# Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators?

We know now that hyperbolic 3-manifolds virtually embed into right-angled Artin groups as quasiconvex subgroups. Also, quasiconvexity depends on the generating set.

I have been constructing a space at infinity for right-angled artin groups where the boundaries of quasiconvex hyperbolic subgroups will embed in a natural way; but to construct it, I've been forced to add all redundant generators of the form $b=a_1...a_k$, where all if the $a_i$ are distinct commuting generators in the standard generating set for the RAAG (they can also be unversed of generators). This is just adding in all 'diagonal' generators, and corresponds in free abelian groups to changing from an $L^1$ metric to an $L^{\infty}$ metric.

My question is, will this make some quasi convex subgroups no longer convex? In the $\mathbb{Z}^2$ case (generated by $a,b$), I think that the subgroup generated by $a$ is quasi convex before adding the generators $ab$, $ab^{-1}$, etc. So my final question is, will non-elementary hyperbolic quasi convex subgroups stay quasi convex after adding in these generators?

Edit: the easiest example of a non-elementary hyperbolic subgroup of a RAAG is the following: take the pentagon RAAG generated by $a,b,c,d,e$. Then the subgroup generated by $ab^{-1},bc^{-1},cd^{-1},de^{-1}$ is a two-holed torus group with standard relations.

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Being quasiconvex in the context of hyperbolic groups, does not depend on the (finite) generating set. This can be seeing a corollary of Morse lemma (quasi-geodesics are uniformly close to geodesics). It also follows from the fact that a subgroup $H$ in a hyperbolic group $G$ is quasiconvex if and only if the embedding $H\to G$ is a quasi-isometric embedding. None of this generalizes to nonhyperbolic groups, like $Z^2$. –  Misha Apr 24 '13 at 22:39
Tiny quibble: the easiest example of a non-elementary hyperbolic subgroup of a RAAG is $F_2\subseteq F_2$. I agree that yours is the easiest interesting example. –  HJRW Apr 25 '13 at 10:57

First, one should decide what quasiconvexity (qc) means in the context of subgroups $H\subset G$, where $G$ is merely a semihyperbolic group, e.g., a RAAG. (I am assuming that generating sets are fixed.) Here are eight competing definitions:

1. $H$ is qc in $G$ if there exists a constant $C$ so that every geodesic in $G$ connecting elements of $H$ is within distance $\le C$ from $H$.

2. $H$ is qc in $G$ if there exists a constant $C$ so that for all $x, y\in H$ there exists a geodesic in $G$ connecting $x, y$, which is within distance $\le C$ from $H$.

3. See e.g. here for a discussion of the following definition: $H$ is quasiconvex with respect to a fixed combing of $G$ if there exists a constant $C$ so that every combing path in $G$ connecting elements of $H$ is within distance $\le C$ from $H$.

4. $H$ is qc in $G$ if $H$ is quasi-isometrically embedded in $G$.

5. $H$ is qc in $G$ if there exists an $L$-Lipschitz retraction $G\to H$ (which is not assumed to be a homomorphism).

6. Assuming that some "reasonable" notions of boundary for $G$ and $H$ are fixed: $H$ is qc in $G$ if there exists an equivariant injective continuous map $\partial H\to \partial G$.

7. Assume that $G$ is a CAT(0) group, so it acts geometrically on a CAT(0) space $X$. Then $H$ is qc in $G$ if its image under the orbit map is qc in $X$ (in the sense of Definition 1).

8. Same setup as in 7, only assume that the convex hull of an orbit of $H$ is within bounded distance from the orbit itself.

All these notions are equivalent if $G$ is hyperbolic (and $X$ above is $CAT(-1)$) but not otherwise. (There are other related notions but I will limit myself to these eight.)

As the question does not specify the notion of quasiconvexity, I will use Definition 1. Here is an example of instability of quasiconvexity for nonelementary hyperbolic subgroup $H\subset G$, where $G$ is a RAAG.

Let $G=F_2\times Z$ and $H=F_2$ embedded in $G$ as the first factor. First, let's give $G$ the obvious generating set $a, b, c$, where $c$ is the central generator and $a, b\in H$. Then, $H$ is qc in $G$. Now, let's add the generator $d=ac$. Then $H$ is no longer qc in $G$.

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Great answer! I am indeed assuming number one. Your example answers the question. It seems to me that quasiconvexity will only be preserved if the subgroup is already generated by diagonal elements. Thanks for the care and time you took in your response! I had never heard of many of these equivalent conditions. –  Brian Rushton Apr 25 '13 at 0:55
Definition 3 is interesting when the combing is regular and part of an automatic structure of $G$. In that case, the combing words that lie in $H$ form a regular set, and it is often not difficult, starting from the automatic structure of $G$, to compute a finite state automaton to recognise membership of combing words $H$. Handling intersections of such subgroups can then be done by standard operations on finite state automata. –  Derek Holt Apr 25 '13 at 12:39
great answer. But what is the difference between 1 and 2 ? :) –  Alexey Kvashchuk Apr 26 '13 at 6:41
@Alexey: Quantifiers! 1 says every geodesic, where 2 says there exists a geodesic. The subgroup $\langle xy \rangle$ of the free abelian group with free generators $x,y$ is qc using 2 but not using 1. –  Derek Holt Apr 26 '13 at 8:39
@Derek: yes, indeed you are right. For hyperbolic groups both notions are equivalent though, since geodesics are "close". –  Alexey Kvashchuk Apr 26 '13 at 9:08

As this is too long for a comment, I'll post it as an answer.

As you seem to be interested in CAT(0) cube complexes, I want to point out that there is, in fact, a ninth type of quasiconvexity, which turns out to be the 'right' notion in this context.

Definition: Suppose that a group $G$ acts properly discontinuously and cocompactly by isometries on a CAT(0) cube complex $X$. A subgroup $H$ is called combinatorially convex cocompact (CCC) if it acts properly discontinuously and cocompactly on a convex subcomplex $Y\subseteq X$.

(Note: this terminology is not standard---people often just write 'quasiconvex'---but I think this is less confusing and more descriptive.)

Haglund proved that this is equivalent to an $H$-orbit's being quasiconvex in the 1-skeleton of $X$ [1]. Therefore, in the special case where $G$ is a RAAG (equipped with its standard generating set) and $X$ is the universal cover of its Salvetti complex, this coincides with Misha's definition 1.

To justify that this is the 'right' definition in preference to definition 1, let me point out that it's clear that a CCC subgroup of a CCC subgroup is again CCC, whereas for definition 1 the transitivity of quasiconvexity doesn't even make sense (since the subgroup isn't equipped with a generating set).

[1] Haglund, Frédéric, 'Finite index subgroups of graph products.', Geom. Dedicata 135 (2008), 167–209.

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