z})$ acts by the dual action on the second factor since the identification by the product with $[-1,1]$ reverses orientation. Hence this gives the sort of homomorphism you seek in this case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.