Journal of Lie Theory Volume 9 (1999) 293–304 ◯^{C}1999 Heldermann Verlag

Adjoint vector fields on the tangent space of semisimple symmetric spaces

T. Levasseur and R. Ushirobira

Communicated by F. Knop

Abstract. Let 𝔤 be a semisimple complex Lie algebra and ϑ ∈ Aut 𝔤 be an involution. If 𝔤 = 𝔨 ⊕ 𝔭 is the decomposition associated to ϑ, define a Lie subalgebra of End 𝔭 by 𝔨̃ = {X : ∀f ∈ S(𝔭^{∗})^{𝔨}, X.f = 0}. We prove that ad_{𝔭}(𝔨) = 𝔨̃ if, and only if, each irreducible factor of rank one of the symmetric pair (𝔤, 𝔨) is isomorphic to (𝔰𝔬(q + 1), 𝔰𝔬(q)).

0. Introduction

Let 𝔤 be a semisimple complex Lie algebra with adjoint group G. Let ϑ ∈ Aut(𝔤) be an involution and set 𝔨 = Ker(ϑ − I), 𝔭 = Ker(ϑ + I), hence 𝔤 = 𝔨 ⊕ 𝔭. The pair (𝔤, ϑ), or (𝔤, 𝔨), will be called a (semisimple) symmetric pair. Let Θ(𝔭) be the Lie algebra of (algebraic) vector fields on 𝔭. Thus Θ(𝔭) identifies with Der_{ℂ} O(𝔭), where O(𝔭) = S(𝔭^{∗}). There exists a Lie algebra homomorphism τ : 𝔤𝔩(𝔭) → Θ(𝔭) defined by (τ(X).f)(v) = \frac{d}{dt}_{|t=0}f(e^{−tX}.v) for v ∈ 𝔭, f ∈ O(𝔭) and X ∈ 𝔤𝔩(𝔭). This applies in particular to ad(X), X ∈ 𝔨, and we still set τ(X) = τ(ad(X)).

Let K be the connected algebraic subgroup of G such that Lie(K) = 𝔨. Recall, cf. [7], that

O(𝔭)^{K} = {f ∈ O(𝔭) : τ(𝔨).f = 0} = ℂ[u_{1}, …, u_{p}]

is a polynomial ring. Here, p is the rank of (𝔤, ϑ), i.e. the dimension of a Cartan subspace 𝔞 ⊂ 𝔭 for (𝔤, ϑ). One defines a Lie subalgebra of 𝔤𝔩(𝔭), containing ad(𝔨), by setting

𝔨̃ = {X ∈ 𝔤𝔩(𝔭) : τ(X).f = 0 for all f ∈ O(𝔭)^{K} }.

The Lie algebra 𝔨̃ has been considered by various authors (see, e.g., [8, 10]), in relation with the description of spherical hyperfunctions, or eigendistributions, on 𝔭. Observe that if 𝔰 ⊂ 𝔨 is an ideal of 𝔤, we have ad(𝔰) = 0 and ad(𝔨) = ad(𝔨/𝔰). We will therefore assume that 𝔨 does not contain a nonzero ideal of 𝔤. Then (𝔤, 𝔨) decomposes as a direct product ∏_{i=1}^{t}(𝔤^{i}, 𝔨^{i}) where each factor (𝔤^{i}, 𝔨^{i}) is irreducible, see [4, VIII.5].

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When p = 1, the invariant u_{1} is (up to a non-zero scalar) the nondegenerate quadratic form on 𝔭 induced by the Killing form B of 𝔤. Then, 𝔨̃ = 𝔰𝔬(𝔭, u_{1}) and 𝔨̃ ⊋ ad(𝔨), unless (𝔤, 𝔨) ≅ (𝔰𝔬(q + 1, ℂ), 𝔰𝔬(q, ℂ)). The main result of this note is the following theorem, which does not seem to have been noticed before.

Theorem. (Main Theorem) Let (𝔤, ϑ) be as above. Then ad(𝔨) = 𝔨̃ if, and only if, each irreducible factor of rank one of (𝔤, 𝔨) is isomorphic to (𝔰𝔬(q+1, ℂ), 𝔰𝔬(q, ℂ)).

The proof of the theorem goes as follows. Let K̃ be the connected algebraic subgroup of GL(𝔭) such that Lie(K̃) = 𝔨̃, we first prove that the representation (K̃ : 𝔭) is polar (see [1, 2]). Now, using the results of [1] one can suppose that there exists a semisimple symmetric pair (𝔤̃, ϑ̃) with associated decomposition 𝔤̃ = 𝔨̃⊕𝔭 and Cartan subspace 𝔞. Then, a case by case examination of the restricted root systems Σ(𝔤, 𝔞) and Σ(𝔤̃, 𝔞) enables us to conclude the proof.

Our interest in this theorem originates in the more general problem of describing the O(𝔭)-module of vector fields on 𝔭 which annihilate O(𝔭)^{K}. Set

ℰ = {d ∈ Θ(𝔭) : d.f = 0 for all f ∈ O(𝔭)^{K} }.

Then, E = O(𝔭)τ(𝔨) ⊂ Ẽ = O(𝔭)τ(𝔨̃) ⊂ ℰ and we conjecture that ℰ = O(𝔭)τ(𝔨̃). The equality ℰ = O(𝔭)τ(𝔨) was established by J. Dixmier [3] in the diagonal case, that is to say when 𝔤 = 𝔤_{1} × 𝔤_{1}, 𝔤_{1} semisimple, ϑ(x, y) = (y, x). It is not difficult to prove that the same conclusion holds when (𝔤, 𝔨) has maximal rank, i.e. p = rk 𝔤 (this is also a very particular case of the results in [13]). Furthermore, the modules E, Ẽ and ℰ are graded O(𝔭)-submodules of Θ(𝔭) whose degree zero parts are given by E_{0} = τ(𝔨), Ẽ_{0} = ℰ_{0} = τ(𝔨̃). Therefore, the Main Theorem indicates in which case one has E ⊊ Ẽ = O(𝔭)ℰ_{0}.

