Convex hyperbolic domains and Clans

Yuncherl Choi

November 2, 1999

Contents

1. Convex hyperbolic cones 1

2. Convex hyperbolic domains 3

3. Algebra of a convex hyperbolic homogeneous domain 5

4. Decomposition of clan 9

5. Examples of clans 13

1. Convex hyperbolic cones

Let V be a finite dimensional real vector space and V^{∗} be the dual vector space of V. Definition. (1) A open subset Ω ⊂ V is called a \underline{cone} if for all x ∈ Ω, tx is contained in Ω where t is a positive real number.

(2) A subset Ω ⊂ V is called \underline{convex}if for any two points x, y ∈ Ω the line segment tx + (1 − t)y, 0 ≤ t ≤ 1 contained in Ω.

(3) A subset Ω ⊂ V is called \underline{hyperbolic} if it doesn’t any straight full line.

(4) For any cone Ω, the \underline{dual cone} Ω^{∗} is the set of elements y ∈ V^{∗} such that y(x) > 0 for all x ∈ Ω̄, x ≠ 0.

Proposition 1 Let Ω be a cone in a vector space V. Then we have followings:

(1) For any nonempty cone Ω, if Ω^{∗} ≠ ∅, then Ω^{∗∗} = Ω under the natural identification of V^{∗∗} and V.

(2) Ω is hyperbolic if and only if Ω ∩ −Ω = ∅.

(3) Ω is hyperbolic if and only if Ω^{∗} ≠ ∅.

Definition. (1) A cone Ω is called \underline{self dual} if Ω = Ω^{∗}.

(2) Aut(Ω) = \{g ∈ GL(V)|gΩ ⊂ Ω\} is called the \underline{automorphism group} of Ω.

(3) A cone Ω is called \underline{homogeneous}if Aut(Ω) acts on Ω transitively.

(4) A cone Ω is called \underline{symmetric}if it is homogeneous and self dual.

Proposition 2 (1) If Ω is a convex hyperbolic cone, then Aut(Ω^{∗}) = Aut(Ω)^{∗}.

(2) For any x ∈ Ω, Aut(Ω)_{x} is compact. Furthermore, if H is a compact subset of Aut(Ω), then there exists an element x ∈ Ω such that H ⊂ Aut(Ω)_{x}.

Definition. Let Ω be a convex hyperbolic cone.

(1) The \underline{characteristic function} ϕ of Ω is defined for x ∈ Ω by

ϕ(x) = ∫_{Ω^{∗}} e^{−y(x)}dy,

where dy denotes the Euclidian measure on V.

(2) For any point x ∈ Ω, define \underline{dual point} x^{∗} as a differential of − log ϕ at x, i.e.

x^{∗} = −D log ϕ(x).

Proposition 3 Let Ω be a convex hyperbolic cone. Then

(1) The characteristic function ϕ is uniformly convergent for x in any compact subset of Ω.

(2) ϕ(gx) = \frac{ϕ(x)}{| deg(g)|}

(3) For any sequence {x_{m}} ⊂ Ω, if {x_{m}} converges to x_{0} as m → ∞ where x_{0} ∈ ∂Ω, then we have:

lim_{m→∞} ϕ(x_{m}) = ∞.

(4) D^{2} log ϕ(x) is positive definite at each point x ∈ Ω.

(5) The map x ↦ x^{∗} is a bijection of Ω onto Ω^{∗}.

(6) x^{∗}(x) = n for all x ∈ Ω where n = dim V.

(7) The map x ↦ x^{∗} has a unique fixed point. Hence we have Ω ∩ Ω^{∗} ≠ ∅.

(8) For all x ∈ Ω and g ∈ Aut(Ω), (gx)^{∗} = (g^{∗})^{−1}x^{∗}.

(9) If Ω is homogeneous, then ϕ(x)ϕ(x^{∗}) = const and (x^{∗})^{∗} = x.

Proposition 4 Let Ω be a symmetric cone. Then (1) Aut(Ω) = Aut(Ω)^{∗}.

(2) If x_{0} be a unique point in Ω such that x_{0}^{∗} = x_{0}, then

Aut(Ω) ∩ O(V) = Aut(Ω)_{x_{0}},

where O(V) is the orthogonal group of V.

