TITLE: When are Ehrhart functions of compact convex sets polynomials?
QUESTION [13 upvotes]: Given a lattice $L$ and a subset $P\subset \mathbb R^d$, we define for each positive integer $t$ $$f_P(L,t)=|(tP\cap L)|$$ the number of lattice points in $tP$. Let's say $P$ is nice if $f_P(L,t)$ is a polynomial. We know that if $P$ is a convex polytope with vertices in $L$ then $P$ is nice and $f_P(L,t)$ is its Ehrhart polynomial. My question is about some converse of this statement.

Are there some mild assumptions (for example convexity etc.) on $P$, under which if $f_P(L,t)$ is a polynomial with respect to at least some lattice $L$ then $P$ must be a convex polytope? Or a weaker question: Is any polynomial arising this way also the Ehrhart polynomial of some polytope?

P.S. I haven't thought much about this question so I apologize if it is well-known or it has an obvious negative answer. Also feel free to retag.

Richard Stanley suggested the following in the comments (edited to take into account a trivial family of counter-examples):

Could the following be true? It seems more in line with the question. Let $P$ be a compact convex $n$-dimensional set in $\mathbb R^n$. Suppose that the Ehrhart function $f_P(t)$ is a polynomial for positive integers $t$. Then $P$ is a translation of a rational polytope.


Edit: I would also be interested in a slightly weaker statement: Suppose a convex set has positive curvature almost everywhere, must the Ehrhart function necessarily be non-polynomial?
For example given an arbitrary lattice, what would be the easiest way to see that a circle doesnt have a polynomial Ehrhart function?

REPLY [3 votes]: I believe that the strong form of the conjecture is false. In lieu of a simple counterexample, let me point you towards a centrally symmetric 10-gon $\hat P$ in arXiv:0801.2812, Figure 6. It is a bit of a mess to explain exactly what it is, but it has something to do with Picard lattice of a toric DM stack. It need not be rational or a translate of rational.
The key properties of $\hat P$ are 
(1) It is centrally symmetric.
(2) The midpoints of all the sides are lattice points. As a result, opposite sides are 
lattice translates of each other.
As a result, generic translates of $\hat P$ have the same number of lattice points. Indeed,
as you move the polytope in a plane along a general curve as soon as a point appears on one side of it, another point exits from the opposite side. This implies that the opposite sides of $\hat P$ glue together to give a "no-gaps" cover of the torus $\mathbb R^2/L$ (preimage of a generic point has the same cardinality $k$). Then if one takes a $t$-multiple of it, one gets a "no-gaps" cover of $\mathbb R^2/tL$, and will thus have $kt^2$ points in $t(\hat P+ c)$ for a generic shift $c$.
I assume that this construction can be simplified to give something more explicit and palatable, so long as the property that the opposite sides are lattice translates of each other is satisfied.
It clearly requires flat sides to be able to glue them together on the torus, so this idea is not going to work for the positive curvature problem.