Spectral Cones in Euclidean Jordan Algebras

Spectral cones in Euclidean Jordan algebras

Juyoung Jeong Department of Mathematics and Statistics University of Maryland, Baltimore County Baltimore, Maryland 21250, USA [email protected]

and

M. Seetharama Gowda Department of Mathematics and Statistics University of Maryland, Baltimore County Baltimore, Maryland 21250, USA [email protected]

August 2, 2016

Abstract

A spectral cone in a Euclidean Jordan algebra V of rank n is of the form K = λ−1(Q), where Q is a permutation invariant convex cone in Rn and λ : V → Rn is the eigenvalue map (which takes x to λ(x), the vector of eigenvalues of x with entries written in the decreasing order). In this paper, we describe some properties of spectral cones. We show, for example, that spectral cones are invariant under automorphisms of V, that the dual of a spectral cone is a spectral cone when V is simple or carries the canonical inner product, and characterize the pointedness/solidness of a spectral cone. We also show that for any spectral cone K in V, dim(K) ∈ {0, 1, m − 1, m}, where dim(K) denotes the dimension of K and m is the dimension of V.

Key Words: Euclidean Jordan algebra, spectral cone, automorphism, dual cone, dimension

AMS Subject Classiﬁcation: 15A18, 15A51, 15A57, 17C20, 17C30, 52A20

1 1 Introduction

Let V be a Euclidean Jordan algebra of rank n and λ : V → Rn denote the eigenvalue map (which takes x to λ(x), the vector of eigenvalues of x with entries written in the decreasing order). A set E in V is said to be a spectral set [1] if there exists a permutation invariant set Q in Rn such that E = λ−1(Q). A function F : V → R is said to be a spectral function [1] if there is a permutation invariant function f : Rn → R such that F = f ◦ λ. Motivated by the works of Baes [1], Sun and Sun [11], Ramirez, Seeger, and Sossa [10], in [7], we presented some new results on spectral sets and functions, speciﬁcally addressing characterization, invariance under automorphisms, Schur- convexity, etc. In the present paper, we study spectral cones, which, by deﬁnition, are spectral sets of the form

K = λ−1(Q), where Q is a permutation invariant convex cone in Rn. The symmetric cone in a Euclidean Jordan n algebra is an important example of a spectral cone as it comes from Q = R+ (the nonnegative orthant in Rn). In this paper, we describe some properties of spectral cones, speciﬁcally looking for properties that are similar to those of symmetric cones. We show, for example, that spectral cones are invariant under automorphisms of V, that the dual of a spectral cone is a spectral cone when V is simple or carries the canonical inner product, and characterize the pointedness/solidness of a spectral cone. We also show that for any spectral cone K in V, dim(K) ∈ {0, 1, m − 1, m}, where dim(K) denotes the dimension of K and m is the dimension of V. The paper is organized as follows. In Section 2, we cover some preliminary material. Section 3 deals with the invariance of spectral sets under automorphisms, establishes the ‘linearity’ of λ−1 over permutation invariant sets in Rn, and describes a duality result. In Section 4, we provide examples of polyhedral and non-polyhedral self-dual permutation invariant cones in Rn. Section 5 deals with a characterization of spectral cones and studies the closure, interior, dual, and orthogonal complement of a spectral cone. The dimensionality of a spectral cone is described in Section 6. Finally, in Section 7, we characterize pointedness/solidness of spectral cones.

2 Preliminaries

For a set S in a real inner product space (H, h·, ·i), the closure, interior, and boundary are denoted, respectively, by S, S◦, and ∂ S. The dual and the orthogonal complement of S are deﬁned by

S∗ = {x ∈ H : hx, yi ≥ 0 ∀ y ∈ S} and S⊥ = S∗ ∩ −S∗.

We say that S is a cone if 0 < α ∈ R, x ∈ S ⇒ α x ∈ S and convex if x, y ∈ S, 0 ≤ α, β ∈ R, α+β = 1 ⇒ α x + β y ∈ S. Also, a convex cone S is pointed if S ∩ −S ⊆ {0} and solid if S◦ 6= ∅. Given a set S, we write conv(S) for its convex hull and cone(S) for its convex conic hull. For sets S1 and

2 S2 and real numbers α1 and α2, we let α1 S1 + α2 S2 := {α1 x + α2 y : x ∈ S1, y ∈ S2}.

2.1 Permutation matrices and majorization in Rn

Vectors in Rn are considered as column vectors and Rn carries the usual inner product. An n × n permutation matrix is a matrix obtained by permuting the rows of an n × n identity matrix. The n set of all n × n permutation matrices is denoted by Σn. A set Q in R is said to be permutation invariant if σ(Q) = Q for all σ ∈ Σn. T n For any u = (u1, u2, . . . , un) in R consider Σn(u) = {σ(u): σ ∈ Σn}, the set of all possible permutations of u. If we look (only) at the ﬁrst components of vectors in this collection, we see ui (for i = 1, 2, . . . , n) appearing exactly (n − 1)!-times. Hence, adding all these ﬁrst components, we get the sum (n − 1)! tr(u), where tr(u) = u1 + u2 + ··· + un. The same sum is obtained when other components are considered. Thus, X σ(u) = (n − 1)! tr(u) 1, (1)

σ∈Σn where 1 denotes the vector in Rn with all entries 1. Given u ∈ Rn, we write u↓ for its decreasing rearrangement: that is the vector obtained by rearranging the coordinates of u in the decreasing order. Likewise, we write u↑ for the increasing rearrangement of u. For two vectors u and v in Rn with their decreasing rearrangements u↓ and v↓, we say that u is majorized by v [9] and write u ≺ v if

k k n n X ↓ X ↓ X ↓ X ↓ ui ≤ vi for 1 ≤ k ≤ n − 1 and ui = vi . i=1 i=1 i=1 i=1 A theorem of Hardy, Littlewood, and P´olya ([2], Theorem II.1.10) says that u ≺ v if and only if u = Av, for some doubly stochastic matrix A (which is a nonnegative matrix with row and column sums one). Also, a theorem of Birkhoﬀ ([2], Theorem II.2.3) says that every doubly stochastic matrix is a convex combination of permutation matrices. The following elementary proposition will be useful.

