IN ITS LIE ALGEBRA

ADJOINT ORBIT TYPES OF COMPACT EXCEPTIONAL LIE GROUP G

IN ITS LIE ALGEBRA

Takashi MIYASAKA

Introduction

A Lie group G naturally acts on its Lie algebra g, called the adjoint action.

In this paper, we determine the orbit types of the compact exceptional Lie group G

in its Lie algebra g

. As results, the group G

has four orbit types in the Lie algebra g

as

G

/G

, G

/(U (1) × U(1)), G

/((Sp(1) × U(1))/Z

), G

/((U (1) × Sp(1))/Z

).

These orbits, especially the last two orbits, are not equivalent, that is, there exists no G

-equivariant homeomorphism among them. Finally, the author would like to thank Professor Ichiro Yokota for his earnest guidance, useful advice and constant encouragement.

0. Preliminaries and notation

(1) For a group G and an element s of G, es denotes the inner automor- phism induced by s:

es(g) = sgs

, g ∈ G,

then G

= {g ∈ G | sg = gs}. Hereafter G

is briefly written by G

. (2) For a transformation group G of a space M , the isotropy subgroup

of G at a point m ∈ M is denoted by G

: G

= {g ∈ G | gm = m}.

(3) As mentioned in the introduction, a Lie group G acts on its Lie algebra g. When G is a compact Lie group, any element X ∈ g is transformed to some element D of a fixed Cartan subalgebra h.

Hence, to determine the conjugate classes of isotropy subgroups G

, it suffices to consider the case of X = D ∈ h.

1. The Cayley algebra C and the group G

Let C be the division Cayley algebra with the canonical basis {e

= 1, e

, . . . , e

} and in C the conjugation x, the inner product (x, y) and the length |x| are naturally defined ([1], [3]). The Cayley algebra C contains

17

naturally the field R of real numbers, furthermore the field C of complex numbers and the field H of quaternion numbers:

R = {x = x1 | x ∈ R}, C = {x

+ x

e

| x

, x

∈ R},

H = {x

+ x

e

+ x

e

+ x

e

| x

, x

, x

, x

∈ R}.

The automorphism group G

of the Cayley algebra C:

G

= {α ∈ Iso

(C) | α(xy) = (αx)(αy)}

is the simply connected compact Lie group of type G

.

Any element x ∈ C can be expressed as x = a + be

, a, b ∈ H, and C is isomorphic to the algebra H ⊕ He

with multiplication

(a + be

)(c + de

) = (ac − db) + (bc + da)e

. We define an R-linear transformation γ of C by

γ(a + be

) = a − be

, a + be

∈ H ⊕ He

= C.

Next, to an element

x = a + m

e

+ m

e

+ m

e

, a, m

, m

, m

∈ C of C, we associate an element

a +



 m

m

m





of the algebra C ⊕ C

with the multiplication

(a + m)(b + n) = (ab − hm, ni) + (an + bm − m × n),

where hm, ni =

mn and m ×n ∈ C

is the exterior product of m, n. Note that C ⊕ C

is a left C-module. Hereafter we identify C with H ⊕ He

and C ⊕ C

:

C = H ⊕ He

, C = C ⊕ C

. We define an R-linear transformation γ

of C by

γ

(a + m) = a + m, a + m ∈ C ⊕ C

= C.

Then, γ, γ

∈ G

, γ

= γ

= 1 and γ, γ

are conjugate in G

([1]).

2. Subgroups (Sp(1) × Sp(1))/Z

and SU (3) of G

Proposition 1 ([3]). (G

)

∼ = (Sp(1)×Sp(1))/Z

, Z

= {(1, 1), (−1, −1)}.

Proof. Let Sp(1) = {p ∈ H | pp = 1} and we define a map ϕ : Sp(1) × Sp(1) → (G

)

by

ϕ(p, q)(a + be

) = qaq + (pbq)e

, a + be

∈ H ⊕ He

= C.

ϕ is well-defined: ϕ(p, q) ∈ (G

)

and ϕ is a homomorphism. We shall show that ϕ is onto. Let α ∈ (G

)

. Note that

C

= {x ∈ C | γx = x} = H, C

= {x ∈ C | γx = −x} = He

and these spaces are invariant under the action of the group (G

)

. Now, since α ∈ (G

)

induces an automorphism of C

= H, there exists q ∈ Sp(1) such that

αa = qaq, a ∈ H.

Let β = ϕ(1, q)

α. Then β ∈ (G

)

and β |H = 1. Since β induces an endomorphism of He

, there exists p ∈ H such that βe

= pe

. From

|p| = |pe

| = |βe

| = |e

| = 1, we see that p ∈ Sp(1). Then

β(a + be

) = βa + (βb)(βe

) = a + b(pe

) = a + (pb)e

= ϕ(p, 1)(a + be

).

Hence, β = ϕ(p, 1) and α = ϕ(1, q)ϕ(p, 1) = ϕ(p, q) which shows that ϕ is onto. ker ϕ = {(1, 1), (−1, −1)} = Z

. Thus, we have (G

)

∼ = (Sp(1) ×

Sp(1))/Z

. ¤

Proposition 2 ([1], [3]). (G

)

∼ = SU (3).

