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Suppose $n$ is an integer greater than 3. Sometimes ago I heard somewhere that it is still not known if there exist complete finite-volume hyperbolic $n$-manifolds having exactly one cusp.

Could someone either confirm that the problem of finding such examples in every dimension is still open, or, preferably, give me a reference for examples of one-cusped hyperbolic manifolds in arbitrary dimension?

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    $\begingroup$ I think this is still open. This is mentioned as an open problem here: ams.org/mathscinet-getitem?mr=1917053 $\endgroup$
    – Ian Agol
    Commented Sep 7, 2010 at 15:31
  • $\begingroup$ Dear Ian, thank you very much for your reply. I also checked the papers that cite Long-Reid's work, and it seems that the problem is still open. $\endgroup$
    – Roberto Frigerio
    Commented Sep 9, 2010 at 10:41
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    $\begingroup$ There are known 1-cusped hyperbolic orbifolds up to dimension 9, I think. See e.g.: dx.doi.org/doi:10.1016/j.jalgebra.2006.12.024 So one could attempt to look for irregular covers of these orbifolds which are manifolds. I don't know what is known about this though. $\endgroup$
    – Ian Agol
    Commented Sep 10, 2010 at 0:24
  • $\begingroup$ Dear Ian, thank you again! this reference sounds quite interesting! I will give a look at it soon. $\endgroup$
    – Roberto Frigerio
    Commented Sep 19, 2010 at 20:49

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Dear Roberto, to add information of Agol's comment, in Theorem 1.3 of this paper it is proved that there aren't one-cusped arithmetic hyperbolic $n$-orbifolds for $n\geq 30$. Moreover, Stover shows one-cusped arithmetic hyperbolic orbifolds in dimensions 10 and 11.

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  • $\begingroup$ Thanks for your answer. I saw Stover's paper on the arxiv some days ago, I hope I will have time to read it soon! $\endgroup$
    – Roberto Frigerio
    Commented Dec 27, 2011 at 17:52
  • $\begingroup$ Are there known non-arithmetic methods for constructing finite-volume hyperbolic manifolds of arbitrary dimension? $\endgroup$
    – Anton Lukyanenko
    Commented Dec 27, 2011 at 20:30
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    $\begingroup$ Anton, there are special constructions of non-arithmetic hyperbolic manifolds in small dimensions >3, coming e.g. from reflection groups or certain special moduli spaces. Otherwise, in high dimensions, the only construction of non-arithmetic lattices available seems to be the Gromov-Piatetskii-Shapiro method of interbreeding, or a slight variation I call inbreeding. It may be possible to take a multicusp arithmetic manifold and interbreed it to construct a one cusp manifold. $\endgroup$
    – Ian Agol
    Commented Dec 27, 2011 at 20:47
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Dear all, today a paper by Kolpakov and Martelli on the arxiv appeared that shows that there exist lots of 4-dimensional cusped hyperbolic manifolds with one cusp. Here is the reference

http://arxiv.org/abs/1303.6122

The general case is still open, I think.

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Indeed, the general case is open. In the case of dimension four, we used the Coxeter $24$-cell, which is an ideal right-angled polytope (ideal means that all of its vertices are on $\partial \mathbb{H}^4$). It is know (a little result of mine) that there are now ideal right angled polytopes in dimensions greater than or equal to $7$ (however, I do not know anything about dimensions $5$ and $6$). Thus, there is no hope to use ideal right-angled polytopes in higher dimensions.

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    $\begingroup$ There are right-angled finite-volume polytopes up to dimension 8: ams.org/mathscinet-getitem?mr=2130566 I believe all the vertices are ideal, although I have to double-check. $\endgroup$
    – Ian Agol
    Commented Sep 2, 2013 at 2:22
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    $\begingroup$ Dear Ian, thank you for your comment! In Potyagailo-Vinberg's construction, there are both ideal and proper vertices. I think that each polytope $P_n$ they construct in dimension n becomes a facet of the next $P_{n+1}$. E.g., each facet of $P_4$ is isometric to $P_3$, e.t.c. However, $P_3$ has already both ideal and proper vertices, thus all $P_n$, $4\leq n \leq 8$, will have vertices of both types (ideal and proper). $\endgroup$
    – SashaKolpakov
    Commented Sep 2, 2013 at 4:24
  • $\begingroup$ However, if you want all the vertices to be ideal, then the dimension is $\leq 6$ (arxiv.org/abs/1211.2944v2) and I know examples only up to $n=4$. I wonder if there are ``totally ideal'' (all vertices ideal) right-angled polytopes of dimension $n=5,6$ or there is a proof that there are none and $n=4$ is a sharp bound. $\endgroup$
    – SashaKolpakov
    Commented Sep 2, 2013 at 7:30

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