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Denote by $F_n$ the free group of rank $n$. We say that an automorphism $\phi\in Aut(F_n)$ is geometric if there exists a surface with boundary $M$ and a homeomorphism $h\colon M\to M$ such that $h$ induces $\phi$ on $\pi_1$.

Every automorphism of $F_2$ is geometric, but in higher rank geometric automorphisms are rare. I have seen several examples of nongeometric automorphisms, but all of which were of infinite order.

Question: Are there any examples of periodic (i.e. finite-order) nongeometric automorphism? Or can one show that periodic automorphisms are always geometric, possibly making use of the fact that every periodic automorphism can be realized as a simplicial graph automorphism? I wouldn't be surprised if we could nicely embed the graph into a suitable surface and extend the automorphism of the graph to a surface homeomorphism.

Observation: Every periodic automorphism $\phi$ fixes some nontrivial conjugacy class (maybe up to inversion), which is a necessary condition for $\phi$ to possibly be induced by a homeomorphism of a surface with connected boundary. But this observation really only plays a role when $n$ is even, as the fundamental group of a surface with connected boundary has even rank.

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When you speak about automorphisms of free groups, do you really mean outer automorphisms? I ask because $h : M \to M$ does not induce a well-defined automorphism of $\pi_1 M$, at best it induces an outer automorphism. –  Lee Mosher May 6 '13 at 21:22
    
The definition of geometric automorphisms also makes sense for actual (i.e. not outer) automorphisms, but basepoints would need to be preserved. If an outer auto $\mathcal{O}$ is geometrically realized by $h$ then any auto $\phi$ representing $\mathcal{O}$ is realized by the basepoint-preserving homeomorphism $p_{\phi}\circ h$, where $p_{\phi}$ is a suitable push map, a surface homeomorphism isotopic to the identity that pushes $h(\mathrm{basepoint})$ back to the basepoint. However, an arbitrary surface homeomorphism does not induce a well-defined actual free group auto, that's correct. –  Sebastian May 7 '13 at 9:54

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up vote 7 down vote accepted

Assuming that you mean outer automorphisms of $F_n$ (as per my comment), the answer is that if $n \ge 3$ then there is a nongeometric finite order element of $Out(F_n)$, for example the outer automorphism class of $a_1 \mapsto a_1^{-1}$ and $a_i \mapsto a_i$ for $2 \le i \le n$.

For the proof, given $h : M \to M$ that induces a nonidentity, finite order, outer automorphism $\phi$ of $\pi_1 M$, I'll show that there does not exist a free factor of $\pi_1 M$ of rank $\ge 2$ whose conjugacy class is fixed by $\phi$ and whose elements all have conjugacy classes fixed by $\phi$, in contrast with the above example having such a free factor of rank $n-1$.

By Nielsen realization, after an isotopy of $h$ we may assume that $h$ is a finite order isometry with respect to some hyperbolic structure on $M$ with totally geodesic boundary. If there is a free factor $G < \pi_1 M$ of rank $\ge 2$ with conjugacy class fixed by $h$ then $h$ lifts to the $G$-cover $\tilde M$, and that lift restricts to an isometry of the convex hull $\tilde h : \mathcal{H}(\tilde M) \to \mathcal{H}(\tilde M)$. Since $rank(G) \ge 2$, the convex hull $\mathcal{H}(\tilde M)$ is itself a compact hyperbolic surface with totally geodesic boundary (if $rank(G)$ were equal to $1$ then the convex hull would just be a closed geodesic, and no contradiction would arise). Since $h$ fixes the conjugacy classes of all the elements of $G$, and since $G$ is a free factor (malnormality of $G$ suffices), it follows that $h$ induces the identity outer automorphism of $\pi_1(G)$. From this it follows in turn that $\tilde h$ is the identity isometry on $\mathcal{H}(\tilde M)$. It follows by projection that $h$ is locally the identity at some points of $M$, which by analytic continuation implies that $h$ is globally the identity. So $h$ induces the identity outer automorphism.

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Thank you very much for this nice example! –  Sebastian May 7 '13 at 9:59
    
You're welcome, Sebastian. –  Lee Mosher May 7 '13 at 13:20

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