[math-fun] Question about compact convex polyhedra in 3-space

2 Feb 2021

      Suppose P is a compact convex polyhedron in 3-space, topologically equivalent to the sphere, with finitely many convex planar faces each bounded by a convex polygon with finitely many straight sides.

Then every point x of P that is not a vertex has a neighborhood that is isometric to a neighborhood in the plane. And each vertex is locally isometric to a neighborhood of the vertex of a cone, with total angle around the vertex* equal to some theta in the open interval (0, pi).

For the jth vertex, call this number theta_j.

If this cone is approximated by a smooth surface then its unit normals translated to the origin will cover an area of the unit sphere equal to 2pi - theta. So if we assign the number 

   2pi - theta_j

to the jth vertex, the Gauss-Bonnet theorem tells us that

   Sum (2pi - theta_j)   =   4pi

or equivalently

   Sum theta_j   =  2pi(V-2)

if V is the number of vertices.**

Now suppose we have a finite number of regions P_j in the plane, each bounded by a polygon with straight sides, and that we are given a pairing of all the edges with edges of equal length so that, if the resulting abstract surface has V vertices, E edges, and F faces, the following two conditions are satisfied:

1) V - E + F  =  2

(The Euler characteristic of the resulting surface is 2, from which we know that it must be a sphere.) 

and

2) The total angle about vertex j is theta_j in the interval (0, pi), such that

   Sum theta_j   =   2pi(V-2).

Question: 
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Does there then necessarily exist a convex polyhedron in 3-space that is isometric to the resulting abstract surface?

—Dan

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* The total angle of a cone can be seen if you slit the cone from the base to the vertex and flatten it out, it will fill a certain fraction f of a disk. Then its total angle is 2pi*f.

** E.g., for a cube V = 8 and each theta_j = 3pi/2, and indeed 8(3pi/2) = 2pi(8-2).

Dan Asimov

Victor Miller

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