n Pn Proposition 2.1 Suppose u (6= 0) and v be vectors in R with decreasing entries and 1 ui = Pn 1 vi = 0. Then, Pk (i) 1 ui > 0 for all k with 1 ≤ k ≤ n − 1, and (ii) v ≺ α u for some positive number α. Pk Pn Pn Proof. Suppose 1 ui ≤ 0 for some k with 1 ≤ k ≤ n − 1. As 1 ui = 0, we have k+1 ui ≥ 0. Since the entries of u are decreasing, we must have uk+1 ≥ 0. This implies that u1 ≥ u2 ≥ Pk · · · ≥ uk ≥ 0. But then, 1 ui ≤ 0 implies that u1 = u2 = ··· = uk = 0. From this we get Pn Pn 0 ≥ uk+1 ≥ uk+2 ≥ · · · ≥ un. As these inequalities imply 0 ≥ k+1 ui, we see that 0 = k+1 ui

3 from which we get 0 = uk+1 = uk+2 = ··· = un. Thus, u = 0, leading to a contradiction. Hence we have (i). Pk Pk Now, because of (i), we can ﬁnd a positive α such that 1 vi ≤ α( 1 ui) for all k with 1 ≤ k ≤ n−1. Pn Pn Since α( 1 ui) = 1 vi = 0, we see that v ≺ α u. This gives (ii).

2.2 Euclidean Jordan algebras

Throughout this paper, V denotes a Euclidean Jordan algebra [3]; for short, we say that V is an algebra. For x, y ∈ V, we denote their inner product by hx, yi and Jordan product by x ◦ y. We let e denote the unit element in V and

V+ := {x ◦ x : x ∈ V} denote the corresponding symmetric cone. Recall that a Euclidean Jordan algebra V is simple if it is not a direct sum of nonzero Euclidean Jordan algebras. It is known, see [3], that any nonzero Euclidean Jordan algebra is, in a unique way, a direct sum/product of simple Euclidean Jordan algebras. Moreover, every simple algebra is isomorphic to one of the following ﬁve algebras:

(i) the algebra Sn of n × n real symmetric matrices,

(ii) the algebra Hn of n × n complex Hermitian matrices,

(iii) the algebra Qn of n × n quaternion Hermitian matrices,

(iv) the algebra O3 of 3 × 3 octonian Hermitian matrices,

(v) the Jordan spin algebra Ln for n ≥ 3.

The Euclidean space Rn is a Euclidean Jordan algebra under componentwise product of vectors and the usual inner product.

We say that an algebra V is essentially simple if it is either simple or isomorphic to Rn.

An element c ∈ V is an idempotent if c2 = c; it is a primitive idempotent if it is nonzero and cannot be written as a sum of two nonzero idempotents. We say a ﬁnite set {e1, e2, . . . , en} of primitive idempotents in V is a Jordan frame if n X ei ◦ ej = 0 if i 6= j and ei = e. i=1 It turns out that the number of elements in any Jordan frame is the same; this common number is called the rank of V.

Proposition 2.2 (Spectral decomposition theorem [3]) Suppose V is a Euclidean Jordan algebra of rank n. Then, for every x ∈ V, there exist uniquely determined real numbers λ1(x), . . . , λn(x)

4 (called the eigenvalues of x) and a Jordan frame {e1, . . . , en} such that

x = λ1(x)e1 + ··· + λn(x)en.

Conversely, given any Jordan frame {e1, . . . , en} and real numbers λ1, λ2, . . . , λn, the sum λ1e1 +

λ2e2 + ··· + λnen deﬁnes an element of V whose eigenvalues are λ1, λ2, . . . , λn.

The trace of any x ∈ V is deﬁned by: n X tr(x) := λi(x). i=1 n T Note that when V = R , for any u = (u1, u2, . . . , un) , tr(u) = u1 + u2 + ··· + un.

From now on, we assume that the eigenvalues of x ∈ V are arranged so that λ1(x) ≥ λ2(x) ≥ · · · ≥

λn(x) and let T λ(x) = λ1(x), λ2(x), . . . , λn(x) .

We note that the mapping x 7→ λ(x), called the eigenvalue map, is well-deﬁned, continuous [1] and positively homogeneous. We say that two elements x, y ∈ V are spectrally equivalent if they have the same eigenvalues; symbolically,

x ∼ y in V ⇔ λ(x) = λ(y) in Rn.

Starting from the given inner product on V, we can deﬁne the canonical inner product by

hx, yic := tr(x ◦ y).

This inner product is compatible with the given Jordan product; idempotents, Jordan frames, spectral decomposition and eigenvalues of an element, etc., remain the same under this new inner product. We note that with respect to the canonical inner product, the norm of any primitive idempotent is one. It is known, see [3], Prop. III,4,1, that in any simple algebra, the inner product is a (positive) multiple of the canonical inner product. Thus, when V is either simple or carries the canonical inner product, all primitive idempotents have the same norm.

Proposition 2.3 (Peirce decomposition theorem [3]) Let V be of rank n and {e1, . . . , en} be a

Jordan frame of V. Then the space V is the orthogonal direct sum of subspaces Vij (i ≤ j), where for i, j ∈ {1, . . . , n},

Vii := {x ∈ V : x ◦ ei = x} = Rei, 1 Vij := {x ∈ V : x ◦ ei = 2 x = x ◦ ej} when i 6= j.

