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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?

Remarks: (1) The answer is no in dimension $2$.

(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.

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Why not take a polygonal prism? Depending on the polygon, you can get twice the number of vertices of the (d-1)- dimensional polygon. Or am I forgetting something? Gerhard "Ask Me About System Design" Paseman, 2012.02.09 –  Gerhard Paseman Feb 9 '12 at 17:59
    
The more I look at the question, the more one thing becomes clear: at least one of us has not had enough coffee this morning. Gerhard "It's Quite Possible Its Me" Paseman, 2012.02.09 –  Gerhard Paseman Feb 9 '12 at 18:08
    
Interestingly, it is known that there are only a finite number of maximal hollow lattice polytopes in a fixed dimension, where "maximal hollow" means "not properly contained in a larger hollow lattice polytope" (Averkov, Wagner, Weismantel; Nill & Ziegler). –  Joseph O'Rourke Feb 9 '12 at 19:07
    
I think that ask me about system design provided an answer to the question: in dimensions ≥3, the number of vertices is not bounded. –  André Henriques Feb 9 '12 at 19:54
    
Joseph, if it is not too much trouble, would you leave a comment enlightening me (and future readers) briefly on hollow polytopes? A rough idea or Wikipedia link would suffice. Gerhard "Enquiring Minds Want To Know" Paseman, 2012.02.09 –  Gerhard Paseman Feb 9 '12 at 21:09
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1 Answer

up vote 4 down vote accepted

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.

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.

Gerhard "Had Enough Coffee This Morning" Paseman, 2012.02.09

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You are absolutely right. –  Roland Bacher Feb 9 '12 at 21:11
    
An interesting follow up question (especiially for d=2) is to ask for the fewest number of lattice points enclosed by a polygon with n vertices, and use this to examine larger dimension polytopes where most of the lattice points are at the vertices. I invite you to ask a well formulated version of this. Gerhard "Ask Me About System Design" Paseman, 2012.02.09 –  Gerhard Paseman Feb 9 '12 at 21:23
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