4
$\begingroup$

Suppose we are given a convex polytope in terms of the face variables.

That is, let $Y = (1,x_1,\dots,x_n)$ and suppose we have vectors $W_a$ in $\mathbb{R}^{n+1}$ such that the locus $W_a \cdot Y \ge 0$ defines a convex polytope in $\mathbb{R}^n$. The vectors $W_a$ are the face variables.

Is it possible to change slightly these $W_a$ in such a way that the locus $W_a \cdot Y \ge 0$ defines a polytope combinatorially equivalent to the original one? (Same facet poset)

In this sense, how "continuous" is the combinatorial structure of a polytope as a function of the face variables?

$\endgroup$
  • 4
    $\begingroup$ If this were the case, you could perturb your polytope to make it rational; but it is well-known that there are some polytopes not combinatorially equivalent to any rational polytope. $\endgroup$ – Sam Hopkins Jul 20 '18 at 21:14
7
$\begingroup$

If you apply an affine, or more generally, a projective transformation to $\mathbb{R}^n$, then the image of the polytope will have the same combinatorics, while the "face variables" will change in a non-trivial way.

In some cases this is the only possibility: there are polytopes whose combinatorics determines them up to a projective transformation, see for example

Adiprasito, Karim A.; Ziegler, Günter M., Many projectively unique polytopes, Invent. Math. 199, No. 3, 581-652 (2015). ZBL1339.52011.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.