Given are two convex bodies $K, L \subset \mathbb{R}^n$ that contain the origin as an interior point. Assume the number of integer points contained in $\lambda K$ equals the number of integer points contained in $\lambda L$ for every $\lambda > 0$. Is $K$ equal to $L$ modulo unimodular transformations?

If not, in what way is $K$ similar to $L$ besides the well-known fact that they must have the same volume?

Addendum: Anton's example shows that $K$ and $L$ are not necessarily equal modulo unimodular transformations even when they are taken to be two integral polygons in the plane and Abhinav Kumar points out in the comments that in the case $K$ is a integral polytope the number of integer points inside $n K$ ($n$ a natural number) is the value of the Ehrhart polynomial of $K$ at $n$.

Added 21/11/2013: For $K$ and $L$ ellipsoids this problem translates to "what can we say about isospectral flat Riemannian tori?". Therefore, by a famous remark by Milnor there exist ellipsoids $K$ and $L$ in $\mathbb{R}^{16}$ that are not unimodularly equivalent and for which the number of integer points contained in $\lambda K$ equals the number of integer points contained in $\lambda L$ for every $\lambda > 0$.

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    $\begingroup$ I would be surprised if this were true even for integer polytopes - that the Erhart polynomial determines the polytope (though I can't seem to find an immediate counterexample by searching online ...). You can certainly do $GL_n(\mathbb{Z})$ transformations without changing the number of integer points. $\endgroup$ Nov 20, 2013 at 12:48
  • $\begingroup$ @AbhinavKumar: yes, I meant equal modulo (integer) unimodular transformations. I'll edit the question. Thanks. $\endgroup$ Nov 20, 2013 at 13:02
  • $\begingroup$ @AbhinavKumar: thanks also for mentioning the Ehrhart polynomial. That's what I was missing. Hopefully this is well known. Are some convex bodies at least known to be determined by their Ehrhart polynomials? $\endgroup$ Nov 20, 2013 at 13:17
  • $\begingroup$ Looking for references on the Ehrhart polynomial I found a really neat book by M. Becks and S. Robins: math.sfsu.edu/beck/papers/noprint.pdf $\endgroup$ Nov 20, 2013 at 13:46
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    $\begingroup$ Erhart polynomial coincides with your function ONLY for integer values of $\lambda$. $\endgroup$ Nov 20, 2013 at 13:46

2 Answers 2


As it was noticed by Abhinav Kumar, you only can hope for equality up to $\mathrm{GL}_n(\mathbb Z)$ transformations.

The following picture shows two $\mathrm{GL}_n(\mathbb Z)$-distinct plane figures which have the same number of integer points after any rescaling.


It still might be true that such bodies are $\mathrm{GL}_n(\mathbb Z)$-equidecomposable.

P.S. This question is related.

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    $\begingroup$ Thanks Anton. I like your guess about the equivariant equidecomposability. By the way, in a precise sense, this question can be rephrased as "How much can you hear in the "shape" of a flat Finsler torus?". $\endgroup$ Nov 20, 2013 at 19:02
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    $\begingroup$ At least the two bodies you pictured are equidecomposable: the parts below the horizontal diagonal are isometric, and the parts above are changed one to another by the map $(x,y)\mapsto (x,y+x)$. $\endgroup$
    – ACL
    Feb 2, 2014 at 6:04
  • $\begingroup$ @ACL, sure, this is the point, I do not see an other way to construct such pairs of sets. $\endgroup$ Feb 2, 2014 at 18:52

In this neat paper by C. Haase & T. McAllister (which appeared in Cont. Math. 451 (2008)) you can find many examples, but as already mentioned in Anton's answer, the question of equidecomposability is open (and in fact, the authors offer a conjecture about this).


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