5 Thus, given any Jordan frame {e1, . . . , en}, we can write any element x ∈ V as n X X x = xiei + xij, i=1 1≤i

{e1, . . . , en}.

Let φ and Ψ be linear transformations on V. We say that

• φ is an algebra automorphism (or ‘automorphism’ for short) if it is invertible and

φ(x ◦ y) = φ(x) ◦ φ(y) ∀ x, y ∈ V.

The set of all automorphisms of V is denoted by Aut(V). We note that automorphisms map Jordan frames to Jordan frames. Thus, eigenvalues of an element remain the same under the action of an automorphism. In particular,

φ(x) ∼ x ∀ x ∈ V, φ ∈ Aut(V).

We say that a set Ω in V is invariant under automorphisms (or automorphism invariant) if

φ(Ω) ⊆ Ω ∀ φ ∈ Aut(V).

• Ψ is doubly stochastic [4] if it is positive, unital, and trace preserving, i.e., Ψ(V+) ⊆ V+, Ψ(e) = e, and tr(Ψ(x)) = tr(x) for all x ∈ V. When V = Rn, such a Ψ is a doubly stochastic matrix (which is an n × n nonnegative matrix with each row/column sum one).

We recall the following two results from [7].

Proposition 2.4 ([3], Theorem IV.2.5, [7], Proposition 3) Let V be essentially simple. If {e1, . . . , en} 0 0 0 and {e1, . . . , en} are any two Jordan frames in V, then there exists φ ∈ Aut(V) such that φ(ei) = ei for all i = 1, . . . , n. In particular, if x ∼ y in V, then there exists φ ∈ Aut(V) such that x = φ(y).

Proposition 2.5 ([7], Proposition 9) Given x, y ∈ V, there exist doubly stochastic matrices A and B such that

λ(x + y) = Aλ(x) + Bλ(y). (2)

When V is simple, we can take A = B.

Our next two results are from [4].

Proposition 2.6 ([4], Proposition 1) Let V1 and V2 be two non-isomorphic simple Euclidean

Jordan algebras. If φ ∈ Aut(V1 × V2), then φ is of the form (φ1, φ2) for some φi ∈ Aut(Vi), i = 1, 2,

6 that is,

φ(x) = (φ1(x1), φ2(x2)), ∀ x = (x1, x2) ∈ V1 × V2.

The result below describes a connection between majorization, automorphisms, and doubly stochastic transformations in the setting of Euclidean Jordan algebras. We deﬁne majorization in V by

x ≺ y in V ⇔ λ(x) ≺ λ(y) in Rn.

Proposition 2.7 ([4], Theorem 6) For x, y ∈ V, consider the following statements:

(a) x = Φ(y), where Φ is a convex combination of automorphisms of V.

(b) x = Ψ(y), where Ψ is doubly stochastic on V.

Then, (a) ⇒ (b) ⇒ (c). Furthermore, when V is essentially simple, reverse implications hold.

3 Spectral sets in Euclidean Jordan algebras

In preparation for a study of spectral cones, in this section, we describe some results on spectral sets. Recall that a set E in V is a spectral set if there exists a permutation invariant set Q in Rn such that E = λ−1(Q).

Proposition 3.1 ([7], Theorem 1) A set E in V is a spectral set if and only if

x ∼ y, y ∈ E ⇒ x ∈ E.

One key observation is the following: Suppose E is a spectral set, x ∈ E with its spectral decom- 0 0 0 position x = λ1e1 + λ2e2 + ··· + λnen. Then for any Jordan frame {e1, e2, . . . , en}, the element 0 0 0 x := λ1e1 + λ2e2 + ··· + λnen ∈ E (as x ∼ x).

Proposition 3.2 ([7], Theorem 3) For a set E in V, consider the following:

(a) E = λ−1(Q), where Q is permutation invariant and convex in Rn.

(b) E is convex, and x ≺ y, y ∈ E ⇒ x ∈ E.

(d) E is convex and invariant under automorphisms.

Then (a) ⇔ (b) ⇔ (c) and (c) ⇒ (d). Furthermore, all statements are equivalent when V is essentially simple.

7 Let E be a spectral set in V so that E = λ−1(Q) for some permutation invariant set Q in Rn. T Then, Σn(λ(E)) ⊆ Q. For any q = (q1, q2, . . . , qn) ∈ Q and any Jordan frame {e1, e2, . . . , en}, Pn ↓ ↓ x := 1 qiei ∈ E because λ(x) = q ∈ Q. Since q = σ(q ) for some σ ∈ Σn, it follows that Q ⊆ Σn(λ(E)); hence

Q = Σn(λ(E)). (3)

This means that Q is uniquely deﬁned for each spectral set E. In this setting, by deﬁning E♦ := Q and Q♦ := E, we have (from [7], Section 6) the metaformula(s)

#(E♦) = [#(E)]♦ and #(Q♦) = [#(Q)]♦, (4) where # is one of the operations of closure, interior, boundary, and convex hull. In particular, closure/interior/boundary/convex hull of a spectral set is a spectral set.

The following result shows that on permutation invariant (convex) sets, λ−1 has linear behavior.

Theorem 3.3 The following statements hold:

(i) Let Q be a permutation invariant set in Rn and α ≥ 0 in R. Then

λ−1(−Q) = −[λ−1(Q)] and λ−1(α Q) = α λ−1(Q).

n (ii) Let Q1 and Q2 be permutation invariant convex sets in R . Then,

−1 −1 −1 λ (Q1 + Q2) = λ (Q1) + λ (Q2).

