Let $X$ be a negatively curved symmetric space. In other words, $X$ is one of the four examples: a hyperbolic space, a complex hyperbolic space, a quaternionic hyperbolic space or the hyperbolic Cayley plane.

Let $B\subset X$ be a compact set (wlog, a large ball). Does there always exist a subgroup $\Gamma$ of the isometry group of $X$ such that $X/\Gamma$ is a compact manifold and the images of $B$ under $\Gamma$ are disjoint (i.e. $B\cap\gamma B=\emptyset$ for every nontrivial $\gamma\in\Gamma$)?

Remark: It would suffice to find just one (for each $X$) residually finite discrete co-compact group of isometries. I am sure there is one but I don't know where to look. (Or maybe they all are residually finite?)

Meta-proof: If the answer is negative, then, for some of the model geometries (in some dimension), all compact manifolds with this local geometry are "uniformly thin" - all injectivity radii are bounded above by some universal constant. Such a wonderful fact would be widely known.

  • 5
    $\begingroup$ Malcev's theorem: Every finitely generated group of matrices is residually finite. $\endgroup$ – Greg Kuperberg May 29 '10 at 22:23
  • $\begingroup$ Thanks! One stupid little thing remains: why do co-compact subgroups exist? $\endgroup$ – Sergei Ivanov May 29 '10 at 23:08
  • $\begingroup$ Sergei: The fact that cocompact lattices exist doesn't just follow from abstract nonsense type arguments, in fact, Margulis showed that they don't exist for some homogeneous spaces $G/H$ with semisimple $G$ and non-compact $H.$ So some version of Borel and Harish-Chandra is necessary (if I'm not mistaken, the mere existence part is due to Borel alone). $\endgroup$ – Victor Protsak May 30 '10 at 0:55

The examples are arithmetic groups, constructed in general by Borel and Harish-Chandra. See also Dave Witte Morris' preliminary book. However, examples in hyperbolic and complex hyperbolic spaces probably go back further to the study of quadratic forms.

For hyperbolic lattices, one can take a quadratic form over a quadratic number field (such as $\mathbb{Q}(\sqrt{2})$), which is Lorentzian at one place, and definite at the other places (such as $x_1^2+\cdots +x_n^2-\sqrt{2} x_{n+1}^2$), and take the group of matrices $\Gamma$ in $GL(n+1,\mathbb{Z}[\sqrt{2}])$ which preserve this quadratic form. Then Mahler's compactness theorem (cf. Witte Morris) implies that the quotient $\mathbb{H}^n/\Gamma$ is compact. Then by Selberg's Lemma and residual finiteness (as Greg points out, Malcev's Theorem), you may find a torsion-free subgroup of finite-index with as large injectivity radius as you like.

For hyperbolic and complex-hyperbolic spaces, there are other examples which don't come from the arithmetic construction (in fact, most hyperbolic surfaces and 3-manifolds are not arithmetic). These are attributable to Gromov and Piatetski-Shapiro in the hyperbolic case in all dimensions, and there are finitely many examples in the complex hyperbolic case (at least for $\mathbb{C}$-dim >1) going back to Deligne and Mostow (see also Thurston's paper). However, Gromov and Schoen have shown that quaternionic and Cayley-hyperbolic lattices are all arithmetic, so the Borel Harish-Chandra construction is complete in these cases.

Addendum: Related to Protsak's comment, there is a simple example of a negatively curved homogeneous space which has no lattice action. This is Thurston's "9th geometry", which is excluded as a geometry precisely for this reason. One can take the double warped product metric $$ dr^2 + e^{2a r} dx^2 + e^{2b r} dy^2, $$ for $a,b >0$. When $a=b$, this gives hyperbolic space. But when $a\neq b$, the sectional curvatures are $-a^2, -b^2, -ab$. This has a solvable transitive group of isometries, so is homogeneous. But using the solvability, one may see that there is no cocompact action. (Remark: when $a=-b$, this gives sol geometry).

  • $\begingroup$ Interesting example, but the isometric restriction is a bit strong. It feels complementary to the result I quoted, where $G$ is semisimple and there is no obvious metric (since $H$ is noncompact). $\endgroup$ – Victor Protsak May 30 '10 at 5:55
  • $\begingroup$ Oh, sorry, I didn't read your comment carefully. I assumed you meant a homogeneous space with a negatively curved metric (because of the nature of the question), and I didn't notice the requirement that G be semisimple. $\endgroup$ – Ian Agol May 30 '10 at 16:14

Let me give an explicit example for the Poincare disk in dimension 2. In the disk model, just take the regular $4g$-gon centered on the origin. Identifying opposite sides, we get a compact surface of genus g. The polygon is a fundamental domain for the group $\Gamma$ generated by $b_1...b_{2g-1}$ with

$b_k=\pmatrix{\cosh(a) & e^{ik\pi\over 2g}\sinh(a)\cr e^{-ik\pi\over 2g}\sinh(a)& \cosh(a)}$

where a is defined by $\cosh(a)=1/\tan(\pi/4g)$. The value of a is in fact equal to the radius of the maximal disk contained in the fundamental domain. This is also the distance from any side of the 4g-gon to the origin. It goes to infinity with g, so we are done.

The Gauss-Bonnet formula tells us that surfaces of a given genus have a bounded volume, hence cannot contains arbitrarily large embedded balls. Hope that helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.