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Gruber P. Convex and Discrete Geometry

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254 <strong>Convex</strong> Polytopes<br />

The Face Lattices of P <strong>and</strong> P ∗ are Anti-Isomorphic<br />

We first present a tool, where<br />

is the polar of P, see Sect. 9.1.<br />

P ∗ = � y : x · y ≤ 1forx ∈ P �<br />

Proposition 14.2. Let P ∈ P such that o ∈ int P. Then the following statements<br />

hold:<br />

(i) P ∗ ∈ P.<br />

(ii) P ∗∗ = P.<br />

Proof. (i) Represent P in the form P = conv{x1,...,xn}. Then<br />

P ∗ = � y : (λ1x1 +···+λnxn) · y ≤ 1forλi ≥ 0, λ1 +···+λn = 1 �<br />

= � y : xi · y ≤ 1fori = 1,...,n} =<br />

n�<br />

{y : xi · y ≤ 1 �<br />

is an intersection of finitely many closed halfspaces. Since o ∈ int P, there is ϱ>0<br />

such that ϱB d ⊆ P. The definition of polarity then easily yields P ∗ ⊆ (ϱB d ) ∗ =<br />

(1/ϱ)B d . Thus P ∗ is bounded. Since P ∗ is the intersection of finitely many halfspaces<br />

<strong>and</strong> is bounded, Theorem 14.2 implies that P ∗ ∈ P.<br />

(ii) To show that P ⊆ P ∗∗ ,letx ∈ P. Then x · y ≤ 1 for all y ∈ P ∗ by the<br />

definition of P ∗ . This, in turn, implies that x ∈ P ∗∗ by the definition of P ∗∗ .To<br />

show that P ⊇ P ∗∗ ,letx ∈ E d \P. The separation theorem 4.4 then provides a point<br />

y ∈ E d such that x · y > 1 while z · y ≤ 1 for all z ∈ P. Hence y ∈ P ∗ .Fromx · y > 1<br />

we then conclude that x �∈ P ∗∗ by the definition of P ∗∗ . The proof that P = P ∗∗ is<br />

complete. ⊓⊔<br />

The following result relates the face lattices of P <strong>and</strong> P ∗ , where dim ∅=−1.<br />

Theorem 14.7. Let P ∈ P such that o ∈ int P. Define a mapping ♦ = ♦P by<br />

i=1<br />

F ♦ = � y ∈ P ∗ : x · y = 1 for x ∈ F � for F ∈ F(P),<br />

where, in particular, ∅ ♦ = P ∗ <strong>and</strong> P ♦ =∅. Then the following statements hold:<br />

(i) ♦ is a one-to-one mapping of F(P) onto F(P ∗ ).<br />

(ii) dim F ♦ + dim F = d − 1 for each F ∈ F(P).<br />

(iii) ♦ is an anti-isomorphism of the lattice � F(P), ∧, ∨ � onto the lattice<br />

� F(P ∗ ), ∧, ∨ � .<br />

By the latter we mean that ♦ is one-to-one <strong>and</strong> onto <strong>and</strong><br />

(F ∧ G) ♦ = F ♦ ∨ G ♦ <strong>and</strong> (F ∨ G) ♦ = F ♦ ∧ G ♦ for F, G ∈ F(P).

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