Notes of Camille Horbez’ Cambridge lecture 9-01-2017

Growth of automorphisms of hyperbolic groups

Joint with Rémi Coulon, Arnaud Hilion and Gilbert Levitt.

1. Introduction

Conjugacy length {\|g\|} is infimum of lengths of conjugates of {g}.

Question. For an outer automorphism {\Phi}, what are the possible behaviours of {\|\Phi^n(g)\|} as {n} tends to infinity, and as {g} varies in {G} ?

Surface group case is Thurston’s theory. In that case, conjugacy length is equivalent to length with respect to some hyperbolic metric on the surface.

Theorem 1 (Thurston)

  • If {\Phi} is pseudo-Anosov, all curves grow at the same exponential speed, given by the stretching factor.
  • If {\Phi} is reducible, some power of {\Phi} preserves a splitting into several pseudo-Anosov’s, growth is again exponential, governed by the maximal stretching factor.
  • If no piece is pseudo-Anosov, {\Phi} is made of commuting Dehn twists, growth is linear, unless it is bounded.

Free group case started by Bestvina-Handel, who defined train tracks for free groups.

Theorem 2 (Bestvina-Handel, Levitt) Growth is exponential times polynomial, with a finite choice of exponents. Mixed exponential/polynomial arises. Polynomial degree can take any value between 0 and {N-1} for {G=F_N}.

2. Results

Theorem 3 (Coulon, Hilion, Horbez, Levitt) Let {G} be torsion free hyperbolic. Then for all {g\in G},

\displaystyle  \begin{array}{rcl}  \|\Phi^n(g)\|^{1/n}\rightarrow \lambda, \end{array}

or, if {\|\Phi^n(g)\|^{1/n}\rightarrow 1},

\displaystyle  \begin{array}{rcl}  \|\Phi^n(g)\|\sim n^p, \end{array}

with only finitely many choices for {\lambda} and {p}, which are algebraic integers.

3. The one-ended case

Theorem 4 (Coulon, Hilion, Horbez, Levitt) If furthermore {G} is one-ended, the growth is either purely exponential or linear, or bounded.

This relies on Bowditch’s canonical JSJ decomposition: splitting with cyclic edge groups. Vertex groups are either elementary (i.e. cyclic) or nonelementary, either surface of rigid. Rigid vertex groups have finite outer automorphism groups. Pseudo-Anosov maps of surface vertex groups are responsible for exponential growth. Symmetries of the decomposition and Dehn-like twists contribute to polynomial growth.

4. The general case

No train tracks available. We use probabilistic arguments instead.

Step 1. Finding the top rate.

There is a nontrivial very small {G}-tree such that maximal exponential growth is encontered precisely by those elements which are loxodromic in that tree.

Step 2. Induction on a measurement of complexity. Possible due to the fact that point stabilizers in {T} are {\Phi}-invariant and simpler. For free group {F_N}, complexity is {N}.

5. Proof of top rate characterization

Consider action on Cayley graph twisted by {\Phi}.

Bestvina-Paulin: renormalize and extract converging sequence in equivariant Gromov-Hausdorff topology, obtain limiting action on tree. Non canonical!

Instead, consider all actions on hyperbolic metric spaces at once, they form a compact space. Use a weak limit of averages of atomic measures based at twisted actions. The assertions are proved for a.e. limiting tree actions, according to this measure.

The top exponent has the following alternative description,

\displaystyle  \begin{array}{rcl}  \log \lambda=\lim\frac{1}{n}d(X,X\circ \Phi^n), \end{array}

where {d} is the equivariant Lipschitz distance on actions.

Indeed, the upper bound {\lambda\leq\limsup \|\Phi^n(g)\|^{1/n}} is general. The lower bound uses Karlsson-Ledrappier’s claim that, for a.e. action on a tree {T},

\displaystyle  \begin{array}{rcl}  \log \lambda=\lim -\frac{1}{n}\log d(T,X\circ \Phi^n). \end{array}

Since

\displaystyle  \begin{array}{rcl}  \|g\|_T\leq d(T,X\circ \Phi^n)\|\Phi^n(g)\|, \end{array}

this shows that loxodromic elements grow at exponential speed {\lambda}.

Exponents arise from actions on trees or from the JSJ decomposition of one-ended pieces of {G}, in either case, they are algebraic integers.

5.1. Polynomial growth

Assume that { \|\Phi^n(g)\|^{1/n}} tends to 1. Levitt-Lusztig show that {\Phi} preserves a free splitting of {G}. Behaviour of { \|\Phi^n(g)\|} under free product is not that obvious.

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