Minkowski Valuations

Transcription

1 Minkowski Valuations Monika Ludwig Abstract Centroid and difference bodies define SL(n) equivariant operators on convex bodies and these operators are valuations with respect to Minkowski addition. We derive a classification of SL(n) equivariant Minkowski valuations and give a characterization of these operators. We also derive a classification of SL(n) contravariant Minkowski valuations and of L p-minkowski valuations AMS subject classification: 52A20 (52B11, 52B45) Centroid, difference, and projection bodies are fundamental notions in the affine geometry of convex bodies. The most important affine isoperimetric inequalities (and open problems) are formulated using these bodies. We show that the operators defined by these bodies together with the identity are basically the only examples of homogeneous, SL(n) equivariant or contravariant Minkowski valuations. The centroid body Γ K of a convex body K R n is a classical notion from geometry (see [5], [16], [36]) that has attracted much attention in recent years (see [4], [6], [8], [20], [21], [25], [27], [31]). If K is o-symmetric, then Γ K is the body whose boundary consists of the locus of the centroids of the halves of K formed when K is cut by hyperplanes through the origin. In general it can be defined in the following way. Let K n denote the set of convex bodies (that is, of compact, convex sets) in R n, and let Ko n denote the set of convex bodies in R n that contain the origin. A convex body K is uniquely determined by its support function h(k, ), where h(k, v) = max{v x : x K}, v R n, and where v x denotes the standard inner product of v and x. The moment body M K of K Ko n is the convex body whose support function is given by h(m K, v) = v x dx. If the n-dimensional volume vol n (K) of K is positive, then the centroid body Γ K of K is defined by 1 Γ K = vol n (K) M K. The fundamental affine isoperimetric inequality for centroid bodies is the Busemann-Petty centroid inequality [32]: Among bodies of given volume precisely for centered ellipsoids the centroid bodies have minimal volume. It is one of the major open problems to determine the reverse inequality (see [31]). The difference body D K of K K n is the Minkowski (or vector) sum of K and its reflection in the origin, that is, K D K = K + ( K). The operation that forms the difference body is essentially that known as central symmetrization and as such finds many applications in geometry and mathematical physics. The fundamental affine isoperimetric inequality for difference bodies is 1

2 the Rogers-Shephard inequality [34]: Among bodies of given volume precisely for simplices the difference bodies have maximal volume. The moment and difference operators are both Minkowski valuations. Here an operator Z is called a Minkowski valuation if Z K 1 + Z K 2 = Z(K 1 K 2 ) + Z(K 1 K 2 ), whenever K 1, K 2, K 1 K 2 K n and addition on K n is Minkowski addition. Valuations on convex bodies are a classical concept. In 1900, Dehn used them for solving Hilbert s third problem on the non-equidecomposability of convex polytopes of equal volume in R 3. Probably the most famous result on valuations is Hadwiger s characterization of rigid motion invariant real valued valuations continuous with respect to the Hausdorff metric as linear combinations of quermassintegrals. See [9], [15], [29], [30] for information on the classical theory and [1] - [3], [11] - [14], [18], [19], [37] for some of the more recent results. An operator Z : K n K n is Minkowski additive if Z(K 1 + K 2 ) = Z K 1 + Z K 2 for K 1, K 2 K n. Note that every Minkowski additive operator is a Minkowski valuation but not vice versa. Continuous Minkowski additive operators that commute with rigid motions are called endomorphisms. Schneider [35] (see also [36]) showed that there is a great variety of these operators. He obtained a complete classification of endomorphisms in K 2 and characterizations of special endomorphisms in K n. These results were further extended by Kiderlen [10]. We show that the moment and difference operators are basically the only examples of homogeneous, SL(n) equivariant Minkowski valuations. Here an operator Z : K n o K n is called SL(n) equivariant, if Z(φK) = φ Z K for φ SL(n), and it is called homogeneous of degree r, r R, if Z(s K) = s r Z K for s 0. Let P n o denote the set of convex polytopes in R n that contain the origin. For n 3, we show that Z : P n o K n is a homogeneous, SL(n) equivariant Minkowski valuation if and only if there are constants c 0 R, c 1, c 2 0 such that Z P = c 0 m(p ) + c 1 M P or Z P = c 1 P + c 2 ( P ) for every P Po n, where m(p ) is the moment vector of P. In particular, this implies that these are all continuous, homogeneous, SL(n) equivariant Minkowski valuations on Ko n. Combined with McMullen s polynomial expansion for translation invariant valuations [28], it implies that Z : P n K n is a translation invariant, SL(n) equivariant Minkowski valuation if and only if there is a constant c 0 such that Z P = c D P for every P P n. We also derive the corresponding results in the context of the L p -Brunn-Minkowski theory and give a characterization of L p -centroid bodies. The projection body Π K of K is the convex body whose support function is given for u S n 1 by h(π K, u) = vol n 1 (K u ), where vol n 1 denotes (n 1)-dimensional volume and K u denotes the image of the orthogonal projection of K onto the subspace orthogonal to u. Projection bodies, which were introduced by Minkowski, are an important tool for studying projections (see [5]). In recent years, projection bodies and their generalizations 2

3 in the L p -Brunn-Minkowski theory have attracted increased attention, see [7], [25], [26], [39]. The fundamental affine isoperimetric inequalities for projection bodies are the Petty projection inequality [33] and the Zhang projection inequality [38]: Among bodies of given volume precisely for ellipsoids the polar projection bodies have maximal volume and precisely for simplices the polar projection bodies have minimal volume. It is a major open problem to determine the corresponding results for the volume of the projection body itself (see [23]). We derive a classification of homogeneous, SL(n) contravariant Minkowski valuations and give a characterization of the projection operator. Here an operator Z : K n o K n is called SL(n) contravariant, if Z(φK) = φ t Z K for φ SL(n), where φ t denotes the inverse of the transpose of φ. This classification generalizes a result in [17] where it is shown that an operator Z : P n P n is a Minkowski valuation that is SL(n) contravariant, translation invariant and homogeneous of degree (n 1) if and only if it is a multiple of the projection operator. Here we obtain that an operator Z : P n K n is a translation invariant, SL(n) contravariant Minkowski valuation if and only if there is a constant c 0 such that Z P = c Π P for every P P n. We also derive the corresponding results in the context of the L p -Brunn-Minkowski theory. 1 Equivariant Minkowski valuations In this section, our main result on the classification of SL(n) equivariant Minkowski valuations is formulated. The important examples are the difference operator, the moment operator and the moment vector. Here the moment vector m(k) of a convex body K Ko n is defined by m(k) = x dx. Note that the moment operator M : K n o K n and the moment vector m : K n o R n commute (up to a determinantal factor) with general linear transformations: M(φK) = det φ φ M K and m(φk) = det φ φ m(k) for φ GL(n). The difference operator and the identity are SL(n) equivariant and homogeneous of degree 1. As we will see, for n 3 these are already all examples of homogeneous, SL(n) equivariant Minkowski valuations on P n o. For n = 2, we have an additional example. Let E o (P ) be the set of edges of P that contain the origin. For a 0, b 0 0 and a i, b i R with a i +b 0 +b 1, a 0 +a 1 +b i 0, i = 1, 2, define Z : P 2 o P 2 o by K Z P = a 0 P + b 0 ( P ) + (a i E i + b i ( E i )), (1) where the sum is taken over E i E o (P ). Here and throughout formulae like (1) have to be read as h(z P, v) = a 0 h(p, v) + b 0 h( P, v) + (a i h(e i, v) + b i h( E i, v)) for v R 2. The notation is only used if h(z P, ) is the support function of a convex body, which is here guaranteed by the conditions on a i, b i. Note that Z : Po 2 3

