MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose a polytope $P$ is contained in its dual polytope $\tilde{P}$. Does there always exist a polytope $Q$ that contains $P$ and is self dual $Q=\tilde{Q}$? Is there any bound on the minimal number of vertices that $Q$ should have given the number of vertices of $P$? Is there any algorithm for finding such a polytope?

The dual of a subset $A$ of a vector space $V$ is defined as $\tilde{A}:=\{\vec{y}\in V |\vec{y}\cdot\vec{x}\geq -c\; \forall \vec{x}\in A\}$, where $c$ is a given positive constant, and $\vec{y}\cdot\vec{x}$ is the scaler product between $\vec{y}$ and $\vec{x}$.

share|cite|improve this question
    
Are there any references where this definition of self-dual is used. I believe the standard definition of self-dual is not this strict. – Stephen Sturgeon Dec 5 '13 at 18:24
    
Self-dual convex compact set containing $P$ does really exist by Zorn lemma (applied to the set of convex sets $X$ contained in $\tilde{P}$ satisfying $y\cdot x\geqslant -c$ for all $x,y\in X$.) What we have to cjeck that we may increase such $X$ a little if $X\ne \tilde{X}$, this is rather easy. – Fedor Petrov Mar 16 at 15:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.