1
$\begingroup$

Edit: Fix a lattice $N = \mathbb{Z}^n$ and let $N_{\mathbb{R}} = N \otimes \mathbb{R}$. Let $C(S)$ be a strongly convex rational cone in $N_{\mathbb{R}}$ generated by a finite set $S \subset N$, with following two properties:

  • No $v \in S$ is in the interior of $C(S)$.
  • $S$ generates $N$ as a $\mathbb{Z}$-module.

I have a vector $v \in N$, which has the property:

$v$ is NOT a nonnegative $\mathbb{Z}$-linear combination of vectors in $S$.

Does it imply that $v \notin C(S)$? If not, is there any simple low-dimensional counterexample?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Counterexample: lattice is $\mathbb Z^2$, $S = \{(1,0),(1,1),(1,3)\}$, $v=(1,2)$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Because of your simple answer, now I recognize that I need to put an additional assumption: What if none of $v \in S$ is in the interior of $C(S)$? $\endgroup$
    – Han-Bom Moon
    Commented Sep 20, 2013 at 22:39
  • $\begingroup$ Still false: $S=\{(1,0,0),(1,1,0),(1,0,1),(1,2,2)\}, v=(1,1,1)$. $\endgroup$
    – Lev Borisov
    Commented Sep 20, 2013 at 22:58
  • $\begingroup$ @Lev : you should perhaps add that in each case, there are only finitely many $v$ that are counterexamples. $\endgroup$
    – BS.
    Commented Sep 21, 2013 at 8:54
  • $\begingroup$ Not true: there are could be infinitely many "missing" $v$ near the boundary of the cone. $\endgroup$
    – Lev Borisov
    Commented Sep 21, 2013 at 11:25
  • $\begingroup$ Even if the cone is rational ? I would like to see an example. $\endgroup$
    – BS.
    Commented Sep 21, 2013 at 11:58

You must log in to answer this question.

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