4 K 2 defined by (1) is SL(2) equivariant, homogeneous of degree 1, and Minkowski additive. The following result is our classification of SL(n) equivariant Minkowski valuations. The proof is given in Section 3. Theorem 1. Let Z : Po n K n, n 3, be a Minkowski valuation which is SL(n) equivariant and homogeneous of degree r. If r = n + 1, then there are constants a 0 R, a 1 0 such that Z P = a 0 m(p ) + a 1 M P for every P P n o. If r = 1, then there are constants a, b 0 such that Z P = a P + b( P ) for every P P n o. In all other cases, Z P = {o} for every P P n o. Let Z : P 2 o K 2 be a Minkowski valuation which is SL(2) equivariant and homogeneous of degree r. If r = 3, then there are constants a 0 R, a 1 0 such that Z P = a 0 m(p ) + a 1 M P for every P P 2 o. If r = 1, then there are constants a 0, b 0 0 and a i, b i R with a i + b 0 + b 1, a 0 + a 1 + b i 0, i = 1, 2, such that Z P = a 0 P + b 0 ( P ) + (a i E i + b i ( E i )) for every P P 2 o, where the sum is taken over E i E o (P ). In all other cases, Z P = {o} for every P P 2 o. Let K n and K n o be equipped with the topology defined by the Hausdorff metric. We have the following simple consequence of Theorem 1. Note that there are further examples of homogeneous, SL(n) equivariant Minkowski valuations on K n o that are not continuous (see [35]). Corollary 1.1. If Z : Ko n K n, n 2, is a continuous, homogeneous, SL(n) equivariant Minkowski valuation, then there are constants a 0 R, a 1, a 2 0 such that Z K = a 0 m(k) + a 1 M K or Z K = a 1 K + a 2 ( K) for every K K n o. For translation invariant valuations, we obtain the following result. Corollary 1.2. If Z : P n K n, n 2, is a translation invariant, SL(n) equivariant Minkowski valuation, then there is a constant c 0 such that Z = c D. Here an operator Z : K n K n is called translation invariant, if Z(K + x) = Z K for x R n. The proof of this corollary is given in Section 7. 2 Equivariant L p -Minkowski valuations For p > 1, the L p -Minkowski sum K 1 + p K 2 of K 1, K 2 K n o is defined by h p (K 1 + p K 2, v) = h p (K 1, v) + h p (K 2, v) for v R n. This notion, which was introduced by Firey in the middle of the last century, is at the heart of the L p -Brunn-Minkowski theory (or Brunn-Minkowski- Firey theory), see [22], [24]. We derive a classification of homogeneous, SL(n) 4

5 equivariant L p -Minkowski valuations. Here an operator Z : Ko n Ko n is called an L p -Minkowski valuation if Z K 1 + p Z K 2 = Z(K 1 K 2 ) + p Z(K 1 K 2 ), whenever K 1, K 2, K 1 K 2 K n o. The important examples of SL(n) equivariant L p -Minkowski valuations are the L p -moment operator and the L p -difference operator. These can be defined in the following way. For 1 τ 1, define M τ p : K n o K n o by h p (M τ p K, v) = K ( v x + τ(v x)) p dx for v R n. For τ = 0, we obtain the L p -moment operator M p. If vol n (K) > 0, then the L p -centroid body Γ p K of K is defined by Γ p K = c n,p vol n (K) M p K, where the constant c n,p is chosen such that Γ p B = B for the unit ball B. L p - centroid bodies were introduced by Lutwak and Zhang [27], for p = 2 they are the Legendre ellipsoids of classical mechanics. Lutwak, Yang, and Zhang [25] obtained the L p -version of the Busemann-Petty centroid inequality, see Campi and Gronchi [4] for a different proof. Note that M τ p(φk) = det φ 1/p φ M τ p K for φ GL(n), and that M τ p is an L p -Minkowski valuation. The L p -difference operator D p, defined by D p K = K + p ( K) for K K n o, and the identity are SL(n) equivariant, homogeneous of degree 1 and L p -Minkowski valuations. We will show that for n 3 these are all examples. For n = 2, there are additional examples. For a 0, b 0 0 and a i, b i R with a 0 + a i, b 0 + b i 0, i = 1, 2, define Z : P 2 o K 2 o by Z P = a 0 P + p b 0 ( P ) + p p(ai E i + p b i ( E i )) for every P Po 2, where the sum is taken over E i E o (P ). Here p denotes L p - Minkowski sum. Then Z : Po 2 Ko 2 is SL(2) equivariant, homogeneous of degree 1, and L p -Minkowski additive. A further operator is obtained in the following way. Let ψ π/2 denote the rotation by an angle π/2 and for 1 τ 1, let ˆΠ τ p be the operator defined in Section 5. Since ˆΠ τ p is SL(2) contravariant, we have (ψ π/2 ˆΠτ p )(φp ) = ψ π/2 φ t ψ 1 π/2 (ψ π/2 ˆΠ τ p)p = φ(ψ π/2 ˆΠτ p )P, that is, ψ π/2 ˆΠτ p is SL(2) equivariant. Since ˆΠ τ p is homogeneous of degree 2/p 1, an L p -Minkowski valuation and not continuous, so is ψ π/2 ˆΠτ p. The following result is our classification of SL(n) equivariant L p -Minkowski valuations. The proof is given in Section 3. Theorem 1 p. Let Z : P n o K n o, n 3, be an L p -Minkowski valuation, p > 1, which is SL(n) equivariant and homogeneous of degree r. If r = n/p + 1, then there are constants a 0 and 1 τ 1 such that Z P = a M τ p P 5

6 for every P P n o. If r = 1, then there are constants a, b 0 such that Z P = a P + p b( P ) for every P P n o. In all other cases, Z P = {o} for every P P n o. Let Z : P 2 o K 2 o be an L p -Minkowski valuation, p > 1, which is SL(2) equivariant and homogeneous of degree r. If r = 2/p + 1, then there are constants a 0 and 1 τ 1 such that Z P = a M τ p P for every P P 2 o. If r = 1, then there are constants a 0, b 0 0 and a i, b i R with a 0 + a i, b 0 + b i 0, i = 1, 2, such that Z P = a 0 P + p b 0 ( P ) + p p(ai E i + p b i ( E i )) for every P P 2 o, where the sum is taken over E i E o (P ). If r = 2/p 1, then there are constants a 0 and 1 τ 1 such that Z P = a ψ π/2 ˆΠτ p P for every P P 2 o. In all other cases, Z P = {o} for every P P 2 o. As a simple consequence we obtain the following corollary. Corollary. If Z : Ko n Ko n, n 2, is a continuous, homogeneous, SL(n) equivariant L p -Minkowski valuation, p > 1, then there are constants a, b 0 and 1 τ 1 such that for every K K n o. Z K = a M τ p K or Z K = a K + p b( K) 3 Proof of Theorems 1 and 1 p In the following, we work in n-dimensional Euclidean space R n with origin o, a fixed orthonormal basis e 1,..., e n, and use coordinates x = (x 1,..., x n ) for x R n. An operator Z : P n o K n is called simple, if Z P = {o} for every P P n o with dim P < n. Here dim P is the dimension of the linear hull, lin P, of P. As a first step in the proof, we show that every operator which is SL(n) equivariant and homogeneous of degree r 1 is simple. Lemma 1. Let Z : P n o K n be an operator which is SL(n) equivariant and homogeneous of degree r. Then Z P lin P. If dim P (n 1) and r 1, then Z P = {o}. Proof. Let P Po n be such that lin P is the k-dimensional subspace with equation x k+1 =... = x n = 0. Since every P Po n with dim P = k is a linear image of such a polytope P and since Z is SL(n) equivariant, it suffices to prove the lemma in this case. Let ( ) I B φ =, 0 A where I is the k k identity matrix, 0 is the (n k) k zero matrix, B is an (n k) k matrix, and A is an (n k) (n k) matrix with determinant 1. Then φ SL(n) and φp = P. (2) 6

