Walls of CAT(0) cube complex sufficiently far apart implies intersection of stabilizers finite - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:13:45Z http://mathoverflow.net/feeds/question/115784 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115784/walls-of-cat0-cube-complex-sufficiently-far-apart-implies-intersection-of-stabi Walls of CAT(0) cube complex sufficiently far apart implies intersection of stabilizers finite K. Bulinski 2012-12-08T08:47:38Z 2012-12-08T18:46:21Z <p>I was reading through <a href="http://arxiv.org/abs/1204.2810" rel="nofollow">Agol's paper on the Virtual Haken Conjecture</a> and I came across a claim whose proof I am after. It seems to boil down to the following claim about the hyperplanes and their stabilizers under the action of a hyperbolic group on a CAT(0) cube complex:</p> <p><strong>Claim:</strong> Suppose $G$ is a (word-)hyperbolic group acting properly discontinously cocompactly and faithfully on a $CAT(0)$ cube complex $X$. Then there exists an $R>0$ so that if two walls(hyperplanes) $W,W' \subset X$ are such that $d(W,W')>R$ then $G_W \cap G_{W'}$ is finite ($G_W \subset G$ denotes the stabilizer of $W$ i.e the elements that send $W$ to itself)</p> <p><strong>Note:</strong> Certain easy examples show that the condition that $G$ be hyperbolic is necessary, e.g $\mathbb{Z}^2$ acts on the cube complex $\mathbb{R}^2$ and the stabilizer of any two horizontal hyperplanes is $\mathbb{Z}$ which is infinite. A similair example shows that cocompactness is also a necessary assumption.</p> http://mathoverflow.net/questions/115784/walls-of-cat0-cube-complex-sufficiently-far-apart-implies-intersection-of-stabi/115822#115822 Answer by HW for Walls of CAT(0) cube complex sufficiently far apart implies intersection of stabilizers finite HW 2012-12-08T18:11:13Z 2012-12-08T18:46:21Z <p>Here's a proof.</p> <p><strong>Lemma:</strong> Suppose $G$ is a (word-)hyperbolic group acting properly discontinously, cocompactly and faithfully on a C⁢A⁢T⁢(0) space $X$. Then there is a uniform bound $R_0$ on the width $R$ of isometrically embedded flat strips $\mathbb{R}\times [0,R]$ in $X$.</p> <p><em>Proof:</em> If not then, by cocompactness, there exist nested flat discs of diameter tending to infinity. Their union is an embedded copy of $\mathbb{R}^2$, which contradicts hyperbolicity. <em>QED</em> </p> <p>Now suppose that walls $W,W'$ have stabilizers with infinite intersection. Then that intersection is an infinite word-hyperbolic group (since it is quasiconvex in $G$) and so contains an element $\gamma$ of infinite order. Because $W,W'$ are convex and so themselves CAT(0), each contains an axis $l_W,l_{W'}$ for $\gamma$. By standard facts about CAT(0) spaces (see Bridson--Haefliger), any two axes bound a flat strip. Therefore, by the lemma, $l_W$ and $l_{W'}$ are at distance at most $R_0$, as claimed.</p>