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. 



Is the number of vertices of a convex $d-$dimensional lattice polytop without interior lattice points bounded?
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


