## Notes of Martin Bridson’s lecture 3

1. Torsion in ${Out(F_n)}$ and applications

Skipping train tracks and mapping tori.

1.1. Finite subgroups

Theorem 1 ${Out(F_n)}$ has finitely many classes of finite subgroups.

question: Does this holds in arbitrary finitely presented linear groups ?

1.2. Proof

Prove a fixed point theorem.

Proposition 2 (Khramstov, Culler, Zimmermann independantly) Every finite subgroup ${Out(F_n)}$ fixes points in the spine ${K_n}$.

Indeed, let ${G\subset Out(F_n)}$ be a finite subgroup, let ${G'}$ be its inverse image in ${Aut(F_n)}$. It is virtually free, so a finite graph of finite groups (Stallings). ${G'}$ acts cocompactly on the Bass-Serre tree ${T}$. Mod out by the kernel, a copy of ${F_n}$ action. The obtained graph is a finite covering of the graph (of groups). So ${G}$ is a group of isometries of a graph of rank ${n}$, whence a fixed point.

Example: The isometry group of a rose is the wreath product ${W_n =({\mathbb Z}/2)\wr{\mathbb Z}}$ (it is the largest finite subgroup in ${Out(F_n)}$).

Example: The isometry group of a cage (aka multitheta) is ${\mathfrak{S}_{n+1}\times{\mathbb Z}/2}$.

Remark 1 Let ${G\subset Out(F_n)}$ be a finite subgroup. ${G}$ acts on some finite graph ${\Gamma}$ of rank ${n}$. If ${G}$ has a fixed point in ${\Gamma}$, then ${G}$ lifts to ${Aut(F_n)}$.

Lifting is not always possible. For instance ${W_n}$ does not lift. Here is an other example.

Example: Consider ${\Gamma=}$ simplicial ${n-1}$-cycle with cycles glued at vertices. Let ${\theta}$ rotate the picture. Then the inverse image of ${\theta}$ in ${Aut(F_n)}$ is torsion free.

1.3. Consequences

1. Natural map ${Aut(F_n)\rightarrow Out(F_n)}$ does not split.
2. Neither does

$\displaystyle \begin{array}{rcl} 0\rightarrow F_n /[F_n ,F_n]\rightarrow Aut(F_n)/[F_n ,F_n]\rightarrow Out(F_n)/[F_n ,F_n]\rightarrow 1. \end{array}$

This implies

Theorem 3 Let ${M=H_1 (\F_n)}$ be the standard ${Out(F_n)}$-module. Then ${H^2 (Out(F_n),M)\not=0}$. However ${H^2 (Out(F_n),M^*)=0}$.

1.4. Trying to construct homomorphisms ${Out(F_n)\rightarrow Out(F_m)}$

Start with a rose. Take a finite Galois covering ${\Gamma}$. Assume the corresponding subgroup is characteristic (i.e. ${Out(F_n)}$-invariant). Individual elements of ${Out(F_n)}$, acting on the base rose, lift to ${\Gamma}$. But this need not provide a lift of the ${Out(F_n)}$ action in general. We look for special cases where it does.

Restrict attention to abelian covers, with groups ${N=[F_n ,F_n]}$ or ${N=[F_n ,F_n]F_n^r}$.

Theorem 4 The action of ${Out(F_n)}$ on the rose lifts to the abelian covering with Galois group ${N}$ iff ${N=[F_n ,F_n]F_n^r}$ with ${(r,n-1)=1}$.

Example: If ${n}$ is even, ${n-1}$ is odd, so coprime to ${2}$, so we get a homomorphism ${Out(F_n)\rightarrow Out(F_m)}$ for ${m=2^n (n-1) +1}$. These examples were discovered earlier by Bogopolski and Puga, with a more algebraic argument, based on presentations.

Proof: The ${Aut\rightarrow Out}$ exact sequence fits in a commutative diagram with the corresponding sequence for groups of homotopy equivalences of graphs. Playing with finite subgroups, we know the former does not split, so the latter does not either.

Theorem 5 (Kielak) A homomorphism ${Out(F_n)\rightarrow Out(F_m)}$ with infinite image can exist only if ${m\geq \begin{pmatrix}n \\ 2\end{pmatrix}}$.

2. ${IA_n}$

Definition 6

$\displaystyle \begin{array}{rcl} IA_n =ker(Out(F_n)\rightarrow Gl(n,{\mathbb Z})). \end{array}$

Gersten’s presentation shows that ${IA_n}$ is generated by ${\lambda_{ij}\rho_{ij}^{-1}}$.

Theorem 7 (Magnus) The subgroup generated by the ${\lambda_{ij}^{-1}\rho_{ij}}$ and the ${a_i[a_j,a_k]}$ is normal. It follows that ${IA_n}$ is finitely generated.

Not very geometric. Unknown wether ${IA_n}$ is finitely presented.

2.1. Maps from higher rank lattices

In some sense, ${IA_n}$ contains the non-linear part of ${Out(F_n)}$.

Theorem 8 (Bridson, Wade) Let ${\Gamma}$ be an irreducible lattive in a higher rank semisimple Lie group. Then every homomorphism ${\Gamma\rightarrow Out(F_n)}$ has finite image.

Theorem holds under a much weaker assumption on ${\Gamma}$ : say ${\Gamma}$ is ${{\mathbb Z}}$-averse if no finite index subgroup

2.2. Proof

Case 1: Assume image ${\phi(\Gamma)}$ contains an iwip. Then apply

Theorem 9 (Dahmani, Guirardel, Osin) If ${\psi}$ is iwip, some power of it normally generates a free group.

Let ${N:=\phi(\Gamma)\cap\ll\psi^N \gg}$. It is a free infinite index normal subgroup of ${\phi(\Gamma)}$. This contradicts Margulis’ normal subgroup theorem.

Case 2: Assume image ${\phi(\Gamma)}$ contains no iwip. Then apply

Theorem 10 (Handel, Mosher) If subgroup ${H\subset Out(F_n)}$ does not contain iwip’s, then, up to finite index, ${H}$ leave invariant some free factor up to conjugacy. In particular, ${H}$‘s action on homology is block triangular.

Then use induction on ${n}$. Prove that every finite index subgroup ${\Gamma_0 \subset \Gamma}$ acts on ${H_1 (F_n)}$ via a finite group.

Step 3: Reduce to the case when ${\phi}$ maps into ${IA_n}$. For this use Lie algebra methods à la Magnus.

Case 4: Assume image ${\phi(\Gamma)\subset IA_n}$. ${IA_n}$ has a filtration by ${G_i}$‘s such that ${G_i /G_{i+1}}$ are free abelian and ${\bigcup G_i}$ is trivial. ${\Gamma}$ maps trivially to all these quotients. The ${G_i}$‘s arise from the lower central series of ${F_n}$. Since ${F_n}$ is residually nilpotent (Magnus), ${\bigcup G_i =\{1\}}$.

Proposition 11 (Andreadakis) ${G_i /G_{i+1}}$ is free abelian.

The corresponding homomorphisms to ${{\mathbb Z}^d}$‘s are unfortunately called Johnson homomorphisms nowadays.