Proof. (i) We ﬁrst observe that −Q is permutation invariant. Let x ∈ λ−1(Q) with its spectral Pn Pn ↓ ↑ decomposition x = 1 λi(x)ei. From −x = 1 [−λi(x)]ei we get λ(−x) = [−λ(x)] = −[λ(x) ]. Since Q is permutation invariant and λ(x) ∈ Q, we have λ(x)↑ ∈ Q and −[λ(x)↑] ∈ −Q. Thus, λ(−x) ∈ −Q and −x ∈ λ−1(−Q). Hence, −[λ−1(Q)] ⊆ λ−1(−Q). Now, let y ∈ λ−1(−Q). Then, by the previous inclusion applied to −Q, −y ∈ λ−1[−(−Q)] = λ−1(Q), or equivalently, y ∈ −[λ−1(Q)]. This proves the inclusion λ−1(−Q) ⊆ −[λ−1(Q)]. Thus we have the ﬁrst part of statement (i). The second part, for α = 0 is obvious; the case α > 0 follows easily from the positive homogeneity of λ. n (ii) Now suppose that Q1 and Q2 are permutation invariant convex sets in R . Then Q1 + Q2 is −1 −1 also permutation invariant and convex. Let x ∈ λ (Q1) and y ∈ λ (Q2) so that λ(x) ∈ Q1 and

λ(y) ∈ Q2. Then, by Proposition 2.5,

λ(x + y) = A λ(x) + B λ(y), where A and B are doubly stochastic matrices on Rn. By a well-known theorem of Birkhoﬀ

([2], Theorem II.2.3), A and B are convex combinations of permutation matrices. As Q1 and

Q2 are permutation invariant convex sets, it follows that Aλ(x) ∈ Q1 and Bλ(y) ∈ Q2. Thus,

8 −1 λ(x + y) ∈ Q1 + Q2. This implies that x + y ∈ λ (Q1 + Q2). Hence,

−1 −1 −1 λ (Q1) + λ (Q2) ⊆ λ (Q1 + Q2).

−1 Pn To see the reverse inclusion, let z ∈ λ (Q1 + Q2) with spectral decomposition z = 1 λi(z)ei. Pn Then, λ(z) ∈ Q1 + Q2. Let λ(z) = u + v, where u ∈ Q1 and v ∈ Q2. Deﬁne x = 1 uiei and Pn ↓ ↓ −1 y = 1 vi ei so that z = x + y. As λ(x) = u ∈ Q1 and λ(y) = v ∈ Q2, we see that x ∈ λ (Q1), −1 −1 −1 −1 −1 −1 y ∈ λ (Q2). Thus, z = x + y ∈ λ (Q1) + λ (Q2). Hence, λ (Q1 + Q2) ⊆ λ (Q1) + λ (Q2). Thus we have (ii).

The following is an easy consequence.

n Corollary 3.4 Let Q1 and Q2 be permutation invariant convex sets in R and α1, α2 ∈ R. Then,

−1 −1 −1 λ (α1 Q1 + α2 Q2) = α1 λ (Q1) + α2 λ (Q2).

Our next result deals with the dual of a spectral set. For a permutation invariant set Q in Rn, we note that its dual (computed with respect to the usual inner product in Rn) is also permutation invariant as

T ∗ hσ(u), qi = hu, σ (q)i ≥ 0 (∀ u ∈ Q , q ∈ Q, and σ ∈ Σn).

The dual of a spectral set is computed with respect to the given inner product in V.

Theorem 3.5 Suppose V is either simple or carries the canonical inner product. Then, for any permutation invariant set Q in Rn,

∗ λ−1(Q) = λ−1(Q∗).

Proof. By assumption, all primitive idempotents in V have the same norm (see the remarks made after the deﬁnition of canonical inner product in Section 2). Let ω := ||c||2 for any primitive −1 ∗ P idempotent c in V. Now, let x ∈ λ (Q) and x = i λi(x)ei be its spectral decomposition. For P ↓ −1 any q ∈ Q, set y = i qiei. As λ(y) = q ∈ Q, we have y ∈ λ (Q) and so hx, yi ≥ 0. Thus, we have * n n + n X X X 2 0 ≤ hx, yi = λi(x)ei, qiei = λi(x)qi keik = ω hλ(x), qi , i=1 i=1 i=1 where hλ(x), qi denotes the usual inner product between the vectors λ(x) and q in Rn. Since ω > 0, hλ(x), qi ≥ 0. As q is arbitrary in Q, we have λ(x) ∈ Q∗ and so x ∈ λ−1(Q∗). This shows λ−1(Q)∗ ⊆ λ−1(Q∗). To prove the reverse implication, let x ∈ λ−1(Q∗) so that λ(x) ∈ Q∗. Now, let y ∈ λ−1(Q) with P its spectral decomposition y = i λi(y)ei. We write the Peirce decomposition of x with respect to

9 the Jordan frame {e1, . . . , en} as n X X x = xiei + xij, i=1 i 0 with αk = 1, and σk ∈ Σn. k=1 k=1 Since Q∗ is permutation invariant and convex, u ∈ Q∗. Since y ∈ λ−1(Q) ⇒ λ(y) ∈ Q, we must have hu, λ(y)i ≥ 0. Hence, from (5), we get hx, yi ≥ 0. As y is arbitrary in λ−1(Q), this proves that x ∈ [λ−1(Q)]∗. Thus, λ−1(Q∗) ⊆ λ−1(Q)∗, completing the proof.

The following example shows that the dual of a spectral set need not always be spectral.

2 Example 1. Consider the algebra V = R , where for x = (x1, x2) and y = (y1, y2), the Jordan and inner products are deﬁned, respectively, by x ∗ y := (x1y1, x2y2) and hx, yi := 2x1y1 + 3x2y2. 2 Since the set (subspace) Q := {(x1, x2): x1 = x2} in R is permutation invariant,

K := λ−1(Q) = R e is a spectral set in our algebra V. With respect to the above inner product,

∗ ⊥ K = K = {(x1, x2) : 2x1 + 3x2 = 0}.

