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Let $F_n$ denote the free group on $n$ generators $x_1,\ldots , x_n$.

Recall that an element $\varphi\in\mathrm{Aut}(F_n)$ is an IA automorphism if it induces the identity on the abelianization $F_n/[F_n,F_n]=\mathbb{Z}^n$. In other words, $\varphi$ is in the kernel of the induced map $\mathrm{Aut}(F_n)\to Gl(n,\mathbb{Z})$. This kernel is denoted $\mathrm{IA}_n$.

Does every $\varphi\in\mathrm{IA}_n$ have a fixed point in the complement of the commutator subgroup $[F_n,F_n]$?

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  • $\begingroup$ Why do you think it might be true for $n\ge 3$? $\endgroup$ – Mark Sapir Jul 31 '13 at 14:19
  • $\begingroup$ @MarkSapir Wishful thinking, mostly. I am interested in any information or references regarding IA automorphisms of free groups. $\endgroup$ – Mark Grant Jul 31 '13 at 14:28
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    $\begingroup$ Did you look at elements of the Torelli subgroup for a punctured surface? $\endgroup$ – Mark Sapir Jul 31 '13 at 14:34
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    $\begingroup$ As Mark says, just take a pseudo-Anosov homeomorphism of a genus g surface with 1 boundary component which lies in Torelli subgroup. Then, even up to conjugation, only the peripheral elements are fixed, and they are in the commutator subgroup. $\endgroup$ – Misha Jul 31 '13 at 15:32
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    $\begingroup$ You can construct examples with no fixed elements at all, out of the examples mentioned by Misha. Let $S$ be of genus $g$ with $1$ boundary component, consider two isomorphisms from $\pi_1(S)$ to $F_n$ taking the boundary component to two nonconjugate elements, push forward two Torelli pseudo-Anosov mapping classes on $S$ under these two isomorphisms respectively to get two $IA_n$ automorphisms of $F_n$ fixing different conjugacy classes, then multiply high powers of those automorphisms together. The resulting $IA_n$ automorphism will fix no conjugacy classes at all. $\endgroup$ – Lee Mosher Jul 31 '13 at 22:02
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Here's a compilation of some of the answers given in comments.

As pointed out by Mark Sapir and Misha, if $S$ is a compact surface with one boundary component whose fundamental group is equipped with an isomorphism $\alpha : \pi_1(S) \to F_n$, and if $f : S \to S$ is any pseudo-Anosov homeomorphism, then the outer automorphism of $\pi_1(S)$ induced by $f$ ($\leftrightarrow$ the mapping class of $f$) pushes forward via $\alpha$ to give an outer automorphism $\phi$ of $F_n$ whose only periodic conjugacy classes are those associated to the boundary component of $S$. So for any automorphism representing $\phi$, all of its fixed points are in the commutator subgroup. And one can easily construct such a pseudo-Anosov $f$ which is in the Torelli subgroup of the mapping class group of $F_n$: apply Penner's recipe, taking two filling, separating curves $c,d$ and composing a positive Dehn twist about $c$ with a negative Dehn twist about $d$. The (outer) automorphism of $F_n$ induced by such an $f$ is in $IA_n$.

But from those examples, one can construct further $IA_n$ outer automorphisms whose only periodic conjugacy class is the identity. To do this, pick two isomorphisms $\alpha_1,\alpha_2 : \pi_1 S \to F_n$ taking the conjugacy class represented by $\partial S$ to two distinct conjugacy classes in $F_n$. Let $\phi_1,\phi_2$ be the two $IA_n$ outer automorphisms of $F_n$ obtained by pushing $f_*$ forward via $\alpha_1,\alpha_2$ respectively, so their fixed conjugacy classes are distinct from each other. Then for high powers of the exponents, the $IA_n$ outer automorphism $\phi_1^i \phi_2^j$ has no periodic conjugacy class at all; this is a consequence of Theorem 1 of this recent preprint of Pritam Ghosh.

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