English

When is a polynomially growing automorphism of $F_n$ geometric ?

Group Theory 2016-05-25 v1

Abstract

The main result of this paper is an algorithmic answer to the question raised in the title, up to replacing the given ϕ^Out(Fn)\hat{\phi} \in Out(F_n) by a positive power. In order to provide this algorithm, it is shown that every polynomially growing automorphism ϕ^\hat \phi can be represented by an iterated Dehn twist on some graph-of-groups G\cal{G} with π1G=Fn\pi_1{\cal{G}} = F_n. One then uses results of two previous papers \cite{KY01, KY02} as well as some classical results such as the Whitehead algorithm to prove the claim.

Keywords

Cite

@article{arxiv.1605.07390,
  title  = {When is a polynomially growing automorphism of $F_n$ geometric ?},
  author = {Kaidi Ye},
  journal= {arXiv preprint arXiv:1605.07390},
  year   = {2016}
}