Now, (2, −3) ∼ (−3, 2) with (−3, 2) ∈ K∗ and (2, −3) 6∈ K∗. Hence, by Proposition 3.1, K∗ is not spectral.

10 4 Some permutation invariant convex cones in Rn

In this section, we describe some permutation invariant convex cones in Rn. Given any nonempty set S in Rn,

Q := cone (Σn(S)), the convex cone generated by Σn(S), is a permutation invariant convex cone. For p ∈ {1, 2, . . . , n}, n the rearrangement cone Qp [5] is deﬁned by n n ↑ ↑ ↑ Qp := {u ∈ R : u1 + u2 + ··· + up ≥ 0},

↑ ↑ ↑ n where we note that u1 + u2 + ··· + up is the sum of the smallest p entries of u ∈ R .

Proposition 4.1 The following statements hold:

n (a) Qp is a permutation invariant polyhedral (closed convex) solid cone; it is pointed when 1 ≤ p ≤ n − 1.

n n n (b) Q1 ⊆ Q2 ⊆ · · · ⊆ Qn. n n (c) Qp is isomorphic to Qn−p for 1 ≤ p ≤ n − 1.

Proof. (a) For any nonempty index set I ⊆ {1, 2, . . . , n}, let |I| denote the cardinality of I and eI be the vector with (eI )i = 1 for i ∈ I and (eI )i = 0 otherwise. Then,

↑ ↑ ↑ u1 + u2 + ··· + up = min hu, eI i , |I|=p hence,

n n ↑ ↑ ↑ Qp = {u ∈ R : u1 + u2 + ··· + up ≥ 0} n = {u ∈ R : minhu, eI i ≥ 0} |I|=p n = {u ∈ R : hu, eI i ≥ 0 for every eI with |I| = p}.

n Since Qp is now the intersection of a ﬁnite number of closed half-spaces, it is a polyhedral closed ↑ ↑ n convex cone. Since σ(u) = u for any permutation matrix σ ∈ Σn, we see that Qp is permutation n n n invariant. That Qp has nonempty interior follows from the inclusion R+ ⊆ Qp . Now let 1 ≤ n p ≤ n − 1. To see the pointedness, suppose both u and −u are in Qp . Then, the inequalities min|I|=p hu, eI i ≥ 0 and min|I|=p h−u, eI i ≥ 0 imply that hu, eI i = 0 whenever |I| = p. This means that sum of any p entries of u is zero. Hence, as p ≤ n − 1, all entries of u are equal, and so, u = 0. n Thus, Qp is pointed when 1 ≤ p ≤ n − 1. n ↑ ↑ ↑ ↑ ↑ (b) Let 1 ≤ p ≤ n − 1. If u ∈ Qp , then u1 + u2 + ··· + up ≥ 0. Since 0 ≤ up ≤ up+1, we get ↑ ↑ ↑ ↑ n u1 + u2 + ··· + up + up+1 ≥ 0. Thus, u ∈ Qp+1. (c) We let 1 ≤ p ≤ n − 1. Let 1 denote the vector in Rn with all entries one, N := 1 1T (the n × n

11 matrix with all entries one), and I denote the identity matrix. Deﬁning 1 M := N − I, p p n we easily verify that Mp Mn−p = I; hence Mp is invertible. Now, let u ∈ Qn−p; this means that the Pn ↓ tr(u) sum of n−p smallest entries of u is nonnegative, that is, p+1 ui ≥ 0. Then, from Mpu = p 1−u ↑ ↓ ↑ tr(u) ↓ and (−u) = − u we have (Mpu) = p 1 − u . Hence, p p n X ↑ X ↓ X ↓ (Mpu)i = tr(u) − ui = ui ≥ 0. i=1 1 p+1 n n n n n n Thus Mpu ∈ Qp , and so Mp(Qn−p) ⊆ Qp . Similarly, Mn−p(Qp ) ⊆ Qn−p. Hence, Mp(Qn−p) = n Qp .

Our next example is a permutation invariant non-polyhedral self-dual cone.

Example 2. In Rn (n ≥ 3), let rn Q = u : tr(u) ≥ ||u|| , 2 where ||u|| denotes the 2-norm of the vector u in Rn. As trace and the 2-norm are permutation invariant, we see that the above set is permutation invariant. We show that Q is the image of the n n Lorentz cone L+ (the symmetric cone in L ) under an orthogonal matrix. Recall  1  " n # 2 n  T X 2  n 1 L+ := u = (u1, u2, . . . , un) : u1 ≥ ui = u ∈ R : hu, c1i ≥ √ ||u||||c1|| ,  2  2 n where c1 denotes the vector in R with one in the ﬁrst slot and zeros elsewhere. It is easily seen n √1 n that Q = U(L+), where U is an orthogonal matrix with Uc1 = n 1. Since L+ is a non-polyhedral self-dual cone and U is orthogonal, it follows that Q is non-polyhedral and self-dual.

5 Spectral cones in Euclidean Jordan algebras

Recall that a spectral cone, by deﬁnition, is of the form K = λ−1(Q), for some permutation invariant convex cone Q in Rn. It is easy to verify, by the positive homogeneity of λ, that the above K is a cone and, by Proposition 3.2, convex. Thus, every spectral cone is a convex cone. The metaformula (4) shows that the closure/interior of a spectral cone is again a spectral cone. For a list of spectral cones in Sn, see [6].

The following result characterizes spectral cones among convex cones in V.

Theorem 5.1 For a convex cone K in V, consider the following statements:

12 (a) K is a spectral cone.