(3) The map x ↦ x^{∗} is an isometry of Riemannian manifold Ω, i.e. Ω is a Riemannian symmetric space.

2. Convex hyperbolic domains

Let U be a finite dimensional affine space. Definition. (1) Any open subset Ω ⊂ U is called a \underline{domain}.

(2) Aut(Ω) = \{g ∈ Aff(U)|gΩ ⊂ Ω\} is called the \underline{automorphism group} of the affine domain Ω.

(3) A domain Ω is called \underline{homogeneous}if Aut(Ω) acts on Ω transitively.

Proposition 5 Let Ω be a convex hyperbolic homegeneous domain in U and G ⊂ Aut(Ω) be a group that acts transitively on Ω. If there exists an affine subspace U_{1} ⊂ U such that it is invariant under the linear parts of all g ∈ G, then (x_{0} + U_{1}) ∩ Ω is a convex hyperbolic homogeneous domain in x_{0} + U_{1} for all x_{0} ∈ Ω.

Proof. Let G_{1} be the subgroup of G formed by the element g ∈ G that leave the affine subspace x_{0} + U_{1} invariant. Since the subspace U_{1} is invariant under the linear part of all g ∈ G, it follows that every transformation g ∈ G for which gx_{0} ∈ x_{0} + U_{1} belongs to G_{1}. It is now obvious that G_{1} acts transitively on (x_{0} + U_{1}) ∩ Ω. □

Definition. Given a convex hyperbolic domain Ω in an n-dimensional affine space U, imbed the space U in the (n + 1)-dimensional vector space V as a hyperplane not passing through 0. Then we can define the cone Ω̃ as following:

Ω̃ = {tx ∈ V |x ∈ Ω, t > 0}.

In this case, the cone Ω̃ is said to be \underline{fitted}onto Ω.

Remark. (1) The cone Ω̃ is clearly hyperbolic and convex.

(2) If Ω is homogeneous, then Ω̃ is also.

(3) The map ϕ = ϕΩ̃|_{Ω} is a characteristic fuction of Ω.

(4) Every nonempty cross section of a convex hyperbolic cone by a hyperplane not passing through 0 is a convex hyperbolic domain.

Proposition 6 Let Ω be a convex hyperbolic homegeneous domain in U and G be a Lie subgroup of Aut(Ω) that acts transitively on Ω. Then

(1) G_{x_{0}} is a maximal compact subgroup of G for all x_{0} ∈ Ω.

(2) The number of connected components of the group G is finite.

(3) The identity component Aut_{0}(Ω) of Aut(Ω) coincides with the idnetity component of its normalizer in the complete affine group Aff(U). Moreove it coincides with the idnetity component of an affine algebraic group.

(4) Let K be the stability subgroup of some point x_{0} ∈ Ω and T be a maximal connected triangular subgroup of (Ω). Then we have

Aut(Ω) = K · T,

where K ∩ T = {id}. The subgroup T acts simply transitively in Ω.

(5) For every convex hyperbolic homogeneous cone Ω̃, there exists a convex hyperbolic homogeneous domain Ω of dimension one less such that Ω̃ is isomorphic to the cone fitted onto Ω.

Proof. (1) Let U_{x_{0}} be the vector space such that x_{0} is zero point. Then G_{x_{0}} ⊂ GL(U_{x_{0}}), and G_{x_{0}} leave the bounded subset (Ω − x_{0}) ∩ (x_{0} − Ω) ⊂ U_{x_{0}}. Hence we conclude that G_{x_{0}} is compact.

Suppose now that K is a compact subgroup of G. We take a bounded open set M ⊂ Ω and consider the set

K · M = {gx|g ∈ K, x ∈ M},

which is invariant under all the transformations in K. Since K is compact, the set K · M is bounded. Its center of gravity x_{1} lies in Ω and is a fixed point for K. Thus K ⊂ G_{x_{1}} and therefore is conjugate to a subgroup K′ of G_{x_{0}}. □

Question. Let Ω be a convex hyperbolic homegeneous domain in U and G be a Lie subgroup of Aut(Ω) that acts transitively on Ω. Then

(1) Is the identity component G_{x_{0}} of G is algebraic?