7 Write x = (x, x ) with x = (x 1,..., x k ) and x = (x k+1,..., x n ) for x R n. Let x Z P. Since Z is SL(n) equivariant, (2) implies that y = φx Z P. Therefore ( ) ( y x + Bx ) = Z P. (3) y Ax This is true for every (n k) k matrix B and every (n k) (n k) matrix A with determinant 1. If x o, this implies that y can be an arbitrary vector. Since Z P is bounded, it follows that x = o. Thus Z P lin P. Now, let r 1 and let φ = ( I 0 0 s where I is the (n 1) (n 1) identity matrix and s 0. Then ), Z(φP ) = s (r 1)/n Z P = Z P. Since this holds for every s 0 and Z P is bounded, this implies that Z P = {o}. Next, we show that we can reduce the proof of Theorem 1 to showing that the corresponding results hold for simplices. The arguments used in the proof of the next lemma are well known from several extension theorems for valuations (cf. [9] or [15]). Lemma 2. Let Z 1, Z 2 : Po n K n be Minkowski valuations. If Z 1 S = Z 2 S for every n-dimensional simplex S Po n, then Z 1 = Z 2. Proof. If T Po n is a simplex with dim T = k < n, then there is a simplex T such that T = T H, where H = lin T and such that T H + and T H are both (k + 1)-dimensional. Here H +, H denote the closed halfspaces bounded by H. Since Z i is a valuation, we have for i = 1, 2 Z i T + Z i T = Z i (T H + ) + Z i (T H ). If Z 1 S = Z 2 S for every (k + 1)-dimensional simplex S, this implies that Z 1 T = Z 2 T. Thus we get by induction that Z 1 T = Z 2 T for every simplex T Po n. For x R n and i, i = 1, 2, fixed, µ(p ) = h(z i P, x) is a real valued valuation. Let P Po n be dissected into n-dimensional polytopes P 1,... P m Po n, that is, P = P 1... P m and the P i s have pairwise disjoint interiors. Induction on the dimension and the number of terms shows that the inclusion-exclusion principle µ(p ) = I ( 1) I 1 µ(p I ) (4) holds, where the sum is taken over all ordered k-tuples I = {i 1,..., i k } such that 1 i 1 <... < i k n and k = 1,..., m. Here I denotes the cardinality of I and P I = P i1... P ik (cf. [15]). Let P Po n. If dim P = n, then we dissect P into n-dimensional simplices S i P n, i = 1,..., m. Then (4) implies that Z 1 P = Z 2 P. If dim P < k, we use the same argument in lin P. The following lemma is used to prove Theorem 1 p. The proof is same as that of Lemma 2. Lemma 2 p. Let Z 1, Z 2 : P n o K n o be L p -Minkowski valuations. If Z 1 S = Z 2 S for every n-dimensional simplex S P n o, then Z 1 = Z 2. 7

8 To prove Theorems 1 and 1 p we determine the value of Z for n-dimensional simplices. Since Z is SL(n) equivariant and homogeneous, it is enough to consider a standard simplex. So let T be the simplex with vertices o, e 1,..., e n. We start with the planar case. Proposition 1. Let Z : P 2 o K 2 be a Minkowski valuation which is SL(2) equivariant and homogeneous of degree r. If r = 3, then there are constants a 0 R, a 1 0 such that Z T = a 0 m(t ) + a 1 M T. If r = 1, then there are constants a 0, b 0 0 and a i, b i R with a i + b 0 + b 1, a 0 + a 2 + b i 0, i = 1, 2, such that Z T = a 0 T + b 0 ( T ) + a 1 [o, e 1 ] + b 1 [o, e 1 ] + a 2 [o, e 2 ] + b 2 [o, e 2 ]. In all other cases, Z T = {o}. Proposition 1 p. Let Z : P 2 o K 2 o be an L p -Minkowski valuation, p > 1, which is SL(2) equivariant and homogeneous of degree r. If r = 2/p + 1, then there are constants a 0 and 1 τ 1 such that Z T = a M τ p T. If r = 1, then there are constants a 0, b 0 0 and a i, b i R with a 0 + a i, b 0 + b i 0, i = 1, 2, such that Z T = a 0 T + p b 0 ( T ) + p a 1 [o, e 1 ] + p b 1 [o, e 1 ] + p a 2 [o, e 2 ] + p b 2 [o, e 2 ]. If r = 2/p 1, then there are constants a 1, a 2 0 such that In all other cases, Z T = {o}. Z T = [ a 1 (e 1 e 2 ), a 2 (e 1 e 2 )]. Proposition 1 is basically a special case of Proposition 1 p. The only difference is the range of the operators. We prove both propositions at the same time. Proof. If p > 1, let Z : Po 2 Ko, 2 and if p = 1, let Z : Po 2 K 2. For 0 < λ < 1, let H λ be the hyperplane through the origin o with normal vector (1 λ) e 1 λ e 2. Then H λ dissects T into two simplices T H + λ and T H λ, where H + λ, H λ are the closed halfspaces bounded by H λ. Since Z is a valuation, we have Z T + p Z(T H λ ) = Z(T H + λ ) + p Z(T H λ ). (5) Set T = T e 1, where e 1 denotes the subspace orthogonal to e 1, and set ( ) ( ) λ 0 1 λ φ λ = and ψ 1 λ 1 λ =. (6) 0 1 λ Then T H λ = ψ λ T, T H + λ = φ λt, and T H λ = ψ λt. Set q = (r + 1)/2. Since Z is SL(2) equivariant and homogeneous of degree 2 q + 1, (5) implies that Z T + p (1 λ) q ψ λ Z T = λ q φ λ Z T + p (1 λ) q ψ λ Z T. (7) 1. Let q 0. By Lemma 1, Z is a simple valuation. Set f(x) = h p (Z T, x). Then (7) implies that f(x) = λ p q f(φ t λx) + (1 λ) p q f(ψ t λx) for 0 < λ < 1, x R 2. (8) We need the following lemma, which is also used for the case q = 0. 8

