Let $X$ be a compact finite dimensional Alexandrov space with curvature bounded below.

Does there exist $\varepsilon_0>0$ (depending on $X$) such that for any $\varepsilon \in (0,\varepsilon_0)$ and any point $x\in X$ the open ball $B(x,\varepsilon)$ is contractible? Is similar statement true for closed balls?

$B(x,\varepsilon)$ denotes the open ball of radius $\varepsilon$ with the center at $x$.


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 $B(x,\varepsilon)$ is an annulus.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.