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Totoro
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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 freely on $S^{n-k-1}$?

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

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 freely on $S^{n-k-1}$?

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

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?

Source Link
Totoro
  • 2.5k
  • 9
  • 19

Alexandrov spaces of zero curvature

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 freely on $S^{n-k-1}$?

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