9 Lemma 3. Let f : R 2 R be a function which is positively homogeneous of degree p and for which (8) holds. Then for x 1 x 2 0 f(x 1, x 2 ) = f( x 1, x 2 ) = q+p xp1 x p q+p 2 (x 1 x 2 ) p q f(e 1 ), (9) q+p xp1 x p q+p 2 (x 1 x 2 ) p q f( e 1 ), (10) for x 2 x 1 0 f(x 1, x 2 ) = f( x 1, x 2 ) = q+p xp2 x p q+p 1 (x 2 x 1 ) p q f(e 2 ), (11) q+p xp2 x p q+p 1 (x 2 x 1 ) p q f( e 2 ), (12) and for x 1, x 2 0 f( x 1, x 2 ) = f(x 1, x 2 ) = x p q+p 2 (x 1 + x 2 ) p q f(e x p q+p 1 2) + (x 1 + x 2 ) p q f( e 1), (13) x p q+p 1 (x 1 + x 2 ) p q f(e x p q+p 2 1) + (x 1 + x 2 ) p q f( e 2). (14) Proof. The vector e 1 is an eigenvector of φ t λ with eigenvalue λ, the vector e 1 + e 2 is an eigenvector with eigenvalue 1. The vector e 2 is an eigenvector of ψλ t with eigenvalue (1 λ), the vector e 1 + e 2 is an eigenvector with eigenvalue 1. Therefore and f(e 1 ) = λ p q f(λ e 1 ) + (1 λ) p q f(e 1 + λ e 2 ) f(e 1 + λ e 2 ) = 1 λp q+p (1 λ) p q f(e 1). Since f is homogeneous of degree p, this can be written as f(x 1, x 2 ) = xp q+p 1 x p q+p 2 (x 1 x 2 ) p q f(e 1 ) for x 1 x 2 0. Similarly, (10) (12) are derived. From (8) we obtain which can be written as f( (1 λ) e 1 + λ e 2 ) = λ p q+p f(e 2 ) + (1 λ) p q+p f( e 1 ), f( x 1, x 2 ) = for x 1, x 2 0. Similarly, (14) is derived. Let q > 1/p. If f(e 1 ) 0, then by Lemma 3 x p q+p 2 (x 1 + x 2 ) p q f(e x p q+p 1 2) + (x 1 + x 2 ) p q f( e 1) lim f(e 1 λ p q+p 1 + λ e 2 ) = lim λ 1 λ 1 (1 λ) p q f(e 1) = ±. Since f(x) = h p (Z T, x) and since Z T is bounded, this is not possible. Thus f(e 1 ) = 0. Similarly, we obtain from Lemma 3 that f( e 1 ) = f(e 2 ) = f( e 2 ) = 0. Therefore Lemma 3 implies that f(x) = 0 for every x R 2, that is, Z T = {o}. 9

10 Let q < 1/p. Then by Lemma 3 we have lim f(e 1 + λ e 2 ) = lim f( e 1 λ e 2 ) = 0. λ 1 λ 1 Thus Z T (e 1 + e 2 ) and there are constants a 1, a 2 R such that f(x) = h p ([ a 1 (e 1 e 2 ), a 2 (e 1 e 2 )], x) for x R 2. If p > 1 and q = 1/p 1, then a 1, a 2 0 and this is just the statement of the proposition. If q 1/p 1, we use (8) with x = ±(1, 1) and obtain that a p i = λp q+p a p i + (1 λ)p q+p a p i. This implies that a 1 = a 2 = 0 and Z T = {o}. Let q = 1/p. Since f(x) = h p (Z T, x) is continuous, we obtain from Lemma 3 that f(e 1 ) = f(e 2 ) and f( e 1 ) = f( e 2 ). Thus f(x 1, x 2 ) = xp+1 1 x p+1 2 f(e 1 ) for x 1 x 2 0, x 1 x 2 f( x 1, x 2 ) = xp+1 1 x p+1 2 f( e 1 ) for x 1 x 2 0, x 1 x 2 (15) f( x 1, x 2 ) = x p+1 2 f(e 1 ) + xp+1 1 f( e 1 ) for x 1, x 2 0 x 1 + x 2 x 1 + x 2 and f(x 1, x 2 ) = f(x 2, x 1 ). Set g(x) = τ h(m(t ), x)+h(m T, x) with τ R for p = 1 and set g(x) = h p (M τ p T, x) with 1 τ 1 for p > 1. Then a simple calculation shows that g(x 1, x 2 ) = (1 + τ)p (x p+1 1 x p+1 2 ) (p + 1) (p + 2) (x 1 x 2 ) g(x 1, x 2 ) = (1 τ)p (x p+1 1 x p+1 2 ) (p + 1) (p + 2) (x 1 x 2 ) g(x 1, x 2 ) = (1 + τ)p x p (1 τ) p x p+1 1 (p + 1) (p + 2)(x 1 + x 2 ) for x 1 x 2 0, for x 1 x 2 0, for x 1, x 2 0 (16) and that g(x 1, x 2 ) = g(x 2, x 1 ). Comparing this with (15) shows that for p = 1 f(x) = a 0 h(m(t ), x) + a 1 h(m T, x), where a 0 R and a 1 0 are suitable constants, and that for p > 1 f(x) = a h(m τ p T, x), where a 0 and 1 τ 1 are suitable constants. 2. Let q = 0. Set f(x) = h p (Z T, x) h p (Z T, x). By Lemma 1, Z T e 1. Therefore φ λ T = T and (5) implies that f(x) = f(φ t λx) + f(ψ t λx) for 0 < λ < 1, x R 2. Thus (8) holds with q = 0 and we can apply Lemma 3. Since Z T e 1, we have Z T = [ s e 2, t e 2 ] with s, t R, s t. Note that if p > 1, then h(z T, ) 0 and s, t 0. Setting a 0 = f(e 1 )+f(e 2 ), a 1 = f(e 2 ), a 2 = 10

11 f(e 1 )+t p and b 0 = f( e 1 )+f( e 2 ), b 1 = f( e 2 ), b 2 = f( e 1 )+s p, we obtain from Lemma 3 that for x 1 x 2 0 for x 2 x 1 0 and for x 1, x 2 0 h p (Z T, (x 1, x 2 )) = x p 1 (a 0 + a 1 ) + x p 2 a 2, h p (Z T, ( x 1, x 2 )) = x p 1 (b 0 + b 1 ) + x p 2 b 2, h p (Z T, (x 1, x 2 )) = x p 1 a 1 + x p 2 (a 0 + a 2 ), h p (Z T, ( x 1, x 2 )) = x p 1 b 1 + x p 2 (b 0 + b 2 ), h p (Z T, ( x 1, x 2 )) = x p 1 (b 0 + b 1 ) + x p 2 (a 0 + a 2 ), h p (Z T, (x 1, x 2 )) = x p 1 (a 0 + a 1 ) + x p 2 (b 0 + b 2 ). As a support function h(z T, x) is convex. Therefore h(z T, e 1 ) + h(z T, e 2 ) h(z T, e 1 + e 2 ), which implies that a 0 0. Similarly, we obtain b 0 0. If p > 1, then a 0 + a 1, a 0 + a 2, b 0 + b 1, b 0 + b 2 0. This is just the statement of the proposition. Since h(z T, e 1 + e 2 ) + h(z T, e 1 + e 2 ) 2 h(z T, e 2 ), we have a 1 + b 0 + b 1 0. Similarly, we obtain a 2 + b 0 + b 1 0 and a 0 + a 1 + b 1, a 0 + a 1 + b 2 0. Thus the proposition holds true. Next, we consider the case n 3. Proposition 2. Let Z : P n o K n, n 3, be a Minkowski valuation which is SL(n) equivariant and homogeneous of degree r. If r = n + 1, then there are constants a 0 R, a 1 0 such that Z T = a 0 m(t ) + a 1 M T. If r = 1, then there are constants a, b 0 such that In all other cases, Z T = {o}. Z T = a T + b( T ). Proposition 2 p. Let Z : P n o K n o, n 3, be an L p -Minkowski valuation, p > 1, which is SL(n) equivariant and homogeneous of degree r. If r = n/p + 1, then there are constants a 0 and 1 τ 1 such that Z T = a M τ p T. If r = 1, then there are constants a, b 0 such that In all other cases, Z T = {o}. Z T = a T + p b( T ). We prove both propositions at the same time. Proof. If p > 1, let Z : Po n Ko n, and if p = 1, let Z : Po n K n. For 0 < λ < 1, i < j, let H λ = H λ (i, j) be the hyperplane through o with normal vector (1 λ) e i λ e j. Then H λ dissects T into two simplices T H + λ and T H λ. Since Z is a valuation, we have Z T + p Z(T H λ ) = Z(T H + λ ) + p Z(T H λ ). (17) 11

