Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with alexandrov-geometry
Search options answers only
not deleted
user 4354
Alexandrov geometry studies non smooth analogues of Riemannian manifolds with curvature bounded from below or above. It includes spaces with curvature bounded below (briefly $\mathrm{CBB}[\kappa]$) and spaces with curvature bounded above (briefly $\mathrm{CAT}[\kappa]$).
5
votes
Accepted
Generalization of Radon's theorem
In dimension 1, Radon's theorem says that for any 3 points on the real line, one of them belongs to the segment between the two others. This becomes false if one replaces the real line by the tripod ( …
- 32.4k
14
votes
Accepted
Tverberg's theorem in CAT(0) spaces
No. Let $X$ be a tripod (three segments with one common endpoint), $d=1$, $r=2$ and $E$ the set of the 3 leaf points.
- 32.4k
9
votes
Accepted
CAT(K) and Busemann
Answering a sensible question appeared in comments: If a geodesic space is Busemann and CAT(1), then it must be CAT(0).
Indeed, CAT(1) implies that the space has well-defined metric angles between ge …
- 32.4k
11
votes
Accepted
Alexandrov geometry techniques for Finsler manifolds.
As Anton mentioned in a comment, a non-Riemannian Finsler manifold cannot be an Alexandrov space. If you found an opposite statement in our book, I would appreciate the page number where it appears (I …
- 32.4k
9
votes
Accepted
Stability of midpoints in CAT(0) spaces
No, even if $X=\mathbb R^2$.
Let $A_1$ be (the convex hull of) 4 points with coordinates $(\pm 1,\pm 1)$. Then $m(A_1)=(0,0)$, as the 4 points are on the circle $S_1$ of radius $\sqrt 2$ centered at …
- 32.4k
9
votes
Accepted
Details of Perelman's example about soul of Alexandrov space
No $X^5$ is not a cone over $CP^2$ and is not compact. The projection has nothing to do with the cone structure. In fact, it's better to forget about the cone structure altogether (until you ask what …
- 32.4k
6
votes
Is there Domain Invariance for Alexandrov spaces?
The discussion in the comments is getting too long, let me sum up the proof. This is community wiki; feel free to correct errors and fill in details.
Let $H^*$ denote the Alexander-Spanier cohomology …