## Polygonal words

Last Friday, Henry Wilton gave a talk at Caltech about his recent joint work with Sang-hyun Kim on polygonal words in free groups. Their work is motivated by the following well-known question of Gromov:

Question(Gromov): Let $G$ be a one-ended word-hyperbolic group. Does $G$ contain a subgroup isomorphic to the fundamental group of a closed hyperbolic surface?

Let me briefly say what “one-ended” and “word-hyperbolic” mean.

A group is said to be word-hyperbolic if it acts properly and cocompactly by isometries on a proper $\delta$-hyperbolic path metric space — i.e. a path metric space in which there is a constant $\delta$ so that geodesic triangles in the metric space have the property that each side of the triangle is contained in the $\delta$-neighborhood of the union of the other two sides (colloquially, triangles are thin). This condition distills the essence of negative curvature in the large, and was shown by Gromov to be equivalent to several other conditions (eg. that the group satisfies a linear isoperimetric inequality; that every ultralimit of the group is an $\mathbb{R}$-tree). Free groups are hyperbolic; fundamental groups of closed manifolds with negative sectional curvature (eg surfaces with negative Euler characteristic) are word-hyperbolic; “random” groups are hyperbolic — and so on. In fact, it is an open question whether a group $G$ that admits a finite $K(G,1)$ is word hyperbolic if and only if it does not contain a copy of a Baumslag-Solitar group $BS(m,n):=\langle x,y \; | \; x^{-1}y^{m}x = y^n \rangle$ for $m,n \ne 0$ (note that the group $\mathbb{Z}\oplus \mathbb{Z}$ is the special case $m=n=1$); in any case, this is a very good heuristic for identifying the word-hyperbolic groups one typically meets in examples.

If $G$ is a finitely generated group, the ends of $G$ really means the ends (as defined by Freudenthal) of the Cayley graph of $G$ with respect to some finite generating set. Given a proper topological space $X$, the set of compact subsets of $X$ gives rise to an inverse system of inclusions, where $X-K'$ includes into $X-K$ whenever $K$ is a subset of $K'$. This inverse system defines an inverse system of maps of discrete spaces $\pi_0(X-K') \to \pi_0(X-K)$, and the inverse limit of this system is a compact, totally disconnected space $\mathcal{E}(X)$, called the space of ends of $X$. A proper topological space is canonically compactified by its set of ends; in fact, the compactification $X \cup \mathcal{E}(X)$ is the “biggest” compactification of $X$ by a totally disconnected space, in the sense that for any other compactification $X \subset Y$ where $Y-X$ is zero dimensional, there is a continuous map $X \cup \mathcal{E}(X) \to Y$ which is the identity on $X$.

For a word-hyperbolic group $G$, the Cayley graph can be compactified by adding the ideal boundary $\partial_\infty G$, but this is typically not totally disconnected. In this case, the ends of $G$ can be recovered as the components of $\partial_\infty G$.

A group $G$ acts on its own ends $\mathcal{E}(G)$. An elementary argument shows that the cardinality of $\mathcal{E}(G)$ is one of $0,1,2,\infty$ (if a compact set $V$ disconnects $e_1,e_2,e_3$ then infinitely many translates of $V$ converging to $e_1$ separate $e_3$ from infinitely many other ends accumulating on $e_1$). A group has no ends if and only if it is finite. Stallings famously showed that a (finitely generated) group has at least $2$ ends if and only if it admits a nontrivial description as an HNN extension or amalgamated free product over a finite group. One version of the argument proceeds more or less as follows, at least when $G$ is finitely presented. Let $M$ be an $n$-dimensional Riemannian manifold with fundamental group $G$, and let $\tilde{M}$ denote the universal cover. We can identify the ends of $G$ with the ends of $\tilde{M}$. Let $V$ be a least ($n-1$-dimensional) area hypersurface in $\tilde{M}$ amongst all hypersurfaces that separate some end from some other (here the hypothesis that $G$ has at least two ends is used). Then every translate of $V$ by an element of $G$ is either equal to $V$ or disjoint from it, or else one could use the Meeks-Yau “roundoff trick” to find a new $V'$ with strictly lower area than $V$. The translates of $V$ decompose $\tilde{M}$ into pieces, and one can build a tree $T$ whose vertices correspond to to components of $\tilde{M} - G\cdot V$, and whose edges correspond to the translates $G\cdot V$. The group $G$ acts on this tree, with finite edge stabilizers (by the compactness of $V$), exhibiting $G$ either as an HNN extension or an amalgamated product over the edge stabilizers. Note that the special case $|\mathcal{E}(G)|=2$ occurs if and only if $G$ has a finite index subgroup which is isomorphic to $\mathbb{Z}$.

Free groups and virtually free groups do not contain closed surface subgroups; Gromov’s question more or less asks whether these are the only examples of word-hyperbolic groups with this property.