(2) Does G have a decomposition as proposition 6(4)?

3. Algebra of a convex hyperbolic homogeneous domain

Let Ω be a convex hyperbolic homogeneous domain in an affine space U. By proposition 6(4), there exists a triangular affine subgroup G of Aut(Ω) such that acts simply transitvely on Ω.

Choose a point x_{0} in Ω such that ϕ(x_{0}) = 1. The map ev_{x_{0}} which is defined by ev_{x_{0}}(g) = g · x_{0} is an affine diffeomorphism from G onto Ω. With this evaluation map, we define the affine flat connection ∇ on G by

∇ = ev_{x_{0}}^{∗}D,

where D is the usual connection on Ω induced from the directional derivative on the affine space U. Since the connection ∇ is left invariant, it induces a multiplication on the Lie algebra 𝔤 = Lie(G):

X · Y = ∇_{X}Y.

This multiplication is compatible with Lie multiplication and left symmetric from the flat condition of connection.

Definition. The LSA A = (𝔤, ·) is called the algebra of the convex hyperbolic homogeneous domain Ω with respact to x_{0} ∈ Ω.

The characteristic function ϕ̃ on G is indeced from the characteristic function ϕ on Ω by the evaluation map: for any g ∈ G,

ϕ̃(g) = (ev_{x_{0}}^{∗}ϕ)(g) = ϕ(g · x_{0}) = \frac{1}{det(g)}ϕ(x_{0}) = \frac{1}{det(g)}.

Hence we have for any X ∈ 𝔤,

log ϕ̃(exp X) = log \frac{1}{det(exp X)} = −trX(= −trλ_{X}).

Now define the 1-form s(X) = trX. Clearly it is a Lie algebra homomorphism, hence we have that for X, Y ∈ 𝔤

s(XY) = s(Y X).

By the way, exp X = id + X + \frac{1}{2!}X^{2} + \frac{1}{3!}X^{3} + · · ·. Hence we have the following by the Taylor expansion:

log ϕ̃(exp X)

= ∇_{X} log ϕ̃(id) + \frac{1}{2!}[∇_{X^{2}} log ϕ̃(id) + ∇_{X}^{2} log ϕ̃(id)]

+[\frac{1}{3!}∇_{X^{3}} log ϕ̃(id) + \frac{1}{2!}∇_{X}∇_{X^{2}} log ϕ̃(id) + \frac{1}{2!}∇_{X^{2}}∇_{X} log ϕ̃(id) + \frac{1}{3!}∇_{X}^{3} log ϕ̃(id)]

+ · · · .

From this, we obtain

s(X) = −∇_{X} log ϕ̃(id)
s(X^{2}) = ∇_{X}^{2} log ϕ̃(id)
s(XY) = ∇_{X}∇_{Y} log ϕ̃(id)
s(X^{3}) = −2∇_{X}^{3} log ϕ̃(id)
⋮  ⋮  .

Definition. A LSA A is called a \underline{clan} if the operator of left multiplication have only real eigenvalue and there exists a Lie algebra homomorphism s : A^{−} → ℝ such that the induced bilinear form <, > defined by < x, y >= s(x · y) is positive definite.

Proposition 7 (1) The algebra of the convex hyperbolic homogeneous domain is a clan.

(2) The algebra of the convex hyperbolic homogeneous cone is a clan with identity.

(3) The algebra of the cone Ω̃ fitted onto the convex hyperbolic homogeneous domain Ω is obtained from the algebra of the domain Ω by adjunction of an identity element.

\xymatrix{𝔤 \ar[d]_{exp} \ar[r]^{−s} & ℜ \ar[d]^{exp} \\ G \ar[r]^{ϕ̃} \ar[d]_{ev_{x0}} & ℜ_{+} \ar[u]^{log} \\ Ω \ar[ur]^{ϕ}}

Conversely, from the clan A we obtain a convex hyperbolic homogeneous domain by evauating the Lie group G which is simply connected and A^{−} = Lie(G) with a point in the affine space U whose dimension is same as A.

Proposition 8 [2] A connected linear Lie group G(resp. linear Lie algebra 𝔤) is triangular if and only if every element of G(resp. 𝔤) has only real eigenvalues.

