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Let $X$ be an $n$-dimensional Alexandrov space with curvature satisfying both $\ge 0$ and $\le 0$. Can we prove that any tangent cone of $X$ must be isometric to $\mathbb R^{k} \times C(S^{n-k-1}/\Gamma)$, where $\Gamma$ is a finite subgroup of $O(n-k)$ acting on $S^{n-k-1}$?

In general, do we have a classification for such space $X$ up to isometry?

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    $\begingroup$ Nikolaev proved that a space with two sided curvature bound is a $C^{1,\alpha}$ Riemannian manifold with almost the same curvature bounds. I think if the curvature is constant, the manifold is actually $C^\infty$ of constant curvature, and this probably goes back to an old result of Alexandrov. The result is stated in theorem 10.10.13 of Burago-Burago-Ivanov's "A course in metric geometry". Thus in your case you get flat Riemannian manifolds, and their tangent cones are Euclidean spaces. $\endgroup$
    – Igor Belegradek
    Commented Mar 16, 2021 at 12:53
  • $\begingroup$ @IgorBelegradek I am confused about the theorem 10.10.13 you mentioned. Clearly, $R^4/\mathbb Z_2$ is an Alexandrov space with zero curvature, but its tangent cone at $0$ is not the Euclidean space. In fact, on page 207 of the book Geometry IV, the author requires the condition of "local extendability of a shortest curve", which excludes the flat cone case. $\endgroup$
    – Totoro
    Commented Mar 16, 2021 at 13:24
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    $\begingroup$ The quotient of a Euclidean space by the antipodal involution does not have nonpositive curvature. In general the Euclidean cone on a space $X$ of curvature $\le 1$ has curvature $\le 0$ if and only if $X$ has injectivity radius $\ge \pi$. In this case $X$ is the projective space. $\endgroup$
    – Igor Belegradek
    Commented Mar 16, 2021 at 14:03
  • $\begingroup$ @Totoro $\mathbb R^4/\mathbb Z_2$ is not $CAT(0)$. It has triangles which are too fat. This is easiest to see if you look at $\mathbb R^2/\mathbb Z_2$ and take a triangle with the cone point inside. Or just look at a biangle in this space. In CAT(0) spaces geodesics are unique and there are no biangles. $\endgroup$
    – Vitali Kapovitch
    Commented Mar 16, 2021 at 16:47

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No, that's too much to ask for. If you take any convex cone in $\mathbb R^n$ (any such cone is both $CAT(0)$ and Alexandrov of $curv\ge 0$) then the tangent space at the origin is just the cone itself. So it need not be any kind of quotient of $\mathbb R^n$. The above picture is general. If an $n$-dimensional space is both locally $CAT(0)$ and Alexandrov of $curv\ge 0$ then the exponential map at any point is an isometry on a small ball and the tangent cone is a convex cone in $\mathbb R^n$.

BTW, regarding Igor's comment about local extendability of geodesics and a result of Nikolaev, in a recent paper with Kell and Ketterer we showed that if a space $X$ is both locally $CAT(K)$ and $CD(k,n)$ (has generalized Ricci curvature bounded below, which is a weaker condition than Alexandrov curvature bounded below) then $X$ is a manifold with boundary and the interior points are exactly the points where all geodesics locally extend.

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    $\begingroup$ The comment you refer to is now deleted, but all it said was that I didn't know if geodesics extend in constant curvature spaces. Now I do, thanks to the paper you link. $\endgroup$
    – Igor Belegradek
    Commented Mar 16, 2021 at 17:27

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