Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 49 (2008), No. 2, 541-547.

The Endomorphisms of the Lattice of Closed Convex Cones

Rolf Schneider ^{∗}

Mathematisches Institut, Albert-Ludwigs-Universität 79104 Freiburg i. Br., Germany e-mail: rolf.schneider@math.uni-freiburg.de

Abstract. The set of closed convex cones in ℝ^{d} (d ≥ 3), with the operations of intersection and closed sum, is a lattice. We determine all endomorphisms of this lattice. As a consequence, we obtain a characterization of the duality of convex cones. MSC 2000: 52A20 Keywords: lattice of convex cones, lattice endomorphism, duality, order preserving mapping

1. Introduction

By ℝ^{d} we denote the d-dimensional real Euclidean vector space, equipped with its standard scalar product ⟨·, ·⟩. We assume throughout that d ≥ 3. Let C^{d} be the set of closed convex cones in ℝ^{d} (including {0} and ℝ^{d}). The set C^{d} together with the two operations ∩ (intersection) and ∨, defined by

C ∨ D := cl conv (C ∪ D) = cl (C + D), C, D ∈ C^{d},

is a lattice. The following theorem determines all endomorphisms of this lattice.

Theorem. Let d ≥ 3. Let φ : C^{d} → C^{d} be a mapping satisfying

φ(C ∩ D) = φ(C) ∩ φ(D),  (1)
φ(C ∨ D) = φ(C) ∨ φ(D)  (2)

∗Partially supported by the European Network PHD, FP6 Marie Curie Actions, RTN, Contract MCRN-2004-511953.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2008 Heldermann Verlag

542 R. Schneider: The Endomorphisms of the Lattice of …

for all C, D ∈ C^{d}. Then either φ is constant, or there exists a linear transformation g ∈ GL(d) such that φ(C) = gC for all C ∈ C^{d}.

This theorem fits into a series of similar results (where the assumptions on the dimension d may differ). Gruber [6] has determined the endomorphisms of the lattice (K^{d}, ∩, ∨), where K^{d} is the set of compact convex sets in ℝ^{d} (including the empty set) and K ∨ L := conv(K ∪ L). Gruber [7] has further classified the endomorphisms of the lattice (B^{d}, ∩, ∨), where B^{d} is the set of unit balls of norms on ℝ^{d}. This was extended in [4] to a determination of the endomorphisms of the lattices (K_{0}^{d}, ∩, ∨) and (K_{(0)}^{d}, ∩, ∨), where K_{0}^{d} and K_{(0)}^{d} denote, respectively, the set of compact convex sets in ℝ^{d} containing 0 or containing 0 in the interior. The latter result was used to answer a question by Vitali Milman, asking for a characterization of the mapping that associates with every convex body in K_{(0)}^{d} its polar body.

That the theorem does not hold for d = 2, can be seen as follows. Let h : ℝ → ℝ be a strictly increasing, continuous function satisfying h(α + π) = h(α) + π for α ∈ ℝ. Using Cartesian coordinates in ℝ^{2}, let R_{α} := {λ(cos α, sin α) : λ ≥ 0}. Define φ(ℝ^{2}) := ℝ^{2} and φ({0}) := {0}. Every cone C ∈ C^{2} \ {ℝ^{2}, {0}} is of the form C = ⋃_{β≤α≤γ} R_{α} with β ≤ γ ≤ β + π, where the angle β is unique mod 2π. If we define φ(C) := ⋃_{h(β)≤α≤h(γ)} R_{α}, then φ is an endomorphism of the lattice (C^{2}, ∩, ∨).

As in [4], the knowledge of the endomorphisms leads to a characterization of duality. For C ∈ C^{d}, the dual cone is defined by

C^{∗} := {x ∈ ℝ^{d} : ⟨x, y⟩ ≤ 0 for all y ∈ C};

it is again in C^{d}. The mapping C ↦ C^{∗} is an involution that interchanges the lattice operations and reverses the order.

In a series of papers, Artstein-Avidan and Milman [1], [2], [3] have distilled the essential properties of an abstract duality from various classical duality operations for functions and convex sets. They have established the surprising fact that the involution property together with the sole condition of order reversion is in several cases sufficient for the essentially unique characterization of a duality. For example, the Legendre transform is, up to obvious linear modifications, the only involution on the set of lower semi-continuous convex functions that reverses the order. Our condition (4) in the subsequent corollary follows their scheme, and this provides yet another example of the phenomenon they discovered.

