2005
DOI: 10.4310/pamq.2005.v1.n3.a6
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Compact Clifford-Klein forms of symmetric spaces - revisited

Abstract: This article discusses the existence problem of a compact quotient of a symmetric space by a properly discontinuous group with emphasis on the non-Riemannian case. Discontinuous groups are not always abundant in a homogeneous space G/H if H is non-compact. The first half of the article elucidates general machinery to study discontinuous groups for G/H, followed by the most update and complete list of symmetric spaces with/without compact quotients. In the second half, as applications of general theory, we prov… Show more

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Cited by 76 publications
(83 citation statements)
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“…Note that L always admits torsion-free uniform lattices by [3]. Kobayashi and Yoshino conjectured that any reductive homogeneous space G/H admitting compact quotients admits standard ones [21,Conj. 3.3.10]; this conjecture remains open.…”
Section: Deformation Of Compact Quotients In the Real Casementioning
confidence: 99%
“…Note that L always admits torsion-free uniform lattices by [3]. Kobayashi and Yoshino conjectured that any reductive homogeneous space G/H admitting compact quotients admits standard ones [21,Conj. 3.3.10]; this conjecture remains open.…”
Section: Deformation Of Compact Quotients In the Real Casementioning
confidence: 99%
“…The aim of this paper is to prove the following. Recently, Tojo [34] proved that irreducible symmetric spaces G/H that admit standard Clifford-Klein forms are exactly those in the list [18]. Thus, there is an overlap with our results in this case.…”
Section: Introductionmentioning
confidence: 63%
“…One interesting problem in differential geometry is to decide whether a given homogeneous space G/I possesses or not a compact quotient. A more general related question is to decide whether there exist compact manifolds locally modelled on (G, G/I ) (see, for instance [1,2,22,24]). …”
Section: Ghys Non Standard Examplesmentioning
confidence: 99%
“…It turns out that compact quotients of S n are known to exist only for n = 1, 3 or 7. We discussed the case n = 3 above, and the existence of a compact quotient of S 7 was proved in [22]. Here, we dare ask with [22]: Conjecture 2.3 S n has no compact quotients, for n = 1, 3, 7.…”
Section: Ghys Non Standard Examplesmentioning
confidence: 99%