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Edit: Addressing Igor's comment I'd like to correct the references I gave. The correct reference for the exact argument I sketch should be the original book by Alexandrov "Intrinsic Geometry of Convex Surfaces"(Chapter 7, section 6, Lemmas 1-3). In the later book by Alexandrov and Zalgaller a similar (but necessarily more complicated) argument is given for the case of surfaces with bounded integral curvature (Theorem 10, page 84). One can reconstruct the original proof in the easier case of curvature bounded below from that one but it requires some work.

This is indeed classical and is due to Alexandrov ( see the book by Alexandrov and Zalgaller "Intrinsic geometry of surfaces").

The idea of general structure the proof is as follows. You take a very fine triangulation of $X$ and substitute the curved triangles by triangles with the same sides in the space form of constant curvature $-1$. This will give you a polyhedral surface of curvature $\ge -1$ (you only need to check that the cone angles at vertices are $\le 2\pi$ which is immediate from the definition of an Alexandrov space). Away from the cone points the metric will be smooth of constant curvature $-1$. The cone points can then be easily smoothed to get a smooth metric of $\sec\ge -1$. There are some hidden technical difficulties here (the The hard part is to show that the resulting polyhedral surface is close to the original space $X$) but they can be overcome.X$. A more interesting result is another theorem of Alexandrov ( "Intrinsic Geometry of Convex Surfaces") that locally any 2-dimensional Alexandrov space of$curv\ge -1$is isometric to a level set of a convex function in$\mathbb H^3$. When the lower bound is 0 he proved an even sharper result that any 2-sphere of curvature$\ge 0$is isometric to the boundary of a convex body in$\mathbb R^3$. BTW, the result you attribute to Perelman that a 2-dimensional Alexandrov space is a topological manifold is actually due to Alexandrov too. 3 added 11 characters in body Edit: Addressing Igor's comment I'd like to correct the references I gave. The correct reference for the exact argument I sketch should be the original book by Alexandrov "Intrinsic Geometry of Convex Surfaces"(Chapter 7, section 6, Lemmas 1-3). In the later book by Alexandrov and Zalgaller a similar (but necessarily more complicated) argument is given for the case of surfaces with bounded integral curvature (Theorem 10, page 84). One can reconstruct the original proof in the easier case of curvature bounded below from that one but it requires some work. This is indeed classical and is due to Alexandrov ( see the book by Alexandrov and Zalgaller "Intrinsic geometry of surfaces"). The idea of the proof is as follows. You take a very fine triangulation of$X$and substitute the curved triangles by triangles with the same sides in the space form of constant curvature$-1$. This will give you a polyhedral surface of curvature$\ge -1$(you only need to check that the cone angles at vertices are$\le 2\pi$which is immediate from the definition of an Alexandrov space). Away from the cone points the metric will be smooth of constant curvature$-1$. The cone points can then be easily smoothed to get a smooth metric of$\sec\ge -1$. There are some hidden technical difficulties here (the hard part is to show that the resulting polyhedral surface is close to the original space$X$) but they can be overcome. A more interesting result is another theorem of Alexandrov ( "Intrinsic Geometry of Convex Surfaces") that locally any 2-dimensional Alexandrov space of$curv\ge -1$is isometric to a level set of a convex function in$\mathbb H^3$. When the lower bound is 0 he proved an even sharper result that any 2-sphere of curvature$\ge 0$is isometric to the boundary of a convex body in$\mathbb R^3$. BTW, the result you attribute to Perelman that a 2-dimensional Alexandrov space is a topological manifold is actually due to Alexandrov too. 2 corrected references Edit: Addressing Igor's comment I'd like to correct the references I gave. The correct reference for the exact argument I sketch should be the original book by Alexandrov "Intrinsic Geometry of Convex Surfaces"(section 6, Lemmas 1-3). In the later book by Alexandrov and Zalgaller a similar (but necessarily more complicated) argument is given for the case of surfaces with bounded integral curvature (Theorem 10, page 84). One can reconstruct the original proof in the easier case of curvature bounded below from that one but it requires some work. This is indeed classical and is due to Alexandrov ( see the book by Alexandrov and Zalgaller "Intrinsic geometry of surfaces"). The idea of the proof is as follows. You take a very fine triangulation of$X$and substitute the curved triangles by triangles with the same sides in the space form of constant curvature$-1$. This will give you a polyhedral surface of curvature$\ge -1$(you only need to check that the cone angles at vertices are$\le 2\pi$which is immediate from the definition of an Alexandrov space). Away from the cone points the metric will be smooth of constant curvature$-1$. The cone points can then be easily smoothed to get a smooth metric of$\sec\ge -1$. There are some hidden technical difficulties here (specifically, the hard part is to show that the resulting polyhedral surface is close to the original space$X$one needs to make sure that the triangles in the polyhedral surface are convex) X$) but they can be overcome.

A more interesting result is another theorem of Alexandrov ( "Intrinsic Geometry of Convex Surfaces") that locally any 2-dimensional Alexandrov space of $curv\ge -1$ is isometric to a level set of a convex function in $\mathbb H^3$. When the lower bound is 0 he proved an even sharper result that any 2-sphere of curvature $\ge 0$ is isometric to the boundary of a convex body in $\mathbb R^3$.

BTW, the result you attribute to Perelman that a 2-dimensional Alexandrov space is a topological manifold is actually due to Alexandrov too.

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