Compact geodesic orbit spaces with a simple isotropy group
aa r X i v : . [ m a t h . DG ] S e p COMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPYGROUP
Z. CHEN, Y. NIKOLAYEVSKY, AND YU. NIKONOROV
Abstract.
Let M = G/H be a compact, simply connected, Riemannian homogeneousspace, where G is (almost) effective and H is a simple Lie group. In this paper, we firstclassify all G -naturally reductive metrics on M , and then all G -geodesic orbit metricson M . Introduction
A Riemannian manifold (
M, g ) is called a geodesic orbit manifold (or a manifold withhomogeneous geodesics, or a GO manifold) if any geodesic of M is an orbit of a 1-parameter subgroup of the full isometry group of ( M, g ) (without loss of generality, onecan replace the full isometry group by its connected identity component). A Riemannianmanifold ( M = G/H, g ), where H is a compact subgroup of the Lie group G and g isa G -invariant Riemannian metric on M , is called a G -geodesic orbit space (or a spacewith G -homogeneous geodesics, or a G -GO space) if any geodesic of M is an orbit of a1-parameter subgroup of the group G . Hence a Riemannian manifold ( M, g ) is a geodesicorbit manifold, if it is a geodesic orbit space with respect to its full isometry group. Thisterminology was introduced in [18] by O. Kowalski and L. Vanhecke who initiated thesystematic study of such spaces.The class of geodesic orbit spaces includes (but is not limited to) symmetric spaces,weakly symmetric spaces [5, 30, 33], normal and generalised normal homogeneous spaces,naturally reductive spaces [9], Clifford-Wolf homogeneous manifolds [7] and δ -homogeneousmanifolds [6]. For the current state of knowledge in the theory of geodesic orbit spacesand manifolds we refer the reader to [2, 3, 10, 11, 23] and the bibliographies therein.Let ( M = G/H, g ) be a homogeneous Riemannian space and let g = h ⊕ p be anAd( H )-invariant decomposition, where g is the Lie algebra of G , h is the Lie algebra of H and p is identified with the tangent space of M at eH . The Riemannian metric g is G -invariant and is determined by an Ad( H )-invariant inner product ( · , · ) on p . Themetric g is called naturally reductive if an Ad( H )-invariant complement p can be chosen Mathematics Subject Classification.
Key words and phrases. homogeneous Riemannian manifolds, geodesic orbit spaces, naturally reduc-tive spaces.Supported by National Natural Science Foundation of China (11931009) and Natural Science Foun-dation of Tianjin (19JCYBJC30600). The second and the third author would like to thank S.S.ChernInstitute of Mathematics and Nankai University (Tianjin, China) for their support and hospitality. in such a way that ([
X, Y ] p , X ) = 0 for all X, Y ∈ p , where the subscript p denotes the p -component. In this case, we say that the (naturally reductive) metric g is generated bythe pair ( p , ( · , · )). For comparison, on the Lie algebra level, g is geodesic orbit if and onlyif for any X ∈ p (with any choice of p ), there exists Z ∈ h such that ([ X + Z, Y ] p , X ) = 0for all Y ∈ p [18, Proposition 2.1]. It immediately follows that any naturally reductivespace is a G -geodesic orbit space; the converse is false when dim M ≥
6. Clearly, theproperty of being naturally reductive depends on the choice of the group G (the choiceof the presentation M = G/H ); both enlarging and reducing G may result in gaining orlosing this property. In this paper, the presentation M = G/H (and hence the group G ) will be fixed, and so “naturally reductive” will always mean “ G -naturally reductive”,unless explicitly stated otherwise.Our setup in this paper is as follows. Let M = G/H be a compact, connected, simplyconnected, Riemannian homogeneous space, with G acting almost effectively (this meansthat any normal subgroup of G contained in H is discrete). We classify all the G -GOmetrics on M , both naturally reductive and not, under the assumption that H is asimple Lie group (that is, any normal proper subgroup of H is discrete). Note that H is then closed (compact) and connected and G is connected. Moreover, the fundamentalgroup of H must be finite, and since G is compact (and hence reductive), with a finitefundamental group (from the exact sequence of the fibration H → G → G/H ), G mustbe a compact semisimple Lie group.We first characterise naturally reductive metrics on G/H . Let g = ⊕ Ni =1 g i , N ≥ g into simple ideals. The inclusion h ֒ → g followed by the linearprojection g → g i relative to this decomposition defines a projection of h to each of g i ,which is a homomorphism of Lie algebras. As h is simple, every such homomorphismis either trivial or injective. Relabel the ideals g i in such a way that g = L N i =1 g i ⊕ L N i = N +1 g i ⊕ L Ni = N +1 g i , where 0 ≤ N ≤ N ≤ N, N < N , and the projection of h to g i is trivial for i = 1 , . . . , N , is injective, but not surjective for i = N + 1 , . . . , N , andis bijective for i = N + 1 , . . . , N (so that g N +1 , . . . , g N are isomorphic to h ). Denote by h· , ·i i minus the Killing form on g i , for i = 1 , . . . , N . For i = N + 1 , . . . , N , denote by h· , ·i i the (negative) multiple of the Killing form on g i normalised in such a way that itsrestriction to the projection of h to g i equals minus the Killing form on h . Theorem 1.
Let M = G/H be a compact, connected, simply connected, Riemannianhomogeneous space, where G is almost effective and H is a simple Lie group. An in-variant metric on M is ( G - ) naturally reductive if and only if it is generated by a pair ( p , ( · , · )) such that, in the above notation, (a) either p = ⊕ i = j g i is an ideal in g , for some j ∈ { N + 2 , · · · , N } ( so that g j is isomorphic to h ) , and ( · , · ) is an ad( p ) -invariant inner product on p , that is, ( · , · ) = P i = j β i h· , ·i i , where β i > . (b) or p is the orthogonal complement to h ⊂ g relative to an ad( g ) -invariant qua-dratic form Q = P Ni =1 γ i h· , ·i i on g and ( · , · ) = Q | p , where OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 3 (i) either γ i > for all i = 1 , . . . , N , (ii) or there exists j ∈ { N + 1 , . . . , N } such that γ j < and γ i > for all i = j ,and P Ni = N +1 γ i < .Remark . Note that all the metrics from (a) are reducible when
N >
2; this is notnecessarily true for metrics in (b). Also note that if g contains no simple ideals isomorphicto h , then any naturally reductive metric is normal (that is, it is a restriction to p ofa bi-invariant metric on g ; a normal metric is always naturally reductive). Theorem 1generalises the result of [22, Theorem 1] for Ledger-Obata spaces . In fact, a Ledger-Obataspace is the homogeneous space
G/H with N = 0 in our notation.The classification of G -GO metrics which are not naturally reductive is given in thefollowing Theorem. Theorem 2.
Let M = G/H be a compact, connected, simply connected, Riemannianhomogeneous space, where G is almost effective and H is a simple Lie group. Suppose M is a G -GO space. Then either M is ( G - ) naturally reductive, or one of the followingis true. (A) If M is an irreducible Riemannian manifold, then G is simple and M belongsto the following list, up to a finite cover ( the corresponding metrics are given inTable 1 ) . (1) SO(9) / Spin(7) ; (2) SO(10) / Spin(7) ; (3) SO(11) / Spin(7) ; (4) E / Spin(10) ; (5) SU( n + p ) / SU( n ) , n ≥ , ≤ p ≤ n − ; (6) SO(2 n + 1) / SU( n ) , n ≥ ; (7) SO(4 n + 2) / SU(2 n + 1) , n ≥ ; (8) Sp( n + 1) / Sp( n ) , n ≥ ; (9) SU(2 n + 1) / Sp( n ) , n ≥ ; (10) Spin(8) / G ; (11) SO(9) / G . (B) If M is reducible, then it is the Riemannian product of one of the spaces in (A) and a compact semisimple Lie group with a bi-invariant metric. Note that the cases in (A) are mutually exclusive. We also note that many of thesespaces already appeared in the literature. For example, the spaces (1), (8) and (5) with p = 1 are spheres with a GO metric [24]; the spaces (1), (4), (9), (10) and (5) with p = 1are weakly symmetric [31]; the spaces (1), (4), (7), (10) and (5) with p = 1 are GO spaceswith exactly two irreducible isotropy components [8]. Moreover, the space (6) with n even (and several others from our list) is fibered over a compact symmetric space, withthe GO metric having the property that its restriction to the tangent space of the fiberis proportional to the restriction of the Killing form on G (so that the tangent space ofthe fiber at eH is an eigenspace of the metric endomorphism — see Section 2.1) [28]. It Z. CHEN, Y. NIKOLAYEVSKY, AND YU. NIKONOROV should be noted that the GO metrics on the spaces (6) with n even and with n odd arevery different — see the details in Table 1; in particular, in the odd case, a nontrivialalgebraic condition has to be satisfied.We note that in the other “extremal” case, when the isotropy subgroup is abelian,any GO metric is naturally reductive by the result of [26] (these two classifications verynicely complement one another; note that some authors include SO(2) in the list ofsimple groups).The paper is organised as follows. In Section 2 we provide the necessary backgroundmaterial and also give a detailed description of the GO metrics on the spaces listed inTheorem 2(A) (see Table 1). In Section 3 we prove Theorem 1. In the rest of the paperwe give the proof of Theorem 2. The proof is based on the study of different types ofsubmodules in the decomposition of p : trivial, large and adjoint modules are consideredin Section 4, and tiny modules, in Section 5 (we refer to Section 2 for unexplainedterminology). 2. Preliminaries
Generalities.
Throughout the paper, we will adopt the assumptions of Theorem 1and Theorem 2 (although some notions and facts below do not require all of them).Namely, we work with a compact, connected, simply connected, Riemannian homoge-neous space M = G/H , where the Lie group G acts almost effectively, and H is a simpleLie group. As we noted above, H is then compact and connected and G is a compact,connected, semisimple Lie group.Let g be the Lie algebra of G and h ⊂ g the Lie algebra of H . Denote by h· , ·i minus the Killing form on g . Throughout the paper, “orthogonal” means “orthogonalrelative to h· , ·i ” unless otherwise is explicitly stated. Let g = h ⊕ p be an Ad( H )-reductive decomposition (one possibility is to take p as the orthogonal complement m to h in g ). Then p can be naturally identified with the tangent space T eH ( G/H ), and theRiemannian metric g is determined by some positive h· , ·i -symmetric Ad( H )-equivariantendomorphism A : p → p by the formula g eH ( X, Y ) = h AX, Y i , for X, Y ∈ p . We call A the metric endomorphism .We have the following fact [1, Proposition 1], [25, Proposition 2]. Lemma 1.
A homogeneous Riemannian manifold ( M = G/H, g ) with a semisimplegroup G and an Ad( H ) -reductive decomposition g = h ⊕ p is a G -geodesic orbit space ifand only if for any X ∈ p there exists Z ∈ h such that (1) [ X + Z, AX ] = 0 . Note that the claim of the Lemma does not depend on a particular choice of p ; inparticular, one can take p to be the orthogonal complement m to h . In the assumptionsof the Lemma, we call any map Z : p → h such that [ X + Z ( X ) , AX ] = 0 for all X ∈ p , a geodesic graph . In general, a geodesic graph may not be unique, but if it at all exists (thatis, if M is a G -GO space), then it can be chosen Ad( H )-equivariant. One obvious, but OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 5 potentially confusing point here is that although for almost all objects (decompositions,modules, inner products, etc.) throughout the paper Ad( H )-invariancy/equivariancy issynonymous with ad( h )-invariancy/equivariancy respectively, this is not necessarily truefor a geodesic graph: an Ad( H )-equivariant geodesic graph does not have to be ad( h )-equivariant (the obvious reason being non-linearity — cf. Lemma 3 in Section 2.3).In the proof of Theorem 2 below we choose and fix p to be the orthogonal complement m to h in g . Let α , . . . , α m > A , and let m , . . . , m m be the corresponding eigenspaces. Each m i is an H -module andthe decomposition m = m ⊕ m ⊕ · · · ⊕ m m is orthogonal and Ad( H )-invariant. Since H is connected, a submodule of m is Ad( H )-irreducible (respectively Ad( H )-invariant)if and only if it is ad( h )-irreducible (respectively ad( h )-invariant).We can further decompose every submodule m i in the decomposition m = ⊕ mi =1 m i intoan orthogonal sum of irreducible modules. Labelling them through we get the orthogonaldecomposition(2) m = ⊕ pr =1 n r into irreducible h -modules n r each of which lies in some m i . Note that at least one ofthe modules n r is nontrivial, as G acts almost effectively (here and below, by a trivialmodule we mean a module on which the group/algebra acts trivially).We call an H -module n ⊂ m large if the principal stationary subgroup of the actionof Ad( H ) on n is discrete. On the level of Lie algebras, n is large if for some X ∈ n (andthen for all X in an open and dense subset of n ), the centraliser z h ( X ) in h is trivial. Inthe context of GO spaces, if m is large (in particular, if one of its submodules is large),then by (1) the geodesic graph Z is uniquely determined on an open, dense subset.An H -module is called small if it is not large. Clearly a trivial module is alwayssmall; the adjoint module is also small. Irreducible small modules for compact simpleand semisimple Lie groups are given in [14, Table 1 and 2]. In the case of simple groups,the list includes the adjoint representations, the standard representations of the classicalgroups, the “defining” representations of the exceptional groups, two infinite series andfive low-dimensional modules of the classical groups. Note that a small module can bethe sum of more than one nontrivial irreducible submodules (see also [12, Section 3]).We call a module tiny if it is small, irreducible, nontrivial and not adjoint; tiny modulesare listed in Table 2.The GO condition imposes strong restrictions on the decompositions of m into theeigenspaces of A and on the decomposition (2). Lemma 2 ([23, Section 5]) . In the assumptions of Theorem 2 ( and in the above notation ) ,we have the following. (a) For any X ∈ m i , Y ∈ m j , with i = j , there exists Z ∈ h such that [ X, Y ] = α i α i − α j [ Z, X ] + α j α i − α j [ Z, Y ] . So for i = j we have [ m i , m j ] ⊂ m i ⊕ m j , and, in thenotation of (2) , if n r ⊂ m i , n s ⊂ m j , then [ n r , n s ] ⊂ n r ⊕ n s . Z. CHEN, Y. NIKOLAYEVSKY, AND YU. NIKONOROV (b)
For any X ∈ m i , Y ∈ m j , with i = j , there exist Z ∈ z h ( X ) , Z ∈ z h ( Y ) suchthat [ X, Y ] = [ Z , X ] + [ Z , Y ] . (c) Consequently, any two large modules m i , m j with i = j commute. Furthermore if m i is large, then for any n r ⊂ m i in the decomposition (2) we have [ m ⊥ i , n r ] ⊂ n r . (d) If X, Y ∈ m i satisfy [ h , X ] ⊥ Y , then [ X, Y ] ∈ m i .Remark . Note that the last inclusion in Lemma 2(a) is a very powerful fact which willbe used in many places in the proofs below (see also Remark 4). Moreover, it defines anAd( H )-equivariant homomorphism n r × n s → n r ⊗ n s , for all r = s . So in particular, ifthe irreducible decomposition of the H -module n r × n s contains no modules isomorphicto either n r or n s , we get [ n r , n s ] = 0. It would substantially simplify our arguments ifa complete or even a partial classification of such pairs of modules would be known (forour purposes, we can assume that at least one of the modules is small), but we were notable to find it in the literature; in many cases in Section 5, we use this condition forindividual pairs of modules.2.2. The table.