Proof. It is enough to show that the connected linear Lie group G whose elements have only real eigenvalues is solvable. If this were not the case, G would contain a nontrivial semisimple subgroup. Since every nontrivial semisimple linear group contains a nontrivial compact subgroup, G contains a nontrivial compact subgroup, which is impossible. □

At first, let A be a clan with an identity e and G be a simply connected Lie group such that its Lie algebra is A^{−}(= 𝔤). We consider the subset A_{0} of A defined by

A_{0} = {x ∈ A|s(x) = 0}.

Then A_{0}^{−} is an ideal of A^{−} since s([x, y]) = 0 for all x, y ∈ A^{−}. Hence the Lie algebra A^{−} is the direct sum of the Lie ideals, i.e.

A^{−} = A_{0}^{−} ⊕ {te}.

From this, we say that the Lie group G is a direct product of the normal subgroups:

G = G_{0} × {t · id}_{t>0},

where the Lie algebra of G_{0} is A_{0}^{−}.

Let Ω = G · e be the orbit of G in A as a vector space and S = G_{0} · e be the orbit of the normal subgroup G_{0} at e. Then we see that

Ω = ⋃_{t>0} tS.

We consider the tangent space T_{e}S, then

T_{e}S = {e + a|s(a) = 0} = {x|s(x) = s(e)}.

If a ∈ A_{0}, then

(exp a) · e = e + a + \frac{1}{2!}a^{2} + \frac{1}{3!}a^{3} + · · · .

From this we obtain the inequality: for surficiently small a ∈ A_{0},

s((exp a)e) ≥ s(e).

This says that S is convex at e and therefore it is convex at all points since S is homogeneous. Hence we conclude that the cone Ω is convex.

From the convecity of S, s(x) ≥ s(e) for all x ∈ S and s(tx) ≥ ts(e) > 0. This shows that for all x ∈ Ω

s(x) > 0.

We define the map from G to A^{∗} the dual vector space of A as the dual evaluation map at s ∈ A^{∗}: for all x ∈ A and g ∈ G

(g^{∗} · s)(x) = s(g · x).

Then the map is regular at identity and the differential of this map is an isomorphism from A^{−} onto A^{∗}. Hence G^{∗} · s fill a neiborhood of s and g^{∗}s is positive on Ω for all g ∈ G. From this we conclude that Ω is hyperbolic.

Now we consider the clan A which does not have an identity. By adjunction of an identity e, we obtain an algebra Ã from A:

Ã = A ⊕ {te}.

Then Ã is a clan:

(1) Ã is an LSA.

(2) The operator of left multiplication have only real eigenvalues.

Let s be a linear form in A such that it is symmetric and the induced 2-form < x, y >= s(xy) is positive definite. Let α be the maximum value of s on the unit sphere s(a^{2}) = 1. Then for all a ∈ A, we have

(s(a))^{2} ≤ α^{2}s(a^{2}).

Now extend the form s to Ã by putting

s(e) = 1 + α^{2}.

(3) s is symmetric in Ã.

(4) The induced 2-form is positive definite in Ã.

The simply connected Lie group G̃ whose Lie algebra Ã^{−} is decomposed of normal subgroups:

G̃ = G × {t id}_{t>0}

where G is the simply connected Lie group and the Lie algebra of G is A^{−}. Then since the orbit of G̃ is a convex hyperbolic cone Ω̃, the orbit of G, e + A ∩ Ω̃, is a convex hyperbolic domain in e + A. Hence we conclude that the orbit of G at 0, G · 0, is a convex hyperbolic domain in A.

Proposition 9 There is a one to one corespondence between the set of clans and the set of convex hyperbolic homogeneous domains. In particular, there is also a one to one corespondence between the set of clans with identity and the set of convex hyperbolic homogeneous cones.

Remark. { Clan’s }/∼⟺^{1−1} { Convex hyperbolic homogeneous domain’s}/ ∼

4. Decomposition of clan

Definition. A LSA A is called a \underline{clan} if the operator of left multiplication have only real eigenvalue and there exists a Lie algebra homomorphism s : A^{−} → ℝ such that the induced bilinear form <, > defined by < x, y >= s(x · y) is positive definite.

