If you want to kill your productivity, move to a new house!

I know things have been extremely slow lately, but I was moving last week (and well, that means packing everything, moving things to the new house, cleaning the old house, unpacking things at the new one… well, you get the picture).

My goal for this week is presenting Aleksandrov’s solution to  the Minkowski problem (see an earlier post I did introducing this problem). So I am going to leave you a problem as a preview, it is a sort of discrete version of the Minkowski problem:

Let $n_1,...,n_k$  be a family of non-coplanar unit vectors in $\mathbb{R}^3$ and let $\alpha_1,...,\alpha_k$ be positive numbers such that

$\sum \limits_{i=1}^k \alpha_in_i=0$

Then, show that there exists a convex closed polyhedron with exactly $k$ faces with normal vectors given by $n_1,...,n_k$ and corresponding areas $\alpha_1,...,\alpha_k$. Plus, this polyhedron is unique up to translation.

This is fact is not surprising (since it is not hard to check that any polyhedron has this property), but the proof is far from trivial. As you may guess, the proof cannot be constructive, it will use a continuity argument to show that there must be at least one such polyhedron. I will present this in my next post.

2 Responses to “If you want to kill your productivity, move to a new house!”

1. August 19, 2009 at 1:44 am

It *can’t* have a constructive solution, or you wouldn’t like it as much?

I was curious if you knew of any variants of the Minkowski problem where you have boundaries. Say you draw a circle on the sphere and only define your function to one side of this circle. Can you find a convex surface with boundary in space subject to the same constraints? Furthermore, can you use this notion along with some sort of gluing scheme to describe, say, a peanut uniquely by specifying only its gaussian curvature in this manner?

2. December 19, 2009 at 9:51 am

Yeah, of course there are variants with boundary, and it is almost the same as the Dirichlet problem. For the boundary you will need some condition, e.g. that the projection of the boundary curve into some plane be a convex curve…