# “Inflating” a closed, defined metric, manifold

Consider a two-dimensional Riemannian manifold homeomorphic to the sphere, with a defined metric.

Since we do not suppose that manifold to have a positive curvature, we are not in the hypotheses of the Pogorelov's uniqueness theorem. As a consequnce, in general, there are many possible isometric embeddings in $$\mathbb R^3$$ for that manifold.

The question is the following: is the maximum volume embedding of that manifold unique, up to rigid transformations?

The problem can be expressed in a more intuitive way: Suppose we have an inflatable object whose walls can be bent but not stretched. It can be seen as a closed manifold of known metric. If I inflate such an object until I reach the maximum volume, is the resulting object rigid? (suppose that there aren't narrow bottlenecks that could divide the object into two or more parts).

• Isn't there a problem with the existence of a maximizer, in this generality? – Pietro Majer Jul 15 '19 at 10:40
• It’s in fact conjectured that any closed smooth surface in $\mathbb R^3$ is rigid. There was a similar conjecture for polyhedra, but Connelly found a counterexample. – Deane Yang Jul 15 '19 at 13:38
• For submetric embeddings (e.g. "inflatable objects" whose walls can compress but not stretch), some related conjectures and problems are discussed in these papers of Pak and Pak and Schlenker: math.ucla.edu/~pak/papers/pillow4 tandfonline.com/doi/abs/10.1142/S140292511000057X – j.c. Jul 15 '19 at 14:44
• @PFerro, I don't remember where I first heard about this, but I did post the question on MathOverflow several years ago and others were able to provide some references. mathoverflow.net/questions/1975/… – Deane Yang Jul 15 '19 at 15:40
• (above I mean "if e.g. $U$ is starshaped). Anyway, the point is, after inflating an object with non-positive curvature, some parts of it may touch each other, no? – Pietro Majer Jul 15 '19 at 16:02