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?



Alexandrov spaces of zero curvature
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](https://arxiv.org/abs/1908.07036) 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.


