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Results tagged with alexandrov-geometry
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user 1441
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 $\mathrm{CAT}[\kappa]$).
16
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Applications of Alexandrov spaces to Riemannian geometry
The two sources of applications come from two sources of examples of Alexandrov spaces:
Limits of Riemannian manifolds with lower curvature bound.
Quotients of Riemannian manifolds by an isometric g …
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10
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Example of non-closed convex hull in a CAT(0) space
There are such examples already in Riemannian world!
In fact in any generic Riemannian manifold of dimension $\ge3$ convex hull of 3 points in general position is not closed.
BUT it is hard to make ex …
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9
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Can Alexandrov surfaces of CAT(0) type be approximated by CAT(0) polyhedra?
I am sure it done somewhere, but I do not know a ref. I did something like this in my "Metric minimizing surfaces", but do not want to claim originality.
You may fix a finite set of points draw all t …
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9
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Metrically singular Alexandrov space.
Suspicious example:
Take a "funny" manifold with sectional curvature $\ge 1$ say $X$;
funny means Cayley flag or Aloff--Walach/Eschenburg/Bazaikin space, (not $S^n$ or $\mathbb{C}\mathrm{P}^n$ or $\ma …
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9
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Convex subcomplexes of CAT(0) cubical complexes
Yes, it is true.
You condition (2) implies that $X$ is locally convex;
this can be proved the same way as the flag condition for $\mathrm{CAT}[0]$-ness.
It remains to note that for $\mathrm{CAT}[0]$ …
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8
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Whether the manifold part of an Alexandrov space is connected?
Yes.
Assume $A$ is an $m$-dimensional Alexandrov space and $\Omega\subset A$ be the maximal open subset which is a topological $m$-manifold and $A'\subset A$ be the subset of all points with tangent …
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8
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Alexandrov angles in Riemannian manifolds
Your equality is two inequalities.
To show the upper bound you can use the triangle inequality --- come closer to $p$ along the geodesic and apply the local estimates.
(This is the "first variation i …
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8
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Hausdorff convergence of submanifolds in Riemannian manifolds
You are right about 1-Lipschitz map, but that is about all you can expect.
In particular the dimension of $X$ might be not integer.
Assume your $X_i$ all isometric to to a flat torus $\mathbb{T}$ and …
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7
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Set of regular points in an Alexandrov space with curvature bounded below
"Yes" to both questions.
For the second, take the projection to the tangent plane and note that its bi-Lipschitz in a small neighborhood of $x$ with constants as close to 1 as you want.
For the firs …
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Can we realize the smooth metric of an Alexandrov space with nonnegative curvature by a Riem...
Yes, smooth distance functions plus Alexandrov means Riemannian,
but you should make all the definitions precise.
After Otsu and Shioya, there was a paper of Perelman "DC structure on Alexandrov spac …
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7
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3-dim positively curved Alexandrov space
I guess you are interested in topological classification (?).
Given a 3-dimensional Alexandrov space $M$,
you can always find an other Alexandrov space $\bar M$ with isometric involution $J$
such tha …
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7
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Is there Domain Invariance for Alexandrov spaces?
The following lemma from Grove--Petersen, A radius sphere theorem does the trick.
Lemma 1. Let $X$ be a compact Alexandrov space without boundary. Then $X$ has a fundamental class in Alexander-Spa …
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Rigidity of triangle comparison in Alexandrov spaces
The question is not stated precisely.
So I'm free to say anything :)
If you are interested in "non-uniqueness" then the anser is "NO". In any such triangle $[x y z]$ there are at least two distinct …
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Contractibility of balls in Alexandrov spaces
Formally speaking the answer is "no".
Take a 2-dimensional cone with small total angle. Then for any $\varepsilon>0$ there is a point $x$ close enuf to the tip of the cone such that $B(x,\varepsilon) …
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Isometric imbedding of a 2-disk into Euclidean 3-space
Take doubling of the disc, we obtain a metric on the sphere.
By Perelman's theorem it had nonnegative curvature in the sense of Alexandrov.
Therefore, by Alexandrov's theorem, it is isometric to a con …
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