1. Generalities

We retain the notation of the introduction. Furthermore, we set n = dim 𝔭, ad(x).y = [x, y] and g.x = Ad(g).x for x, y ∈ 𝔤, g ∈ G. If V ⊂ 𝔤, we denote by V^{x} the subset of elements of V which commute with x.

By [7], dim 𝔭 − dim 𝔨 = dim 𝔭^{v} − dim 𝔨^{v} for all v ∈ 𝔭. Define the set of regular elements in 𝔭 by

𝔭^{reg} = {v ∈ 𝔭 : dim K.v = n − p} = {v ∈ 𝔭 : dim 𝔭^{v} = p} .

Then, cf. [7], one has p = min_{v∈𝔭} dim 𝔭^{v} = dim 𝔞 and max_{v∈𝔭} dim K.v = dim 𝔭 − p. One can write 𝔞 = 𝔭^{x} for a generic element x, i.e. x ∈ 𝔭^{reg} and x semisimple in 𝔤.

Recall (see [4, Proposition X.1.4] and [5, Lemma III.4.1]) that the symmetric pair (𝔤, ϑ) is the complexification of a real symmetric pair (𝔤_{0}, ϑ_{0}) where 𝔤_{0} = 𝔨_{0} ⊕𝔭_{0} is a Cartan decomposition of the real form 𝔤_{0} of 𝔤. Thus 𝔨_{0} is a compactly embedded subalgebra of 𝔤_{0} and the restriction of B to 𝔭_{0} is a 𝔨_{0}-invariant scalar product. We then have 𝔨 = 𝔨_{0} ⊗_{ℝ} ℂ, 𝔭 = 𝔭_{0} ⊗_{ℝ} ℂ, ϑ = ϑ_{0} ⊗_{ℝ} 1 and

S(𝔭_{0}^{∗})^{𝔨_{0}} ⊗_{ℝ} ℂ = S(𝔭^{∗})^{𝔨} = O(𝔭)^{K} = ℂ[u_{1}, …, u_{p}].

It follows that S(𝔭_{0}^{∗})^{𝔨_{0}} is a polynomial ring in p variables and that we may choose the generators u_{1}, …, u_{p} in S(𝔭_{0}^{∗}), the first invariant u_{1} being the nondegenerate quadratic form on 𝔭_{0} induced by the restriction of B. We have 𝔤𝔩(𝔭) = 𝔤𝔩(𝔭_{0}) ⊗_{ℝ} ℂ = 𝔤𝔩(𝔭_{0}) ⊕ i𝔤𝔩(𝔭_{0}) and, if X ∈ 𝔤𝔩(𝔭_{0}), the vector field τ(X) is a derivation of the polynomial ring S(𝔭_{0}^{∗}). Notice that 𝔰_{0} = {X ∈ 𝔤𝔩(𝔭_{0}) : τ(X).u_{1} = 0} is the orthogonal Lie algebra 𝔰𝔬(𝔭_{0}, u_{1}) ≅ 𝔰𝔬(n, ℝ).

Define [8, §4] a closed subgroup of GL(𝔭_{0}) by

K′_{0} = {g ∈ GL(𝔭_{0}) : g.u_{j} = u_{j} for all j = 1, …, p} .

Since K′_{0} ⊂ SO(𝔭_{0}, u_{1}), K′_{0} is a compact Lie group. Denote by K̃_{0} its identity component and set 𝔨̃_{0} = Lie(K′_{0}) = Lie(K̃_{0}). We have

𝔨̃_{0} = {X ∈ 𝔤𝔩(𝔭_{0}) : τ(X).f = 0 for all f ∈ S(𝔭_{0}^{∗})^{𝔨_{0}}}

and ad(𝔨_{0}) ⊂ 𝔨̃_{0} ⊂ 𝔰_{0}. Let K̃ = (K̃_{0})_{ℂ} ⊂ GL(𝔭) be the complexification of K̃_{0} (see [11, Chap. 5, Theorem 12]). Then, K̃ is a reductive algebraic group and is the unique connected reductive subgroup of GL(𝔭) such that Lie(K̃) = 𝔨̃_{0} ⊗_{ℝ} ℂ. One verifies easily that 𝔨̃ = 𝔨̃_{0} ⊗_{ℝ} ℂ. It will be convenient to denote the K̃_{0}-module 𝔭_{0} by 𝔭̃_{0}.

Recall that the pair (𝔤, 𝔨) is said to be irreducible if (𝔤_{0}, 𝔨_{0}) is irreducible in the following sense [5, VIII.5]: 𝔨_{0} does not contain a nonzero ideal of 𝔤_{0} and the K_{0}-module 𝔭_{0} is simple. Decompose (𝔤_{0}, 𝔨_{0}) as a finite direct sum of irreducible symmetric pairs (𝔤_{0}^{i}, 𝔨_{0}^{i}), 1 ⩽ i ⩽ t. We can then define, in a similar way, 𝔨̃_{0}^{i} ⊂ 𝔤𝔩(𝔭_{0}^{i}), K̃^{i} ⊂ GL(𝔭^{i}) etc., for each i = 1, …, t.

Lemma 1.1. We have 𝔨̃_{0} = 𝔨̃_{0}^{1} × ⋯ × 𝔨̃_{0}^{t} and K̃_{0} = K̃_{0}^{1} × ⋯ × K̃_{0}^{t}.

