Let F=\langle a,b\rangle be the free group on two generators, and let \phi:F \to F be the endomorphism defined on generators by \phi(a)=ab and \phi(b)=ba. We define Sapir’s group C to be the ascending HNN extension

F*_\phi:=\langle a,b,t\; | \; a^t=ab,b^t=ba\rangle

This group was studied by Crisp-Sageev-Sapir in the context of their work on right-angled Artin groups, and independently by Feighn (according to Mark Sapir); both sought (unsuccessfully) to determine whether C contains a subgroup isomorphic to the fundamental group of a closed, oriented surface of genus at least 2. Sapir has conjectured in personal communication that C does not contain a surface subgroup, and explicitly posed this question as Problem 8.1 in his problem list.

After three years of thinking about this question on and off, Alden Walker and I have recently succeeded in finding a surface subgroup of C, and it is the purpose of this blog post to describe this surface, how it was found, and some related observations. By pushing the technique further, Alden and I managed to prove that for a fixed free group F of finite rank, and for a random endomorphism \phi of length n (i.e. one taking the generators to random words of length n), the associated HNN extension contains a closed surface subgroup with probability going to 1 as n \to \infty. This result is part of a larger project which we expect to post to the arXiv soon.

The context of this problem is Gromov’s notorious question:

Question(Gromov): Does every 1-ended hyperbolic group contain a surface subgroup?

Actually, it is not at all clear if Gromov really asked this question, or what sort of answer he expected. There is a discussion of this in the introduction to a recent paper by Henry Wilton. A positive answer to this question is known in only a few special cases, including

(this list is not exhaustive). One strategy to find a surface subgroup is to define a class \mathcal{G} of groups with the property that every one-ended hyperbolic group contains a subgroup in the class \mathcal{G}, and then to show that every group in this class further contains a surface subgroup. A reasonable candidate for the class \mathcal{G} is the class of one-ended graphs of free groups. The logic behind this choice is that it is very easy to produce many free subgroups of a one-ended hyperbolic group (in fact, this is more or less the only kind of subgroup one knows how to produce) by Klein’s pingpong argument, and one could perhaps argue that because there are so many such subgroups, that intersect in quite rich and interesting ways, a sufficiently rich collection is one-ended while at the same time has the structure of a graph of groups. On the other hand, the structure of a graph of free groups is similar in some ways to the structure of a Haken 3-manifold, and one knows enough about the components of the graph (i.e. the free factors) that one can try to build a surface subgroup by amalgamating surface-with-boundary subgroups along cyclic subgroups of the edge groups.

Anyway, this is more philosophy than mathematics, but it does partly explain why the class \mathcal{G} has been widely studied by geometric group theorists interested in Gromov’s question. One important class of graphs of groups are the HNN extensions, whose underlying graphs consist of a single vertex and a single edge joining this vertex to itself. An (injective) endomorphism \phi:F \to F of a free group thus gives rise to an HNN extension F *_\phi in the class \mathcal{G}.

Now, suppose S is a map from a surface subgroup to F *_\phi. There is a homomorphism F*_\phi \to \mathbb{Z} sending F to 0 and the conjugating element t to 1. The kernel intersected with the image of S will determine an infinite cyclic cover \widetilde{S} of S, and one would like to determine whether this map is injective. We can think of \widetilde{S} as an infinite union of subsurfaces S_i with boundary, where each S_i is attached to S_{i-1} and S_{i+1}, and contained in a conjugate F^{t^i} of the subgroup F. If we identify each F^{t^i} with \phi^i(F) \subset F for i\ge 0, then we can think of S_i = \phi^i(S_0). Let \widetilde{S}^+ denote the union of the S_i with i\ge 0. Evidently it is sufficient to show that the inclusion of \widetilde{S}^+ to F is injective, since any loop in the kernel of \widetilde{S} is conjugate into \widetilde{S}^+. This is convenient, since we can discuss surface-with-boundary subgroups of a fixed free group, and essentially ignore the endomorphism \phi.

The first thing to check is that each separate inclusion S_i \to F is injective. Each S_i may be represented by a certain kind of diagram, called a fatgraph. Basically, a fatgraph is a graph in the usual sense, together with a choice of cyclic ordering of the edges incident to each vertex. A fatgraph Y embeds canonically as the spine of some surface S(Y) which itself deformation retracts back to Y, in such a way that the cyclic order on edges inherited from the embedding agrees with the fatgraph structure. The oriented edges of S(Y) are labeled with reduced words in F in such a way that the labels on opposite sides of an edge of Y are inverse in F. In this way, a fatgraph “represents” a surface-with-boundary mapping to F. Here is an example of a (disconnected) fatgraph, whose underlying surface is homeomorphic to the union of two 4-punctured spheres:

Now, the fundamental group of every (component of every) fatgraph is free, but the map to F is not necessarily injective. Stallings gave a celebrated criterion for a simplicial map from a graph to a rose (i.e. a standard graph with fundamental group F) to be injective, namely that the map should be folded, or equivalently, that the map should be an immersion on the link of every vertex. In terms of fatgraphs, this means that there should be at most one incoming edge at each vertex with each label. The graph pictured above is folded in this sense. Notice if every boundary word is reduced, a 2- or 3-valent vertex is necessarily (locally) folded.

