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Let $F$ be a fixed free group of finite rank. If $H$ is a finitely generated subgroup of $F$ and $A$ is a basis for $F$, then we can form the Stallings graph $\Gamma_A(H)$ for $H$. It is the unique smallest ($|A|$-labelled) subgraph of the covering space of a bouquet of $A$-circles corresponding to $H$ that contains the base point and carries the fundamental group.

It is easy to see that if $\Gamma_A(H)$ embeds in $\Gamma_A(K)$, then $H$ is a free factor in $K$. If one fixes the basis $A$ then there are free factors of $F$ itself that are not induced by graph embeddings.

Let us say that a finitely generated subgroup $H$ of $F$ is a geometric free factor of a subgroup $K$ if there exists some basis $B$ for $F$ such that $\Gamma_B(H)$ embeds as a labelled graph in $\Gamma_B(K)$.

Question. Is every free factor $H$ of a finitely generated subgroup $K$ of a free group $F$ a geometric free factor?

History. I came upon this question in some joint work with Auinger on finite semigroups. To solve some problem we needed all geometric free factors of open subgroups to be closed in the pro-C topology in order for nice things to happen. We would have liked to have been able to drop the word "geometric." Enric Ventura has independently considered this question.

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2 Answers 2

up vote 11 down vote accepted

If you take a free group on two generators to be $F=\langle a,b\rangle$, and take an index 2 subgroup to be $K$, then the subgroup $H$ generated by $[a,b]$ is not a geometric free factor in $K$ for any choice of generators for $F$ (but is a free factor). One may verify this for all three 2-generator subgroups of $F$ with respect to the generators $a,b$. Changing to another set of generators (applying an automorphism of $F$) sends the commutator to a conjugate of commutator of generators, which is therefore not geometric with respect to any generating set.

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I should have though to try a finite index subgroup since the Stallings graph is the whole covering space. Do you know examples where both subgroups have infinite index? –  Benjamin Steinberg May 2 '12 at 13:04
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@Benjamin: I think just wedge this example with another loop. –  Ian Agol May 2 '12 at 14:50

Here is what I think will be an infinite index example, although I'm missing some details of proof. Take $F = \langle a,b \rangle$, identified with the fundamental group of a rose $R$ with two petals labelled $a$, $b$. Let $\Theta$ be the rank 2 graph with oriented edges $e_1,e_2,e_3$ having the same initial and terminal endpoints, labelled by three reduced words $w_1,w_2,w_3$ in the letters $a,b,a^{-1},b^{-1}$. Make sure the initial three letters of these words are all distinct, as are the terminal three letters, so the induced map $\Theta \to R$ is an immersion and the $\pi_1$-image of $\Theta$ is an infinite index, rank 2 subgroup $B$ in $F$. Make sure that the words $w_1,w_2,w_3$ are really long and random, for instance at the very least each of them should obtain every possible length 10 subword of $F$. Now start applying the standard generators of $Aut(F)$, for example $a \to a$, $b \to ab$, and you will see that the induced self-map on the graph $\Theta$ is homotopic to a homeomorphism. I believe that this property continues to be true as you apply any word in the generators of $Aut(F)$. As a consequence, the only geometric subgroups of $B$ are those represented by the three loops in $\Theta$. Roughly speaking the reason for this behavior is that any element of $Aut(F)$ can be represented by an equivariant Stallings fold sequence on the universal cover of $R$, and if two segments of $R$ are identified under this fold sequence then those two segments never have the property that they contain every length 10 word of $F$, so the only identifications induced on $\Theta$ are short initial segments of its three edges.

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