(b) If x ≺ y and y ∈ K, then x ∈ K.

(d) For every doubly stochastic map Ψ on V, Ψ(K) ⊆ K.

(e) For every algebra automorphism φ on V, φ(K) ⊆ K.

Then, (a) ⇔ (b) ⇔ (c) and (b) ⇒ (d) ⇒ (e). Moreover, when V is essentially simple, all the above statements are equivalent.

Proof. (a) ⇒ (b) ⇒ (c): When (a) holds, K is a spectral convex set. For such a set, (b) and (c) hold by Proposition 3.2. (c) ⇒ (a): When (c) holds, from Proposition 3.2, we can write K = λ−1(Q), where Q is permutation invariant and convex. Since K is given to be a cone, we see that Q (which equals Σn(λ(K)) by (3)) is a cone. Thus, K is a spectral cone. (b) ⇒ (d): Assume that K is spectral cone and let Ψ be a doubly stochastic map on V. For y ∈ K, let x = Ψ(y). Then by Proposition 2.7, x ≺ y. It follows that x ∈ K. (d) ⇒ (e): This is obvious, as every algebra automorphism is doubly stochastic. Now we prove (e) ⇒ (b) assuming V is essentially simple. Then, Proposition 3.2 shows that K is a spectral set that corresponds to a permutation invariant convex set Q in Rn. As K is also a cone, we see that Q is a cone. Hence, K is a spectral cone.

Remark. When V is not essentially simple, the reverse implications in the above theorem may not 2 2 hold. For instance, consider V = R × S , K = R+ × S , and

1 0 −1 0 x = − 1, 0 2 , y = 1, 0 2 . From Proposition 2.6, we see that K is invariant under automorphisms of V. However, K is not a spectral set as x ∼ y, y ∈ K, and x∈ / K. Thus, (e) holds, but not (a).

The above example shows that the Cartesian product of two spectral cones need not be spectral. However, as we see below, certain projections of a spectral cone are spectral.

Corollary 5.2 Suppose V1, V2,..., VN are essentially simple algebras and V = V1 × V2 × · · · × VN .

Let Πj : V → Vj denote the projection map and K be a spectral cone in V. Then Πj(K) is a spectral cone in Vj for all j = 1, 2 ...,N.

Proof. By the linearity of Πj,Πj(K) is a convex cone in Vj. To show that Πj(K) is a spectral cone in the essentially simple algebra Vj, we show that it is invariant under automorphisms of Vj and apply Theorem 5.1. Now, without loss of generality, let j = 1, x1 ∈ Π1(K) and φ1 ∈ Aut(V1).

Then, there exists xi ∈ Vi, i = 2, 3,...,N such that x := (x1, x2, . . . , xN ) ∈ K. Let φi be the

13 identity transformation on Vi, i = 2,...,N. Then, φ := (φ1, φ2, φ3, . . . , φN ) ∈ Aut(V), where the action of φ on V1 × V2 × · · · × VN is given by φ(z1, z2, . . . , zn) := (φ(z1), φ(z2), . . . , φ(zn)). Since

φ(x) ∈ K by Theorem 5.1, we see that φ1(x1) ∈ Π1(K). This completes the proof.

Remarks. (1) The automorphisms of Rn are just permutation matrices. Since Rn is also essentially simple, spectral cones in Rn are just permutation invariant convex cones. (2) Let V be essentially simple. For any z ∈ V, consider

K = cone{φ(z): φ ∈ Aut(V)}, the convex cone generated by {φ(z): φ ∈ Aut(V)}. Then, by Theorem 5.1, K is a spectral cone.

Theorem 5.3 Let K1, K2, and K be spectral cones in V and α1, α2 ∈ R. Then, the following hold:

◦ (a) α1 K1 + α2 K2, K, and K are spectral cones. (b) When V is simple or carries the canonical inner product, K∗ and K⊥ are spectral cones. −1 n Proof. Let Ki = λ (Qi), where Qi is a permutation invariant convex cone in R for i = 1, 2. n Then, α1 Q1 + α2 Q2 is a permutation invariant convex cone in R and from Corollary 3.4, α1 K1 + −1 −1 α2 K2 = λ (α1 Q1 + α2 Q2). Thus, α1 K1 + α2 K2 is a spectral cone. Let K = λ (Q), where Q is a permutation invariant convex cone in Rn. It is easy to see that Q, Q◦,Q∗ and Q⊥ (deﬁned with respect to the usual inner product in Rn) are permutation invariant convex cones in Rn. (Note that Q⊥ = Q∗ ∩ −Q∗.) That K and K◦ are spectral cones follow from the metaformula (4). Thus we have (a). To see (b) suppose that V is simple or carries the canonical inner product. Then, Theorem 3.5 shows that K∗ is a spectral cone. Finally, the equality

K⊥ = K∗ ∩ −K∗ = λ−1(Q∗) ∩ λ−1(−Q∗) = λ−1(Q∗ ∩ −Q∗) = λ−1(Q⊥) (6) shows that K⊥ is a spectral cone.

Remark. In [6], Lemma 3, it is shown that the dual of a spectral cone in Sn is a spectral cone. Item (b) in the above theorem is a generalization. However, it does not hold in a general algebra: The set K in Example 1 is a spectral cone (actually, a subspace) while K∗ (which is K⊥) is not even a spectral set.

6 A dimensionality result

Inspired by a result of Lahtonen and its proof in [8], we present the following dimensionality result for spectral cones. Recall that the dimension of a convex cone K is the dimension of the subspace

14 K − K. Throughout this section, for the Euclidean Jordan algebra V, we let

m := dim(V).

Theorem 6.1 For any spectral cone K in V,

dim(K) ∈ {0, 1, m − 1, m}.

