5
$\begingroup$

My question has an easily formulated generalization, which I will state first. Let $\sigma \subseteq \mathbf{R}^n$ be a full-dimensional strongly convex polyhedral cone. For each lattice point $m \in \sigma^o \cap \mathbf{Z}^n$, minimally generating inside the interior cone $\sigma^o$, let $S(m) \subseteq \sigma^{\vee} \cap \mathbf{Z}^n$ denote the set of lattice points $u$ with $\langle u,m \rangle = 1$. The generalized question is:

Does $S(m) = S(m') \not = \varnothing$ imply that $m = m'$?

UPDATE: As Minseon Shin pointed out, there was a $2$-dimensional counter-example to the previous formulation of the above.


For my main question, as a special case of the above, assume that $\sigma$ is the nef cone of a smooth projective toric variety $X_{\Sigma}$. Then my question amounts to the following:

Let $D_1$ and $D_2$ be two ample divisors, minimally generating inside the ample cone. Suppose that there exists two effective curves $C_1,C_2$ such that $D_1 \cdot C_1 = D_2 \cdot C_2 = 1$. Then does $D_1 \cdot C = 1 \Leftrightarrow D_2 \cdot C = 1$ for all effective curves $C$ imply that $D_1 = D_2$?

$\endgroup$
  • 1
    $\begingroup$ Is the following a counterexample to the generalization? Let $n = 2$, let $s \ge 4$ be an integer, let $\sigma \subseteq \mathbf{R}^{2}$ be the cone generated by $(s,1)$ and $(0,1)$, let $m = (1,1)$ and $m' = (2,1)$. Then $S(m) = S(m') = \{(0,1)\}$. $\endgroup$ – Minseon Shin Aug 4 at 23:47
  • $\begingroup$ Yes, thanks - I'll update my question. $\endgroup$ – Mellon Aug 5 at 9:19

Your Answer

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

Browse other questions tagged or ask your own question.