Solutions to some equations in a free group

Question

Let $F$ be the free group on (say) two generators, $a$ and $b$. Let $A$ and $B$ be (freely reduced) elements of $F$. Let $W(X, Y)$ denote a word on the words $X, Y$.

-Is it ever true that the equation $W(a, b) = W(A, B)$, has finitely many non-conjugate solutions (by conjugate solution I mean there exists a word $V$ such that $V^{-1}AV = A^{\prime}$ and $V^{-1}BV=B^{\prime}$)?

For example, take $W(a, b) = a^{-1}b^nab^m$. We therefore want to find $A$, $B$ such that $a^{-1}b^nab^m = A^{-1}B^nAB^m$. We can take $A=b^ia$ and $B=b$ for all $i$, and so this equation has infinitely many (non-conjugate) solutions.

In fact, $a\mapsto b^ia$, $b \mapsto b$ defines an automorphism of $F$ (as free groups are Hopfian). Further, different $i$'s give different coset representatives of Out(F), and so a related question would be,

-Does there exist a word $W \in F$ such that there are only finitely many outer automorphisms $\phi$ such that $W\phi = W$?

I cannot seem to get anywhere with this. The only examples I can find are, essentially, trivial. For example, $W(a, b) = a$. However, this doesn't quite work, as then $b$ can be whatever we want (essentially, exclude this because its boring).

Any help/ideas of papers to look at would be greatly appreciated.

Answer 1

Let $G=\langle x,y,c\mid w(x,y)=c\rangle$ (which, in this case, happens to be a free group). The set of solutions to the system that you are looking for corresponds precisely to the set of homomorphisms $f:G\to F$ satisfying $f(c)=w(a,b)$. The basic idea of Sela's approach to the study of equations over $F$ (see here, et seq.) is to study the splittings of $G$ as an amalgamated free product or HNN extension. Specifically, we have the following. (I'm not sure this exact statement is written down anywhere, but I think I've got it correct. To really do this right, you should probably intone the magic words 'graded limit group' somewhere.)

Theorem (Sela): There are infinitely many conjugacy classes of solutions if and only if $G$ splits (as an amalgamated free product or HNN extension) over a cyclic group, with infinite centraliser, relative to $c$. (That is, $c$ should be conjugate into one of the vertex groups of the splitting.)

Sketch proof: If there are infinitely many solutions then one can take a Gromov--Hausdorff limit and get an action on a real tree. The Rips machine promotes this to an action on a simplicial tree.

Conversely, if $G$ splits as hypothesised then a Dehn twist in this splitting gives one infinitely many outer automorphisms that preserve $w$, as required. QEDish

So, in order to write down an example of the form you want, you simply need a $w$ such that $F$ has no cyclic splittings relative to $w$. This is easier said than done, though in principle any `sufficiently generic' $w$ should do. In a recent paper, Cashen and Macura described an algorithm for finding cyclic splittings of $F$ relative to $w$. (In fact, they describe their algorithm as finding cut pairs in a certain decomposition space. But these turn out to correspond to cyclic splittings, as Cashen has shown.)

According to Theorem 6.3 of Cashen and Macura, any word $w$ for which the Whitehead graph is complete will be `rigid', ie $F$ does not split cyclically relative to $w$. For instance, I think the word

$w=aba^{-1}b^{-2}ab^{-1}a^{-2}$

works.

Answer 2

The free group on two generators is exceptional: the group of outer automorphisms of $F(a,b)$ is $GL(2, \mathbb Z)$. That's because every automorphism preserves the commutator $[a,b]$ up to conjugacy. Every outer automorphism of the fundamental group of a surface with boundary that permutes the boundary components up to conjugacy is represented by a homeomorphism of the surface, by a theorem I believe of Nielsen, so outer automorphisms of $F(a,b) \leftrightarrow $ isotopy classes of homeomorphisms of a torus minus an open disk $\leftrightarrow GL(2,\mathbb Z)$ acting on $\mathbb R^2/\mathbb Z^2$ (you can remove the lattice points to make it a punctured torus).

A free homotopy class of curves $\alpha$ on a surface fills the surface is for any other non-peripheral homotopy class $\beta$, it is impossible to homotope $\alpha$ and $\beta$ so they don't intersect. Surfaces of negative Euler characteristic such as the punctured torus have complete Riemannian metrics of constant negative curvature (hyperbolic structures), where every free homotopy class is represented by a unique geodesic. In a hyperbolic structure $\alpha$ fills the surface if and only if every component of the surface minus $\alpha$ is either simply-connected, or it retracts to an arbitrarily small neighborhood of a puncture.

The rubber band theorem of Steve Kerckhoff says that for a homotopy class $\alpha$ that fills a surface $S$ of finite type and negative Euler characteristic, there is a unique hyperbolic structure for which the length of $\alpha$ is minimized: a rubber band following $\alpha$ pulls the surface into a specific shape. (The length of $\alpha$ is a convex function, in a certain sense, in Teichmüller space: this uses either my earthquake theorem, or Scott Wolpert's geometric analysis of the Weil-Peterson metric. This was the key step in Kerckhoff's proof of the Nielsen fixed point theorem which had an infamous history).

In particular, any outer automorphism of the free group that fixes such a conjugacy class is homotopic to an isometry of a particular hyperbolic structure, which implies the group of all such elements is finite. For an $\alpha$ that doesn't fill the surface, a Dehn twist about any non-intersecting simple closed curve generates an infinite group of automorphisms, as in your example. In the figure, the arcs labeled a* and b* keep track of a homorphism to the free group on two generators.

alt text http://dl.dropbox.com/u/5390048/UnfilledTorus.png

It's easy to invent lots of homotopy classes like this on the punctured torus, just as you wrap a ribbon around a present. The curve illustrated below cuts the punctured torus into an open annulus around the puncture, and an open disk, so it qualifies. I leave the word description of this curve as an exercise.

alt text http://dl.dropbox.com/u/5390048/gift-torus.png

For free groups on more generators, the theory is trickier, but there is a beautiful theory developed by Bestvina and Handel of train track maps representing automorphisms of free groups, that gives a method to determine what elements are fixed up to conjugacy by a particular automorphism. I believe this will give examples in the general case, but I think it takes some work and I haven't thought through the details. A related subject: there's an elaborate theory of when word-hyperbolic groups can have infinite automorphism groups, via the theory of group actions on $\mathbb R$-trees; and your question is close to a special case of this, for one-relator groups.

Answer 3

These equations have been considered here: Hmelevskiĭ, Ju. I. Systems of equations in a free group. I, II. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 35 (1971), 1237–1268; ibid. 36 (1972), 110–179. All solutions have been described there (see Theorem 3).

Update. See also more recent Touikan, Nicholas W. M.(3-MGL) The equation $w(x,y)=u$ over free groups: an algebraic approach. J. Group Theory 12 (2009), no. 4, 611–634.