Proof. Let SU (3) = {A ∈ M(3, C) | A

A = E, det A = 1} and we define a map ψ : SU (3) → (G

)

by

ψ(A)(a + m) = a + Am, a + m ∈ C ⊕ C

= C.

ψ is well-defined: ψ(A) ∈ (G

)

. ψ is injective and a homomorphism. We shall show that ψ is onto. Let α ∈ (G

)

. Note that α induces a C-linear transformation of C

. Now let

αe

= a

, αe

= a

, αe

= a

and consider a matrix A = (a

, a

, a

) ∈ M(3, C). From (αe

)(αe

) = α(e

e

) = −α(e

), we have a

a

= −a

, namely, −ha

, a

i−a

× a

= −a

, then

ha

, a

i = 0, a

= a

× a

.

Similarly, we have ha

, a

i = ha

, a

i = 0. Moreover, from (αe

)(αe

) =

α(e

e

) = α( −1) = −1, we have ha

, a

i = 1, hence, A ∈ U(3) = {A ∈

M (3, C) | A

A = E }. Finally, det A = (a

, a

×a

) = (a

, a

) = ha

, a

i =

1 (where (a, b) is the usual inner product in C

: (a, b) =

ab). Hence, we

have A ∈ SU(3) and ψ(A) = α which shows that ψ is onto. Thus, we have

SU (3) ∼ = (G

)

. ¤

3. Adjoint orbit types of G

in the Lie algebra g

The Lie algebra g

of the group G

is given by

g

= {D ∈ Hom

(C) | D(xy) = (Dx)y + x(Dy)}.

The adjoint action of the group G

on the Lie algebra g

is given by µ : G

× g

→ g

, µ(α, D) = αDα

.

Now, we shall determine the adjoint orbit types of the group G

in g

. Since the group G

contains the subgroup SU (3) (Proposition 2), the Lie algebra g

also contains the subalgebra su(3) = {D ∈ M(3, C) | D

+ D = 0, tr(D) = 0 }. We choose a Cartan subalgebra h of su(3) as

h = (

D(r, s, t) =



 re

0 0 0 se

0 0 0 te



 ¯¯

¯¯ ¯ r, s, t ∈ R, r + s + t = 0 )

,

which is also a Cartan subalgebra of g

. The order of r, s, t of D(r, s, t) is arbitrarily exchanged by the action of G

.

Note that between the induced mappings ϕ

: sp(1) ⊕ sp(1) → g

and ψ

: su(3) → g

of ϕ and ψ, there exist the following relations:

ϕ

(e

, 0) = ψ

(diag(0, e

, −e

)), ϕ

(0, e

) = ψ

(diag(2e

, −e

, −e

)).

Theorem 3. The orbit types in g

through D(r, s, t) ∈ h under the adjoint action of the group G

are as follows:

(1) In the case: r = s = t = 0, the orbit through D(0, 0, 0) is G

/G

.

(2) In the case: r, s, t are non-zero and distinct, the orbit through D(r, s, t) is

G

/(U (1) × U(1)).

(3) In the case: r is non-zero, the orbit through D(2r, −r, −r) is G

/((Sp(1) × U(1))/Z

).

(4) In the case: r is non-zero, the orbit through D(0, r, −r) is G

/((U (1) × Sp(1))/Z

).

Proof. (1) trivial.

(2) Let U (1) = {a ∈ C | aa = 1} and S(U(1) × U(1) × U(1)) be the

diagonal subgroup of SU (3) and ψ : S(U (1) × U(1) × U(1)) → (G

)

be the restriction map ψ of Proposition 2. Then, ψ is well-defined and

injective. We shall show that ψ is onto. For this purpose, we first show that for α ∈ (G

)

, we have

αe

= ±e

.

Indeed, let αe

= a + m ∈ C ⊕ C

. From the condition αD(r, s, t) = D(r, s, t)α,

0 = α0 = αD(r, s, t)e

= D(r, s, t)αe

= D(r, s, t)(a + m) = D(r, s, t)m, we have m = 0. So αe

= a. Since |a| = |αe

| = |e

| = 1 and α1 = 1, we have αe

= ±e

. In the case of αe

= −e

, consider γ

∈ G

. Then αγ

e

= e

, so αγ

= ψ(A), A = (a

) ∈ SU(3) (Proposition 2). Hence, α = ψ(A)γ

. Then

Aγ

D(r, s, t)



 1 0 0



 = Aγ



 re

0 0



 = A



 −re

0 0



 =



 −ra

e

−ra

e

−ra

e



 .

On the other hand, D(r, s, t)Aγ



 1 0 0



 = D(r, s, t)A



 1 0 0



 = D(r, s, t)



 a

a

a



 =



 re

a

se

a

te

a



 .

From the condition αD(r, s, t) = D(r, s, t)α, we have

t

( −ra

e

, −ra

e

, −ra

e

) =

(re

a

, se

a

, te

a

).