In the above notation, we give an explicit description of the GO metricsin Theorem 2(A) in the table below (for the proofs, see Lemma 5(g) and Section 5).In the table, the direct sum always means the orthogonal direct sum; note that in manycases, the modules m i (the eigenspaces of the metric endomorphism A ) are reducible.The condition on the eigenvalues of A in the last column in all but one case simply saysthat for any positive α , . . . , α m , the resulting metric is GO, and that it is naturallyreductive only when A is a multiple of the identity (so that the metric is normal); theonly exception is case (6 ): the eigenvalues of the metric endomorphism A of a GO metricon the space SO(2 n + 1) / SU( n ) where n ≥ ), A has three eigenspaces). Theexceptions are case (9) where the dimension is 3 and A can have two or three eigenspaces,and case (10) where A has two eigenspaces which are isomorphic as h -modules, and sothe dimension of the space of GO metrics is again 3 (any invariant metric is GO [33]).2.3. Natural reductivity.
It is easy to see that in the assumptions of Lemma 1, thespace ( M = G/H, g ) is naturally reductive if there exists an Ad( H )-reductive decompo-sition g = h ⊕ p such that [ X, AX ] = 0. It follows that any G -GO space is naturallyreductive; the converse is true when dim M ≤
4, but is false starting from dimension 5.More precisely, in [18] the authors constructed examples of G -GO spaces of dimension 5which are not naturally reductive but can made be such by choosing a different transitiveisometry group acting on M ; further, in dimension 6 there are examples of GO spaceswhich are not naturally reductive, for any choice of a transitive isometry group.The following fact, which is a stronger version of [18, Proposition 2.10] (see also [27]),will be useful to detect whether a GO space is naturally reductive, without the necessityto produce a specific p . OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 7 H ⊂ G Modules m i Condition(1) Spin(7) ⊂ SO(8) ⊂ SO(9) m ⊕ so (8) = so (9) , m = R , m ⊕ so (7) = so (8), m is spin α = α (2) Spin(7) ⊂ SO(8) ⊂ SO(10) m ⊕ so (8) = so (10) , m = 2 R ⊕ R , m ⊕ so (7) = so (8), m is spin α = α (3) Spin(7) ⊂ SO(8) ⊂ SO(11) m ⊕ so (8) = so (10) , m = 3 R ⊕ so (3), m ⊕ so (7) = so (8), m is spin α = α (4) Spin(10) ⊂ Spin(10)SO(2) ⊂ E m ⊕ so (10) ⊕ R = e , m = R α = α (5) SU( n ) ⊂ S(U( n ) × U( p )) ⊂ SU( n + p ) n ≥ , ≤ p ≤ n − m = su ( p ) ⊕ p C n , m = R α = α (6 ) SU( n ) ⊂ U( n ) ⊂ SO(2 n ) ⊂ SO(2 n + 1) n ≥ , n is odd m ⊕ su ( n ) ⊕ R = so (2 n ), m = C n , m = R nα − =( n − α − + α − , α = α (6 ) SU( n ) ⊂ U( n ) ⊂ SO(2 n ) ⊂ SO(2 n + 1) n ≥ , n is even m ⊕ su ( n ) = so (2 n ), m = C n α = α (7) SU(2 n + 1) ⊂ U(2 n + 1) ⊂ SO(4 n + 2) n ≥ m ⊕ su (2 n + 1) ⊕ R = so (4 n + 2), m = R α = α (8) Sp( n ) ⊂ Sp( n ) × Sp(1) ⊂ Sp( n + 1) n ≥ m = H n , m = sp (1) , m is trivial α = α (9) Sp( n ) ⊂ SU(2 n ) ⊂ SU(2 n + 1) n ≥ m ⊕ sp ( n ) = su (2 n ), m = H n = C n , m = R not( α = α = α )(10) G ⊂ Spin(7) ⊂ Spin(8) m , m are any two isomorphic 7-di-mensional g -modules in g ⊥ ⊂ so (8) α = α (11) G ⊂ SO(7) ⊂ SO(9) m ⊕ g = so (7), m ⊕ so (7) = so (9) , m = 2 R ⊕ R α = α Table 1.
GO metrics on the spaces in Theorem 2(A)
Lemma 3.
Suppose a homogeneous Riemannian manifold ( M = G/H, g ) is a G -geodesicorbit space. Then ( M, g ) is naturally reductive if and only if for some ( and then for any )Ad( H ) -reductive decomposition g = h ⊕ p , there is a geodesic graph Z : p → h which islinear.Proof. The claim will follow from [18, Proposition 2.10] if we can show that the existenceof a linear geodesic graph implies the existence of a linear Ad( H )-equivariant geodesicgraph. This is similar to the fact that a geodesic graph can be always chosen Ad( H )-equivariant (see the paragraph after Lemma 1 in Section 1). Z. CHEN, Y. NIKOLAYEVSKY, AND YU. NIKONOROV
Suppose we have a linear geodesic graph Z : p → h . It is easy to see that for any h ∈ H ,the map X Ad h Z (Ad h − X ) is also a geodesic graph. But a convex linear combinationof geodesic graphs is also a geodesic graph. Integrating the latter expression over H bythe Haar measure µ such that µ ( H ) = 1 we obtain an Ad( H )-equivariant geodesic graph X Z ′ ( X ) := R H Ad h Z (Ad h − X ) dµ ( h ). Note that Z ′ is linear as Z is such. (cid:3) Naturally reductive spaces. Proof of Theorem 1
In this section we prove Theorem 1 following the approach in [22, Section 3] (whichcorresponds, in our notation, to the special case N = 0).Suppose p is an ad( h )-invariant complement to h in g . Then the space q = [ p , p ] + p is an ideal in g .By [15, Theorem 4], if a naturally reductive metric is generated by a pair ( p , ( · , · )),then there is a (unique) ad( q )-invariant, non-degenerate quadratic form Q on q such that(3) Q ( p , q ∩ h ) = 0 and Q | p = ( · , · ) . The converse is also true: if Q is an ad( q )-invariant, non-degenerate quadratic form whichsatisfies the first equation of (3) and whose restriction to p is positive definite, then thatrestriction defines a naturally reductive metric; this follows from ad( q )-invariancy of Q and from the fact that q is complemented in g by an ideal.We clearly have p + h = g , and so in the notation of Section 1, there are only twopossible cases.In the first case, we have q = ⊕ N − i =1 g i (up to relabelling the modules g N +1 , . . . , g N ),and then the (linear) projection of h ⊂ g to g N is an isomorphism (so in particular, N < N ). Then q ∩ h = 0 and g = q ⊕ h , and so p = q , an ideal. Furthermore, Q = ( · , · )by the second equation of (3) and has the form given in (a).In the second case, q = g . The quadratic form Q is ad( g )-invariant, and so we have Q = P Ni =1 γ i h· , ·i i , with γ i = 0. Then by (3) the space p is the Q -orthogonal com-plement to h in g . Note that N − N ≥
2, for if N = N + 1, we get h = g N andthen p = q = ⊕ N − i =1 g i . We need to choose γ i in such a way that the restriction of Q to p is positive definite. Clearly, for i = 1 , . . . , N we have g i ⊂ p , and so γ i > γ i > i = N + 1 , . . . , N , as every such g i contains anonzero vector Q -orthogonal to h . The restriction of Q to p will obviously be positivedefinite if γ i > i = N + 1 , . . . , N ; then Q itself is positive definite and so themetric ( · , · ) is normal; this gives case (b)(i). Suppose Q is indefinite. If γ j , γ k < j, k ∈ { N + 1 , . . . , N } , j = k , then Q is negative definite on the Q -orthogonalcomplement to h in g j ⊕ g k . It therefore remains to consider the case when γ j < j ∈ { N + 1 , . . . , N } . Up to relabelling we can assume that j = N and so γ i > i < N . Identify the images of the (linear) projections of h to g i , i = N + 1 , . . . , N , with h (recall that each of these projections is an isomorphismon its image and that every inner product h· , ·i i is normalised in such a way that thecorresponding projection is a linear isometry). Then the restriction of Q to p is positive OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 9 definite if and only if for any Y N +1 , . . . , Y N ∈ h such that P Ni = N +1 γ i Y i = 0 and notall Y i are zeros, we have P Ni = N +1 γ i k Y i k > h ). Equivalently, P N − i = N +1 γ i k Y i k + γ − N k P N − i = N +1 γ i Y i k > Y i is nonzero. Taking Y N +1 = · · · = Y N − = 0 we obtain a nec-essary condition γ N + P N − i = N +1 γ i <
0. But this condition is also sufficient, as from P N − k = N +1 γ k < − γ N we obtain ( P N − k = N +1 γ k )( P N − i = N +1 γ i k Y i k + γ − N k P N − i = N +1 γ i Y i k ) ≥ ( P N − k = N +1 γ k )( P N − i = N +1 γ i k Y i k ) − k P N − i = N +1 γ i Y i k ≥
0, by the Cauchy-Schwarz inequal-ity, with the equality only possible when Y N +1 = · · · = Y N − and P N − i = N +1 γ i Y i = 0, thatis, when all Y i are zeros. This gives the condition in (b)(ii) and completes the proof. Remark . Using Theorem 1 one can easily write down the inner product and the metricendomorphism A on the orthogonal complement m to h in g . Note that first every γ i , i = N + 1 , . . . , N , has to be re-scaled by the ratio of the restriction of the Killingform of g i to the projection of h to g i and the Killing form of h .4. G -GO spaces. Trivial, large and adjoint modules In this section, we study trivial, large and adjoint modules in the decomposition of m .In the next section, we will complete the proof of Theorem 2 by a case-by-case study oftiny modules. Throughout both sections ( in particular, in all the Lemmas and the Propositions ) ,we adopt the assumptions of Theorem 2 and we use the terminology and the notationintroduced in Section 2. Recall that M = G/H is a compact, connected, simply connected, Riemannian homo-geneous space, with G acting almost effectively and H being a simple Lie group. Then G is compact, connected and semi-simple and H is compact and connected. We denote by h· , ·i minus the Killing form on g and consider the orthogonal decomposition g = h ⊕ m ,where the h -module m is identified with the tangent space of M at eH . The metric on M is defined by the metric endomorphism A whose eigenspaces m i , i = 1 , . . . , m , areorthogonal h -modules with corresponding eigenvalues α i >
0. We will use Lemmas 1 and3 to check when M is a G -GO space and when M is not naturally reductive respectively.Our strategy is to consider the decomposition of m = ⊕ mi =1 m i into the eigenspaces m i of A together with the “finer” decomposition (2). In this section, we will first study thetrivial modules which may occur in (2). Next we show that if at least one module in thedecomposition (2) is either large or adjoint, then any G -GO metric is naturally reductive(with two exceptions, SU(3) / SU(2) and Sp(2) / Sp(1)). Our main tools will be Lemma 1,Lemma 2 and Lemma 3. It will therefore follow that a G -GO space may be not naturallyreductive only when all the modules in the decomposition (2) are either trivial or tiny(recall that this means that the module is irreducible, nontrivial, not adjoint, and thatits generic element has a nontrivial centraliser in h ). We will then use the classificationof such modules given in [14, Table 1] to complete the proof of Theorem 2. We can assume that m >
1, as otherwise the metric is normal and hence naturallyreductive. Another easy but useful observation is as follows.
Lemma 4.
In the assumptions of Theorem 2, suppose that one of the eigenspaces of A contains a nonzero ideal of g . Then M is the Riemannian product of a compact,simply connected Lie group with a bi-invariant metric and a compact, connected, simplyconnected homogeneous space ˆ M = ˆ G/H . Moreover, M is a G -GO space ( respectively G -naturally reductive ) if and only if ˆ M is a ˆ G -GO space ( respectively ˆ G -naturally reduc-tive ) .Proof. We can assume that in the presentation M = G/H both G and H are simplyconnected (by replacing, if necessary, G by its universal cover and H by the identitycomponent of its full preimage under the covering map). It is sufficient to prove thelemma when the ideal is simple. Let g = ⊕ Nl =1 g l be the decomposition of g into simpleideals and suppose g k ⊂ m i for some i = 1 , . . . , m . Denote ˆ g = g ⊥ k and ˆ m = g ⊥ k ∩ m . Wecompute the linear holonomy algebra of M using the construction in [16]. Extend themetric endomorphism A to the operator C on g which is symmetric relative to h· , ·i and isdefined by C | m = A and C | h = 0. For Z ∈ g define D Z : m → m by D Z ( Y ) = [ Z, Y ] m , for Y ∈ m , where the subscript m denotes the m -component. Then by [16, Theorem 2.3], thelinear holonomy algebra of M = G/H is the Lie algebra generated by all the operatorson m of the form Γ Z = D Z + C − D Z C − C − D CZ , Z ∈ g . (note that C − is only applied to elements of m ). As g k ⊂ g is an ideal and is C -invariant we obtain that Γ Z ( g k ) ⊂ g k , for all Z ∈ g , and so the linear holonomy algebrapreserves the orthogonal decomposition m = ˆ m ⊕ g k . As M is simply connected, it is theRiemannian product ˆ M × M k , where the tangent spaces to ˆ M and M k at eH are ˆ m and g k respectively. Now g k is an ideal orthogonal to (and hence commuting with) h andthe restriction of A to g k is a multiple of the identity. It follows that M k is the simplyconnected, compact Lie group with the Lie algebra g k and with a bi-invariant metric.Furthermore, h ⊂ ˆ g and so H ⊂ ˆ G , where ˆ G is the compact, simply connected Lie groupwith the Lie algebra ˆ g . Then ˆ M = ˆ G/H , with the metric defined (relative to minus theKilling form of ˆ g ) by the metric endomorphism ˆ A which is the restriction of A to ˆ m .Now if M is a G -GO space, then ˆ M is ˆ G -GO (and the converse is also true). One wayto see that is to define the geodesic graph ˆ Z : ˆ m → h (see Lemma 1) by restricting ageodesic graph Z : m → h to ˆ m and using the fact that [ h , g k ] = [ ˆ m , g k ] = 0. By Lemma 3this also implies that if M is naturally reductive, then ˆ M also is (relative to ˆ G ). (cid:3) Trivial modules.
