Gruber P. Convex and Discrete Geometry

19 Lattice Polytopes 313
1. Hence vd+1 = α1 +···+αd+1 < d + 1. Thus, being integer, vd+1 ≤ d, which
shows that q has degree at most d. The proof of (1) is complete.
Second, we show Proposition (i) for simplices.
(6) Let S ∈ P Z d p be a simplex. Then L(nS) = pS(n) for n ∈ N, where pS is a
polynomial of degree d, with leading coefficient V (S) and constant term 1.
To show (6), note that according to (1), q(t) = a0+a1t +···+adt d . By the definition
of q in (5), the coefficient ai is the number of points v ∈ F ∩ Z d+1 with vd+1 = i.
Thus, in particular,
(7) a0 = 1 and a0 + a1 +···+ad = h,
see (2). Denote the k-dimensional volume by Vk. Wehave,
� �
Vd+1(F) = (d + 1)!Vd+1 conv(T ∪{o})
= (d + 1)!
d + 1 Vd(T ) = d !V (S).
Since h = Vd+1(F) by (2), it follows that
(8) V (S) = h
d ! .
Inserting
q(t) = a0 + a1t +···+adt d
into (1) and comparing the coefficients of t n in the two power series, yields
� �
d + n
L(nS) =
d
a0 +
� �
d + n − 1
d
a1 +···+
�
n
�
ad
d
= a0 +···+ a0 +···+ad
n
d !
d = 1 +···+V (S) n d = pS(n), say,
for n ∈ N
by (7) and (8). The proof of (6) is complete.
Using (6), we now show the following related statement.
(9) Let R ∈ P Z d be a simplex with c = dim R < d. Then L(nR) = pR(n)
for n ∈ N, where pR is a polynomial of degree less than d with constant
term 1.
Embed E c into E d as usual (first c coordinates). There is an integer unimodular d ×d
matrix U ∈ U such that UR is a simplex in PZ c p. Now apply (6) to UR and note
that L(nR) = L(nU R) to get (9).
Third, the following proposition will be shown.
(10) Let S ∈ P Z d p be a simplex. Then L(n int S) = qS(n) for n ∈ N, where
qS is a polynomial of degree d with leading coefficient V (S) and constant
term (−1) d .