7
$\begingroup$

Let $K \subset \mathbb{R}^3$ be a bounded convex set such that the points with integer coordinates in $K$ are not all coplanar. Is it true that $|K \cap \mathbb{Z}^3| \leq 6{\rm Vol}(K) + 3$?

$\endgroup$
5
  • 1
    $\begingroup$ What makes you think this is correct? $\endgroup$ Mar 8, 2015 at 17:54
  • 3
    $\begingroup$ Theorem 7.2.1 in Han Duong's Minimal Volume K-point Lattice D-simplices (I googled for it) shows a roughly similar bound for lattice polyhedra. So the question seems to be plausible, at least. I would be interested to hear a bit more in the way of background... $\endgroup$ Mar 8, 2015 at 19:23
  • 3
    $\begingroup$ By replacing $K$ with the convex hull of its lattice points, we may assume $K$ is a lattice polyhedron, so the theorem found by @PeteL.Clark applies. If $k \ge 1$ is the number of lattice points in the interior of $K$ and $b$ is the number on the boundary, it says$$ 6 \mathrm{Vol}(K) + 3 \ge 2b + 3k - 4 .$$ By assumption, the lattice points in $K$ are not all coplanar, so $K$ must have at least $4$ vertices, which means $b \ge 4$. Hence, $$ 2b + 3k - 4 \ge b + 3k \ge b + k = |K \cap \mathbb{Z}^3| .$$ However, this theorem only applies if $k \ge 1$. What about the remaining case? $\endgroup$ Mar 8, 2015 at 19:57
  • $\begingroup$ With some possibly other constants definitely yes. $\endgroup$ Mar 8, 2015 at 20:04
  • $\begingroup$ In their paper Lattice points in convex bodies (Israel J. Math, Vol. 74, Nos. 2-3, 1991), H. Gillet and C. Soulé prove a similar inequality in all dimensions: Under your assumptions, $\lvert K\cap \mathbf Z^n\rvert \leq 6^n/\mathop{\rm Vol}(K^*)$, where $K^*$ is the convex body dual to $K$ (Prop. 3). They eventually combine this inequality with Bourgain-Milman's one. $\endgroup$
    – ACL
    Mar 8, 2015 at 20:45

1 Answer 1

3
$\begingroup$

It suffices to prove the bound for (non-planar) polytopes $K$ with integer vertices. Let $v$ be a vertex of $K$. By triangulating faces of $K$ which does not contain $v$ and considering the corresponding simplexes with vertex $v$ based at these triangles one can obtain a decomposition of $K$ to simplexes with integer vertices.For such simplexes one can easily prove that the number of lattice points inside and on the faces except the vertices is at most six times the volume of the simplex minus 1:

Proof. By a well known result the number of lattice points in the parallelepiped generated by independent integer vectors $v_1,v_2,v_3$: $$\{\lambda_1 v_1+\lambda_2 v_2 + \lambda_3 v_3 : 0\leq \lambda_1,\lambda_2,\lambda_3<1 \}$$ is equal to the volume of parallelepiped which is six times the volume of the simplex with vertices $0,v_1,v_2,v_3$ and the origin with other lattice points of the simplex except vertices are in this parallelepiped.

So the number of all integer points in $K$ is at most $6 \mathrm{Vol(K)}$-$F$(the number of triangles = simplexes) + 1 (for $v$) + $V$ (the number of other vertices of $K$). But one can consider the triangulation as a connected triangulation in plane. By easy induction (adding triangles one by one) $E$ (the number of edges) is at most $2F+1$. So by Euler's formula: (we don't consider the unbounded face) $$V+F = E+1 \leq 2F+2 \Rightarrow V \leq F +2.$$

Therefore the total number of integer points of $K$ is $\leq 6\mathrm{Vol}(K)+3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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