Corollary 1. Let d ≥ 3. Let ψ : C^{d} → C^{d} be a mapping satisfying

ψ(ψ(C)) = C,  for all C ∈ C^{d}  (3)

and

C ⊂ D ⇒ ψ(C) ⊃ ψ(D)  for all C, D ∈ C^{d}.  (4)

Then there exists a selfadjoint linear transformation g ∈ GL(d) such that ψ(C) = gC^{∗} for all C ∈ C^{d}.

As one of many outcomes of their approach, Artstein-Avidan and Milman [3, Theorem 13] have also arrived at the following unexpected result. Let C_{0}^{d} denote the system of closed convex sets in ℝ^{d} containing 0. If φ : C_{0}^{d} → C_{0}^{d} is a bijective map satisfying K_{1} ⊂ K_{2} ⇔ φ(K_{1}) ⊂ φ(K_{2}) for all K_{1}, K_{2} ∈ C_{0}^{d}, then there exists a linear transformation g ∈ GL(d) such that φ(K) = gK for K ∈ C_{0}^{d}. Our theorem implies a similar result for convex cones.

Corollary 2. Let d ≥ 3. Let φ : C^{d} → C^{d} be a bijective mapping satisfying

C_{1} ⊂ C_{2} ⇔ φ(C_{1}) ⊂ φ(C_{2})  for all C_{1}, C_{2} ∈ C^{d}.  (5)

Then there exists a linear transformation g ∈ GL(d) such that φ(C) = gC for C ∈ C^{d}.

In fact, as will be clear from the proof, analogous assertions are true (for d ≥ 2) for each of the systems K^{d}, B^{d}, K_{0}^{d}, K_{(0)}^{d}, since in each case the lattice endomorphisms are known to be induced either by an affine transformation (in the first case, according to Gruber [6]) or by a linear transformation (in the other three cases). Only in the case of C_{0}^{d} treated by Artstein-Avidan and Milman, no corresponding classification of the lattice endomorphisms is known, so that their proof proceeds in a different way.

2. Proof of the theorem

The main line of reasoning is similar to that in [4], but there are some characteristic differences (which is already seen from the fact that the two-dimensional case is exceptional).

Let φ : C^{d} → C^{d} satisfy the assumptions (1) and (2). Then φ is inclusion preserving, that is, C ⊂ D for C, D ∈ C^{d} implies φ(C) ⊂ φ(D).

We write {0} =: ¯0. A ray is a set R^{x} := {λx : λ ≥ 0}, where x ∈ ℝ^{d} \ {0}. By R we denote the set of all rays in ℝ^{d}.

Following the procedure in [6], we distinguish several cases.

Case 1: φ(R) = φ(¯0) for all R ∈ R. We choose d + 1 rays R_{1}, …, R_{d+1} with R_{1} ∨ ⋯ ∨ R_{d+1} = ℝ^{d}. For any C ∈ C^{d} we have ¯0 ⊂ C ⊂ R_{1} ∨ ⋯ ∨ R_{d+1} and hence φ(¯0) ⊂ φ(C) ⊂ φ(R_{1}) ∨ ⋯ ∨ φ(R_{d+1}) = φ(¯0), hence

φ(C) = φ(¯0) for C ∈ C^{d}.  (6)

Case 2: φ(R) = φ(¯0) for some R ∈ R, but not for all rays R.

Let A := {R ∈ R : φ(R) = φ(¯0)}, and let A be the union of all rays in A. If S, U ∈ A, T ∈ R and T ⊂ S ∨ U, then φ(¯0) ⊂ φ(T) ⊂ φ(S) ∨ φ(U) = φ(¯0), hence T ∈ A; thus A is a convex cone. Since A ≠ ℝ^{d}, the closure of A is contained in a closed halfspace with 0 in the boundary. We can choose rays R ∈ A and T, U ∈ R \ A with U ∩ T = ¯0 and T ⊂ R ∨ U. Then φ(¯0) ⊂ φ(T) ⊂ [φ(R) ∨ φ(U)] ∩ φ(T) = φ(U) ∩ φ(T) = φ(¯0), hence φ(T) = φ(¯0), which is a contradiction.

Thus, Case 2 cannot occur, and we are left with the situation where φ(R) ≠ φ(¯0) for all R ∈ R. This implies

dim φ(R) ≥ 1 for R ∈ R.

