# Parabolic subgroups of relatively hyperbolic and CAT(0) groups

Let $G$ be a finitely generated group. We say that $G$ is CAT(0) if it acts properly and co-compactly by isometries on a CAT(0) space. We say it is hyperbolic relative to a collection $\Omega$ of subgroups if it acts properly by isometries on a Gromov-hyperbolic space $X$ such that limit points are either conical or bounded parabolic and the stabilizers of the parabolic points are precisely the elements of $\Omega$. Those subgroups are called (maximal) parabolic subgroups or peripheral subgroups.

Question: Assume that $G$ is both CAT(0) and relatively hyperbolic. Are the peripheral subgroups CAT(0) themselves ?

• As you expected, the second "more general" question has a negative answer. For instance, a product of two nonabelian free groups has plenty of undistorted subgroups that are not CAT(0) (see for instance my answer mathoverflow.net/a/107199/14094) – YCor Mar 16 '18 at 13:27
• Maybe a replacement for the "more general question" would be to ask about subgroups that are retracts in the large scale Lipschitz category. – YCor Mar 16 '18 at 14:18
• @YCor, I read the second question as asking for something slightly stronger: a quasiconvex subgroup that's not itself CAT(0). Undistorted and quasiconvex coincide in the hyperbolic case, but in the CAT(0) case the natural meaning of quasiconvex is surely that some orbit in the CAT(0) space is quasiconvex. It seems unlikely that the examples you cite have this property. – HJRW Mar 16 '18 at 15:04
• @HJRW I indeed considered undistorted and quasiconvex as synonyms. The OP will clarify if he has another meaning in mind. – YCor Mar 16 '18 at 16:02
• Thank you both for your comments. I was more thinking of undistorded subgroups too, so Ycor's answer is fine. However, as HJRW suggests, it would be more reasonnable to think of quasi-convex subgroups in his sense. Actually, is it obvious/true that a parabolic subgroup is quasi-convex then ? – M. Dus Mar 17 '18 at 9:11

Peripheral subgroups of relatively hyperbolic CAT(0) groups are indeed CAT(0) themselves. In fact, more is true: Morse subgroups of CAT(0) groups are CAT(0) themselves.

Definition. Given a finitely generated group $G$, $H \subset G$ is a Morse subgroup if, for every $A>0$ and $B \geq 0$, there exists some constant $K \geq 0$ such that any $(A,B)$-quasigeodesic between any two points of $H$ stays in the $K$-neighborhood of $H$ (in some fixed Cayley graph constructed from a finite generating set; the definition does not depend on this choice).

In his paper Quasi-convexity of hyperbolically embedded subgroups, Sisto proved that hyperbolically embedded subgroups are Morse subgroups, so in particular peripheral subgroups of relatively hyperbolic groups are Morse subgroups.

Now, I claim that Morse subgroups in CAT(0) groups are always convex-compact. (I already proved this statement for groups acting geometrically on CAT(0) cube complexes in my article Hyperbolicities in CAT(0) cube complexes (Proposition 4.2).)

Proposition: Let $X$ be a complete CAT(0) space and $S \subset X$ a Morse subset. The Hausdorff distance between $S$ and its convex hull is finite.

Proof. Let $x$ be a point in the convex hull of $S$, and fix a point $y \in S$. The first observation is that, for every $\epsilon>0$, there exists some $z \in S$ whose projection onto $[x,y]$, say $p$, is at distance less than $\epsilon$ from $x$. Otherwise, if $m$ denotes the point of $[x,y]$ at distance $\epsilon$ from $x$ and $\pi : X \to [x,y]$ the projection onto $[x,y]$, then $\pi^{-1}([m,y])$ would be a convex subspace(*) containing $S$ but not $x$, contradicting the fact that $x$ belongs to the convex hull of $S$.

Now, I claim that $[y,p] \cup [p,z]$ is a $(\sqrt{2},0)$-quasigeodesic. The only point to verify is that, if $a \in [p,y]$ and $b \in [p,z]$, then $d(a,b) \geq \frac{1}{\sqrt{2}} (d(a,p)+d(p,z))$.