12 Define φ λ = φ λ (i, j), ψ λ = ψ λ (i, j) by φ λ e i = λ e i + (1 λ) e j, φ λ e k = e k for k i, (18) ψ λ e j = λ e i + (1 λ) e j, ψ λ e k = e k for k j. (19) Then T H + λ = φ λt and T H λ = ψ λt. Set T = T e i. Then T H λ = ψ λ T. Set q = (r 1)/n. Since Z is SL(n) equivariant and homogeneous of degree n q + 1, (17) implies that Z T + p (1 λ) q ψ λ Z T = λ q φ λ Z T + p (1 λ) q ψ λ Z T. (20) We need the following symmetry relations. Let i j, k l and let α SL(n) be such that α t e i = e k, α t e j = e l and αt = T. Since Z is SL(n) equivariant, Z T = α Z T. This implies that for s, t R h(z T, s e i + t e j ) = h(z T, s e k + t e l ). (21) 1. Let q 0. By Lemma 1, Z is a simple valuation. Set g(x) = h p (Z T, x). Then (20) implies that g(x) = λ p q g(φ t λx) + (1 λ) p q g(ψ t λx) for 0 < λ < 1, i < j, x R n. (22) For k i, j, we have g(e k ) = λ p q g(e k ) + (1 λ) p q g(e k ). Let q 1/p. Then this implies that g(e k ) = 0. Similarly, we obtain that g( e k ) = 0. Since (22) implies that Lemma 3 holds for f(s, t) = g(s e i + t e j ), we obtain that g(s e i + t e j ) = 0 for (s, t) R 2. Combined with the following lemma this shows that Z T = {o} for q 1/p. Lemma 4. Let g 1, g 2 : R n R be functions for which (22) holds. If g 1 (x) = g 2 (x) for every x R n where all but two coordinates vanish, then g 1 = g 2. Proof. Assume that g 1 (x) = g 2 (x) holds true for every x R n where at most k, 2 k n 1, coordinates are 0. We show that then g 1 (x) = g 2 (x) for every x R n where at most (k + 1) coordinates are 0. So let x = (x 1,..., x n ), n 3, be such that at most k coordinates are 0. Since at least two coordinates of x have the same sign and (22) holds for every pair i < j, we may assume that x 1, x 2 > 0 or x 1, x 2 < 0 and use (22) for i = 1, j = 2. First, let 0 < x 2 < x 1 or x 1 < x 2 < 0. By (22), we have for 0 < λ < 1 Since g i (ψ t λ x) = λp q g i (φ t λψ t x) + (1 λ)p q g i (x). λ ψ t λ x = (x 1, λ 1 λ x λ x 2, x 3,..., x n ), φ t λψ t λ x = (x 2, λ 1 λ x λ x 2, x 3,..., x n ), we set λ = x 2 /x 1 and obtain 0 < λ < 1. In ψ t λ x and φt λ ψ t λ x at most k coordinates are 0. Now, let 0 < x 1 < x 2 or x 2 < x 1 < 0. By (22), we have g i (φ t λ x) = λp q g i (x) + (1 λ) p q g i (ψ t λφ t λ x). Since φ t λ x = ( 1 λ x 1 1 λ λ x 2, x 2,..., x n ), 12

13 ψλφ t t λ x = ( 1 λ x 1 1 λ λ x 2, x 1, x 3,..., x n ), we set λ = 1 x 1 /x 2 and obtain 0 < λ < 1. As before we obtain g 1 (x) = g 2 (x). In φ t λ x and ψt λ φ t x at most k coordinates are 0. λ Let q = 1/p. Note that for x = s e 1 + t e 2, s, t R, h(m T, x) is a multiple of h(m T 2, (s, t)), where T 2 is the 2-dimensional standard simplex, and that corresponding statements hold for h(m(t ), x) and h(m τ p T, x). Set f(s, t) = g(s e 1 +t e 2 ). Since (22) implies that Lemma 3 holds for f(s, t) = g(s e 1 + t e 2 ), we obtain from (15) and (16) that for p = 1 and x = s e 1 + t e 2, g(x) = h(z T, x) = a 0 h(m(t ), x) + a 1 h(m T, x), (23) where a 0 R and a 1 0 are suitable constants. Similarly, we obtain that for p > 1 and x = s e 1 + t e 2, g(x) = h p (Z T, x) = a h(m τ p T, x), (24) where a 0 and 1 τ 1 are suitable constants. Since Z as well as m, M, M τ p are SL(n) equivariant, we obtain from (21) and the same argument applied to m, M, M τ p that (23) and (24) hold for every x R n where all but two coordinates vanish. Thus applying Lemma 4 shows that the propositions hold true for q = 1/p. 2. Let q = 0. Let i = 1, j = 2. Set g(x) = h p (Z T, x) h p (Z T, x). By Lemma 1, Z T e 1. Therefore φ λ T = T and (17) implies that g(x) = g(φ t λx) + g(ψ t λx) for 0 < λ < 1, x R n. Thus (22) holds with q = 0. Set a = h p (Z T, e 2 ) and b = h p (Z T, e 2 ). We apply Lemma 3 with f(x 1, x 2 ) = h p (Z T, x) h p (Z T, x) for x = x 1 e 1 + x 2 e 2 and obtain that for x 1 x 2 0 for x 2 x 1 0 and for x 1, x 2 0 h p (Z T, (x 1, x 2, 0,..., 0)) = (x p 1 xp 2 ) f(e 1) + x p 2 a, h p (Z T, ( x 1, x 2, 0,..., 0)) = (x p 1 xp 2 ) f( e 1) + x p 2 b, (25) h p (Z T, (x 1, x 2, 0,..., 0)) = (x p 2 xp 1 ) f(e 2) + x p 2 a, h p (Z T, ( x 1, x 2, 0,..., 0)) = (x p 2 xp 1 ) f( e 2) + x p 2 b, (26) h p (Z T, ( x 1, x 2, 0,..., 0)) = x p 2 f(e 2) + x p 1 f( e 1) + x 2 a, h p (Z T, (x 1, x 2, 0,..., 0)) = x p 1 f(e 1) + x p 2 f( e 2) + x 2 b. (27) Note that (21), (25), and (26) imply that f(e 1 ) = f(e 2 ) + a. (28) Since there is a β SL(n) such that βe 2 = e 3, βe 3 = e 2, and βt = T and since Z is SL(n) equivariant, we have h(z T, e 2 ) = h(z T, e 3 ). (29) Since e 3 and e 1 + e 2 are eigenvectors with eigenvalue 1 of φ t λ and ψt λ, it follows from (20) that h(z T, e 3 ) = h(z T, e 3 ) and h(z T, e 1 + e 2 ) = h(z T, e 1 + e 2 ). (30) 13