Proof. We assume without loss of generality that K is nonempty and that V carries the canonical inner product so that the norm of any primitive idempotent is one. (See the remarks made after Proposition 2.2.) Now we show that either every element of K is a scalar multiple of e or every element of K⊥ is a scalar multiple of e. Assuming the contrary, let x ∈ K and y ∈ K⊥ have at least two distinct eigenvalues. Writing their spectral decompositions,

0 0 0 x = x1e1 + x2e2 + ··· + xnen and y = y1e1 + y2e2 + ··· + ynen, we assume, without loss of generality, x1 6= x2 and y1 6= y2. Now, deﬁne x, x by 0 0 0 0 x := x1e1 + x2e2 + x3e3 + ··· + xnen, 0 0 0 0 x := x2e1 + x1e2 + x3e3 + ··· + xnen. We note that x ∼ x and x ∼ x. Since K is a spectral set and x ∈ K, by Theorem 5.1, x, x ∈ K; hence, x, x ⊥ y. Then, 0 = hx, yi − x, y = x − x, y 0 0 0 0 0 = (x1 − x2)(e1 − e2), y1e1 + y2e2 + ··· + ynen

= (x1 − x2)y1 − (x1 − x2)y2 = (x1 − x2)(y1 − y2) 6= 0. We reach a contradiction. This shows that either K or K⊥ must contain just multiples of e. Thus, the dimension of K or K⊥ is at most 1. As dim(K) + dim(K⊥) = dim(V) = m, we deduce that the possible values for dim(K) are 0, 1, m − 1, and m.

It is easy to see that a nonempty (spectral) cone has dimension zero if and only if it is {0}. Also, it is of dimension m if and only if it has nonempty interior. We now describe nonempty spectral cones in V with dimensions 1 and m − 1.

Theorem 6.2 For a nonempty spectral cone K in V, the following statements hold.

(a) dim(K) = 1 if and only if K is a nonzero cone contained in R e.

(b) dim(K) = m − 1 if and only if K = {x ∈ V : tr(x) = 0}.

We note that in V, there are ﬁve nonzero cones contained in R e, namely, R+ e, R++ e, −R+ e,

−R++ e, and R e.

15 Proof. (a) As the ‘if’ part is obvious, We prove the ‘only if’ part. Suppose dim(K) = 1. Let K = λ−1(Q), where Q is a permutation invariant convex cone in Rn. We claim that Q has dimension one. If, on the contrary, u, v are linearly independent vectors in Q, then for any Jordan Pn Pn frame {e1, e2, . . . , en}, x := 1 uiei and y := 1 viei are two linearly independent elements in K contradicting dim(K) = 1. Thus, Q has dimension one; let Q be contained in the span of a nonzero vector w. As Q is permutation invariant, this w must be a multiple of 1. Then K = λ−1(Q) is contained in R e, as {e} = λ−1({1}). (b) If K = {x ∈ V : tr(x) = 0}, then it has dimension m − 1 as x 7→ tr(x) from V to R is linear with null space K. Now suppose (without loss of generality) V carries the canonical inner product and K has dimension m − 1. Letting K = λ−1(Q), we see, from (6), λ−1(Q⊥) = K⊥ = (K − K)⊥ has dimension one. As in Item (a), we can show that Q⊥ is spanned by the vector 1 in Rn; hence, tr(v) = hv, 1i = 0 for all v ∈ Q. This means that Q ⊆ M, where M = {v ∈ Rn : tr(v) = 0}. Now, take any vector v ∈ M and a nonzero u ∈ Q. As M and Q are permutation invariant, we may assume that the entries of Pn Pn u and v are decreasing. Note that i=1 vi = i=1 ui = 0. Now, from Proposition 2.1, v ≺ α u for some α > 0. Since Q is a permutation invariant convex cone, α u ∈ Q and by Proposition 3.2, v ∈ Q. This proves that Q = M. From this, we get K = λ−1(Q) = {x ∈ V : tr(x) = 0}. This proves the result.

Corollary 6.3 Suppose K is a spectral cone in V such that {0} ⊆ K ⊆ {x ∈ V : tr(x) = 0}. Then, either K = {0} or K = {x ∈ V : tr(x) = 0}.

Proof. For the speciﬁed K, let K = λ−1(Q), where Q is a permutation invariant convex cone in Rn. Then,

{0} ⊆ λ−1(Q) ⊆ λ−1(M),

n −1 where M = {v ∈ R : tr(v) = 0}. Now, from (3), P = Σn(λ(λ (P ))) for any permutation invariant set P in Rn; thus,

{0} ⊆ Q ⊆ M.

The proof of Item (b) in the above theorem shows that either Q = {0} or Q = M. Then, K, which is λ−1(Q), is either {0} or {x ∈ V : tr(x) = 0}.

Remark. Suppose K is a nonempty spectral cone diﬀerent from {0} and {x ∈ V : tr(x) = 0}. Then, either e ∈ K or −e ∈ K. This is seen as follows. Let K = λ−1(Q), where Q is a permutation invariant convex cone in Rn. Then, Q is nonempty and diﬀerent from {0} and {u ∈ Rn : tr(u) = 0}. Let u be a nonzero element of Q with tr(u) 6= 0. As Q is permutation invariant, from (1), X (n − 1)! tr(u) 1 = σ(u) ∈ Q.

σ∈Σn

16 Since Q is a cone, we see that either 1 ∈ Q or −1 ∈ Q. From this, we see that either e ∈ K or −e ∈ K.

7 Pointed/solid spectral cones

The following result is a generalization of a similar result stated in the setting of V = Sn (see [5], Theorem 3.3).

Theorem 7.1 Let K = λ−1(Q), where Q is a permutation invariant convex cone in Rn. Then

(i) K is pointed if and only if Q is pointed,

(ii) K is solid if and only if Q is solid.

