Nov 28, 2017 - We prove that for various classes of M-spaces the only g.o. metric is ... sufficient conditions, so that a G-invariant metric on G/K1 i...

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arXiv:1610.01278v4 [math.DG] 28 Nov 2017

ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO Abstract. We investigate homogeneous geodesics in a class of homogeneous spaces called M -spaces, which are defined as follows. Let G/K be a generalized flag manifold with K = C(S) = S × K1 , where S is a torus in a compact simple Lie group G and K1 is the semisimple part of K. Then the associated M -space is the homogeneous space G/K1 . These spaces were introduced and studied by H.C. Wang in 1954. We prove that for various classes of M -spaces the only g.o. metric is the standard metric. For other classes of M -spaces we give either necessary, or necessary and sufficient conditions, so that a G-invariant metric on G/K1 is a g.o. metric. The analysis is based on properties of the isotropy representation m = m1 ⊕ · · · ⊕ ms of the flag manifold G/K (as Ad(K)-modules) and corresponding decomposition n = s ⊕ m1 ⊕ · · · ⊕ ms of the tangent space of the M -space G/K1 (as Ad(K1 )-modules).

2010 Mathematical Subject Classification. Primary 53C25. Secondary 53C30.

Key words. Generalized flag manifold; isotropy representation; M -space; t-roots; homogeneous geodesic; geodesic vector; g.o. space.

Introduction Let (M, g) be a homogeneous Riemannian manifold, i.e. a connected Riemannian manifold on which the largest connected group G of isometries acts transitively. Then M can be expressed as a homogeneous space (G/K, g), where K is the isotropy group at a fixed pointed o of M, and g is a G-invariant metric. A geodesic γ(t) through the origin o of M = G/K is called homogeneous if it is an orbit of a one-parameter subgroup of G, that is γ(t) = exp(tX)(o), t ∈ R, (1) where X is a non zero vector of g. A homogeneous Riemannian manifold M = G/K is called a g.o. space, if all geodesics are homogeneous with respect to the largest connected group of isometries Io (M). A G-invariant metric g on M is called G-g.o. if all geodesics are homogeneous with respect to the group G ⊆ Io (M). Of course a G-g.o. metric is a g.o. metric, but the converse is not true in general. In this paper we only consider G-g.o. metrics, which we also call them g.o. metrics. Naturally reductive spaces, symmetric spaces and weekly symmetric spaces are g.o. spaces ([10], [13], [21], [31]). In [22] O. Kowalski, F. Pr¨ ufer and L. Vanhecke gave an explicit classification of all naturally reductive spaces up to dimension five. In [24] O. Kowalski and L. Vanhecke gave a classification of all g.o. spaces, which are in no way naturally reductive, up to dimension six. In [20] C. Gordon described g.o. spaces which are nilmanifolds, and in [29] H. Tamaru classified homogeneous g.o. spaces which are fibered over irreducible symmetric spaces. In [15] and [17] O. Kowalski and Z. Duˇsek investigated homogeneous geodesics in Heisenberg groups and some H-type groups. Examples of g.o. spaces in dimension seven were obtained by Duˇsek, O. Kowalski and S. Nikˇcevi´c in [18]. In [2] the first author and D.V. Alekseevsky classified generalized flag manifolds which are g.o. spaces. Recently, the first two authors classified generalized 1

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Wallach spaces which are g.o. spaces ([9]). Also, in [14] Z. Chen and Yu. Nikonorov classified compact simply connected g.o. spaces with two isotropy summands. Other interesting results about g.o. spaces can be found in [3], [8], [11], [16], [19], [23], [25], [26], [27], [28] and [32]. Finally, the notion of homogeneous geodesics can be extended to geodesics which are orbits of a product of two exponential factors (cf. [6], [7]). The general problem of classification of compact homogeneous Riemannian manifolds (M = G/K, g) with homogeneous geodesics remains open. The object of the present paper is to study homogeneous geodesics in M-spaces. These spaces were introduced and studied by H.C. Wang in [30] and are defined as follows: Let G/K be a generalized flag manifold with K = C(S) = S × K1 , where S is a torus in a compact simple Lie group G and K1 is the semisimple part of K. Then the corresponding M-space is the homogeneous space G/K1 . Let g and k be the Lie algebras of the Lie groups G and K respectively. Let g = k ⊕ m be an Ad(K)-invariant reductive decomposition of the Lie algebra g, where m∼ = To (G/K). This is orthogonal with respect to B = −Killing from on g. Assume that m = m1 ⊕ · · · ⊕ ms (2)

is a B-orthogonal decomposition of m into pairwise inequivalent irreducible ad(k)modules. Let G/K1 be the corresponding M-space and s and k1 be the Lie algebras of S and K1 respectively. We denote by n the tangent space To (G/K1 ), where o = eK1 , it follows that n = s ⊕ m. A G-invariant metric g on G/K1 induces a scalar product h·, ·i on n which is Ad(K1 )-invariant. Such an Ad(K1 )-invariant scalar product h·, ·i on n can be expressed as hx, yi = B(Λx, y) (x, y ∈ n), where Λ is an Ad(K1 )-equivariant positive definite symmetric operator on n. Conversely, any such operator Λ determines an Ad(K1 )-invariant scalar product hx, yi = B(Λx, y) on n, which in turn determines a G-invariant Riemannian metric g on n. We say that Λ is the operator associated to the metric g, or simply the associated operator. A Riemannian metric generated by the inner product B(·, ·) is called standard metric. An M-space G/K1 with the standard metric is g.o., since it is naturally reductive. The main results of the paper are the following: Theorem 1. Let G/K be a generalized flag manifold with s ≥ 3 in the decomposition (2). Let G/K1 be the corresponding M-space. If (G/K1 , g) is a g.o. space, then g = h·, ·i = Λ |s +λB(·, ·) |m1 ⊕m2 ⊕···⊕ms , (λ > 0), where Λ is the operator associated to the metric g. Corollary 1. Let G/K be a generalized flag manifold with s ≥ 3 in the decomposition (2). Let G/K1 be the corresponding M-space. If dim s = 1 and there exists some j ∈ {1, . . . , s} such that mj is reducible as an Ad(K1 )-module, then (G/K1 , g) is a g.o. space if and only if g is the standard metric. For a generalized flag manifold with two isotropy summands we always assume that it satisfies [m1 , m1 ] ⊆ k ⊕ m2 . So it does not occur dim m1 = dim m2 = 2, since G is simple. If there exists i ∈ {1, 2} such that dim mi = 2, we obtain that i = 2 (see also convention on [Ar-Ch]). Theorem 2. Let G/K be a generalized flag manifold with s = 2 in the decomposition (2). Let G/K1 be the corresponding M-space.