14 By (29), (21), and (30), we obtain a = h p (Z T, e 2 ) = h p (Z T, e 1 ) = f(e 1 ) and therefore by (28) that f(e 2 ) = 0. Similarly, we get b = f( e 1 ). Thus it follows from (25) (27) that h p (Z T, x) = a h p (T, x) + b h p ( T, x) (31) for every x = (x 1, x 2, 0,..., 0). Since Z T is convex, we have a, b 0. The operators Z and P a P + p b( P ) are SL(n) equivariant. Therefore (31) holds for every x R n where all but two coordinates vanish. We apply Lemma 4 with g 1 (x) = h p (Z T, x) h p (Z T, x) and g 2 (x) = a(h p (T, x) h p (T, x)) + b(h p ( T, x) h p ( T, x)) and obtain that for every x R n where h p (Z T, x) = a h p (T, x) + b h p ( T, x) + w(x), w(x) = h p (Z T, x) a h p (T, x) b h p ( T, x). We identify e 1 with R n 1 and use induction on the dimension to obtain Z T. The operator Z restricted to convex polytopes in e 1 is a homogeneous, SL(n 1) equivariant L p -Minkowski valuation. Let n = 3. Then we obtain from Proposition 1 and 1 p that there are constants such that Z T = a 1 T + p b 1 ( T ) + p a 2 [o, e 2 ] + p b 2 [o, e 2 ] + p a 3 [o, e 3 ] + p b 3 [o, e 3 ]. We have and h p (Z T, e 2 + e 3 ) = a + w(e 2 + e 3 ) = h p (Z T, e 2 + e 3 ) = a 1 + a 2 + a 3 h p (Z T, e 1 + e 3 ) = a + w(e 1 + e 2 ) = h p (Z T, e 2 ) = a 1 + a 2. Thus (21) implies that a 3 = 0. Similarly, we obtain a 2 = b 2 = b 3 = 0. This shows that w(x) = 0 for n = 3 and Z T = a T + p b( T ). Let n 4. Then we get by induction that Z T = a T + p b ( T ). Therefore w(x) = 0 also in this case. Thus Z T = a T + p b( T ) holds for n 4. 4 Contravariant Minkowski valuations In this section, our main result on the classification of SL(n) contravariant Minkowski valuations is formulated. The most important example is the projection operator Π : K n K n. The following transformation rule for Π, which is due to Petty (cf. [5]), shows that Π is SL(n) contravariant and homogeneous of degree (n 1): Π(φK) = det φ φ t Π K for φ GL(n). On Po n we define an additional operator with the same transformation behaviour. Note that for a polytope P, Π P can be written in the following way (cf. [5]). A vector v is a scaled facet normal of P, if v is an outer normal to a facet of P and 14

15 if the length of v is equal to the (n 1)-dimensional volume of the corresponding facet. Then Π P = [o, v], v V(P ) where V(P ) is the set of scaled facet normals of P. Define the operator Π o : Po n K n by Π o P = [o, v], v V o(p ) where V o (P ) is the set of scaled facet normals of P that correspond to facets that contain the origin. We set Π o P = {o} if P contains the origin in its interior. Note that Π o is a Minkowski valuation and that it is not continuous. For a 1 0, a 2, a 3 R with a 1 + a 2 + a 3 0, the operator Z : Po n K n defined by Z P = a 1 Π P + a 2 Π o P + a 3 ( Π o P ) is a Minkowski valuation which is SL(n) contravariant and homogeneous of degree (n 1). As we will see, for n 3 these are already all examples of homogeneous, SL(n) contravariant Minkowski valuations on Po n. For n = 2, we have additional examples. Let ψ π/2 denote the rotation by an angle π/2. If Z : Po 2 K 2 is SL(2) equivariant, then (ψ π/2 Z)(φP ) = ψ π/2 φψ 1 π/2 (ψ π/2 Z)P = φ t (ψ π/2 Z)P, that is, ψ π/2 Z is SL(2) contravariant. If Z is a Minkowski valuation and homogeneous of degree r, so is ψ π/2 Z. This implies that the rotated versions of the operators from Theorem 1 are homogeneous, SL(2) contravariant Minkowski valuations. The following result is our classification of SL(n) contravariant Minkowski valuations. The proof is given in Section 6. Theorem 2. Let Z : Po n K n, n 3, be a Minkowski valuation which is SL(n) contravariant and homogeneous of degree r. If r = n 1, then there are constants a 1 0, a 2, a 3 R with a 1 + a 2 + a 3 0 such that Z P = a 1 Π P + a 2 Π o P + a 3 ( Π o P ) for every P Po n. In all other cases, Z P = {o} for every P Po n. Let Z : Po 2 K 2 be a Minkowski valuation which is SL(2) contravariant and homogeneous of degree r. If r = 3, then there are constants a 0 R and a 1 0 such that ( Z P = ψ π/2 a0 m(p ) + a 1 M P ) for every P P 2 o. If r = 1, then there are constants a 0, b 0 0 and a i, b i R with a i + b 0 + b 1, a 0 + a 1 + b i 0, i = 1, 2, such that Z P = ψ π/2 ( a0 P + b 0 ( P ) + (a i E i + b i ( E i )) ) for every P P 2 o, where the sum is taken over E i E o (P ). In all other cases, Z P = {o} for every P P 2 o. The following simple consequence of this theorem holds. Corollary 2.1. If Z : Ko n K n, n 3, is a continuous, homogeneous, SL(n) contravariant Minkowski valuation, then there is a constant a 0 such that Z K = a Π K 15

16 for every K K n o. If Z : K 2 o K 2 is a continuous, homogeneous, SL(2) contravariant Minkowski valuation, then there are constants a 0 R and a 1, b 1 0 such that Z K = ψ π/2 ( a0 m(k) + a 1 M K ) or Z K = ψ π/2 ( a1 K + b 1 ( K) ) for every K K 2 o. For translation invariant valuations we obtain the following result. The proof is given in Section 7. Corollary 2.2. Let Z : P n K n, n 2, be a translation invariant, SL(n) contravariant Minkowski valuation. Then there is a constant c 0 such that Z = c Π. 5 Contravariant L p -Minkowski valuations In [25] Lutwak, Yang, and Zhang introduced the L p -projection operator Π p and obtained an L p -version of the Petty projection inequality. For K K n o which contain the origin as an interior point, Π p is defined by h p (Π p K, v) = c n,p S n 1 v u p ds p (K, u) for v R n, where c n,p is chosen such that Π p B = B for the unit ball B. Here S p (K, ) is the L p -surface area measure of K, that is S p (K, ) = h(k, ) 1 p S(K, ), where S(K, ) is the surface area measure of K. It is proved in [25] that Π p (φk) = det φ 1/p φ t Π p K for φ GL(n). Note that Π p is an L p -Minkowski valuation on convex bodies that contain the origin in their interiors and that it is not bounded on Ko n. We need the following generalization of this operator. For 1 τ 1 and K Ko n which contain o as an interior point, define Π τ p K by h p (Π τ p K, v) = c n,p S n 1 ( v u + τ(v u)) p ds p (K, u). Then Π τ p is an SL(n) contravariant L p -Minkowski valuation on convex bodies that contain the origin in their interiors. For P Po n, set h p (ˆΠ τ p P, v) = c n,p ( v u + τ(v u)) p ds p (P, u), S n 1 \ω o(p ) where ω o (P ) is the set of outer unit normal vectors to facets of P that contain the origin. Note that ˆΠ τ p is bounded but not continuous on Ko n and that it is an L p -Minkowski valuation. As we will show, these operators are the only examples of homogeneous, SL(n) contravariant L p -Minkowski valuations for n 3. For n = 2, the rotated version of the operators from Theorem 1 p are additional examples. The proof of the following result is given in Section 6. Theorem 2 p. Let Z : P n o K n, n 3, be an L p -Minkowski valuation, p > 1, which is SL(n) contravariant and homogeneous of degree r. If r = n/p 1, then there are constants a 0 and 1 τ 1 such that Z P = a ˆΠ τ p P 16

