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?

  • 1
    $\begingroup$ Regarding the 'related question': M = dim(P) -1 suffices (Theorem 2.2.12, Toric Varieties, Cox-Little-Schenck). $\endgroup$ – auniket Mar 10 '14 at 23:51

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.

| cite | improve this answer | |
  • $\begingroup$ 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 Mar 10 '14 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 Mar 10 '14 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.