Is it possible to classify explicitly compact 2-dimensional Alexandrov spaces with curvature bounded below (either with or without boundary)?

If yes, a reference would be helpful.

EDIT: If the question is too general, may be one can classify somehow non-negatively curved compact 2-dimensional Alexandrov spaces, e.g. as pieces of convex surfaces (see my comment below). In general I would be interested to know as precise result as possible.

  • $\begingroup$ Classify up to what? isometry? $\endgroup$ – ThiKu Nov 13 '15 at 7:35
  • $\begingroup$ Yes. For example all non-negatively curved metrics on the 2-sphere can be realized as convex (possibly degenerate) surfaces in Euclidean 3-space (Alexandrov's theorem). May be in general there is no such a nice description. Then I am asking on the most precise result available. $\endgroup$ – MKO Nov 13 '15 at 7:54
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    $\begingroup$ You can get something local in the same vein : if $X$ is an alexandrov surface with curvature $\geq\kappa$, then every point of $X$ has a neighborhood isometric ta a small piece of the boundary of a convex subset in the space form of dimension 3 and curvature $\kappa$. $\endgroup$ – Thomas Richard Nov 13 '15 at 13:08

Applying doubling theorem, we can get rid of boundary. Then pass to the universal cover. The obtained space is isometric to convex surface in the model space with curvature $\kappa$ and any motion of your original space can be extended as a motion of this surface.

So your space is a quotient of convex surface by a discrete group of motions of the model space. In particular, if your space had a boundary it will be lifted to plane curves in the model space.

In general, the surface is not unique (up to congruence), so it is not exactly a classification, but close.

  • $\begingroup$ 1) The universal cover may not be compact. Can it still be realized as convex surface in the model space? 2) If one has a convex closed surface in the model space, and one chooses a domain with nice boundary in it, under what conditions the closure of the domain is an Alexandrov space with curvature bounded below? This is necessary to describe Alexandrov 2-spaces with boundary. $\endgroup$ – MKO Nov 14 '15 at 8:12
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    $\begingroup$ @sva (1) Right, it can be noncompact, but still it can be realized as a convex surface, the proof by approximation by closed surfaces. (2) It has to be bounded by curve with positive turn, but as I said, to describe the surface you may assume that it bounded by a plane curve (this is the plane of reflection which preserves your surface). $\endgroup$ – Anton Petrunin Nov 14 '15 at 13:26

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