## Notes of Arnaud Hilion’s talk

Automorphisms of free groups and free Burnside groups

Joint work with Rémi Coulon

1. Rémi Coulon’s results on free Burnside groups

The free Burnside group of rank ${r}$ and exponent ${n}$, ${B_r (n)}$, is the group on ${r}$ generators, with relators all the ${n}$-th powers. Every automorphism of the free group defines an automorphism of ${B_r (n)}$. Converse is false. For instance, ${B_1 (n)={\mathbb Z}/n{\mathbb Z}}$ has arithmetically defined automorphisms which do not arise in this manner. But we shall ignore these.

Theorem 1 (Coulon, Hilion) Let ${\phi\in Aut(F_r)}$. The following are equivalent.

1. ${\phi}$ has exponential growth.
2. ${\exists n\in{\mathbb N}}$ such that the induced automorphism of ${B_r (n)}$ has infinite order.
3. ${\exists n_0 \in{\mathbb N}}$ such that for all odd ${n\geq n_0}$, the induced automorphism of ${B_r (n)}$ has infinite order.

Recall from Cashen’s talk that exponential growth means that

$\displaystyle \begin{array}{rcl} \sup_{g\in G}\limsup_{k\rightarrow\infty}\frac{\log|\phi^k (g)|}{k}>0. \end{array}$

Theorem 2 (Coulon) Let ${\phi\in Aut(F_r)}$ have polynomial growth. Then for all ${n\in{\mathbb N}}$, the induced automorphism of ${B_r (n)}$ has finite order.

Proof by induction on ${r}$. Indeed, either ${\phi}$ preserves a free splitting or ${F_r=F\star\langle t\rangle}$, ${\phi(F)=F}$ and ${\phi(t)=tu}$ for some ${u\in F}$. I illustrate the rest of the proof on an example, which is in fact an older theorem.

Theorem 3 (Cherepanov 2005) Let ${\phi}$ map ${a\mapsto ab}$, ${b\mapsto a}$. Then ${\phi}$ has exponential growth equal to the golden ratio. There exists ${n\in{\mathbb N}}$ such that the induced automorphism of ${B_r (n)}$ has infinite order.

Note that in the sequence ${\phi^k (a)}$, each word is a prefix of the next one, there are squares but no fourth powers (this is due to the fact that the matrix ${\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}}$ is primitive).

Next use Novikov-Adian’s criterion for non vanishing of ${B_r (n)}$, ${n}$ large: if ${w\in F_r}$ is a reduced word which does not contain fourth powers, then ${w\not=1}$ in ${B_r (n)}$ for odd ${n>10000}$.

Here is an other example. Let ${\phi}$ map ${a\mapsto a}$, ${b\mapsto ba}$, ${c\mapsto cbd}$, ${d\mapsto dcbd}$. This preserves a free factor ${\langle a,b \rangle}$, and has polynomial growth on that factor. On the other hand, there is expansion on ${c}$ and ${d}$, and growth is exponential. Words contain high powers of ${a}$. Kill ${a}$, and Novikov-Adian’s criterion applies.

It is easy to produce examples where the criterion does not apply to any quotient, so we need an improved criterion.

2. New criterion

One can locate the source of exponential growth via a relative train track map. If growth is exponential, there is (at least) one exponential stratum. One can assume that it is a top stratum. Then growth is polynomial on the other strata. We color edges of the top stratum in red, and edges of the other strata in yellow. If ${e}$ is a red edge, ${f(e)}$ is a path consisting of yellow and red edges, begining and ending by red edges. The set ${\Gamma}$ of all the maximal yellow subpaths of ${f(e)}$ (${e}$ describing all the red edges) is finite. Thus, there exists ${n_1}$ such that for odd ${n>n_1}$, loops in ${\Gamma}$, viewed as element of ${F_r}$, are non trivial in ${B_r (n)}$.

After collapsing all yellow subpaths in ${f^k(e)}$, get a word in alphabet consisting of red edges. This word is, a priori, not reduced. The fact that the transition matrix of the red stratum is primitive implies that this word does not contain any ${n_2}$-th powers, for some ${n_2}$.

Theorem 4 (Coulon) Let ${w\in F_r}$ be a reduced word. Assume ${w}$ is trivial in ${B_r (n)}$. Then there exists a finite sequence of elementary moves which sends ${w}$ to ${1}$, where an elementary move is the following: if you see some power ${u^p}$ with ${p>n/4}$ in ${w}$, then replace ${u^p}$ by ${u^{n-p}}$ in ${w}$.

This is some kind of “Dehn semi-algorithm”. Semi because the length may increase. But after a few steps, length decreases. Coulon uses a geometric method (mesoscopic curvature according to Delzant-Gromov) for the proof.

Taking ${n>n_2}$ implies that if ${\bar{e}f^k(e)}$ is trivial in ${B_r (n)}$, elementary moves can only occur in the yellow part of ${\bar{e}f^k(e)}$.

Taking ${n>n_1}$ implies that no maximal yellow subpath of ${\bar{e}f^k(e)}$ can be removed by an elementary move. Indeed, such a path is some ${f^q(\gamma)}$ for some ${\gamma)\in\Gamma)}$. If it is trivial in ${B_r (n)}$, then ${\gamma}$ has also to be trivial in ${B_r (n)}$ since ${f}$ represent an automorphism.

Thus the induced automorphism of ${B_r (n)}$ has infinite order, for ${n>max(n_1,n_2)}$.