# A.D. Alexandrov imbedding theorem for metrics with symmetry

A well known theorem due to A.D. Alexandrov says that any metric on the 2-sphere $$S^2$$ with curvature at least -1 (in the sense of Alexandrov) can be isometrically realized either as convex surface in the hyperbolic 3-space or as a doubled 2-dimensional convex set there (I omit the precise description of the latter notion).

REMARK: I do not know if the uniquness of the isometric imbedding holds up to isometries of the hyperbolic space. For isometric imbeddings into Euclidean space such a uniqueness was proven by Pogorelov.

Assume that the metric as above on $$S^2$$ is invariant under the antipodal involution $$x\mapsto -x$$. Is it true that there exists an isometric imbedding into the hyperbolic space which intertwines the antipodal involution and the reflection of the hyperbolic space with respect to some fixed point? In particular the convex surface in the image would have center of symmetry.

## 1 Answer

The existence of a symmetric embedding can be proved as follows.

Approximate the metric by a sequence of symmetric hyperbolic cone-metrics (i. e. locally hyperbolic with cone points of angles $$< 2\pi$$). Each of these metrics has a unique, and therefore centrally symmetric, realization as the boundary of a convex polytope in the hyperbolic space. From this sequence of polytopes one can choose (by Blaschke's selection principle) a converging subsequence. The limit convex body is centrally symmetric, and its boundary is isometric to our given metric.

• Why these polyhedral metrics have unique realization in hyperbolic space? In Euclidean space this is known, but in hyperbolic space I did not see a reference (but I am not a specialist). Thank you. – makt Aug 10 '19 at 8:00
• The rigidity of convex hyperbolic polyhedra can be proved in exactly the same way as that of Euclidean polyhedra: compare the dihedral angles in two realizations, then there are at least four sign changes around every vertex, then this contradicts the Euler formula. – Ivan Izmestiev Aug 10 '19 at 8:23
• Thank you. I am wondering if there is a reference to the hyperbolic case. – makt Aug 10 '19 at 8:26
• There is a more artificial but elegant way to prove this: the Pogorelov map. It associates to any pair of isometric hyperbolic polyhedra a pair of isometric Euclidean polyhedra. – Ivan Izmestiev Aug 10 '19 at 8:30
• Yes, from the last section of my article "Projective background of the infinitesimal rigidity of frameworks" or Jean-Marc Schlenker's "Hyperbolic manifolds with convex boundary". – Ivan Izmestiev Aug 10 '19 at 8:42