For every eigenvalue α i of A , denote by t i the maximal h -trivialsubmodule of m i , and m ′ i its orthogonal complement in m i . Let t = ⊕ mi =1 t i and m ′ = ⊕ mi =1 m ′ i . We will need the following Lemma (note that statement (d) is well known [20]). Lemma 5.
In the assumptions of Theorem 2, suppose the metric is GO. In the abovenotation, we have the following.
OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 11 (a)
The submodules t i are commuting, reductive ideals of the subalgebra t = z g ( h ) ⊂ g . (b) For any i = 1 , . . . , m , we have [ t , m ′ i ] ⊂ m ′ i . The restrictions of ad t and ad h commute on every m ′ i . Moreover, ad t i respects any irreducible decomposition of m ′ j if j = i , and the decomposition of m ′ i into the sums of isomorphic modules. (c) If an irreducible submodule n ⊂ m ′ is ad T -invariant for T ∈ t , then either [ T, n ] =0 or (ad T ) | n = λ id | n for some λ < . In the latter case, [ n , n ] ⊥ n . (d) The Lie algebra of skew-symmetric operators on an irreducible module n whichcommute with the restriction of ad h to n is either trivial, or is isomorphic to oneof so (2) or so (3) ; then the module n is said to be of real , complex or quaternionictype , respectively. In particular, the adjoint module is of real type.Suppose additionally that no eigenspace m i contains a nonzero ideal of g ( cf. Lemma 4 ) .Then we have the following. (e) [ T, m ′ ] = 0 , for any nonzero T ∈ t . (f) If no irreducible module n r in the decomposition (2) is adjoint, then g is simple. (g) If no more than one irreducible module n r in the decomposition (2) is nontrivial,then either the metric is normal, or M is one of the following spaces: SO(4 n + 2) / SU(2 n + 1) , n ≥ , SU( n + 1) / SU( n ) , n ≥ , E / Spin(10) , Sp( n + 1) / Sp( n ) , n ≥ . The spaces in (g) are spaces of cases (7), (5) with p = 1, (4) and (8) of Theorem 2(A)respectively (cf. [8, Theorem 2]). Remark . As a side remark we note that assertion (b) combined with Lemma 2(a)imposes further restrictions on the brackets of nontrivial irreducible submodules of m (similar to the second statement in (c)). For example, if n and n are irreduciblesubmodules lying in different eigenspaces m i and m j of A and ad t acts nontrivially on n ⊕ n , then either [ n , n ] ⊂ n or [ n , n ] ⊂ n . Proof. (a) Clearly t is the centraliser of h and hence is a subalgebra in g . Furthermore,by Lemma 2(b) we have [ t i , t j ] = 0 for i = j and by Lemma 2(d), [ t i , t i ] ⊂ m i ∩ t = t i .(b) By Lemma 2(a) we have [ t i , m ′ j ] ⊂ m ′ j for i = j , and moreover, irreducible h -submodules of m ′ j are ad( t i )-invariant. Furthermore, [ m ′ i , t i ] ⊂ m ′ i by Lemma 2(d) andfrom the fact that t i is a subalgebra. The restrictions of ad t and ad h clearly commute onevery m ′ i ; then by Schur’s Lemma, ad t i preserves the isotopic components of m ′ i .(c) The first statement is obvious, as n is irreducible and (ad T ) | n is a symmetric opera-tor commuting with ad h by (b). To prove the second statement, consider the three-form ω ∈ Λ n defined by ω ( X, Y, Z ) = h [ X, Y ] , Z i for X, Y, Z ∈ n . Then Ad(exp( π ( − λ ) − / T ))acts on n as − id | n and leaves ω invariant, so ω = 0.(d) Both statements are well known.(e) Suppose the centraliser k of m ′ in t is nontrivial. Then by (b), k is an ideal in t and hence also in g , as [ k , h ⊕ m ′ ] = 0. Then k ∩ t i , i = 1 , . . . , m , is an ideal of g by (a). As k ∩ t i ⊂ m i , it must be zero by our assumption. It follows that [ k , t i ] = 0, for all i = 1 , . . . , m , and so k lies in the centre of g contradicting the fact that g is semisimple.(f) Let g = ⊕ Nl =1 g l be the decomposition of g into simple ideals. As h is simple, theprojection of h to each of them is either a monomorphism or trivial, and it is nontrivialfor at least one l , say for l = N . If the projection to any other ideal g l , l < N , isalso nontrivial, then m contains an adjoint module, and hence so does (2). Otherwise, h ⊂ g N and so g ′ = ⊕ N − l =1 g l lies in m and is a trivial h -module. Then g ′ ⊂ t by (a) andso m ′ ⊂ g N . It follows that [ g ′ , m ′ ] = 0, and therefore g ′ = 0 by (e).(g) If all the modules in (2) are trivial (that is, if m is a trivial module), then say t = m is an ideal of g by (a), a contradiction. Let n = n ⊂ m be the only nontrivialirreducible module. Then m = n ⊕ t , and so by (b), (e) and (d), we have dim t ∈ { , , } .If t = 0, then m = n = m and so the metric is normal. If dim t = 1, the module t cannot lie in m (otherwise the metric is normal), so m has exactly two irreduciblesubmodules, m = n and m = t = R . By (e) and (d), the module n is not adjoint,and so g is simple by (f). Consulting the classification in [8, Theorem 2] we find thatthe spaces M for which h is simple and one of the irreducible submodules in m is one-dimensional are SO(4 n + 2) / SU(2 n + 1) for n ≥
2, SU( n + 1) / SU( n ) for n ≥ / Spin(10).Now suppose dim t = 3. Then by (b), (e) and (d), t is a subalgebra isomorphic to so (3) and so by (a) it is a single ideal t i . That ideal cannot lie in m (as otherwise themetric is normal) and so we have m = m ⊕ m , where m = n and m = t . Similar tothe above, g is simple and t acts on n nontrivially. But then h ′ = h ⊕ t is a subalgebraof g and ( g , h ′ ) is a symmetric pair by (c). Examining the list in [28, Theorem 4.1] wefind that M = Sp( n + 1) / Sp( n ) , n ≥ (cid:3) Large modules.
Recall that a module (not necessarily irreducible) is said to belarge if it contains an element whose centraliser in h is trivial. It is clear that the setof such elements in a large module is open and dense and that a module is large ifits submodule is large. Moreover, from (1) it follows that a geodesic graph is uniquelydefined on an open, dense subset of a large module. As usual, we adopt the assumptionsof Theorem 2 and use the notation of Section 2.We start with the following technical fact. Lemma 6.
Suppose a module n := n ⊂ m i in the decomposition (2) is nontrivial andthat its orthogonal complement n ′ = ⊕ pr =2 n r in m is a large module. Denote by U ⊂ n ′ the ( open and dense ) set of those elements whose centraliser in h is trivial. Then therestriction of a geodesic graph Z to the subset U × n is uniquely determined and thereexist a unique linear map L : n ′ → h and a unique map Ω :
U →
Lin( n , h ) such that forall X ∈ U , Y ∈ n , we have (4) Z = Z ( X + Y ) = LX + Ω( X ) Y. OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 13
Moreover, for all X ∈ U and Y ∈ n , we have [Ω( X ) Y, Y ] = 0 , (5) [ AX, X ] = [
LX, X ] , (6) α i [ LX, Y ] + [Ω( X ) Y, AX ] = [ AX − α i X, Y ] . (7) Proof.
Let X ∈ U ⊂ n ′ , Y ∈ n . Applying equation (1) to X + Y we get(8) [ Z ( X + Y ) , AX ] + α i [ Z ( X + Y ) , Y ] + [ X, AX ] + [
Y, AX − α i X ] = 0 . Note that the geodesic graph X + Y Z ( X + Y ) is uniquely defined on U × n . Let F be the restriction of Z to U . Taking Y = 0 in (8) we obtain(9) [ F ( X ) , AX ] + [ X, AX ] = 0 , for X ∈ U , and so [ Z ( X + Y ) − F ( X ) , AX ]+ α i [ Z ( X + Y ) , Y ]+[ Y, AX − α i X ] = 0 from (8).Projecting this equation to n ′ we obtain [ Z ( X + Y ) − F ( X ) , AX ] + π n ′ [ Y, AX − α i X ] =0, where π n ′ : m → n ′ is the orthogonal projection (the fact that [ Y, AX − α i X ] ∈ m follows from Lemma 2(a), as AX − α i X ⊥ m i ). From the fact that the secondterm is linear in Y and that Z ( X + Y ) and F ( X ) are uniquely determined we findthat the element Z ( X + Y ) − F ( X ) ∈ h depends linearly on Y , for every X ∈ n ′ .Therefore there exists a map Ω : U →
Lin( n , h ) such that for all X ∈ U , Y ∈ n ,we have Z ( X + Y ) = F ( X ) + Ω( X ) Y . Substituting into (8) and using (9) we obtain[Ω( X ) Y, AX ]+ α i [ F ( X )+Ω( X ) Y, Y ]+[
Y, AX − α i X ] = 0. Considering the left-hand side,for every fixed X ∈ U , as a polynomial in Y we obtain (5). To prove (6) and (7) it remainsto show that the map F : U → h is in fact linear. Projecting the latter equation to n wefind [ F ( X ) , Y ] + π n [ Y, α − i AX − X ] = 0, and so [ F ( X + X ) − F ( X ) − F ( X ) , Y ] = 0,for all Y ∈ n and for all X , X ∈ U such that X + X ∈ U . But the centraliser of n in h is an ideal which must be trivial, as h is simple and n is a nontrivial module. It followsthat F ( X + X ) = F ( X ) + F ( X ), for an open, dense set of pairs ( X , X ) ∈ n ′ × n ′ .The fact that F is homogeneous of degree 1 in X ∈ U follows from (9). Therefore thereexists a linear map L : n ′ → h whose restriction to U coincides with F . (cid:3) Note that from (7) it follows that the map Ω :
U →
Lin( n , h ) is analytic on U : relativeto some bases for n and h , the entries of its matrix are given by rational functions of X ∈ U [17]. We also note that as Z is unique, it is Ad( H )-equivariant, which implies that L is a homomorphism of h -modules. In particular, if n ′ contains no adjoint submodules,then L = 0 by Schur’s Lemma, and then [ AX, X ] = 0, for all X ∈ n ′ , by (6) (this impliesthat all the modules m i ∩ n ⊥ and m j , j = i pairwise commute).Furthermore, we have the following useful fact. Lemma 7.
Suppose the decomposition (2) contains nontrivial modules n = n whoseorthogonal complements are both large. Then the metric is naturally reductive.Proof. Denote q = ⊕ pr =3 n r . Let U r ⊂ n r ⊕ q , r = 1 ,
2, be the sets of elements whosecentraliser in h is trivial. Each of the subsets U r is open and dense in n r ⊕ q . For r = 1 , let U ′ r be the set of those elements X ∈ q for which there exists an open and dense subset V r,X ⊂ n r such that for all Y r ∈ V r,X we have Y r + X ∈ U r . Note that both U ′ and U ′ are open and dense in q , as also is the set U ′ = U ′ ∩ U ′ . By Lemma 6, for r = 1 , L r : n r ⊕ q → h and maps Ω r : U r → Lin( n r , h ) such that for any X ∈ U ′ and any Y ∈ V ,X , Y ∈ V ,X , the geodesic graph is given by Z = L ( X + Y ) + Ω ( X + Y ) Y = L ( X + Y ) + Ω ( X + Y ) Y , and(10) α [ L ( X + Y ) , Y ] + [Ω ( X + Y ) Y , A ( X + Y )] = [( A − α id)( X + Y ) , Y ] , (11)where the latter equations follows from (7) and we assume that n ⊂ m and n ⊂ m j (note that we can have j = 1). Projecting (11) to q we get [Ω ( X + Y ) Y , AX ] = π q [( A − α id) X, Y ] (note that π q [( A − α id) Y , Y ] = ( α j − α ) π q [ Y , Y ] = 0, by Lemma 2 (a)).Furthermore, by (5) [Ω ( X + Y ) Y , Y ] = 0. From the last two equations it followsthat for any X ∈ U ′ , Y ∈ V ,X and any Y ′ , Y ′′ ∈ V ,X we have [Ω ( X + Y ′ ) Y − Ω ( X + Y ′′ ) Y , A ( X + Y )] = 0. As X + Y ∈ U r and ad h commutes with A we obtainΩ ( X + Y ′ ) Y = Ω ( X + Y ′′ ) Y . Therefore there exists a map Ψ : U ′ → Lin( n , h ) suchthat Ω ( X + Y ) Y = Ψ ( X ) Y , for all X ∈ U ′ and all Y ∈ V ,X , Y ∈ V ,X . Similarly,there exists a map Ψ : U ′ → Lin( n , h ) such that Ω ( X + Y ) Y = Ψ ( X ) Y , for all X ∈ U ′ and all Y ∈ V ,X , Y ∈ V ,X . Substituting into (10) we get Z = L X + L Y +Ψ ( X ) Y = L X + L Y + Ψ ( X ) Y which now holds for all X ∈ U ′ and all Y ∈ n , Y ∈ n . Thus L X = L X and (Ψ ( X ) − L ) Y = 0, for all X ∈ U ′ and all Y ∈ n . Therefore we have Z = L X + L Y + L Y , for all X + Y + Y ∈ m , and so the metric is naturally reductiveby Lemma 3. (cid:3) The following Proposition effectively reduces the list of possible irreducible moduleswhich may appear in the decomposition (2) to a finite number of candidates, for everygiven group H . Proposition 1.
