Ehrhart–Macdonald reciprocity

Remarks

One proof triangulates $P$ into lattice simplices and writes $P$ and $P^\circ$ as disjoint unions of half-open simplices. For a single simplex, the integer points of all dilates are counted with a rational generating function over the cone above the simplex. The generating functions of a cone and of its interior are related by $z \mapsto 1/z$, and this symmetry gives the sign $(-1)^d$ and the change from $P$ to $P^\circ$.

Example: for $P = [0,1]^2$, $E_P(t) = (t+1)^2$, so $E_P(-t) = (t-1)^2$. The open square $(0,t)^2$ contains exactly the integer points $(a,b)$ with $1 \le a, b \le t-1$, and there are $(t-1)^2$ of them. For the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$, $E_P(-t) = (t-1)(t-2)/2$, which counts the integer points with $a, b \ge 1$ and $a + b \le t - 1$.

Taking $t = 1$ gives the number of interior lattice points of $P$ as $(-1)^d E_P(-1)$. The statement extends to rational polytopes, where $E_P$ is replaced by the Ehrhart quasi-polynomial.

Going deeper

For a student

Reciprocity says that the Ehrhart polynomials of $P$ and of $P^\circ$ are tied by $t \mapsto -t$. It is one instance of a family of combinatorial reciprocity laws, which also includes reciprocity for order polynomials and for chromatic polynomials. Together with $E_P(0) = 1$ and the leading coefficient being the volume, it constrains the coefficients of $E_P$. For example, a lattice polygon $P$ with area $A$ and $b$ lattice points on its boundary has $E_P(t) = A t^2 + \frac{b}{2} t + 1$. Reciprocity at $t = 1$ then gives $A - \frac{b}{2} + 1$ interior lattice points, which is Pick's theorem.