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I've found in the literature these facts:

  • Any closed flat manifold is virtually (i.e. finitely covered by) a torus, and any finite-volume real hyperbolic manifold has virtually (i.e. is finitely covered by a manifold with) only torus cusp sections.

  • Any closed Nil 3-manifold is virtually a circle bundle over a 2-torus, and any finite-volume complex hyperbolic surface has virtually only such cusp sections.

How to complete this picture for higher-dimensional complex hyperbolic manifolds? And for quaternionic hyperbolic manifolds and octonionic hyperbolic surfaces?

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    $\begingroup$ Check out this paper msp.org/agt/2004/4-2/p03.xhtml $\endgroup$ Commented Apr 21 at 2:44
  • $\begingroup$ So I didn't google it well enough! Does the paper answer the second sentence of each item? i.e., is every finite-volume X-hyperbolic manifold finitely covered by a manifold whose cusps are all topologically of the simplest form (like tori for X = R)? (From a very quick glance, it looks like it answers at least the question with some in place of all). If yes, where is it stated and proved? Thanks! $\endgroup$ Commented Apr 21 at 11:07
  • $\begingroup$ (I edited the question by adding "only" for clarity) $\endgroup$ Commented Apr 21 at 11:14
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    $\begingroup$ I guess it might depend on what you mean by simplest form? McReynolds makes a distinction between infranil and nil almost flat manifolds. The infranil ones have a finite-sheeted cover which is nil, and this persists under finite sheeted covers. By Theorem 1.4, such groups are separable. So for each cusp, one can pass to a cover where the preimage of the cusp is nil. Do this for each cusp, then pass to a common regular cover, and all of the cusps will be nil. Not sure if this answers your question. $\endgroup$ Commented Apr 22 at 23:04
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    $\begingroup$ Yes, this answers my question except for the complex case. But thanks to you I searched better, and found the answer: the Corollary here ams.org/journals/proc/1998-126-08/S0002-9939-98-04289-0/… , where an "elementary geometric proof" of this fact (previously proven by Garland-Raghunathan using a result of Borel) is given $\endgroup$ Commented Apr 23 at 7:26

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