If one of the irreducible modules in the decomposition (2) is large, theneither the metric is naturally reductive or M is SU(3) / SU(2) or Sp(2) / Sp(1) .Proof.
By Lemma 4 we can assume that no m i contains a nonzero ideal of g . Otherwise,that ideal would be a trivial module, and so by factoring it out we would not lose a largemodule in decomposition (2). By Lemma 5(g) we can assume that at least two modulesin the decomposition (2) are nontrivial (one can easily check that all the irreduciblemodules in the corresponding decompositions are small except for in the two cases givenin the proposition). Furthermore, by Lemma 7, we can assume that the decomposition(2) has exactly two nontrivial modules, one of them being large and another one, small.Denote by them n and n ′ respectively.We first suppose n ′ is the adjoint module. Then by Lemma 6 applied to n ′ we obtainthat for an open, dense subset U ⊂ ( n ′ ) ⊥ ∩ m , a map Ω : U →
Lin( n ′ , h ) and a linearmap L : U → h , the geodesic graph is given by Z = LX + Ω( X ) Y for all X ∈ U , Y ∈ n ′ .Moreover, by (5) we have [Ω( X ) Y, Y ] = 0, for all X ∈ U , Y ∈ n ′ . As n ′ is the adjointmodule, there exists a linear isomorphism ι : h → n ′ such that for all V , V ∈ h we OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 15 have [ V , ιV ] = ι [ V , V ] (see 4.3). For X ∈ U define an endomorphism P X ∈ End( h ) by P X V = Ω( X ) ιV . Then for all V ∈ h we have [ P X V, V ] = 0, and so by [19, Theorem 5.28], P X commutes with all ad V , V ∈ h . As the adjoint module is of real type, P X = f ( X )id h ,for some function f : U → R , so that Ω( X ) Y = f ( X ) ι − Y , for all Y ∈ n ′ . Choosing X , X ∈ U such that X + X ∈ U and the intersection of Span( X , X ) with the trivialsubmodule of m is zero (the set of such pairs ( X , X ) is open end dense in U × U ) wefind from (7) that the function f is locally a constant, say c ∈ R . Then Z = LX + cι − Y on an open subset of m , hence on the whole m and so the metric is naturally reductiveby Lemma 3.We can therefore assume that n ′ is a tiny module. As m contains no adjoint modules,the algebra g is simple by Lemma 5(f). Furthermore, assuming that both n and n ′ lie in the same m i and applying Lemma 6 to n ′ we obtain that on the right-hand sideof (7), the vector AX − α i X lies in a trivial submodule of m orthogonal to m i , and so[ AX − α i X, Y ] ∈ n ′ , by Lemma 2(a). Projecting (7) to ( n ′ ) ⊥ we obtain [Ω( X ) Y, AX ] = 0,and so Ω( X ) Y = 0, for all X ∈ U ⊂ ( n ′ ) ⊥ and all Y ∈ n ′ . Then Z = LX which impliesthat the metric is naturally reductive (note that in fact Z = L = 0 — see the commentbefore Lemma 7).We can therefore assume that n and n ′ lie in different eigenspaces of A . We have m = n ⊕ n ′ ⊕ t , where t is trivial, and so h ⊕ t is a subalgebra of g having exactly two irreducibleisotropy modules, n and n ′ (by Lemma 5(b)). Moreover, by Lemma 1, the restriction of A to n ⊕ n ′ gives a GO metric on the space G/ ( HK ), where K is the centraliser of H in G (the Lie algebra of K is t ). That metric is not normal, as n and n ′ lie in differenteigenspaces of A . Since G is simple, by examining the list in [8, Theorem 2] we get thefollowing candidates for M (we have omitted the spaces in Lemma 5(g), as for them onlyone submodule in m is nontrivial): Spin(8) / G , SO(9) / G , SO(2 n + 1) / SU( n ) ( n ≥ / Spin(7) , SU( n + p ) / SU( n ) ( n ≥ , ≤ p ≤ n − , SU(2 n + 1) / Sp( n ) ( n ≥ m are tiny (and are listed in Table 2). This contradicts theassumption that n is large. (cid:3) Adjoint modules.
For any adjoint module s ⊂ m , there is a well-defined linearbijections ι : h → s such that for all U, V ∈ h ,(12) [ U, ιV ] = ι [ U, V ] . Lemma 8.
The direct sum of the adjoint module and a nontrivial module is a largemodule.Proof.
Suppose s is the adjoint module and n is an irreducible, nontrivial module. Identi-fying s with h via ι as in (12) we see that it is sufficient to find two elements X ∈ h , Y ∈ n whose centralisers have trivial intersection. This latter condition means that the rank ofthe linear system [ Z, X ] = [
Z, Y ] = 0 in the variable Z is maximal (equals dim h ); theset of pairs ( X, Y ) for which it is not is Zariski closed in the complexification h C × n C , and so it is sufficient to construct X ∈ h C , Y ∈ n C whose centralisers in h C have trivialintersection. To do that, take X to be regular and denote c ⊂ h C the Cartan subalgebradefined by X . Let γ be the dominant weight of n C . Then every element of its orbitunder the action of the Weyl group W of h C on c ∗ is also a weight of n C . Furthermore,the orbit W ( γ ) spans c ∗ as h C is simple. Take Y to be a linear combination of nonzerovectors Y g ∈ V g ( γ ) , for all g ∈ W , where V g ( γ ) is the weight space corresponding to theroot g ( γ ). Now the centraliser of X is c , but no nonzero vector from c centralises Y . (cid:3) Furthermore, we have the following proposition.
Proposition 2.
If one of the irreducible modules in the decomposition (2) is adjoint,then the metric is naturally reductive.Proof.
Suppose the decomposition (2) contains an adjoint module s . In the assump-tion that the metric is GO but not naturally reductive, by Lemma 8, Lemma 7 andLemma 5(g) we can assume that exactly one other module n in (2) is nontrivial, so that m = s ⊕ n ⊕ t , where t is trivial. By Proposition 1 we can assume that such n is small.Furthermore, we can assume that m contains no simple ideals of g . For if g a ⊂ m isa simple ideal of g , then h lies in the sum of other ideals of g , and so g a is a trivial h -module. But then by Lemma 5(a) it entirely lies in one of the eigenspaces m i and wecan factor it out by Lemma 4.We first assume that both s and n lie in the same eigenspace m of A . Then m is anonzero, trivial module. Take a nonzero T ∈ m . As s is of real type we have [ T, s ] = 0.Take X = S + N + T , where S ∈ s , N ∈ n . By (1), there exists Z ∈ h such that0 = [ Z + X, AX ] = α [ Z, S ] + α [ Z, N ] + ( α − α )[ N, T ] , As [ T, n ] ⊂ n by Lemma 5(b) the latter equation gives [ Z, S ] = 0 and [
Z, N ] = (1 − α − α )[ T, N ]. Let c ⊂ h be a Cartan subalgebra and let V ∈ c be a regular vector.Taking S = ιV we obtain that Z ∈ c by (12). By Lemma 5(e), (c) we can assume, up toscaling, that the restriction of (1 − α − α ) ad T to n is an almost Hermitian structure, andthen by Lemma 5(b), the restriction of ad h to n is a subalgebra of su ( n ) (it lies in u ( n ),the centraliser of (ad T ) | n and hence in su ( n ) as h is simple). Then (ad c ) | n is an abeliansubalgebra of su ( n ) which lies in a Cartan subalgebra c ′ of su ( n ). But then choosing aunitary basis for n we find that the equation [ Z, N ] = (1 − α − α )[ T, N ] is equivalentto the fact that for x ∈ C n (2 n = dim n ), there is a real, diagonal n × n matrix D withTr D = 0 such that i Dx = i x , which is false for a generic x ∈ C n , a contradiction.Now suppose s and n lie in different eigenspaces of A . The homogeneous space ˆ M = G/ ( HK ) where K is the connected subgroup of G whose Lie algebra is t has exactly twoirreducible components in its isotropy representations (note that t acts separately on s and on n by Lemma 5(b)) and moreover, the restriction of A to s ⊕ n defines a GO metricon ˆ M which is not normal. By [8, Proposition 1] we can have one of three cases (note that g must be semisimple and no ideal of it is allowed to be orthogonal to h ). In the first case, g = h ⊕ h ⊕ h and h ⊂ g is the diagonal (then t = 0). Then M is a Ledger-Obata space OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 17 and the metric is naturally reductive by [8, Proposition 3] (see also [22, Proposition 1];in fact, any invariant metric on the Ledger-Obata space H /H is naturally reductive,even without imposing the GO condition). In the second case, the algebra g is simple.Examining the cases in [8, Theorem 2], we find that in neither of them m contains anadjoint module. In the third case, we have g = g ⊕ g , where g , g are simple ideals,with both projections h and h from h to g and to g respectively being isomorphic to h ,and with g = h . Then g = h ⊕ n ⊕ t , the algebra h ⊂ g is the diagonal in h ⊕ h , andthe adjoint module s is its orthogonal complement in h ⊕ h . Note that then [ s , n ] ⊂ n ,and moreover, the action of s on n coincides with that of h , that is, [ S, N ] = [ ι − S, N ],for S ∈ s , N ∈ n (note that ι is defined up to scaling and we can take it to be a linearisometry). Up to relabelling, we have s ⊂ m , n ⊂ m . Suppose T ∈ t j , the trivialsubmodule of m j , j = 1 , , . . . , m . Then (1) with X = S + N + T, S ∈ s , N ∈ n , givesthat there exists Z ∈ h such that α [ Z, S ]+ α [ Z, N ]+( α − α )[ S, N ]+( α − α j )[ T, N ] = 0(we used the fact that [ T, s ] = 0). Set V = ι − S ∈ h . Then [ S, N ] = [
V, N ], and so theGO condition is equivalent to(13) [
Z, V ] = 0 , [ α Z + ( α − α ) V, N ] = ( α j − α )[ T, N ] . If ( α j − α ) T = 0, we argue as in the previous paragraph: by Lemma 5(e), (c), ( α j − α )(ad T ) | n is a nonzero multiple of an almost Hermitian structure on n , and then for aregular V ∈ h , from the first equation of (13), Z lies in the Cartan subalgebra defined by V which lies in a Cartan subalgebra of su ( n ). But then the second equation of (13) cannotbe satisfied with a generic N ∈ n , a contradiction. It now follows from Lemma 5(a)that t = t ⊂ m , and then (13) is satisfied with Z ( S + N + T ) = ( α α − − V =( α α − − ι − S . So the metric is naturally reductive by Lemma 3. (cid:3) G -GO spaces. Tiny modules Now we are in a position to complete the proof of Theorem 2.In the assumptions of Theorem 2 we assume that the GO metric is not naturallyreductive . Summarising the results of the previous sections we can additionally assumethe following: • all nontrivial modules in the decomposition (2) are tiny, and there are at leasttwo of them (by Propositions 1 and 2 and Lemma 5(g)); • there are not “too many” of them: there is no more than one nontrivial modulewhose complement is large (by Lemma 7); • no m i contains an ideal of g by Lemma 4; • g is simple (by Lemma 5(f)); • and finally, note that no GO metric constructed below is naturally reductive,unless it is normal (by Propositions 2 and Remark 1).The list of tiny modules from [14, Table 1] is given in Table 2. Note that some simplegroups (e.g. SU(2) and E ) have no tiny representations, while some others may have upto three. In the second column, for representations coming from the s -representations of Group Representation dim type StationarySO( n ) , n ≥ R n n r SO( n − n ) , n ≥ C n n c SU( n − n ) , n ≥ H n n q Sp( n − n ) , n ≥ s : SO(2 n ) / U( n ) n ( n −
1) c SU(2) [ n ] Sp( n ) , n ≥ s : SU(2 n ) / Sp( n ) ( n − n + 1) r Sp(1) n SU(6) s : E / SU(6)SU(2) 40 q T Spin(7) spin 8 r G Spin(9) spin, s : F / Spin(9) 16 r Spin(7)Spin(10) spin, s : E / Spin(10)SO(2) 32 c SU(4)Spin(12) spin, s : E / Spin(12)SU(2) 64 q SU(2) G standard, O ∩ ⊥ s : E /F
26 r Spin(8)E s : E / E SO(2) 54 c Spin(8)E s : E / E SU(2) 112 q Spin(8)
Table 2.