Kim and Wilton study Gromov’s question in a very, very concrete case, namely that case that $G$ is the double of a free group $F$ along a word $w$; i.e. $G = F *_{\langle w \rangle } F$ (hereafter denoted $D(w)$). Such groups are known to be one-ended if and only if $w$ is not contained in a proper free factor of $F$ (it is clear that this condition is necessary), and to be hyperbolic if and only if $w$ is not a proper power, by a result of Bestvina-Feighn. To see that this condition is necessary, observe that the double $\mathbb{Z} *_{p\mathbb{Z}} \mathbb{Z}$ is isomorphic to the fundamental group of a Seifert fiber space, with base space a disk with two orbifold points of order $p$; such a group contains a $\mathbb{Z}\oplus \mathbb{Z}$. One might think that such groups are too simple to give an insight into Gromov’s question. However, these groups (or perhaps the slightly larger class of graphs of free groups with cyclic edge groups) are a critical case for at least two reasons:

1. The “smaller” a group is, the less room there is inside it for a surface group; thus the “simplest” groups should have the best chance of being a counterexample to Gromov’s question.
2. If $G$ is word-hyperbolic and one-ended, one can try to find a surface subgroup by first looking for a graph of free groups $H$ in $G$, and then looking for a surface group in $H$. Since a closed surface group is itself a graph of free groups, one cannot “miss” any surface groups this way.

Not too long ago, I found an interesting construction of surface groups in certain graphs of free groups with cyclic edge groups. In fact, I showed that every nontrivial element of $H_2(G;\mathbb{Q})$ in such a group is virtually represented by a sum of surface subgroups. Such surface subgroups are obtained by finding maps of surface groups into $G$ which minimize the Gromov norm in their (projective) homology class. I think it is useful to extend Gromov’s question by making the following

Conjecture: Let $G$ be a word-hyperbolic group, and let $\alpha \in H_2(G;\mathbb{Q})$ be nonzero. Then some multiple of $\alpha$ is represented by a norm-minimizing surface (which is necessarily $\pi_1$-injective).

Note that this conjecture does not generalize to wider classes of groups. There are even examples of $\text{CAT}(0)$ groups $G$ with nonzero homology classes $\alpha \in H_2(G;\mathbb{Q})$ with positive, rational Gromov norm, for which there are no $\pi_1$-injective surfaces representing a multiple of $\alpha$ at all.

It is time to define polygonal words in free groups.

Definition: Let $F$ be free. Let $X$ be a wedge of circles whose edges are free generators for $F$. A cyclically reduced word $w$ in these generators is polygonal if there exists a van-Kampen graph $\Gamma$ on a surface $S$ such that:

1. every complementary region is a disk whose boundary is a nontrivial (possibly negative) power of $w$;
2. the (labelled) graph $\Gamma$ immerses in $X$ in a label preserving way;
3. the Euler characteristic of $S$ is strictly less than the number of disks.

The last condition rules out trivial examples; for example, the double of a single disk whose boundary is labeled by $w^n$. Notice that it is very important to allow both positive and negative powers of $w$ as boundaries of complementary regions. In fact, if $w$ is not in the commutator subgroup, then the sum of the powers over all complementary regions is necessarily zero (and if $w$ is in the commutator subgroup, then $D(w)$ has nontrivial $H_2$, so one already knows that there is a surface subgroup).

Condition 2. means that at each vertex of $\Gamma$, there is at most one oriented label corresponding to each generator of $F$ or its inverse. This is really the crucial geometric property. If $\Gamma,S$ is a van-Kampen graph as above, then a theorem of Marshall Hall implies that there is a finite cover of $X$ into which $\Gamma$ embeds (in fact, this observation underlies Stallings’s work on foldings of graphs). If we build a $2$-complex $Y$ with $\pi_1(Y)=D(w)$ by attaching two ends of a cylinder to suitable loops in two copies of $X$, then a tubular neighborhood of $\Gamma$ in $S$ (i.e. what is sometimes called a “fatgraph” ) embeds in a finite cover $\tilde{Y}$ of $Y$, and its double — a surface of strictly negative Euler characteristic — embeds as a closed surface in $\tilde{Y}$, and is therefore $\pi_1$-injective. Hence if $w$ is polygonal, $D(w)$ contains a surface subgroup.

Not every word is polygonal. Kim-Wilton discuss some interesting examples in their paper, including:

1. suppose $w$ is a cyclically reduced product of proper powers of the generators or their inverses (e.g a word like $a^3b^7a^{-2}c^{13}$ but not a word like $a^3bc^{-1}$); then $w$ is polygonal;
2. a word of the form $\prod_i a^{p_{2i-1}}(a^{p_{2i}})^b$ is polygonal if $|p_i|>1$ for each $i$;
3. the word $abab^2ab^3$ is not polygonal.

