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. 



Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators?
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](http://berstein.wordpress.com/2011/02/24/well-behaved-subgroups-of-non-positively-curved-groups/) 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 $G$. 


