most recent 30 from http://mathoverflow.net 2013-05-23T20:05:15Z http://mathoverflow.net/feeds/question/88012 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88012/is-the-number-of-vertices-of-a-convex-d-dimensional-lattice-polytop-without-int Roland Bacher 2012-02-09T17:14:01Z 2012-02-09T20:58:09Z

<p>The lattice polytop $[0,n_1]\times[0,n_2]\times\dots\times[0,n_{d-1}]\times[0,1]$ contains $(n_1+1)(n_2+1)\cdots(n_{d-1}+1)2$ integral points on the boundary and no integral points in its interior. Its number of vertices, $2^d$, is however bounded by a function depending only on its dimension $d$. Does there exist a sequence of convex $d-$dimensional lattice-polytops without interior lattice points and more and more vertices? </p> <p>Remarks: (1) The answer is no in dimension $2$.</p> <p>(2) This question is motivated by a result of Lagarias-Ziegler who showed that the volume (and thus the number of vertices) of a convex $d-$dimensional lattice polytop is bounded if it contains exactly $k>0$ interior lattice points. If no sequence as above exist, then the condition on the existence of $k$ interior lattice points can perhaps be modified into a condition on the number of integral vertices (which has to be sufficiently large) and integral boundary points. </p>

http://mathoverflow.net/questions/88012/is-the-number-of-vertices-of-a-convex-d-dimensional-lattice-polytop-without-int/88034#88034 Gerhard Paseman 2012-02-09T20:58:09Z 2012-02-09T20:58:09Z

<p>Encouraged by Andre Henriques' comment (and not seeing a response which puts more constraints on the problem), I shall promote my comment to an answer.</p> <p>Consider an arbitrary convex polygon P in R^2 which has n vertices for your favorite sufficiently large positive integer n. Then P x [0,1] (or an appropriate represntation) is a convex polytope in R^3 with no interior lattice points and 2n vertices. It should be easy to extend this example to higher dimensions. Thus a sequence of such polytopes with unbounded number of vertices exists in R^d for any fixed d with d > 2.</p> <p>Gerhard "Had Enough Coffee This Morning" Paseman, 2012.02.09</p>