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Let $M^n$ be a smooth manifold of dimension $n$. Let $M$ given a metric with curvature bounded below in the sense of Alexandrov which induces the original topology of $M$.

It is true that the dimension of $M$ is equal to $n$ in the sense of the theory of Alexandrov spaces, namley for any $\delta>0$ there exist $(n,\delta)$-strained points, and for some $\delta>0$ there are no $(n+1,\delta)$-strained points.

At the moment I am particularly interested in the case $n=2$ (the case $n=1$ is trivial).

Remark. Burago, Gromov, and Perelman in their paper "Alexandrov spaces with curvature bounded below" mentioned that they did not know at the time (1992) an answer to a more general question, see Remark 6.6 there. I think my question is more specific.

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    $\begingroup$ Corollary 10.8.23 in chapter 10 of Burago-Burago-Ivanov says that in an n-dimensional Alexandrov space the set of n-strained points form an open dense set, which is an n-dimensional manifold. Thus the answer to your question is yes. $\endgroup$
    – Igor Belegradek
    Commented Sep 7, 2019 at 16:12
  • $\begingroup$ @IgorBelegradek: I do not see why the corollary is applicable. The problem is that there might be sets of k-strained points for any k. So the corresponding Alexandrov space is infinite dimensional. $\endgroup$
    – asv
    Commented Sep 7, 2019 at 16:33
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    $\begingroup$ An $(m, \epsilon)$-strained point with small enough $\epsilon$ has a neighborhood homeomorphic to $\mathbb R^m$ (Proposition 10.8.15 in BBI), and if $m>n$, then there is no such neighborhood in an $n$-dimensional manifold. $\endgroup$
    – Igor Belegradek
    Commented Sep 7, 2019 at 16:44
  • $\begingroup$ This is true provided m is the maximal possible number of strained points. But if the number of strained points might be arbitrarily large, the argument does not work. $\endgroup$
    – asv
    Commented Sep 7, 2019 at 16:55
  • $\begingroup$ What do you mean by a "maximal possible number of strained points"? Are you using the same definition of a strained point as in BBI? $\endgroup$
    – Igor Belegradek
    Commented Sep 7, 2019 at 17:11

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If $n$ is defined, then the statement has already been proved in the paper by Burago, Gromov, and Perelman. However, there might be no such $n$; in other words, the space has infinite dimension in this sense.

In this case, it is sufficient to show that our Alexandrov space has infinite topological dimension. The latter follows from 15.6 in our book; see also the references therein.

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  • $\begingroup$ What do you mean by “if $n$ is defined”? It is assumed from the vary beginning that $M$ is a topological manifold of dimension $n$. $\endgroup$
    – asv
    Commented Feb 17 at 9:23
  • $\begingroup$ @asv look at your definition of $n$ as maximum... --- in principle, there might be no such maximum (even if $M$ is finite-dimesional manifold). $\endgroup$
    – Anton Petrunin
    Commented Feb 17 at 20:46

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