**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 and exponent , , is the group on generators, with relators all the -th powers. Every automorphism of the free group defines an automorphism of . Converse is false. For instance, has arithmetically defined automorphisms which do not arise in this manner. But we shall ignore these.

Theorem 1 (Coulon, Hilion)Let . The following are equivalent.

- has exponential growth.
- such that the induced automorphism of has infinite order.
- such that for all odd , the induced automorphism of has infinite order.

Recall from Cashen’s talk that exponential growth means that

Theorem 2 (Coulon)Let have polynomial growth. Then for all , the induced automorphism of has finite order.

Proof by induction on . Indeed, either preserves a free splitting or , and for some . I illustrate the rest of the proof on an example, which is in fact an older theorem.

Theorem 3 (Cherepanov 2005)Let map , . Then has exponential growth equal to the golden ratio. There exists such that the induced automorphism of has infinite order.

Note that in the sequence , 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 is primitive).

Next use Novikov-Adian’s criterion for non vanishing of , large: if is a reduced word which does not contain fourth powers, then in for odd .

Here is an other example. Let map , , , . This preserves a free factor , and has polynomial growth on that factor. On the other hand, there is expansion on and , and growth is exponential. Words contain high powers of . Kill , 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 is a red edge, is a path consisting of yellow and red edges, begining and ending by red edges. The set of all the maximal yellow subpaths of ( describing all the red edges) is finite. Thus, there exists such that for odd , loops in , viewed as element of , are non trivial in .

After collapsing all yellow subpaths in , 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 -th powers, for some .

Theorem 4 (Coulon)Let be a reduced word. Assume is trivial in . Then there exists a finite sequence of elementary moves which sends to , where an elementary move is the following: if you see some power with in , then replace by in .

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 implies that if is trivial in , elementary moves can only occur in the yellow part of .

Taking implies that no maximal yellow subpath of can be removed by an elementary move. Indeed, such a path is some for some . If it is trivial in , then has also to be trivial in since represent an automorphism.

Thus the induced automorphism of has infinite order, for .