Proof. We write the proof for t = 2, the general case being similar. Let {e_{i}, x_{i} = e_{i}^{∗}}_{i} and {f_{i}, y_{i} = f_{i}^{∗}}_{i} be orthonormal coordinate systems (w.r.t. the Killing forms) on 𝔭_{0}^{1} and 𝔭_{0}^{2}. Thus, S(𝔭_{0}^{∗})^{𝔨_{0}} = S((𝔭_{0}^{1})^{∗})^{𝔨_{0}^{1}} ⊗_{ℝ} S((𝔭_{0}^{2})^{∗})^{𝔨_{0}^{2}}. Let X ∈ End 𝔭_{0} and write X = ( \begin{matrix}A & B \\ C & D \end{matrix} ) with A = [a_{ij}] ∈ End 𝔭_{0}^{1}, B = [b_{ij}] ∈ L(𝔭_{0}^{2}, 𝔭_{0}^{1}), C = [c_{ij}] ∈ L(𝔭_{0}^{1}, 𝔭_{0}^{2}), D = [d_{ij}] ∈ End 𝔭_{0}^{2}. Then,

τ(X) = ∑_{s}(A_{s}(x) + B_{s}(y))\frac{∂}{∂x_{s}} + ∑_{q}(C_{q}(x) + D_{q}(y))\frac{∂}{∂y_{q}}

where A_{s}(x) = − ∑_{u} a_{su}x_{u}, B_{s}(y) = − ∑_{u} b_{su}y_{u}, C_{q}(x) = − ∑_{u} c_{qu}x_{u}, D_{q}(y) = − ∑_{u} d_{qu}y_{u}. Suppose that X ∈ 𝔨̃_{0} and let f(x) ∈ S((𝔭_{0}^{1})^{∗})^{𝔨_{0}^{1}}. Then, from τ(X).f = 0 we deduce that

∑_{s} A_{s}(x)\frac{∂f(x)}{∂x_{s}} = − ∑_{s} B_{s}(y)\frac{∂f(x)}{∂x_{s}},

which forces ∑_{s} A_{s}(x)\frac{∂f(x)}{∂x_{s}} = ∑_{s} B_{s}(y)\frac{∂f(x)}{∂x_{s}} = 0. Similarly,

∑_{s} C_{s}(x)\frac{∂g(y)}{∂y_{s}} = ∑_{s} D_{s}(y)\frac{∂g(y)}{∂y_{s}} = 0

for all g(y) ∈ S((𝔭_{0}^{2})^{∗})^{𝔨_{0}^{2}}. Now, taking f(x) = ∑_{s} x_{s}^{2} we obtain ∑_{s} B_{s}(y)x_{s} = 0 and therefore B_{s}(y) = 0. Hence B = 0 and, similarly, C = 0 (use g(y) = ∑_{q} y_{q}^{2}). This proves that X = A × D with A ∈ 𝔨̃_{0}^{1}, D ∈ 𝔨̃_{0}^{2}. The second assertion follows easily.

Remark 1.2. The previous lemma shows that ad(𝔨_{0}) = 𝔨̃_{0} if and only if ad(𝔨_{0}^{i}) = 𝔨̃_{0}^{i} for all i. Therefore, to prove the theorem of the introduction, we may assume that the symmetric pair (𝔤, 𝔨) is irreducible.

Lemma 1.3. Suppose that (𝔤, 𝔨) is irreducible and p = 1. Then, ad(𝔨_{0}) = 𝔨̃_{0} if and only if (𝔤, 𝔨) is isomorphic to (𝔰𝔬(n + 1, ℂ), 𝔰𝔬(n, ℂ)).

Proof. Note that 𝔨_{0} \underrightarrow{∼} ad(𝔨_{0}) ⊂ 𝔨̃_{0} = 𝔰_{0} with 𝔰_{0} ⊗_{ℝ} ℂ ≅ 𝔰𝔬(n, ℂ). Assume that (𝔤, 𝔨) ≅ (𝔰𝔬(n+1, ℂ), 𝔰𝔬(n, ℂ)); then, dim 𝔨_{0} = dim_{ℂ} 𝔨 = dim 𝔰_{0} and therefore ad(𝔨_{0}) = 𝔰_{0}. Conversely, if ad(𝔨_{0}) = 𝔰_{0}, we obtain that 𝔨 ≅ 𝔰𝔬(n, ℂ) acting naturally on 𝔭 ≅ ℂ^{n}. It follows that (𝔤, 𝔨) ≅ (𝔰𝔬(n + 1, ℂ), 𝔰𝔬(n, ℂ)).

Recall (for completeness) the following lemma, cf. [8, Corollary 4.4] for a proof in the analytic case.

Lemma 1.4. Let V ⊂ 𝔭 be an affine open subset and f ∈ O(V), then

{∀ X ∈ 𝔨̃, τ(X).f = 0} ⟺ {∀ X ∈ 𝔨, τ(X).f = 0}.

In particular, O(𝔭)^{K} = O(𝔭)K̃ and S(𝔭_{0}^{∗})^{K_{0}} = S(𝔭_{0}^{∗})K̃_{0} .

Proof. Let X ∈ 𝔨̃ and let f ∈ O(V) be such that τ(𝔨).f = 0. By [9, Lemma 4.9] (or the proof of [8, Lemma 4.3]), there exists 0 ≠ ψ ∈ O(𝔭) such that ψτ(X) ∈ O(𝔭)τ(𝔨). Hence (ψτ(X)).f = 0, forcing τ(X).f = 0. The converse is obvious and the last assertions follow easily by taking V = 𝔭.

Corollary 1.5. Let v ∈ 𝔭_{0}. Then K_{0}.v = K̃_{0}.v.

Proof. By Lemma 1.4, the invariant functions u_{j} separate both the K_{0}-orbits and the K̃_{0}-orbits, see e.g. [12, (0.4)]. We clearly have K_{0}.v ⊂ K̃_{0}.v. Suppose that y ∈ K̃_{0}.v \ K_{0}.v. Since K_{0}.y ≠ K_{0}.v, we get that u_{j}(y) ≠ u_{j}(v) for some j. But this yields K̃_{0}.v ≠ K̃_{0}.y and a contradiction.

Let (L : E) be a finite dimensional representation of a compact group L. Fix an L-invariant scalar product B on E and set 𝔩 = Lie(L). Recall [1] that v ∈ E is said to be L-regular if dim L.v is maximal. The representation (L : E) is called polar if, whenever v, v′ ∈ E are regular, there exists k ∈ L such that 𝔞_{v} = k.𝔞_{v′}, where 𝔞_{v} is the orthogonal of 𝔩.v with respect to B. A subspace of the form 𝔞_{v}, v regular, is called a Cartan subspace for (L : E) and we define the rank of (L : E) to be rk(L : E) = dim 𝔞_{v}. We then have max_{v∈E} dim L.v = dim E − rk(L : E).

