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Gruber P. Convex and Discrete Geometry

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Let φn,µn be as in the proof of Proposition 19.3. Then<br />

19 Lattice Polytopes 331<br />

L(nP) = φn(P) = µn(P + H d ) = µn(a0S0 +···+ad Sd + H d )<br />

= a0µn(S0 + H d ) +···+adµn(Sd + H d )<br />

� � � �<br />

n + 0<br />

n + d<br />

= a0φn(S0) +···+adφn(Sd) = a0 +···+ad<br />

0<br />

d<br />

= b0 + b1n +···+bdn d for P ∈ P Z d , n ∈ N,<br />

by Proposition 19.3, where a0,...,ad <strong>and</strong> thus b0,...,bd are suitable integers,<br />

depending only on P. Thus we may write,<br />

(14) L(nP) = L0(P) + L1(P)n +···+Ld(P)n d for P ∈ P Z d , n ∈ N.<br />

It is easy to check that the mappings Li : P Z d → R are integer unimodular invariant<br />

valuations: let P, Q ∈ P Z d be such that P ∪ Q, P ∩ Q ∈ P Z d . Then L � n(P ∪ Q) � +<br />

L � n(P ∩ Q) � = L(nP) + L(nQ). Represent each of these four expressions as a<br />

polynomial in n according to (14) <strong>and</strong> compare coefficients. By (14),<br />

Fixing n ∈ N, this shows that<br />

L(mnP) = L0(nP) + L1(nP)m +···+Ld(nP)m d<br />

= L0(P) + L1(P)mn +···+Ld(P)m d n d<br />

for P ∈ P Z d , m, n ∈ N.<br />

Li(nP) = n i Li(P) for P ∈ P Z d , n ∈ N, i = 0,...,d.<br />

Since L(nSi) = ( n+i<br />

i ) is a polynomial in n of degree i, we obtain<br />

Li(Si) �= 0, Li+1(Si) =···= Ld(Si) = 0fori = 0,...,d.<br />

This readily implies that L0,...,Ld are linearly independent. The proof of (13) is<br />

complete.<br />

The theorem finally follows from (12) <strong>and</strong> (13). ⊓⊔<br />

Ehrhart’s Lattice Point Enumerators Theorem<br />

As simple corollaries of the theorem of Betke <strong>and</strong> Kneser we get again the<br />

theorem 19.1 of Ehrhart [292, 293] on lattice point enumerators, but without the<br />

information that the constant term in the polynomial is 1.<br />

Corollary 19.1. Let P ∈ P Z d p . Then the following hold:<br />

(i) L(nP) = pP(n) for n ∈ N, where pP is a polynomial of degree d with leading<br />

coefficient V (P).<br />

(ii) L o (nP) = (−1) d pP(−n) for n ∈ N.<br />

Proof. (i) By the Betke–Kneser theorem,

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