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1) If both m1 and m2 are irreducible as Ad(K1 )-modules, then (G/K1 , g) is a g.o. space if and only if for every V ∈ s and for every X = X1 + X2 ∈ m (Xi ∈ mi , i = 1, 2), there exists k ∈ k1 such that [µ1 k + (µ1 − µ)V + (µ1 − µ2 )X2 , X1 ] = 0 and

[µ2 k + (µ2 − µ)V, X2 ] = 0. Here g = h·, ·i = µ |s +µ1 B(·, ·) |m1 +µ2 B(·, ·) |m2 is any G-invariant metric on G/K1 . 2) If there exists j ∈ {1, 2} such that mj is reducible as an Ad(K1 )-module and mi (i 6= j) is irreducible as an Ad(K1 )-module, and (G/K1 , g) is a g.o. space, then g = h·, ·i = µB(·, ·) |s+mj +µi B(·, ·) |mi . (3) 3) If both m1 and m2 are reducible as Ad(K1 )-modules, then (G/K1 , g) is a g.o. space if and only if g is the standard metric. Corollary 2. Let G/K be a generalized flag manifold with s = 2 and dim m2 = 2. Then the corresponding M-space (G/K1 , g), where g is given by (3), is a g.o. space for every µ 6= µ1 if and only if for every V ∈ s and for every X1 ∈ m1 , X2 ∈ m2 there exists k ∈ k1 such that [k + V + X2 , X1 ] = 0. (4) Theorem 3. Let G/K be a generalized flag manifold with s = 1 in the decomposition (2). Let G/K1 be the corresponding M-space. 1) If m is irreducible as Ad(K1 )-module, then g.o. metrics on G/K1 have been studied by Z.Chen and Yu.Nikonorov (cf. [14]). 2) If m is reducible as Ad(K1 )-module, then (G/K1 , g) is a g.o. space if and only if g is the standard metric. The paper is organised as follows: In Section 1 we recall certain Lie theoretic properties of a generalized flag manifold G/K and corresponding M-space G/K1 . Then we investigate Ad(K1 )-irreducible submodules (cf. Lemma 2, 3). In Section 2, 3 we recall basic facts about g.o. spaces. In Sections 4, 5, 6 we give the proofs of Theorems 1, 2, 3 and Corollaries 1, 2. Acknowledgements. The first author was supported by Grant # E.037 from the research committee of the University of Patras (Programme K. Karatheodori). The second author is supported by NSFC 11501390, and by funding of Sichuan University of Science and Engineering grant Grant No. 2015RC10. The third author is supported by NSFC 11571242. 1. Generalized flag manifolds and M-spaces Let G/K = G/C(S) be a generalized flag manifold, where G is a compact semisimple Lie group and S is a torus in G, here C(S) denotes the centralizer of S in G. Let g and k be the Lie algebras of the Lie groups G and K respectively, and gC and kC be the complexifications of g and k respectively. Let g = k ⊕ m be a reductive decomposition with respect to B = −Killing form on g with [k, m] ⊂ m. Let T be a maximal torus of G containing S. Then this is a maximal torus in K. Let a be the Lie algebra of T and aC its complex. Then aC is a Cartan subalgebra of gC . Let R be a root system of gC with respect to aC and Π = {α1 , . . . , αl }, (l = dimC aC ) be a system of simple roots of R, and {Λ1 , . . . , Λl } be the fundamental weights of gC corresponding to Π, that is 2B(Λi ,αj ) = δij , (1 ≤ i, j ≤ l). We can identify (aC )∗ with aC as follows: For every B(αj ,αj )

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α ∈ (aC )∗ it corresponds to hα ∈ aC by the equation B(H, hα ) = α(H) for all H ∈ aC . Let X gC = aC ⊕ gCα (5) α∈R

be the root space decomposition, where

gCα = {X ∈ gC : [H, X] = α(H)X ∀H ∈ aC }.

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Since kC contains aC , there is a subset RK of R such that kC = aC ⊕ α∈RK gCα . We choose a system of simple roots ΠK of RK and a system of simple roots Π of R so that + ΠK ⊂ Π. We choose an ordering in R+ . Then there is a natural ordering in RK , so P + + C C that RK ⊂ R . Set RM = R \ RK (complementary roots). Then m = α∈RM g . P

+ + Definition 1. An invariant ordering RM in RM is a choice of a subset RM ⊂ RM such that + − − + (i) R = RK ⊔ RM ⊔ RM , where RM = {−α : α ∈ RM }, + + (ii) If α, β ∈ RM and α + β ∈ RM , we have α + β ∈ RM , + + + (iii) If α ∈ RM , β ∈ RK and α + β ∈ R, we have α + β ∈ RM . + + For any α, β ∈ RM we define α > β if and only if α − β ∈ RM .

We choose a Weyl basis {Eα , Hα : α ∈ R} in gC with B(Eα , E−α ) = 1, [Eα , E−α ] = Hα and ( 0, if α + β 6∈ R and α + β 6= 0, [Eα , Eβ ] = (7) Nα,β Eα+β , if α + β ∈ R, where the structural constants Nα,β (6= 0) satisfy Nα,β = −N−α,−β and Nβ,α = −Nα,β . Then we have that X X gCα , (8) gCα ⊕ gC = aC ⊕ α∈RM

α∈RK

C

and {Eα : α ∈ RM } is a basis of m . It is well known that X X √ (RAα + RBα ), R −1Hα ⊕ gu =

(9)

α∈R+

α∈R+

√ where Aα = Eα − E−α , Bα = −1(Eα + E−α ) (α ∈ R+ ) is a compact real form of gC . Hence we can identify g with gu . In fact g = gu is the fixed point set of the √ √ √ conjugationPX + −1Y → 7 X + −1Y = X − −1Y in gC so that Eα = −E−α . √ P + + Hence k = α∈R+ R −1Hα ⊕ α∈R+ (RAα + RBα ). We set RM = R+ \ RK . Then K X m= (RAα + RBα ). (10) α∈R+ M

The next lemma gives us information about the Lie algebra structure of g. √ Lemma 1. The Lie bracket among the elements of {Aα , Bα , −1Hβ : α ∈ R+ , β ∈ Π} of g are given by √ [ −1Hα , Aβ ] = β(Hα)Bβ , [Aα , Aβ ] = Nα,β Aα+β + N−α,β Aα−β (α 6= β), √ [ −1Hα , Bβ ] = −β(Hα )Aβ , [Bα , Bβ ] = −Nα,β Aα+β − Nα,−β Aα−β (α 6= β), √ [Aα , Bα ] = 2 −1Hα , [Aα , Bβ ] = Nα,β Bα+β + Nα,−β Bα−β (α 6= β), where Nα,β are the structural constants in (7).

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An important invariant of a generalized flag manifold G/K is the set Rt of t-roots. Their importance arises from the fact that the knowledge of Rt gives us crucial information about the decomposition of the isotropy representation of the flag manifold G/K. From now on we fix a system of simple roots Π = {α1 , . . . , αr , φ1 , . . . , φk } of R, so that ΠK = {φ1 , . . . , φk } is a basis of the root system RK and ΠM = Π \ ΠK = {α1 , . . . , αr } (r + k = l). We consider the decomposition R = RK ∪ RM and let √ √ t = z(kC ) ∩ −1a = {X ∈ −1a : φ(X) = 0, for all φ ∈ RK }, (11)

where z(kC ) is the center of kC . Consider the restriction map κ : (aC )∗ → t∗ defined by κ(α) = α |t , and set Rt = κ(R) = κ(RM ). Note that κ(RK ) = 0 and κ(0) = 0. + + The elements of Rt are called t-roots. For an invariant ordering RM = R+ \ RK + + − + + in RM , we set Rt = κ(RM ) and Rt = −Rt = {−ξ : ξ ∈ Rt }. It is obvious that − Rt− = κ(RM ), thus the splitting Rt = Rt− ∪ Rt+ defines an ordering in Rt . A t-root ξ ∈ Rt+ (respectively ξ ∈ Rt− ) will be called positive (respectively negative). A t-root is called simple if it is not a sum of two positive t-roots. The set Πt of all simple t-roots is called a t-basis of t∗ , in the sense that any t-root can be written as a linear combination of its elements with integer coefficients of the same sign. Definition 2 ([2]). (1) Two t-roots ξ, η ∈ Rt are called adjacent if one of the following occurs: (i) If η is a multiple of ξ, then η 6= ±2ξ and ξ 6= ±2η. (ii) If η is not a multiple of ξ, then ξ + η ∈ Rt or ξ − η ∈ Rt . (2) Two t-roots ξ, η ∈ Rt are called connected if there is a chain of t-roots ξ = ξ1 , ξ2 , . . . , ξk = η

such that ξi , ξi+1 are adjacent (i = 1, . . . , k − 1). We remark that ξ and ±ξ are connected, and if ξ, 2ξ are the only positive t-roots, then these are not connected. We define the relation ξ ∼ η ⇔ ξ, η are connected.