Tiny modulescompact symmetric spaces, we give those spaces. The fourth column indicates the type:real, complex or quaternionic; the fifth, the principal stationary subgroup.In the rest of the proof, we consider the groups in Table 2 one-by-one. Our strategy,for every individual group, will be first to consider all the possible decompositions (2);there will be a finite number of them: the nontrivial submodules are controlled by theabove assumptions, and the trivial ones, by Lemma 5. Some of those cases will bethen sorted out by the dimension count, as g must be simple. The remaining ones,when there are only two nontrivial modules, can be reduced to the classification in [8,Theorem 2] (in particular, if the trivial submodule t ⊂ m is zero, this classification appliesdirectly). For the small number of remaining cases, we consider possible “distributions”of the modules in the decomposition (2) among the eigenspaces m i , i = 1 , . . . , m , of themetric endomorphism A (note that m ≥
2) using Lemma 2, the decompositions of thetensor products and the external squares into irreducible modules and the classificationof compact irreducible symmetric spaces. If no contradiction is reached up to this point,we apply the GO criterion from Lemma 1 to determine the GO metric; then we identifythe corresponding space from the list in Theorem 2(A) (and in Table 1).Throughout this section we use the notation introduced earlier (in Sections 2 and 4.1);the direct sum of a ≥ n is abbreviated to a n .5.1. Types B and D: H = SO ( n ) , Spin ( n ) , n ≥ OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 19 n ) , n ≥ . From Table 2, there is only one tiny module, R n , the standard one.It follows that in the decomposition (2), all the modules are either standard or trivial.Let m = a R n ⊕ t , m = a R n ⊕ t be two eigenspaces of A , where a , a ≥ t , t are trivial. We have [ t , t ] = 0 by Lemma 5(a), and then for any n = R n ⊂ m , n = R n ⊂ m , we have [ t , n ] = [ t , n ] = 0 by Lemma 5(b) and (d) and [ n , n ] = 0 as theirreducible decomposition of R n ⊗ R n contains no module R n . It follows that [ m , m ] = 0.Therefore all the modules m i pairwise commute, and so the metric is naturally reductive(we can take Z = 0 in (1)).5.1.2. Spin(7) . From Table 2, there are only two tiny modules, the 8-dimensional module s for the spin representation and the 7-dimensional module R for the standard repre-sentation of SO(7). If there are no spin modules, we have a representation of SO(7)and then any GO metric must be naturally reductive as we have shown above. Wetherefore assume that there is at least one spin module. We claim that the sum of anyfour modules each of which is either spin or standard (and at least one is spin) is large.Indeed, if we have four spin modules, then choosing a generic element in one of themwe get the stationary subgroup G represented on R as the automorphism group of thealgebra of octonions O . As any three non-associating octonions generate O we obtainthat the stationary subgroup of a generic quadruple of elements is trivial. Next, supposewe have one standard module R and three spin modules. Then the stationary subgroupof a nonzero element from R is Spin(6) = SU(4), and its representation on each ofthe spin modules is the standard representation of SU(4) on C = R . The stationarysubgroup of a generic triple of elements is trivial. Next, suppose we have two standard R modules and two spin modules. Then the stationary subgroup of a generic pair ofelements from R is Spin(5) = Sp(2), and its representation on each of the spin mod-ules is the standard representation of Sp(2) on H = R . The stationary subgroup of ageneric pair of elements is again trivial. Finally, if we have three standard R modulesand one spin module, the stationary subgroup of a generic triple of elements from R isSpin(4) = Sp(1) × Sp(1), and its representation on the spin module is the sum of thetwo standard representations of Sp(1) on H = R . The stationary subgroup of a genericelement in R (the sum of two elements from each copy of H ) is again trivial.Up to relabelling, the decomposition (2) takes the form m = ⊕ qr =1 n r ⊕ t , where t isa trivial module, and from among the modules n r , for r = 1 , . . . , q , we have s ≥ a ≥ R modules, with s + a = q . By Lemma 7 and the argumentsabove we can assume that q ≤
4, and by Lemma 5(g), that q ≥
2. We will considerseveral cases depending on the value of q ∈ { , , } and the “distribution” of nontrivialmodules among the eigenspaces m i . For i = 1 , . . . , m , we have m i = s i s ⊕ a i R ⊕ t i ,where t i are trivial modules. We have s = P mi =1 s i , a = P mi =1 a i , with s ≥ ≤ q (= s + a ) ≤
4. Note that m > t i commutes with all m j , j = i , and may not commute with m ′ i only when m ′ i contains atleast two isomorphic modules. Then by Lemma 5(e) we obtain that for no i = 1 , . . . , m , the module m i can be trivial (so that a i + s i >
1, for all i = 1 , . . . , m ), and that t i canonly be nonzero when either s i > a i >
1, and in that case, t i is isomorphic to asubalgebra of so ( s i ) ⊕ so ( a i ).The above argument shows that if q = 2, then m contains no trivial submodules (forif both nontrivial submodules lie in the same m , then m = m ). Then by the result of[8, Theorem 2], we get the GO space M = Spin(9) / Spin(7) in Theorem 2(A)(1), with m = m ⊕ m , where m is the spin module and m is the standard module.Assume that q = 3 or q = 4. Furthermore, we have the following irreducible decom-positions of Spin(7) modules:(14) s ⊗ s = R ⊕ R ⊕ so (7) ⊕ . . . , R ⊗ R = R ⊕ so (7) ⊕ . . . , s ⊗ R = s ⊕ . . . , where so (7) is the adjoint module and the dots denote the sums of irreducible largemodules (these modules cannot occur in the decomposition of g viewed as the Spin(7)module). It then follows from Lemma 2(a) that any two spin modules lying in differenteigenspaces m i , m j , i = j , commute. Therefore if m contains no standard submodules R , then any two different eigenspaces m i commute and hence the metric is normal. Wecan therefore further assume that a ≥ g must be simple, from the dimension count and the above conditions we obtainthe following list of candidates (where s j are spin modules).(i) q = 4 , g = f , and n = s , n = s , n = s , n = R .(ii) q = 4 , g = f , and m = s ⊕ s ⊕ t , n = R , n = R , where t = so (2).(iii) q = 4 , g = f , and m = R ⊕ R ⊕ t , n = s , n = s , where t = so (2).(iv) q = 4 , g = so (11) or g = sp (5), and m = s ⊕ s ⊕ s ⊕ t , m = R , where t = so (3).(v) q = 3 , g = so (10), and m = s ⊕ s ⊕ t , m = R , where t = so (2).Note that in cases (i, ii) and (iii) the irreducible submodules n r may lie either in thesame or in different eigenspaces m i .By Lemma 2(a), (d) and from (14) we find that the sum h ′ of h = so (7) and all thetrivial and all the standard submodules of m is a subalgebra of g (not necessarily simple)and that its orthogonal complement p (which is the sum of all the spin submodules of m )satisfies [ p , p ] ⊂ h ′ . It follows that ( g , h ′ ) is a symmetric pair, and additionally dim( g / h ′ )is a multiple of 8.In particular, if g = f , the classification in [13] shows that there is only one such pair,( g , h ′ ) = ( f , so (9)), which corresponds to the Cayley projective plane. We immediatelysee that case (i) is not possible by the dimension count. Case (ii) is also not possible,because from Lemma 5(b), t would lie in the centre of h ′ = so (9) which is trivial. Incase (iii) we have h ′ = h ⊕ R ⊕ R ⊕ t = so (9) (note that t acts nontrivially on R ⊕ R ).Moreover, the modules s and s commute — this follows from Lemma 2(a) and (14) ifthey lie in two different eigenspaces of the metric automorphism A , and from Lemma 2(d)and (14) if they lie in the same eigenspace. But this is a contradiction as no two linearindependent vectors in the tangent space of the Cayley projective plane may commute OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 21 (as elements of f ), since otherwise the sectional curvature of the two-plane spanned bythem would equal zero.We now separately consider two remaining cases.(iv) In this case, h ⊕ R = so (8) and t lies in the centre of h ′ . There is no symmetricpair ( g , h ′ ) = ( sp (5) , so (8) ⊕ so (3)), and so g = so (11) giving the symmetric pair ( g , h ′ ) =( so (11) , so (8) ⊕ so (3)). The corresponding homogeneous space is SO(11) / Spin(7), whereSpin(7) ⊂ SO(8) ⊂ SO(8) × SO(3). It is an S -fibration over the Stieffel manifoldSO(11) / SO(8) with a normal metric (the construction is similar to the that in [28,Section 2], but with the non-symmetric base).By Lemma (1), the GO condition is equivalent to the fact that for any X r ∈ s r , r =1 , ,
3, and T ∈ t , Y ∈ R , there exists Z ∈ h such that 0 = [ X + X + X + T + Y + Z, α ( X + X + X + T ) + α Y ] = ( α − α )[ Y, X + X + X + T ] + α [ Z, Y ] + α [ Z, X + X + X + T ] = α [ Z + (1 − α α − ) Y, X ] + α [ Z + (1 − α α − ) Y, X ] + α [ Z +(1 − α α − ) Y, X ] + α [ Z, Y ]. By (14) and Lemma 2(a), the four terms on the right-handside belong to s , s , s and R respectively, and so the GO condition is equivalent to theexistence of Z ∈ h such that(15) [ Z, Y ] = 0 , [ Z + (1 − α α − ) Y, X r ] = 0 , for r = 1 , , . Now if Y = 0, one can take Z = 0. If Y = 0, then from the first equation, Z belongs tothe stationary subalgebra so (6) = su (4) ⊂ so (7) of Y . Identifying three modules s , s , s with a single spin module s via an isomorphism we see that relative to some unitary basisfor s = C , the operator (ad Y ) | s is proportional to the multiplication by i, and the actionof su (4) commutes with it. Then (15) is equivalent to [ Z, X r ] = µ i X r , r = 1 , ,
3, where µ ∈ R , µ = 0. It is sufficient to show that a required Z ∈ su (4) exists for X , X , X unitary orthonormal. Extending { X , X , X } to a unitary basis { X , X , X , X } wecan take Z ∈ su (4) such that (ad Z ) | s = diag( µ i , µ i , µ i , − µ i) relative to this basis. Itfollows that the metric is GO; it is not naturally reductive unless it is normal. This givesspace (3) in Theorem 2(A).(v) In this case, h ⊕ R = so (8) and t lies in the centre of h ′ . We obtain the sym-metric pair ( g , h ′ ) = ( so (10) , so (8) ⊕ so (2)). The corresponding homogeneous space isSO(10) / Spin(7), where Spin(7) ⊂ SO(8) ⊂ SO(8) × SO(2), which is an S -fibration overthe Stieffel manifold SO(10) / SO(8) with a normal metric. The fact that the metric isGO, can be established by repeating the arguments for the previous case, with obviousmodifications. This is the space (2) in Theorem 2(A).5.1.3. Spin(9) . From Table 2, there are only two tiny modules, the 16-dimensional mod-ule s for the spin representation and the 9-dimensional module R for the standardrepresentation of SO(9). Then in the decomposition (2), all the modules n r are eitherspin or standard or trivial. We have the following irreducible decompositions of Spin(9)modules:(16) s ⊗ s = R ⊕ R ⊕ so (9) ⊕ . . . , Λ s = so (9) ⊕ . . . , R ⊗ R = R ⊕ so (9) ⊕ . . . , s ⊗ R = s ⊕ . . . , where so (9) is the adjoint module and dots denote the sums of irreducible large modules(these modules cannot occur in the decomposition of g viewed as the Spin(9) module).Let m and m be two eigenspaces of the metric endomorphism A and let t i , i = 1 ,
2, bethe (maximal) trivial submodule of m i . Suppose s i ⊂ m i , i = 1 ,
2, are spin submodulesand ( R ) i ⊂ m i , i = 1 ,
2, are standard submodules. Then by Lemma 2(a) and (16) wehave [ s , s ] = 0 and [( R ) , ( R ) ] = 0, and by Lemma 5(a), [ t , t ] = 0. Furthermore,as both the spin and the standard modules are of real type, Lemma 5(b) gives [ t , s ⊕ ( R ) ] = [ t , s ⊕ ( R ) ] = 0. Finally, from Lemma 2(a) and (16) we get [( R ) , s ] ⊂ s ,which gives a homomorphism from ( R ) to Λ s . But this must be trivial by (16), whichimplies that [( R ) , s ] = 0 and similarly, [( R ) , s ] = 0. It follows that [ m , m ] = 0.Therefore all the modules m i pairwise commute, and so the metric is naturally reductive.5.1.4. Spin(10) . From Table 2, there are only two tiny modules, the 32-dimensionalmodule s for the spin representation and the 10-dimensional module R for the standardrepresentation of SO(10). If m contains no spin modules, we can take H = SO(10) andthen any GO metric must be naturally reductive by 5.1.1. We can therefore assume that m contains at least one spin module, and moreover, at least one other nontrivial module,by Lemma 5(g).We first show that the sum of two spin modules is large. The spin representation comesfrom the s -representation: it is the representation of Spin(10) on the tangent space ofthe symmetric space Q = E / Spin(10)SO(2). This symmetric space has rank 2, withthe restricted root system of type BC ; there are 6 roots: ε and ε of multiplicity 8each, ε ± ε of multiplicity 6 each and 2 ε and 2 ε of multiplicity 1 each [29, Table 1].The stationary subalgebra k (0) of a regular element of a maximal abelian subalgebra a ⊂ q = T o Q is su (4) ⊕ so (2). By [21, Lemma 2.25(a)], for every 6-dimensional rootspace q ( λ ) , λ = ε ± ε , the subalgebra spanned by [ q ( λ ) , q ( λ )] is an ideal of k (0)isomorphic to so (6), and it acts as the standard representation of so (6) on q ( λ ). By [21,Corollary 2.26(a)], for every 8-dimensional root space q ( λ ) , λ = ε , ε , the subalgebraspanned by [ q ( λ ) , q ( λ )] is an ideal of k (0) isomorphic to su (4), and it acts as the standardrepresentation of su (4) on q ( λ ). It is easy to see that the stationary subalgebra in k (0)of the element X + + X − + Y + Y , where X ± ∈ q ( ε ± ε ) and Y r ∈ q ( ε r ) , r = 1 ,
2, aregeneric vectors, is trivial. Therefore the sum of two copies of s is a large module.We next show that the sum of s and three copies of R is large. Indeed, the stationarysubgroup of a triple of linear independent elements of R is Spin(7) ⊂ Spin(10). Then s is the sum of four 8-dimensional irreducible Spin(7) submodules, and so is large by theargument in 5.1.2.We can therefore assume by Lemma 7 that the decomposition (2) takes either the form m = s ⊕ s ⊕ t , where t is a trivial module, or the form m = s ⊕ a R ⊕ t , where t is atrivial module and 1 ≤ a ≤ s is of complex type, we find by Lemma 5 that t is isomorphic toeither a subalgebra of u (2) if s and s lie in the same eigenspace m i , or of u (1) ⊕ u (1), OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 23 if they lie in different eigenspaces. The dimension count shows that, in any case, 109 ≤ dim g ≤ a = 1, as s is of complex type and R is of real type, weobtain by Lemma 5 that t is isomorphic to a subalgebra of u (1). Then dim g ∈ { , } .Similarly, if a = 3, we obtain that t is isomorphic to a subalgebra of u (1) ⊕ so (3), andso 107 ≤ dim g ≤ a = 2, t is isomorphic to a subalgebra of u (1) ⊕ so (2), and from thedimension count, the only candidate for g is su (10). But Spin(10) cannot be a subgroupof SU(10) as Spin(10) has no faithful real representation on R .So in the case H = Spin(10), any GO metric is naturally reductive.5.1.5. Spin(12) . From Table 2, there are only two tiny modules, the 64-dimensionalspin module s and the 12-dimensional module R for the standard representation ofSO(12). We can assume that m contains at least one spin module by 5.1.1, and atleast one other nontrivial module by Lemma 5(g). We claim that the sum of the spinmodule and either of the spin or the standard module is large. Indeed, if the secondmodule is the standard module R , the stationary subgroup of a nonzero element of it isSpin(11) ⊂ Spin(12). Its representation on s is still irreducible, with the trivial principalstationary subgroup (no Spin(11) entry in Table 2). Now suppose the second module isalso spin. The spin representation comes from the s -representation for the symmetricspace Q = E / Spin(12)SU(2). The symmetric space Q has rank 4, with the restrictedroot system of type F ; there are 12 roots of multiplicity 1 and 12 roots of multiplicity 4[29, Table 1]. The stationary subalgebra k (0) of a regular element of a maximal abeliansubalgebra a ⊂ q = T o Q is so (3) ⊕ so (3) ⊕ so (3). By [21, Lemma 2.25(a)], for every4-dimensional root space q ( λ ), the subalgebra spanned by [ q ( λ ) , q ( λ )] is an ideal of k (0) isomorphic to so (4), and it acts as the standard representation of so (4) on q ( λ ).Take three 4-dimensional root spaces q ( λ ) such that [ q ( λ ) , q ( λ )] span the same ideal so (4) ⊂ k (0) and choose three generic vectors, X , X , X , one in each of them. Then thestationary subalgebra of X + X + X in k (0) is the “remaining” ideal so (3). Now choosea nonzero vector Y in a 4-dimensional root space q ( λ ) such that [ q ( λ ) , q ( λ )] contains thatideal. Then the stationary subalgebra of X + X + X + Y in k (0) is trivial. Thereforethe sum of two copies of s is a large module.We can therefore assume that the decomposition (2) takes either the form m = s ⊕ s ⊕ t , where t is a trivial module, or the form m = s ⊕ R ⊕ t , where t is a trivial module.In the second case, both s and R are ad( t )-invariant by Lemma 5(b). As s is ofquaternionic type and R is of real type, t is a subalgebra of sp (1), by Lemma 5(e),(d).It follows that 142 ≤ dim g ≤ g whose dimension liesin this range is su (12), but Spin(12) cannot be a subgroup of SU(12) as Spin(12) has nofaithful real representation on R . Similarly, in the first case, t must be a subalgebra of sp (2), which gives 194 ≤ dim g ≤ g whose dimensionlies in this range is su (15), but Spin(12) is not a subgroup of SU(15) as Spin(12) has nofaithful real representation on R . So in the case H = Spin(12), any GO metric is naturally reductive.5.2. Type A: H = SU ( n ) , n ≥ From Table 2, we can have the following tiny SU( n )modules. For all n ≥
3, we have the standard module of dimension 2 n , and for all n ≥ p of dimension n ( n −
1) coming from the s -representation for thesymmetric space Q = SO(2 n ) / U( n ) (note that for n = 3 this module is standard, andfor n = 4 it is reducible). In addition, for n = 4 we have the tiny module of dimension6 coming from the standard representation of SO(6) = SU(4) / Z , and for n = 6, thereis a tiny module q of dimension 40 from the s -representation for the symmetric space Q = E / SU(6)SU(2).Before considering various cases we prove the following lemma. Consider the homo-geneous space
G/H = SO(2 n + 1) / SU( n ), where n ≥ H = SU( n ) ⊂ U( n ) ⊂ SO(2 n ) ⊂ SO(2 n + 1) = G . At the level of Lie algebras, we have h = su ( n ) ⊂ u ( n ) ⊂ so (2 n ) ⊂ so (2 n + 1) = g . We have an orthogonal decomposition into h -modules: g = h ⊕ t ⊕ n ⊕ s , where t is the one-dimensional trivial module which is the orthogonalcomplement to h in u ( n ), the module n is the orthogonal complement to u ( n ) in so (2 n )and the module s is the standard 2 n -dimensional module which is the orthogonal com-plement to so (2 n ) in so (2 n + 1) (note that the module n is reducible when n = 4 and isstandard when n = 3). Lemma 9.
In the above notation, suppose the metric on
SO(2 n + 1) / SU( n ) , n ≥ , isdefined by a metric automorphism A such that the modules s , n , t lie in the eigenspaces of A with the eigenvalues α , α , α > respectively. Then if n is even, the metric is GO ifand only if α = α ; if n is odd, the metric is GO if and only if nα − = ( n − α − + α − .Proof. We have [ n , s ] = [ t , s ] = s and [ t , n ] = n . For X ∈ s , Y ∈ n , T ∈ t , the GOcondition (1) gives [ Z + X + Y + T, α X + α Y + α T ] = 0, which is equivalent to[ Z + σT, Y ] = 0 , [ Z + ρY + τ T, X ] = 0 , where σ = 1 − α α − , ρ = 1 − α α − , τ = 1 − α α − . Choose an almost Hermitianstructure in s = C n in such a way that (ad t ) | s is a real multiple of the multiplication by i . Then for w = X ∈ C n we have [ Z, X ] =
M w , where M is an n × n skew-Hermitianmatrix with trace zero, [ T, X ] = i tw , where t ∈ R , and [ Y, X ] =
N Cw , where N is an n × n complex skew-symmetric matrix (depending on Y ) and C is the componentwisecomplex conjugation. The above GO condition is then equivalent to the following: forany w ∈ C n , t ∈ R and N ∈ so ( n, C ), there exists an n × n traceless, skew-Hermitianmatrix M such that(17) ( M + σt i I n ) N + N ( M + σt i I n ) t = 0 , ( M + ρN C + τ t i ) w = 0 . We consider two cases depending on the parity of n .Suppose n is even; let n = 2 k . Take N in (17) to be generic (that is, rk N = n and allthe eigenvalues of N are pairwise distinct). We can choose a unitary basis for C n relativeto which N = diag( µ J, µ J, . . . , µ k J ), where J = ( − ) and ± µ , ± µ , . . . , ± µ k ∈ C arenonzero and pairwise distinct. Then from the first equation of (17) we obtain M + σt i I n = OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 25 diag( F , F , . . . , F k ), where F , F , . . . , F k ∈ su (2). It follows that Tr( M + σt i I n ) = 0 andso σ = 0. This gives α = α and τ = ρ .Now the second equation of (17) gives F j w j + ρµ j J w j + ρt i w j = 0 , j = 1 , , . . . , k, where the coordinates of the vector w j = (cid:16) w j w j (cid:17) ∈ C are the (2 j − j -thcoordinates of w respectively. If w j = 0, the latter equation is trivially satisfied, with anarbitrary F j . Otherwise, a direct calculation gives F j = (cid:16) a j i z j − z j − a j i (cid:17) , where a j ∈ R and z j ∈ C are given by( | w j | + | w j | ) (cid:18) a j i z j (cid:19) = − ρt i (cid:18) | w j | − | w j | w j w j (cid:19) + ρ (cid:18) i Im( µ j w j w j ) µ j ( w j ) + µ j ( w j ) (cid:19) . As the right-hand side is continuous in µ j , the entries of N , we deduce that a tracelessskew-Hermitian matrix M which satisfies (17) exists for all N ∈ so ( n, C ) (and all t ∈ R and w ∈ C n ). Hence the metric so defined is GO.Now suppose n is odd; let n = 2 k +1. The proof is similar to that in the even case withsome modifications. We again take a generic N and choose a unitary basis for C n relativeto which N = diag( µ J, µ J, . . . , µ k J, J = ( − ) and ± µ , ± µ , . . . , ± µ k ∈ C are nonzero and pairwise distinct. The first equation of (17) gives M + σt i I n =diag( F , F , . . . , F k , c ), where F , F , . . . , F k ∈ su (2) and c ∈ C . Comparing the traces wefind that c = nσt i and so M + σt i I n = diag( F − σt i I , F − σt i I , . . . , F k − σt i I , kσt i ).Then the second equation of (17) gives( F j + ( τ − σ ) t i I ) w j + ρµ j J w j = 0 , j = 1 , , . . . , k, kσ + τ = 0 , where the coordinates of the vector w j = (cid:16) w j w j (cid:17) ∈ C are the (2 j − j -thcoordinates of w respectively. From the last equation we obtain nα − = ( n − α − + α − .From the first k equations we find, provided w j = 0, that the entries of the matrix F j = (cid:16) a j i z j − z j − a j i (cid:17) ∈ su (2) , a j ∈ R , z j ∈ C are given by( | w j | + | w j | ) (cid:18) a j i z j (cid:19) = nσt i (cid:18) | w j | − | w j | w j w j (cid:19) + ρ (cid:18) i Im( µ j w j w j ) µ j ( w j ) + µ j ( w j ) (cid:19) . Similar to the even case this proves that the metric is GO. (cid:3) n ) , n ≥ ; all modules standard. We first consider the case when n ≥ n − n − p defined above and the one-dimensional trivial module. Suppose we have two standard modules s ⊂ m i and s ⊂ m j , i = j . Then by Lemma 2(a) [ s , s ] ⊂ s ⊕ s ,and hence s and s commute; in particular, [[ s , s ] , s ] = 0. On the other hand, [ s , s ]is contained in the subalgebra which is the direct sum of h and a one-dimensional trivialmodule, and moreover, is an ideal in that subalgebra. As s is nontrivial we obtain h ⊂ [ s , s ] which contradicts the fact that [[ s , s ] , s ] = 0, as s is also nontrivial.It follows that all the standard modules s r lie in the same eigenspace, and so up torelabelling we have m = ⊕ pr =1 s r ⊕ t and m i = t i for i = 2 , . . . , m , where t , t , . . . , t m aretrivial modules. Note that 2 ≤ p ≤ n − p = 1 we obtain the space SU( n + 1) / SU( n )from Lemma 5(g)) and m ≥
2, so that t = 0. Moreover, the module t must alsobe nonzero, for if t = 0 we obtain that [ s , s ] = 0 by Lemma 2(d) and then repeatthe argument above. Furthermore, the pair ( g , h ′ ), where h ′ = h ⊕ t , is a symmetricpair (as the bracket of standard modules contains no standard modules and hence liesin h ′ , and m ′ := ⊕ pr =1 s r is ad( t )-invariant, by Lemma 5(b)). As g is simple and h ′ is the sum of two ideals, su ( n ) and t , with dim t ≥
2, from the classification [13] wefind that ( g , h ′ ) = ( su ( n + p ) , su ( n ) ⊕ su ( p ) ⊕ R ). From Lemma 5(e), (a) we obtainthat m = 2 and t ⊕ t = su ( p ) ⊕ R . Note that t = su ( p ) as ad t preserves everyindividual module s r , r = 1 , . . . , p , by Lemma 5(b); therefore t = su ( p ) , t = R . Itfollows that m = ⊕ pr =1 s r ⊕ su ( p ) and m = R . The corresponding homogeneous spaceis SU( n + p ) / SU( n ) , ≤ p ≤ n −
1, where SU( n ) ⊂ SU( n ) × SU( p ) ⊂ S(U( n ) × U( p ));this is the space (5) in Theorem 2(A) (with p ≥ n ≥ X r ∈ s r , r =1 , . . . , p , and T ∈ t , T ∈ t , there exists Z ∈ h such that [ P pr =1 X r + T + T + Z , α ( P pr =1 X r + T ) + α T ] = 0. Note that t commutes with both h and t , and preserveseach of the s r ’s. Therefore the GO condition is equivalent to the existence of Z ∈ h suchthat(18) [ Z + (1 − α α − ) T, X r ] = 0 , for r = 1 , . . . , p. If T = 0, we can take Z = 0. If T = 0, then identifying the modules s r with a singlestandard module n via an isomorphism we see that relative to some unitary basis for n = C n , the operator (ad T ) | s is real proportional to the multiplication by i, and the actionof su ( n ) commutes with it. Then (18) is equivalent to [ Z, X r ] = µ i X r , r = 1 , . . . , p , where µ ∈ R , µ = 0. It is sufficient to show that a required Z ∈ su ( n ) exists when X r areunitary orthonormal. Extending { X , . . . , X p } to a unitary basis { X , . . . , X p , . . . , X n } (recall that p ≤ n −
1) we can take Z ∈ su ( n ) such that (ad Z ) | s relative to this basis isgiven by the diagonal matrix whose first p entries are µ i and the remaining n − p , are − p ( n − p ) − µ i. It follows that the metric is GO.5.2.2. SU( n ) , n ≥ and n = 6 . From Table 2, there are two tiny modules, the standardmodule s = C n and the n ( n − p , and so in the decomposition(2), every nontrivial submodule is isomorphic to either s or p . There are at least twoof them, and we can assume that at least one is isomorphic to p by 5.2.1. Note thatboth p ⊕ p and p ⊕ s are large modules. Indeed, for a generic element of p we have the OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 27 set of [ n/
2] mutually unitary orthogonal subspaces C ⊂ C n such that the stationarysubalgebra is the sum of [ n/
2] copies of su (2) each acting in its own C . To see that theprincipal stationary subalgebra of p ⊕ s is trivial we take an element in s = C n all whosecomponents in the subspaces C are nonzero; for p ⊕ p , we take two generic elements of p sharing no common subspaces C in the corresponding sets.It follows that the only possible cases for the decomposition (2) are m = p ⊕ p ⊕ t and m = p ⊕ s ⊕ t , where t is a trivial module.We will need the irreducible decompositions of the tensor products of s and p :(19) s ⊗ s = 2 R ⊕ p ⊕ su ( n ) ⊕ . . . , for n = 5 or n = 7 , p ⊗ p = 2 R ⊕ su ( n ) ⊕ . . . , s ⊗ p = s ⊕ . . . , for n ≥ , p ⊗ p = 2 R ⊕ su (5) ⊕ s ⊕ . . . , s ⊗ p = s ⊕ p ⊕ . . . , for n = 5 , where R is the trivial module, su ( n ) is the adjoint module and dots denote the sums ofirreducible large modules.We now consider the case m = p ⊕ p ⊕ t . From Lemma 5 we find t ⊂ u (2). Furthermore,by (19), p ⊗ p does not contain p , and so ( g , h ′ = su ( n ) ⊕ t ) is a symmetric pair, with thecorresponding symmetric space of dimension 2 n ( n − n ≥
5, and so this case is not possible.Next consider the case m = p ⊕ s ⊕ t . From Lemma 5 we have t ⊂ R . First suppose that n ≥
7. Then by (19), ( g , h ′ = su ( n ) ⊕ p ⊕ t ) is a symmetric pair, with the correspondingsymmetric space of dimension 2 n and with 2 n − n − ≤ dim h ′ ≤ n − n + 1 andrk h ′ ≥ n −
1. From the classification in [13] we find that there is only one such pair:( g , h ′ ) = ( so (2 n + 1) , so (2 n )). Then we get dim t = 1 and su ( n ) ⊕ p ⊕ t = so (2 n ) and weobtain a family of GO metrics as in Lemma 9.The last case to consider is m = p ⊕ s ⊕ t and n = 5. We again have dim t ≤ g ∈ { , , } . Then we get dim t = 1 and g = sp (5)or g = so (11). The first case is not possible, as the only way to realise the algebra u (5) = su (5) ⊕ t as a subalgebra of sp (5) is the one corresponding to the symmetric pair( sp (5) , u (5)). But as both p and s are ad( t )-invariant, the 30-dimensional irreducible u (5)-module which is the tangent space of the corresponding symmetric space remainsirreducible for the subalgebra su (5). In the case g = so (11) there is again, the only wayto realise u (5) = su (5) ⊕ t as a subalgebra of so (11): we have u (5) ⊂ so (10) ⊂ so (11),and then this case is completed by application of Lemma 9.Thus we obtain the spaces in Theorem 2(A)(6) with n ≥ n = 5.5.2.3. SU(3) . From Table 2, there is only one tiny module, the standard one, of dimen-sion 6. Note that the sum of two of them is a large module, and so by Lemma 7 andLemma 5(g) we can assume that the decomposition (2) takes the form m = s ⊕ s ⊕ t ,where s and s are standard and t is a trivial module. As the standard module is ofcomplex type and as there are two of them, from Lemma 5 we find that t is a subal-gebra of u (2). It follows that 20 ≤ dim g ≤
24. This gives three possibilities: either g = sp (3) , dim t = 1, or g = so (7) , dim t = 1, or g = su (5) , dim t = 4. In the first case, there is only one way, up to conjugation, to realise the algebra u (3) = h ⊕ t as a subalgebra of g = sp (3), namely the one corresponding to the symmetricpair ( sp (3) , u (3)). As we have at least two different eigenspaces m i of A , and dim t = 1,we obtain by Lemma 5(b) that both s and s are ad( t )-invariant. But then they are u (3)-invariant which contradicts the fact that the representation of u (3) on the tangentspace of the corresponding symmetric space is irreducible.In the second case, there is again only one way, up to conjugation, to realise thealgebra u (3) = h ⊕ t as a subalgebra of g = so (7): we have u (3) ⊂ so (6) ⊂ so (7), wherethe first inclusion corresponds to the symmetric pair ( so (6) , u (3)). We have thereforethe following modules in the decomposition (2): the standard module s which is theorthogonal complement to so (6) in so (7), the standard module s which is the orthogonalcomplement to u (3) in so (6), and the one-dimensional module t , the centre of u (3). ByLemma 9 we get GO metrics on the space SO(7) / SU(3) from Theorem 2(A)(6) with n = 3.In the third case, from Lemma 5 we obtain that t can have dimension 4 only if, upto relabelling, the eigenspaces of A are given by m = s ⊕ s ⊕ t , m = t , where t is isomorphic to su (2) and dim t = 1. Repeating the arguments in the last twoparagraphs of 5.2.1, we obtain a family of GO metrics on SU(5) / SU(3) (the space (5) inTheorem 2(A) with n = 3 and p = 2).5.2.4. SU(4) . From Table 2, there are two tiny modules, the standard module s = C and the module n = R coming from the standard representation of SO(6) = SU(4) / Z (note that the orthogonal complement to u (4) in so (8) is reducible and is the direct sumof two copies of n ).The cases when all nontrivial modules in the decomposition (2) are isomorphic to n or all are standard have already been considered in 5.1.1 and 5.2.1 respectively. Wecan therefore assume that at least one module in (2) is standard and at least one isisomorphic to n . Note that the module 2 n ⊕ s is large. Indeed, the stationary subgroupof a nonzero element from s = C is SU(3), and the restriction of the representation ofSU(4) on n to SU(3) is the standard representation of SU(3) on n = R = C . Then thestationary subgroup of a pair of linear independent vectors from C is trivial. Similarly,the module n ⊕ s is large, as the stationary subgroup of a nonzero element from n isSp(2) = Spin(5) ⊂ Spin(6) = SU(4), and so s can be viewed as the standard module forSp(2). Then the stationary subgroup of a pair of linear independent vectors from s = H is trivial.It therefore follows that the decomposition (2) takes the form m = q s ⊕ r n ⊕ t , where t is a trivial module and ( q, r ) = (1 , , (2 , , (1 , s is of complex type and n is of real type. Then in the case ( q, r ) = (1 , t ≤
1, and so dim g ∈ { , } , which is a contradiction as there are no simple groupsof these dimensions. Similarly, in the case ( q, r ) = (2 ,
1) we obtain that t is a subalgebraof u (2), and so 37 ≤ dim g ≤
41, which again leads to a contradiction. In the case( q, r ) = (1 , t ≤
2, and from the dimension count we obtain one of
OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 29 the following three cases: either t = 0 and g = su (6), or dim t = 1 and then g = sp (4) or g = so (9).We have the following irreducible decompositions of SU(4) modules:(20) s ⊗ s = 2 R ⊕ n ⊕ su (4) ⊕ . . . , n ⊗ n = R ⊕ su (4) ⊕ . . . , s ⊗ n = s ⊕ . . . , where su (4) is the adjoint module and dots denote the sums of irreducible large modules.It follows that h ′ = h ⊕ n ⊕ t is a subalgebra of g and that ( g , h ′ ) is a symmetric pair.The corresponding symmetric space must be of dimension 8, and from the classificationin [13] we find that the only possible case is g = so (9) , h ′ = so (8), and then s is theorthogonal complement to so (8) is so (9), t is the one-dimensional centraliser of su (4) in so (8) and 2 n is the orthogonal complement to u (4) = su (4) ⊕ t in so (8). Denote by n , n the two copies of n . If n and n lie in different eigenspaces m i , then by Lemma 2(a) and(20) we get [ n , n ] = 0, and so [[ n , n ] , n ] = 0 which contradicts the fact that both n and n are irreducible nontrivial modules. Therefore n ⊕ n lie in the same eigenspace m i of A . By a similar argument using Lemma 2(d) we obtain that t ⊂ m i . It followsthat the metric endomorphism A has two eigenspaces, m = n ⊕ n ⊕ t and m = s . ByLemma 9, we get the space SO(9) / SU(4) in Theorem 2(A)(6) with n = 4.5.2.5. SU(6) . We have three tiny modules, the standard module s of dimension 12, themodule p of dimension 30 and the module q of dimension 40 which comes from the s -representation for the symmetric space Q = E / SU(6)SU(2). As above, we can assumethat in the decomposition (2) we have at least two nontrivial modules and that at leastone of them is not standard. We claim that the sum of any two modules each of whichis isomorphic to s , p or q and at least one is not standard is a large module. For themodules p ⊕ p and p ⊕ s the argument is similar to that in the first paragraph of 5.2.2.In the remaining cases, one of the modules is q with the stationary subalgebra R (seeTable 2) which is a subalgebra of a Cartan subalgebra of su (6). It follows that the sumof q and any nontrivial module is large, by the argument in the proof of Lemma 8.Therefore the decomposition (2) takes the form m = n ⊕ n ⊕ t , where n , n ∈ { s , p , q } and at least one of them is not standard, and t is trivial. Note that s and p are of complextype and q of quaternionic type. We consider all possible cases.Suppose m = 2 q ⊕ t . Then by Lemma 5 we have t ⊂ sp (2), and by the dimension countwe get dim g = 115 + dim t , so that 115 ≤ dim g ≤ su (11) and so (16), both of dimension 120. But sp (2) contains nosubalgebras of dimension 5, so this case is not possible.Suppose m = p ⊕ q ⊕ t . Then t ⊂ sp (1) ⊕ R , and dim g = 105 + dim t , so that105 ≤ dim g ≤ sp (7) and so (15), both of dimension 105. But then t = 0 and so m = p ⊕ q which implies that anyGO metric (if one at all exists) is normal, by [8, Theorem 2].Suppose m = s ⊕ q ⊕ t . Then again t ⊂ sp (1) ⊕ R , and so dim g = 87 + dim t whichgives the only candidate so (14). Then t = sp (1) ⊕ R . But the only faithful representationof su (6) of dimension at most 14 is the standard representation on R . It follows that su (6) ⊂ so (12) ⊂ so (14), and so the centraliser of su (6) in so (14) is R (cannot be aslarge as t = sp (1) ⊕ R ). This contradiction shows that this case is also not possible.Next suppose m = 2 p ⊕ t . By Lemma 5 we have t ⊂ u (2), and by the dimensioncount, the only possible case is g = su (10) , t = u (2). But the only realisation of su (6)as a subalgebra of su (10) comes from the inclusion C ⊂ C , and then the centraliser of su (6) is u (4) which is much bigger than t . So this case is also not possible.Finally let m = p ⊕ s ⊕ t . Then t ⊂ R and the dimension count gives dim t = 1and dim g = 78, so that g is either e or sp (6) or so (13). But the case g = e is notpossible, as the only su (6) subalgebra in e is the one coming from the symmetric pair( e , su (6) ⊕ su (2)), and its centraliser contains su (2) in contradiction with dim t = 1.Furthermore, if g = sp (6), then the only su (6) subalgebra is the one coming from thesymmetric pair ( sp (6) , u (6)). But then, as both p and s are ad( t )-invariant, they arealso u (6)-invariant which contradicts the fact that the representation of u (6) on thetangent space of the corresponding symmetric space is irreducible. The last remainingcase is g = so (13). The only way to realise u (6) = su (6) ⊕ t as a subalgebra of so (13)is u (6) ⊂ so (12) ⊂ so (13), where ( so (12) , u (6)) is a symmetric pair. By Lemma 9 weobtain the space SO(13) / SU(6) in Theorem 2(A)(6) with n = 6.5.3. Type C: H = Sp ( n ) , n ≥ From Table 2, we can have two tiny Sp( n ) modules:the standard module s of dimension 4 n and the module p of dimension ( n − n + 1)obtained from the s -representation for the symmetric space Q = SU(2 n ) / Sp( n ) (note thatfor n = 2, the module p comes from the standard representation of SO(5) = Sp(2) / Z on R ).We first consider the case when all the nontrivial modules in decomposition (2) arestandard. The sum of n such modules is a large module, and so we can assume that m = ⊕ pi =1 s i ⊕ t , where s i are standard modules, t is trivial and 2 ≤ p ≤ n . As the tensorsquare of the standard module contains no standard modules, we find that ( g , h ′ = h ⊕ t )is a symmetric pair, with the corresponding symmetric space of dimension 4 pn , and with t being a nonzero ideal in h ′ (for if t = 0, the eigenspaces m i are the sums of the standardmodules and hence pairwise commute by Lemma 2(a); this gives a naturally reductivemetric: we can take Z = 0 in (1)). From the classification in [13], there are only twocases. In the first one, we have n = 2 and ( g , h ′ ) = ( so (5 + q ) , so (5) ⊕ so ( q )); but then8 p = 5 q , from the dimension count, and so p ≥
5, a contradiction. In the second case, n ≥ g , h ′ ) = ( sp ( n + q ) , sp ( n ) ⊕ sp ( q )). It follows that q = p and that t = sp ( p ), and so by Lemma 5(a), t entirely lies in a single eigenspace m i . But then fromLemma 5(b), we can have t so big only when all the standard modules s , . . . , s p also liein the same eigenspace which implies that A is the multiple of the identity and so theGO metric is normal (note that this argument does not work when p = 1, in which casewe obtain the space Sp( n + 1) / Sp( n ) from Lemma 5(g) which carries a GO metric whichis not naturally reductive).We can therefore assume that the decomposition (2) contains the module p and at leastone other nontrivial module. Note that p ⊕ s is a large module. Indeed, the stationary OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 31 subalgebra of a generic element of p is n sp (1), which acts on s = H n by acting as sp (1) on n orthogonal copies of H ; taking an element of s whose component in each of these copiesis nonzero we find that the stationary subalgebra is trivial. Assume that all nontrivialmodules in (2) are isomorphic to p . If n = 2, we have the standard representation of so (5) on R , and so the sum of four copies of p is large, but the sum of three is still small.If n ≥
3, the sum of two copies of p is already large. To see this, we identify the module p with the tangent space q of the symmetric space Q = SU(2 n ) / Sp( n ) and choose ageneric element X ∈ q . Its centraliser in q is a maximal abelian subalgebra a ⊂ q ofdimension n − q ∩ a ⊥ into the orthogonal sum of n ( n −
1) root spaces q ( λ ) of dimension 4, with the root system of type A n − [29, Table 1].The stationary subalgebra of X is k (0) = n sp (1) = ⊕ ni =1 ( sp (1)) i , with every root space q ( λ ) being k (0)-invariant, and moreover, for the root λ = ± ( ε i − ε j ) , ≤ i < j ≤ n , thesubalgebra ( sp (1)) i ⊕ ( sp (1)) j ⊂ k (0) acts on q ( λ ) as the standard representation of so (4),and all the other ( sp (1)) k , k = i, j , act on q ( λ ) trivially [21, Lemma 2.25]. It followsthat if we choose Y ∈ q whose component in each of the root spaces q ( λ ) is nonzero, itsstationary subalgebra in k (0) will be trivial.Suppose all the nontrivial modules in the decomposition (2) are isomorphic to p . If n = 2 we have the standard representation of so (5) on R ; this case has been analysedin 5.1.1. If n ≥ m contains exactly two copies of p . As p isof real type, no module m i can be trivial by Lemma 5(e), (d) and (b), and so the twocopies of p must lie in different eigenspaces m i , and moreover, neither of these eigenspacescontains a trivial submodule. It follows that the only possibility is m = m ⊕ m , with m = p , m = p , where p and p are isomorphic to p . But then from [8, Theorem 2]we find that any GO metric is normal, hence naturally reductive.The last remaining case to consider is when the decomposition (2) has the form m = p ⊕ s ⊕ t , where t is a trivial module. Note that p is of real type and so by Lemma 5 wehave [ t , p ] = 0 , [ t , s ] ⊂ s and t ⊂ sp (1). From [8, Theorem 2] we also find that t = 0.Moreover, the irreducible decomposition of the tensor product p ⊗ s does not contain p , and so [ p , s ] ⊂ s and [ p , p ] ⊥ s . Also, the irreducible decomposition of the tensorproduct s ⊗ s does not contain s . It follows that ( g , h ′ = h ⊕ p ⊕ t ) is a symmetric pair,with t a nonzero ideal in h ′ . The dimension of the corresponding symmetric space is 4 n ,and we also know that dim h ′ ∈ { n , n + 2 } , rk h ′ ≥ n + 1 and that h ′ contains anonzero ideal t ⊂ sp (1). From the classification in [13] we find that the only possibilityis ( g , h ′ ) = ( su (2 n + 1) , su (2 n ) ⊕ R ). Then s is the orthogonal complement to su (2 n ) in su (2 n + 1) and p is the orthogonal complement to h in su (2 n ) (note that the only way,up to automorphism, to realise sp ( n ) as a subalgebra of su (2 n ) is the one correspondingto the symmetric pair ( su (2 n ) , sp ( n ))).We now consider the GO condition. Suppose s , p and t lie in the eigenspaces of A with the eigenvalues α , α and α respectively (some of them, but not all three, may beequal). Then (1) is equivalent to the fact that for any X ∈ s , Y ∈ p and T ∈ t , there exists Z ∈ h such that(21) [ Z, Y ] = 0 , [ Z + ρY + τ T, X ] = 0 , where ρ = 1 − α α − , τ = 1 − α α − . We can now identify s with C n in such away that the action of ad t is the multiplication by a real multiple of i and choose aunitary basis in such a way that su (2 n ) is the space of complex matrices of the form (cid:0) K L − L ∗ K (cid:1) , where K and K are skew-Hermitian with Tr K + Tr K = 0 and L is anarbitrary n × n complex matrix, and sp ( n ) ⊂ su (2 n ) is the space of complex matricesof the form (cid:0) K S − S K (cid:1) , where K is skew-Hermitian and S is an n × n symmetric complexmatrix. Then p is the space of matrices of the form (cid:0) K LL − K (cid:1) , where K is skew-Hermitianwith Tr K = 0 and L ∈ so ( n, C ). Choose a generic Y ∈ p . We can specify the basisfurther (conjugate by an element of Sp( n )) in such a way that the maximal abeliansubalgebra of p containing Y is the space ( i D i D ) where D is a real diagonal matrixwith Tr D = 0. Taking Y of this form with D = diag( d , . . . , d n ) such that d i ∈ R arenonzero and pairwise distinct we obtain from the first equation of (21) that Z = (cid:0) K S − S K (cid:1) ,where K = diag( i x , . . . , i x n ) , S = diag( z , . . . , z n ) , x j ∈ R , z j ∈ C . We also havead T = i tI n , t ∈ R . Then the second equation of (21) splits into n pairs of equations ofthe form i ( x j + ρd j + τ t ) w j + z j w n + j = 0 , − z j w j + i ( − x j + ρd j + τ t ) w n + j = 0 , j = 1 , . . . , n, where w j , w n + j ∈ C are the corresponding coordinates of X ∈ s relative to the chosenbasis. Now if w j = w n + j = 0 we can choose x j and z j arbitrarily. If not, a direct calcula-tion gives x j = ( | w n + j | − | w j | )( | w n + j | + | w j | ) − ( ρd j + τ t ) , z j = − i w j w n + j ( | w n + j | + | w j | ) − ( ρd j + τ t ). As these expressions are continuous in d j , the equations (21) have asolution Z ∈ h for all Y ∈ p , X ∈ s and T ∈ t . Hence the metric so defined is GO whichgives the space SU(2 n + 1) / Sp( n ) , n ≥
2, in Theorem 2(A)(9).5.4.
Exceptional groups. . From Table 2, there is only one tiny module, the space of imaginary octonions O ′ := O ∩ ⊥ which is the space of the defining representation for G , and so thedecomposition (2) takes the form m = q O ′ ⊕ t , where t is a trivial module. As anythree non-associating imaginary octonions generate the whole algebra of octonions O ,the sum of three copies of O ′ is a large module (one can check that the sum of two isstill a small module, with the principal stationary subgroup Sp(1)), and so by Lemma 7and Lemma 5(g) we have q ∈ { , } . As the module O ′ is of real type, Lemma 5(b) and(e) imply that no m i can be trivial (and hence not all the modules O ′ lie in a single m i )and that m i may contain a nonzero trivial submodule only if it also contains at least twocopies of O ′ . If q = 2 these conditions imply that m = m ⊕ m , where both m and m are isomorphic to O ′ , so that m is the sum of two irreducible submodules. As g is simple,by [8, Theorem 2] we obtain M = Spin(8) / G , the space (10) in Theorem 2(A).Suppose q = 3. If the three modules O ′ lie in three different eigenspaces of A , thensimilar to the above, we get m = 3 O ′ . Then dim g = 35, and so g = su (6). But g is OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 33 not a subalgebra of su (6) (as the complexification of g has no nontrivial representationon C ), a contradiction. So up to relabelling we have m = m ⊕ m , where m = n ⊕ n ⊕ t , m = n , where the modules n , n and n are isomorphic to O ′ and t isa trivial module. If t = 0, we get a contradiction by the dimension count as above.Furthermore, by Lemma 5(b) we get [ t , n ] = 0, and so by Lemma 5(b) and (e), forevery nonzero T ∈ t , the restriction of ad T to n ⊕ n is nonzero and commutes withad h , and so we obtain that t = R T , where (ad T ) n ⊕ n is given by the matrix (cid:0) I − I (cid:1) ,relative to bases for n , n which correspond via isomorphism. Then g is a simple algebraof dimension 36 and h ′ = h ⊕ t is its subalgebra; moreover, the modules n ⊕ n and n are irreducible for h ′ and the metric on ( n ⊕ n ) ⊕ n obtained by the restrictionof A is GO. Then the pair ( g , h ′ ) corresponds to [8, Theorem 2(2)], and so we get thespace M = SO(9) / G in Theorem 2(A)(11). To see that the metric is GO we note that[ t , n ] = n and [ t , n ] = n . By Lemma 2(a) we have [ n , n i ] ⊂ n ⊕ n i for i = 1 , t we get [ n , n i ] ⊂ n i for i = 1 ,
2, as [ t , n ] = 0. Nowfor X = N + N + T ′ + N , where N i ∈ n i for i = 1 , , T ′ ∈ t , equation (1)gives α [ Z, N ] + α [ Z, N ] + α [ Z, N ] + ( α − α )[ N , N ] + ( α − α )[ N , N ] = 0 andso projecting to the modules n i , i = 1 , ,
3, we get(22) [
Z, N ] = 0 , [ Z, N i ] = ( α − α − N , N i ] , for i = 1 , . If N = 0 we can take Z = 0. Suppose N = 0. Identify all three modules n , n , n with a copy of O ′ via isomorphisms and denote x i , i = 1 , ,
3, the image of N i underthis identification. The stationary subalgebra of x is su (3) acting by the standardrepresentation on C = O ′ ∩ x ⊥ (see Table 2; the stationary subalgebras of all nonzeroelements are principal). Moreover, the action of ad N on O ′ commutes with the actionof su (3) and so it is trivial on x and is a multiplication by µ i on C for some µ ∈ R .To satisfy equation (22) it is sufficient, for any x , x ∈ C , to find M ∈ su (3) such that M x = ρ i x , M x = ρ i x , where ρ = µ ( α − α − ∈ R . We can assume that x and x are be complex orthonormal. Then, relative to a unitary basis { x , x , y } for C we cantake M = diag( ρ i , ρ i , − ρ i } .5.4.2. F . From Table 2, there is only one tiny module, of dimension 26, which comesfrom the s -representation for the symmetric space Q = E / F . The sum of three suchmodules is a large module (the sum of two is still small, by the dimension count 52 =dim F < n of dimension 26can be viewed as the space of 3 × h = f can be represented as the direct sum of two subspaces, the Lie subalgebra g whoseelements act on a matrix N ∈ n componentwise, and the subspace so (3 , O ) of the 3 × N ∈ n as amatrix commutator [4, Theorem 5]. Now take N ∈ n to be diagonal (then it is real), withpairwise non-equal entries. The corresponding stationary subalgebra of h is the directsum of g and the subspace { diag( a , a , a ) | a , a , a ∈ O ′ , a + a + a = 0 } ⊂ so (3 , O )(note that the stationary subalgebra is so (8), as per Table 2). Take N , N ∈ n defined by N = , N = x yx ∗ zy ∗ z ∗ , where the octonions x, y, z are non-associating. Note that the action of g on N istrivial as all the entries of N are real, and that a matrix Q = diag( a , a , a ) with a , a , a ∈ O ′ , a + a + a = 0, commutes with N only when Q = 0. It follows that thestationary subalgebra of the pair of matrices ( N , N ) is the subalgebra g ⊂ h acting onthe matrices from n componentwise. But as the entries x, y, z of the matrix N generatethe whole algebra of octonions O , the only element of g which maps all of them to zerois zero. It follows that the stationary subalgebra of the triple ( N , N , N ) is trivial.Now by the arguments similar to those in (5.4.1) we get that the decomposition (2)takes the form m = q n ⊕ t , where t is a trivial module and q = 2 ,
3. As the module n is of real type, Lemma 5(b) and (e) imply that no m i can be trivial and that m i maycontain a nonzero trivial submodule only if it also contains at least two copies of n . Thenfor q = 2, the only possibility is m = m ⊕ m and m = n , m = n , where n , n areisomorphic to n , which is not possible by [8, Theorem 2]. Suppose q = 3. If the modules n , n , n isomorphic to n lie in three different eigenspaces of A , then m = n ⊕ n ⊕ n ,and so dim g = 130, a contradiction, as there is no simple algebra of this dimension.Otherwise, up to relabelling, we have m = m ⊕ m , where m = n ⊕ n ⊕ t , m = n ,where t is a trivial module. As n is of real type, by Lemma 5(b) we get [ t , n ] = 0, andso by Lemma 5(b) and (e) t ⊂ so (2) acting on n ⊕ n . This is a contradiction, as thereis no simple Lie algebra whose dimension lies in { , } . It follows that in the case H = F , any GO metric is naturally reductive.5.4.3. E . From Table 2, there is only one tiny module n , of dimension 54, which comesfrom the s -representation for the symmetric space Q = E / E SO(2). The sum of threesuch modules is a large module. Indeed, the symmetric space Q has rank 3, with therestricted root system of type C ; there are three roots of multiplicity 1 and six roots ofmultiplicity 8 [29, Table 1]. The stationary subalgebra of a regular element of a maximalabelian subalgebra a ⊂ q = T o Q is so (8) ⊂ e , and by [21, Lemma 2.25(a)], it acts asthe standard representation of so (8) on each of the six 8-dimensional root spaces (andacts trivially on the three 1-dimensional root spaces). If we choose a generic 6-tuple ofvectors, one in each of the six 8-dimensional root spaces, the stationary subalgebra oftheir sum will be so (2) ⊂ so (8) ⊂ e , but if we choose two such 6-tuples, the stationarysubalgebra will be trivial. We now argue as in the previous cases. The decomposition (2)takes the form m = q n ⊕ t , where t is a trivial module and q ∈ { , } . In the case q = 2,as n is of complex type, the centraliser of ad h in so ( n ⊕ n ) (where n , n are isomorphiccopies of n ) is u (2), and so by Lemma 5(b), (e) and (d), the maximal dimension of t is4. Then 186 ≤ dim g ≤ g whose dimension satisfiesthis inequality is so (20), but e is not a subalgebra of so (20) as it has no nontrivial realrepresentation on R . Suppose q = 3. If all three submodules n , n and n isomorphic OMPACT GEODESIC ORBIT SPACES WITH A SIMPLE ISOTROPY GROUP 35 to n lie in different m i , then by Lemma 5(b), each of them is ad( t )-invariant, and sofrom Lemma 5(e) and (d), we obtain dim t ≤
3. If (up to relabelling) m ⊃ n ⊕ n and m ⊃ n we get ad T ∈ u (2) ⊕ u (1), for all T ∈ t and so dim t ≤
5. Finally, if m = n ⊕ n ⊕ n ⊕ t , where t is trivial, then all the other modules m i , < i ≤ m , aretrivial. By [8, Theorem 2] we can assume that m ≥
3, and then by Lemma 5, t = ⊕ mi =1 t i is a subalgebra of u (3), with t i being commuting ideals and t , t = 0. It follows thatdim t ≤
5. Therefore in all three cases, we have 240 ≤ dim g ≤ H = E , anyGO metric is naturally reductive.5.4.4. E . From Table 2, there is only one tiny module n , of dimension 112, which comesfrom the s -representation for the symmetric space Q = E / E SU(2). We show thatthe sum of two such modules is a large module. The symmetric space Q has rank 4,with the restricted root system of type F ; there are 12 roots of multiplicity 1 and 12roots of multiplicity 8 [29, Table 1]. The stationary subalgebra of a regular element ofa maximal abelian subalgebra a ⊂ q = T o Q is so (8) ⊂ e , and by [21, Lemma 2.25(a)],it acts as the standard representation of so (8) on each of the twelve 8-dimensional rootspaces. If we choose a generic 12-tuple of vectors, one in each of the twelve 8-dimensionalroot spaces, the stationary subalgebra will be trivial. We can therefore assume that thedecomposition (2) takes the form m = 2 n ⊕ t , where t is a trivial module. But now fromLemma 5 we get t ⊂ sp (2). It follows that 357 ≤ dim g ≤ g whose dimension lies in this range is su (19), but e cannot be its subalgebra as e has no nontrivial real representation on R . So in the case H = E , any GO metricmust be naturally reductive.This completes the proof of Theorem 2. References [1] D. V. Alekseevsky, A. Arvanitoyeorgos,
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