In fact, since ¯0 ⊂ R, the equality φ(R) = ¯0 would imply φ(¯0) ⊂ φ(R) = ¯0 and hence φ(R) = φ(¯0), a contradiction.

Case 3: φ(R) ≠ φ(¯0) for all R ∈ R, φ(¯0) ≠ ¯0.

Case 4: φ(R) ≠ φ(¯0) for all R ∈ R, φ(¯0) = ¯0; there exists a ray P ∈ R with dim φ(P) ≥ 2.

With the same proof as in [4] we have: (P1) In an n-dimensional affine space, let M be a fixed closed convex set and let F be a family of n-dimensional closed convex sets such that K ≠ M for all K ∈ F and K_{1} ∩ K_{2} = M whenever K_{1}, K_{2} ∈ F and K_{1} ≠ K_{2}. Then F is at most countable.

By assumption, d ≥ 3. We show simultaneously that Case 3 and Case 4 both lead to a contradiction. In Case 3, we put P := ¯0.

Let B := φ(P) and b := dim B, then b ≥ 1 in Case 3 and b ≥ 2 in Case 4. A ray R is called free if R ⊄ lin P (so every ray is free in Case 3). Let R

be a free ray. Then φ(R) ∩ φ(P) = φ(R ∩ P) = φ(¯0). If φ(R ∨ P) = B, then φ(R) ∨ φ(P) = φ(P), thus φ(R) ⊂ φ(P) and hence φ(R) ∩ φ(P) = φ(R) ≠ φ(¯0), a contradiction. This shows that φ(R ∨ P) ≠ B.

By a sheet we understand a set R ∨ lin P with a free ray R. A sheet is called bad if it contains a free ray R with φ(P ∨ R) ⊂ lin B. If R, S are free rays in different sheets, then (R ∨ P) ∩ (S ∨ P) = P, hence φ(R ∨ P) ∩ φ(S ∨ P) = B. Now it follows from (P1) (applied in lin B) that there are at most countably many bad sheets. The other sheets are called good.

Suppose that b ≥ d − 1. Let the set S contain precisely one free ray from every good sheet, and no other elements, and put F := {φ(R ∨ P) : R ∈ S}. If R ∈ S, then φ(R ∨ P) ⊄ lin B, hence dim φ(R ∨ P) = d. It follows from (P1) that the family F is countable. This is a contradiction, since there are uncountably many good sheets. This proves that b ≤ d − 2.

Let k ∈ {1, …, d − b}. A set {x_{1}, …, x_{k}} of points in ℝ^{d} \ lin P, briefly a k-set, is called full if dim φ(R^{x_{1}} ∨ ⋯ ∨ R^{x_{k}} ∨ P) ≥ b + k. A k-flat E ⊂ ℝ^{d} is called general if 0 ∉ E in Case 3, and if dim aff(E ∪ lin P) = k + 2 in Case 4.

If x_{1} is contained in a good sheet, then B ⊂ φ(R^{x_{1}} ∨ P) ⊄ lin B, hence {x_{1}} is a full 1-set. We assert the following.

(P2) Let k ∈ {2, …, d − b}. In every general (k − 1)-flat E ⊂ ℝ^{d} there is a full k-set.

The proof of this proposition can be taken verbally from [4], if expressions like x̄_{1} ∨ ⋯ ∨ x̄_{k} ∨ p̄ are replaced by R^{x_{1}} ∨ ⋯ ∨ R^{x_{k}} ∨ P. It need, therefore, not be repeated.

Let k = d−b, and choose a general k-flat F ⊂ ℝ^{d}. Let E be the family of (k − 1)-flats contained in F and parallel to a fixed (k − 1)-flat. Since E ∈ E is general, by (P2) there exists a full k-set {x_{1}, …, x_{k}} in E; put C_{E} := R^{x_{1}} ∨ ⋯ ∨ R^{x_{k}} ∨ P. Then dim φ(C_{E}) = d and C_{E_{1}} ∩ C_{E_{2}} = P for E_{1}, E_{2} ∈ E with E_{1} ≠ E_{2}, hence φ(C_{E_{1}}) ∩ φ(C_{E_{2}}) = B. By (P1), this is a contradiction, since E is uncountable.

Since Cases 3 and 4 cannot occur, we are left with

Case 5: φ(¯0) = ¯0, and dim φ(R) = 1 for all R ∈ R.

