**On the difficulty of inverting automorphisms of free groups**

Joint work with Pedro Silva and Manuel Ladra

**1. Motivation **

I will give an application of the asymetry of the metric on Outer space.

**Question [Myasnikov]**: Can you find a group where multiplication is easy but inverting is difficult ?

Multiplying in is very easy. Inversion does not seem to be.

**Example**:

Then inverse involves a tower of power operations.

We shall formalize the problem, and see that inversion is not that hard, in fact.

** 1.1. Inversion growth **

** 1.2. Notation **

Definition 1In a group generated by , define norms for automorphisms as follows,

Definition 2In a group , let be the worst norm of the inverse of automorphisms of length .

**Question**: How does this function grow ?

We shall concentrate on the free group . One uses cyclic length (min of lengths of cyclic conjugates), although it does not satisfy triangle inequality. One uses a third length, where one maximizes over conjugations (where the same conjugation is applied to all generators). This gives three inversion growth functions, , and , which are not equivalent.

**2. Lower bounds for free groups **

** 2.1. Rank case **

Theorem 3is quadratic, and are linear.

** 2.2. Rank case **

Theorem 4, is at most polynomial and .

I expect that all three functions are exactly polynomial of nearby degrees.

** 2.3. Proof of the lower bound on **

Use the obvious generalization of the example above. Generators are sent to cyclically reduced elements, so cyclic lengths are easy to calculate.

** 2.4. Proof of the lower bound on **

One wins a factor of by composing the previous example with conjugation by a large element.

**3. Upper bounds **

We use the asymetric distance on Outer space (see Martino’s talk).

Definition 5For , the -thick part of Outer space is the set of metric graphs in which all cycles have length .

Theorem 6 (Bestvina, Algom-Kfir)On , .

Corollary 7All three inversion functions are bounded above by .

Indeed, given , consider a rose . Then equal up to constants. . Both and belong to .

**4. Case **

Theorem 8For ,

Proof uses a decomposition of automorphisms. Every can be written

where is inner, are letter permutations and is positive.

**5. Fixed subgroups **

Fixators of automorphisms can be complicated: for instance, generated by unexpectedly long elements (Stallings).

Theorem 9 (Bestvina-Handel)Fixed point sets of arbitrary automorphisms have rank .

They introduced train tracks for that purpose. Later extensions to endomorphisms.

** 5.1. Inertia **

Definition 10Say a subgroup is inert if for every subgroup .

Theorem 11 (Dicks, Ventura 1996)Fixed point sets of sets of monomorphisms are inert.

Does this hold for general endomorphisms ?

Theorem 12 (Martino-Ventura, Ciobanu-Dicks)There exist subgroups which are not fixed point sets.

** 5.2. Algorithmic issues **

Theorem 13 (Maslakova 2003, corrected by Dicks 2011)Fixed subgroups of endomorphisms are computable.

Theorem 14 (Ventura 2010)Given a subgroup, one can decide wether it is the fixed point set of a set of endomorphisms (resp. endomorphisms).

But I cannot decide if it is the fixed point set of a single endomorphism (resp. endomorphism).