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Let $v_i=a_ie_i\in\mathbb R^d$ and $w_i=b_ie_i\in\mathbb R^d$ for $i=1,\dots,d$ where $e_i$'s are unit vetcors and $a_i,b_i$ are positive integers. Let $$S=conv\{0,rv_i+sw_i:i=1,\dots,d\}.$$ Can we express the cardinality of the set $S\cap\mathbb N_{\geq 0}^d$ as a bigraded polynomial of $r$ and $s$?

I am almost sure that it is a polynomial. But I can not prove it.

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  • $\begingroup$ Are $a_i$ and $b_i$ integers? $\endgroup$ Commented Jun 2, 2015 at 23:20
  • $\begingroup$ Of course, it's a polynomial. For any lattice polytopes $P$ and $Q$ the number of lattice points in $rP+sQ$ is a polynomial in $r$ and $s$. $\endgroup$ Commented Jun 3, 2015 at 1:52
  • $\begingroup$ @LevBorisov: I don't think that $S$ is of the form $rP+sQ$. $\endgroup$ Commented Jun 3, 2015 at 14:01
  • $\begingroup$ Sorry, my mistake, see answer below. $\endgroup$ Commented Jun 3, 2015 at 17:36

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Let's see what happens in dim 2. You have $conv((0,0),(ra_1+sb_1,0),(0,ra_2+sb_2))$. The number of points in the closed triandle $(0,0),(A,0),(0,B)$ is $(A+1)(B+1)/2$ plus half the number of points on the diagonal, which is $gcd(A,B)+1$. So up to a polynomial in $r$ and $s$ you get $gcd(ra_1+sb_1,ra_2+sb_2)$. I am pretty sure it is not polynomial in $r,s$. For example, take $a_1=1$, $b_1=0$, $a_2=0$, $b_2=1$, so you get $gcd(r,s)$.

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  • $\begingroup$ $a_i$and $b_i$ must be positive. $\endgroup$ Commented Jun 3, 2015 at 18:39
  • $\begingroup$ Makes no difference, really. Just make a linear change of variables that sends $(1,0)$, $(0,1)$ to $(1,1)$, $(2,1)$. $\endgroup$ Commented Jun 4, 2015 at 2:27
  • $\begingroup$ @LevBorisov: Thank you for the answer. $\endgroup$
    – Cusp
    Commented Jun 4, 2015 at 4:04

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