6
$\begingroup$

An integral polytope $P$ is normal if all lattice points inside the integer dilation $kP$ can be expressed as $p_1+p_2+\dots+p_k$, where $p_i \in P$ are lattice points.

I am looking for an example $P$ for which the above is true for $k=2$, but fails for higher $k$.

A related question: is there a number $M$, that only depends on the dimension of $P$, such that if the above holds for $k\leq M$, then $P$ is integral?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Regarding the 'related question': M = dim(P) -1 suffices (Theorem 2.2.12, Toric Varieties, Cox-Little-Schenck). $\endgroup$
    – pinaki
    Commented Mar 10, 2014 at 23:51

1 Answer 1

Reset to default
12
$\begingroup$

Examples of this sort (and worse) are given, for dimension 5 and up, in http://scholar.google.com/scholar?cluster=14055290405510744870&hl=en&oi=scholarr David Handelman [me], Effectiveness of an affine invariant for indecomposable integral polytopes, J Pure and Applied Algebra 66 (1990) 165–184, section 3, pp 16ff.

Specifically, this gives integral polytopes $K$ such that $e(K \cap Z^d) = eK \cap Z^d$ [the first denotes the sum of $e$ points in $K \cap Z^d$) for $e \leq d/2$, but for no $e > d/2$, and when $e \geq d/2$, $eK$ is projectively faithful (that is, its set of lattice points generates the standard copy of $Z^d$ as an abelian group) inside Euclidean space of dimension $d$. When the dimension is 6 or more, we can also assume $K$ itself is projectively faithful.

Towards the related question, any $M \geq d-1$ will do, as in my answer to Lattice points in dilated polytopes and sumsets.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Ah! Nice examples! Is it also possible to construct such counter-examples, even when the polytope contains the origin and all the points $(1,0,...,0)$, to $(0,0,...,0,1)$? $\endgroup$
    – Per Alexandersson
    Commented Mar 10, 2014 at 19:47
  • 1
    $\begingroup$ Hi. I've modified the Answer, to remove a minor inaccuracy. When $d \geq 6$, then the relevant $K$ is projectively faithful, hence we can translate it so that the origin is a vertex, and the lattice points in $K$ contain a basis for the standard copy of $Z^d$. Then there exists an element of SL(d,Z) mapping the basis to the standard basis as you wanted. (When $d = 5$, the construction in the paper does not give $K$ projectively faithful, so it's not clear how to proceed. When $d=4$, I don't know if it is possible to obtain the weaker property.) $\endgroup$
    – David Handelman
    Commented Mar 10, 2014 at 22:46

You must log in to answer this question.

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