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Let $X$ be a CAT(0) space and $G$ its group of isometries. Then $X$ is said to be cocompact, if there exists a compact set $K\subset X$ with $X=G.K$. The space $X$ is called periodic, if there exists a locally isometric covering $X\to C$ where $C$ is a compact metric space. If $X$ is periodic, then $X$ is the universal covering of $C$ and the group of deck transformations is included in $G$, so $X$ is cocompact.

My question is for the converse: If $X$ is cocompact, is it true that $X$ is periodic? This means, is it true that there exists a subgroup $\Gamma$ of $G$ which acts properly discontinuously such that the quotient $\Gamma\backslash X$ is compact?

If this is not true for general CAT(0) spaces, what if $X$ is a building?

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  • $\begingroup$ Take countably many copies of an isosceles triangle and glue them together in a chain along the equal sides. By Reshetnyak's theorem, you get a $\mathrm{CAT}[0]$-space, say $X$. $X$ admits a cocompact action of $\mathbb{Z}$, but all cocompact actions have a fixed point. Therefore they do not come from a covering map $X\to C$, so it is not periodic. $\endgroup$ Aug 24, 2015 at 12:15
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    $\begingroup$ @AntonPetrunin: Thanks a lot. What if $X$ is locally compact? …if it is an affine building? $\endgroup$
    – user1688
    Aug 24, 2015 at 12:20

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2 examples: a manifold and a tree:

1) Consider the group $G_s=\mathbf{R}^2\rtimes\mathbf{R}$, where $s>1$ and the action is given by $t.(x,y)=(e^tx,e^{st}y)$. Endow it with a negatively curved left-invariant Riemannian metric (it exists, by Heintze); let $X$ be the resulting Riemannian manifold, it is transitive.

I claim that $H=\mathrm{Isom}(X)$ has no lattice (so $X$ is not periodic). Indeed, otherwise $H$ is unimodular; also $H$ is Gromov-hyperbolic and is not focal because it is unimodular, and $H_0$ is not compact, from which it follows by standard results that if $W$ is the maximal compact normal subgroup in $H$, $H/W$ is an open subgroup in the isometry group of some rank 1 symmetric space of noncompact type, and since the Gromov boundary of $H$ is 2-dimensional, this can only be the 3-dimensional real hyperbolic space. But $G_s$ does not embed cocompactly into $\mathrm{Isom}(\mathbb{H}^3(\mathbf{R}))$, contradiction.

2) Start with a trivalent regular tree; first endow it with the usual distance, and fix a Busemann function $b$. Let the length of and edge $e$ be 1,2,3 according to whether $b(e)$ equals $[0,1]$, $[1,2]$, or $[2,3]$ modulo 3. If $T$ is the resulting tree with the new distorted distance, $\mathrm{Isom}(T)$ is cocompact on $T$ and fixed a point at infinity, hence is not unimodular, so has no lattice, thus $T$ is cocompact and not periodic.

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