# Reciprocity (Ehrhart-style) for real polytopes?

Is there some sense in which the well-known Ehrhart reciprocity law for rational, convex, polytopes can be extended to any convex polytope with arbitrary real vertices?

In other words, given any real closed (and convex) polytope $P$, is the function that enumerates integer points in dilates of $P$ somehow simply related to the function that enumerates integer points in the dilates of the interior of $P$?

Here's one possible direction, given that we no longer have the usual quasi-polynomial behavior for the integer point enumerator. Does there exist some piecewise smooth function $L(t)$, of a real variable $t$, which satisfies the following properties:

(1) $L(t)$ agrees with the integer point enumerator #{${\mathbb Z}^d \cap tP$} at all positive integer values of $t$.

(2) $L(-t)$ counts the integer points in the interior of $tP$, for all positive integers $t$.

(3) The Fourier transform of $L$ is supported at a countable set of points.

Intuitively, for all real and positive $t$, such a smooth function $L(t)$ would approximate the piecewise constant function #{${\mathbb Z}^d \cap tP$}). Property (3) above was suggested by Allen Knutson.

(ADDED) Let's try to see what happens in dimension one. So we let $P := [0, \alpha]$, a one-dimensional polytope, with $\alpha$ a positive irrational number. We let $L_P(n)$ be the number of integer points in the dilation $nP$, so that here $L_P(n) = [n\alpha]+1$. For the interior $P^o$, we have $L_{P^o}(n) = [n\alpha]$. Using {x} $= \frac{1}{2} + \sum_{k\in \mathbb Z - \{0\}} \frac{1}{k}e^{2\pi i k x}$, and $L_P(n)= n\alpha -$ {$n\alpha$}+1, we have: $$L_P(n) = [n\alpha]+1 = n\alpha + \frac{1}{2} - \sum_{k \in \mathbb Z - \{0\} } \frac{1}{k}e^{2\pi i k n \alpha},$$ and let's use this last expression as the extension of $L_P(n)$ to all real values of $n$. We now check for reciprocity, using this new definition of our extended function $L_P(n)$: $$L_P(-n) = -n\alpha + \frac{1}{2} - \sum_{k \in \mathbb Z - \{0\} } \frac{1}{k}e^{-2\pi i k n \alpha}$$ $$= -n\alpha + \frac{1}{2} + \sum_{k \in \mathbb Z - \{0\} } \frac{1}{k}e^{2\pi i k n \alpha} \ \ (\text{replacing } k \in \mathbb Z \ by \ -k \in \mathbb Z)$$ $$= - [ n\alpha]= - L_{P^o}(n),$$ which is a reciprocity law.

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I don't understand the question. Say we have two functions $h(n),h^\circ(n)$ enumerating points in $nP,nP^\circ$. Then in the case of integer vertices, we can say "these are polynomials! so it makes sense to plug in negative values" and observe that $h^\circ(n) = (-1)^n h(n)$. But for a general $P$, how do you extend to negative argument? – Allen Knutson Nov 8 '12 at 2:48
You'll need some additional constraint on your interpolating function, since any values at the integers can be interpolated by an entire function. – Andreas Blass Nov 9 '12 at 0:36
I think there should be some Fourier transform statement, in terms of distributions on circles. Polynomial $\iff$ supported at the identity, quasipolynomial $\iff$ supported at some finite set of elements of finite order. Maybe given $h(>0)$ there's a unique extension to ${\mathbb Z}$ such that the Fourier transform is supported on a finite set? Or is a countable sum that converges especially quickly? – Allen Knutson Nov 9 '12 at 3:46
Let $\hat f$ be the delta function on the circle supported at $2\pi x \in [0,2\pi)$. I think the Fourier transform is $f(n) = \exp(nx)$, so, periodic iff $x$ is rational. More to the point, if $x$ is rational then $f$ is quasipolynomial. Apply derivatives and such to $\hat f$ to get more complicated distributions at $2\pi x$; meanwhile $f$ becomes $\exp(nx) p(n)$. Now take finite linear combinations, and you exactly get quasipolynomials. – Allen Knutson Nov 9 '12 at 14:50
(I meant $\exp(inx)$, of course) – Allen Knutson Nov 9 '12 at 14:57