Whether the manifold part of an Alexandrov space is connected?

Question

The title is my question.

Alexandrov space here means finite dimensional Alexandrov space with curvature bounded below ,denoted by CBB.

Let $\gamma$ be a simple curve in a $n$ dimensional CBB $M$ with two end points in the manifold part of $M$.

Analysis. $\gamma$ can be covered by a finite collection $\mathcal U$ of cone neighborhoods, where the cone neighborhood here means a subspace of $M$ homeomorphic to $\mathbb{R}^m\times cone$. Since every cone nbhd in $\mathcal{U}$ must contain two manifold points. If we are able to find a new path contained in the manifold part of every cone nbhd in $\mathcal{U}$, then we done. So my question is reduced to the the restriction of the cone nbhd. However, since I don't know the the structure of the $cone$ part in the cone nbhd very well. I don't know how to get such a new path. Maybe the density of the manifold points in $M$ would help, but I don't know how to use this condition.

Answer 1

Yes.

Assume $A$ is an $m$-dimensional Alexandrov space and $\Omega\subset A$ be the maximal open subset which is a topological $m$-manifold and $A'\subset A$ be the subset of all points with tangent space isometric to Euclidean space.

From Perelmans paper "Beginning of Morse...", we get that $A'\subset \Omega$.

The set $A'$ is a dense convex subset in $A$, this follows from Burago--Gromov--Perelman paper and from my paper on parallel translation. (Convexity means that any geodesic with the ends in $A'$ lies completely in $A'$.) In particular $A'$ is connected; hence the result follows.