# Mapping Out(F_n) to the mapping class group

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.

• 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.
– YCor
Jun 1, 2022 at 20:11
• @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. Jun 1, 2022 at 20:41
• If you have any homomorphism with infinite image, you get an infinite subgroup of MCG with Property T, and this is unknown too.
– YCor
Jun 1, 2022 at 20:51
• Very curious about the long story if you can say a bit! Jun 2, 2022 at 0:34
• 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… Jun 2, 2022 at 13:06

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.
• The restriction map $Mod(V_g) \rightarrow Mod(S_g)$ is not surjective (but it is injective). Jun 2, 2022 at 6:54
• 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 Jun 2, 2022 at 21:41
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.