In this definition, since s is Lie algebra homomorphism, the bilinear form <, > is symmetric.

Let {e_{1}, ⋯, e_{n}} be an orthornormal basis with respect to <, > and put u =

∑_{i=1}^{n} s(e_{i})e_{i}, then we have

s(ux) = s(x) = s(xu),  (1)

for all x ∈ A. Furthermore, u is an unique element of A satisfying (1) because the bilinear form <, > is nondegenerate.

Proposition 10 (1) ρ_{u} is self-adjoint with respect to <, >.

(2) u is an idempotent, i.e. u · u = u. In this case u ∈ A is called principal idempotent.

(3) ρ_{u}^{2} = ρ_{u}.

Proof. (1) For any x, y ∈ A,

< x, ρ_{u}y > − < y, ρ_{u}x > = s(x(yu)) − s(y(xu))
= s([λ_{x}, λ_{y}](u))
= s(λ_{[x,y]}(u))
= s([x, y]u) = s([x, y]) = 0.

This proves.

(2) For any x ∈ A, we have

< uu, x >=< ρ_{u}(u), a >=< u, ρ_{u}(x) >= s(u(xu)) = s(xu) =< u, x > .

Since the bilinear form <, > is nondegenerate, we prove.

(3) Let P := ρ_{u} − ρ_{u}^{2} = ρ_{uu} − ρ_{u}^{2} = [λ_{u}, ρ_{u}]. Since ρ_{u} is self-adjoint, P is also. Moreover, P is commute with ρ_{u}. Now,

trP^{2} = tr[λ_{u}, ρ_{u}]P = tr(λ_{u}ρ_{u}P − ρ_{u}λ_{u}P)
= trλ_{u}ρ_{u}P − trρ_{u}λ_{u}P = trλ_{u}ρ_{u}P − trλ_{u}P ρ_{u}
= tr(λ_{u}ρ_{u}P − λ_{u}P ρ_{u}) = trλ_{u}[ρ_{u}, P]
= 0.

Thereofre we conclude that P = 0. □

Proposition 10(3) says that ρ_{u} is an orthogonal projector onto a subspace A_{0} ⊂ A, hence we have an orthogonal decomposition

A = A_{0} ⊕ K,

where K is the kernel of ρ_{u}. And it is called \underline{principal decomposition} of A.

Remark. If A has an identity e, then by the uniqueness of pricipal idempotent, we have u = e. Hence we conclude that A_{0} = A.

Proposition 11 (1) λ_{u} + λ_{u}^{∗} = ρ_{u} + id.

(2) λ_{u} = \frac{1}{2}(ρ_{u} + id), i.e. λ_{u} = id on A_{0} and λ_{u} = \frac{1}{2}id on K.

Proof. (1) For any x, y ∈ A, we have

< (λ_{u} + λ_{u}^{∗})(x), y > = < λ_{u}x + λ_{u}^{∗}x, y >=< ux, y > + < x, uy >
= s((ux)y + x(uy)) = s(u(xy) + (xu)y)
= s(xy) + s((xu)y) =< x, y > + < ρ_{u}x, y >
= < (ρ_{u} + id)(x), y > .

Hence by the nondegneracy of <, >, we prove.

(2) Put h = \frac{1}{2}(ρ_{u} + id) − λ_{u}, then h is skew-symmetric from the proposition 10. It means that h has only pure imaginary eigenvalues. However, we have [ρ_{u}, λ_{u}] = 0 from proposition 10, hence [ρ_{u} + id, λ_{u}] = 0, and ρ_{u} + id, λ_{u} have real eigen values. By the Lie Theorem, h has real eigen values. Therefore we conclude that h = 0. □

Lemma 1 Let A be a LSA and suppose that

λ_{z}x = t_{x}x, λ_{z}y = t_{y}y, ρ_{z}x = s_{x}x,

where x, y, z ∈ A and t_{x}, t_{y}, s_{x} ∈ ℝ. Then we have

λ_{z}(xy) = (t_{x} + t_{y} − s_{x})xy.

Proof.