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One can easily check that this is an equivalence relation. Let Ri be the equivalent classes consisting of mutually connected t-roots. Then the set Rt is decomposed into a disjoint union Rt = R1 ∪ · · · ∪ Rr . (13) Definition 3. The set of t-roots Rt is called connected if r = 1. Also, if G is simple, then for s ≥ 3 in the decomposition (2), the set of t-roots is connected (cf. [2]). Proposition 1 ([4]). There is one-to-one correspondence between t-roots and complex irreducible ad(kC )-submodules mξ of mC . This correspondence is given by X Rt ∋ ξ ↔ mξ = CEα . α∈RM :κ(α)=ξ

Thus mC =

mξ . Moreover, these submodules are inequivalent as ad(kC )-modules. √ √ Since the complex conjugation τ : gC → gC , X + −1Y 7→ X − −1Y (X, Y ∈ g) of gC with respect to the compact real form g interchanges the root spaces, i.e. τ (Eα ) = P

ξ∈Rt

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E−α and τ (E−α ) = Eα , a decomposition of the real Ad(K)-module m = (mC )τ into real irreducible Ad(K)-submodule is given by X m= (mξ ⊕ m−ξ )τ , (14) + ξ∈R+ t =κ(RM )

where V τ denotes the set of fixed points of the complex conjugation τ in a vector subspace V ⊂ gC . If, for simplicity, we set Rt+ = {ξ1 , . . . , ξs }, then according to (14) each real irreducible Ad(K)-submodule mi = (mξi ⊕ m−ξi )τ (1 ≤ i ≤ s) corresponding to the positive t-roots ξi , is given by X (RAα + RBα ). (15) mi = α∈R+ M :κ(α)=ξi

Definition 4 ([30]). Let G/K be a generalized flag manifold with G simple and K = C(S) = S × K1 , where S is a torus in G and K1 is the semisimple part of K, here C(S) denotes the centralizer of S in G. The homogeneous space G/K1 is called the corresponding M-space. The following Lemma is important in our study. Lemma 2. Assume that an Ad(K)-irreducible submodule mi (i ∈ {1, . . . , s}) is Ad(K1 )reducible. Then we have a decomposition mi = ni1 ⊕ ni2 , where ni1 and ni2 are equivalent irreducible Ad(K1 )-invariant submodules. Proof. Let Rt+ = {ξ1, . . . , ξs } be a set of positive t-roots for the generalized flag manifold G/K. Set + Rj+ = {α ∈ RM : α |t = ξj }, j = 1, . . . , s. (16) Set αj |t = αj , (j = 1, . . . , r), where ΠM = Π \ ΠK = {α1 , . . . , αr }. Then Πt = {α1 , . . . , αr } is a t-basis of t∗ . Therefore, there exist αi1 , . . . , αip , i1 , . . . , ip ∈ {1, . . . , r} and positive integers b1 , . . . , bp such that α |t = b1 αi1 + · · · + bp αip for any α ∈ Ri+ . It follows that B(Λij , α) 6= 0 for any α ∈ Ri+ and j ∈ {1, . . . , p}. Then we have that √ √ (17) [ −1hΛij , Aα ] = B(Λij , α)Bα , [ −1hΛij , Bα ] = −B(Λij , α)Aα

for any α ∈ Ri+ . Assume that an Ad(K)-irreducible submodule mi (i ∈ {1, . . . , s}) is Ad(K1 )-reducible and set mi = ni1 ⊕ · · · ⊕ nili , where nik (k = 1, . . . , li ) are irreducible Ad(K1 )-invariant √ √ submodules. By (17) it is easy to check that [ −1hΛi1 , ni1 ] = [ −1hΛiq , ni1 ] for √ ′ any q ∈ {2, . . . , p}. Set n = [ −1hΛi1 , ni1 ]. Then by (17) it follows that ni1 = √ √ √ √ ′ ′ [ −1hΛi1 , [ −1hΛi1 , ni1 ]] = [ −1hΛi1 , n ] = [ −1hΛiq , n ] for any q ∈ {2, . . . , p}. Then ′ we have that ni1 and n are irreducible Ad(K1 )-invariant subspaces which are isomorphic. = √ Since ΠM =√{α1 , . . . , αr }, this implies that s = {H ∈ a i: B(H, Πi K ) = 0} i R −1hΛ1 +· · ·+R −1hΛr . Since k = s⊕k1 , it follows that [s⊕k1 , n1 ] = [s, n1 ]+[k1 , n1 ] ⊂ ′ ′ ′ ′ ′ ′ n + ni1 and [s ⊕ k1 , n ] = [s, n ] + [k1 , n ] ⊂ ni1 + n , so ni1 + n is an Ad(K)-invariant sub′ space of mi . Since mi is irreducible as an Ad(K)-module, we obtain that mi = ni1 + n . ′ ′ This implies that ni2 ⊕ · · · ⊕ nili ⊆ n . Since ni1 and n are isomorphic, it follows that dim ni1 ≥ dim ni2 . In the same way we obtain that dim ni2 ≥ dim ni1 . Hence we have ′ n = ni2 , √ which implies that li = 2. It is obvious that ni1 and ni2 are equivalent by the ′ map ad( −1hΛi1 ) : ni1 → n = ni2 .

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Remark 1. There are several choices for the equivalent map from submodule ni1 to √ i n2 in Lemma 2, it is ad( −1hΛiq ) : ni1 → ni2 . for any q ∈ {1, . . . , p}. Here it is α |t = a1 αi1 + · · · + ap αip for any α ∈ Ri+ , where i1 , . . . , ip ∈ {1, . . . , r} and αj |t = αj , (j = 1, . . . , r). Remark 2. If there exists j ∈ {1, . . . , s} such that dim mj = 2, then mj is Ad(K1 )reducible, and mj = nj1 ⊕ nj2 , where nj1 = RAα and nj2 = RBα , (α ∈ Rj+ ). With regard to the set (16) let Ri− = −Ri+ and let Ri = Ri+ ⊔ Ri− . Also, we fix a Cartan subalgebra a1 of the Lie algebra k1 . Definition 5. A root α ∈ Ri is called the lowest (resp. highest) root in Ri+ if α−γ ∈ /R + (resp. α + γ ∈ / R) for any γ ∈ RK .

Remark 3. If α, β ∈ Ri+ are the lowest and highest roots respectively, α = β if and only if dim mi = 2.

The following Propositions and Lemma are very important for the proofs of Theorems 1, 2, 3 and Corollaries 1, 2. Proposition 2. Let α , β be the lowest and highest root respectively in Ri+ , for some i = 1, . . . , s. If there exist α, β ∈ Ri such that such that α|a1 = β|a1 and α(h) = −β(h) ′ ′ ′ ′ for any h ∈ s, then we have α |a1 = −β |a1 and α (h) = β (h) for any h ∈ s. Also, we ′ ′ have that α = β if and only if β = −α. ′

′

Proof. Since α|a1 = β|a1 , it follows that α + δ ∈ R if and only if β + δ ∈ R for any δ ∈ RK . Since α(h) = −β(h) for any h ∈ s, it follows that one of α, β belongs to Ri+ , the other belongs to Ri− . If α ∈ Ri+ and α is the highest root, it follows that −β is the lowest root. We set ′ ′ α = α and β = −β, it is easy to check that the conclusion holds. + If α ∈ Ri+ and α is not the highest root, then there exists γ ∈ RK such that + α + γ ∈ Ri . Since (α + γ)(h) = α(h) and (β + γ)(h) = β(h) for any h ∈ s, we have (α + γ)(h) = −(β + γ)(h). Since α + γ ∈ Ri+ , it follows that β + γ ∈ Ri− . Set α + γ = β1 and β + γ = ω1 . If β1 is not the highest root, then there exists γ1 ∈ RK such that β1 + γ1 ∈ Ri+ and ω1 + γ1 ∈ Ri− . We do this several times, until we get ′ ′ + βk = βk−1 + γk−1 , where βk ∈ Ri+ and βk + γ ∈ / R for any γ ∈ RK . This implies + that βk is the highest root in Ri . Then it follows that ωk = (ωk−1 + γk−1) ∈ Ri− ′ ′ + and ωk + γ ∈ / R for any γ ∈ RK . This implies that −ωk ∈ Ri+ is the lowest root. Since βk = α + γ + γ1 + · · · + γk−1 and ωk = β + γ + γ1 + · · · + γk−1, it follows that βk |a1 = ωk |a1 . Since βk (h) = (α + γ + γ1 + · · · + γk−1)(h) = α(h) and ωk (h) = (β + γ + γ1 + · · · + γk−1 )(h) = β(h) for any h ∈ s, it follows that βk (h) = −ωk (h) for any h ∈ s. ′ ′ Next we prove that α = β if and only if α = −β. If α = −β, since α |a1 = β |a1 , it follows that α |a1 = β |a1 = 0, which implies that ′ ′ dim mi = 2. Then it follows that α = β . ′ ′ If α = β , by Remark 3 we have dim mi = 2. Since α(h) = −β(h) for any h ∈ s, it follows that α = −β. Lemma 3. An Ad(K)-irreducible submodule mi (i ∈ {1, . . . , s}) is Ad(K1 )-reducible given by mi = ni1 ⊕ ni2 if and only if α|a1 = −β|a1 and α(h) = β(h) for any h ∈ s, where α, β ∈ Ri+ are the lowest and highest roots respectively, and a1 is the Cartan subalgebra of k1 . Moreover, if dim mi 6= 2 and mi is reducible as an Ad(K1 )-submodule, we have

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mi = ni1 ⊕ ni2 and

( ni1 = U(k1 )(Aα + A−β ), ni2 = U(k1 )(Aα − A−β ),

(18)

where U(k1 ) is the universal enveloping algebra of k1 and the action of U(k1 ) on ni1 and ni2 is determined by the adjoint representation of g. Proof. We prove the necessity of Lemma 3 first. If mi is reducible as Ad(K1 )-module and dim mi = 2, Remark 3 implies that β = α. It is obvious that the results hold. Let mi be reducible as Ad(K1 )-module and dim mi 6= 2. Assume that α|a1 6= β|a1 for any α 6= β ∈ Ri . Then it follows that U(k1 )(X) = mi for any nonzero vector X ∈ mi , which means that mi is irreducible as Ad(K1 )-module and this is a contradiction. Hence we have there that exist α 6= β ∈ Ri such that α|a1 = β|a1 . Since α, β ∈ Ri , it follows that α(h) = β(h) or α(h) = −β(h) for any h ∈ s. Since α 6= β we have α(h) = −β(h) ′ ′ ′ ′ for any h ∈ s. By Proposition 2 we have α |a1 = −β |a1 and α (h) = β (h) for any ′ ′ h ∈ s, where α and β are the lowest root and highest root respectively. Conversely, assume that α|a1 = −β|a1 and α(h) = β(h) for any h ∈ s, where α, β ∈ Ri+ are the lowest and highest roots respectively. If α = β, by Remark 3 we obtain that dim mi = 2, and Remark 2 implies that mi is reducible as Ad(K1 )-module. If α 6= β, we set ( ni1 = U(k1 )(Aα + A−β ), (19) ni2 = U(k1 )(Aα − A−β ). + Since α|a1 = −β|a1 , it follows that α + γ ∈ R if and only if β − γ ∈ R, where γ ∈ RK . By Lemma 1 we have √ [ −1hγ , Aα ± A−β ] = (α(hγ )(Bα ± B−β ), √ −1hγ , Bα ± B−β ] = −α(hγ )(Aα ± A−β ), [ (20) [Aγ , Aα ± A−β ] = Nγ,α Aγ+α ± Nγ,−β Aγ−β , [Bγ , Aα ± A−β ] = (−Nα,γ Bα+γ ) ± (−N−β,γ B−β+γ ), [B , B ± B ] = (−N A ) ± (−N A ), γ

α

γ,α

−β

γ+α

γ,−β

γ−β

+ where γ ∈ RK . If α + γ ∈ / R, we have Nα,γ = 0, and if β − γ ∈ / R, we have i i Nβ,−γ = 0. By (20) it is√ easy to check that n√ 1 and n2 are Ad(K1 )-invariant irreducible i i i submodules, and n2 = [ −1hΛiq , n1 ], n1 = [ −1hΛiq , ni2 ] for any q ∈ {1, . . . , p}. Here α |t= a1 αi1 + · · · + ap αip for any α ∈ Ri+ , where i1 , . . . , ip ∈ {1, . . . , r} and αj |t = αj , (j = 1, . . . , r). Then it follows that ni1 + ni2 is Ad(K)-invariant submodule. Since ni1 , ni2 ⊂ mi and mi is irreducible Ad(K)-module, this implies that mi = ni1 ⊕ ni2 . ′

′

′

′

′

′

Proposition 3. Let α |a1 = −β |a1 and α (h) = β (h) for any h ∈ s, where α , β ∈ Ri are the lowest root and highest root respectively. Then for any α ∈ Ri+ there exists β ∈ Ri+ such that α|a1 = −β|a1 and α(h) = β(h) for any h ∈ s. ′

′

′

Proof. First we assume that α 6= β . Any α ∈ Ri can be expressed as α = α +γ0 +γ1 + ′ ′ · · · + γk , and (α + γ0 + γ1 + · · · + γj ) ∈ Ri+ , (j ≤ k). If k = 0, this means that α = α .

RIEMANNIAN M -SPACES WITH HOMOGENEOUS GEODESICS ′

′

′

9

′

Since α |a1 = −β |a1 , it follows that α +δ ∈ R if and only if −β +δ ∈ R for any δ ∈ RK . ′ ′ This implies that β = −(−β + γ0 + γ1 + · · · + γk ) = (β − γ0 − γ1 − · · · − γk ) ∈ Ri+ . Now we prove that α|a1 = −β|a1 and α(h) = β(h) for any h ∈ s. ′ ′ Since α = α + γ0 + γ1 + · · · + γk , it follows that α(h) = (α + γ0 + γ1 + · · · + γk )(h) = ′ ′ α (h) + (γ0 + γ1 + · · · + γk )(h) for any h ∈ a1 . Since β = β − γ0 − γ1 − · · · − γk , it ′ ′ follows that β(h) = (β − γ0 − γ1 + · · · − γk )(h) = β (h) − (γ0 + γ1 + · · · + γk )(h) for ′ ′ ′ ′ any h ∈ a1 . Since α |a1 = −β |a1 , it follows that α (h) = −β (h) for any h ∈ a1 , hence we have α(h) = −β(h) for any h ∈ a1 , which means that α|a1 = −β|a1 . Since γ(h) = 0 ′ ′ ′ and α (h) = β (h) for any h ∈ s, it follows that α(h) = (α + γ0 + γ1 + · · · + γk )(h) = ′ ′ ′ α (h) = β (h) = (β − γ0 − γ1 − · · · − γk )(h) = β(h) for any h ∈ s. ′ ′ ′ If α = β , we have α = β = α . 2. Invariant metrics on M-spaces

√ √ Let n be the tangent space of a M-space G/K1 with n = s⊕m. Let { −1hΛ1 , . . . , −1hΛr } be a basis of s with respect to ΠM = {α1 , . . . , αr }. There are two cases for Ad(K1 )invariant irreducible decomposition of the tangent space n, based on properties of the isotropy representation m = m1 ⊕ · · · ⊕ ms of the flag manifold G/K. Then we consider G-invariant metrics on G/K1 which are Ad(K1 )-invariant corresponding to the Ad(K1 )-irreducible decomposition of n. Case A. Assume that mi for any i ∈ {1, . . . , s} in the decomposition (2) is irreducible as an Ad(K1 )-submodule, so we get that √ √ n = R −1hΛ1 ⊕ · · · ⊕ R −1hΛr ⊕ m1 ⊕ · · · ⊕ ms (21) is the Ad(K1 )-irreducible decomposition. Let h·, ·i = B(Λ·, ·) be an Ad(K1 )-invariant scalar product on n, where Λ is the associated operator. Therefore, G-invariant metrics on G/K1 which are Ad(K1 )-invariant are defined by (22) h·, ·i = Λ|s + λ1 B(·, ·)|m1 + · · · + λs B(·, ·)|ms , where A|s is positive definite symmetry matrix. Case B. Assume that there exists an r ′ ∈ {1, . . . , s} such that mi , i = 1, . . . , r ′ are Ad(K1 )-reducible submodules, and mi are Ad(K1 )-irreducible submodules for i = r ′ + 1, . . . , s. Set mi = ni1 ⊕ ni2 , i = 1, . . . , r ′, where ni1 and ni2 are equivalent and irreducible Ad(K1 )-submodules. It follows that √ √ ′ ′ n = R −1hΛ1 ⊕ · · · ⊕ R −1hΛr ⊕ (n11 ⊕ n12 ) ⊕ · · · ⊕ (nr1 ⊕ nr2 ) ⊕ mr′ +1 ⊕ · · · ⊕ ms (23)

is the Ad(K1 )-irreducible decomposition. Let h·, ·i = B(Λ·, ·) be an Ad(K1 )-invariant scalar product on n, where Λ is the √ + } adapted associated operator. We fix basis B = { −1hΛi , Aα , Bα , i = 1, . . . , r, α ∈ RM to the decomposition (23). Let A and A|p be the matrix representation of Λ and Λ|p respectively, where p denotes a subspace of n. Then G-invariant metrics on G/K1 which are Ad(K1 )-invariant are defined by A|s 0 ... 0 0 A|m1 , (24) A= . . .. .. ... 0 . . . . . . A|ms where A|s is a positive definite symmetric matrix, and A|mi , i = 1, . . . , r ′ has the form

10

ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO

A|mi =

µi1 Id|ni1 Ai21 Ai12 µi2 Id|ni2

, µi1, µi2 > 0.

(25)

The block matrices Ai12 and Ai21 correspond to Ad(K1 )-equivariant maps φ1 : ni1 → ni2 and φ2 : ni2 → ni1 respectively. Moreover, the symmetry of Λ implies that Ai12 = Ai21 . Consequently, for any vector Xj ∈ nij ⊂ mi , j = 1, 2, it is ΛXj = Λ|mi Xj = µij Xj + Aijk Xj ,

Aijk : nij → nik , j 6= k ∈ {1, 2}.

(26)

Since mp , p = r ′ + 1, . . . , s are irreducible as Ad(K1 )-modules, it follows that Λ|mp = µp Id|mp , for some µp > 0. 3. Riemannian g.o. spaces Let (M = G/K, g) be a homogeneous Riemannian manifold with G a compact connected semisimple Lie group. Let g and k be the Lie algebras of G and K respectively and g = k ⊕ m be a reductive decomposition. Definition 6. A nonzero vector X ∈ g is called a geodesic vector if the curve (1) is a geodesic. Lemma 4 ([24]). A nonzero vector X ∈ g is a geodesic vector if and only if h[X, Y ]m , Xm i = 0

(27)

for all Y ∈ m. Here the subscript m denotes the projection into m. A useful description of homogeneous geodesics (1) is provided by the following : Proposition 4 ([2]). Let (M = G/K, g) be a homogeneous Riemannian manifold and Λ be the associated operator. Let a ∈ k and x ∈ m. Then the following are equivalent: (1) The orbit γ(t) = expt(a+x)·o of the one-parameter subgroup expt(a+x) through the point o = eK is a geodesic of M. (2) [a + x, Λx] ∈ k. (3) h[a, x], yi = hx, [x, y]m i for all y ∈ m. (4) h[a + x, y]m , xi = 0 for all y ∈ m. An important corollary of Proposition 4 is the following: Corollary 3 ([2]). Let (M = G/K, g) be a homogeneous Riemannian manifold. Then (M = G/K, g) is a g.o. space if and only if for every x ∈ m there exists an a(x) ∈ k such that [a(x) + x, Λx] ∈ k. (28) For later use we recall the following: Proposition 5. ([3, Proposition 5]) Let (M = G/H, g) be a compact g.o. space with associated operator Λ. Let X, Y ∈ m be eigenvectors Λ with different eigenvalues λ, µ. Then λ µ [X, Y ] = [h, X] + [h, Y ] (29) λ−µ λ−µ for some h ∈ h.

RIEMANNIAN M -SPACES WITH HOMOGENEOUS GEODESICS

11

Proposition 6. ([20], [29]) Let G be a connected semisimple Lie group and H ⊃ K be compact Lie subgroups in G. Let MF and MC be the tangent spaces of F = H/K and C = G/H respectively. Then the metric ga,b = aB |MF +bB |MC , (a, b ∈ R+ ) is a g.o. metric on G/K if and only if for any vF ∈ MF , vC ∈ MC there exists X ∈ k such that [X, vF ] = [X + vF , vC ] = 0. 4. Proof of Theorem 1 and Corollary 1 Let G/K be a generalized flag manifold with K = C(S) = S × K1 , where S is a torus in the simple compact Lie group G and K1 is the semisimple part of K. Then the corresponding M-space is G/K1 . We denote by g and k the Lie algebras of G and K respectively. Let B = −Killing form. Then the module m decomposes into a direct sum of Ad(K)-invariant irreducible submodules pairwise orthogonal with respect to B (cf. (2)). Proof of Theorem 1. Case 1. Assume that mi for any i ∈ {1, . . . , s} in the decomposition (2) is irreducible as an Ad(K1 )-submodule. Then the tangent space n ∼ = To (G/K1 ) is decomposed into irreducible Ad(K1 )-invariant submodules: √ √ n = R −1hΛ1 ⊕ · · · ⊕ R −1hΛr ⊕ m1 ⊕ m2 ⊕ · · · ⊕ ms . (30) √ √ Here s = R −1hΛ1 ⊕ · · · ⊕ R −1hΛr . Therefore, an Ad(K1 )-invariant inner product h·, ·i = B(Λ·, ·) is expressed as h·, ·i = Λ|s + λ1 B(·, ·)|m1 + · · · + λs B(·, ·)|ms ,

(31)

where Λ is the associate operator on n. Let Rt+ = {ξ1 , · · · , ξs } be the set of positive t-roots of the generalized flag manifold G/K with s ≥ 3. Since Rt+ is connected, for any ξ, η ∈ Rt+ there exists (without loss of generality) a chain of positive t-roots ξ = ζ1 , ζ2 , . . . , ζk = η, where ζi , ζi+1 are adjacent (i = 1, . . . , k − 1). We define the subset {mi1 , mi2 , . . . , mik } of {m1 , m2 , . . . , ms } by X miq = (RAα + RBα ), (q = 1, . . . , k).

(32)

(33)

α∈R+ M :κ(α)=ζq

Since ζq , ζq+1 (q = 1, . . . , k −1) are adjacent, then either ζq +ζq+1 ∈ Rt or ζq+1 −ζq ∈ Rt . Therefore, we have X X [miq , miq+1 ] ⊆ (RAα + RBα ) ⊕ (RAα + RBα ) . α∈R+ M :κ(α)=ζq +ζq+1

α∈RM :κ(α)=ζq+1 −ζq

P

If ζq + ζq+1 ∈ / Rt , then α∈R+ :κ(α)=ζq+1 +ζq (RAα + RBα ) = {0}. If ζq+1 − ζq ∈ / Rt , then M P α∈RM :κ(α)=ζq+1 −ζq (RAα + RBα ) = {0}. Since ζq + ζq+1 6= ±ζq , ζq + ζq+1 6= ±ζq+1 and ζq − ζq+1 6= ±ζq , ζq − ζq+1 6= ±ζq+1 , it follows that X X (RAα + RBα )∩ miq ⊕ miq+1 = {0}. (RAα + RBα ) ⊕ α∈R+ M :κ(α)=ζq +ζq+1

α∈RM :κ(α)=ζq+1 −ζq

ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO

12

Therefore we get that [miq , miq+1 ] ∩ (miq ⊕ miq+1 ) = {0}.

(34)

Also, since ζq , ζq+1 are adjacent (q = 1, . . . , k − 1) in (32), there exist X ∈ miq , Y ∈ miq+1 eigenvectors of Λ such that [X, Y ] 6= 0. If we had that λiq 6= λiq+1 , then Proposition 5 implies that [X, Y ] ⊂ miq ⊕ miq+1 , which contradicts (34), hence λiq = λiq+1 , (q = 1, . . . , k − 1). Since this is true for any ξ, η ∈ Rt+ we obtain that λ1 = λ2 = · · · = λs , and the conclusion follows. Case 2. Assume that there exists an r ′ ∈ {1, . . . , s} such that mi (i = 1, . . . , r ′ ) are reducible as Ad(K1 )-submodules, and mi are irreducible as Ad(K1 )-submodules for i = r ′ + 1, . . . , s. By Lemma 2 we have mi = ni1 ⊕ ni2 , (i = 1, . . . , r ′ ), where ni1 , ni2 are equivalent and irreducible as Ad(K1 )-submodules. It follows that √ √ ′ ′ n = R −1hΛ1 ⊕ · · · ⊕ R −1hΛr ⊕ (n11 ⊕ n12 ) ⊕ · · · ⊕ (nr1 ⊕ nr2 ) ⊕ mr′ +1 ⊕ · · · ⊕ ms (35) is the Ad(K1 )-irreducible decomposition. Therefore G-invariant metrics on G/K1 are defined by (24). Since Rt+ = {ξ1 , · · · , ξs } is connected, for any ξ, η ∈ Rt+ there exists a chain of positive t-roots ξ = ζ1 , ζ2 , . . . , ζk = η, (36) where ζi , ζi+1 are adjacent (i = 1, . . . , k − 1). We define the subset {mi1 , mi2 , . . . , mik } of {m1 , m2 , . . . , ms } by X miq = (RAα + RBα ), (q = 1, . . . , k).

(37)

α∈R+ M :κ(α)=ζq

Since ζq , ζq+1 (q = 1, . . . , k − 1) are adjacent, it is either ζq + ζq+1 ∈ Rt or ζq+1 − ζq ∈ Rt . There are three possibilities for miq , miq+1 as Ad(K1 )-modules: (a) Both of miq , miq+1 are irreducible as Ad(K1 )-modules. Since miq , miq+1 are irreducible as Ad(K1 )-modules, it follows that the associated operator Λ|miq = µiq Id|miq , Λ|miq+1 = µiq+1 Id|miq+1 , (µiq , µiq+1 > 0). As in the proof in Case 1 we obtain that µiq = µiq+1 . (b) One of miq , miq+1 is irreducible as Ad(K1 )-module and the other is reducible as Ad(K1 )-module. (b1) Assume that miq is reducible as Ad(K1 )-module. i i i i By Lemma 2 it follows that miq = n1q ⊕ n2q , where n1q , n2q are equivalent and irreducible Ad(K1 )-modules. Thus Λ|miq has the form ! i i µ1q Id|niq A21q i i 1 A|miq = (38) , µ1q , µ2q > 0, iq iq A12 µ2 Id|niq 2

i Ajkq

i njq

i nkq ,

: j 6= k ∈ {1, 2} is an Ad(K1 )-equivalent map, and ΛX = where → i iq iq Λ|miq X = µj X + Ajk X for any X ∈ njq ⊂ miq , j ∈ {1, 2}. Since miq+1 is irreducible as Ad(K1 )-module, it follows that Λ|miq+1 = µiq+1 Id|miq+1 for some µiq+1 > 0. Since G/K1 is a g.o. space, by Corollary 3 it follows that for any X ∈ n there exists i a k ∈ k1 such that [k + X, ΛX] ∈ k1 . We choose non zero vectors X1 ∈ n1q ⊂ miq and

RIEMANNIAN M -SPACES WITH HOMOGENEOUS GEODESICS

13

X2 ∈ miq+1 , with [k + X1 + X2 , Λ(X1 + X2 )] ∈ k1 . Since Λ(X1 + X2 ) = Λ|miq X1 + i i Λ|miq+1 X2 = µ1q X1 + A12q X1 + µiq+1 X2 , we obtain that i

i

[k + X1 + X2 , Λ(X1 + X2 )] = [k + X1 + X2 , µ1q X1 + A12q X1 + µiq+1 X2 ] i

i

i

i

i

= [k, µ1q X1 + A12q X1 ] + [k, µiq+1 X2 ] + (µiq+1 − µ1q )[X1 , X2 ] + [X1 , A12q X1 ] + [X2 , A12q X1 ].

Also, since ζq + ζq+1 6= ±ζq , ζq + ζq+1 6= ±ζq+1 and ζq − ζq+1 6= ±ζq , ζq − ζq+1 6= ±ζq+1 , it follows that i

i

i

[k, µ1q X1 + A12q X1 ] ∈ miq , [k, µiq+1 X2 ] ∈ miq+1 , [X1 , A12q X1 ] ∈ k ⊕ mj1 , (iq 6= j1 6= iq+1 ), i

i

[X2 , A12q X1 ] ∈ mj2 ⊕ mj3 , (µiq+1 − µ1q )[X1 , X2 ] ∈ mj2 ⊕ mj3 , (iq 6= j2 , j3 6= iq+1 ). Then it follows that i

i

i

i

i

([k, µ1q X1 +A12q X1 ]+[k, µiq+1 X2 ])∩([X1 , A12q X1 ]+[X2 , A12q X1 ]+(µiq+1 −µ1q )[X1 , X2 ]) = {0}. i

i

Next, we prove that A12q = 0 and µ1q = µiq+1 . √ √ i i By Remark 1 we have that there exists −1hΛl ∈ s such that n2q = [ −1hΛl , n1q ] √ i i i i and n1q = [ −1hΛl , n2q ]. Assume that Ajkq 6= 0, j 6= k ∈ {1, 2} and Ajkq is the matrix representation of the following map √ i i ad( −1hΛl ) : njq → nkq , j 6= k ∈ {1, 2}. Since miq is reducible as Ad(K1 )-module, by Lemma 3 and Proposition 3 we obtain that there exist α, β ∈ Ri+q such that α |a1 = −β |a1 and α(h) = β(h) for any h ∈ s. Since ζq , ζq+1 are adjacent, it follows that either ζq +ζq+1 ∈ Rt or ζq+1 −ζq ∈ Rt . Assume + that ζq + ζq+1 ∈ Rt , this implies that there exists γ ∈ RM such that κ(γ) = ζq+1 and κ(α+γ) 6= 0. Hence we have that there exists j 6= q, q +1 such that κ(α+γ) = ξj ∈ Rt+ . i We now distinguish two cases and we will get a contradiction to Ajkq 6= 0. (i) Assume that α = β. Since α = β and α |a1 = −β |a1 , it follows that α(h) = β(h) = 0 for any h ∈ a1 , and i i this implies that dim miq = 2. By Remark 2 we have n1q = RAα , n2q = RBα . We choose X1 = Aα and X2 = Aγ . Then we have √ i i i [X1 , A12q X1 ] + [X2 , A12q X1 ] + (µ1q − µiq+1 )[X1 , X2 ] = [Aα , [ −1hΛl , Aα ] √ i +[Aγ , [ −1hΛl , Aα ]] + (µiq+1 − µ1q )[Aα , Aγ ]. It is easy to get that

√ i [X1 , A12q X1 ] = 2 −1α(hΛl )hα ∈ s, i

[X2 , A12q X1 ] = α(hΛl )(Nγ,α Bγ+α + Nγ,−α Bγ−α ), i

i

(µiq+1 − µ1q )[X1 , X2 ] = (µiq+1 − µ1q )(Nγ,α Aγ+α + Nγ,−α Aγ−α ).

Therefore we have that i

i

i

i

([X1 , A12q X1 ] + [X2 , A12q X1 ]) ∩ (µ1q − µiq+1 )[X1 , X2 ] = {0}. i

i

(39)

Since ([X1 , A12q X1 ] + [X2 , A12q X1 ]) ∈ / k1 and (µ1q − µiq+1 )[X1 , X2 ] ∈ / k1 , by (39) it follows iq iq i that (µiq+1 −µ1 )[X1 , X2 ] = 2(µiq+1 −µ1 )(Nγ,α Aγ+α +Nγ,−α Aγ−α ) = 0 and [X1 , A12q X1 ] = i i 0. It is easy to check that (µiq+1 −µ1q )[X1 , X2 ] = 2(µiq+1 −µ1q )(Nγ,α Aγ+α +Nγ,−α Aγ−α ) =

ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO

14

√ i i 0 if and only if µiq+1 = µ1q . It is obvious that [X1 , A12q X1 ] = 2 −1α(hΛl )hα 6= 0, which i i is a contradiction, so we have A12q = A21q = 0. (ii) Assume that α 6= β. We will prove that (γ + β) 6= (γ ± α) 6= (γ − β). Assuming that (γ + β) = (γ + α), it follows that (γ + β)(h) = (γ + α)(h) for any h ∈ a1 . This implies that β(h) = α(h) for any h ∈ a1 . Since −β(h) = α(h) for any h ∈ a1 , it follows that β(h) = α(h) = 0 for any h ∈ a1 , this implies that dim miq = 2. By Remark 3 and Proposition 2 we have β = −α. But β(h) = α(h) for any h ∈ s, this implies that α, β ∈ Ri+q , which is a contradiction. Hence we have (γ + β) 6= (γ + α). Assuming that (γ + β) = (γ − α), it follows that (γ + β)(h) = (γ − α)(h) for any h ∈ s. This implies that β(h) = −α(h) for any h ∈ s. But β(h) = α(h) for any h ∈ s, which is a contradiction. Hence we have (γ + β) 6= (γ − α). We prove that (γ ± α) 6= (γ − β) by the same method as above. We choose X1 = Aα + A−β and X2 = Aγ . Since i

i

(µiq+1 − µ1q )[X1 , X2 ] = (µiq+1 − µ1q )(Nγ,α Aγ+α + Nγ,−α Aγ−α + Nγ,−β Aγ−β + Nγ,β Aγ+β ), √ i [X1 , A12q X1 ] = (2 −1α(hΛl )(hα − h−β ) − 2α(hΛl )Nα,−β Bα−β ) ∈ k, i

[X2 , A12q X1 ] = α(hΛl )(Nγ,α Bγ+α + Nγ,−α Bγ−α − Nγ,−β Bγ−β − Nγ,β Bγ+β ) ∈ / k,

it follows that

i

i

i

(µiq+1 − µ1q )[X1 , X2 ] ∩ ([X1 , A12q X1 ] + [X2 , A12q X1 ]) = {0}.

(40) i

Since (γ + β) 6= (γ ± α) 6= (γ − β), it is easy to check that (µiq+1 − µ1q )[X1 , X2 ] = i (µiq+1 − µ1q )(Nγ,α Aγ+α + Nγ,−α Aγ−α + Nγ,−β Aγ−β + Nγ,β Aγ+β ) ∈ k1 if and only if i µ1q = µiq+1 . i i Since [X1 , A12q X1 ] ∈ k and [X2 , A12q X1 ] ∈ / k, it follows that i

i

i

[X1 , A12q X1 ] ∩ [X2 , A12q X1 ] = {0}.

(41)

This implies that [X2 , A12q X1 ] ∈ k1 if and only if i

[X2 , A12q X1 ] = α(hΛl )(Nγ,α Bγ+α + Nγ,−α Bγ−α − Nγ,−β Bγ−β − Nγ,β Bγ+β ) = 0.

(42)

Since (γ + β) 6= (γ ± α) 6= (γ − β) and Nγ,α 6= 0, it follows that i

[X2 , A12q X1 ] = α(hΛl )(Nγ,α Bγ+α + Nγ,−α Bγ−α − Nγ,−β Bγ−β − Nγ,β Bγ+β ) 6= 0, i

which contradicts with (42). Hence we get A12q = 0. i i We prove µ2q = µiq+1 by the same method. Consequently, we have Ajkq = 0, for i i j 6= k ∈ {1, 2} and µ1q = µ2q = µiq+1 .

(b2) Assume that miq+1 is reducible as Ad(K1 )-module. i i i i By Lemma 2 it follows that miq+1 = n1q+1 ⊕ n2q+1 , where n1q+1 , n2q+1 are equivalent and irreducible Ad(K1 )-modules. Thus Λ|miq+1 has the form ! i i µ1q+1 Id|niq+1 A21q+1 i i 1 A|miq+1 = , µ1q+1 , µ2q+1 > 0, (43) iq+1 iq+1 i A12 µ2 Id|n q+1 2

RIEMANNIAN M -SPACES WITH HOMOGENEOUS GEODESICS i

i

15

i

where Ajkq+1 : njq+1 → nkq+1 , j 6= k ∈ {1, 2} is an Ad(K1 )-equivalent map. We can prove i i i i that A21q+1 = A12q+1 = 0 and µ1q+1 = µ2q+1 = µiq by the same method as above, where Λ |miq = µiq Id |miq . (c) Both miq , miq+1 are reducible as Ad(K1 )-modules. i i i i i i By Lemma 2 it follows that miq = n1q ⊕ n2q and miq+1 = n1q+1 ⊕ n2q+1 , where n1q , n2q i i are equivalent and irreducible Ad(K1 )-modules, and n1q+1 , n2q+1 are equivalent and irreducible Ad(K1 )-modules. Thus Λ|miq has the form ! i i µ1q Id|niq A21q i i 1 A|miq = , µ1q , µ2q > 0, (44) iq iq µ2 Id|niq A12 2

and Λ|miq+1 has the form i

A|miq+1 =

i

µ1q+1 Id|niq+

A21q+1

i

µ2q+1 Id|niq+1

1

A12q+1

i

2

!

i

i

, µ1q+1 , µ2q+1 > 0.