17 for every P Po n. In all other cases, Z P = {o} for every P Po n. Let Z : Po 2 K 2 be an L p -Minkowski valuation, p > 1, which is SL(2) contravariant and homogeneous of degree r. If r = 2/p + 1, then there are constants a 0 and 1 τ 1 such that Z P = a ψ π/2 M τ p P If r = 1, then there are constants a 0, b 0 0 and a i, b i R with a 0 + a i, b 0 + b i 0, i = 1, 2, such that Z P = ψ π/2 ( a0 P + p b 0 ( P ) + p p(ai E i + p b i ( E i )) ) for every P P 2 o, where the sum is taken over E i E o (P ). If r = 2/p 1, then there are constants a 0 and 1 τ 1 such that Z P = a ˆΠ τ p P for every P P 2 o. In all other cases, Z P = {o} for every P P 2 o. As a simple consequence we obtain the following result. Corollary. If Z : Ko n K n, n 3, is a continuous, homogeneous, SL(n) contravariant L p -Minkowski valuation, p > 1, then Z K = {o} for every K K n o. If Z : K 2 o K 2 is a continuous, homogeneous, SL(2) contravariant L p -Minkowski valuation, p > 1, then there are constants a, b 0 and 1 τ 1 such that for every K K 2 o. Z K = a ψ π/2 M τ p K or Z K = ψ π/2 (a K + p b( K)) 6 Proof of Theorems 2 and 2 p As a first step in the proof, we show that every operator which is SL(n) contravariant and homogeneous of degree r n 1 is simple. Lemma 5. Let Z : Po n K n be an operator which is SL(n) contravariant and homogeneous of degree r. If dim P < (n 1), then Z P = {o}. If dim P = (n 1), then Z P (lin P ). If dim P = (n 1) and r n 1, then Z P = {o}. Proof. Let P Po n be such that lin P is the k-dimensional subspace with equation x k+1 =... = x n = 0. Since every P Po n with dim P = k is a linear image of such a polytope P and since Z is SL(n) contravariant, it suffices to prove the lemma in this case. Let ( ) I B φ =, 0 A where I is the k k identity matrix, 0 is the (n k) k zero matrix, B is an (n k) k matrix, and A is an (n k) (n k) matrix with determinant 1. Then φ SL(n), ( ) φ t I 0 = C A t with C = A t B t, and φp = P. (32) 17

18 Write x = ( ) x x with x = (x 1,..., x k ) and x = (x k+1,..., x n ) for x R n. Let x Z P. Since Z is SL(n) contravariant, (32) implies that y = φ t x Z P, that is, ( ) ( y x ) = Cx + A t x Z P. (33) y This is true for every (n k) k matrix B and every (n k) (n k) matrix A with determinant 1. If x o, this implies that y can be an arbitrary vector. Since Z P is bounded, it follows that x = o. Thus Z P (lin P ). If k = (n 1) and q = 1, this proves the lemma. Let k < (n 1). Then x = o and (33) holds for every (n k) (n k) matrix A with determinant 1. Since Z P is bounded and (n k) 2, this implies that x = o. Let k = (n 1), r n 1, and let φ = ( I 0 0 s where I is the (n 1) (n 1) identity matrix and s 0. Then ), Z(φP ) = s (r (n 1))/n Z P = Z P. Since this holds for every s 0 and Z P is bounded, this implies that Z P = {o}. By Lemmas 2 and 2 p, it suffices to determine the value of Z for n-dimensional simplices to prove Theorems 2 and 2 p. Since Z is SL(n) contravariant and homogeneous, it is enough to determine Z T, where T is the simplex with vertices o, e 1,..., e n. We start with the planar case. Proposition 3. Let Z : P 2 o K 2 be a Minkowski valuation which is SL(2) contravariant and homogeneous of degree r. If r = 3, then there are constants a 0 R and a 1 0 such that Z T = ψ π/2 ( a0 m(t ) + a 1 M T ). If r = 1, then there are constants a 0, b 0 0 and a i, b i R with a i + b 0 + b 1, a 0 + a 2 + b i 0, i = 1, 2, such that Z T = ψ π/2 (a 0 T + b 0 ( T ) + a 1 [o, e 1 ] + b 1 [o, e 1 ] + a 2 [o, e 2 ] + b 2 [o, e 2 ]). In all other cases, Z T = {o}. Proposition 3 p. Let Z : P 2 o K 2 o be an L p -Minkowski valuation, p > 1, which is SL(2) contravariant and homogeneous of degree r. If r = 2/p + 1, then there are constants a 0 and 1 τ 1 such that Z T = a ψ π/2 M τ p T. If r = 1, then there are constants a 0, b 0 0 and a i, b i R with a 0 + a i, b 0 + b i 0, i = 1, 2, such that Z T = ψ π/2 (a 0 T + p b 0 ( T ) + p a 1 [o, e 1 ] + p b 1 [o, e 1 ] + p a 2 [o, e 2 ] + p b 2 [o, e 2 ]). If r = 2/p 1, then there are constants a 1, a 2 0 such that In all other cases, Z T = {o}. Z T = [ a 1 (e 1 + e 2 ), a 2 (e 1 + e 2 )]. 18

19 That these propositions hold true can be seen in the following way. Let Z be as in Proposition 3 or 3 p. Then ψ π/2 Z is an L p -Minkowski valuation which is SL(2) equivariant and homogeneous of degree r. Thus Propositions 1 and 1 p immediately imply that the above propositions hold true. Next, we consider the case n 3. Proposition 4. Let Z : Po n K n, n 3, be a Minkowski valuation, which is SL(n) contravariant and homogeneous of degree r. If r = n 1 then there are constants a 1 0, a 2, a 3 R with a 1 + a 2 + a 3 0 such that In all other case, Z T = {o}. Z T = a 1 Π T + a 2 Π o T + a 3 ( Π o T ). Proposition 4 p. Let Z : P n o K n o, n 3, be an L p -Minkowski valuation, p > 1, which is SL(n) contravariant and homogeneous of degree r. If r = n/p 1, then there are constants a 1, a 2 0, such that In all other cases, Z T = {o}. Z T = [ a 1 (e e n ), a 2 (e e n )]. We prove both propositions at the same time. Proof. If p > 1, let Z : P n o K n o, and if p = 1, let Z : P n o K n. For 0 < λ < 1, i < j, let H λ = H λ (i, j) be the hyperplane through o with normal vector (1 λ) e i λ e j. Then H λ dissects T into the two simplices φ λ T and ψ λ T, where φ λ and ψ λ are defined by (18). Set T = T e i. Then T H λ = ψ λ T. Set q = (r + 1)/n. Since Z is SL(n) contravariant, homogeneous of degree n q 1 and an L p -Minkowski valuation, we have Z T + p (1 λ) q ψ t λ Z T = λ q φ t λ Z T + p (1 λ) q ψ t λ Z T. (34) Denote by α π/2 = α π/2 (i, j) the rotation by an angle π/2 in the plane lin{e i, e j }, that is, α π/2 e i = e j, α π/2 e j = e i, α π/2 e k = e k for k i, j. Then we obtain that (α π/2 Z)(φ λ T ) = λ (r+1)/n α π/2 φ t λ α 1 π/2 (α π/2 Z)T = λ (r 1)/n φ λ (α π/2 Z)T (35) and (α π/2 Z)(ψ λ T ) = λ (r 1)/n ψ λ (α π/2 Z)T. (36) 1. Let q 1. By Lemma 1, Z is a simple valuation. Set g(x) = h p (α π/2 Z T, x). Then (34) (36) imply that g(x) = λ p q g(φ t λx) + (1 λ) p q g(ψ t λx) for 0 < λ < 1, i < j, x R n. (37) Therefore we have for k i, j g(e k ) = λ p q g(e k ) + (1 λ) p q g(e k ). If q 1/p, this implies that g(e k ) = 0. Similarly, we obtain that g( e k ) = 0. Since (37) implies that Lemma 3 holds for f(x 1, x 2 ) = g(x 1 e i + x 2 e j ), we obtain that g(x) = 0 for x = x 1 e i + x 2 e j. Thus Lemma 4 implies that Z T = {o}. If q = 1/p, then (37) implies that g(e i e j ) = λ 1 p g(e i e j ) + (1 λ) 1 p g(e i e j ). 19