OK, this is a criterion that will certify that an individual S_i might be injective, when represented as a fatgraph. What about the dynamics of \phi? Notice that the endomorphism \phi has a particularly nice property: if we think of it as representing a self-map f of the standard rose X to itself, then the map f is an immersion, in the sense of Stallings. This means that if each of the surfaces S_i is represented by a folded fatgraph Y_i, then each f^i(Y_0)=Y_i will be folded if Y_0 is. This suggests the following definition:

Definition. A fatgraph Y with associated surface S(Y) is f-folded if there is a decomposition of its boundary into \partial^- and \partial^+ in such a way that f(\partial^-) = \partial^+ (with the opposite orientation), and satisfying the following properties:

  1. The graph Y is Stallings folded
  2. Every f-vertex in \partial^+ (i.e. the images under f of the vertices of \partial^-) is associated to a 2-valent vertex of Y
  3. No vertex of Y is associated to more than one f-vertex in \partial^+
  4. No vertex of Y is associated to more than one vertex in \partial^-

When we talk about a vertex of Y being “associated” to a vertex of \partial S(Y) we mean that the vertex of \partial S(Y) maps to the given vertex of Y under the deformation retraction of S(Y) to (Y) (this deformation retraction is simplicial when restricted to \partial S(Y)).

Now, suppose Y is f-folded. We can glue S(Y) to S(f(Y)) by gluing \partial^+ to f(\partial^-). Condition 4 implies that the resulting surface is S(Y_1) where Y_1 = Y \cup f(Y). In a similar way we can define

Y_n:=Y\cup f(Y)\cup \cdots \cup f^n(Y)

and S_n:=S(Y_n). Now, conditions 2 and 3 imply that every vertex of Y_n is obtained by gluing some vertex of f^i(Y) to a sequence of 2-valent vertices in various f^j(Y) with j>i. In particular, since every vertex of Y is locally folded, the same is true of every vertex of f^i(Y), and therefore also of Y_n. Hence Y_n is folded, and thus injective. Since \cup_n S(Y_n) = \widetilde{S}^+ as above, it follows that the suspension of an f-folded surface is injective in *_\phi.

The definition of f-folded can be modified for an endomorphism \phi which is not an immersion of X. One of the main theorems Alden and I prove is that a “random” endomorphism admits many f-folded surfaces in this sense, and therefore the associated HNN extension has (many) surface subgroups. For a random endomorphism of length n, the genus of these surfaces will typically be of order at least O(n), but the number will grow at least like g^{Cg} for genus g\gg n.

Now, Sapir’s group C is certainly not random in any sense; nevertheless, it is possible to search for an f-folded surface. A priori finding an f-folded surface with given boundary seems to require trying exponentially many gluings, and is apparently impractical. However, Alden and I are able to show that the search for such a surface can be reduced to a linear programming problem, and thus becomes eminently practical. Sure enough, a computer search rapidly found the following example of an f-folded surface in Sapir’s group:

In a bit of detail: the picture above is a fatgraph whose boundary decomposes into three components labeled ababABAB and one component labeled \phi^4(babaBABA^3). There is a 3-fold cover whose boundary decomposes into \partial^- consisting of three copies of ababABAB^3 and \partial^+ consisting of three components labeled \phi^4(babaBABA^3). The components of \partial^- are the ones indicated by the blue circles, and one can see that they are embedded, satisfying condition 4. The red dots are the f-vertices, and one can check that they are distinct and on 2-valent vertices of the fatgraph. Finally, one can check that the surface is folded in the usual sense of Stallings. It follows that the suspension is an injective surface in Sapir’s group, of genus 31.

(added Thursday, February 21, 2013): Jack Button has just posted a paper to the arXiv making the observation that a random HNN extension of a free group (in the sense of Alden and I, as above) will satisfy the small cancellation condition C'(\lambda) for any \lambda>0 as n \to \infty, with probability \to 1, and therefore will be the fundamental group of a special cube complex, by a result of Wise. This is good to know, and underlines the extent to which such HNN extensions resemble 3-manifold groups.