The representation (K_{0} : 𝔭_{0}) is known to be polar and is called a symmetric space representation, see [1]. In this case a Cartan subspace is provided by a maximal abelian Lie subalgebra 𝔞_{0} contained in 𝔭_{0}; then, 𝔞 = 𝔞_{0} ⊗_{ℝ}ℂ is a Cartan subspace for (𝔤, ϑ).

Proposition 1.6. The representation (K̃_{0} : 𝔭̃_{0}) is polar.

Proof. By Corollary 1.5, v_{0} ∈ 𝔭 is K_{0}-regular if and only if it is K̃_{0}-regular and we have 𝔨_{0}.v_{0} = 𝔨̃_{0}.v_{0}. Set 𝔞_{0} = 𝔞_{v_{0}} = (𝔨_{0}.v_{0})^{⊥}. Let v ∈ 𝔭_{0} be regular, we then have 𝔞_{0} = k.𝔞_{v} = k.(𝔨_{0}.v)^{⊥} = k.(𝔨̃_{0}.v)^{⊥} for some k ∈ K_{0}. This implies that (K̃_{0} : 𝔭̃_{0}) is polar with 𝔞_{0} as Cartan subspace.

We need to recall a few facts from the theory of symmetric spaces [4, VI.3]. Let 𝔞_{0} be a Cartan subspace for (K_{0} : 𝔭_{0}) and let λ ∈ 𝔞_{0}^{∗}. One sets:

𝔤_{0}^{λ} = {x ∈ 𝔤_{0} : [a, x] = λ(a)x for all a ∈ 𝔞_{0}}
Σ = {α ∈ 𝔞_{0}^{∗} : α ≠ 0 and 𝔤_{0}^{α} ≠ 0}
𝔪_{0} = 𝔤_{0}^{0} ∩ 𝔨_{0} = cent_{𝔨_{0}}(𝔞_{0})

Then Σ is a root system, possibly non reduced; we fix a choice Σ^{+} of positive roots. Define the reduced associated root system by

Σ_{red} = {λ ∈ Σ : λ ∉ 2Σ}.

(If Σ is reduced we have Σ_{red} = Σ; otherwise, in the irreducible case, Σ is of type (BC)_{p} and Σ_{red} ≅ B_{p}.)

If V is a real vector space we denote by V_{ℂ} its complexification and if 𝔩_{0} is a subspace of 𝔤_{0}, we set 𝔩 = (𝔩_{0})_{ℂ}. With this notation the decomposition 𝔤_{0} = ⊕_{λ∈Σ∪{0}}𝔤_{0}^{λ} yields

𝔤 = 𝔨 ⊕ 𝔭 = _{⨁λ∈Σ∪{0}}𝔤^{λ}
𝔪 = cent_{𝔨}(𝔞)

Recall that the multiplicity of λ ∈ Σ is m_{λ} = dim_{ℂ} 𝔤^{λ} = dim 𝔤_{0}^{λ}. Let λ ∈ Σ^{+} and set

𝔨_{0}^{λ} = {X ∈ 𝔨_{0} : ad(a)^{2}.X = λ(a)^{2}X for all a ∈ 𝔞_{0}}
𝔭_{0}^{λ} = {v ∈ 𝔭_{0} : ad(a)^{2}.v = λ(a)^{2}v for all a ∈ 𝔞_{0}}.

Then, 𝔨_{0} = 𝔪_{0} ⊕ (⊕_{λ∈Σ^{+}}𝔨_{0}^{λ}), 𝔭_{0} = 𝔞_{0} ⊕ (⊕_{λ∈Σ^{+}}𝔭_{0}^{λ}). Furthermore, see [5, III.4], 𝔤^{λ} ⊕ 𝔤^{−λ} = 𝔨^{λ} ⊕ 𝔭^{λ}. Let v ∈ 𝔞 be generic, i.e. λ(v) ≠ 0 for all λ ∈ Σ_{red}, then ad(v) induces an isomorphism 𝔭^{λ} \underrightarrow{∼} 𝔨^{λ}. It follows in particular that m_{λ} = dim 𝔤^{λ} = dim 𝔨^{λ} = dim 𝔭^{λ}.

Denote the set of generic elements in 𝔞 by

𝔞′ = {v ∈ 𝔞 : α(v) ≠ 0 for all α ∈ Σ}

and let 𝔞^{sing} = 𝔞 \ 𝔞′ be the set of singular elements. We recall, for completeness, the following lemma.

Lemma 1.7. Let x ∈ 𝔞. Then

(i) 𝔨^{x} = 𝔪 ⊕ (⊕_{{λ∈Σ^{+}:λ(x)=0}}𝔨^{λ})

(ii) x generic ⟺ 𝔨^{x} = 𝔪 ⟺ dim 𝔨^{x} is minimal ⟺ dim 𝔨^{x} = dim 𝔭−p.

Proof. (i) follows from 𝔨 = 𝔪 ⊕ (⊕_{λ∈Σ^{+}}𝔨^{λ}) and Ker ad(a)^{2} = Ker ad(a) for a ∈ 𝔞 (since a is semisimple).

(ii) is consequence of (i) and the definitions.

For α ∈ Σ_{red}^{+} we set 𝔞_{α} = Ker α = {a ∈ 𝔞 : α(a) = 0}. Therefore,

𝔞^{sing} = ⋃_{α∈Σ_{red}^{+}}𝔞_{α}  (1)

and the 𝔞_{α} are pairwise distinct hyperplanes. Set

𝔞′_{0} = 𝔞′ ∩ 𝔞_{0}, 𝔞_{0}^{sing} = 𝔞_{0} ∩ 𝔞^{sing}, 𝔞_{0,α} = 𝔞_{0} ∩ 𝔞_{α}.

Since dim K_{0}.x = dim_{ℂ} K.x for all x ∈ 𝔞_{0}, it follows from Lemma 1.7 that 𝔞′_{0} is the set of regular elements in 𝔞_{0}.

2. Proof of ad(𝔨) = 𝔨̃

We continue with the notation of the previous sections. Recall that the proof of the Main Theorem reduces to the case when (𝔤_{0}, 𝔨_{0}) is irreducible, see Remark 1.2. From now on, we assume that this hypothesis holds. Since ad : 𝔨_{0} → 𝔤𝔩(𝔭_{0}) is injective, we will identify 𝔨_{0} with the Lie subalgebra ad(𝔨_{0}) of 𝔨̃_{0}, therefore 𝔨 is identified with ad(𝔨). Note that the representations (K_{0} : 𝔭_{0}) and (K̃_{0} : 𝔭̃_{0}) are irreducible and faithful.

From the classification of irreducible polar representations one can deduce the following result, see [1, Theorem 9, Theorem 10 and Proposition 6].

Proposition 2.1. Let (L_{0} : V_{0}) be an irreducible faithful polar representation of a compact Lie group L_{0}. Then, there exists a semisimple symmetric pair (\overline{𝔤}_{0}, \overline{𝔨}_{0}) such that (with obvious notation):

(i) \overline{𝔤}_{0} = \overline{𝔨}_{0} ⊕ V_{0} is the associated Cartan decomposition;

(ii) L_{0} ⊂ \overline{K}_{0} and (L_{0} : V_{0}) is the restriction of (\overline{K}_{0} : V_{0});

(iii) S(V_{0}^{∗})^{\overline{K}^{0}} = S(V_{0}^{∗})^{L_{0}} .

Corollary 2.2. The representation (K̃_{0} : 𝔭̃_{0}) is an irreducible symmetric space representation.

Proof. By Proposition 1.6 and Proposition 2.1, there exists a semisimple symmetric pair (\overline{𝔤}_{0}, \overline{𝔨}_{0}) such that 𝔭_{0} = 𝔭̃_{0} = \overline{𝔭}_{0} (as vector spaces), 𝔨_{0} ⊂ 𝔨̃_{0} ⊂ \overline{𝔨}_{0} and S(𝔭_{0}^{∗})^{K_{0}} = S(𝔭_{0}^{∗})^{\overline{K}^{0}} . It follows then from the definition of 𝔨̃_{0} that 𝔨̃_{0} = \overline{𝔨}_{0}.

Remark. B. Kostant has informed us that Corollary 2.2 can also be deduced from the results contained in [6].

From the previous corollary we may suppose now that (K̃_{0} : 𝔭̃_{0}) is coming from a semisimple symmetric pair (𝔤̃_{0}, 𝔨̃_{0}). Without lost of generality we can assume that 𝔤̃_{0} has Cartan decomposition 𝔤̃_{0} = 𝔨̃_{0} ⊕ 𝔭̃_{0} and that, if [ , ]^{∼} is the bracket on 𝔤̃_{0}, [X, v] = [X, v]^{∼}, [X, Y] = [X, Y]^{∼} for all X, Y ∈ 𝔨_{0}, v ∈ 𝔭_{0} = 𝔭̃_{0}. Notice that if 𝔩_{0} ⊂ 𝔨̃_{0} ⊂ End 𝔭̃_{0} is an ideal of 𝔤̃_{0}, then 𝔩_{0}.𝔭̃_{0} = [𝔩_{0}, 𝔭̃_{0}]^{∼} ⊂ 𝔨̃_{0} ∩ 𝔭̃_{0} = 0 and therefore 𝔩_{0} = 0. Thus the symmetric pair (𝔤̃_{0}, 𝔨̃_{0}) is also irreducible. Recall that we have fixed the Cartan subspace 𝔞_{0} and that we can take 𝔞̃_{0} = 𝔞_{0} as Cartan subspace for (𝔤̃_{0}, 𝔨̃_{0}), see Proposition 1.6. The associated Weyl groups will be denoted by W and W̃.

The notation given in §1 for 𝔨_{0}, 𝔭_{0}, 𝔞_{0}, 𝔤_{0}, etc. can be introduced for 𝔨̃_{0}, 𝔭̃_{0}, 𝔞̃_{0}, 𝔤̃_{0}, etc. If an object x is defined relatively to (𝔤_{0}, 𝔨_{0}) we denote by x̃ the corresponding one, relatively to (𝔤̃_{0}, 𝔨̃_{0}). Since there is only one degree two invariant in S(𝔭_{0}^{∗})^{K_{0}} = S(𝔭̃_{0}^{∗})K̃_{0} , the scalar product B on 𝔭_{0} is a positive scalar multiple of the scalar product B̃ on 𝔭̃_{0} and we will suppose in the sequel that they are actually equal.

Proposition 2.3. (1) There exists a bijection t : Σ_{red}^{+} → Σ̃_{red}^{+}, α ↦ α̃, such that 𝔞_{0,α} = 𝔞̃_{0,α̃}.

(2) W = W̃.

(3) Let α ∈ Σ_{red}^{+} and w ∈ W be such that w.α ∈ Σ_{red}^{+}. Then t(w.α) = ±w.t(α).

(4) There exist c_{1}, c_{2} ∈ ℝ^{∗} such that

α̃ = \begin{cases}±c_{1}α & if α short, \\ ±c_{2}α & if α long. \end{cases}

Proof. (1) By Corollary 1.5 we have 𝔞_{0}^{sing} = 𝔞̃_{0}^{sing} , hence we get from (1):

⋃_{α∈Σ_{red}^{+}} 𝔞_{0,α} = ⋃_{β∈Σ̃_{red}^{+}} 𝔞̃_{0,β}

Since the hyperplanes occuring in each side of the previous equality are pairwise distinct, we obtain that

∀ α ∈ Σ_{red}^{+}, ∃! t(α) ∈ Σ̃_{red}^{+}, 𝔞_{0,α} = 𝔞̃_{0,t(α)}.

It is then clear that α ↦ t(α) = α̃ gives the required bijection. Notice that Ker α = Ker α̃ (in 𝔞_{0}) implies that α̃ = c_{α}α for some c_{α} ∈ ℝ^{∗}.

(2) Recall that W is generated by the reflections r_{α}, α ∈ Σ_{red}^{+}, and that the reflecting hyperplane of r_{α} is 𝔞_{0,α}. Thus r_{α} = rα̃ and it follows that W = W̃.

(3) We have Ker w.α = w(Ker α), thus w(𝔞_{0,α}) = w(𝔞̃_{0,α̃}) is equivalent to Ker w.α = Ker w.α̃. Let ϵ = ±1 such that ϵw.α̃ ∈ Σ̃_{red}^{+}. Then Ker w.α̃ = 𝔞̃_{0,ϵw.α̃} = 𝔞_{0,w.α} and, by definition of t(w.α), we obtain that t(w.α) = ϵw.α̃.

(4) Let α, β ∈ Σ_{red}^{+} having the same length and w ∈ W be such that β = w.α. By (3), β̃ = ±w.α̃ and, therefore, β̃ = c_{β}β = ±c_{α}w.α = ±c_{α}β. Hence c_{β} = ±c_{α}. The assertion then follows easily (with the convention that all the roots are short when there is only one root length in Σ).

Corollary 2.4. (1) If Σ_{red} ∉ {B_{p}, C_{p}}, then Σ_{red} and Σ̃_{red} are of the same type.

(2) If Σ_{red} ∈ {B_{p}, C_{p}}, then Σ̃_{red} ∈ {B_{p}, C_{p}}.

Proof. Recall that the Weyl group distinguishes irreducible root systems which are not of type B_{p} or C_{p} and that the Weyl groups of B_{p} and C_{p} are the same. The claims are therefore consequences of Proposition 2.3(2).

Observe that it could happen that Σ_{red} ≅ B_{p} and Σ̃_{red} ≅ C_{p}, the bijection t being given by t(α) = 2α, α short, t(α) = α, α long. (Similarly, Σ_{red} ≅ C_{p} and Σ̃_{red} ≅ B_{p} could occur.) In case Σ = Σ_{red} ≅ F_{4} (resp. G_{2}) we must have Σ̃ ≅ F_{4} (resp. G_{2}) but it possible that t interchanges the short and long roots. In summary, we have the following possibilities for the pair (Σ_{red}, Σ̃_{red}):

• (A_{p}, A_{p}), (D_{p}, D_{p}), (E_{p}, E_{p});

• (F_{4}, F_{4}), (G_{2}, G_{2});

• (B_{p}, B_{p}), (C_{p}, C_{p}), (B_{p}, C_{p}), (C_{p}, B_{p}).

For all λ ∈ Σ_{red}^{+} we set 𝔪_{λ} = cent_{𝔨}(𝔞_{λ}) = {x ∈ 𝔨 : [x, 𝔞_{λ}] = 0}. If, similarly, 𝔪̃λ̃ = cent𝔨̃(𝔞̃λ̃) we obtain from 𝔞_{λ} = 𝔞̃λ̃ that

𝔪_{λ} = 𝔪̃λ̃ ∩ 𝔨.  (2)

The Lie algebra 𝔪_{λ} is described by the following well known lemma.

Lemma 2.5. Let λ ∈ Σ_{red}^{+}. Then, 𝔪_{λ} = 𝔪 ⊕ 𝔨^{λ} ⊕ 𝔨^{2λ} (with the convention that 𝔨^{2λ} = 0 if 2λ ∉ Σ).

Proof. Let X ∈ 𝔨 and set X = X_{0} + ∑_{α∈Σ^{+}} X_{α}, X_{0} ∈ 𝔪, X_{α} ∈ 𝔨^{α}. Thus X ∈ 𝔪_{λ} if and only if ∑_{α∈Σ^{+}}[a, X_{α}] = 0 for all a ∈ 𝔞_{λ}. But, since [a, X_{α}] ∈ 𝔭^{α}, this is equivalent to [a, X_{α}] = 0 for all α ∈ Σ^{+} and a ∈ 𝔞_{λ}. Hence,

X ∈ 𝔪_{λ} ⟺ ∀α ∈ Σ^{+}, ∀a ∈ 𝔞_{λ}, X_{α} ∈ Ker ad(a) = Ker ad(a)^{2}
⟺ ∀α ∈ Σ^{+}, ∀a ∈ 𝔞_{λ}, α(a) = 0 or X_{α} = 0.

Therefore, if X_{α} ≠ 0, 𝔞_{λ} = Ker λ ⊂ Ker α; thus Ker λ = Ker α and α = λ or 2λ. Conversely, if X ∈ 𝔨^{λ} or 𝔨^{2λ} we have X ∈ Ker ad(a)^{2} = Ker ad(a) for all a ∈ 𝔞_{λ}. Hence X ∈ cent_{𝔨}(𝔞_{λ}).

Let λ ∈ Σ_{red}^{+}; set
𝔰_{λ} = 𝔨^{λ} ⊕ 𝔨^{2λ}, s_{λ} = dim 𝔰_{λ} = m_{λ} + m_{2λ}

(with m_{2λ} = 0 if 2λ ∉ Σ). Notice that s_{λ} = dim(𝔭^{λ} ⊕ 𝔭^{2λ}).

Lemma 2.6. One has s_{λ} = s̃λ̃ for all λ ∈ Σ_{red}^{+}.

Proof. It follows from Lemma 2.5 and (2) that 𝔪 ⊕ 𝔰_{λ} ⊂ 𝔪̃ ⊕ 𝔰̃λ̃. Let ϕ : 𝔪̃λ̃ ↠ 𝔰̃λ̃ be the projection afforded by the decomposition 𝔪̃λ̃ = 𝔪̃ ⊕ 𝔰̃λ̃. By composing ϕ with the inclusions 𝔰_{λ} ↪ 𝔪_{λ} ↪ 𝔪̃λ̃, we obtain a linear map φ : 𝔰_{λ} → 𝔰̃λ̃. Suppose that φ(x) = 0, then x ∈ 𝔪̃ ∩ 𝔰_{λ} = 𝔪̃ ∩ 𝔨 ∩ 𝔰_{λ} = 𝔪 ∩ 𝔰_{λ} = 0. Thus φ is injective and, consequently, s_{λ} ⩽ s̃λ̃. Now, recall that

𝔭 = 𝔭̃ = 𝔞 ⊕ (⊕_{λ∈Σ_{red}^{+}}𝔭^{λ} ⊕ 𝔭^{2λ}) = 𝔞 ⊕ (⊕_{λ̃∈Σ̃_{red}^{+}}𝔭̃λ̃ ⊕ 𝔭̃^{2λ̃}).

Therefore ∑_{λ∈Σ_{red}^{+}} s_{λ} = ∑_{λ̃∈Σ̃_{red}^{+}} s̃λ̃ and, since s_{λ} ⩽ s̃λ̃, we obtain that s_{λ} = s̃λ̃ for all λ ∈ Σ_{red}^{+}.

Remark. One has 𝔭^{λ} ⊕ 𝔭^{2λ} = 𝔭̃λ̃ ⊕ 𝔭̃^{2λ̃} for all λ ∈ Σ_{red}^{+}. This can be shown as follows. Let v ∈ 𝔞′, then ad(v) induces an isomorphism 𝔨^{α} \underrightarrow{∼} 𝔭^{α} for all α ∈ Σ^{+}. Recall that if X ∈ 𝔨, [v, X]^{∼} = [v, X]. Thus ad𝔤̃(v) restricted to 𝔪_{λ} coincides with ad(v). It follows that

𝔭^{λ} ⊕ 𝔭^{2λ} = ad𝔤̃(v).𝔰_{λ} ⊂ ad𝔤̃(v).(𝔪̃ ⊕ 𝔰̃λ̃) = 𝔭̃λ̃ ⊕ 𝔭̃^{2λ̃}.

Since s_{λ} = s̃λ̃, we get that 𝔭^{λ} ⊕ 𝔭^{2λ} = 𝔭̃λ̃ ⊕ 𝔭̃^{2λ̃}.

We now set:

s_{1} = s_{λ} if λ ∈ Σ_{red}^{+} is short,
s_{2} = s_{λ} if λ ∈ Σ_{red}^{+} is long,
s_{2} = 0 if all λ ∈ Σ^{+} are short.

(3)

Hence, we can associate to the Lie algebra 𝔤_{0} two ordered pairs (s_{1}, s_{2}) and (s_{2}, s_{1}). It is shown in Appendix A that these pairs almost determine 𝔤_{0}. A similar definition holds for the pair 𝔤̃_{0} and gives the pairs (s̃_{1}, s̃_{2}), (s̃_{2}, s̃_{1}). We now compare the s_{i} and s̃_{j}.

Lemma 2.7. (1) Assume that Σ is simply laced. Then, (s_{1}, s_{2}) = (s̃_{1}, s̃_{2}). (2) Assume that Σ has two root lengths. Then,

(s_{1}, s_{2}) = \begin{cases}(s̃_{1}, s̃_{2}) or (s̃_{2}, s̃_{1}) & if (Σ_{red}, Σ̃_{red}) = (F_{4}, F_{4}), (G_{2}, G_{2}), (B_{2}, B_{2}), \\ (s̃_{1}, s̃_{2}) & if (Σ_{red}, Σ̃_{red}) = (B_{p}, B_{p}), (C_{p}, C_{p}), p ⩾ 3, \\ (s̃_{2}, s̃_{1}) & if (Σ_{red}, Σ̃_{red}) = (B_{p}, C_{p}), (C_{p}, B_{p}), p ⩾ 3. \end{cases}

Proof. Observe first that s_{α} = s_{β} if α, β have the same length; then Lemma 2.6 yields s̃α̃ = s̃β̃ = s̃_{1} or s̃_{2}, depending on the length of α̃.

(1) is clear.

(2) Recall that if Σ_{red} has two root lengths, then the number of short roots is equal to the number of long roots if, and only if, Σ_{red} is of type B_{2} = C_{2}, F_{4} or G_{2}. The assertion then follows from Lemma 2.6 and Proposition 2.3(4).

Theorem 2.8. Assume that p ⩾ 2. Then, 𝔤_{0} ≅ 𝔤̃_{0} and, therefore, 𝔨_{0} = 𝔨̃_{0}.

Proof. By Corollary 2.4 and Lemma 2.7, the hypothesis (h.j), j = 1, …, 4, of Appendix A hold. Thus, by Theorem 2.9, if 𝔤_{0} ≅ ˜𝔤̸_{0} we are in one of the following cases. Case 1: Diagonal case with Σ, Σ̃ ∈ {B_{p}, C_{p}}. Then, dim 𝔨_{0} = dim 𝔤_{0} = dim 𝔤̃_{0} = dim 𝔨̃_{0} and 𝔨_{0} ⊂ 𝔨̃_{0} force 𝔨_{0} = 𝔨̃_{0} and, consequently, 𝔤_{0} ≅ 𝔤̃_{0}. Case 2: 𝔤_{0} and 𝔤̃_{0} are of type BI(p, p + 1) or CI(p). This implies that 𝔨_{0} and 𝔨̃_{0} are isomorphic to 𝔰𝔬(p)×𝔰𝔬(p+1) or 𝔲(p), which are both of dimension p^{2}. Since 𝔨_{0} ⊂ 𝔨̃_{0}, this implies 𝔨_{0} = 𝔨̃_{0}. But 𝔰𝔬(p) × 𝔰𝔬(p + 1) ≅ 𝔲(p) only happens when p = 2 (see [4, p. 519]), in which case 𝔤_{0} ≅ 𝔤̃_{0} ≅ 𝔰𝔬(2, 3).

Proof of the Main Theorem. As noticed in Remark 1.2, we may assume that (𝔤, 𝔨) is irreducible. Now, the assertion follows from Lemma 1.3 if (𝔤, 𝔨) has rank one and from Theorem 2.8 if this rank is ⩾ 2.