In this case, the φ-image of a ray is either a ray or a line. We show first that it is always a ray.

Let x_{1}, …, x_{d+1} ∈ ℝ^{d} be the vertices of a simplex containing 0 in its interior. For each i ∈ {1, …, d + 1}, choose a point x′_{i} ∈ φ(R^{x_{i}}) \ {0}. Now the argument used in [4], at the beginning of the treatment of Case 5 (with x̄_{i} and ā replaced by R^{x_{i}} and R^{a}, respectively) shows that each φ(R), R ∈ R, has 0 as an endpoint and hence is a ray.

Let E ⊂ ℝ^{d} be a two-dimensional linear subspace. Let R, S ⊂ E be different rays. If φ(S) = −φ(R), we choose a ray T ⊂ E different from R and S. Then φ(T) ≠ −φ(R), since φ(T) ∩ φ(S) = ¯0. Hence, we can assume from the beginning that φ(S) ≠ −φ(R). Let E′ be the two-dimensional subspace spanned by φ(R) and φ(S). In the following, A, B, Z are rays in E. If Z ⊂ A∨B and φ(A), φ(B) ⊂ E′, then φ(Z) ⊂ φ(A) ∨ φ(B) ⊂ E′. This yields φ(Z) ⊂ E′ for all Z ⊂ R ∨ S. Let Z ⊂ (R − S) \ (−S). Then R ⊂ Z ∨ S, hence φ(R) ⊂ φ(Z) ∨ φ(S), which implies φ(Z) ⊂ E′. Similarly, Z ⊂ (S − R) \ (−R) implies φ(Z) ⊂ E′. Finally, if Z ⊂ −R − S, we can choose rays U ⊂ R − S and V ⊂ S − R with Z ⊂ U ∨ V, which gives φ(Z) ⊂ E′. We have proved that φ(R) ⊂ E′ for every ray R ⊂ E.

Let R ∈ R. We assert that φ(−R) = −φ(R). For the proof, choose S ∈ R with S ≠ ±R. Let A, B, C be any three of the rays R, −R, S, −S. Then (A ∨ B) ∩ C = ¯0, hence (φ(A) ∨ φ(B)) ∩ φ(C) = ¯0 and thus φ(C) ⊄ φ(A) ∨ φ(B). Since this holds for all choices of A, B, C from {R, −R, S, −S} and since φ(R), φ(−R), φ(S), φ(−S) lie in a two-dimensional subspace, the set {φ(R), φ(−R), φ(S), φ(−S)} must be of the form {±U, ±V } with two rays U, V satisfying U ≠ ±V. Suppose that φ(−R) = −φ(S). Choose a ray T ⊂ S −R with T ≠ S, −R. Then φ(T) ⊂ φ(S) ∨ φ(−R) and φ(T) ∩ φ(−R) = φ(T) ∩ φ(S) = ¯0, a contradiction. Thus, φ(−R) = −φ(R) is the only possibility.

Let L be the set of one-dimensional linear subspaces of ℝ^{d}. For L ∈ L, there are two rays R, −R with L = R ∨ (−R); we call them the generating rays of L. Since φ(−R) = −φ(R), the line f(L) := φ(R) − φ(R) does not depend on the choice of R. Thus, we have defined a map f : L → L. Let L_{1}, L_{2} ∈ L, L_{1} ≠ L_{2}. Let R_{i} be a generating ray of L_{i} (i = 1, 2). If f(L_{1}) = f(L_{2}), then either φ(R_{1}) = φ(R_{2}) or φ(R_{1}) = −φ(R_{2}). But R_{1} ∩ R_{2} = R_{1} ∩ (−R_{2}) = ¯0, hence φ(R_{1}) ∩ φ(R_{2}) = φ(R_{1}) ∩ (−φ(R_{2})) = ¯0, a contradiction. Thus, the map f is injective.

Suppose that L_{1}, L_{2}, L_{3} ∈ L are different and coplanar, that is, contained in a two-dimensional linear subspace of ℝ^{d}. Then we can choose generating rays R_{i} of L_{i} (i = 1, 2, 3) such that R_{2} ⊂ R_{1} ∨ R_{3}. Then φ(R_{2}) ⊂ φ(R_{1}) ∨ φ(R_{2}). Therefore, the lines f(L_{1}), f(L_{2}), f(L_{3}) are coplanar.