But, it is easy to see that this is false. Hence, we have αe

= e

, so α = ψ(A), A ∈ SU(3). From the condition αD(r, s, t) = D(r, s, t)α again, we have α = ψ(A), A ∈ S(U(1) × U(1) × U(1)), which shows that ψ is onto. Thus we have (G

)

= S(U (1) × U(1) × U(1)) ∼ = U (1) × U(1).

(3) Since

exp(πD(2, −1, −1)) = exp(πψ

(diag(2e

, −e

, −e

)))

= exp(πϕ

(0, e

)) = γ,

if α ∈ G

commutes with D(2, −1, −1), then α also commutes with γ.

Hence, α ∈ (G

)

= ϕ(Sp(1) × Sp(1)) (Proposition 1). So there ex- ist p, q ∈ Sp(1) such that α = ϕ(p, q). Again from the commutativity with exp

³ π

2 D(2, −1, −1) ´

= exp

³ π

2 ψ

(diag(2e

, −e

, −e

))

´

= ϕ(1, e

),

we have ϕ(p, q)ϕ(1, e

) = ϕ(1, e

)ϕ(p, q), hence, ϕ(p, qe

) = ϕ(p, e

q). So

qe

= e

q, therefore q ∈ C ∩ Sp(1) = U(1). Conversely, α = ϕ(p, q)

(p ∈ Sp(1), q ∈ U(1)) commutes with ϕ(1, e

), t ∈ R, so α also com-

mutes with ϕ

(0, e

) = D(2, −1, −1). Thus, we have (G

)

=

ϕ(Sp(1) × U(1)) = (Sp(1) × U(1))/Z

.

(4)

exp(πD(0, 1, −1)) = exp(πψ

(diag(0, e

, −e

)))

= exp(π(ϕ

(e

, 0)) = γ.

Hence, (G

)

= (U (1) × Sp(1))/Z

is proved in a similar way to (3)

above. ¤

Proposition 4. If r and s are non-zero, then the groups (G

)

∼ = (U (1) × Sp(1))/Z

and (G

)

∼ = (Sp(1) × U(1))/Z

are not conju- gate in the group G

.

Proof. We shall prove that D(0, r, −r) and D(2s, −s, −s) are not conjugate under the action of the group G

. Suppose that there exists α ∈ G

such that

α(D(0, r, −r)) = (D(2s, −s, −s))α.

Let αe

= a

+ m ∈ C ⊕ C

, m =

(m

, m

, m

). Then, from α



 0 0 0

0 re

0 0 0 −re







 1 0 0



 =



 2se

0 0

0 −se

0

0 0 −se







a

+



 m

m

m







 ,

we have



 0 0 0



 =



 2sm

e

−sm

e

−sm

e



, hence, m

= m

= m

= 0, so

αe

= a

∈ C.

From (αe

)(αe

) = α(e

e

) = α( −1) = −1, we have a

a

= −1, hence, αe

= a

= ±e

. Similarly, we have αe

= ±e

. Then αe

= (αe

)(αe

) = ( ±e

)( ±e

) = ±1, which is a contradiction. ¤ Theorem 5. If r and s are non-zero, then the orbit spaces

X = {α(D(0, r, −r))α

| α ∈ G

} and Y ={α(D(2s, −s, −s))α

| α ∈ G

} are not equivalent.

Proof. Suppose that there exists a G

-equivariant homeomorphism h : X → Y . Then, there exists α ∈ G

such that

(1) h(D(0, r, −r)) = α(D(2s, −s, −s))α

. For any δ ∈ (G

)

, we have

δ(h(D(0, r, −r)))δ

= δ(α(D(2s, −s, −s))α

)δ

.

Since h is a G

-equivariant homeomorphism, we have e δh = he δ. Hence, we see

(2) h(D(0, r, −r)) = δ(α(D(2s, −s, −s))α

)δ

.

From (1) and (2), we have

δ(α(D(2s, −s, −s))α

)δ

= α(D(2s, −s, −s))α

,

that is, α

(δ(α(D(2s, −s, −s))α

)δ

)α = D(2s, −s, −s). This implies α

δα ∈ (G

)

. Thus, we have

α

((G

)

)α ⊂ (G

)

.

Since dim((G

)

) = dim((G

)

) (Theorem 3 (3) and (4)), the inclusion above must be equal, that is,

α

((G

)

)α = (G

)

.

This means that the groups (G

)

and (G

)

are conjugate

in G

, which contradicts Proposition 4. ¤

References

[1] I. Yokota, Realizations of involutive automorphisms σ and G^σ of exceptional linear Lie groups, Part I, G = G2, F4 and E6, Tsukuba J. Math. 14(1990), 185–223.

[2] I. Yokota, Simple Lie groups of classical type (in Japanese), Gendai-Sugakusya, 1990.

[3] I. Yokota, Simple Lie groups of exceptional type (in Japanese), Gendai-Sugakusya, 1992.

Takashi Miyasaka Takatoh High School Obara, Takatoh 396-0293, Japan

e-mail address: [email protected] (Received July 13, 2001 )