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Gruber P. Convex and Discrete Geometry

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The Steiner Point<br />

13 The Space of <strong>Convex</strong> Bodies 239<br />

The Steiner point or curvature centroid sC of a convex body C is the point<br />

sC = 1<br />

�<br />

hC(u)udσ(u),<br />

κd<br />

Sd−1 where the integral is to be understood componentwise. If C is a proper convex body<br />

of class C 2 , it can be shown that sC is the centroid of mass distributed over bd C with<br />

density equal to the Gauss curvature of bd C. A nice application of the Steiner point<br />

to approximate matching of shapes was given by Aichholzer, Alt <strong>and</strong> Rote [5].<br />

Homomorphisms of 〈C, +〉 into 〈E d , +〉 <strong>and</strong> a Characterization of the Steiner<br />

Point<br />

The following result was proved by Schneider [898].<br />

Theorem 13.6. Let s : C → E d be a mapping. Then the following statements are<br />

equivalent:<br />

(i) s(C) = sC for C ∈ C.<br />

(ii) s satisfies the properties:<br />

(a) s(C + D) = s(C) + s(D) for C, D ∈ C<br />

(b) s(mC) = ms(C) for C ∈ C <strong>and</strong> all rigid motions m : E d → E d<br />

(c) s(·) is continuous<br />

Proof. (i)⇒(ii) This is a simple exercise.<br />

(ii)⇒(i) The first step is to show the following:<br />

(1) Let H : Hd → Ed be a linear mapping such that H(rh) = rH(h) for<br />

h ∈ Hd <strong>and</strong> each rotation r : Ed → Ed . Then<br />

�<br />

H(h) = β h(u)udσ(u) for h ∈ H d ,<br />

S d−1<br />

where β is a real constant depending only on d <strong>and</strong> H <strong>and</strong> the integral is<br />

considered componentwise.<br />

To prove this, let h = h0 + h1 +···+hm where hn ∈ Hd n . From Proposition 13.1(ii)<br />

<strong>and</strong> the fact that each coordinate ui of u ∈ Sd−1 is a spherical harmonic of degree 1<br />

in Hd 1 by Proposition 13.1(iii), it follows that<br />

�<br />

�<br />

h(u)udσ(u) = h1(u)udσ(u).<br />

S d−1<br />

S d−1<br />

For the proof of (1) it is thus sufficient to show that<br />

�<br />

(2) H(hn) = β h1(u)udσ(u) for n = 1 <strong>and</strong> = o for n �= 1.<br />

S d−1

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