Gruber P. Convex and Discrete Geometry

328 Convex Polytopes
(7) ψ|H d = 0.
To see this, note that for the generating elements of H d ,
ψ(P − UP− u) = ψ(P) − ψ(UP+ u) = φ(P) − φ(UP+ u) = 0,
⎛
ψ ⎝P − �
Pi + �
⎞
Pi ∩ Pj −··· ⎠
i
i< j
= ψ(P) − �
ψ(Pi) + �
ψ(Pi ∩ Pj) −···
i
i< j
= φ(P) − �
φ(Pi) + �
φ(Pi ∩ Pj ) −···=0,
i
i< j
where we have used the facts that φ is integer unimodular invariant by assumption
and satisfies the inclusion–exclusion formula by Theorem 19.5. Since this holds for
the generating elements of H d , we obtain (7). Since ψ : G d → A is a homomorphism,
(7) shows that it gives rise to a homomorphism of G d /H d → A. Wealso
denote it by ψ. Clearly ψ(P + H d ) = ψ(P) = φ(P) for P ∈ P Z d . To conclude the
proof of (6) we have to show that ψ is unique. Let µ be another homomorphism of
G d /H d → A satisfying (5). Then
ψ(P + H d ) = φ(P) = µ(P + H d ) for P ∈ P Z d .
Since the cosets P + H d : P ∈ P Z d generate G d /H d and ψ, µ are both homomorphisms,
ψ = µ, concluding the proof of (6).
The second step is to prove the following reverse statement:
(8) Let ψ : G d /H d → A be a homomorphism. Then the mapping φ : P Z d →
A, defined by (5), is a integer unimodular invariant valuation.
We first show that φ is a valuation. Let P, Q ∈ P Z d , where P ∪ Q, P ∩ Q ∈ P Z d .
Then
φ(P ∪ Q) = ψ(P ∪ Q + H d )
= ψ � P ∪ Q − (P ∪ Q − P − Q + P ∩ Q) + H d�
= ψ(P + Q − P ∩ Q + H d )
= ψ(P + H d ) + ψ(Q + H d ) − ψ(P ∩ Q + H d )
= φ(P) + φ(Q) − φ(P ∩ Q)
by the definitions of φ and H d , since ψ is a homomorphism. To see that φ is integer
unimodular invariant, let P ∈ P Z d , U ∈ U, u ∈ Z d . Then
φ(UP+ u) = ψ(UP+ u + H d ) = ψ(UP+ u + (P − UP− u) + H d )
= ψ(P + H d ) = φ(P)
by the definitions of φ and H d , concluding the proof of (8).
Having shown (6) and (8), the proof of the proposition is complete. ⊓⊔