λ_{z}(xy) = z(xy) = (zx)y + x(zy) − (xz)y
= t_{x}xy + t_{y}xy − s_{x}xy = (t_{x} + t_{y} − s_{x})xy

□

By applying the lemma, we have the followings:

λ_{u}(xy) = \{ \begin{matrix}xy & if x, y ∈ A_{0} \\ \frac{1}{2}xy & if x ∈ A_{0}, y ∈ K \\ \frac{3}{2}xy & if x ∈ K, y ∈ A_{0} \\ xy & if x, y ∈ K \end{matrix}

Hence the multiplcation table is given as followings:

\begin{matrix}· & A_{0} & K \\ A_{0} & A_{0} & K \\ K & 0 & A_{0} \end{matrix}  \begin{matrix}[,] & A_{0}^{−} & K^{−} \\ A_{0}^{−} & A_{0}^{−} & K^{−} \\ K^{−} & K^{−} & 0 \end{matrix}

In fact, the multiplication restricted to K is commutative since for any x, y ∈ K we have

xy = ρ_{u}(xy) = (xy)u = x(yu) + (yx)u − y(xu) = (yx)u = ρ_{u}(yx) = yx.

Hence we conculude that the associated Lie algebra K^{−} of K is abelian.

Corollary Let A be a clan and assume its associated Lie algebra A^{−} is unimodula, then the principal idempotent u is an identity of A.

Proof. From the principal decomposition, we have

\frac{1}{2} dim K = trλ_{u}|_{K} = trρ_{u}|_{K} = 0.

Hence dim K = 0, i.e. ρ_{u} = λ_{u} = id. □

Proposition 12 Let A be a clan with the identity e. Then there exists a subalgebra A_{1} and an idempotent v ∈ A such that

A = A_{1} + {tv},

and u = e − v is the principal idempotent of the clan A_{1}.

Proof. From the proposition 6(5), the cone Ω̃ obtained from A is isomorphic to the cone fitted onto a convex hyperbolic domain Ω. If A_{1} is the algebra of Ω, then A is obtained from A_{1} by adjunction of an identity. Let u be the principal idempotent of the clan A_{1}. We put v = e − u. Then the decomposition follows. □ Lemma 2 Let L be a unimodular Lie algebr and L_{0} be an ideal of L. Then L_{0} is also unimodular.

Proof. For any x ∈ L_{0}, trad_{x}|_{L_{0}} = trad_{x}. □

Corollary Let A be a clan and assume its associated Lie algebra A^{−} is unimodula, then there exist the set of idempotents {e_{1}, ⋯, e_{n}} where n is the dimension of A and satisfies

A = {t_{1}e_{1}} + ⋯ + {t_{n}e_{n}}.

Proposition 13 Let A be a clan with an identity. Then (1) A has a \underline{normal decomposition}, i.e.

A = ∑_{i≤j} A_{ij} (i, j = 1, ⋯, m)

where (i) A_{ii} is 1-dimensional and generated by idempotent e_{i} for all i = 1, ⋯, m.

(ii) the operaters λ_{i} = λ_{e_{i}}, ρ_{i} = ρ_{e_{i}} leave the space A_{jk} invariant with the following table

A_{ii}  A_{ji}(j ≠ i) A_{ik}(k ≠ i) A_{jk}(j, k ≠ i)
λ_{i}  1  \frac{1}{2}  \frac{1}{2}  0
ρ_{i}  1  1  0  0

(2) The multiplication law satisfies the followings:

A_{ij}A_{kl} = 0 ifj ≠ k, l
A_{ij}A_{jk} ⊂ A_{ik}
A_{ij}A_{kj} ⊂ A_{ik} or A_{ki}.

(3) The any idempotent v of A can be expressed in the form

v = e_{i_{1}} + ⋯ + e_{i_{k}},

where i_{1} < ⋯ < i_{k}.

