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 $P$ f_P(L,t)$is nice a polynomial with respect to at least some lattice$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$is a translation of a rational polytope. 4 added 335 characters in body Given a lattice$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.) under which if$P$is nice 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: 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 rational polytope. 3 added 38 characters in body 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.) under which if$P$is nice 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?