Since d ≥ 3, then it follows from the Fundamental Theorem of Projective Geometry (see, e.g., Faure [5] for a proof of a general version), together with the injectivity of f, that there exists a linear transformation g of ℝ^{d} such that f(L) = gL for all L ∈ L.

Let R ∈ R. Then φ(R) and φ(−R) = −φ(R) are rays with gR − gR = g(R − R) = f(R − R) = φ(R) − φ(R). It follows that either φ(R) = gR or φ(R) = −gR. Let A_{±} := {R ∈ R : φ(R) = ±gR}, and let A_{±} be the union of the rays in A_{±}. Then A_{+} and A_{−} are convex cones. If both are non-empty, they have a common boundary ray S and we can choose rays R_{+} ∈ A_{+} and R_{−} ∈ A_{−} such that S ⊂ R_{+} ∨ R_{−}. Then φ(S) ⊂ φ(R_{+}) ∨ φ(R_{−}), hence either gS ⊂ gR_{+} − gR_{−} or −gS ⊂ gR_{+} − gR_{−}, which is a contradiction. It follows that A_{+} or A_{−} is empty, and we can assume (replacing g by −g if necessary) that A_{−} = ∅. Then φ(R) = gR for all R ∈ R.

Let C ∈ C^{d} and R ∈ R. If R ⊂ C, then gR = φ(R) ⊂ φ(C), hence gC ⊂ φ(C). Let S be a ray with S ⊄ gC. Then g^{−1}S ⊄ C, hence g^{−1}S ∩ C = ¯0. This gives S ∩ φ(C) = g(g^{−1}S) ∩ φ(C) = φ(g^{−1}S) ∩ φ(C) = ¯0 and thus S ⊄ φ(C). We have obtained gC = φ(C), which completes the proof of the theorem.

3. Proof of the corollaries

First we prove Corollary 2. Let φ satisfy the assumptions of the corollary. For the reader’s convenience, we modify (to sets) an argument that is due to Artstein-Avidan and Milman [2, Lemma 4]. Let C_{1}, C_{2} ∈ C^{d}. Then (5) gives φ(C_{1} ∩ C_{2}) ⊂ φ(C_{1}) ∩ φ(C_{2}). If this inclusion is strict, then by the bijectivity of φ there exists B ∈ C^{d} with φ(C_{1} ∩C_{2}){⊂ \atop ≠}φ(B) ⊂ φ(C_{1})∩φ(C_{2}), and (5) gives C_{1} ∩C_{2} ⊂ B ⊂ C_{1} ∩ C_{2}, a contradiction. In the same way it follows that φ(C_{1} ∨ C_{2}) = φ(C_{1}) ∨ φ(C_{2}), thus φ is a lattice endomorphism. Now the assertion of Corollary 2 follows from the theorem.

For the proof of Corollary 1, we note that the duality mapping C ↦ C^{∗} reverses inclusions. Therefore, if ψ satisfies the assumptions of Corollary 1, then the map φ defined by φ(C) := ψ(C)^{∗} satisfies the assumptions of Corollary 2. Hence, there exists a linear transformation g ∈ GL(d) such that φ(C) = gC for C ∈ C^{d}. Then ψ(C) = (gC)^{∗} = g^{−t}C^{∗}, and (3) implies g^{t} = g. This proves Corollary 1.

References

[1] Artstein-Avidan, S.; Milman, V.: The concept of duality in asymptotic geometric analysis, and the characterization of the Legendre transform. Preprint.

[2] Artstein-Avidan, S.; Milman, V.: The concept of duality for measure projections of convex bodies. Preprint.

[3] Artstein-Avidan, S.; Milman, V.: Characterization of the abstract concept of duality. Preprint.

[4] Böröczky, K. J.; Schneider. R.: A characterization of the duality mapping for convex bodies. Geom. Funct. Anal. (to appear). http://home.mathematik.uni-freiburg.de/rschnei [5] Faure, C.-A.: An elementary proof of the fundamental theorem of projective

geometry. Geom. Dedicata 90 (2002) 145–151. Zbl 0996.51001 [6] Gruber, P.: The endomorphisms of the lattice of convex bodies. Abh. Math.

Semin. Univ. Hamburg 61 (1991) 121–130. Zbl 0754.52006 [7] Gruber, P.: The endomorphisms of the lattice of norms in finite dimensions.

Abh. Math. Semin. Univ. Hamburg 62 (1992) 179–189. Zbl 0779.52006

Received September 30, 2007