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Let $X^n$ be an $n$-dimensional complete Alexandrov space with curvature bounded below (or a smooth Riemannian manifold, possibly with boundary). Let $U\subset X$ be an open dense subset with the following convexity property: Any shortest path connecting any two points of $U$ is contained in $U$.

Question 1. Is the Hausdorff dimension of $X\backslash U$ at most $n-1$?

Question 2. If the answer to question 1 is yes, whether the corresponding Hausdorff measure of $X\backslash U$ is locally finite?

The answer is not known to me even in the case of Riemannian manifolds.

A reference would be helpful.

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In the Riemannian case $U$ has to be whole manifold.

In Alexandrov space it is expected that $U$ has full measure, if true it would be a partial answer to your question.

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  • $\begingroup$ Thank you. But having the full measure is weaker than what I asked. Does one expect having the stronger statement as I asked? Are there counter-examples to it? $\endgroup$
    – asv
    Nov 10, 2015 at 18:24
  • $\begingroup$ @sva The main example of $U$ is the set of points with almost Euclidean tangent cone; in this case the answer to your question is "yes" and it is written already in Burago-Gromov-Perelman. Your question is a generalization for which I do not see an application. $\endgroup$
    – Anton Petrunin
    Nov 10, 2015 at 20:27
  • $\begingroup$ The situation I have in mind is when $U$ is the complement to the boundary of an Alexandrov space. As you told me privately, in this case the answer to my question is 'yes'. I am trying to reconstruct a proof. $\endgroup$
    – asv
    Nov 11, 2015 at 5:02

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