Consider a comparison triangle $\Delta = \Delta(\bar{a},\bar{b},\bar{p})$ for $[a,p]\cup[p,b] \cup [a,b]$. Notice that $\angle_{\bar{p}}(\bar{a},\bar{b}) \geq \angle_p(a,b) \geq \pi/2$. Therefore, $d(\bar{a},\bar{b})$ greater or equal to the length of the hypothenuse of a right-angled triangle whose small sides have lengths $\alpha := d(\bar{a},\bar{p})$ and $\beta :=d(\bar{b},\bar{p})$. One has $$\begin{array}{lcl} d(a,b) & = & d(\bar{a},\bar{b}) \geq \sqrt{\alpha^2+\beta^2} = (\alpha + \beta) \cdot \frac{\sqrt{1+(\alpha/\beta)^2}}{1+ \alpha/\beta} \\ & \geq & \frac{1}{\sqrt{2}} (\alpha+\beta) = \frac{1}{\sqrt{2}} ( d(\bar{a},\bar{p})+d(\bar{b},\bar{p}) \\ & \geq & \frac{1}{\sqrt{2}} ( d(a,p)+d(b,p) \end{array}.$$ Because $S$ is Morse, there exists a universal constant $K$ such that $d(p,S) \leq K$. Consequently, $d(x,S) \leq d(x,p)+d(p,S) \leq K+ \epsilon$.

So the conlusion is: if $K$ is such that any $(\sqrt{2},0)$-quasigeodesic between any two points of $S$ stays in the $K$-neighborhood of $S$, then the Hausdorff distance between $S$ and its convex hull is at most $K$. $\square$

Corollary: Let $G$ be a group acting geometrically on a CAT(0) space $X$. If $H \subset G$ is a Morse subgroup, then $H$ is convex-cocompact, ie., there exists a convex subspace $Y \subset X$ on which $H$ acts cocompactly.

Proof. The map $\left\{ \begin{array}{ccc} G & \to & X \\ g & \mapsto & g \cdot x_0 \end{array} \right.$, where $x_0 \in X$ is a fixed basepoint, is a quasi-isometry, so the orbit $H \cdot x_0$ is a Morse subspace. It follows from the previous proposition that $H$ acts cocompactly on the convex hull of $H \cdot x_0$. $\square$

(*) The assertion is not true. The pre-image of a single point under a projection onto a geodesic is not necessarily convex in a CAT(0) space. So the proof above is not correct. I do not erase the argument in case it would lead to a correct proof.

• I accepted the answer. However, I've read the proof again and there is one thing I don't actually understand. Why is $\pi^{-1}([m,x])$ a convex subspace in the proof of the proposition ? – M. Dus May 6 '18 at 22:38
• The property should be the following: Let $X$ be a CAT(0) space and $x,y,a,b \in X$ four points. If $p : X \to [a,b]$ denotes the projection onto $[a,b]$, then $p(z) \in [p(x),p(y)]$ for every $z \in [x,y]$. In fact, it seems reasonable to think that the image of a convex subspace under a projection is convex as well. However, I don't see a good argument yet. I will think about this problem. – AGenevois May 13 '18 at 14:10
• Yes, this was my first expectation, but I could not prove the property you claim (and I even found somewhere on the Internet that it was an open question, although I'm not sure it is). Now, for the second claim, it depends on what you project. On a geodesic it should be fine, I agree, but on a convex set it's not (consider the projection of a line onto a ball, where the image is not convex because it lies on the sphere). – M. Dus May 14 '18 at 6:05
• @M. Dus: It turns out that this "obvious" lemma is wrong. There are easy counter-examples in products of trees. So my proof is not correct: you should unaccept my answer. An alternative approach would be to use a characterisation of being relatively hyperbolic due to Drutu and Sapir: being asymptotically tree-graded. But I don't see yet how to write a complete argument. – AGenevois Jun 5 '18 at 9:34
• Thanks for your interest in this question. Maybe you are right about the Drutu Sapir approach, but maybe this is not true in general. However, using Hruska and Kleiner paper Hadamard spaces with isolated flats, we should be able to prove it's true when the parabolic subgroups are virtually abelian. – M. Dus Jun 5 '18 at 9:48