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What is the classification of 3-dim positively curved Alexandrov space? And if a 3-dim positively curved Alexandrov space has a totally (quasi)geodesic subset,then the classification?

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I guess you are interested in topological classification (?).

Given a 3-dimensional Alexandrov space $M$, you can always find an other Alexandrov space $\bar M$ with isometric involution $J$ such that

  • $M$ is isometric to $\bar M/J$
  • $\bar M$ is topological manifold without boundary.

This is almost an answer to your first question.

If $M$ is compact and simply connected then $M$ has to be homeomorphic to one of the following (I might miss something):

  • $\mathbb S^3$
  • spherical suspension over $\mathbb R\mathrm P^2$
  • $\mathbb D^3$
  • ball in the cone over $\mathbb R\mathrm P^2$.

If noncompact then you get in addition $\mathbb R^3$ and cone over $\mathbb R\mathrm P^2$.

If the space has a totally quasigeodesic surface then cutting along this surface should give you an Alexandrov space with boundary. Since the curvature is positive it should be disc or ball in cone over $\mathbb R\mathrm P^2$. So you original space is described by an isometric involution of the boundary of one of these spaces.

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So you original space is described by an isometric involution of the boundary of one of these spaces,you mean(?) the totally quasigeodesic surface is the boundary of disk or ball in cone over S(RP^2).Then except S^3,S(RP^2), S^3/J,S(RP^2)/J,what else? Someone guess 3-dim positively curved Alexandrov space is S^3/J,right?In my previous question ,if we change 3 by n,then the result will be? – jiangsaiyin Aug 30 at 6:55
I have read some of your book "Alexandrov geometry",but topic on this is so limit,can you recommend some other books or papers? – jiangsaiyin Aug 30 at 7:19

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