20 Therefore g(e i e j ) = 0. Similarly, we obtain that g(e j e i ) = 0. Thus we have for i < j, Z T (e i e j ). This shows that Z T = [ a 1 (e e n ), a 2 (e e n )] with suitable constants a 1, a Let q = 1. Lemma 5 implies that Z T lin{e i }. Since there is a β SL(n) such that βe i = e i and βt = T and since Z is SL(n) contravariant, we obtain that Z T = [ a e i, a e i ] with a 0. By (34) we have for k i, j h p (Z T, e k )+(1 λ) p h p (Z T, e k ) = h p (Z T, e k ) = λ p h p (Z T, e k )+(1 λ) p h p (Z T, e k ). If p > 1, this implies that h(z T, e k ) = 0. Similarly, we obtain that h(z T, e k ) = 0. Thus Z T = {o}. So let p = 1. Setting f(x 1, x 2 ) = h p (Z T, x) h p (Z T, x) for x = x 1 e i + x 2 e j, we see that (37) holds. Therefore we can apply Lemma 3. Since for i j, k l there is an α SL(n) such that αe i = e k, αe j = e l and αt = T and since Z is SL(n) contravariant, we see that f does not depend on i, j and that for x 1, x 2 R h(z T, x 1 e i + x 2 e j ) = h(z T, x 1 e k + x 2 e l ). (38) Thus we obtain for x 1 x 2 0 and for x 1, x 2 0 h(z T, x 1 e i x 2 e j ) = x 1 (f(e 1 ) + a) x 2 f(e 1 ), h(z T, x 1 e i + x 2 e j ) = x 1 (f( e 1 ) + a) x 2 f( e 1 ), h(z T, x 1 e i + x 2 e j ) = x 1 (f(e 1 ) + a) + x 2 f(e 2 ), h(z T, x 1 e i x 2 e j ) = x 1 (f( e 1 ) + a) + x 2 f( e 2 ). It follows from (38) that f(e 1 ) + a = f(e 2 ) and f( e 1 ) a = f( e 2 ). Setting a 1 = f(e 1 ) + f( e 1 ) + a, a 2 = f( e 1 ), a 3 = f(e 1 ) and using (38) shows that h(z T, x) = a 1 h(π T, x) + a 2 h(π o T, x) + a 3 h( Π o T, x) (39) for x = x 1 e i + x 2 e j. Thus by Lemma 4 the proposition is proved. 7 Proof of Corollaries 1.2 and 2.2 The main tool is the following result by McMullen. Theorem ([28]). Let µ : P n R be a translation invariant valuation. Then for s Q, s 0, n µ(s P ) = s i µ i (P ). i=0 The coefficient µ i (P ) (which is independent of s) is a translation invariant valuation on P n, which is homogeneous of degree i. Let Z : P n K n be a translation invariant Minkowski valuation. Then for x R n fixed, P h(zp, x) is a translation invariant real valued valuation. Thus for s Q, s 0, there is a polynomial expansion h(z(s P ), x) = n s i µ i (P, x), i=0 20

21 where for every x R n the coefficient µ i (, x) is a translation invariant valuation on P n, which is homogeneous of degree i. First, we consider Corollary 1.2, that is, Z is SL(n) equivariant. Then we have for φ SL(n), x R n, h(z(φp ), x) = Thus for i = 1,..., n, n s i µ i (φp, x) = h(z P, φ t x) = i=0 n s i µ i (P, φ t x). µ i (φp, x) = µ i (P, φ t x) for φ SL(n), x R n. (40) Note that a function f : R n R which is homogeneous of degree 1 is the support function of a convex body if and only if f is sublinear, that is, f(x+y) f(x)+f(y) for x, y R n (cf. [36]). Since h(z(s P ), ) is sublinear for s > 0, h(z(s P ), ) µ n (P, ) = lim s s n is also sublinear. Thus µ n (P, ) is a support function. In view of (40) and the homogeneity of µ n, we can apply Theorem 1 and obtain that µ n (P, ) = 0. Therefore n 1 h(z(s P ), ) = s i µ i (P, ). Using induction on the degree of homogeneity and the same arguments as above, we obtain that for i = n 1,..., 2, µ i (P, ) is a support function and that by Theorem 1 µ i (P, ) = 0 for i = n 1,..., 2. By Lemma 1 we have Z{o} = {o} and therefore µ 0 (P, ) = 0. This implies that µ 1 (P, ) = h(z P, ). Thus applying Theorem 1 shows that Z = c D with a suitable constant c 0 and Corollary 1.2 is proved. Now, we consider Corollary 2.2, that is, Z is SL(n) contravariant. Then we have for i = 1,..., n, i=0 µ i (φp, x) = µ i (P, φ 1 x) for φ SL(n), x R n. (41) Since µ n (P, ) = lim s h(z(s P ), )/s n is sublinear, it is a support function. In view of (41) and the homogeneity of µ n, we can apply Theorem 2 and obtain that µ n (P, ) = 0. By Lemma 5 we have Z{o} = {o} and therefore µ 0 (P, ) = 0. This implies that n 1 h(z(s P ), ) = s i µ i (P, ). Since µ 1 (P, ) = lim s 0 h(z(s P ), )/s is sublinear, it is a support function. We apply Theorem 2 and obtain that µ 1 (P, ) = 0. Using induction on the degree of homogeneity and the same arguments as above, we obtain that for i = 1,..., n 2, µ i (P, ) is also a support function and that by Theorem 2 µ i (P, ) = 0 for i = 1,..., n 2. This implies that h(z P, ) = µ n 1 (P, ). Thus applying Theorem 2 shows that Z = c Π with a suitable constant c 0 and Corollary 2.2 is proved. References [1] S. Alesker, Continuous rotation invariant valuations on convex sets, Ann. of Math. (2) 149 (1999), [2] S. Alesker, On P. McMullen s conjecture on translation invariant valuations, Adv. Math. 155 (2000), i=1 i=0 21

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