Let $M$ be an $n$-dimensional Alexandrov space with curvature $\geqslant k$. A point $p\in M$ is said to be an $(n,\delta)$ strained point if there are $n$ pairs of points $a_i, b_i$ such that $$ \tilde{\angle}a_i p b_i >\pi-\delta, \quad\tilde{\angle}a_i p a_j >\frac{\pi}{2}-\delta, $$ $$ \tilde{\angle}a_i pb_j>\frac{\pi}{2}-\delta,\quad \tilde{\angle}b_ipb_j>\frac{\pi}{2}-\delta. $$ We know if $\delta<c(n)$, then the map $\varphi(q)=(|a_1q|,...,|a_nq|)$ is a bi-Lipschitz homeomorphism between a certain neighborhood of $p$ and a domain in $\mathbb{R}^n$.

I want to know how big is $c(n)$? In the paper "A.D. Alexandrov space...", Theorem 5.4 says that if $\delta<\frac{1}{2n}$, then $\varphi$ is a $\frac{(1-2n\delta)}{\sqrt{n}}$-open Lipschitz map. So is the sharp constant $c(n)\leqslant \frac{c_1}{n}$? Can someone give an example that if $\delta>\frac{c_2}{n}$, then a neighborhood of $p$ is not homeomorphic to a domain in $\mathbb{R}^n$, $n\geqslant 3$.


1 Answer 1


Your question is almost equivalent to following problem:

Find the minimal value $\delta_n$ such that $\mathbb{S}^{n-1}$ contains $2\cdot(n+1)$ points on the distance at least $\tfrac\pi2-\delta_n$ from each other.

I guess it is not hard to solve, if the oprimal configuration is centrally symmetric, then $\delta_n$ is the bound you are looking for.

On the other hand, there is another version of strainers where the sharp bound is obvious. The condition is $$\tilde\measuredangle(p\,{}^{a_i}_{a_j})>\tfrac\pi2+\delta$$ for a point array $a_0,\dots,a_n$. In this case the distance map $M\to \mathbb{R}^n$ defined by $$x\mapsto (|a_1-x|,\dots,|a_n-x|)$$ is bi-Lipschitz in a neighborhood of $p$ (note that we did not use $a_0$).

This is not true if $\delta\le 0$, even for $n$-dimensional Euclidean space.

One has to work bit more to introduce this type of strainers, but they provide extra flexibility. In particular they are essential in the proof stability theorem.


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