2005
DOI: 10.1016/j.top.2004.01.010
|View full text |Cite
|
Sign up to set email alerts
|

Cocompact CAT(0) spaces are almost geodesically complete

Abstract: Let M be a Hadamard manifold, that is, a complete simply connected riemannian manifold with non-positive sectional curvatures. Then every geodesic segment α : [0, a] → M from α(0) to α(a) can be extended to a geodesic ray α : [0, ∞) → M. We say then that the Hadamard manifold M is geodesically complete. Note that, in this case, all geodesic rays are proper maps.CAT(0) spaces are generalizations of Hadamard manifolds. For a CAT(0) space X, all geodesic rays α : [0, ∞) → X are proper maps but, in general, X is n… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1
1

Citation Types

0
58
0

Year Published

2009
2009
2022
2022

Publication Types

Select...
6
1

Relationship

0
7

Authors

Journals

citations
Cited by 35 publications
(58 citation statements)
references
References 16 publications
0
58
0
Order By: Relevance
“…The proof of this theorem relies on the study of topology at infinity of systolic complexes. It is similar to the proof of an analogous theorem for CAT (0) spaces . The main difference is that our proof uses the notion of connectedness at infinity, whereas the one in uses cohomology with compact supports.…”
Section: Introductionmentioning
confidence: 88%
See 1 more Smart Citation
“…The proof of this theorem relies on the study of topology at infinity of systolic complexes. It is similar to the proof of an analogous theorem for CAT (0) spaces . The main difference is that our proof uses the notion of connectedness at infinity, whereas the one in uses cohomology with compact supports.…”
Section: Introductionmentioning
confidence: 88%
“…It is similar to the proof of an analogous theorem for CAT (0) spaces . The main difference is that our proof uses the notion of connectedness at infinity, whereas the one in uses cohomology with compact supports. The key fact is that in the setting above, the complex X is not 1‐connected at infinity (see ).…”
Section: Introductionmentioning
confidence: 88%
“…By [19], X has almost extendable geodesics, and this implies that there is q ∈ ∂X with d T (n, q) = π, and by π-convergence g i (q) → p. It follows that there are translates a 1 , b 1 of a, b respectively so that q is separated from D by {a 1 , b 1 }. Further we claim that g i {a 1 , b 1 } = {a, b} for all i big enough.…”
Section: Maximal Inseparable Subsets Of ∂Xmentioning
confidence: 98%
“…We recall the fact (see [19]) that geodesics in X are 'almost extendable' i. Let p : X → L be the projection map from X to L. We claim that there is an M > 0 such that the M-neighborhood of L separates X in at least two components Y 1 , Y 2 such that Y i is not contained in any finite neighborhood of L (i = 1, 2).…”
Section: 3mentioning
confidence: 99%
“…A proof has also been written up by P. Ontaneda in [24]. Bestvina asks in [3] (also in [2]) if all boundaries of a given CAT(0) group also satisfy the stronger condition of being celllike equivalent.…”
mentioning
confidence: 99%