5. Examples of clans

< 2-dimensional Case >

(1) dim A_{0} = 1, A_{0} =< e_{1} >, u = e_{1}

\begin{aligned}& e_{1} & e_{2} \\ e_{1} & e_{1} & \frac{1}{2}e_{2} \\ e_{2} & 0 & te_{1} \end{aligned}  \begin{matrix}[,] & e_{1} & e_{2} \\ e_{1} & 0 & \frac{1}{2}e_{2} \\ e_{2} & −\frac{1}{2}e_{2} & 0 \end{matrix}

For x = ae_{1} + be_{2}, λ_{x} = ( \begin{matrix}a & tb \\ 0 & \frac{1}{2}a \end{matrix} ) , ρ_{x} = ( \begin{matrix}a & tb \\ \frac{1}{2}b & 0 \end{matrix} ), p(x) = 1 + a − \frac{1}{2}tb^{2}

For x = a_{1}e_{1} + b_{1}e_{2}, y = a_{2}e_{1} + b_{2}e_{2}, xy = (a_{1}a_{2} + tb_{1}b_{2})e_{1} + \frac{1}{2}a_{1}b_{2}e_{2} s_{λ}(xy) = \frac{3}{2}(a_{1}a_{2} + tb_{1}b_{2}), s_{ρ}(xy) = a_{1}a_{2} + tb_{1}b_{2}

(2) dim A_{0} = 2, A_{0} = A, u = e_{1} + e_{2}, A_{11} =< e_{1} >, A_{22} =< e_{2} >

\begin{aligned}& e_{1} & e_{2} \\ e_{1} & e_{1} & 0 \\ e_{2} & 0 & e_{2} \end{aligned} , abelian

For x = ae_{1} + be_{2}

λ_{x} = ρ_{x} = ( \begin{matrix}a & 0 \\ 0 & b \end{matrix} ) , p(x) = (1 + a)(1 + b)

For x = a_{1}e_{1} + b_{1}e_{2}, y = a_{2}e_{1} + b_{2}e_{2}, xy = a_{1}a_{2}e_{1} + b_{1}b_{2}e_{2} s_{λ}(xy) = s_{ρ}(xy) = a_{1}a_{2} + b_{1}b_{2}

< 3-dimensional Case >

(1) dim A_{0} = 1, A_{0} =< e_{1} >, u = e_{1}

\begin{aligned}& e_{1} & e_{2} & e_{3} \\ e_{1} & e_{1} & \frac{1}{2}e_{2} & \frac{1}{2}e_{3} \\ e_{2} & 0 & re_{1} & se_{1} \\ e_{3} & 0 & se_{1} & te_{1} \end{aligned}  \begin{matrix}[,] & e_{1} & e_{2} & e_{3} \\ e_{1} & 0 & \frac{1}{2}e_{2} & \frac{1}{2}e_{3} \\ e_{2} & −\frac{1}{2}e_{2} & 0 & 0 \\ e_{3} & −\frac{1}{2}e_{3} & 0 & 0 \end{matrix}

For x = ae_{1} + be_{2} + ce_{3},

λ_{x} = ( \begin{matrix}a & rb + sc & sb + tc \\ 0 & \frac{1}{2}a & 0 \\ 0 & 0 & \frac{1}{2}a \end{matrix} ) , ρ_{x} = ( \begin{matrix}a & rb + sc & sb + tc \\ \frac{1}{2}b & 0 & 0 \\ \frac{1}{2}c & 0 & 0 \end{matrix} ),

p(x) = 1 + a − \frac{1}{2}(rb^{2} + 2sbc + tc^{2})

For x = a_{1}e_{1} + b_{1}e_{2} + c_{1}e_{3}, y = a_{2}e_{1} + b_{2}e_{2} + c_{2}e_{3}

s_{λ}(xy) = 2(a_{1}a_{2} + rb_{1}b_{2} + sb_{1}c_{2} + sc_{1}b_{2} + tc_{1}c_{2}) s_{ρ}(xy) = a_{1}a_{2} + rb_{1}b_{2} + sb_{1}c_{2} + sc_{1}b_{2} + tc_{1}c_{2}

rt − s^{2} ≠ 0 ⇒ nondegenerate r > 0, rt − s^{2} > 0 ⇒ positive definit (2) dim A_{0} = 2, A_{0} =< e_{1}, e_{2} >, u = e_{1} + e_{2}

\begin{aligned}& e_{1} & e_{2} & e_{3} \\ e_{1} & e_{1} & 0 & \frac{1}{2}e_{3} \\ e_{2} & 0 & e_{2} & 0 \\ e_{3} & 0 & 0 & se_{1} \end{aligned}  \begin{matrix}[,] & e_{1} & e_{2} & e_{3} \\ e_{1} & 0 & 0 & \frac{1}{2}e_{3} \\ e_{2} & 0 & 0 & 0 \\ e_{3} & −\frac{1}{2}e_{3} & 0 & 0 \end{matrix}

