3
$\begingroup$

Suppose $P$ is a convex lattice polytope in $Z^3$ without interior lattice points, and we require the interior lattice points of each facet(i.e. dimensional 2 faces) are neither too much nor too few, say, they are less or equal to 2 but bigger or equal to 1.

Is there any classification results on such polytopes? Are there finite such polytopes, or is there any known method can be used to deal with such polytopes? Any suggestions, remarks, references for the problem or similar questions are great appreciated!

$\endgroup$

1 Answer 1

3
$\begingroup$

This is something you might be aware of already (and which does not seem to be exactly the situation you are formulating), but Reeve's tetrahedron is a 3-dimensional integral convex polytope without lattice-points in the interior. Furthermore, the only lattice points are the vertices, so there are four of them.

EDIT: I happened to attend a talk by Benjamin Nill a few weeks ago, and part of his talk seemed related to this kind of question. Here is a paper he wrote with Ziegler: http://arxiv.org/abs/arXiv:1101.4292.

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .