SYMMETRIC SPACES AND STRONGLY ISOTROPY

IRREDUCIBLE SPACES

McKenzie Wang and Wolfgang Ziller

A strongly isotropy irreducible homogeneous space is a connected effective homogeneous space L/H where H is a compact subgroup of a Lie group L such that the identity component H_{0} of H acts irreducibly on T_{eH}(L/H). If we assume instead that H, but not necessarily H_{0}, acts irreducibly on T_{eH}(L/H), then L/H is called an isotropy irreducible homogeneous space. These spaces are interesting because by Schur’s Lemma they all admit an L-invariant Einstein metric, and because they include the irreducible Riemannian symmetric spaces as a sub-family. The non-symmetric strongly isotropy irreducible homogeneous spaces were classified independently by O.V. Manturov [M1,2,3], J.A. Wolf [Wo1], and later by M. Krämer [Kr], with some minor omissions in the first two references. In addition, in [Wo 1] a detailed study of the geometric properties of strongly isotropy irreducible spaces was carried out. The isotropy irreducible spaces which are not strongly isotropy irreducible have recently been classified in [WZ3].

In a different direction, we present in this paper a direct proof of a beautiful observation of C.T.C. Wall concerning a correspondence between the compact simply connected irreducible symmetric spaces on the one hand and the compact strongly isotropy irreducible quotients of the classical groups on the other hand (cf. the remarks added in proof in [Wo1] pp. 147-148). This correspondence has a number of exceptions: certain Grassmannians and the isotropy irreducible space SO(7)/G_{2}, which is diffeomorphic to ℝP^{7}. These exceptions appear at first sight to be even more mysterious than the correspondence. But our main result is that there is an explanation, using general principles, of this correspondence as well as all its exceptions. In particular, this means that the classification of the non-symmetric strongly isotropy irreducible quotients of the classical groups may be read off from that of the irreducible symmetric spaces.

Consider a compact simply connected irreducible symmetric space G/K, where G is the identity component of the isometry group, and K is the isotropy group, which is necessarily connected. Then K is locally of the form H · U(1) or H · Sp(1), or it is semisimple. In the first case the symmetric space is hermitian and the isotropy representation is of the form [π ˆ⊗ϕ]_{ℝ}, where ϕ is a non-trivial 1-dimensional representation of U(1). In the second case the symmetric space is quaternionic and the isotropy representation is of the form π ˆ⊗ν_{2}, where ν_{2} is the two-dimensional representation of Sp(1). In the third case the isotropy group is simple, unless G/K is a real or quaternionic Grassmannian. We hasten to point

Typeset by AMS-TEX

The first author is partially supported by the Natural Sciences and Engineering Research Council of Canada. The second author is partially supported by a grant from the National Science Foundation

out that these cases are not mutually exclusive. The precise correspondence can now be formulated as follows:

Theorem A. Let G/K be a compact, simply connected, effective, irreducible symmetric space.

(a) If G/K = G/H · U(1) is Hermitian symmetric of real dimension 2n with isotropy representation [π ˆ⊗ϕ]_{ℝ}, then π: H → SU(n) is a faithful unitary representation and the homogeneous manifold SU(n)/π(H) is strongly isotropy irreducible. Conversely, if SU(n)/π(H) is strongly isotropy irreducible and non-symmetric with H compact and connected, then [π ˆ⊗ϕ]_{ℝ} is the isotropy representation of a simply connected, irreducible Hermitian symmetric space.

(b) If G/K = G/H ·Sp(1) is quaternionic symmetric of real dimension 4n with isotropy representation π ˆ⊗ν_{2}, then π: H → Sp(n) is a faithful irreducible quaternionic representation and the homogeneous manifold Sp(n)/π(H) is strongly isotropy irreducible. Conversely, if Sp(n)/π(H) is strongly isotropy irreducible and nonsymmetric with H compact and connected, then π ˆ⊗ν_{2} is the isotropy representation of a simply connected, irreducible quaternionic symmetric space.

(c) If K is simple or G/K is a quaternionic projective space Sp(n+1)/ Sp(n) Sp(1), then the isotropy representation π: K → SO(n) is a faithful irreducible orthogonal representation and the homogeneous manifold SO(n)/π(K) is strongly isotropy irreducible. Conversely, if SO(n)/π(K) is strongly isotropy irreducible, nonsymmetric and not equal to SO(7)/ G_{2} and with K compact and connected, then π is the isotropy representation of a simply connected, irreducible symmetric space.

The exceptions preventing the above correspondence from being one-to-one are described in detail in the comments after Theorem (3.1).

The forward correspondence can also be uniformly formulated as follows. If G/K is a compact simply connected irreducible symmetric space of dimension n, then either K is maximal in SO(n) and then SO(n)/K is strongly isotropy irreducible, or K ⊂ L ⊂ SO(n) and L/K is strongly isotropy irreducible. Here L can be either U(n) if G/K is hermitian symmetric or Sp(n) · Sp(1) if G/K is quaternionic symmetric and hence L/K, after being made effective, becomes a quotient of SU(n) or Sp(n). The only exception to this rule are the Grassmannians SO(p + q)/ SO(p) SO(q), p ≥ q ≥ 3, p ≠ 4, q ≠ 4, and Sp(p + q)/ Sp(p) Sp(q), p ≥ q ≥ 2, in which case K is maximal in SO(n), but the isotropy representation of SO(n)/K has as many irreducible sub-representations as there are simple factors in K.

These strongly isotropy irreducible spaces also have the following geometric intepretation. If G/K is a symmetric space as in part (c) of Theorem A and of dimension n, then it gives rise to a curvature tensor on ℝ^{n} which uniquely characterises the symmetric space. The set of all curvature tensors on ℝ^{n} which give rise to the given symmetric space is then equal to SO(n)/π(K) since any two of them are orthogonally equivalent and since the isotropy group of a given one is precisely equal to K. Similarly in the case of SU(n)/π(H) and Sp(n)/π(H) if we consider curvature tensors compatible with the complex or the quaternionic structure.

The statement of the forward correspondence can be found in Theorem 3.1 whose proof involves only general principles. The irreducible symmetric spaces which do not satisfy the hypothesis of Theorem 3.1 give the exceptions in the forward correspondence. To enumerate these we use results in 2.10. In proving the forward correspondence, we need to study carefully some properties of the isotropy representation of a symmetric space. It turns out that the dominant weight of the isotropy representation satisfies a very simple, remarkable condition, which admits further refinements in the hermitian and quaternionic cases. It can be formulated as follows:

Theorem B. Let G/K be a compact, irreducible symmetric space and B the negative of the Killing form of G.

(a) If G/K is not a sphere SO(n + 1)/ SO(n) or a quaternionic projective space Sp(n + 1)/ Sp(n) Sp(1) and has isotropy representation π ⊗ ℂ = π_{λ} or π_{λ} ⊕ π_{λ}^{∗}, where λ is the dominant weight, then B(λ, λ) = 2B(λ, α), where α is any simple root of K with B(λ, α) ≠ 0.

(b) If G/H·U(1) is Hermitian symmetric other than the complex projective spaces, with isotropy representation [π_{λ}]_{ℝ} = [π_{λ′} ˆ⊗ϕ]_{ℝ}, then B(λ, λ^{∗}) = 0 and if α is any simple root of H such that B(λ′, α) ≠ 0, we also have B(λ′, λ′) + B(λ′, λ′^{∗}) = 2B(λ′, α).

(c) If G/H · Sp(1) is quaternionic symmetric other than the quaternionic projective spaces, with isotropy representation π ⊗ ℂ = π_{λ} = π_{λ′} ˆ⊗ν_{2} or π ⊗ ℂ = π_{λ} ⊕ π_{λ}^{∗} = [π_{λ′} ⊕ π_{λ′}^{∗}] ˆ⊗ν_{2}, then B(λ′, λ′) = \frac{3}{2}B(λ′, α), where α is any simple root of H such that B(λ′, α) ≠ 0.

Theorem B is proved in §2 from general principles, see 2.5 and 2.7. It turns out that the identities in Theorem B actually characterize the isotropy representations of symmetric spaces, among all orthogonal representations of compact Lie groups. The precise formulation is as follows:

Theorem C. Let π_{λ}: H → Gl(V) be a faithful irreducible complex representation of a semisimple, compact, connected Lie group H with dominant weight λ, and let < , > be a bi-invariant metric on 𝔥.

(a) If π_{λ} is orthogonal and < λ, λ > = 2 < λ, α > for every simple root α such that < λ, α > ≠ 0, then (H, π_{λ}) is the isotropy representation of a compact, simply connected, irreducible symmetric space, except in the case (G_{2}, ϕ_{7}), where ϕ_{7} is the 7-dimensional representation of G_{2}.

(b) If π_{λ} ≠ π_{λ}^{∗}, and < λ, λ > + < λ, λ^{∗} > = 2 < λ, α > for every simple root α such that < λ, α > ≠ 0, then (H · U(1), [π_{λ} ˆ⊗ϕ]_{ℝ}) is the isotropy representation of a compact, simply connected, hermitian symmetric space.

(c) If π_{λ} is symplectic and < λ, λ > = \frac{3}{2} < λ, α > for every simple root α such that < λ, α > ≠ 0, then (H · Sp(1), π_{λ} ˆ⊗ν_{2}) is the isotropy representation of a compact, simply connected, quaternionic symmetric space.

These identities turn out to be very strong conditions on the dominant weights. Hence, as a byproduct of our study, we obtain a short classification of the simply connected irreducible symmetric spaces in terms of their isotropy representation (which coincides with the holonomy representation). This classification is presented in §2, and may be regarded as a modern version of E. Cartan’s first method of classifying symmetric spaces - that of analysing the holonomy representation (cf. pp. 216-217, 233-263 of [Ca]). In fact, Cartan also used an identity for the weights of the isotropy representation, but it did not involve the dominant weight alone.

The construction of irreducible symmetric spaces from the strongly isotropy irreducible quotients of the classical groups occupies §4. The general idea is to analyse the space of invariant curvature tensors on the candidate for the isotropy representation of the corresponding symmetric space. This candidate is constructed from the given strongly isotropy irreducible space. We show the existence of a non-trivial invariant curvature operator taking values in the appropriate holonomy algebra, and using this curvature operator to define Lie brackets, we recover the corresponding irreducible symmetric pair.

At the end of this paper we compiled a table giving the correspondence and exceptions. As an application of our results, we mention a proof of the fact that the second sym-

metric power of the (1, 0)- part of the isotropy representation of an irreducible hermitian symmetric space other than complex projective space contains exactly two irreducible subrepresentations (see Theorem (5.2)). This fact was proved by invoking the classification theory of symmetric spaces in [CV] and [Bo, Lemma 3.3], and was used in the computations of [CV], [Bo] and in [Mo], p.184. Other consequences of this paper were used in [WZ2]. Lemma II.2.1 in [WZ2] is just (2.2), and the facts about symmetric spaces on pp. 585-589 in [WZ2] are proved in detail here.

Lastly, we would like to mention that similar techniques allow us to give a proof, along general principles, of Dynkin’s classification of the maximal subalgebras of the classical Lie algebras [Dy]. This will be the content of a forth-coming paper.

§1 Preliminaries.

Recall that a connected effective homogeneous space L/H is said to be isotropy irreducible if H is compact and acts irreducibly on T_{eH}(L/H) via the isotropy representation χ. L/H is said to be strongly isotropy irreducible if the identity component H^{0} of H also acts irreducibly on T_{eH}(L/H). If L/H is non-compact, one easily shows (see [WZ3], Theorem 2.1) that L/H is a symmetric space. Clearly, L/H is strongly isotropy irreducible iff L^{0}/(H ∩ L^{0}) is strongly isotropy irreducible, and hence we can assume that L is connected. If L/H is not simply connected but L is connected, then a covering group L^{∗} of L acts transitively on the universal cover of L/H with connected isotropy group H^{∗} and L/H is strongly isotropy irreducible iff L^{∗}/H^{∗} is strongly isotropy irreducible. Hence we can assume, whenever necessary, that L/H is a compact, simply connected, effective homogeneous space with L and H connected.

Let L/H be an arbitrary homogeneous space with isotropy representation χ. If we choose a reductive decomposition 𝔩 = 𝔥 ⊕ 𝔭, we can identify 𝔭 with T_{eK}(L/H) in the usual manner, sending X ∈ 𝔭 to X^{∗}(eH) = \frac{d}{dt}|_{t=0}[exp(tX)H]. Under this identification the isotropy representation χ becomes the adjoint representation of H on 𝔭 and hence is given by

Ad_{L}|H = Ad_{H} ⊕ χ.

Recall that for L = U(n), SU(n), Sp(n), SO(n) respectively, Ad_{L} ⊗ ℂ is given by μ_{n} ⊗ μ_{n}^{∗} , μ_{n} ⊗ μ_{n}^{∗}− 1 , S^{2}ν_{2n} , Λ^{2}ρ_{n} , where μ_{n}, ν_{2n}, ρ_{n} are respectively the standard representations of L on ℂ^{n}, ℂ^{2n}, and ℂ^{n}. Hence, if H is a subgroup of such an L, and if the inclusion is given via a representation π, which must be unitary, symplectic, or orthogonal respectively, then the isotropy representation is determined by the equations π⊗π^{∗} = Ad_{H} ⊕χ , π⊗π^{∗} = 1⊕Ad_{H} ⊕χ , S^{2}π = Ad_{H} ⊕χ , or Λ^{2}π = Ad_{H} ⊕χ respectively.

Now let (M, g) be a compact, irreducible Riemannian symmetric space. Then from the general theory of symmetric spaces (see, e.g. [Wo2], pp. 231-247, 257-263) we can write M = G/K, where G is the identity component Isom^{0}(M, g) of the isometry group of g and K is the isotropy group of a point. Furthermore, K^{0} coincides with the identity component of the linear holonomy group. G is simple, except when M = (K × K)/ΔK, where K is simple, and ΔK is the diagonal subgroup of K × K. Therefore, we may asume that g is the normal homogeneous metric induced by B, the negative of the Killing form of G. The curvature tensor of g is given by R_{X^{∗},Y^{∗}}(Z^{∗}) = −[[X, Y], Z], where X, Y, Z ∈ 𝔭, 𝔤 = 𝔨 ⊥ 𝔭. Notice that [X, Y] ∈ 𝔨, so that [[X, Y], Z] = π([X, Y])(Z) is the infinitesimal isotropy action. Conversely, if 𝔤 = 𝔨⊕𝔭 with [𝔭, 𝔭] ⊂ 𝔨, then there exists a symmetric space G/K, where 𝔤 is the Lie algebra of G and 𝔨 is the Lie algebra of K. Here K is given by the group which makes the infinitesimal isotropy action of 𝔨 on 𝔭 given by π(X)a = [X, a] into an effective representation.

Since (M, g) is irreducible, the isotropy representation π of G/K is irreducible. Furthermore, π ⊗ ℂ is reducible (as a complex representation) iff (M, g) is Hermitian symmetric, in which case K = H · U(1) with H semisimple. Then π ⊗ ℂ = π_{λ} ⊕ π_{λ}^{∗}, where π_{λ} is an irreducible unitary representation of K with dominant weight λ and π_{λ} ≅ π̸_{λ}^{∗}. Furthermore, π_{λ} = π_{λ′} ˆ⊗ϕ, where ϕ is a non-trivial 1-dimensional representation of U(1) and π_{λ′}: H → SU(n) is a unitary representation of H with dominant weight λ′.

G/K is called a quaternionic symmetric space if K = H · Sp(1), and if the isotropy representation satisfies π = σ ˆ⊗ν_{2}, where σ is a quaternionic representation of H. A quaternionic symmetric space is automatically irreducible and hence π is irreducible, which means that σ ⊗ ℂ = π_{λ} or σ ⊗ ℂ = π_{λ} ⊕ π_{λ}^{∗}. Notice that the second case can indeed occur, since the symmetric space SU(n + 2)/S(U(n) U(2)) is both hermitian symmetric and quaternionic symmetric.

Furthermore, the (infinitesimal) isotropy action determines the symmetric space in the following sense (see e.g. [He], p.427, Theorem 5.11) : If G_{i}/K_{i} are two simply connected irreducible symmetric spaces with isotropy representations π_{i} then any isomorphism ϕ of 𝔨_{1} onto 𝔨_{2} such that π_{2} ∘ ϕ is equivalent to π_{1} can be extended to an isomorphism of 𝔤_{1} onto 𝔤_{2} and hence induces an isometry of G_{1}/K_{1} onto G_{2}/K_{2}.

Also notice that if (K, π) is the isotropy representation of a symmetric space G/K, then so is (K, π ∘ σ) where σ is any automorphism of K. Of course, the two symmetric spaces are isometric by the previous result. Indeed, on the level of Lie algebras we can define 𝔤′ = 𝔨 ⊕ 𝔭 with Lie brackets [X, Y]′ = [X, Y], [X, a]′ = [σ(X), a] and [a, b]′ = σ^{−1}([a, b]) where X, Y ∈ 𝔨 and a, b ∈ 𝔭. 𝔤′ is of course isomorphic to 𝔤 via the isomorphism which sends X ∈ 𝔨 to σ^{−1}(X) ∈ 𝔨 and is the identity on 𝔭. The isotropy representation of the corresponding symmetric space G′/K is by construction equal to π ∘ σ. For example, the spin representation Δ_{8}^{±} can be viewed as the isotropy representation of the sphere SO(9)/ SO(8).

Although we do not want to assume the classification of symmetric spaces, we will need the isotropy representation of the Grassmannians (and in particular the corresponding rank 1 symmetric spaces) which are easily determined to be as follows: (G/K, π) = (SO(p + q)/ SO(p) SO(q), ρ_{p} ˆ⊗ρ_{q}), (SU(p + q)/S(U(p) U(q)), [μ_{p} ˆ⊗μ_{q}^{∗} ˆ⊗ϕ]_{ℝ}), (Sp(p + q)/ Sp(p) Sp(q), ν_{p} ˆ⊗ν_{q}) where ϕ is a 1-dimensional representation of a 1-dimensional torus. We will use the fact that any of these representations π is the isotropy representation of a symmetric space, and by the above uniqueness statement, must be the isotropy representation of one of the Grassmannians.

§2. On the Isotropy Representation of a Symmetric Space.

In this chapter we characterize the isotropy representation of a symmetric space in terms of their dominant weights and apply it to give a quick classification of the irreducible symmetric spaces.

Our proofs will use some consequences of the classification of simple Lie algebras, which we collect in the following Lemma. The first part consists of standard facts about the adjoint representation. For the second part see e.g. [He], p.523, Excercise 3. We will use the convention that for each simple root α of a semisimple Lie algebra 𝔥 we have τ_{α} = \frac{2}{<α,α>}H_{α}, where H_{α} is given by α(H) = < H_{α}, H > for all H in the chosen Cartan subalgebra 𝔱 of 𝔥 ⊗ ℂ. If λ_{α} is the basis of 𝔱^{∗} dual to the basis τ_{α} of 𝔱, then λ_{α} are the dominant weights of the basic representations of 𝔥. Furthermore, for the exceptional Lie groups we use the notation ϕ_{k} for an irreducible representation of dimension k.

(2.1) Lemma. Let 𝔥 be a complex simple Lie algebra of finite dimension.

(a) The dominant weight of the adjoint representation, i.e. the maximal root μ, satifies one of the following:

(i) 𝔥 = 𝔰u(n), n ≥ 3, with μ = λ_{β} + λ_{β}^{∗} where β is one of the simple roots at the end of the Dynkin diagram. (ii) 𝔥 = 𝔰p(n), n ≥ 1, with μ = 2λ_{β} where β is the short root at the end of the Dynkin diagram. (iii) 𝔥 = 𝔰o(n), n ≥ 7, with μ = λ_{β} where β is a long simple root not at the end of the Dynkin diagram. (iv) 𝔥 is an exceptional Lie algebra and μ = λ_{β} where β is a long simple root at the end of the Dynkin diagram.

(b) The root system consists of at most 2 orbits under the Weyl group since all roots of the same length are equivalent under the Weyl group, and there are two orbits when 𝔥 is of type B_{n}, C_{n}, G_{2} or F_{4}. In each Weyl orbit there is a unique dominant form. Hence the only roots that are also dominant are the maximal root and, in the case of the above Lie algebras, there exists a unique short root. In the latter case the dominant form is a basic weight and corresponds to the representations (𝔥, π) = (𝔰o(2n + 1), ρ_{2n+1}), n ≥ 2 ; (𝔰p(n), Λ^{2}ν_{2n} − 1), n ≥ 3 ; (𝔤_{2}, ϕ_{7}) ; and (𝔣_{4}, ϕ_{26}). □

Next we collect some facts from representation theory.

(2.2) Lemma. Let 𝔥 be a compact Lie algebra and V a finite dimensional complex vector space. Let < , > be a bi-invariant metric on 𝔥.

(a) Let π_{λ}: 𝔥 → 𝔤l(V) be a faithful irreducible complex representation of 𝔥 with dominant weight λ. If 𝔥 is not semisimple and hence has a 1-dimensional center, we have π_{λ} ⊗ π_{λ}^{∗} = ad(𝔥)_{ℂ} ⊕ χ with χ = 0 iff 𝔥 = 𝔲(n), n ≥ 1, and π_{λ} = μ_{n} or μ_{n}^{∗}. If 𝔥 is semisimple, then π_{λ} ⊗ π_{λ}^{∗} = [ad(𝔥)_{ℂ} + 1] ⊕ χ with χ = 0 iff 𝔥 = 𝔰u(n), n ≥ 2, and π_{λ} = μ_{n} or μ_{n}^{∗}. In both cases π_{λ+λ^{∗}} occurs with multiplicity 1 in π_{λ} ⊗ π_{λ}^{∗} and lies in χ unless (i) χ = 0, (ii) 𝔥 = 𝔰p(n), n ≥ 2, with π_{λ} = ν_{2n}, or (iii) 𝔥 = 𝔰p(n) ⊕ ℝ, n ≥ 2, with π_{λ} = ν_{2n} ˆ⊗ϕ.

(b) Let π_{λ}: 𝔥 → 𝔰p(V, ω) be a faithful irreducible complex representation of 𝔥 on V with invariant symplectic form ω. Then S^{2}π_{λ} = ad(𝔥)_{ℂ} ⊕ χ with χ = 0 iff 𝔥 = 𝔰p(n) and π_{λ} = ν_{2n}. π_{2λ} occurs with multiplicity 1 in S^{2}π_{λ} and if χ ≠ 0 then π_{2λ} ⊂ χ.

(c) Let π_{λ}: 𝔥 → 𝔰o(V, b) be a faithful irreducible complex representation of 𝔥 on V with invariant symmetric bilinear form b. Then Λ^{2}π_{λ} = ad(𝔥)_{ℂ} ⊕ χ with χ = 0 iff 𝔥 = 𝔰o(n) and π_{λ} = ρ_{n} or 𝔥 = 𝔰o(8) and π_{λ} is one of the two half-spin representations Δ_{8}^{±}. Let α be a simple root of 𝔥 such that < λ, α > ≠ 0. Then π_{2λ−α} occurs with multiplicity 1 in Λ^{2}π_{λ}. Furthermore, π_{2λ−α} ⊂ χ except when (i) χ = 0, (ii) 𝔥 = G_{2}, π_{λ} = ϕ_{7}, (iii) 𝔥 = 𝔰o(7), π_{λ} = spin representation Δ_{7} , or (iv) 𝔥 = 𝔰p(n) ⊕ 𝔰p(1) , π_{λ} = ν_{2n} ⊗ ν_{2} and α is the simple root of 𝔰p(1).

Proof. (a) If 𝔥 is not semisimple π_{λ}(𝔥) ⊂ 𝔲(n) and hence π_{λ} ⊗ π_{λ}^{∗} = ad(𝔥)_{ℂ} ⊕ χ where χ is the isotropy representation of the corresponding homogeneous space U(n)/π_{λ}(H). If χ = 0 this means that π_{λ}(𝔥) = 𝔲(n) and hence 𝔥 = 𝔲(n) with π_{λ} = μ_{n} or μ_{n}^{∗}. Similarly if 𝔥 is semisimple, since in this case π_{λ}(𝔥) ⊂ 𝔰u(n). If v_{λ} is a weight vector of π_{λ} with weight λ, then v_{λ}^{∗} is a weight vector of π_{λ}^{∗} with weight λ^{∗}, and hence v_{λ} ⊗ v_{λ}^{∗} is a dominant weight vector in π_{λ} ⊗ π_{λ}^{∗} with weight λ + λ^{∗}, which clearly has multiplicity one. Hence π_{λ+λ^{∗}} occurs in π_{λ} ⊗ π_{λ}^{∗} with multiplicity one. Notice that π_{λ+λ^{∗}} is a representation of 𝔥_{s} , the semisimple part of 𝔥. Now if π_{λ+λ^{∗}} ⊂ ad(𝔥)_{ℂ}, then λ + λ^{∗} is the maximal root of a simple ideal of 𝔥. Since π_{λ} is faithful, 𝔥 has only one simple ideal. (If 𝔥 has no simple ideal, then 𝔥 = 𝔲(1) and our assertions are trivial.) Since both λ and λ^{∗} are dominant (2.1a) implies that 𝔥_{s} = 𝔰u(n) or 𝔰p(n) with π_{λ}|_{𝔥_{s}} = μ_{n}, μ_{n}^{∗}, or ν_{2n}. If 𝔥 is semisimple, we obtain (i) and (ii) and if 𝔥 is not semisimple we obtain (i) and (iii). Notice that in the case of π_{λ} = ν_{2n} or π_{λ} = ν_{2n} ˆ⊗ϕ we have χ = 0 iff n = 1.

(b) By assumption 𝔥 must be semisimple. As in (a) it follows that S^{2}π_{λ} = ad(𝔥)_{ℂ} ⊕ χ where χ is the isotropy representation of Sp(n)/π_{λ}(H) and hence if χ = 0 we have 𝔥 = 𝔰p(n) and π_{λ} = ν_{2n}. If v_{λ} is a weight vector in π_{λ}, then v_{λ} ∘ v_{λ} is a dominant weight vector in S^{2}π_{λ} with weight 2λ which clearly has multiplicity one. Suppose π_{2λ} ⊂ ad(𝔥)_{ℂ}, then 2λ is the maximal root of a simple ideal of 𝔥. By faithfulness, 𝔥 is simple, and by (2.1a) 𝔥 = 𝔰p(n), n ≥ 1, with π_{λ} = ν_{2n}.

(c) 𝔥 must also be semisimple in this case. Λ^{2}π_{λ} = ad(𝔥)_{ℂ} ⊕ χ follows as before. But this time χ = 0 can also occur if 𝔥 = 𝔰o(8) with π_{λ} = Δ_{8}^{±} since in this case π_{λ} gives rise to an isomorphism of 𝔰o(8) onto itself. If < λ, α > ≠ 0, then λ − α is a weight in π_{λ} with multiplicity one and if v_{λ} and v_{λ−α} are corresponding weight vectors, then v_{λ} ∧ v_{λ−α} is a dominant weight vector in Λ^{2}π_{λ} with weight 2λ − α which clearly has multiplicity one. Suppose that π_{2λ−α} ⊂ ad(𝔥)_{ℂ}. Then 2λ − α is the maximal root of a simple ideal of 𝔥. If 𝔥 is not simple, faithfulness implies that α is a root of a different simple ideal of 𝔥 and that these are the only simple ideals in 𝔥. We can therefore assume that λ = λ_{1} +λ_{2} where λ_{i} is a dominant weight of 𝔥_{i}, i = 1, 2, α is a root of 𝔥_{2} and 2λ − α is the maximal root μ of 𝔥_{1}.

Then α = 2λ_{2} and μ = 2λ_{1}. Hence α is a dominant weight of 𝔥_{2} and also a simple root, which is only possible if 𝔥_{2} = 𝔰p(1) and hence λ_{2} = ν_{2}. The condition μ = 2λ_{1} implies by (2.1a) that 𝔥_{1} = 𝔰p(n) and λ_{1} = ν_{2n}. This gives case (iv).

Assume next that 𝔥 is simple and hence 2λ − α = μ. If 𝔥 = 𝔰u(n), n ≥ 3, then μ = λ_{β} + λ_{β}^{∗} where λ_{β} is the basic weight corresponding to the simple root β at one end of the Dynkin diagram of 𝔥. Notice that we must have α = α^{∗} since λ^{∗} = λ and μ^{∗} = μ. This means that n is even and α is the middle vertex of the Dynkin diagram. Since 2λ = μ+α is dominant, this is only possible if n = 3. Furthermore 2λ(τ_{β}) = (λ_{β} + λ_{β^{∗}})(τ_{β}) + α(τ_{β}) = 0 and 2λ(τ_{α}) = (λ_{β} + λ_{β^{∗}})(τ_{α}) + α(τ_{α}) = 2. Similarly λ(τ_{β^{∗}}) = 0 and hence λ = λ_{α} or 𝔥 = 𝔰o(6) and π_{λ} = ρ_{6}, which implies χ = 0.

If 𝔥 = 𝔰p(n), n ≥ 1, then μ = 2λ_{β}, where β is the short root at the end of the Dynkin diagram. Since 2λ = μ + α is dominant, we must have n = 2 if α ≠ β and n = 1 if α = β. In the first case 2λ(τ_{β}) = 2λ_{β}(τ_{β}) + α(τ_{β}) = 0 and 2λ(τ_{α}) = 2λ_{β}(τ_{α}) + α(τ_{α}) = 2. Hence λ = λ_{α} or 𝔥 = 𝔰p(2) = 𝔰o(5), π_{λ} = ρ_{5}. If n = 1 , 2λ(τ_{β}) = 2λ_{β}(τ_{β}) + β(τ_{β}) = 4 and hence λ = 2λ_{β} or 𝔥 = 𝔰o(3), π_{λ} = ρ_{3}. In both cases χ = 0.

In the remaining cases, μ = λ_{β} for some simple root β. The fact that 2λ = μ + α is dominant implies that α cannot have any adjacent simple roots except for β, in particular α ≠ β. Hence α must be at the end of the Dynkin diagram and β right next to it. (2.1a) now implies that 𝔥 = 𝔰o(n) or 𝔤_{2}. If 𝔥 = 𝔰o(n) it follows that 2λ(τ_{β}) = λ_{β}(τ_{β}) + α(τ_{β}) = 0 and 2λ(τ_{α}) = 2λ_{β}(τ_{α}) + α(τ_{α}) = 2. Hence λ = λ_{α} which implies π_{λ} = ρ_{n} or n = 7 with π_{λ} = Δ_{7} or n = 8 with π_{λ} = Δ_{8}^{±}. Similarly if 𝔥 = 𝔤_{2} it follows that λ = λ_{α} and hence π_{λ} = ϕ_{7}. □

Remark. (2.2) proves Lemma II.2.1 of [WZ2]. Note, however, the correction for the statement of the unitary case. This omission does not effect the rest of [WZ2].

Later we will also need a somewhat more general version of (2.2c).

(2.3) Lemma. Let π_{λ}: 𝔥 → 𝔰o(V, b) be a faithful irreducible complex representation of 𝔥 on V with invariant symmetric bilinear form b and let α be a simple root of 𝔥 with < λ, α > ≠ 0. Then 2λ − α is not a root of 𝔥 except in the following cases:

(a) 𝔥 = 𝔰o(n), π_{λ} = ρ_{n}

(b) 𝔥 = 𝔰o(8) , π_{λ} = Δ_{8}^{±}

(c) 𝔥 = 𝔰p(n) ⊕ 𝔰p(1), π_{λ} = ν_{2n} ˆ⊗ν_{2} and α is the simple root of 𝔰p(1).

(d) 𝔥 = G_{2} , π_{λ} = ϕ_{7}

(e) 𝔥 = 𝔰o(7) , π_{λ} = Δ_{7}

Proof. If 𝔥 is not simple we can argue as in (2.2c). If 𝔥 is simple, we observe that 2λ − α is a dominant form and hence if 2λ − α is equal to a root β, β must be dominant also. By (2.1b) there are only two possibilities. If β is a long root it must be equal to the maximal root, and in this case the proof of (2.2c) implies the possibilities (a)-(e). If β is a short root, (2.1b) implies that β = λ_{γ} for some simple root γ. The fact that 2λ = λ_{γ} + α is dominant, again implies that α must be at the end of the Dynkin diagram and γ right next to it. By (2.1b) this is only possible if (𝔥, π_{λ_{γ}}) = (𝔰p(n), Λ^{2}ν_{2n} − 1), n ≥ 2 or (𝔤_{2}, ϕ_{7}). In the first case 2λ(τ_{γ}) = λ_{γ}(τ_{γ}) + α(τ_{γ}) = 0 and 2λ(τ_{α}) = λ_{γ}(τ_{α}) + α(τ_{α}) = 2 and hence λ = λ_{α} = ν_{2n} . But this representation is excluded since ν_{2n} is not orthogonal. In the second case 2λ(τ_{γ}) = λ_{γ}(τ_{γ}) + α(τ_{γ}) = −2 implies that λ is not dominant. □

(2.4) Lemma. The following cannot be the isotropy representation of a symmetric space: (a) 𝔨 = 𝔰p(n) ⊕ 𝔲(1), π = [ν_{2n} ˆ⊗ϕ]_{ℝ}, n ≥ 2

(b) 𝔨 = G_{2}, π = ϕ_{7}

(c) 𝔨 = 𝔰o(7), π = Δ_{7}

Proof. (a) In this case the corresponding simply connected irreducible symmetric space G/K would have holonomy group K = Sp(n) · U(1). But in general the holonomy group can be either Sp(n) or Sp(n) · Sp(1), but not Sp(n) · U(1), see [Be], 10.66 and 14.46.

In the case of (b) and (c) one easily sees that the representation π of 𝔨 acts transitively on the set of 2-frames. Hence if this were the isotropy representation of a symmetric space M, M would have to have constant curvature. But in this case the holonomy group of M would be bigger than K. □

We now come to the main result of this chapter.

(2.5) Theorem. Let (M = G/K, g) be a compact symmetric space with G = Isom^{0}(M, g) and without Euclidean factors. Let B be the negative of the Killing form of G and λ be a dominant weight of its (complex) isotropy representation π. Suppose that α is any root of 𝔨 such that B(λ, α) ≠ 0 and that 2λ − α is not a root of 𝔨. Then B(λ, λ) = 2B(λ, α).

Proof. We write 𝔤 = 𝔨⊥𝔭 and select a maximal abelian subalgebra 𝔱 ⊂ 𝔨_{s}, as well as a system of simple roots {α_{i}} for (𝔨_{s} ⊗ ℂ, 𝔱 ⊗ ℂ). As is well-known, there is a basis of 𝔨 ⊗ ℂ of the form \{z_{i}, H_{j}, X_{α}, X_{−α}\} where \{z_{i}\} spans the center of \sqrt{−1}𝔨, \{H_{j} = H_{α_{j}}\} spans \sqrt{−1}𝔱, and [X_{α}, X_{−α}] = H_{α} ∈ \sqrt{−1}𝔱, where H_{α} is dual to α with respect to (B|_{𝔨}) ⊗ ℂ. Next let {v_{ρ}} be a basis of weight vectors for π⊗ℂ such that B(v_{λ}, v_{−λ}) = 1 and π(X_{−α})v_{λ} = v_{λ−α}. Notice that v_{λ−α} ≠ 0 since B(λ, α) ≠ 0. Since B([v_{λ}, v_{−λ}], H) = −B([v_{λ}, H], v_{−λ}) = B(λ(H)v_{λ}, v_{−λ}) = λ(H) it follows that [v_{λ}, v_{−λ}] = H_{λ} ∈ \sqrt{−1}𝔱, the element dual to λ.

By the Jacobi identity (or equivalently the Bianchi identity for the curvature tensor), we have

[v_{λ}, [v_{−λ}, v_{λ−α}]] + [v_{−λ}[v_{λ−α}, v_{λ}]] + [v_{λ−α}[v_{λ}, v_{−λ}]] = 0.

[v_{λ−α}, v_{λ}] = 0 since in general [v_{ρ_{1}}, v_{ρ_{2}}] = 0 unless ρ_{1} + ρ_{2} is a root of 𝔨 ⊗ ℂ. Therefore,

[v_{λ}, [v_{−λ}, [X_{−α}, v_{λ}]]] + [v_{λ−α}, H_{λ}] = 0.

By the Jacobi identity again,

B(λ, λ − α)v_{λ−α} = (λ − α)(H_{λ})v_{λ−α} = [H_{λ}, v_{λ−α}] = [v_{λ}, [v_{−λ}, [X_{−α}, v_{λ}]]]
= −[v_{λ}, [X_{−α}, H_{λ}]] + [v_{λ}, [v_{λ}, π(X_{−α})v_{−λ}]] = −[v_{λ}, [X_{−α}, H_{λ}]]
= B(λ, α)v_{λ−α}

which yields the identity. □

Remark. E.Cartan first classified the symmetric spaces by classifying their holonomy representation, which coincides with their isotropy representation. Geometrically this approach is especially pleasing in a general discussion of holonomy, in view of the Berger-Simons theorem. However, Cartan never published all the technical details of this classification. The general method is described in [Ca] , see 216–217, 233–263. The key lemma seems to be the following: If λ and μ are two weights of the isotropy representation of a symmetric space such that λ + μ and λ − μ are not roots of 𝔨, then B(λ, μ) = 0. This condition can be deduced by applying the Jacobi identity to v_{λ}, v_{−λ}, and v_{μ} as in the proof of (2.5). It implies the condition in (2.5) only if B(λ, α) ≥ B(α, α) (which by (2.11) is almost never the case) since then λ − 2α is also a weight of π_{λ}. In the following we shall see that (2.5) can be used quite easily to give a shorter proof of Cartan’s first classification.

(2.6) Corollary. Let M = G/K be a compact irreducible symmetric space which is not a sphere SO(n + 1)/ SO(n) or a quaternionic projective space Sp(n + 1)/ Sp(n) Sp(1) and with isotropy representation π ⊗ ℂ = π_{λ} or π_{λ} ⊕ π_{λ}^{∗} . If B is the negative of the Killing form of G and α a simple root of 𝔨 with B(λ, α) ≠ 0, then B(λ, λ) = 2B(λ, α).

Proof. If π⊗ℂ = π_{λ}, then this follows from (2.3), (2.4), and (2.5). If π⊗ℂ = π_{λ}⊕π_{λ}^{∗}, then G/K is hermitian symmetric and hence 𝔨 has a one dimensional center. By effectiveness λ is nonzero on this center and hence 2λ − α is not a root of 𝔨 and (2.5) applies. □

For Hermitian and quaternionic symmetric spaces we obtain a further refinement:

(2.7) Theorem.

(a) Let G/K be a compact irreducible Hermitian symmetric space other than the complex projective spaces, with K = H · U(1) and isotropy representation π = [π_{λ}]_{ℝ} = [π_{λ′} ˆ⊗ϕ]_{ℝ}. Then B(λ, λ^{∗}) = 0 and if α is any simple root of H such that B(λ′, α) ≠ 0, we also have B(λ′, λ′) + B(λ′, λ′^{∗}) = 2B(λ′, α) .

(b) Let G/K be a compact quaternionic symmetric space other than the quaternionic projective spaces, with K = H · Sp(1) and isotropy representation π ⊗ ℂ = π_{λ} = π_{λ′} ˆ⊗ν_{2} or π ⊗ ℂ = π_{λ} ⊕ π_{λ}^{∗} = [π_{λ′} ⊕ π_{λ′}^{∗}] ˆ⊗ν_{2}. If α is any simple root of H such that B(λ′, α) ≠ 0, then B(λ′, λ′) = \frac{3}{2}B(λ′, α).

Proof. (a) In the Hermitian case, λ = λ′ + x, where x is the weight of ϕ. There are weight vectors v_{λ} of π_{λ}, v_{λ^{∗}} and v_{−λ} of π_{λ^{∗}} such that [v_{λ}, v_{−λ}] = H_{λ} ∈ 𝔨. Furthermore, [v_{λ^{∗}}, v_{−λ}] = 0 since λ^{∗} − λ cannot be a root of 𝔨 (it is nonzero on 𝔲(1)). By the Jacobi identity, we have

[[v_{λ}, v_{−λ}], v_{λ^{∗}}] + [[v_{−λ}, v_{λ^{∗}}], v_{λ}] + [[v_{λ^{∗}}, v_{λ}], v_{−λ}] = 0
or B(λ, λ^{∗})v_{λ^{∗}} + [[v_{λ^{∗}}, v_{λ}], v_{−λ}] = 0

If λ + λ^{∗} is not a root of 𝔨, then [v_{λ^{∗}}, v_{λ}] = 0 and so B(λ, λ^{∗}) = 0, as desired. If λ + λ^{∗} is a root of a simple ideal of 𝔨, since it is dominant and cannot be a basic weight on that simple ideal, it must be the maximal root. Then (2.1a) implies that 𝔥 = 𝔰u(n) ⊕ 𝔲(1) or 𝔰p(n) ⊕ 𝔲(1), with π_{λ} = μ_{n} ˆ⊗ϕ and ν_{2n} ˆ⊗ϕ respectively. In the first case we have the isotropy representation of complex projective space, which is excluded by hypothesis. The second case is ruled out by (2.4a).

Since λ^{∗} = λ′^{∗} − x , B(λ, λ^{∗}) = 0 implies B(λ′, λ′^{∗}) = B(x, x). Furthermore, by Corollary (2.6) 2B(λ′, α) = 2B(λ, α) = B(λ, λ) = B(λ′, λ′) + B(x, x) = B(λ′, λ′) + B(λ′, λ′^{∗}).

(b) In the quaternionic case λ = λ′ + \frac{1}{2}β, where β is the simple root of Sp(1). (2.6) implies that

B(λ, λ) = 2B(λ, α) = 2B(λ′, α) = B(λ′, λ′) + \frac{1}{4}B(β, β)
and B(λ, λ) = 2B(λ, β) = B(β, β).

Hence 2B(λ′, α) = B(λ′, λ′)+\frac{1}{2}B(λ, β) = B(λ′, λ′)+\frac{1}{2}B(λ′, α), which implies our claim. □

We now come to our first application of (2.5), which is actually the only one we need in the forward correspondence in §2.

(2.8) Theorem. Let G/K be a compact irreducible symmetric space with isotropy representation π and dominant weight λ. If G/K ≠ [SU(n) × SU(n)]/Δ SU(n) then for every simple ideal 𝔥 ⊂ 𝔨 there exists a unique simple root α of 𝔥 such that B(λ, α) ≠ 0.

Proof. Assume that α and β are simple roots of a simple ideal 𝔥 of 𝔨 such that B(λ, α) ≠ 0 and B(λ, β) ≠ 0. Then there is a minimal chain of simple roots α_{1} = α, α_{2}, …, α_{r} = β connecting α and β. Let γ = α_{1} + ⋯ + α_{r}. Now γ is a root of 𝔥 and B(γ, λ) ≠ 0. Since our claim is certainly true for the spheres and quaternionic projective spaces, (2.6) implies that B(λ, λ) = 2B(λ, α) = 2B(λ, β). We claim that 2λ − γ is a root of 𝔨. If not, we have in addition B(λ, λ) = 2B(λ, γ) = 2B(λ, α_{1}) + ⋯ + 2B(λ, α_{r}) ≤ 2B(λ, α) + 2B(λ, β), which is a contradiction.

Since 2λ−γ is a root of 𝔨, 𝔨 must be semisimple. We claim next that 𝔨 is simple. Suppose not, then let 𝔥′ be a second simple ideal in 𝔨. Since π is faithful, there exists a simple root α′ of 𝔥′ such that B(λ, α′) ≠ 0. Now B(2λ − γ, α′) = 2B(λ, α′) ≠ 0, so 2λ − γ|_{𝔥} ≡ 0. On the other hand, B(2λ − γ, α) = 2B(λ, α) − B(α, α) − B(α_{2}, α) ≠ 0, a contradiction to the assumption that 2λ − γ is a root.

Therefore, 𝔨 = 𝔥 is simple. As one can verify directly, 2λ − γ is a dominant weight of 𝔨. Moreover, B(2λ − γ, α) ≠ 0, B(2λ − γ, β) ≠ 0 so 2λ − γ is not a basic weight. It is also a root of 𝔨 , so by (2.1), 𝔨 = 𝔰u(n), n ≥ 3 and 2λ − γ is the maximal root. It follows that α and β must be at the end of the Dynkin diagram and hence λ = γ = maximal root and G/K = [SU(n) × SU(n)]/Δ SU(n). □

We will now show that the conditions on the isotropy representation in (2.6) and (2.7) are so strong that one can easily derive a classification of the symmetric spaces via their isotropy representation. This will also enable us to prove a converse to (2.6) and (2.7). The results in the remainder of this section will mostly not be needed in the proof of the C.T.C.Wall correspondence, except in a minor way in the proof of (3.4) and (4.4). We first prove the following Lemma.

(2.9) Lemma. Let 𝔥 be a compact simple Lie algebra with ad(𝔥)-invariant inner product < , > and π_{λ}: 𝔥 → 𝔤l(V) an irreducible faithful complex representation of 𝔥.

(a) Suppose that α is a simple root of 𝔥 such that < λ, α > ≠ 0 and < λ, λ > ≤ < λ, α >. Then up to an automorphism of 𝔥 one of the following holds:

(i) λ = λ^{∗}, (𝔥, π) = (𝔰o(n), ρ_{n}), n ≥ 3, or (𝔰p(n), ν_{2n}), n ≥ 2, or (𝔰p(1), ν_{2}). In the first two cases \frac{<λ,λ>}{<λ,α>} = 1 and in the third case \frac{<λ,λ>}{<λ,α>} = \frac{1}{2}.

(ii) λ ≠ λ^{∗}, (𝔥, π) = (𝔰u(n), μ_{n}), n ≥ 3, and \frac{<λ,λ>}{<λ,α>} = \frac{n−1}{n}.

(b) If α is a simple root of 𝔥 such that < λ, α > ≠ 0 and < λ, λ > + < λ, λ^{∗} > ≤ < λ, α >, then (𝔥, π) = (𝔰u(n), μ_{n}), n ≥ 2, with < λ, λ > + < λ, λ^{∗} > = < λ, α >.

Proof. (a) Without loss of generality, we may choose < , > to be the multiple of the Killing form of 𝔥 such that the longest root has length \sqrt{2}. If rank(𝔥) = 1, then < λ, λ > ≤ < λ, α > implies that π = ν_{2} or ρ_{3} with \frac{<λ,λ>}{<λ,α>} = \frac{1}{2} or 1 respectively. Hence we will assume that rank(𝔥) ≥ 2 in the following.

Let λ = ∑_{β} n_{β}λ_{β}, where the sum is over all simple roots and n_{β} are the non-negative integers \frac{2<λ,β>}{<β,β>}. We can also write λ as ∑_{β} r_{β}β where β runs over all simple roots of 𝔥 and r_{β} are positive rational numbers. We first observe that if we write λ_{α} = ∑_{β} r_{αβ}β , then

r_{αβ} = \frac{2 < λ_{α}, λ_{β} >}{< β, β >} = \frac{< λ_{α}, λ_{β} >}{< λ_{β}, β >}.

Since 1 ∈ π_{λ} ⊗ π_{λ^{∗}} , this has the particular consequence that \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = r_{αα} must be

an integer or half integer if λ_{α} is orthogonal or symplectic and that \frac{<λ_{α},λ_{α}>+<λ_{α},λ_{α}^{∗}>}{<λ_{α},α>} = r_{αα} + r_{αα^{∗}} must be an integer in general.

We will first show that if (2.9) holds for basic representations of 𝔥 then it holds for all irreducible representations of 𝔥. Suppose that there is a simple root β other than α for which < λ, β > ≠ 0. Then < λ, λ > > n_{α}^{2} < λ_{α}, λ_{α} > +n_{β}^{2} < λ_{α}, λ_{β} > since < λ_{α}, λ_{β} > > 0. Dividing throughout by < λ, α > = n_{α} < λ_{α}, α >, we obtain \frac{n_{α}<λ_{α},λ_{α}>}{<λ_{α},α>} < \frac{<λ,λ>}{<λ,α>} ≤ 1. By assumption, (2.9) holds for λ_{α}, so n_{α} = 1 and 𝔥 = 𝔰u(n), n ≥ 3. Since all roots of 𝔰u(n) have the same length, < λ_{α}, α > = < λ_{β}, β >. Hence

\frac{< λ_{α}, λ_{α} >}{< λ_{α}, α >} + \frac{n_{β}^{2} < λ_{β}, λ_{β} >}{< λ_{β}, β >} < \frac{< λ, λ >}{< λ, α >} ≤ 1.

Since \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = \frac{n−1}{n} , \frac{<λ_{β},λ_{β}>}{<λ_{β},β>} < \frac{1}{n_{β}^{2}n} < \frac{1}{2}, a contradiction to the assumption that (2.9)

holds for basic weights. Therefore, the inequality \frac{<λ,λ>}{<λ,α>} ≤ 1 holds only if λ is a multiple of a basic weight. Next assume λ = kλ_{α} with k ≥ 2. Then λ − α must be a dominant and hence < λ, λ − α > > 0 which contradicts the assumption that \frac{<λ,λ>}{<λ,α>} ≤ 1.

Hence \frac{<λ,λ>}{<λ,α>} ≤ 1 can only hold for a basic weight λ_{α} and in this case \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = r_{αα}. We therefore need to determine which simple roots satisfy r_{αα} ≤ 1. For this purpose we make the observation that \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = \frac{2<λ_{α},λ_{α}>}{<α,α>} are the diagonal entries in the inverse of the Cartan matrix. Indeed, since λ_{α} is dual to τ_{α}, < λ_{α}, λ_{β} > is the inverse matrix of < τ_{α}, τ_{β} > = \frac{4<α,β>}{<α,α> <β,β>} and hence \frac{2<λ_{α},λ_{β}>}{<β,β>} is the inverse matrix of the Cartan matrix g_{αβ} = \frac{2<α,β>}{<β,β>}. See eg. [Dy] p.117 Table 2 for the inverse of the Cartan matrix (but notice that Dynkin uses a different normalization). One can also easily obtain the diagonal entries of the inverse Cartan matrix by using the following observation. If H is a compact, simply connected, simple Lie group with center Z(H) and Cartan matrix g_{αβ}, then det g_{αβ} = |Z(H)| (see e.g. [He], X.3.31). Hence it follows that if α is a simple root of H and if we delete the vertex corresponding to α from the Dynkin diagram of H, we obtain several connected Dynkin diagrams corresponding to simply connected Lie groups H_{i} and r_{αα} = g^{αα} = (∏_{i} |Z(H_{i})|) /|Z(H)|. In either case one easily shows that the only cases where r_{αα} ≤ 1 are the representations listed in (a).

(b) The condition in (b) implies that \frac{<λ,λ>}{<λ,α>} < 1 and hence we can apply (a). If π_{λ} = ρ_{n} or ν_{2n}, then we have λ = λ^{∗} and hence 𝔥 = 𝔰p(1) = 𝔰u(2) with π_{λ} = ν_{2} = μ_{2}. If π_{λ} = μ_{n} one easily checks that \frac{<λ,λ>+<λ,λ^{∗}>}{<λ,α>} = 1. □

Remarks. 1) One can prove a Lemma analogous to (2.9) if 𝔥 is only compact but not necessarily simple if one requires that \frac{<λ,λ>}{<λ,α>} ≤ 1 for all simple roots α with < λ, α > ≠ 0 and if one includes (𝔲(n), μ_{n}) as one of the possibilities. Indeed if λ = ∑_{i} λ_{i} + x with π_{λ} = π_{λ_{1}} ˆ⊗ ⋯ ˆ⊗π_{λ_{k}} ˆ⊗ϕ, then there exist simple roots α_{i} with < λ, α_{i} > ≠ 0 and hence < λ, λ > = ∑_{j} < λ_{j}, λ_{j} > + < x, x > ≤ < λ, α_{i} > = < λ_{i}, α_{i} >. If k ≥ 2, this implies that \frac{<λ_{i},λ_{i}>}{<λ_{i},α_{i}>} ≤ \frac{1}{2} for some i. Applying (2.9) we see that this is only possible if 𝔥 = 𝔰p(1) ⊕ 𝔰p(1) = 𝔰o(4) with π = ν_{2} ˆ⊗ν_{2} = ρ_{4}. If k = 1 and x ≠ 0, (2.9) implies that 𝔥 = u(n) with π = μ_{n}.

2) (2.9) and Remark 1) generalizes [Ko1], Theorem 2.5 where he showed that if an irreducible representation has a weight μ such that μ is orthogonal to all weights σ ≠ ±μ, then it must be ν_{2n}, ρ_{n}, or μ_{n}. This easily reduces to the case where μ is the dominant weight λ. If α is a simple root with < λ, α > ≠ 0, then λ − α is a weight and hence our (2.9) shows that it is sufficient to assume that λ and λ − α are orthogonal.

(2.10) Proposition.

(a) Let G/K be a compact irreducible Hermitian symmetric space with K = H · U(1). Then H is simple unless (𝔨, π) = (𝔰u(m)⊕𝔰u(n)⊕𝔲(1), [μ_{m} ˆ⊗μ_{n}^{∗} ⊗ϕ]_{ℝ}), m ≥ n ≥ 2. In general π = [π_{λ′} ˆ⊗ϕ]_{ℝ} with λ′ ≠ λ′^{∗} unless G/K = SO(n + 2)/ SO(n) SO(2).

(b) Let G/K be a quaternionic symmetric space with K = H ·Sp(1). Then H is simple unless (𝔨, π) = (𝔰o(n) ⊕ 𝔰o(4) = [𝔰o(n) ⊕ 𝔰p(1)] ⊕ 𝔰p(1), ρ_{n} ˆ⊗ρ_{4} = [ρ_{n} ˆ⊗ν_{2}] ˆ⊗ν_{2}) or (𝔨, π) = (𝔰u(2)⊕𝔰u(n)⊕𝔲(1) = [𝔰u(n)⊕𝔲(1)]⊕𝔰p(1), [ν_{2} ˆ⊗μ_{n} ˆ⊗ϕ]_{ℝ} = [μ_{n} ˆ⊗ϕ]_{ℝ} ˆ⊗ν_{2}). In particular the only quaternionic symmetric space that is also hermitian symmetric is the Grassmannian SU(n + 2)/S(U(n) U(2)).

(c) Let G/K be a compact irreducible symmetric space with 𝔨 = 𝔨_{1} ⊕ ⋯ ⊕ 𝔨_{s} , s ≥ 2, where the 𝔨_{i} are simple non-abelian ideals. Then s ≤ 4 and (𝔨, π) belongs to one of the following:

(i) s = 2 and (k, π) = (𝔰o(m) ⊕ 𝔰o(n), ρ_{m} ˆ⊗ρ_{n}), m ≥ n ≥ 3, m ≠ 4, n ≠ 4.

(ii) s = 2 and (𝔨, π) = (𝔰p(m) ⊕ 𝔰p(n), ν_{2m} ˆ⊗ν_{2n}), m ≥ n ≥ 1.

(iii) s = 2 and (𝔨, π) = (𝔥 ⊕ 𝔰p(1), π_{λ′} ˆ⊗ν_{2}) with 𝔥 simple and π_{λ′} symplectic. (iv) s = 3 and (𝔨, π) = (𝔰o(n) ⊕ 𝔰o(4), ρ_{n} ˆ⊗ρ_{4}), n ≥ 3, n ≠ 4.

(v) s = 4 and (𝔨, π) = (𝔰o(4) ⊕ 𝔰o(4), ρ_{4} ˆ⊗ρ_{4}).

Proof. We first prove (c) and start by assuming that s = 2. Let λ be the dominant weight of π. Then λ = λ_{1} + λ_{2}, where λ_{i} are dominant weights of 𝔨_{i}. We denote by α_{i} a simple root of 𝔨_{i} such that B(λ, α_{i}) ≠ 0. Assume also that (𝔨, π) ≠ (𝔰p(n) ⊕ 𝔰p(1), ν_{2n} ˆ⊗ν_{2}). By (2.6) B(λ, λ) = 2B(λ, α_{1}) = 2B(λ, α_{2}). This implies that

\frac{B(λ_{1}, λ_{1})}{B(λ_{1}, α_{1})} + \frac{B(λ_{2}, λ_{2})}{B(λ_{2}, α_{2})} = 2.

By relabelling 𝔨_{1} and 𝔨_{2}, we may assume further that \frac{B(λ_{1},λ_{1})}{B(λ_{1},α_{1})} ≤ 1. Since π_{λ} is orthogonal,

π_{λ_{i}} are self-contragredient, so by (2.9) either \frac{B(λ_{i},λ_{i})}{B(λ_{i},α_{i})} = 1, in which case (𝔨, π) = (𝔰o(m) ⊕

𝔰o(n), ρ_{m} ˆ⊗ρ_{n}), m ≥ n ≥ 3 or (𝔰p(m) ⊕ 𝔰p(n), ν_{2m} ˆ⊗ν_{2n}) m ≥ n ≥ 2, or else \frac{B(λ_{1},λ_{1})}{B(λ_{1},α_{1})} = \frac{1}{2}, in which case 𝔨_{1} = 𝔰p(1), π_{λ_{1}} = ν_{2}. In this latter situation, 𝔨_{2} = 𝔥 is simple and π_{λ_{2}} = π_{λ′} is symplectic. Notice that B(λ_{2}, λ_{2}) = \frac{3}{2}B(λ_{2}, α_{2}) holds.

If s > 2, we proceed analogously and obtain ∑_{i=1}^{s} \frac{B(λ_{i},λ_{i})}{B(λ_{i},α_{i})} = 2. If one term equals 1, by (2.9) it must correspond to (𝔰o(n), ρ_{n}), n ≠ 4. Hence s = 3, with the other two terms corresponding to (𝔰p(1), ν_{2}), i.e. (𝔨, π) = (𝔰o(n) ⊕ 𝔰o(4), ρ_{n} ˆ⊗ρ_{4}). If no term equals 1, then all equal \frac{1}{2} and correspond to (𝔰p(1), ν_{2}), i.e. (𝔨, π) = (𝔰o(4) ⊕ 𝔰o(4), ρ_{4} ˆ⊗ρ_{4}).

Next we prove (a). Let 𝔥 = 𝔥_{1}⊕⋯⊕𝔥_{s} where 𝔥_{i} are simple non-abelian ideals and s ≥ 2. Then λ = λ_{1} + ⋯ + λ_{s} + x, where x is the weight on 𝔲(1). Let α_{i} be a simple root of 𝔥_{i} such that B(λ, α_{i}) ≠ 0. By (2.6), we have ∑_{i=1}^{s} B(λ_{i}, λ_{i}) + B(x, x) = 2B(λ, α_{i}) for each i. Since G/K is not complex projective space, (2.7) implies that ∑_{i=1}^{s} B(λ_{i}, λ_{i}^{∗})−B(x, x) = 0. Therefore, ∑_{i=1}^{s} (\frac{B(λ_{i},λ_{i})+B(λ_{i},λ_{i}^{∗})}{B(λ_{i},α_{i})}) = 2. If s ≥ 2, it follows that there is some i for which

\frac{B(λ_{i},λ_{i})+B(λ_{i},λ_{i}^{∗})}{B(λ_{i},α_{i})} ≤ 1. By (2.9b) 𝔨_{i} = 𝔰u(n) and π_{λ_{i}} = μ_{n} with \frac{B(λ_{i},λ_{i})+B(λ_{i},λ_{i}^{∗})}{B(λ_{i},α_{i})} = 1. Hence s = 2 and (𝔨, π) = (𝔰u(m) ⊕ 𝔰u(n) ⊕ 𝔲(1), [μ_{m} ˆ⊗μ_{n} ⊗ ϕ]_{ℝ}).

If we assume that λ = λ′ + x with λ′ = λ′^{∗}, then the above shows that s = 1 and then (2.7a) implies that B(λ′, λ′) = B(λ′, α) and by (2.9) and (2.4a) we are in the case of the complex quadric SO(n + 2)/ SO(n) SO(2).

(b) follows by combining (a) and (c). □

(2.8) and (2.10) reduces the study of irreducible symmetric spaces to the following three cases:

(A) M = (G/H · U(1), π) with H simple and π = [π_{λ} ˆ⊗ϕ]_{ℝ}, λ = kλ_{α} ≠ λ^{∗} and < λ, λ >

+ < λ, λ^{∗} > = 2 < λ, α >. (B) M = (G/H · Sp(1), π) with H simple and π ⊗ ℂ = π_{λ} ˆ⊗ν_{2}, λ = kλ_{α} and < λ, λ >

= \frac{3}{2} < λ, α >. (C) M = (G/K, π) with K simple and π ⊗ℂ = π_{λ}, λ = kλ_{α} and < λ, λ > = 2 < λ, α >. In the above three cases the equations hold with respect to any inner product < , >

since the group in question is simple. We first show that in most cases k = 1.

(2.11) Proposition. If in the cases (A)–(C) above we have k > 1, then one of the following holds:

(A) k = 2 with (𝔨 ⊕ 𝔲(1), π) = (𝔲(n), S^{2}μ_{n}) which is the isotropy representation of the

symmetric space Sp(n)/ U(n). (B) k = 3 with (𝔨 ⊕ 𝔰p(1), π) = (𝔰p(1) ⊕ 𝔰p(1), S^{3}ν_{2} ˆ⊗ν_{2}) which is the isotropy represen-

tation of the symmetric space G_{2} / SO(4).

(C) k = 2 with (𝔨, π) = (𝔰o(n), S^{2}ρ_{n} − 1), n ≥ 5, which is the isotropy representation of the symmetric space SU(n)/ SO(n); or (𝔨, π) = (𝔰p(n), S^{2}ν_{2n}), n ≥ 1, which is the isotropy representation of the symmetric space [Sp(n) × Sp(n)]/Δ Sp(n). Otherwise, k = 4 with (𝔨, π) = (𝔰o(3), S^{4}ν_{2}), which is the isotropy representation of the symmetric space SU(3)/ SO(3).

Proof. In case (A) we have 2 = \frac{<λ,λ>+<λ,λ^{∗}>}{<λ,α>} = k \frac{<λ_{α},λ_{α}>+<λ_{α},λ_{α}^{∗}>}{<λ_{α},α>} and hence (2.9b) implies that 𝔨 = 𝔰u(n) with λ_{α} = μ_{n} and k = 2.

In case (B) we have k \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = \frac{3}{2} and since kλ_{α} must be symplectic, (2.9a) implies that 𝔨 = 𝔰p(1) with λ_{α} = ν_{2} and k = 3.

In case (C) we have 2 = \frac{<λ,λ>}{<λ,α>} = k \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} . (2.9) together with the fact that kλ_{α} must be orthogonal implies that 𝔨 = 𝔰o(n) with λ_{α} = ρ_{n}, n ≥ 5, and k = 2, or 𝔨 = 𝔰p(n) with λ_{α} = ν_{n}, n ≥ 2, and k = 2, or 𝔨 = 𝔰p(1) with λ_{α} = ν_{2} and k = 4. □

The classification of symmetric spaces is now reduced to the cases (A)–(C) with λ a basic weight λ_{α}. We can easily obtain a complete list by using the observation, which we made in the proof of (2.9), that r_{αα} = \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} are the diagonal entries in the inverse of the Cartan matrix. Now, as in the proof of (2.9), one immediately enumerates the possibilities where r_{αα} = \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} ≤ 2 and easily checks that they belong to one of the following cases:

(A) If λ_{α} ≠ λ_{α}^{∗} then \frac{<λ_{α},λ_{α}>+<λ_{α},λ_{α}^{∗}>}{<λ_{α},α>} = 2 or \frac{<λ_{α},λ_{α}>+<λ_{α},λ_{α}^{∗}>}{<λ_{α},α>} = 1. In the first case one obtains the entries (2), (4), and (5) in our Table at the end of this paper and in the second case λ_{α} = μ_{n}, corresponding to entry (7).

(B) If λ_{α} is symplectic then \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = \frac{3}{2} or \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = 1 or \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = \frac{1}{2}. In the first case one obtains the entries (10)-(13) in our Table, in the second case λ_{α} = ν_{2n}, n ≥ 2, and in the third case λ_{α} = ν_{2}, both corresponding to entry (15).

(C) If λ_{α} is orthogonal then \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = 2 or \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = \frac{3}{2} or \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} = 1. In the first case one obtains the entries (17) and (19)-(24) in our Table as well as (𝔨, λ_{α}) = (𝔤_{2}, ϕ_{7}) which by (2.4) cannot be the isotropy representation of a symmetric space. In the second case (𝔨, λ_{α}) = (𝔰o(7), Δ_{7}) which cannot be the isotropy representation of a symmetric space either. In the third case λ_{α} = ρ_{n} which corresponds to entry (25).

This finishes the classification of symmetric spaces. These arguments can be easily modified to obtain the following converse to (2.6) and (2.7).

(2.12) Theorem. Let π_{λ}: 𝔥 → 𝔤l(V) be a faithful irreducible representation of a compact semisimple Lie algebra 𝔥 on a complex vector space V, and let < , > be a bi-invariant metric on 𝔥.

(a) If π_{λ} ≠ π_{λ^{∗}}, and < λ, λ > + < λ, λ^{∗} > = 2 < λ, α > for every simple root α such that < λ, α > ≠ 0, then (𝔥 ⊕ 𝔲(1), [π_{λ} ˆ⊗ϕ]_{ℝ}) is the isotropy representation of a hermitian symmetric space.

(b) If π_{λ} is symplectic and < λ, λ > = \frac{3}{2} < λ, α > for every simple root α such that < λ, α > ≠ 0, then (𝔥⊕𝔰p(1), π_{λ} ˆ⊗ν_{2}) is the isotropy representation of a quaternionic symmetric space.

(c) If π_{λ} is orthogonal and < λ, λ > = 2 < λ, α > for every simple root α such that < λ, α > ≠ 0, then (𝔥, π_{λ}) is the isotropy representation of an irreducible symmetric space, except in the case (𝔤_{2}, ϕ_{7}).

Proof. Notice that the proof of (2.10) and (2.11) and the classification after (2.11) only uses the condition on the dominant weights in (a)–(c) and hence carries over directly. But we need to find a different proof of (2.8) since we do not assume the stronger condition in (2.5).

But this can be easily done using (2.9) as in the proof of (2.10). For example, in case (c), if α ≠ β are two roots in the same simple ideal 𝔥 of 𝔨 with < λ, α > ≠ 0 and < λ, β > ≠ 0, then \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} + \frac{<λ_{β},λ_{β}>}{<λ_{β},β>} < 2 since < λ_{α}, λ_{β} > > 0. Hence, after

relabeling α and β, we can assume \frac{<λ_{α},λ_{α}>}{<λ_{α},α>} < 1 which implies 𝔥 = 𝔰u(n), λ_{α} = μ_{n}. Since < λ, α > = < λ^{∗}, α^{∗} > = < λ, α^{∗} > ≠ 0 we can assume that β = α^{∗} and hence λ_{β} = μ_{n}^{∗}. If λ = kλ_{α} + kλ_{α}^{∗} + ⋯, then < λ, λ > ≥ k^{2} < λ_{α} + λ_{β}, λ_{α} + λ_{β} > = 2k^{2}(< λ_{α}, λ_{α} > + < λ_{α}, λ_{α}^{∗} >) = 2k^{2} < λ_{α}, α > ≥ 2 < λ, α >, which implies that λ = λ_{α} + λ_{β} or π_{λ} = Ad_{SU(n)}. This proves (2.8) under the assumption (c). Similarly in the case of (a) and (b). □

Remarks. 1) One can also formulate a similar theorem if 𝔥 is compact but not semisimple. If [π_{λ}]_{ℝ} is an irreducible orthogonal representation of such a Lie algebra with < λ, λ > = 2 < λ, α > for every simple root α with < λ, α > ≠ 0 and if in addition < λ, λ^{∗} > = 0, then [π_{λ}]_{ℝ} is the isotropy representation of a hermitian symmetric space. Indeed, the center of 𝔥 is 1-dimensional and if λ = λ′ + x, then the two conditions imply that < λ′, λ′ > + < λ′, λ′^{∗} > = 2 < λ′, α > and hence the claim follows from (2.12a).

2) Our proof of Theorem (2.12) is not ideal since it involves case by case arguments. It indirectly uses the classification of symmetric spaces since we need to know that the resulting representations are indeed isotropy representations of symmetric spaces. For this purpose we use the list of the isotropy representations of all irreducible symmetric spaces in [Wo] (8.11.2) and (8.11.5). It would be much more desirable to have a direct proof of this theorem that does not use the classification of symmetric spaces. We hope to return to this problem in a separate paper.

§3 Strongly isotropy irreducible spaces from irreducible symmetric spaces.

We first formulate the correspondence precisely.

(3.1) Theorem (Forward Correspondence).

(a) Let G/K = G/H · U(1) be a compact, simply connected, effective, irreducible Hermitian symmetric space of real dimension 2n with isotropy representation π ⊗ ℂ = π_{λ} ⊕π_{λ}^{∗} and π_{λ} = π_{λ′} ˆ⊗ϕ. Then π_{λ′}: H → SU(n) is a faithful unitary representation and the homogeneous manifold V = SU(n)/π_{λ′}(H) is strongly isotropy irreducible.

(b) Let G/K = G/H · Sp(1) be a compact, simply connected, effective, quaternionic symmetric space of real dimension 4n with isotropy representation π ⊗ ℂ = σ ˆ⊗ν_{2}, where σ is a faithful irreducible quaternionic representation σ: H → Sp(n) of H. Then the homogeneous manifold V = Sp(n)/σ(H) is strongly isotropy irreducible.

(c) Let G/K be a compact, simply connected, effective, irreducible symmetric space of dimension n and assume that either K is simple or that G/K is a quaternionic projective space. Then the isotropy representation π ⊗ ℂ = π_{λ}: K → SO(n) is a faithful, absolutely irreducible orthogonal representation and the homogeneous manifold V = SO(n)/π(K) is strongly isotropy irreducible.

By applying this construction to the Grassmannians, we can see some complications in this correspondence:

(1) The rank 1 symmetric spaces (a) SU(n+1)/S(U(n) U(1)), (b) Sp(n+1)/ Sp(n) Sp(1), and (c) SO(n + 1)/ SO(n) give rise to the space V consisting of a single point. It follows from (2.2) that these are actually the only cases in which V can be a point.

(2) The Grassmannian SU(n+2)/S(U(n) U(2)) is both hermitian symmetric and quaternionic symmetric, and we obtain respectively the strongly isotropy irreducible spaces SU(2n)/ SU(n) SU(2) and Sp(n)/ U(n), the latter also being a symmetric space. By (2.10b) this complex Grassmannian is actually the only symmetric space which is both hermitian symmetric and quaternionic symmetric.

(3) Besides the above example there are two more in which V is again a symmetric space. Namely if G/K is the hermitian symmetric space SO(n + 2)/ SO(n) SO(2), then V is the symmetric space SU(n)/ SO(n). If G/K is the real symmetric space Sp(3)/ Sp(2) Sp(1), then V is the symmetric space SO(8)/[Sp(2) Sp(1)]/Δℤ_{2} = (SO(8)/ℤ_{2})/ SO(5) SO(3). We will be able to prove this directly in (3.4).

(4) On the other hand, real or quaternionic Grassmannians which are not hermitian symmetric or quaternionic symmetric are excluded in the correspondence. This fact will be explained after the proof of (3.1), where we will show that in these cases the isotropy representation of SO(n)/π(K) has as many irreducible sub-representations as there are simple factors in K.

(5) But we cannot totally exclude the Grassmannians from the correspondence either since we need the complex Grassmannian in (a), the real Grassmannian SO(n + 4)/ SO(n) SO(4) in (b), and the quaternionic Grassmannian Sp(n+1)/ Sp(n) Sp(1) in case (c).

(6) As we shall see, the only strongly isotropy irreducible quotient of a classical group that is excluded from the range of the correspondence is SO(7)/G_{2}. Notice though that (G_{2}, ϕ_{7}) was also an exception in several of our results in §2.

(7) There is one strongly isotropy irreducible space which can arise in two distinct ways from a symmetric space. Namely the symmetric space SU(3)/ SO(3) gives rise to the strongly isotropy irreducible space SO(5)/ SO(3) and the quaternionic symmetric space G_{2} / SO(4) gives rise to the strongly isotropy irreducible space Sp(2)/ Sp(1) = SO(5)/ SO(3). As we will see, this is the only case where this can occur.

It is amusing to observe that it is precisely these familiar spaces that are the exceptional cases in the correspondence.

The first step in proving Theorem (3.1) is the following fact about Casimir operators. Our convention for the Casimir operator of a representation π is that C_{π} = − ∑_{i} π(x_{i}) ∘ π(x_{i}) where x_{i} is an orthonormal basis of 𝔨 with respect to a given biinvariant symmetric bilinear form. Furthermore, if we use a positive definite inner product on 𝔨, then Casimir constants are non-negative. For further details, see [WZ2], pp. 568-571.

(3.2) Proposition. Let G/K be a compact symmetric space with no Euclidean factors, 𝔤 = 𝔨 ⊥ 𝔭 and π be its isotropy representation. Suppose that B is the negative of the Killing form of G and C_{π}, C_{Λ^{2}π} are the Casimir operators with respect to B|_{𝔨}. Let R̂ be the curvature operator of the normal metric (defined by B) acting on Λ^{2}𝔭. Then C_{π} = \frac{1}{2}Id and C_{Λ^{2}π} = Id − 2R̂.

Proof. By assumption G is semisimple and hence B is positive definite. Let {x_{i}} be an orthonormal basis for 𝔨 and {e_{p}} an orthonormal basis for 𝔭.

The fact that C_{π} = \frac{1}{2}Id follows easily from [WZ2], Corollaries I.1.6 and I.1.7, but for the convenience of the reader we give here a direct proof. Indeed, for X, Y ∈ 𝔭 we have:

B(X, Y) = − tr[adXadY] = − ∑_{i} B([X, [Y, x_{i}]], x_{i}) − ∑_{p} B([X, [Y, e_{p}]], e_{p})

= ∑_{i} B([X, x_{i}], [Y, x_{i}]) + ∑_{p} B([X, e_{p}], [Y, e_{p}])

= ∑_{i} B([x_{i}, X], [x_{i}, Y]) + ∑_{p} ∑_{i} B([X, e_{p}], x_{i})B([Y, e_{p}], x_{i})

= − ∑_{i} B([x_{i}, [x_{i}, X]], Y) + ∑_{i} ∑_{p} B([[X, x_{i}], e_{p})B([Y, x_{i}], e_{p})

= B(C_{π}(X), Y) + ∑_{i} B([X, x_{i}], [Y, x_{i}]) = 2B(C_{π}(X), Y).

Hence

C_{Λ^{2}π}(X ∧ Y) = − ∑_{i}(ad(x_{i})^{2}X) ∧ Y − ∑_{i} X ∧ (ad(x_{i})^{2}Y) − 2 ∑_{i}(ad(x_{i})X) ∧ (ad(x_{i})Y)

= X ∧ Y − 2 ∑_{i}(ad(x_{i})X) ∧ (ad(x_{i})Y).

On the other hand

∑_{i}(ad(x_{i})X) ∧ (ad(x_{i})Y)

= ∑_{i} ∑_{p,q} B([x_{i}, X], e_{p})B([x_{i}, Y], e_{q})e_{p} ∧ e_{q}

= ∑_{p,q} B([X, e_{p}], [Y, e_{q}])e_{p} ∧ e_{q}

= − ∑_{p,q} B(e_{p}, [X, [Y, e_{q}]])e_{p} ∧ e_{q}

= ∑_{p,q} B(e_{p}, [Y, [e_{q}, X]] + [e_{q}, [X, Y]])e_{p} ∧ e_{q}by the Jacobi identity

= ∑_{p,q} B([e_{q}, X], [e_{p}, Y])e_{p} ∧ e_{q} + ∑_{p,q} B([e_{p}, e_{q}], [X, Y])e_{p} ∧ e_{q}

= − ∑_{i}(ad(x_{i})X) ∧ (ad(x_{i})Y) − ∑_{p,q} B(R_{X,Y} (e_{p}), e_{q})e_{p} ∧ e_{q}.

Hence

∑_{i} ad(x_{i})X ∧ ad(x_{i})Y = − ∑_{p<q} B(R_{X,Y} (e_{p}), e_{q})e_{p} ∧ e_{q}
= R̂(X ∧ Y)

where our convention is chosen so that sectional curvature is given by K(e_{i} ∧ e_{j}) = < R_{e_{i},e_{j}}(e_{j}), e_{i} > and R̂ = Id for the standard sphere. The formula for C_{Λ^{2}π} follows immediately. □

(3.3) Corollary.

(a) Let G/K be a compact symmetric space with no euclidean factors and with isotropy representation π: K → SO(𝔭). Then Λ^{2}π = Ad_{K} ⊕ χ where χ is the isotropy representation of SO(𝔭)/π(K) and C_{χ} = Id, where the Casimir operator is taken with respect to the negative of the Killing form of G restricted to K.

(b) Let G/K be a compact symmetric space as in (3.1a) or (3.1b) and let χ be the isotropy representation of the corresponding homogeneous space V = SU(n)/π_{λ′}(H) or V = Sp(n)/σ(H) respectively. Then C_{χ} = Id where the Casimir operator is taken with respect to the negative of the Killing form of G restricted to H.

Proof. It follows from §1 that Λ^{2}π = Ad_{K} ⊕ χ and that χ is the isotropy representation of SO(𝔭)/π(K). The embedding ad(𝔨) ⊂ Λ^{2}(𝔭) ≅ 𝔰o(𝔭) is given by ad(X) → ∑_{p<q} B(ad(X)e_{p}, e_{q})e_{p} ∧ e_{q} and hence R̂(X ∧ Y) is equal to the image of −ad[X, Y]. We can choose the representation space 𝔪 for χ ⊂ Λ^{2}π to be perpendicular to ad(𝔨) and since R̂ is a symmetric operator, it keeps the decomposition Λ^{2}(𝔭) = ad(𝔨) ⊕ 𝔪 invariant and must be 0 on 𝔪. This together with (3.2) proves (a).

To prove (b) we obseve that in the case of G/H · U(1) we have H · U(1) ⊂ SU(n) U(1) ⊂ SO(2n) and hence the isotropy representation of SU(n)/π_{λ′}(H) is contained in the isotropy representation of SO(2n)/π(H · U(1)). Similarly for the case of G/H · Sp(1) and hence (b) follows from (a). □

Remarks. 1) Notice that in (3.3a) the representation π and hence χ may be reducible, but the Casimir constants with respect to B|_{𝔨} of the irreducible sub-representations of χ are all equal. But this does not say that SO(𝔭)/π(K) is normal homogeneous Einstein (see [WZ2], I.1.7) since the Casimir constants with respect to B_{SO(p)}|_{𝔨} could be different. See also (5.4).

2) There is a converse to (3.3), which we proved in [WZ2], III.3.4, III.3.6, III.3.8, assuming all the results of this paper. I states that if SO(n)/π(H), SU(n)/π(H), or Sp(n)/π(H) is a homogeneous space with isotropy representation χ such that C_{χ} = a·Id for some choice of biinvariant metric on H and some constant a, then, up to a few exceptions, π, [π ˆ⊗ϕ]_{ℝ}, π ˆ⊗ν_{2} resp. are the isotropy representation of a symmetric space.

Proof of Theorem (3.1). We begin with the Hermitian symmetric case. If G/K is the complex projective space, then V = point. So we assume that this is not the case. By (2.2a) and (2.4a) we have π_{λ′+λ′^{∗}} ⊂ χ ⊂ π_{λ′} ⊗ π_{λ′}^{∗} and by (3.3b) C_{χ} = Id. Any other irreducible summand in π_{λ′} ⊗ π_{λ′}^{∗} and hence in χ has dominant weight of the form λ′ + λ′^{∗} − (∑_{α} n_{α}α) with n_{α} ≥ 0 and some n_{α} > 0. Now in general one easily shows, see e.g. [WZ2], II.2.2, that if μ and ρ are two dominant weights with ρ = μ − (∑_{α} n_{α}α) and n_{α} ≥ 0 with some n_{α} > 0, then the Casimir constant for π_{ρ} must be strictly smaller than the one for π_{μ}. Hence χ = π_{λ′+λ′^{∗}} and V is strongly isotropy irreducible.

In the case of a quaternionic symmetric space we have σ = π_{λ′} or σ = π_{λ′} ⊕ π_{λ′}^{∗}. In the first case and excluding the quaternionic projective space, we have π_{2λ′} ⊂ χ ⊂ S^{2}π_{λ′} and C_{χ} = Id. Any other summand in χ is of the form 2λ′ − (∑_{α} n_{α}α) and hence has smaller Casimir constant. Thus χ = π_{2λ′} and V is strongly isotropy irreducible. In the second case G/K must also be a hermitian symmetric space. We could now use (2.10b) to conclude that π_{λ′} = μ_{n} and hence V = Sp(n)/ U(n) which is strongly isotropy irreducible. We can also argue directly as follows. We have χ ⊂ S^{2}σ = S^{2}(π_{λ′} ⊕π_{λ′}^{∗}) = S^{2}π_{λ′} ⊕S^{2}π_{λ′}^{∗} ⊕π_{λ′} ⊗π_{λ′}^{∗} and since π_{λ′} ⊗π_{λ′}^{∗} = Ad_{H} ⊕ψ we get χ = S^{2}π_{λ′} ⊕S^{2}π_{λ′}^{∗} ⊕ψ. Since π_{2λ′} ⊂ S^{2}π_{λ′} and since any other summand in S^{2}π_{λ′} would have smaller Casimir constant, it follows that S^{2}π_{λ′} = π_{2λ′} and S^{2}π_{λ′}^{∗} = π_{2λ′}^{∗}. If π_{λ′} ≠ μ_{n} then π_{λ′+λ′^{∗}} ⊂ ψ ⊂ χ and hence must have the same Casimir constant as π_{2λ′}. But these Casimir constants are equal to < λ′ + λ′^{∗}, λ′ + λ′^{∗} + 2δ > and < 2λ′, 2λ′ + 2δ > respectively, where 2δ is the sum of all positive roots of 𝔥. Hence if they are equal it follows that < λ′, λ′ > = < λ′, λ′^{∗} > = < λ′^{∗}, λ′^{∗} > and thus λ′ = λ′^{∗} which would contradict the assumption that G/K is hermitian symmetric. Hence π_{λ′} = μ_{n} and thus χ = π_{2λ′} ⊕ π_{2λ′}^{∗} which means that V is strongly isotropy irreducible.

In the case of (3.1c) assume first that G/K is not a sphere or a quaternionic projective space or the symmetric space [SU(n) × SU(n)]/Δ SU(n). Then, if α is a simple root of K with < λ, α > ≠ 0, (2.2c) and (2.4) imply that π_{2λ−α} ⊂ χ ⊂ Λ^{2}π and by (3.3a) C_{χ} = Id. Since K is simple, (2.8) says that there exists a unique simple root α with < λ, α > ≠ 0. Hence any other irreducible summand in Λ^{2}π and hence in χ must have dominant weight 2λ − α − (∑_{β} n_{β}β) and hence has smaller Casimir constant. Thus χ = π_{2λ−α} and V is strongly isotropy irreducible.

If G/K = [SU(n) × SU(n)]/Δ SU(n), then there exist two simple roots α and α^{∗} with < λ, α > ≠ 0 and < λ, α^{∗} > ≠ 0. Hence π_{2λ−α} and π_{2λ−α^{∗}} ⊂ χ and the same argument as above shows that χ = π_{2λ−α} ⊕ π_{2λ−α^{∗}}. But since (2λ − α)^{∗} = 2λ − α^{∗} this says that χ is irreducible as a real representation.

If G/K = Sp(n + 1)/ Sp(n) Sp(1), then there exist precisely two simple roots α and β with < λ, α > ≠ 0 and < λ, β > ≠ 0 where α is a root in Sp(n) and β is the root of Sp(1). Then by (2.2c), π_{2λ−α} ⊂ χ and π_{2λ−β} ⊂ ad(𝔨). Any other irreducible summand in χ must have dominant weight 2λ − α − (∑_{γ} n_{γ}γ) and hence has smaller Casimir constant. Thus χ = π_{2λ−α} and V is strongly isotropy irreducible.

This finishes the proof of (3.1). □

Remarks. 1) We can see from the above proof why the construction (3.1c) fails to yield isotropy irreducible spaces when G/K is a quaternionic Grassmannian Sp(p+q)/ Sp(p) Sp(q), p ≥ q ≥ 2 or a real Grassmannian SO(p + q)/ SO(p) SO(q), p ≥ q ≥ 3, p ≠ 4, q ≠ 4. This is because K contains at least two simple ideals and the proof of (3.1) shows that there are as many irreducible summands of the form π_{2λ−α} in χ ⊗ ℂ as there are simple ideals in 𝔨 and these all have the same Casimir constant. The proof also shows that Sp(n+1)/ Sp(n) Sp(1) is an exception in (3.1c), since in this case π_{2λ−β} ⊂ ad(𝔨).

2) Notice also that the proof of (3.1) gives a simple description of the isotropy representation χ of the strongly isotropy irreducible spaces V. Indeed, assuming that V is not a point nor a symmetric space, it follws that χ = π_{λ′+λ′^{∗}} in (3.1a), χ = π_{2λ′} in (3.1b), and in (3.1c) χ = π_{2λ−α} or, in the case of G/K = [SU(n) × SU(n)]/Δ SU(n), χ = [π_{2λ−α}]_{ℝ}.

We finally determine when V is again a symmetric space.

(3.4) Proposition. Let G/K be a symmetric space as in (3.1) and V the corresponding strongly isotropy irreducible homogeneous space. Then V is an irreducible symmetric space iff

(a) G/K is the hermitian symmetric space SO(n + 2)/ SO(n) SO(2) with V = SU(n)/ SO(n).

(b) G/K is the quaternionic symmetric space SU(n + 2)/S(U(n) U(2)) with V = Sp(n)/ U(n).

(c) G/K = Sp(3)/ Sp(2) Sp(1) regarded as a real symmetric space with V = SO(8)/[Sp(2) Sp(1)]/Δℤ_{2} = (SO(8)/ℤ_{2})/ SO(5) SO(3)

Proof. If G/K is hermitian symmetric, the proof that V is isotropy irreducible shows that the isotropy representation χ of V is equal to π_{λ′+λ′^{∗}}. If λ′ ≠ λ′^{∗} and if χ is the isotropy representation of a symmetric space, then (2.8) shows (since H must be semisimple) that π_{λ′} = μ_{n}. But in this case V = point. If λ′ = λ′^{∗} then by (2.10a) π_{λ′} = ρ_{n} and hence V = SU(n)/ SO(n), which is a symmetric space.

Now let G/K be quaternionic symmetric. If σ = π_{λ′} ⊕ π_{λ′}^{∗}, then by (2.10b) we have π_{λ′} = μ_{n} and hence V = Sp(n)/ U(n), which is a symmetric space. If σ = π_{λ′}, then the proof of (3.1) shows that χ = π_{2λ′}. If H is not simple, then (2.10b) shows that π_{λ′} = ρ_{n} ˆ⊗ν_{2} but (2.10) also shows that π_{2λ′} cannot be the isotropy representation of a symmetric space.

If H is simple, then we can use (2.6) since all biinvariant inner products on 𝔥 must be a multiple of each other. Since < λ′, λ′ > = \frac{3}{2} < λ′, α > we have < 2λ′, 2λ′ > = 3 < 2λ′, α > and hence (2.6) implies that π_{2λ′} cannot be the isotropy representation of a symmetric space.

We now deal with the case where G/K is neither hermitian symmetric nor quaternionic symmetric. If G/K is the quaternionic projective space, then π_{λ} = ν_{n} ˆ⊗ν_{2} and the proof of (3.1) implies that χ = π_{2λ−α_{1}} where ν_{n} = π_{λ_{α_{1}}} and hence χ = π_{λ_{α_{2}}} ˆ⊗ρ_{3} where α_{2} is adjacent to α_{1}. But this cannot be the isotropy representation of a symmetric space by (2.10) if n ≥ 3. If n = 2 it is the isotropy representation of the symmetric space which is the unoriented Grassmannian (SO(8)/ℤ_{2})/ SO(5) SO(3).

If G/K = [SU(n) × SU(n)]/Δ SU(n), then the proof of (3.1) implies that χ ⊗ ℂ = π_{2λ−α} ⊕ π_{2λ−α^{∗}}. But this cannot be the isotropy representation of a symmetric since it would have to be hermitian symmetric with K not semisimple.

In all other cases K is simple and the proof of (3.1) implies that χ ⊗ ℂ = π_{2λ−α} and that α is the unique simple root with < λ, α > ≠ 0. Hence λ = kλ_{α}. If rank 𝔥 = 1, it follows from (2.6) that k = 4 and hence 2λ − α = 3α = 6λ_{α} which cannot be the isotropy representation of a symmetric space by (2.6). Hence we can assume that rank 𝔥 > 1 and there must therefore be a root β adjacent to α. Since < 2λ − α, β > ≠ 0, if χ were the isotropy representation of a symmetric space, (2.6) would imply that < 2λ − α, 2λ − α > = 2 < 2λ−α, β > or \frac{2<α,β>}{<α,α>} = −(2k +1). Hence k = 1, 𝔨 = 𝔤_{2}, and α must be the short root of 𝔤_{2}, i.e. π_{λ} = ϕ_{7}. But ϕ_{7} cannot be the isotropy representation of a symmetric space by (2.4). □

Remark. As mentioned earlier, there is one other phenomena which keeps the correspondence from being one to one. Namely the quaternionic symmetric space G_{2} / SO(4) and the symmetric space SU(3)/ SO(3) give rise to the same strongly isotropy irreducible space SO(5)/ SO(3). One easily sees that this is the only case where this can happen. Indeed, if two distinct symmetric spaces are of the same type (3.1a), (3.1b), or (3.1c), then the corresponding spaces V must also be distinct since the symmetric spaces are determined by their isotropy representation. If two symmetric spaces belong to two different categories, then the corresponding spaces V can be the same only due to the low dimensional isomorphisms 𝔰o(6) ≅ 𝔰u(4), and 𝔰o(5) ≅ 𝔰p(2). By listing the possible symmetric spaces that can give rise to spaces V in these low dimensions, one sees that the above is the only case where this can happen.

§4 Symmetric Spaces from Strongly Isotropy Irreducible Quotients of the Classical Groups.

Let V be a strongly isotropy irreducible quotient of a classical group. The embedding of the isotropy subgroup in the classical group is described by a faithful representation π′ : H → SU(N), π′ : H → Sp(N), or π′ : H → SO(N). Since G/H is strongly isotropy irreducible iff G/H^{0} is, we may as well assume that H is connected. In the case of SU(N) and Sp(N) this is equivalent to requiring that V is simply connected. In the case of SO(N), V may or may not be simply connected.

In view of the construction in (1.2), the isotropy representation π of the corresponding symmetric space G/K should be given by [π′ ⊗ ϕ]_{ℝ}, π′ ⊗ ν_{2}, or π′ respectively, with the corresponding 𝔨 = 𝔥 ⊕ 𝔲(1), 𝔥 ⊕ 𝔰p(1), or 𝔥. Let 𝔭 be the representation space for the above, so that 𝔭 = ℝ^{2N}, ℝ^{4N}, or ℝ^{N} respectively. To recover the symmetric space we let 𝔤 = 𝔨 ⊕ 𝔭. One’s first impulse is to define a suitable Lie bracket for 𝔤 such that [𝔭, 𝔭] ⊂ 𝔨. Indeed, we already have a Lie bracket in 𝔨, and if X ∈ 𝔨, v ∈ 𝔭, we can define [X, v] = [−v, X] as π(X)(v). For u, v ∈ 𝔭 we can define [u, v] = −[v, u] = Cpr_{𝔨}(u∧v), where C is a suitable constant and pr_{𝔨} is the orthogonal projection associated to the embedding π: 𝔨 → 𝔰o(𝔭) ≅ Λ^{2}(𝔭) with respect to the Killing form on 𝔰o(𝔭). The difficulty with this approach is in proving the Jacobi identity. We do not know whether this approach can be successful or not. Instead, we take advantage of the following lemma of Cartan, which says that a symmetric space can be recovered from its isotropy representation and its curvature tensor. To formulate it, let R be the space of algebraic curvature tensors on 𝔭, i.e. the space of all tensors R of type (4, 0) on 𝔭 which are skew symmetric in the first two and last two arguments and which satisfy the Jacobi identity R(x_{1}, x_{2}, x_{3}, x_{4}) + R(x_{2}, x_{3}, x_{1}, x_{4}) + R(x_{3}, x_{1}, x_{2}, x_{4}) = 0. For each R ∈ R, denote by R̂ the corresponding curvature operator, i.e. the symmetric endomorphism R̂: Λ^{2}(𝔭) → Λ^{2}(𝔭), which can also be regarded as an element in S^{2}Λ^{2}(𝔭).

(4.1) Lemma (Cartan). Let π: 𝔨 → 𝔰o(𝔭) be an orthogonal representation of a compact Lie algebra 𝔨. Suppose that 𝔭 admits a non-trivial 𝔨-invariant algebraic curvature tensor R with values in π(𝔨) ⊂ 𝔰o(𝔭) ≅ Λ^{2}(𝔭). Then 𝔤 = 𝔨 ⊕ 𝔭 can be made into a Lie algebra by defining [X, v] = −[v, X] = π(X)(v) for X ∈ 𝔨, v ∈ 𝔭 and [u, v] = Y ∈ 𝔨 with π(Y) = R(u, v) for u, v ∈ 𝔭. Furthermore (𝔤, 𝔨) is then a symmetric pair. □

See [Ca], p. 225 or [Si], pp. 215-6 for a more modern proof.

Our strategy is therefore to show that for the π: 𝔨 → 𝔰o(𝔭) we constructed from the strongly isotropy irreducible quotients of the classical groups, there exists a non-trivial invariant curvature with values in π(𝔨). To do this, recall that S^{2}Λ^{2}(𝔭) = Λ^{4}(𝔭)⊕R. From the decomposition of Λ^{2}(𝔭) into 𝔨-irreducible summands we can deduce the dimension of 𝔨-invariant elements in S^{2}Λ^{2}(𝔭). An invariant curvature tensor on p is a symmetric endomorphism of Λ^{2}(𝔭) and if sufficiently many invariant curvature tensors exist, a suitable linear combination will yield a non-trivial one taking values in π(𝔨) in Λ^{2}(𝔭). Our principal task is therefore to analyse the invariant 4-forms on p using the isotropy irreducibility condition given. From the result of this analysis, we will be able to determine the dimension of the space of invariant curvatures on 𝔭. From Lemma (4.1) and the general theory of symmetric spaces it then follows that there exists a simply connected symmetric space G′/K′ where G′ is simply connected with Lie algebra 𝔤 and K′ is the connected subgroup with Lie algebra 𝔨. G′/K′ is not yet effective. By construction, the infinitesimal isotropy representation of G′/K′ is equal to π and since π is an effective representation of K, the effective version of the symmetric space must have isotropy group K, which will then finish the proof.

After finishing this paper, we were kindly informed by one of the referees that the characterization of the isotropy representations of irreducible symmetric spaces in terms of invariant 4-forms on the tangent spaces has been studied by Kostant , see [Ko2].

A. The Orthogonal Case.

(4.2) Theorem. Let π: K → SO(N) be a faithful, orthogonal representation of a compact connected Lie group K such that SO(N)/π(K) is strongly isotropy irreducible, nonsymmetric, and not SO(7)/ G_{2}. Then (K, π) is the isotropy representation of a compact, simply connected, irreducible symmetric space G/K. Furthermore, K is a simple Lie group unless K = Sp(n) Sp(1) with π = ν_{n} ˆ⊗ν_{2}.

Proof. It follows from §1 that Λ^{2}π = Ad_{K} ⊕ χ and our assumptions imply that χ is irreducible over ℝ. If π is not irreducible, then π = π_{1} ⊕ π_{2}. But π_{1} and π_{2} must both be orthogonal (of dimension p and q) and hence K ⊂ SO(p)×SO(q) ⊂ SO(N) with N = p+q. But then the isotropy representation of [SO(p) × SO(q)]/K is contained in χ and hence K = SO(p) × SO(q) and π = ρ_{p} ˆ⊗1 ⊕ 1 ˆ⊗ρ_{q}. In this case V is the Grassmannian and hence already symmetric.

If π is not absolutely irreducible, then π⊗ℂ = σ⊕σ^{∗} where σ is a unitary n-dimensional representation. But then K ⊂ U(n) ⊂ SO(N) with N = 2n and as before it follows that K = U(n) and hence V is a symmetric space.

We can therefore assume that π is absolutely irreducible and hence also that K is semisimple. Let 𝔭 be the representation space on which π acts and denote by b the inner product on 𝔭, invariant under π. Then π: 𝔨 → 𝔰o(𝔭, b). Furthermore, 𝔰o(𝔭, b) carries a natural symmetric bilinear form given by < A, B > = −\frac{1}{2} tr(AB) which induces an inner product on 𝔨, again denoted by < , >, via the embedding π. Let λ be the dominant weight of π ⊗ ℂ and α be a simple root of 𝔨 such that < λ, α > ≠ 0. It follows from (2.2c) that π_{2λ−α} ⊂ χ ⊗ ℂ unless (i) χ = 0, (ii) 𝔥 = G_{2}, π_{λ} = ϕ_{7}, (iii) 𝔥 = 𝔰o(7), π_{λ} = Δ_{7} , or (iv) 𝔥 = 𝔰p(n) ⊕ 𝔰p(1) , π_{λ} = ν_{2n} ⊗ ν_{2} and α is the simple root of 𝔰p(1). In case (i), V is only a point. Case (ii) is excluded by assumption and in case (iii) V = SO(8)/ Spin(7) = (SO(8)/{±Id})/ SO(7) = ℝP^{7} which is a symmetric space and hence also excluded. In case (iv) π is already the isotropy representation of the quaternionic symmetric space ℍP^{n}. Notice also that it follows from (3.1c) that SO(4n)/π(Sp(n) Sp(1)) is indeed strongly isotropy irreducible.

Hence we can assume that π_{2λ−α} ⊂ χ ⊗ ℂ. Isotropy irreducibility implies that either χ ⊗ ℂ = π_{2λ−α} or χ ⊗ ℂ = π_{2λ−α} ⊕ π_{2λ−α}^{∗}. Hence we may assume that α and α^{∗} are the only simple roots β of 𝔨 such that < λ, β > ≠ 0. By faithfulness, 𝔨 is simple.

Consider now S^{2}Λ^{2}π = S^{2}(ad(𝔨) ⊕ χ) = S^{2}(ad(𝔨)) ⊕ S^{2}(χ) ⊕ [ad(𝔨) ⊗ χ]. Since σ ⊗ ρ contains a trivial summand iff σ = ρ^{∗}, and since 2λ − α is not a root of 𝔨, it follows that ad(𝔨) ⊗ χ does not contain a trivial summand. But since ad(𝔨) and χ are both orthogonal, S^{2}(ad(𝔨)) and S^{2}(χ) both contain a trivial summand coming from the metric. Hence dim(S^{2}Λ^{2}𝔭)^{𝔨} = 2. But S^{2}Λ^{2}𝔭 = Λ^{4}𝔭 ⊕ R. Each R ∈ R is a symmetric endomorphism of ad(𝔨) ⊕ χ. If R is 𝔨-invariant, then it is a diagonal operator with one eigenvalue for each summand of ad(𝔨) ⊕ χ. Suppose that dim(Λ^{4}𝔭)^{𝔨} = 0, then there are two linearly independent invariant curvature tensors R_{1} and R_{2} in R. It follows that up to a constant there is a unique linear combination of R_{1} and R_{2} that yields a non-trivial invariant curvature tensor with values in ad(𝔨). By (4.1) we would be done. So in the following we will prove that dim(Λ^{4}𝔭)^{𝔨} = 0.

Since π is absolutely irreducible we may work over the complex numbers whenever convenient. As before, we denote by < , > the natural metric on 𝔰o(𝔭), which induces a metric on 𝔨 via the embedding π. We also denote by b the extension of b to a symmetric bilinear form on 𝔭 ⊗ ℂ. b induces in the usual way an inner product on Λ^{2}𝔭 and the identification Λ^{2}𝔭 ≅ 𝔰o(𝔭), defined by < v ∧ w, A > = b(A(v), w) for all v, w ∈ 𝔭, A ∈ 𝔰o(𝔭), becomes a linear isometry. The invariance of b easily implies that b(v_{ρ_{1}}, v_{ρ_{2}}) = 0 for all weight vector v_{ρ_{1}}, v_{ρ_{2}}, unless ρ_{1} + ρ_{2} = 0.

We also choose the usual Chevalley basis {H_{α}, X_{α}, X_{−α}} of 𝔨⊗ℂ so that < X_{α}, X_{−α} >= 1 and [X_{α}, X_{−α}] = H_{α}. Then choose weight vectors v_{λ}, v_{−λ} such that b(v_{λ}, v_{−λ}) = 1 and define v_{λ−α} = π(X_{−α})v_{λ}, v_{α−λ} = π(X_{α})v_{−λ}. We then have

b(v_{λ−α}, v_{α−λ}) = b(π(X_{−α})v_{λ}, π(X_{α})v_{−λ}) = −b(v_{λ}, π(X_{−α})π(X_{α})v_{−λ})
= −b(v_{λ}, π([X_{−α}, X_{α}])v_{−λ}) = b(v_{λ}, π(H_{α})v_{−λ}) = − < λ, α > .

We denote by pr_{𝔨} the orthogonal projection from Λ^{2}𝔭 to 𝔨 via the embedding π. If v_{ρ_{1}}, v_{ρ_{2}} are weight vectors with respect to a maximal torus 𝔱 ⊗ ℂ ⊂ 𝔨 ⊗ ℂ, we claim that pr_{𝔨}(v_{ρ_{1}} ∧ v_{ρ_{2}}) has weight ρ_{1} + ρ_{2}, i.e. lies in 𝔱 ⊗ ℂ if ρ_{1} + ρ_{2} = 0, is 0 if ρ_{1} + ρ_{2} is not a root and is a multiple of X_{ρ_{1}+ρ_{2}} if it is a root. Indeed, [π(X), pr_{𝔨}(v_{ρ_{1}} ∧ v_{ρ_{2}})] = pr_{𝔨} (Λ^{2}π(X)(v_{ρ_{1}} ∧ v_{ρ_{2}})) = (ρ_{1}(X) + ρ_{2}(X))pr_{𝔨}(v_{ρ_{1}} ∧ v_{ρ_{2}}) for all X ∈ 𝔱 ⊗ ℂ. Hence one easily shows that

pr_{𝔨}(v_{λ} ∧ v_{−λ}) = H_{λ}
pr_{𝔨}(v_{λ−α} ∧ v_{α−λ}) = − < λ, α > H_{λ−α}
pr_{𝔨}(v_{λ} ∧ v_{α−λ}) = − < λ, α > X_{α}
pr_{𝔨}(v_{λ−α} ∧ v_{−λ}) =< λ, α > X_{−α}
pr_{𝔨}(v_{λ} ∧ v_{λ−α}) = 0

For example, the first equation follows from < pr_{𝔨}(v_{λ} ∧ v_{−λ}), X > = < v_{λ} ∧ v_{−λ}, π(X) > = b(π(X)v_{λ}, v_{−λ}) = λ(X).

Now let η be a 𝔨-invariant 4-form on 𝔭 and let η̂ denote the corresponding symmetric

endomorphism of Λ^{2}(𝔭). Furthermore, assume that η̂ = ( \begin{matrix}a_{1}1 & 0 \\ 0 & a_{2}1 \end{matrix} ) with respect to

Λ^{2}π_{λ} = ad(𝔨) ⊕ χ.

Since η is a 4-form, we have

η(v_{λ}, v_{λ−α}, v_{−λ}, v_{α−λ}) = η(v_{λ}, v_{α−λ}, v_{λ−α}, v_{−λ}) = η(v_{λ}, v_{−λ}, v_{α−λ}, v_{λ−α}).

Let us compute each of these explicitly. First

η(v_{λ}, v_{λ−α}, v_{−λ}, v_{α−λ}) = < η̂(v_{λ} ∧ v_{λ−α}), v_{−λ} ∧ v_{α−λ} >
= a_{2} < v_{λ} ∧ v_{λ−α}, v_{−λ} ∧ v_{α−λ} >
= a_{2}(b(v_{λ}, v_{−λ})b(v_{λ−α}, v_{α−λ}) − b(v_{λ}, v_{α−λ})b(v_{λ−α}, v_{−λ}))
= a_{2}(− < λ, α >).

Furthermore

η(v_{λ}, v_{α−λ}, v_{λ−α}, v_{−λ}) = < η̂(v_{λ} ∧ v_{α−λ}), v_{λ−α} ∧ v_{−λ} >
= < −a_{1} < λ, α > X_{α} + a_{2}y_{1} , < λ, α > X_{−α} + z_{1} >
= −a_{1} < λ, α >^{2}< X_{α}, X_{−α} > +a_{2} < y_{1}, z_{1} >
= −a_{1} < λ, α >^{2} +a_{2} < y_{1}, z_{1} >

where y_{1}, z_{1} are in χ_{ℂ}.

To evaluate < y_{1}, z_{1} >, consider < v_{λ} ∧ v_{α−λ}, v_{λ−α} ∧ v_{−λ} > = − < λ, α >^{2} + < y_{1}, z_{1} > = b(v_{λ}, v_{λ−α})b(v_{α−λ}, v_{−λ}) − b(v_{λ}, v_{−λ})b(v_{α−λ}, v_{λ−α}) =< λ, α >, since 2λ − α is not a root. So < y_{1}, z_{1} > = < λ, α > (1+ < λ, α >).

Finally

η(v_{λ}, v_{−λ}, v_{α−λ}, v_{λ−α}) = < η̂(v_{λ} ∧ v_{−λ}), v_{α−λ} ∧ v_{λ−α} >
= < a_{1}H_{λ} + a_{2}y_{2} , < λ, α > H_{λ−α} + z_{2} >, y_{2}, z_{2} ∈ χ ⊗ ℂ
= a_{1} < λ, α > < λ, λ − α > +a_{2} < y_{2}, z_{2} > .

To evaluate < y_{2}, z_{2} > consider

< v_{λ} ∧ v_{−λ}, v_{α−λ} ∧ v_{λ−α} > = < λ, α > < λ, λ − α > + < y_{2}, z_{2} >

= b(v_{λ}, v_{α−λ})b(v_{−λ}, v_{λ−α}) − b(v_{λ}, v_{λ−α})b(v_{−λ}.v_{α−λ}) = 0

So η(v_{λ}, v_{−λ}, v_{α−λ}, v_{λ−α}) = (a_{1} − a_{2}) < λ, α > < λ, λ − α >.

Combining these computations, we obtain the following linear system that has to be satisfied by a_{1} and a_{2}:

a_{1} < λ, α > −a_{2}(2+ < λ, α >) = 0
a_{1} < λ, λ − α > +a_{2}(1− < λ, λ − α >) = 0

The determinant is 2 < λ, λ > − < λ, α > = < λ, 2λ − α >. Since λ and 2λ − α are both dominant weights, the determinant is non-zero. Hence a_{1} = a_{2} = 0 and dim(Λ^{4}𝔭)^{𝔨} = 0, as claimed. □

Remark. For the isotropy irreducible space SO(7)/ G_{2} = ℝP^{7} we have π_{λ} = ϕ_{7}. But 2λ − α is precisely the maximal root of G_{2} and Λ^{2}π_{λ} = ad(G_{2}) ⊕ π_{λ}, and so the above argument fails. Furthermore, one knows that there actually exists a unique 4-form on ℝ^{7} invariant under ϕ_{7} and hence dim(Λ^{4}𝔭)^{𝔨} = 1, contrary to the conclusion in the above proof.

B. The Quaternionic Case.

(4.3) Theorem. Let π′: H → Sp(N) be a faithful, quaternionic representation of a compact connected Lie group H such that Sp(N)/π′(H) is strongly isotropy irreducible and non-symmetric. Then (K = H · Sp(1), π = π′ ⊗ ν_{2}) is the isotropy representation of a compact, simply connected, irreducible symmetric space G/K. Moreover, either H is simple or H = SO(N) Sp(1) with π′ = ρ_{N} ˆ⊗ν_{2}, N ≥ 3.

Proof. In this case we have S^{2}π′ = Ad_{H} ⊕ χ with χ irreducible over ℝ. If π′ is reducible, it follows as in (4.2) that H = Sp(p) × Sp(q) and π′ = ν_{p} ˆ⊗1 ⊕ 1 ˆ⊗ν_{q}, N = p + q, which implies that Sp(N)/π′(H) is symmetric. Similarly, if π′ ⊗ ℂ = σ ⊕ σ^{∗}, then H = U(n), N = 2n, and again Sp(N)/π′(H) is symmetric.

Hence we can assume that H is semisimple and that π′ is absolutely irreducible with dominant weight λ′. If H is not simple, then π′ = π_{1} ˆ⊗π_{2} and we can assume that π_{1} is orthogonal and π_{2} symplectic. Then S^{2}(π_{1} ˆ⊗π_{2}) = S^{2}π_{1} ˆ⊗S^{2}π_{2} ⊕ Λ^{2}π_{1} ˆ⊗Λ^{2}π_{2}. Since π_{1} is orthogonal, we have S^{2}π_{1} = 1 ⊕ α and Λ^{2}π_{1} = ad(𝔥_{1}) ⊕ β, and since π_{2} is symplectic, we have S^{2}π_{2} = ad(𝔥_{2}) ⊕ γ and Λ^{2}π_{2} = 1 ⊕ δ. In order for S^{2}(π_{1} ˆ⊗π_{2}) to be equal to ad(𝔥) ⊕ χ , the only possibility is that Λ^{2}π_{1} = ad(𝔥_{1}) , S^{2}π_{2} = ad(𝔥_{2}) , and Λ^{2}π_{2} = 1 and hence χ = α ˆ⊗ad(𝔥_{2}). Using (2.2) this implies that, up to an automorphism, π_{1} = ρ_{N} and π_{2} = ν_{2} and hence H = SO(N) Sp(1) with π′ = ρ_{N} ˆ⊗ν_{2}. But [π′ ˆ⊗ν_{2}]_{ℝ} is indeed the isotropy representation of the symmetric space G/K = SO(N + 4)/ SO(N) SO(4).

We can therefore assume from now on that H is simple and π′ = π_{λ′}. The outline of our proof in this case is the same as before, but further technicalities arise since we now have to consider curvature tensors and 4-forms on 𝔭 , a real form of the representation space of π ⊗ ℂ = π′ ⊗ ν_{2}. Now Λ^{2}(π_{λ′} ˆ⊗ν_{2}) = Λ^{2}π_{λ′} ˆ⊗ad(sp(1))_{ℂ} ⊕ S^{2}π_{λ′} ˆ⊗1, and since π_{λ′} is symplectic, Λ^{2}π_{λ′} = 1⊕π̃, while isotropy irreducibility implies that S^{2}π_{λ′} = ad(𝔥)_{ℂ}⊕χ with χ irreducible. Therefore, it is important to decide about the nature of the representation π̃. We do this below. For this purpose we introduce some more notation. Let V′ be the representation space of π′ ⊗ ℂ and ω′ the 𝔥-invariant skew symmetric bilinear form on V′. Hence π′ ⊗ ℂ : 𝔥 ⊗ ℂ → 𝔰p(V′, ω′). On 𝔰p(V′, ω′) we have the natural inner product < A, B > = −\frac{1}{2}tr(AB) which induces an inner product on 𝔥 ⊗ ℂ, again denoted by < , >, via the embedding π′ ⊗ ℂ. ω′ induces a symmetric bilinear form on S^{2}V′, which we also denote by < , > and with respect to this form the equation < v ∘ w, A > = ω′(A(v), w) induces a natural isomorphism S^{2}V′ ≅ 𝔰p(V′, ω′) which also respects the natural inner products on both sides.

(4.4) Proposition. With the same assumptions as in (4.3) and the assumption that H is simple, we have

(a) There is a unique simple root α′ of 𝔥 such that < λ′, α′ > ≠ 0. Furthermore, < λ′, λ′ > = \frac{3}{2} < λ′, α′ > = −3 dim H/ dim Sp(N).

(b) Λ^{2}π_{λ′} = 1 ⊕ π_{2λ′−α′}.

Proof of (a). As in the beginning of the proof of (4.3), it follows that π′ is absolutely irreducible with dominant weight λ′. Also observe that by (2.2b) it follows that χ = π_{2λ′}. Furthermore, since < λ′, α′ > ≠ 0, 2λ′ − α′ is a weight in π_{2λ′} and hence does not lie in ad(𝔥) ⊗ ℂ.

There is a decomposition S^{2}S^{2}V′ = S^{4}V′ ⊕ R′ invariant under Sp(V′, ω′) analogous to the decomposition for the orthogonal group. Here R′ is the space of algebraic curvature tensors on V′, the only difference being that V′ is a complex vector space. If R ∈ R′, we

let R^{∘} denote the corresponding symmetric endomorphism of S^{2}V′.

Corresponding to π′ ⊗ ℂ : 𝔥 ⊗ ℂ → 𝔰p(V′, ω′) and < , > is the projection map pr_{𝔥} from S^{2}V′ to 𝔥⊗ℂ. As before we define weight vectors v_{λ′}, v_{−λ′}, v_{λ′−α′} = π′(X_{−α′})v_{λ′}, v_{α′−λ′} = π′(X_{α′})v_{−λ′} of π′ such that ω′(v_{λ′}, v_{−λ′}) = 1 and it follows again that ω′(v_{λ′−α′}, v_{α′−λ′}) = − < λ′, α′ >. We also have that ω′(v_{ρ_{1}}, v_{ρ_{2}}) = 0 unless ρ_{1} + ρ_{2} = 0 and that pr_{𝔥}(v_{ρ_{1}} ∘ v_{ρ_{2}}) has weight ρ_{1} + ρ_{2}. Hence one easily shows that

pr_{𝔥}(v_{λ′} ∘ v_{−λ′}) = H_{λ′}
pr_{𝔥}(v_{λ′−α′} ∘ v_{α′−λ′}) = − < λ′, α′ > H_{λ′−α′}
pr_{𝔥}(v_{λ′} ∘ v_{α′−λ′}) = − < λ′, α′ > X_{α′}
pr_{𝔥}(v_{−λ′} ∘ v_{λ′−α′}) = < λ′, α′ > X_{−α′}
pr_{𝔥}(v_{λ′} ∘ v_{λ′−α′}) = 0

Define

η(v_{1}, v_{2}, v_{3}, v_{4}) = ω′(pr_{𝔥}(v_{1} ∘ v_{2})v_{3} + pr_{𝔥}(v_{2} ∘ v_{3})v_{1} + pr_{𝔥}(v_{3} ∘ v_{1})v_{2}, v_{4})

Using the fact that

ω′(pr_{𝔥}(v_{1}∘v_{2})v_{3}, v_{4}) = < v_{3}∘v_{4}, pr_{𝔥}(v_{1}∘v_{2}) > = < pr_{𝔥}(v_{1}∘v_{2}), v_{3}∘v_{4} > = ω′(pr_{𝔥}(v_{3}∘v_{4})v_{1}, v_{2})

it easily follows that η ∈ S^{4}V′. Furthermore η is 𝔥-invariant and non-trivial since η(v_{λ′}, v_{−λ′}, v_{λ′}, v_{−λ′}) = 2 < λ′, λ′ >. One can also check that any element in S^{4}V′ considered as a symmetric endomorphism of S^{2}V′ has trace 0. Since η is 𝔥-invariant, η^{∘} has diagonal form with respect to the decomposition S^{2}V′ = [ad(𝔥) ⊗ ℂ] ⊕ [χ ⊗ ℂ]. Since η^{∘} is nonzero and tr(η^{∘}) = 0, its two eigenvalues are nonzero. By renormalizing η, we may assume

that η^{∘}|_{ad(𝔥)⊗ℂ} = 1 and η^{∘}|_{χ⊗ℂ} = a1. We have η(v_{λ′}, v_{−λ′}, v_{λ′}, v_{−λ′}) = η(v_{λ′}, v_{λ′}, v_{−λ′}, v_{−λ′})

and η(v_{λ′}, v_{α′−λ′}, v_{−λ′}, v_{λ′−α′}) = η(v_{λ′}, v_{λ′−α′}, v_{−λ′}, v_{α′−λ′}).

Let v_{λ′} ∘v_{−λ′} = H_{λ′} +x and v_{λ′−α′} ∘v_{α′−λ′} = − < λ′, α′ > H_{λ′−α′} +y, where x, y ∈ χ⊗ℂ. We then have

η(v_{λ′}, v_{−λ′}, v_{λ′}, v_{−λ′}) = < η^{∘}(v_{λ′} ∘ v_{−λ′}), v_{λ′} ∘ v_{−λ′} >
= < H_{λ′} + ax, H_{λ′} + x > =< λ′, λ′ > +a < x, x >
η(v_{λ′}, v_{λ′}, v_{−λ′}, v_{−λ′}) = < η^{∘}(v_{λ′} ∘ v_{λ′}), v_{−λ′} ∘ v_{−λ′} > = a < v_{λ′} ∘ v_{λ′}, v_{−λ′} ∘ v_{−λ′} >
= a(ω(v_{λ′}, v_{−λ′})^{2} + ω(v_{λ′}, v_{−λ′})^{2}) = 2a .

To determine < x, x >, we use

< v_{λ′} ∘ v_{−λ′}, v_{λ′} ∘ v_{−λ′} > = ω′(v_{λ′}, v_{λ′})ω′(v_{−λ′}, v_{−λ′}) + ω′(v_{λ′}, v_{−λ′})ω′(v_{−λ′}, v_{λ′})
= −1 = < λ′, λ′ > + < x, x >

Combining these equations, we obtain a = \frac{<λ′,λ′>}{3+<λ′,λ′>}. On the other hand,

η(v_{λ′}, v_{α′−λ′}, v_{−λ′}, v_{λ′−α′}) = < η^{∘}(v_{λ′} ∘ v_{α′−λ′}), v_{−λ′} ∘ v_{λ′−α′} >
= < − < λ′, α′ > X_{α′} + ax′ , < λ′, α′ > X_{−α′} + y′ >
= − < λ′, α′ >^{2} +a < x′, y′ >
η(v_{λ′}, v_{λ′−α′}, v_{−λ′}, v_{α′−λ′}) = < η^{∘}(v_{λ′} ∘ v_{λ′−α′}), v_{−λ′} ∘ v_{α′−λ′} >
= a < v_{λ′} ∘ v_{λ′−α′}, v_{−λ′} ∘ v_{α′−λ′} >
= aω′(v_{λ′}, v_{−λ′})ω′(v_{λ′−α′}, v_{α′−λ′})
= −a < λ′, α′ >

In addition,

< v_{λ′} ∘ v_{α′−λ′}, v_{−λ′} ∘ v_{λ′−α′} > = − < λ′, α′ >^{2} + < x′, y′ >

= ω′(v_{λ′}, v_{−λ′})ω′(v_{α′−λ′}, v_{λ′−α′})

= < λ′, α′ >

Combining these equations, we get a = \frac{<λ′,α′>}{2+<λ′,α′>}. Equating the expressions for a, we

get < λ′, λ′ > = \frac{3}{2} < λ′, α′ >. Finally, since tr(η^{∘}) = 0, dim H + a dim(Sp(N)/π′(H)) = 0, or \frac{<λ′,λ′>}{3+<λ′,λ′>} = \frac{− dim H}{dim Sp(N)−dim H}. Hence < λ′, λ′ > = \frac{−3 dim H}{dim Sp(N)}.

In the above, α′ is any simple root such that < λ′, α′ > ≠ 0. If another simple root has this property, we obtain a contradiction as in the proof of (2.12). This finishes the proof of (a). □

Before we continue with the proof of (4.4b), we first prove the following general Lemma:

(4.5) Lemma. If π is a unitary representation of a compact Lie algebra 𝔥, then tr C_{Λ^{2}π} = (dim π − 2) tr C_{π}, where the Casimir operators are taken with respect to any biinvariant metric < , > on 𝔥.

Proof. Remember that C_{π} = − ∑ π(X_{i})π(X_{i}), where X_{i} is an orthonormal basis of 𝔥 ⊗ ℂ with respect to < , >. If g is the invariant metric on the representation space for π with orthonormal basis e_{i}, then it induces a metric on the representation space of Λ^{2}π with orthonormal basis e_{i} ∧ e_{j}, i < j. We then have

tr C_{Λ^{2}π} = ∑_{i<j} g(C_{Λ^{2}π}(e_{i} ∧ e_{j}), e_{i} ∧ e_{j})

= −\frac{1}{2} ∑_{i,j} ∑_{k} g(Λ^{2}π(X_{k})Λ^{2}π(X_{k})(e_{i} ∧ e_{j}), e_{i} ∧ e_{j})

= −\frac{1}{2} ∑_{i,j} ∑_{k} g([π(X_{k})π(X_{k})e_{i}] ∧ e_{j} + e_{i} ∧ [π(X_{k})π(X_{k})e_{j}]

+ 2π(X_{k})e_{i} ∧ π(X_{k})e_{j}, e_{i} ∧ e_{j})

= \frac{1}{2} ∑_{i,j} g([C_{π}e_{i}] ∧ e_{j} + e_{i} ∧ [C_{π}e_{j}] − 2 ∑_{k} π(X_{k})e_{i} ∧ π(X_{k})e_{j}, e_{i} ∧ e_{j})

= (dim π − 1) ∑_{i} g(C_{π}e_{i}, e_{i})

− ∑_{i,j} ∑_{k}{g(π(X_{k})e_{i}, e_{i})g(π(X_{k})e_{j}, e_{j}) − g(π(X_{k})e_{i}, e_{j})g(π(X_{k})e_{j}, e_{i})}

= (dim π − 1) tr C_{π} + ∑_{i,j} ∑_{k} g(π(X_{k})e_{i}, e_{j})g(π(X_{k})e_{j}, e_{i})

where we have used the fact that π(H) ⊂ SU(dim π) and hence tr π(X_{k}) = 0. The last term is equal to ∑_{k,i} g(π(X_{k})π(X_{k})e_{i}, e_{i}) = − tr C_{π} which implies our claim. □

Proof of(4.4b). Since π_{λ′} is symplectic, it follows that 1 ∈ Λ^{2}π_{λ′} and since π_{λ′} is irreducible, 1 occurs only once in Λ^{2}π_{λ′}. Furthermore π_{2λ′−α′} ∈ Λ^{2}π_{λ′} since < λ′, α′ > ≠ 0. It follows from (4.5) that

tr C_{Λ^{2}π_{λ′}} = (2N − 2) tr C_{π_{λ′}} = −(2N − 2)(2N) < λ′, λ′ + 2δ′ > .

Any irreducible summand in Λ^{2}π_{λ′} besides 1 and π_{2λ′−α′} has dominant weight of the form 2λ′ − α′− (sum of simple roots) and hence has a smaller Casimir constant. Hence, if there are any other irreducible summands, we have

tr C_{Λ^{2}π_{λ′}} < − (\binom{2N}{2} − 1) < 2λ′ − α′, 2λ′ − α′ + 2δ′ >
= − (2N + 1)(N − 1) [2 < λ′, λ′ + 2δ′ > +2 < λ′, λ′ − α′ >
− 2 < λ′, α′ > + < α′, α′ > −2 < δ′, α′ >]

Using the fact that δ′ is also the sum of all basic weights, we see that < α′, α′ > = 2 < δ′, α′ >. Using this and 2 < λ′, λ′ > = 3 < λ′, α′ > we obtain

−2N(2N − 2) < λ′, λ′ + δ′ > < −(2N + 1)(N − 1) [2 < λ′, λ′ + 2δ′ > − < λ′, α′ >] ,

or

2 < λ′, λ′ + 2δ′ > < (2N + 1) < λ′, α′ > .

Since − < λ′, λ′ + 2δ′ > is the Casimir constant of C_{π_{λ′}} we obtain immediately that −2N < λ′, λ′ + 2δ′ > = tr C_{π_{λ′}} = − ∑_{i} tr(π_{λ′}(X_{i})π_{λ′}(X_{i})) = 2 ∑_{i} < X_{i}, X_{i} > = 2 dim H. Using the expression for < λ′, α′ > in (a), we finally obtain

\frac{2 dim H}{−N} < \frac{−2 dim H}{N} ,

a contradiction. Therefore, there cannot be any further irreducible summands in Λ^{2}π_{λ′}. □

Proof of (4.3) continued. Using (4.4), we see that

Λ^{2}(π ⊗ ℂ) = Λ^{2}(π_{λ′} ˆ⊗ν_{2}) = Λ^{2}π_{λ′} ˆ⊗ad(sp(1))_{ℂ} ⊕ S^{2}π_{λ′} ˆ⊗1
= ad(𝔥)_{ℂ} ˆ⊗1 ⊕ 1 ˆ⊗ad(𝔰p(1))_{ℂ} ⊕ [π_{2λ′−α′} ˆ⊗ad(𝔰p(1))_{ℂ}] ⊕ [π_{2λ′} ˆ⊗1]

Every 𝔨-invariant element of S^{2}Λ^{2}(𝔭) is determined by its four eigenvalues a_{1}, a_{2}, b_{1}, b_{2} corresponding to the above decomposition for Λ^{2}(π_{λ′} ˆ⊗ν_{2}). Since the four irreducible summands of Λ^{2}(π_{λ′} ˆ⊗ν_{2}) are mutually inequivalent and absolutely irreducible, we see that dim(S^{2}Λ^{2}𝔭)^{𝔨} = dim_{ℂ}[S^{2}Λ^{2}(π_{λ′} ˆ⊗ν_{2})]^{𝔨} = 4. Because the symmetric space we are attempting to construct would be quaternionic symmetric, we should expect that dim[Λ^{4}(𝔭)]^{𝔨} = 1. In that event, the space of invariant curvature tensors on 𝔭 would have dimension 3. Hence there would be a linear combination of the three basis elements of R that gave an invariant curvature tensor with values in 𝔨. By (4.1) we would be done. Accordingly, we will prove that dim[Λ^{4}(𝔭)]^{𝔨} = 1.

Next we describe the various bilinear forms that we will use in the calculations below. First, just as π_{λ′} has the representation space V′ with an 𝔥-invariant skew-symmetric form ω′, ν_{2} also has a representation space V′′ with an 𝔰p(1)-invariant skew form ω′′. V′ ⊗V′′ is then the representation space of π_{λ′} ˆ⊗ν_{2} and it has a 𝔨-invariant non-degenerate symmetric bilinear form q given by q(x′ ⊗ x′′, y′ ⊗ y′′) = ω′(x′, y′)ω′′(x′′, y′′). The space 𝔭 is a Kinvariant real form of V′ ⊗V′′ on which q is positive definite. q in turn induces a 𝔨-invariant symmetric bilinear form on Λ^{2}(V′ ⊗ V′′), to be denoted by ( , )_{Λ^{2}(V′⊗V′′)}. On the other hand, there is an isomorphism θ: Λ^{2}(V′ ⊗ V′′) ≅ S^{2}V′ ⊗ Λ^{2}V′′ ⊕ Λ^{2}V′ ⊗ S^{2}V′′ where (x′ ⊗ x′′) ∧ (y′ ⊗ y′′) is mapped to ((x′ ∘ y′) ⊗ (x′′ ∧ y′′), (x′ ∧ y′) ⊗ (x′′ ⊗ y′′)). S^{2}V′, Λ^{2}V′, S^{2}V′′, Λ^{2}V′′ all have invariant non-degenerate symmetric bilinear forms induced by ω′ and ω′′. These in turn induce a non-degenerate symmetric bilinear form on S^{2}V′ ⊗ Λ^{2}V′′ ⊕ Λ^{2}V′ ⊗ S^{2}V′′. Under the above isomorphism, one can check that ( , )_{Λ^{2}(V′⊗V′′)} corresponds to \frac{1}{2} times the form < , >_{S^{2}V′} ⊗ < , >_{Λ^{2}V′′} ⊥ < , >_{Λ^{2}V′} ⊗ < , >_{S^{2}V′′} on the right.

Let λ′′ be the dominant weight of ν_{2} and α′, α′′ be respectively the unique simple root of 𝔥 and 𝔰p(1) such that < λ′, α′ > ≠ 0, < λ′′, α′′ > ≠ 0. Note that 2λ′′ = α′′. Let λ = λ′ +λ′′ be the dominant weight of π_{λ′} ⊗ ν_{2}. Next we choose weight vectors v_{λ′}, v_{−λ′}, v_{λ′−α′}, v_{α′−λ′} of π_{λ′} and v_{λ′′}, v_{−λ′′} of ν_{2} with the following properties:

ω′(v_{λ′}, v_{−λ′}) = 1 , ω′′(v_{λ′′}, v_{−λ′′}) = 1
v_{λ′−α′} = π(X_{−α′})v_{λ′} , v_{α′−λ′} = π(X′_{α})v_{λ′}
ω′(v_{λ′−α′}, v_{α′−λ′}) = − < λ′, α′ > .

Since λ′′ − α′′ = −λ′′, we define v_{λ′′−α′′} = v_{−λ′′} and v_{α′′−λ′′} = v_{λ′′} and hence

ω′′(v_{λ′′−α′′}, v_{α′′−λ′′}) = −1.

As before it follows that

pr_{𝔥}(v_{λ′} ∘ v_{−λ′}) = H_{λ′}
pr_{𝔥}(v_{λ′−α′} ∘ v_{α′−λ′}) = − < λ′, α′ > H_{λ′−α′}
pr_{𝔥}(v_{λ′} ∘ v_{α′−λ′}) = − < λ′, α′ > X_{α′}
pr_{𝔥}(v_{λ′−α′} ∘ v_{−λ′}) =< λ′, α′ > X_{−α′}
pr_{𝔥}(v_{λ′′} ∘ v_{−λ′′}) = H_{λ′′}
pr_{𝔥}(v_{λ′′−α′′} ∘ v_{α′′−λ′′}) = H_{λ′′}

We denote by pr_{1} the projection of Λ^{2}V′ = 1 ⊕ π_{2λ′−α′} onto the trivial summand. Let Ω′ and Ω′′ respectively be the generators of the trivial representations in Λ^{2}V′, Λ^{2}V′′ subject to the normalization condition

ω′(x′, y′) = < Ω′, x′ ∧ y′ >_{Λ^{2}V′} , ω′′(x′′, y′′) =< Ω′′, x′′ ∧ y′′ >_{Λ^{2}V′′} .

It follows from this that < Ω′, Ω′ > = N and < Ω′′, Ω′′ > = 1.

Now one easily shows

pr_{1}(v_{λ′} ∧ v_{−λ′}) = \frac{1}{N} Ω′
pr_{1}(v_{α′−λ′} ∧ v_{λ′−α′}) = \frac{1}{N} < λ′, α′ > Ω′
pr_{1}(v_{λ′} ∧ v_{α′−λ′}) = 0
pr_{1}(v_{λ′−α′} ∧ v_{−λ′}) = 0
v_{λ′′} ∧ v_{−λ′′} = Ω′′
v_{α′′−λ′′} ∧ v_{λ′′−α′′} = Ω′′

Let η ∈ Λ^{4}(π_{λ′} ˆ⊗π_{λ′′})^{𝔨} and define v_{λ−α′} = v_{λ′−α′} ⊗ v_{λ′′} and v_{λ−α′′} = v_{λ′} ⊗ v_{λ′′−α′′}, and so forth. Since η is a 4-form, we have for β = α′ or α′′.

η(v_{λ}, v_{λ−β}, v_{−λ}, v_{β−λ}) = η(v_{λ}, v_{−λ}, v_{β−λ}, v_{λ−β}) = η(v_{λ}, v_{β−λ}, v_{λ−β}, v_{−λ}).

These give rise to 4 homogeneous linear equations in a_{1}, a_{2}, b_{1}, b_{2}. We will show that this system has rank 3, from which our claim follows. For this purpose we produce a 3 × 3 minor with non-vanishing determinant.

(a) η(v_{λ}, v_{λ−α′}, v_{−λ}, v_{α′−λ}) = (η̂(v_{λ} ∧ v_{λ−α′}), v_{−λ} ∧ v_{α′−λ})_{Λ^{2}(V′⊗V′′)}. Now

θ(v_{λ} ∧ v_{λ−α′}) = θ((v_{λ′} ⊗ v_{λ′′}) ∧ (v_{λ′−α′} ⊗ v_{λ′′}))
= (v_{λ′} ∧ v_{λ′−α′}) ⊗ (v_{λ′′} ∘ v_{λ′′}) ∈ π_{2λ′−α′} ˆ⊗ad(𝔰p(1))_{ℂ}

Therefore,

η(v_{λ}, v_{λ−α′}, v_{−λ}, v_{α′−λ}) = b_{1}(v_{λ} ∧ v_{λ−α′}, v_{−λ} ∧ v_{α′−λ})_{Λ^{2}(V′⊗V′′)}
= b_{1}(q(v_{λ}, v_{−λ})q(v_{λ−α′}, v_{α′−λ}) − q(v_{λ}, v_{α′−λ})q(v_{λ−α′}, v_{−λ}))
= b_{1}(ω′(v_{λ′−α′}, v_{α′−λ′}) − 0) = −b_{1} < λ′, α′ >

(b) In an analogous fashion, η(v_{λ}, v_{λ−α′′}, v_{−λ}, v_{α′′−λ}) = −b_{2}.

(c) η(v_{λ}, v_{−λ}, v_{α′−λ}, v_{λ−α′}) = (η̂(v_{λ} ∧ v_{−λ}), v_{α′−λ} ∧ v_{λ−α′})_{Λ^{2}(V′⊗V′′)}. Now

θ(v_{λ} ∧ v_{−λ}) = θ((v_{λ′} ⊗ v_{λ′′}) ∧ (v_{−λ′} ⊗ v_{−λ′′}))

= (v_{λ′} ∘ v_{−λ′}) ⊗ (v_{λ′′} ∧ v_{−λ′′}) + (v_{λ′} ∧ v_{−λ′}) ⊗ (v_{λ′′} ∘ v_{−λ′′})

= (H_{λ′} + s_{1}) ⊗ Ω′′ + (\frac{1}{N} Ω′ + x_{1}) ⊗ H_{λ′′}

with s_{1} ∈ π_{2λ′} and x_{1} ∈ π_{2λ′−α′}.

So θ(η̂(v_{λ} ∧ v_{−λ})) = a_{1}H_{λ′} ⊗ Ω′′ + \frac{a_{2}}{N} Ω′ ⊗ H_{λ′′} + b_{1}x_{1} ⊗ H_{λ′′} + b_{2}s_{1} ⊗ Ω′′. Next,

θ(v_{α′−λ} ∧ v_{λ−α′}) = θ((v_{α′−λ′} ⊗ v_{−λ′′}) ∧ (v_{λ′−α′} ⊗ v_{λ′′}))

= −(v_{α′−λ′} ∘ v_{λ′−α′}) ⊗ (v_{λ′′} ∧ v_{−λ′′}) + (v_{α′−λ′} ∧ v_{λ′−α′}) ⊗ (v_{λ′′} ∘ v_{−λ′′})

= (< λ′, α′ > H_{λ′−α′} − s_{2}) ⊗ Ω′′ + (\frac{1}{N} < λ′, α′ > Ω′ + x_{2}) ⊗ H_{λ′′}

with s_{2} ∈ π_{2λ′} and x_{2} ∈ π_{2λ′−α′}. Hence

η(v_{λ},v_{−λ}, v_{α′−λ}, v_{λ−α′}) = \frac{1}{2} < θ(η̂(v_{λ} ∧ v_{−λ})), θ(v_{α′−λ} ∧ v_{λ−α′}) >

= \frac{1}{2}\{a_{1} < λ′, α′ > < λ′ − α′, λ′ > −b_{2} < s_{1}, s_{2} >_{Λ^{2}V′} +\frac{a_{2}}{N} < λ′, α′ > < λ′′, λ′′ >

+ b_{1} < x_{1}, x_{2} >_{Λ^{2}V′}< λ′′, λ′′ >}

In order to compute < s_{1}, s_{2} >_{S^{2}V′} and < x_{1}, x_{2} >_{Λ^{2}V′}, we examine < v_{λ′}∧v_{−λ′}, v_{α′−λ′}∧ v_{λ′−α′} >_{Λ^{2}V′}= 0 and < v_{λ′} ∘ v_{−λ′}, v_{α′−λ′} ∘ v_{λ′−α′} >_{S^{2}V′}= 0 and obtain < x_{1}, x_{2} >_{Λ^{2}V′}= −\frac{<λ′,α′>}{N} and < s_{1}, s_{2} >_{S^{2}V′}= < λ′, α′ > < λ′, λ′ − α′ >. Substituting these in the above, we get

η(v_{λ}, v_{−λ}, v_{α′−λ}, v_{λ−α′}) = \frac{1}{2} < λ′, α′ > \{a_{1} < λ′ − α′, λ′ > − b_{2} < λ′ − α′, λ′ >

+ \frac{a_{2}}{N} < λ′′, λ′′ > −\frac{b_{1}}{N} < λ′′, λ′′ >\}.

(d) η(v_{λ}, v_{−λ}, v_{α′′−λ}, v_{λ−α′′}) = (η̂(v_{λ} ∧ v_{−λ}), v_{α′′−λ} ∧ v_{λ−α′′})_{Λ^{2}(V′⊗V′′)}

= \frac{1}{2} < θ(η̂(v_{λ} ∧ v_{−λ})), θ(v_{α′′−λ} ∧ v_{λ−α′′}) >

The first argument has been computed in (c). Now

θ(v_{α′′−λ} ∧ v_{λ−α′′}) = θ((v_{−λ′} ⊗ v_{α′′−λ′′}) ∧ (v_{λ′} ⊗ v_{λ′′−α′′}))

= −(v_{λ′} ∘ v_{−λ′}) ⊗ (v_{λ′′−α′′} ∧ v_{α′′−λ′′}) − (v_{λ′} ∧ v_{−λ′}) ⊗ (v_{λ′′−α′′} ∘ v_{α′′−λ′′})

= (H_{λ′} + s_{1}) ⊗ Ω′′ − (\frac{1}{N} Ω′ + x_{1}) ⊗ H_{λ′′}

Also, < v_{λ′} ∧ v_{−λ′}, v_{λ′} ∧ v_{−λ′} >_{Λ^{2}V′}= 1 and < v_{λ′} ∘ v_{−λ′}, v_{λ′} ∘ v_{−λ′} >_{S^{2}V′}= −1 from which we get < x_{1}, x_{1} >_{Λ^{2}V′}= 1 − \frac{1}{N} and < s_{1}, s_{1} >_{S^{2}V′}= −1− < λ′, λ′ >. Therefore,

η(v_{λ}, v_{−λ}, v_{α′′−λ}, v_{λ−α′′}) = \frac{1}{2}\{a_{1} < λ′, λ′ > −b_{2}(1+ < λ′, λ′ >) − \frac{a_{2}}{N} < λ′′, λ′′ >

− b_{1}(1 − \frac{1}{N}) < λ′′, λ′′ >\}.

(e) A similar computation shows that

η(v_{λ}, v_{α′−λ}, v_{λ−α′}, v_{−λ}) = \frac{< λ′, α′ >}{2}\{−a_{1} < λ′, α′ > −b_{1} < λ′′, λ′′ > +b_{2}(1+ < λ′, α′ >)\}.

Combining these computations and using the fact that < λ′, λ′ > = \frac{3}{2} < λ′, α′ > and setting c′ = \frac{1}{2} < λ′, α′ >, c′′ = < λ′′, λ′′ > we obtain the following linear system for a_{1}, a_{2}, b_{1}, b_{2} :

( \begin{matrix}c′ & \frac{c′′}{N} & 2 − \frac{c′′}{N} & −c′ \\ 2c′ & 0 & c′′ − 2 & −(1 + 2c′) \\ 3c′ & −\frac{c′′}{N} & −(1 − \frac{1}{N})c′′ & 1 − 3c′ \\ ∗ & ∗ & ∗ & ∗ \end{matrix} ) ( \begin{matrix}a_{1} \\ a_{2} \\ b_{1} \\ b_{2} \end{matrix} ) = 0

It easily follows that the 3 × 3 minor consisting of the first three rows and columns 1, 2, and 4 has non-zero determinant and hence dim(Λ^{4}(𝔭)^{𝔨}) = dim_{ℂ}[Λ^{4}(V′ ⊗ V′′)]^{𝔨} ≤ 1. But equality holds since we may define η(u_{1} ⊗ v_{1}, u_{2} ⊗ v_{2}, u_{3} ⊗ v_{3}, u_{4} ⊗ v_{4}) by

ω′(u_{1}, u_{2})ω′′(v_{3}, v_{4}) + ω′(u_{1}, u_{4})ω′′(v_{2}, v_{3}) − ω′(u_{1}, u_{3})ω′′(v_{2}, v_{4})
+ω′(u_{3}, u_{4})ω′′(v_{1}, v_{2}) + ω′(u_{2}, u_{3})ω′′(v_{1}, v_{4}) − ω′(u_{2}, u_{4})ω′′(v_{1}, v_{3})

for u_{i} ∈ V′ , v_{i} ∈ V′′ and one can check that η is 𝔨-invariant since ω′, ω′′ are invariant, and η(v_{λ′} ⊗ v_{λ′′}, v_{−λ′} ⊗ v_{−λ′′}, v_{−λ′} ⊗ v_{λ′′}, v_{λ′−α} ⊗ v_{−λ′′}) = 1 so that η ≠ 0. This completes the proof of (4.3). □

C. The Hermitian Case.

(4.6) Theorem. Let π′ = H → SU(N) be a faithful, unitary representation of a compact connected Lie group H such that SU(N)/π′(H) is strongly isotropy irreducible and nonsymmetric. Then (K = H ·U(1), π = [π′ ⊗ϕ]_{ℝ}) is the isotropy representation of a compact, simply connected, irreducible Hermitian symmetric space. Moreover, either H is simple or else H = SU(p) · SU(q), p ≥ q ≥ 2, π′ = μ_{p} ˆ⊗μ_{q}^{∗}.

Proof. In this case we have π′⊗π′^{∗} = 1⊕Ad_{H} ⊕χ with χ irreducible over ℝ. Furthermore, π′ is already a complex representation. If it is not irreducible, then it follows as in (4.2) that H = S(U(p) × U(q)) and π′ = μ_{p} ˆ⊗1 ⊕ 1 ˆ⊗μ_{q}, N = p + q, which implies that SU(N)/π′(H) is symmetric. Hence π′ is an irreducible complex representation with dominant weight λ′. Since π′ is faithful, this also implies that H must be semisimple. If π′ ≅ π′^{∗}, then π′ would be orthogonal or symplectic and hence H ⊂ SO(n) ⊂ SU(n) or H ⊂ Sp(n) ⊂ SU(2n). Isotropy irreducibility then implies that H = SO(n) or H = Sp(n) and hence SU(N)/π′(H) is symmetric. We can therefore also assume that λ′ ≠ λ′^{∗}. If H is not simple, it follows as in the proof of (4.3) that up to an automorphism H = SU(p)SU(q), p ≥ q ≥ 2 and π′ = μ_{p} ˆ⊗μ_{q}^{∗}. But then [π′ ˆ⊗ϕ]_{ℝ} is indeed the isotropy representation of the symmetric space SU(p + q)/S(U(p) U(q)). We can therefore assume from now on that H is simple and π′ = π_{λ′} with λ′ ≠ λ′^{∗}.

As in the previous proofs, we have to consider curvature tensors on

Λ^{2}(π_{λ′} ˆ⊗ϕ ⊕ π_{λ′}^{∗} ˆ⊗ϕ^{∗}) = Λ^{2}π_{λ′} ˆ⊗ϕ^{2} ⊕ Λ^{2}π_{λ′}^{∗} ˆ⊗(ϕ^{∗})^{2} ⊕ 1 ˆ⊗1 ⊕ ad(𝔥)_{ℂ} ˆ⊗1 ⊕ χ_{ℂ} ˆ⊗1.

We again must first analyse Λ^{2}π_{λ′}. For this purpose we need the following general Lemma.

(4.7) Lemma. Let π_{λ}: H → SU(N) be an irreducible unitary representation of a compact connected simple Lie group H. Then S^{2}π_{λ} is irreducible iff H = SU(N) and π_{λ} = μ_{N} or μ_{N}^{∗} or H = Sp(n) and π_{λ} = ν_{2n} with N = 2n.

Proof. Let V ≅ ℂ^{N} denote the representation space of π_{λ} and < , > an ad(𝔥)-invariant inner product on 𝔥. Furthermore, let {H_{1}, ⋯, H_{r}} and {H_{1}^{∗}, ⋯, H_{r}^{∗}} be dual bases of 𝔱 ⊗ ℂ with respect to < , >, where 𝔱 is a maximal abelian subalgebra of 𝔥, and let {H_{i}, X_{α}, X_{−α}} form a Weyl-Chevalley basis of 𝔥 ⊗ ℂ. Then the Casimir operator C_{π_{λ}} with respect to < , > is − ∑_{i=1}^{r} π_{λ}(H_{i})π_{λ}(H_{i}^{∗}) − ∑_{a>0}(π_{λ}(X_{α})π_{λ}(X_{−α}) + π_{λ}(X_{−α})π_{λ}(X_{α})), where α ranges over all positive roots of 𝔥 ⊗ ℂ.

Consider v_{λ} ∘ v_{−λ^{∗}} ∈ S^{2}V. If S^{2}π_{λ} is irreducible, then

C_{S^{2}π_{λ}}(v_{λ} ∘ v_{−λ^{∗}}) = − < 2λ, 2λ + 2δ > (v_{λ} ∘ v_{−λ^{∗}}).

On the other hand,

C_{S^{2}π_{λ}}(v_{λ} ∘ v_{−λ^{∗}}) = C_{π_{λ}}(v_{λ}) ∘ v_{−λ^{∗}} + v_{λ} ∘ C_{π_{λ}}(v_{−λ^{∗}})

+ ∑_{i}[λ(H_{i})λ^{∗}(H_{i}^{∗}) + λ^{∗}(H_{i})λ(H_{i}^{∗})]v_{λ} ∘ v_{−λ^{∗}}

− 2 ∑_{α>0} π_{λ}(X_{−α})v_{λ} ∘ π_{λ}(X_{α})v_{−λ^{∗}}

= (−2 < λ, λ + 2δ > +2 < λ, λ^{∗} >)v_{λ} ∘ v_{−λ^{∗}}

− 2 ∑_{α>0} π_{λ}(X_{−α})v_{λ} ∘ π_{λ}(X_{α})v_{−λ^{∗}}

It follows that ∑_{α>0} π_{λ}(X_{−α})v_{λ}∘π_{λ}(X_{α})v_{−λ^{∗}} = a(v_{λ}∘v_{−λ^{∗}}) with a = < λ, λ+λ^{∗} > < 0. Let ϕ_{−λ}, ϕ_{λ^{∗}} ∈ V^{∗} be dual to v_{λ} and v_{−λ^{∗}}. Applying ϕ_{−λ} ∘ ϕ_{λ^{∗}} to ∑_{α>0} π_{λ}(X_{−α})v_{λ} ∘ π_{λ}(X_{α})v_{−λ^{∗}} we get 0 ≠ a = ∑_{α>0} ϕ_{λ^{∗}}(X_{−α}v_{λ})ϕ_{−λ}(X_{α}v_{−λ^{∗}}) since ϕ_{ρ_{1}}(v_{ρ_{2}}) = 0 unless ρ_{1} + ρ_{2} = 0. In particular this also means that there is a positive root α such that λ + λ^{∗} = α. By (2.1) we conclude that H = SU(N) and π_{λ} = μ_{N} or μ_{N}^{∗} or H = Sp(n) and π_{λ} = ν_{2n}. □

(4.8) Proposition. With the same assumptions as in (4.6) and the assumption that H is simple we have

(a) Λ^{2}π′ is irreducible and S^{2}π′ consists of exactly two irreducible inequivalent subrepresentations.

(b) There exists a unique simple root α of 𝔥 with < λ′, α > ≠ 0 and hence Λ^{2}π′ = π_{2λ′−α}.

Proof. As in the beginning of the proof of (4.6), it follows that π′ = π_{λ′} is irreducible with λ′ ≠ λ′^{∗}. Since SU(N)/π′(H) is not a point, we can also assume that π′ ≠ μ_{n}, μ_{n}^{∗}. Hence (4.7) implies that S^{2}π_{λ′} is reducible. Consider

S^{2}(π_{λ′} ⊗ π_{λ′}^{∗}) = S^{2}π_{λ′} ⊗ S^{2}π_{λ′}^{∗} ⊕ Λ^{2}π_{λ′} ⊗ Λ^{2}π_{λ′}^{∗}.

This is also given by

S^{2}(1 ⊕ ad(𝔥)_{ℂ} ⊕ χ_{ℂ}) = 1 ⊕ S^{2}(ad(𝔥)_{ℂ}) ⊕ S^{2}χ_{ℂ} ⊕ ad(𝔥)_{ℂ} ⊕ χ_{ℂ} ⊕ [ad(𝔥)_{ℂ} ⊗ χ_{ℂ}].

By (2.2a), χ_{ℂ} = π_{λ′+λ′^{∗}} ≠ ad(𝔥)_{ℂ}. Hence there are exactly 3 trivial summands in S^{2}(π_{λ′} ⊗ π_{λ′}^{∗}). This forces S^{2}π_{λ′} to have 2 irreducible inequivalent summands and Λ^{2}π_{λ′} to be irreducible. (b) follows from (a) since for every simple root α with < λ′, α > ≠ 0 the representation π_{2λ′−α} is a subrepresentation of Λ^{2}π′ □ Proof of (4.6) continued. Let V′ ≅ ℂ^{N} be the representation space of π′ and V′′ ≅ ℂ be that for ϕ and define V = V′ ⊗ V′′. Let λ′ be the dominant weight of π′ and λ′′ that for ϕ. Hence λ = λ′ + λ′′ is the dominant weight for π = π′ ˆ⊗ϕ. From the natural pairing between V^{∗} and V one obtains a non-degenerate symmetric bilinear form q on V ⊕ V^{∗} given by q((v_{1}, ϕ_{1}), (v_{2}, ϕ_{2})) = ϕ_{1}(v_{2})+ϕ_{2}(v_{1}). Then 𝔭 is a K-invariant real form of V ⊕V^{∗} on which q is positive definite.

It follows from (4.8) that

Λ^{2}(𝔭) = [ad(𝔥) ˆ⊗1] ⊕ 1 ˆ⊗1 ⊕ [π_{λ′+λ′^{∗}} ˆ⊗1]_{ℝ} ⊕ [π_{2λ′−α} ˆ⊗ϕ^{2}]_{ℝ}
= ad(𝔨) ⊕ [π_{λ+λ^{∗}}]_{ℝ} ⊕ [π_{2λ−α}]_{ℝ}.

Every element of S^{2}Λ^{2}(𝔭) that is 𝔨-invariant has diagonal form with respect to this decomposition of Λ^{2}(𝔭). Let a_{1}, a_{2}, b_{1}, b_{2} be respectively the eigenvalues. Clearly, dim[S^{2}Λ^{2}(𝔭)]^{𝔨} = 4. Hence, as in the quaternionic case, we will show that dim[Λ^{4}(𝔭)]^{𝔨} = 1, and we will be done.

We first establish some notation for future use. There is a non-degenerate symmetric bilinear form < , > on V ⊗ V^{∗} given by < v_{1} ⊗ ϕ_{1}, v_{2} ⊗ ϕ_{2} > = −ϕ_{1}(v_{2})ϕ_{2}(v_{1}). Under the isomorphism V ⊗ V^{∗} ≈ 𝔤l(V) where v ⊗ ϕ ↦ A_{v⊗ϕ} with A_{v⊗ϕ}(x) = ϕ(x)v, < , > corresponds to the form on 𝔤l(V) given by − tr(AB). Furthermore, q induces a symmetric bilinear form ( , )_{Λ^{2}(V ⊕V^{∗})} on Λ^{2}(V ⊕ V^{∗}). The isomorphism θ: Λ^{2}(V ⊕ V^{∗}) ≅ Λ^{2}V ⊕ Λ^{2}V^{∗} ⊕V ⊗V^{∗} given by (v_{1}, ϕ_{1})∧(v_{2}, ϕ_{2}) ↦ (v_{1} ∧v_{2}, ϕ_{1} ∧ϕ_{2}, v_{1} ⊗ϕ_{2} −v_{2} ⊗ϕ_{1}) becomes an isometry if we choose as our invariant symmetric bilinear form q̂⊥ < , > where q̂ comes from the duality of Λ^{2}V and Λ^{2}V^{∗} as q does.

Using the embedding π: 𝔨 = 𝔥 ⊕ 𝔲(1) → 𝔤l(V) ≅ V ⊗ V^{∗}, the inner product on V ⊗ V^{∗} induces an inner product on 𝔨, again denoted by < , > and a projection pr_{𝔨} from V ⊗ V^{∗} to 𝔨. We now select weight vectors v_{λ}, v_{−λ^{∗}}, v_{λ−α} = π(X_{−α})V_{λ} in V and ϕ_{λ^{∗}}, ϕ_{−λ}, ϕ_{α−λ} = π(X_{α})ϕ_{−λ} in V^{∗} such that

ϕ_{−λ}(v_{λ}) = 1 = ϕ_{λ^{∗}}(v_{−λ^{∗}})
ϕ_{α−λ}(v_{λ−α}) = − < λ, α >
pr_{𝔨}(v_{λ} ⊗ ϕ_{α−λ}) = < λ, α > X_{α}
pr_{𝔨}(v_{λ−α} ⊗ ϕ_{−λ}) = − < λ, α > X_{−α}
pr_{𝔨}(v_{λ} ⊗ ϕ_{−λ}) = −H_{λ}
pr_{𝔨}(v_{λ−α} ⊗ ϕ_{α−λ}) = < λ, α > H_{λ−α}
pr_{𝔨}(v_{−λ^{∗}} ⊗ ϕ_{λ^{∗}}) = H_{λ^{∗}}

Let η ∈ [Λ^{4}(𝔭)]^{𝔨} and consider its complexification. Let η̂ be the corresponding symmetric endomorphism of Λ^{2}(𝔭). Then we have

η(v_{λ}, v_{λ−α}, ϕ_{−λ}, ϕ_{α−λ}) = η(v_{λ}, ϕ_{−λ}, ϕ_{α−λ}, v_{λ−α}) = η(v_{λ}, ϕ_{α−λ}, v_{λ−α}, ϕ_{−λ})
η(v_{λ}, ϕ_{−λ}, ϕ_{λ^{∗}}, v_{−λ^{∗}}) = η(v_{λ}, ϕ_{λ^{∗}}, v_{−λ^{∗}}, ϕ_{−λ}) = η(v_{λ}, v_{−λ^{∗}}, ϕ_{−λ}, ϕ_{λ^{∗}})

which yields a homogeneous linear system of order 4 in a_{1}, a_{2}, b_{1}, b_{2}. To obtain the system, we compute as follows:

(a)  η(v_{λ}, v_{λ−α}, ϕ_{−λ}, ϕ_{α−λ}) = (η̂(v_{λ} ∧ v_{λ−α}), ϕ_{−λ} ∧ ϕ_{α−λ})_{Λ^{2}(V ⊕V^{∗})}
= < θ(η̂(v_{λ} ∧ v_{λ−α})), θ(ϕ_{−λ} ∧ ϕ_{α−λ}) >

Now θ(v_{λ} ∧ v_{λ−α}) = v_{λ} ∧ v_{λ−α} ∈ π_{2λ−α} and θ(ϕ_{−λ} ∧ ϕ_{α−λ}) = ϕ_{−λ} ∧ ϕ_{α−λ} ∈ π_{2λ−α}^{∗}. So

η(v_{λ}, v_{λ−α}, ϕ_{−λ}, ϕ_{α−λ}) = b_{2}q̂(v_{λ} ∧ v_{λ−α}, ϕ_{−λ} ∧ ϕ_{α−λ})
= b_{2}(ϕ_{−λ} ∧ ϕ_{α−λ})(v_{λ} ∧ v_{λ−α})
= −b_{2} < λ, α >

(b) Likewise, η(v_{λ}, v_{−λ^{∗}}, ϕ_{−λ}, ϕ_{λ^{∗}}) = b_{2} and η(v_{λ}, ϕ_{λ^{∗}}, v_{−λ^{∗}}, ϕ_{−λ}) = −b_{1}, where we have used the fact that θ(v_{λ} ∧ ϕ_{λ^{∗}}) = v_{λ} ⊗ ϕ_{λ^{∗}} which lies in π_{λ+λ^{∗}}.

(c)  η(v_{λ}, ϕ_{−λ}, ϕ_{α−λ}, v_{λ−α}) = (η̂(v_{λ} ∧ ϕ_{−λ}), ϕ_{α−λ} ∧ v_{λ−α})_{Λ^{2}(V ⊕V^{∗})}
= < θ(η̂(v_{λ} ∧ ϕ_{−λ})), θ(ϕ_{α−λ} ∧ v_{λ−α}) > .

We have θ(v_{λ} ∧ ϕ_{−λ}) = v_{λ} ⊗ ϕ_{−λ} = −H_{λ} + x_{1} and θ(ϕ_{α−λ} ∧ v_{λ−α}) = −v_{λ−α} ⊗ ϕ_{α−λ} = − < λ, α > H_{λ−α} + x_{2}, with x_{1}, x_{2} ∈ π_{λ+λ^{∗}}. Furthermore, < v_{λ} ⊗ ϕ_{−λ}, v_{λ−α} ⊗ ϕ_{α−λ} > = −ϕ_{−λ}(v_{λ−α})ϕ_{α−λ}(v_{λ}) = 0 = < λ, α > < λ, λ − α > + < x_{1}, x_{2} >. Hence < x_{1}, x_{2} > = − < λ, α > < λ, λ − α >. It follows that

η(v_{λ},ϕ_{−λ}, ϕ_{α−λ}, v_{λ−α})

=< −a_{1}H_{λ′} − a_{2}H_{λ′′} + b_{1}x_{1}, − < λ, α > H_{λ′−α}− < λ, α > H_{λ′′} + x_{2} >

= a_{1} < λ, α > < λ′, λ′ − α > +a_{2} < λ, α > < λ′′, λ′′ > −b_{1} < λ, α > < λ, λ − α >

(d) A similar computation shows that

η(v_{λ}, ϕ_{α−λ}, v_{λ−α}, ϕ_{−λ}) = −a_{1} < λ, α >^{2} +b_{1} < λ, α > (1+ < λ, α >).

Combining the above calculations we obtain the following linear system :

( \begin{matrix}0 & 0 & 1 & 1 \\ < λ′, λ′ − α > & < λ′′, λ′′ > & − < λ, λ − α > & 1 \\ − < λ, α > & 0 & 1+ < λ, α > & 1 \\ ∗ & ∗ & ∗ & ∗ \end{matrix} ) ( \begin{matrix}a_{1} \\ a_{2} \\ b_{1} \\ b_{2} \end{matrix} ) = 0

Since the 3 × 3 minor consisting of the first three rows and columns has non-zero determinant, we have dim[Λ^{4}(𝔭)]^{𝔨} ≤ 1. Finally, observe that Λ^{4}(V ⊕ V^{∗}) ⊃ Λ^{2}(V) ⊗ Λ^{2}(V^{∗}), which contains a trivial summand, and so dim[Λ^{4}(𝔭)]^{𝔨} = 1. This completes the proof of (4.6). □

§5 Some Applications.

We end our paper with some applications to the isotropy representation of quaternionic and hermitian symmetric spaces that are of independent interest.

(5.1) Theorem. Let G/K = G/H · Sp(1) be a compact, simply connected, irreducible quaternionic symmetric space with isotropy representation π = π′ ˆ⊗ν_{2} and such that H is simple. Then Λ^{2}π′ = 1 ⊕ σ with σ irreducible. If H is not simple, Λ^{2}π′ − 1 consists of 2 or 3 irreducible summands.

Proof. If H is not simple, then (2.10b) implies that (𝔥, π′) = (𝔰o(n) ⊕ 𝔰p(1), ρ_{n} ˆ⊗ν_{2}) or (𝔲(n), μ_{n}). In both cases one easily shows that Λ^{2}π′−1 consists of 2 irreducible summands, unless n = 4 in the first case, where it consists of 3 irreducible summands. If G/K is the quaternionic projective space, we have (𝔥, π′) = (𝔰p(n), ν_{2n}) and it follows that Λ^{2}π′ − 1 is irreducible. If G/K is not the quaternionic projective space, it follows from (3.1b) that Sp(N)/π′(H), 4N = dim G/K, is isotropy irreducible and positive dimensional. Assuming that and that H is simple, a proof identical to the one in (4.4) shows that Λ^{2}π′ − 1 is irreducible. One only needs to observe, which is needed in the proof of (4.4), that π′ is absolutely irreducible, since G/K is an irreducible symmetric space. □

(5.2) Theorem. Let G/K = G/H · U(1) be a compact, simply connected, irreducible hermitian symmetric space with isotropy representation π = [π′ ˆ⊗ϕ]_{ℝ}. Then we have

(a) If G/K is not a projective space, then S^{2}π′ contains exactly two irreducible inequivalent sub-representations; otherwise S^{2}π′ = S^{2}μ_{n} is irreducible.

(b) If H is simple, i.e. G/K is not a complex Grassmannian SU(p + q)/S(U(p) U(q)), p ≥ q ≥ 2, then Λ^{2}π′ is irreducible; otherwise it contains exactly two irreducible summands.

Proof. If G/K is SU(n + 1)/S(U(n) U(1)), then π′ = μ_{n} and so S^{2}π′ and Λ^{2}π′ are irreducible. If not, (3.1a) implies that SU(N)/π′(H) is positive dimensional and isotropy irreducible, where N = \frac{1}{2} dim G/K. If H is not simple, then (2.10a) implies that (𝔥, π′) = (𝔰u(p)⊕𝔰u(q), μ_{p} ˆ⊗μ_{q}) with p ≥ q ≥ 2, up to an automorphism of 𝔥. In this case both S^{2}π′ and Λ^{2}π′ contain exactly two irreducible sub-representations. In all other cases a proof identical to the one in (4.8) shows that S^{2}π′ contains precisely 2 irreducible summands and Λ^{2}π′ is irreducible. One only needs to observe the following two facts, which are needed in the proof of (4.8). First, π′ must be absolutely irreducible, since G/K is an irreducible symmetric space. Second, (4.7) implies that S^{2}π′ must be reducible, since π′ = μ_{n} was excluded above and π′ = ν_{2n} is excluded since [ν_{2n} ˆ⊗ϕ]_{ℝ} cannot be the isotropy representation of a symmetric space by (2.4). □

Remarks. 1) The fact in (5.2a), that S^{2}π′ consists of exactly two irreducible representations was first observed in [Bo], Lemma 3.3, where a proof was given by verifying it in each case separately. To our knowledge, the above is the first direct proof of this fact. It is used in various vanishing theorems, see e.g. [CV], [Bo], [Mo]. This fact also has the consequence that the curvature operator on S^{2}π′ has at most 2 different eigenvalues. A direct proof of this consequence can also be found in [Gu].

2) The statements in (5.1) and (5.2) should be seen as complementary to the statements that S^{2}π′ in (5.1) and π′ ⊗ π′^{∗} − 1 in (5.2) consist of 2 irreducible summands, as follows from (3.1). There is no corresponding fact for irreducible symmetric spaces which are not hermitian symmetric or quaternionic symmetric. In this case it follows from (3.1c) that Λ^{2}π consists of 2 irreducible summands. But S^{2}π can consist of several irreducible summands.

3) In our forthcoming paper on maximal subgroups we will see that (5.1) and (5.2b) characterize hermitian and quaternionic symmetric spaces in the following sense. If π is a unitary representation such that Λ^{2}π − 1 is irreducible, then π ˆ⊗ν_{2} is the isotropy representation of a quaternionic symmetric space and if π is a unitary representation such that Λ^{2}π is irreducible, then [π ˆ⊗ϕ]_{ℝ} is the isotropy representation of a hermitian symmetric space.

The following result was used extensively in [WZ2], p.585–589. The fact about the Killing forms in (a) was also proved in [DZ], Theorem 11.

(5.3) Theorem. Let G/K be a compact irreducible symmetric space and B_{G}, B_{K} the negative of the Killing forms of G and K respectively. As in (3.1) we let χ be the isotropy representation of the corresponding strongly isotropy irreducible space.

(a) If K is simple and n = dim G/K, then B_{K} = cB_{G}|_{𝔨} with c = 1 − \frac{n}{2 dim K}. Furthermore, C_{χ} = \frac{2 dim K}{n(n−2)} Id, where the Casimir operator is computed with respect to the inner product B_{SO(n)}|_{𝔨}.

(b) If G/K = G/H · U(1) is hermitian symmetric of dimension 2n with H simple, then B_{H} = cB_{G}|_{𝔥} with c = 1 − \frac{n−1}{dim H}. Furthermore, C_{χ} = \frac{dim H}{n(n−1)}Id, where the Casimir operator is computed with respect to B_{SU(n)}|_{𝔥}.

(c) If G/K = G/H · Sp(1) is quaternionic symmetric of dimension 4n with H simple, then B_{H} = cB_{G}|_{𝔥} with c = 1 − \frac{n(2n+1)}{(n+2) dim H}. Furthermore, C_{χ} = \frac{(n+2) dim H}{n(n+1)(2n+1)}Id, where the Casimir operator is computed with respect to B_{Sp(n)}|_{𝔥}.

Proof. (a) If 𝔤 = 𝔨 ⊕ 𝔭 and X ∈ 𝔨 we have

B_{G}(X, X) = − tr_{𝔤} adXadX = B_{K}(X, X) − tr_{𝔭} adXadX

or (1 − c)B_{G}(X, X) = − tr_{𝔭} adXadX = − tr π(X)π(X). Hence, if H_{i} is an orthonormal basis of 𝔨 with respect to B_{G}|_{𝔨}, then

(1 − c) dim K = (1 − c) ∑_{i} B_{G}(H_{i}, H_{i}) = − ∑_{i} tr π(H_{i})π(H_{i}) = tr C_{π}

and the first claim follows from (3.2).

Since B_{SO(n)}(A, B) = −(n − 2) tr AB we have for X ∈ 𝔨

B_{SO(n)}(X, X) = −(n − 2) tr π(X)π(X) = (n − 2)(B_{G} − B_{K})(X, X) = \frac{n(n − 2)}{2 dim H}B_{G}|_{𝔨}(X, X)

and hence our second claim follows from (3.3).

(b) Letting X = i ∈ 𝔲(1), we get B_{G}(X, X) = − tr_{𝔭} adXadX = − tr π(X)π(X) = 2n. Hence we can choose an orthonormal basis H_{i} of 𝔨 with H_{1} = X/\sqrt{2n} and obtain together with (3.2)

(1−c) dim H = (1−c) ∑_{i≥2} B_{G}(H_{i}, H_{i}) = − ∑_{i≥2} tr π(H_{i})π(H_{i}) = tr C_{π}+tr π(H_{1})π(H_{1}) = n−1

which implies our first claim.

Since B_{SU(n)}(A, B) = −2n tr AB we have for X ∈ 𝔥

B_{SU(n)}(X, X) = −2n tr π_{λ′}(X)π_{λ′}(X) = −n tr π(X)π(X)

where we have used the fact that π|_{𝔥} = π_{λ′} ⊕ π_{λ′}^{∗}. Hence B_{SU(n)}|_{𝔥} = n(B_{G} − B_{H}) = \frac{n(n−1)}{dim H}B_{G}|_{𝔥} which implies our second claim.

(c) If X is equal to i, j, or k ∈ 𝔰p(1), we get B_{G}(X, X) = B_{Sp(1)}(X, X)−tr π(X)π(X) = 8 + 4n = 4(n + 2). Hence we can choose an orthonormal basis H_{i} of 𝔨 with 2\sqrt{n + 2}H_{1} = i, 2\sqrt{n + 2}H_{2} = j, 2\sqrt{n + 2}H_{3} = k. Then

(1 − c) dim H = (1 − c) ∑_{i≥4} B_{G}(H_{i}, H_{i}) = tr C_{π} + ∑_{i=1}^{3} tr π(H_{i})π(H_{i}) = 2n − \frac{3n}{n + 2}

from which our first claim follows.

B_{Sp(n)}(A, B) = −2(n + 1) tr AB where we regard A, B as complex matrices via the embedding 𝔰p(n) ⊂ 𝔰u(2n). Hence, for X ∈ 𝔥 we have

B_{Sp(n)}(X, X) = −2(n + 1) tr π_{λ′}(X)π_{λ′}(X) = −(n + 1) tr π(X)π(X)

where we have used the fact that π|_{𝔥} = π_{λ′} ⊕ π_{λ′}. Hence B_{Sp(n)}|_{𝔥} = (n + 1)(B_{G} − B_{H}) = \frac{n(n+1)(2n+1)}{(n+2) dim H}B_{G}|_{𝔥} which implies our second claim. □

We finally mention the following consequences of (2.8), (2.10), and (2.11), which were proved without using the classification of symmetric spaces.

(5.4) Theorem. Let G/K be a compact irreducible symmetric space with isotropy representation π.

(a) K is a simple Lie group, unless G/K is hermitian symmetric, quaternionic symmetric, or a real or quaternionic Grassmannian.

(b) If G/K is a hermitian symmetric space G/H · U(1) or a quaternionic symmetric space G/H · Sp(1), then H is simple unless G/K is a complex or quaternionic Grassmannian.

(c) Let 𝔥 ⊂ 𝔨 be a simple non-abelian ideal such that π|_{𝔥} is not a basic representation. Then (𝔨, π) must be one of the following:

(i) (𝔰o(n)⊕𝔰o(3), ρ_{n} ˆ⊗ρ_{3}) corresponding to the symmetric space SO(n+3)/ SO(n) SO(3). (ii) (𝔰u(n), ad) corresponding to the symmetric space [SU(n) × SU(n)]/△ SU(n). (iii) (𝔰p(n), ad) corresponding to the symmetric space [Sp(n) × Sp(n)]/△ Sp(n). (iv) (𝔰o(n), S^{2}ρ_{n} − 1) corresponding to the symmetric space SU(n)/ SO(n).

(v) (𝔲(n), S^{2}μ_{n}) corresponding to the symmetric space Sp(n)/ U(n).

(vi) (𝔰p(1)⊕𝔰p(1), S^{3}ν_{2} ˆ⊗ν_{2}) corresponding to the symmetric space G_{2} / SO(4). □

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Department of Mathematics and Statistics, McMaster University, Hamilton, Ontario, Canada, L8S 4K1

E-mail address: wang@mcmaster.bitnet

Department of Mathematics, University of Pennsylvania, Philadelphia, Pennsylvania 19104 U.S.A.

E-mail address: wziller@math.upenn.edu