Gruber P. Convex and Discrete Geometry

(t) = adt +···+a1t d + a0t d+1
19 Lattice Polytopes 315
and we obtain, from (13)
L o � � � �
� �
d + n − 1 d + n − 2
n − 1
(nS) =
ad +
ad−1 +···+ a0
d
d
d
= 1 � �
(d + n − 1) ···(n + 1)nad +···+(n − 1) ···(n − 1 − d + 1)a0
d !
= (−1)d � �
(−n + d) ···(−n + 1)a0 +···+(−n) ···(−n − d + 1)ad
d !
= (−1) d
�� � � � �
−n + d
−n
a0 +···+ ad = (−1)
d
d
d pS(−n).
This, together with the corresponding results in lower dimensions, yields the following
counterpart of (6) and (9).
(14) Let S ∈ P Z d be a simplex with c = dim S ≤ d. Then
L o (nS) = (−1) c pS(−n) for n ∈ N.
In the final step we extend (14) to proper convex lattice polytopes.
(15) Let P ∈ P Z d p . Then L o (nP) = (−1) d pP(−n) for n ∈ N.
Represent P as the disjoint union of the relative interiors of the simplices of a complex
of lattice simplices as in the proof of (12). Then
(16) pP(n) = L(nP) = �
L
S
o (nS) + �
L
R
o (nR) +···
= �
(−1) d pS(−n) + �
(−1) d−1 pR(−n) +··· for n ∈ N,
S
R
where S ranges over all proper simplices of this complex, R over all (d − 1)dimensional
simplices, etc. Since pP, pS,..., all are polynomials and since (16)
holds for all n ∈ N, it holds for all t ∈ R in place of n ∈ N. Thus it holds in
particular for all −n where n ∈ N. Hence
(17) (−1) d pP(−n) = �
pS(n) −
S
�
pR(n) +···
R
= �
L(nS) − �
L(nR) +··· for n ∈ N,
S
R
by (i). To finish the proof of (15,) it suffices to show that the last line in (17) is
equal to L o (nP). Letl ∈ Z d . The contribution of l to the last line in (17) equals the
expression
# � {S : l ∈ nS} � − # � {R : l ∈ nR} � +−...
If l �∈ nP, this expression is 0. If l ∈ bd(nP), this expression may be interpreted
(up to the sign) as the Euler characteristic of (a complex of convex polyhedral cones
with apex l and union equal to) the polyhedral support cone of P at l. Thus it is 0.
If l ∈ int nP, this expression is (up to the sign) equal to the Euler characteristic of
(a complex of convex polyhedral cones with apex l and union equal to) E d . Thus it
is 1. The last line in (17) thus equals L o (nP), concluding the proof of (15) and thus
of (ii). ⊓⊔