17
votes
Accepted
Gluing hexagons to get a locally CAT(0) space
The third example is a spine of the Gieseking manifold. The existence of a spine of this sort follows from a result of Iain Aitchison and the fact that the Gieseking is make of a single regular ideal ...
- 68.9k
16
votes
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 ...
- 45k
14
votes
Accepted
Homeomorphism/ homotopy types of non-negatively curved manifolds
As mentioned in comments, the first dimension where an infinite family of pairwise non-homeomorphic closed nonnegatively curved manifolds occurs is $3$ (the lens spaces). The question becomes more ...
- 29.1k
14
votes
Accepted
Translation lengths in CAT(0) spaces
Consider the following two transformations of $\mathbb{R}^3$: $a:(x,y,z)\mapsto (x+1,y,z)$ and $b:(x,y,z)\mapsto (-x,-y,z+1)$. The translation axis for $a^nb$ is the line $x=n/2$, $y=0$ and $a^nb$ ...
- 3,451
10
votes
Accepted
CAT(0) groups that does not act on CAT(0) cubical complex
Many CAT(0) groups cannot act geometrically on CAT(0) cube complexes. For instance:
CAT(0) groups satisfying Kazhdan's property (T), eg. uniform lattices in simple Lie groups of higher rank or in ...
- 8,401
10
votes
When is a extension of $\mathbb{Z}$ by a free group a CAT(0) group?
Seems worth recording here: there are some recent new examples of CAT(0) free-by-cyclic groups due to Rylee Lyman, see https://arxiv.org/abs/1909.03097v3. As she explains, "These are the first ...
- 3,740
10
votes
Accepted
Hausdorff vs Gromov-Hausdorff convergence of convex hypersurfaces
The reference is Lemma 10.2.7 in A course of metric geometry by Burago-Burago-Ivanov. They do it in 3d but it does not matter. The main point is that if two convex bodies are Hausdorff close, then ...
- 29.1k
9
votes
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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 ...
- 45k
9
votes
Amalgamated product acting on CAT(0) cube complex
To extend the gluing result from Bridson--Haefliger to non-positively curved cube complexes, it is important to work in the correct category.
If we want the result to also be a non-positively curved ...
- 25k
7
votes
Accepted
Can we realize the smooth metric of an Alexandrov space with nonnegative curvature by a Riemannian structure?
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 ...
- 45k
7
votes
Accepted
Estimate of number of boundary components of a compact Riemannian 2-surface
I think it follows from Gauss-Bonnet. Suppose $X$ has genus $g$ and $n$ boundary components. Gauss-Bonnet says that
$$\int_X K\;dA+\int_{\partial X}k\;ds=2\pi\chi(X)=2\pi(2-2g-n),$$
where $K$ is ...
- 1,100
7
votes
Bishop-Gromov volume comparison on manifolds with negligible negative Ricci curvature
If $K\subset B(x,R)$ then
$$ \frac{\text{Vol}(B(p, r))}{(r-R)^n}$$
is monotonic for $r>R$.
The proof is the same and it is about maximum one could expect.
- 45k
7
votes
Accepted
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,\...
- 45k
7
votes
Accepted
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 ...
- 45k
7
votes
Accepted
Metrics of constant Gauss curvature on 2-cylinder
An equidistant curve to a line in hyperbolic plane has constant geodesic curvature. So pick a line $L$ in $\mathbb H^2$ and pick $r>0$ such that the $r$-equidistant curves have the desired geodesic ...
- 208
6
votes
Accepted
Rademacher type theorem for Alexandrov spaces
I assume you are interested in the finite-dimensional case.
You can write this function in a distance chart (these charts cover almost all points). Apply the standard Rademacher theorem and notice ...
- 45k
6
votes
Accepted
Harmonic maps are light
As your question operates with $f(\partial D)$, I assume that $f$ is continuous in
$\overline{D}$, though you do not mention this explicitly. Then the answer is yes, the zero set of $f$ is discrete (...
- 91.8k
6
votes
Accepted
What are the extremal CAT(0) metrics?
Let me describe a 6-point counterexample.
Let $K$ be a 2-dimensional cone with total angle $\theta=2{\cdot}\pi+\varepsilon$, where $\varepsilon$ is small and positive (any $0<\varepsilon<\tfrac\...
- 45k
6
votes
Accepted
Estimate of area of 2-dimensional surface
Yes, there is a bound; your problem can be reduced to the case of convex boundary using the following trick.
Without loss of generality, we can assume that $\lambda=-\tfrac1{10}$ and $\kappa=-1$.
In ...
- 45k
6
votes
Accepted
Alexandrov spaces of zero curvature
No, that's too much to ask for. If you take any convex cone in $\mathbb R^n$ (any such cone is both $CAT(0)$ and Alexandrov of $curv\ge 0$) then the tangent space at the origin is just the cone itself....
- 7,842
6
votes
Accepted
Completion of an Alexandrov space
Yes, the conclusion is exactly the main result of Petrunin's paper "A globalization for non-complete but geodesic spaces", Mathematische Annalen volume 366, pages387–393(2016).
- 1,441
5
votes
Accepted
Connected sum in Alexandrov spaces
Yes. In the case of 3-dimensional Alexandrov spaces, a notion of the connected sum has been introduced recently in the preprint "DECOMPOSITIONS OF THREE-DIMENSIONAL ALEXANDROV SPACES" (arXiv:...
- 506
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
5
votes
Accepted
Is any geodesic in the tangent cone of an Alexandrov space a limit geodesic?
The answer is "yes" assuming you are interested in minimizing geodesics.
Moreover the statement holds in the collapsing case as well.
Fix two points $p$ and $q$ in the limit space $X$.
If there is ...
- 45k
5
votes
Discrete approximations of Riemannian manifolds
The cigar gives both a positive answer for 1 and a negative answer for an upper bound in 2.
If it is thin enough, a sequence of aligned points along it is a model and it can be embedded in $\mathbb{R}...
- 686
5
votes
Accepted
Does geometrization of Alexandrov 3-spaces follow from that of 3-orbifolds?
In principle, yes. Note, however, that topologically singular Alexandrov 3-spaces are homeomorphic to non-orientable orbifolds. We could not find an appropriate reference for the geometrization in the ...
5
votes
Accepted
Is there Brownian motion on Alexandrov spaces?
Yes, there is a natural Brownian motion on an Alexandrov space.
In the following paper:
Kuwae, Kazuhiro; Machigashira, Yoshiroh; Shioya, Takashi, Sobolev spaces, Laplacian, and heat kernel on ...
- 29.8k
4
votes
Isoperimetric inequality in CAT(0) surfaces
1) That the disk of maximal area on a smooth surface (if the maximum is attained...) has constant geodesic curvature, was stated by Steiner and proved by Minding, see a historical account on page 1200 ...
- 6,337
4
votes
Convex subcomplexes of CAT(0) cubical complexes
In addition to Anton Petrunin's answer, I would like to mention that a more combinatorial argument is possible. Indeed, in a CAT(0) cube complex $X$, a full subcomplex $Y$ (i.e. a subcomplex which ...
- 8,401
4
votes
Accepted
Multidimensional gluing theorem for Riemannian manifolds
Just learned that the answer is positive, at least its main part saying that if the sum of second fundamental forms is non-negative then the curvature of $M$ is at least $\kappa$. The answer is ...
- 21.8k
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