For x = ae_{1} + be_{2} + ce_{3}

λ_{x} = ( \begin{matrix}a & 0 & sc \\ 0 & b & 0 \\ 0 & 0 & \frac{1}{2}a \end{matrix} ) , ρ_{x} = ( \begin{matrix}a & 0 & sc \\ 0 & b & 0 \\ \frac{1}{2}c & 0 & 0 \end{matrix} ),

p(x) = (b + 1)(a + 1 − \frac{1}{2}sc^{2})

For x = a_{1}e_{1} + b_{1}e_{2} + c_{1}e_{3}, y = a_{2}e_{1} + b_{2}e_{2} + c_{2}e_{3}

s_{λ}(xy) = \frac{3}{2}(a_{1}a_{2} + sc_{1}c_{2}) + b_{1}b_{2}

s_{ρ}(xy) = a_{1}a_{2} + b_{1}b_{2} + sc_{1}c_{2}

if s = 0, then the inner product is degenerate.

\begin{aligned}& e_{1} & e_{2} & e_{3} \\ e_{1} & e_{1} & 0 & \frac{1}{2}e_{3} \\ e_{2} & 0 & e_{2} & \frac{1}{2}e_{3} \\ e_{3} & 0 & e_{3} & te_{1} \end{aligned}  \begin{matrix}[,] & e_{1} & e_{2} & e_{3} \\ e_{1} & 0 & 0 & \frac{1}{2}e_{3} \\ e_{2} & 0 & 0 & −\frac{1}{2}e_{3} \\ e_{3} & −\frac{1}{2}e_{3} & \frac{1}{2}e_{3} & 0 \end{matrix}

For x = ae_{1} + be_{2} + ce_{3},

λ_{x} = ( \begin{matrix}a & 0 & tc \\ 0 & b & 0 \\ 0 & c & \frac{1}{2}a + \frac{1}{2}b \end{matrix} ) , ρ_{x} = ( \begin{matrix}a & 0 & tc \\ 0 & b & 0 \\ \frac{1}{2}c & \frac{1}{2}c & b \end{matrix} )

p(x) = (1 + b)((1 + b)(1 + a) − \frac{1}{2}tc^{2})

For x = a_{1}e_{1} + b_{1}e_{2} + c_{1}e_{3}, y = a_{2}e_{1} + b_{2}e_{2} + c_{2}e_{3}

s_{λ}(xy) = \frac{3}{2}(a_{1}a_{2} + b_{1}b_{2} + tc_{1}c_{2})

s_{ρ}(xy) = a_{1}a_{2} + 2b_{1}b_{2} + tc_{1}c_{2} (4) dim A_{0} = 3, A_{0} = A, u = e_{1} + e_{2} + e_{3},

A_{11} =< e_{1} >, A_{22} =< e_{2} > A_{33} =< e_{3} >

\begin{aligned}& e_{1} & e_{2} & e_{3} \\ e_{1} & e_{1} & 0 & 0 \\ e_{2} & 0 & e_{2} & 0 \\ e_{3} & 0 & 0 & e_{3} \end{aligned} abelian

For x = ae_{1} + be_{2} + ce_{3},

λ_{x} = ρ_{x} = ( \begin{matrix}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix} ) , p(x) = (1 + a)(1 + b)(1 + c)

For x = a_{1}e_{1} + b_{1}e_{2} + c_{1}e_{3}, y = a_{2}e_{1} + b_{2}e_{2} + c_{2}e_{3},

s_{λ}(xy) = s_{ρ}(xy) = a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2}

References

[1] Jacques Faraut and Adam Korányi. Analysis on symmetric cones. The Clarendon Press Oxford University Press, New York, 1994. Oxford Science Publications.

[2] E. B. Vinberg. The Morozov-Borel theorem for real Lie groups. Soviet Math. Dokl., 2:1416–1419, 1961.

[3] E.B. Vinberg. The theory of convex homogeneous cones. Trans. of the Moscow Math. Soc., 12:340–403, 1963.