A. Appendix

Let 𝔤_{0} be a real semisimple Lie algebra. We adopt the notation of §§1 and 2. In particular, we fix a Cartan decomposition 𝔤_{0} = 𝔨_{0} ⊕ 𝔭_{0} and a Cartan subspace 𝔞_{0} ⊂ 𝔭_{0} of dimension p. Let 𝔤̃_{0} be another semisimple Lie algebra with Cartan decomposition 𝔤̃_{0} = 𝔨̃_{0} ⊕𝔭̃_{0}. Any object x defined relatively to 𝔤_{0} has an analogue for 𝔤̃_{0} and it will be denoted by x̃.

We will assume that the pairs (𝔤_{0}, 𝔨_{0}) and (𝔤̃_{0}, 𝔨̃_{0}) are both irreducible and that the following hypothesis hold.

(h.1) p ⩾ 2.

(h.2) Σ_{red} ∈ {B_{p}, C_{p}} if, and only if, Σ̃_{red} ∈ {B_{p}, C_{p}}.

(h.3) Σ_{red} ≅ Σ̃_{red} when Σ_{red} is not of type B_{p} or C_{p}.

(h.4) The pairs (s_{1}, s_{2}), (s̃_{1}, s̃_{2}) being defined as in (3), one has

(s_{1}, s_{2}) = \begin{cases}(s̃_{1}, s̃_{2}) & if Σ is simply laced, \\ (s̃_{1}, s̃_{2}) or (s̃_{2}, s̃_{1}) & if (Σ_{red}, Σ̃_{red}) = (F_{4}, F_{4}), (G_{2}, G_{2}), (B_{2}, B_{2}), \\ (s̃_{1}, s̃_{2}) & if (Σ_{red}, Σ̃_{red}) = (B_{p}, B_{p}), (C_{p}, C_{p}), p ⩾ 3, \\ (s̃_{2}, s̃_{1}) & if (Σ_{red}, Σ̃_{red}) = (B_{p}, C_{p}), (C_{p}, B_{p}), p ⩾ 3. \end{cases}

Observe that the hypothesis are symmetric in 𝔤_{0} and 𝔤̃_{0}.

The notation for the classification of irreducible symmetric pairs, i.e. of semisimple real Lie algebras, will be (almost) as in [4, X.6]; in particular, we adopt the notation of [4, pp. 532-534]. For instance, if 𝔤_{0} = 𝔰𝔬(p, q), 𝔨_{0} = 𝔰𝔬(p) × 𝔰𝔬(q), p ⩽ q, p + q even, we say that 𝔤_{0} is of type DI(p, q).

Suppose that 𝔤_{0} = 𝔤_{1}^{ℝ} for some complex simple Lie algebra 𝔤_{1}. Define an involution ϑ on 𝔤 = 𝔤_{0} ⊗_{ℝ} ℂ ≅ 𝔤_{1} × 𝔤_{1} by ϑ(x, y) = (y, x). Then the symmetric pair (𝔤, 𝔨) is isomorphic to (𝔤_{1} × 𝔤_{1}, 𝔤_{1}). This case will be called the diagonal case and (𝔤, ϑ) is said to be of diagonal type.

Theorem 2.9. Up to symmetry between 𝔤_{0} and 𝔤̃_{0}, the following (exclusive) possibilities hold.

(i) 𝔤_{0} ≅ 𝔤̃_{0}.

(ii) (𝔤, ϑ) and (𝔤̃, ϑ̃) are of diagonal type, Σ ≅ B_{p}, Σ̃ ≅ C_{p}.

(iii) 𝔤_{0} is of type BI(p, p+1), 𝔤̃_{0} is of type CI(p), p ⩾ 3 (thus 𝔨_{0} ≅ 𝔰𝔬(p)×𝔰𝔬(p+ 1), 𝔨̃_{0} ≅ 𝔲(p)).

Proof. The proof is a case by case analysis using [4, X, Table VI]: One computes the pairs (s_{1}, s_{2}) for each type of irreducible symmetric pair (𝔤_{0}, 𝔨_{0}) and, then, one notes that the hypothesis (h.i), i = 1, …, 4, yield the desired result. We will simply make a few remarks in order to explain the method and the appearance of cases (i), (ii), (iii).

If (𝔤, ϑ) is of diagonal type with 𝔤 ≅ 𝔤_{1} × 𝔤_{1}, 𝔤_{1} complex simple of type T_{p} (T = A, B, C, D, E, F, G), then Σ ≅ T_{p} and (s_{1}, s_{2}) = (2, 0) or (2, 2). Then, the (h.i)’s show that only cases (i) or (ii) may occur.

If 𝔤_{0} of type AIII(p, p), then (s_{1}, s_{2}) = (2, 1), Σ ≅ C_{p}. The only possibility for 𝔤̃_{0} and (s̃_{1}, s̃_{2}) = (s_{1}, s_{2}) or (s_{2}, s_{1}) occurs when 𝔤̃_{0} is of type DI(p, p + 2). In this case Σ̃ ≅ B_{p}. When p = 2 we find the isomorphism DI(2, 2 + 2) ≅ AIII(2, 2), see [4, p. 519]. When p ⩾ 3, the hypothesis (h.4) forces (s_{1}, s_{2}) = (2, 1) = (s̃_{2}, s̃_{1}) = (1, 2), hence a contradiction.

If 𝔤_{0} is of type BI(p, 2ℓ + 1 − p), then (s_{1}, s_{2}) = (2ℓ − 2p + 1, 1), Σ ≅ B_{p}. From s_{2} = 1 and s_{1} odd, it follows that the only possibility for 𝔤̃_{0} may occur in type CI(p), where (s̃_{2}, s̃_{1}) = (1, 1), Σ ≅ C_{p}. But this forces 2ℓ − 2p + 1 = 1, i.e. ℓ = p. Recalling that BI(2, 3) ≅ CI(2), see [4, p. 519], this yields case (iii).

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Département de Mathématiques Université de Poitiers F-86022 Poitiers, France levasseu@mathlabo.univ-poitiers.fr Département de Mathématiques Université de Poitiers F-86022 Poitiers, France rosane@mathlabo.univ-poitiers.fr

Received December 17, 1997