Proof. (i) From Theorem 3.3, K ∩ −K = λ−1(Q ∩ −Q). Thus, K ∩ −K ⊆ {0} if and only if Q ∩ −Q ⊆ {0}. This proves (i). (ii) This follows from the fact that K◦ = λ−1(Q◦), see (4).

In the next several results, we characterize pointedness and solidness of spectral cones in Rn and V.

Lemma 7.2 Let Q be a nonempty nonzero permutation invariant convex cone in Rn. Then Q is pointed if and only if exactly one of the following conditions holds:

(i) tr(u) > 0 for all nonzero u ∈ Q (and 1 ∈ Q).

(ii) tr(u) < 0 for all nonzero u ∈ Q (and −1 ∈ Q).

Proof. Suppose Q is pointed. We ﬁrst show that tr(u) 6= 0 for every nonzero u ∈ Q. Suppose, on the contrary, there is a nonzero vector u ∈ Q such that tr(u) = 0. Then by (1), 0 = (n−1)! tr(u) 1 = P σ(u). This implies σ∈Σn " # X X v := σ(u) = σ(u) − I(u) = −I(u) = −u,

σ∈Σn, σ6=I σ∈Σn where I is the identity matrix. However, as both u and v are in Q, we reach a contradiction to the pointedness of Q. Now, if there exist nonzero u, v ∈ Q such that tr(u) > 0 and tr(v) < 0, then for a suitable convex combination w of u and v, we have tr(w) = 0. As Q is convex, w ∈ Q. From what has been proved earlier, w = 0. This contradicts the pointedness of the convex cone Q. Hence, either (a) tr(u) > 0 for all nonzero u ∈ Q or (b) tr(u) < 0 for all nonzero u ∈ Q. Now, in the ﬁrst case, by (1), X (n − 1)! tr(u) 1 = σ(u) ∈ Q.

σ∈Σn

17 It follows (by scaling) that 1 ∈ Q. This is Item (i). Similarly, when (b) holds, we get Item (ii). By the pointedness of Q, both (i) and (ii) cannot hold simultaneously. To see the reverse implication, suppose without loss of generality (i) holds so that tr(u) > 0 for every nonzero u ∈ Q. By the linearity of the trace function, −u cannot be in Q for any nonzero u ∈ Q. Thus, Q is pointed.

Theorem 7.3 Let K be a nonempty, nonzero spectral cone in V. Then K is pointed if and only if exactly one of the following conditions holds:

(i) tr(x) > 0 for all nonzero x ∈ K (and e ∈ K).

(ii) tr(x) < 0 for all nonzero x ∈ K (and −e ∈ K).

Proof. Let K = λ−1(Q) where Q is a permutation invariant convex cone. Then, by an earlier result, K is pointed if and only if Q is pointed. Thus, by the above lemma, K is pointed if and only if either tr(u) > 0 for all u ∈ Q (and 1 ∈ Q) or tr(u) < 0 for all u ∈ Q (and −1 ∈ Q). As tr(λ(x)) = tr(x), and ± 1 ∈ Q if and only if ± e ∈ K, we get the stated results from K = λ−1(Q).

We say that a vector in Rn is a nonconstant vector if it is not a multiple of 1.

Lemma 7.4 Let Q be a permutation invariant convex cone in Rn. Then the following are equivalent:

(a) Q is solid.

(b) 1 ∈ Q◦ or −1 ∈ Q◦.

Proof. (a) ⇒ (b): Suppose Q is solid, that is, Q◦ 6= ∅. As Q◦ cannot be contained in the subspace {u ∈ Rn : tr(u) = 0}, there exists a nonzero vector u ∈ Q◦ with tr(u) 6= 0. As Q◦ is permutation ◦ ◦ invariant, σ(u) ∈ Q for every σ ∈ Σn. Now, as Q is a convex cone, by (1), X (n − 1)! tr(u) 1 = σ(u) ∈ Q◦.

σ∈Σn Since tr(u) 6= 0, by scaling, we get (b). (b) ⇒ (c): This is obvious. T (c) ⇒ (a): Let n ≥ 2. Assume that there is a nonconstant vector u = (u1, u2, . . . , un) in Q and (without loss of generality) 1 ∈ Q. Suppose, if possible, Q is not solid. Then Q − Q 6= Rn and ⊥ ⊥ P so there exists a nonzero v ∈ (Q − Q) = Q . As 1 ∈ Q, we must have i vi = h1, vi = 0, T where v = (v1, v2, v2 . . . , vn) . Now, as u has at least two distinct components, we may assume (by permuting the coordinates) that u1 6= u2. Letu ¯ be the vector in Q obtained from u by interchanging

18 u1 and u2. Since u, u¯ ⊥ v, we get n X hu, vi = u1v1 + u2v2 + uivi = 0, and i=3 n X hu, vi = u2v1 + u1v2 + uivi = 0. i=3

From these, we get 0 = hu − u,¯ vi = (u1 − u2)(v1 − v2). As u1 − u2 6= 0, we must have v1 = v2. Now, we can permute u so that the resulting vector has diﬀerent entries in i and j slots, i 6= j.

Then, the above argument can be repeated to get vi = vj. Hence, v is a multiple of 1. However, as P i vi = 0, we must have v = 0, a contradiction. Thus (a) holds.

Theorem 7.5 Let K be a spectral cone in V. Then the following are equivalent:

(a) K is solid.

(b) e ∈ K◦ or −e ∈ K◦.

Proof. Let K = λ−1(Q) where Q is a permutation invariant convex cone. As K◦ = λ−1(Q◦) and λ(e) = 1, we see that the items listed above are equivalent to the similar ones in the previous lemma. This completes the proof.

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