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Does there exist an infinite order element $\phi\in Out(F_n)$, for some or all $n\geq 3$, which is not iwip but has finite index in its centralizer? How about an element such that all its non-zero powers have this property?

Motivation: iwip elements are Morse (i.e., roughly speaking, all quasi-geodesics connecting points on a given orbit stay close to it), and a standard way of proving that an element is not Morse is showing that it has infinite index in its centralizer, or one of its powers do. An element in the mapping class group of a closed surface is Morse if and only if it is pseudo-Anosov.

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How do you prove this for pseudo-Anosovs? – Igor Rivin Dec 7 '11 at 20:43
Can you please recall what iwip means? – Alain Valette Dec 7 '11 at 21:22
Behrstock proved an equivalent property in this paper (Theorem 6.5): Another proof can be found here: The idea of this last proof is best explained pretending that curve complexes are trees. If this was true then all paths connecting markings projecting on distant simplices will have to cross the pre-images in the marking complex of certain specified simplices. An orbit of a pA does this "as fast as possible", and one can show that a path staying far from such orbit can't "go straight" (acylindricity) and so it is much longer. – Alessandro Sisto Dec 7 '11 at 21:31
An element of $Out(F_n)$ is irreducible if it does not fix (the conjugacy class of) some free factor $F<F_n$ (i.e. $F_n=F*G$ for some other subgroup $G$). It is irreducible with irreducible powers (iwip) if its powers have the same property as well. – Alessandro Sisto Dec 7 '11 at 21:34
I would guess, that such automorphisms exist, because there is a natural embedding of, say, $Aut(F_{3})$ into $Out(F_4)$, where the image one acts normally on the subgroup, generated by the first three free generators, and trivially on the fourth generator. So, taking an arbitrary automorphism of $F_3$ and extending it this way to an automorphism of $F_4$ could produce the needed example... Am I wrong here? – Ashot Minasyan Dec 8 '11 at 12:30
up vote 6 down vote accepted

There is a Nielsen-Thurston type method due to Feighn and Handel which is useful for approaching this question. The method is laid out in the papers arXiv:math/0612702 and arXiv:math.GR/0612705 by Feighn and Handel, "The recognition theorem for $Out(F_n)$ and "Abelian subgroups of $Out(F_n)$". It is an outgrowth of the train track and relative train track machinery developed in earlier papers of Bestvina, Feighn, and Handel, particularly math.GT/9712217, part I of their series on the Tits alternative for $Out(F_n)$.

As Ashot suggests, take $F_4 = \langle a,b,c,d \rangle$ and take $\Phi \in Aut(F_4)$ to preserve $\langle a,b,c\rangle $ so that the action on $\langle a,b,c\rangle $ has no periodic conjugacy classes. This implies that $\Phi(d) = u d^{\pm 1} v$ for words $u,v$ in $a,b,c$. One could take $\Phi(d)=d$ as Ashot suggests but I prefer $u$ and/or $v$ to be nontrivial for reasons explained below. Any such $\Phi$ is a "principal automorphism", which is guaranteed by having a fixed subgroup (when $u$, $v$ are trivial) or by having no fixed subgroup but having three or more attracting periodic points in the boundary of the free group (when $u$ and/or $v$ are nontrivial). Furthermore, if you throw in the condition that the action of $\Phi$ on the $a,b,c$ rose is a train track map and has no periodic Nielsen paths, then $\Phi$ is the ONLY principal automorphism in its outer automorphism class $\phi \in Out(F_n)$ up to conjugation by inner automorphism.

The centralizer $C(\phi)$ in $Out(F_n)$ has to act on the "principal data" of $\phi$, meaning that it has to permute the conjugacy classes of fixed subgroups, and it has to permute orbits of attracting periodic points in the boundary. Furthermore, $\phi$ has an attracting lamination $\Lambda$ that is supported in the $a,b,c$ subgroup, and there is an association between certain attracting periodic points in the boundary and certain rays of the lamination $\Lambda$, and this association implies that $C(\phi)$ preserves $\Lambda$ and so has a well-defined stretch factor homomorphism for $\Lambda$. The gist of the Recognition Theorem in this situation is that an element of $C(\phi)$ is more-or-less determined by all of this "principal data". The situation is a little more complicated to analyze when there is a nontrivial fixed subgroup, so for that reason I actually prefer the opposite case where $u$ and/or $v$ is nontrivial, which for the class of examples under consideration implies that there is no fixed subgroup for any automorphism representing $\phi$ (this is one place where $Out(F_n)$ departs from $MCG(S)$, for if a mapping class is not pseudo-Anosov then after passing to a power there MUST be a principal automorphism having nontrivial fixed subgroup). So in that case, $C(\phi)$ acts on the set of orbits of attracting periodic points, this is a finite set, and so the action has a finite index kernel; the kernel of that action has a stretch factor homomorphism to $Z$, and the kernel of THAT homomorphism is trivial.

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The mapping class group question has a solution using Nielsen-Thurston theory, which goes like this. If $\phi \in MCG(S)$ is pseudo-Anosov and if $\Lambda$ is the stable or unstable lamination of $\phi$ then the subgroup of $MCG(S)$ that preserves $\Lambda$ and has stretch factor 1 on $\Lambda$ is a finite group (there is a proof in "Travaux de Thurston sur les surfaces" MR1134426). The centralizer of $\phi$ must fix the projective class of $\Lambda$, and so it has a homomorphism to $Z$ (the stretch factor homomorphism) with finite kernel.

By the way, the same proof outline basically works for elements of $Out(F_n)$ that are fully irreducible (a.k.a. i.w.i.p.), which means that each conjugacy class of proper nontrivial free factors of $F_n$ is nonperiodic. The proof is given in the Bestvina, Feighn, Handel paper MR1445386, which is essentially Part 0 of their Tits alternative series. Any such outer automorphism $\phi$ has an attracting lamination $\Lambda$, the centralizer of $\phi$ fixes $\Lambda$ (as an abstract lamination), the stabilizer of $\Lambda$ has a well-defined stretch homomorphism, and the kernel of that homomorphism is finite.

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