To see 3, suppose there were a van-Kampen diagram with more disks than Euler characteristic. Then there must be some vertex of valence at least $3$. Since $w$ is positive, the complementary regions must have boundaries which alternate between positive and negative powers of $w$, so the degree of the vertex must be even. On the other hand, since $\Gamma$ must immerse in a wedge of two circles, the degree of every vertex must be at most $4$, so there is consequently some vertex of degree exactly $4$. Since each $a$ is isolated, at least $2$ edges must be labelled $b$; hence exactly two. Hence exactly two edges are labelled $a$. But one of these must be incoming and one outgoing, and therefore these are adjacent, contrary to the fact that $w$ does not contain a $a^{\pm 2}$.

1 above is quite striking to me. When $w$ is in the commutator subgroup, one can consider van-Kampen diagrams as above without the injectivity property, but with the property that every power of $w$ on the boundary of a disk is positive; call such a van-Kampen diagram monotone. It turns out that monotone van-Kampen diagrams always exist when $w \in [F,F]$, and in fact that norm-minimizing surfaces representing powers of the generator of $H_2(D(w))$ are associated to certain monotone diagrams. The construction of such surfaces is an important step in the argument that stable commutator length (a kind of relative Gromov norm) is rational in free groups. In my paper scl, sails and surgery I showed that monomorphisms of free groups that send every generator to a power of that generator induce isometries of the $\text{scl}$ norm; in other words, there is a natural correspondence between certain equivalence classes of monotone surfaces for an arbitrary word in $[F,F]$ and for a word of the kind that Kim-Wilton show is polygonal (Note: Henry Wilton tells me that Brady, Forester and Martinez-Pedroza have independently shown that $D(w)$ contains a surface group for such $w$, but I have not seen their preprint (though I would be very grateful to get a copy!)).

In any case, if not every word is polygonal, all is not lost. To show that $D(w)$ contains a surface subgroup is suffices to show that $D(w')$ contains a surface subgroup, where $w$ and $w'$ differ by an automorphism of $F$. Kim-Wilton conjecture that one can always find an automorphism $\phi$ so that $\phi(w)$ is polygonal. In fact, they make the following:

Conjecture (Kim-Wilton; tiling conjecture): A word $w$ not contained in a proper free factor of shortest length (in a given generating set) in its orbit under $\text{Aut}(F)$ is polygonal.

If true, this would give a positive answer to Gromov’s question for groups of the form $D(w)$.

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### 6 Responses to Polygonal words

1. Mark Sapir says:

I have asked the following question several years ago. Does the group contain a hyperbolic surface subgroup? It is known (Minsyan, that the group is hyperbolic). I conjectured that the answer is “no”, that would be an answer to Gromov’s question.

2. Mark Sapir says:

Sorry, the group did not show in the comment. Some strange picture appeared instead. The group is generated by x,y, t subject to x^t=xy, y^t=yx (mapping torus of the endomorphism phi(x)=xy, phi(y)=yx of the free group of rank 2.

3. kimsh says:

Mark// One can at least show that the relator set $\{ x^t(xy)^{-1}, y^t(yx)^{-1}\}$ of the group $G$ is polygonal; concretely, there exists a van Kampen diagram on a closed non-orientable surface $S$ of the Euler characteristic -2 such that $S$ consists of 4 pentagons (2 for each word) and such that $S$ satisfies the condition 1,2,3 of this post; so, the corresponding double (now we have two edge groups) contains a surface group. I do not know if this van Kampen diagram is $\pi_1$-injectively, though. The relator of an isomorphic (to $G$) one-relator group is also polygonal.

4. kimsh (Sang-hyun Kim) says:

Still more concretely, the surface group $H=\langle a,b,c,d \vert\; aca^b(c^{-1})^d\rangle$ maps (possibly non-injectively) into $G$ by $a\mapsto (yx^{-1})^t, b\mapsto t^{-1}x^{-1}t^{-1}, c\mapsto (yx^{-1})^{x^{-1}}, d\mapsto x t^{-1}$.

5. Danny Calegari says:

There are indeed many interesting maps of surfaces into Mark’s group. If $C$ is any homologically trivial chain in $\langle x,y\rangle$, then $C-\phi(C)$ bounds an (scl-norm minimizing) incompressible surface in $\langle x,y\rangle$ which can be tubed together with a collection of annuli (from $C$ to $\phi(C)$) to make a map of a closed surface.

I looked at a couple of simple examples, but the resulting surfaces were not injective. I can’t tell by looking whether Kim’s example is of a similar kind.

6. Sang-hyun Kim says:

Here is a (belated) update. Tiling conjecture is true for rank two:

http://arxiv.org/abs/1009.3820

Hence, one-ended doubles of rank-two free groups always contain hyperbolic surface groups.