(45)

Since G/K1 is a g.o. space, by Corollary 3 it follows that for any X ∈ n there exists i a k ∈ k1 such that [k + X, ΛX] ∈ k1 . We choose non zero vectors X1 ∈ n1q ⊂ miq i and X2 ∈ n1q+1 ⊂ miq+1 , with [k + X1 + X2 , Λ(X1 + X2 )] ∈ k1 . Since Λ(X1 + X2 ) = i i i i Λ|miq X1 + Λ|miq+1 X2 = µ1q X1 + A12q X1 + µ1q+1 X2 + A12q+1 X2 , we obtain that i

i

i

i

[k + X1 + X2 , Λ(X1 + X2 )] = [k + X1 + X2 , µ1q X1 + A12q X1 + µ1q+1 X2 + A12q+1 X2 ] i

i

i

i

i

i

i

= [k, µ1q X1 + A12q X1 ] + [k, µ1q+1 X2 + A12q+1 X2 ] + (µ1q+1 − µ1q )[X1 , X2 ] + [X1 , A12q X1 ] i

i

i

+[X1 , A12q+1 X2 ] + [X2 , A12q+1 X2 ] + [X2 , A12q X1 ].

Also, since ζq + ζq+1 6= ±ζq , ζq + ζq+1 6= ±ζq+1 and ζq − ζq+1 6= ±ζq , ζq − ζq+1 6= ±ζq+1 , it follows that i

i

i

[k, µ1q X1 + A12q X1 ] ∈ miq , [k, µiq+1 X2 ] ∈ miq+1 , [X1 , A12q X1 ] ∈ k ⊕ mj1 , (iq 6= j1 6= iq+1 ), i

[X2 , A12q+1 X2 ] ∈ k ⊕ mj2 , (iq 6= j2 6= iq+1 , j2 6= j1 ), i

i

i

i

([X2 , A12q X1 ] + [X1 , A12q+1 X2 ] + (µ1q+1 − µ1q )[X1 , X2 ]) ∈ mj3 ⊕ mj4 , (iq 6= j3 , j4 6= iq+1 ). Then it follows that i

i

i

i

i

([k, µ1q X1 + A12q X1 ] + [k, µiq+1 X2 ]) ∩ ((µ1q+1 − µ1q )[X1 , X2 ] + [X1 , A12q X1 ] i

i

i

+[X1 , A12q+1 X2 ] + [X2 , A12q+1 X2 ] + [X2 , A12q X1 ]) = {0}.

Since miq is reducible as Ad(K1 )-module, by Lemma 3 and Proposition 3 we have there exist α1 , β1 ∈ Ri+q such that α1 |a1 = −β1 |a1 and α1 (h) = β1 (h) for any h ∈ s. Since miq+1 is reducible as Ad(K1 )-module, by Lemma 3 and Proposition 3 we have that there exist α2 , β2 ∈ Ri+q+1 such that α2 |a1 = −β2 |a1 and α2 (h) = β2 (h) for any + h ∈ s and κ(α1 + α2 ) 6= 0. It follows that α1 + α2 ∈ RM . iq+1 iq+1 iq iq Next, we will show that A12 = A12 = 0 and µ1 = µ1 . i Case (c1). Assume that A12q = 0. iq+1 i i We will prove that A12 = 0 and µ1q = µ1q+1 . i i Assume that Ajkq+1 6= 0, j 6= k ∈ {1, 2} and Ajkq+1 is the matrix representation of the following map √ i i ad( −1hΛl ) : njq+1 → nkq+1 , j 6= k ∈ {1, 2}.

ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO

16

(i) Assume that α1 = β1 , α2 6= β2 . We choose X1 = Aα1 and X2 = Aα2 + A−β2 . Then we have i

i

i

i

i

i

(µ1q+1 − µ1q )[X1 , X2 ] + [X1 , A12q X1 ] + [X1 , A12q+1 X2 ] + [X2 , A12q+1 X2 ] + [X2 , A12q X1 ] i

i

i

i

= (µ1q+1 − µ1q )[X1 , X2 ] + [X1 , A12q+1 X2 ] + [X2 , A12q+1 X2 ] √ i i = (µ1q+1 − µ1q )[Aα1 , Aα2 + A−β2 ] + [Aα1 , [ −1hΛl , Aα2 + A−β2 ]] √ +[Aα2 + A−β2 , [ −1hΛl , Aα2 + A−β2 ]] and i

i

i

i

(µ1q+1 − µ1q )[X1 , X2 ] = (µ1q+1 − µ1q )(Nα1 ,α2 Aα1 +α2 + N−α1 ,α2 Aα1 −α2 + Nα1 ,−β2 Aα1 −β2 +N−α1 ,−β2 Aα1 +β2 ), i

[X1 , A12q+1 X2 ] = α2 (hΛl )(Nα1 ,α2 Bα1 +α2 + Nα1 ,−α2 Bα1 −α2 − Nα1 ,−β2 Bα1 −β2 − Nα1 ,β2 Bα1 +β2 ), √ i [X2 , A12q+1 X2 ] = 2 −1α2 (hΛl )(hα2 − h−β2 ) − 2α2 (hΛl )Nα2 ,−β2 Bα2 −β2 . It follows that i

i

i

i

(µ1q+1 − µ1q )[X1 , X2 ] ∩ ([X1 , A12q+1 X2 ] + [X2 , A12q+1 X2 ]) = {0}.

(46)

Since α2 6= β2 , by the same method as in (b) we prove that (α1 ± α2 ) 6= (α1 ± β2 ). i i i i It follows that (µ1q+1 − µ1q )[X1 , X2 ] ∈ k1 if and only if µ1q+1 = µ1q . √ i Since [X2 , A12q+1 X2 ] = (2 −1α2 (hΛl )(hα2 − h−β2 ) − 2α2 (hΛl )Nα2 ,−β2 Bα2 −β2 ) ∈ k and i [X1 , A12q+1 X2 ] ∈ / k, it follows that i

i

[X2 , A12q+1 X2 ] ∩ [X1 , A12q+1 X2 ] = {0}, i

this implies that [X1 , A12q+1 X2 ] ∈ k1 if and only if i

[X1 , A12q+1 X2 ] = 0. i

But (α1 ± α2 ) 6= (α1 ± β2 ), it follows that [X1 , A12q+1 X2 ] 6= 0, which is a contradiction. i Hence we have Ajkq+1 = 0. i i i i Therefore if Ajkq = 0, we conclude that Ajkq+1 = 0 and µ1q+1 = µ1q . Also, we prove i i i i i µ1q+1 = µ2q and µ2q+1 = µ1q = µ2q by the same method. (ii) Assume that α1 6= β1 , α2 = β2 . i i We choose X1 = Aα1 + A−β1 and X2 = Aα2 . If Ajkq = 0, we conclude that Ajkq+1 = 0 i i i i i and µ1q+1 = µ1q by the same method as in 1). Also, we prove µ1q+1 = µ2q and µ2q+1 = iq iq µ1 = µ2 by the same method. (iii) Assume that α1 6= β1 , α2 6= β2 .

RIEMANNIAN M -SPACES WITH HOMOGENEOUS GEODESICS

17

We choose X1 = Aα1 + A−β1 and X2 = Aα2 + A−β2 . Then we have i

i

i

i

(µ1q+1 − µ1q )[X1 , X2 ] = (µ1q+1 − µ1q )(Nα1 ,α2 Aα1 +α2 + N−α1 ,α2 Aα1 −α2 + Nα1 ,−β2 Aα1 −β2 +N−α1 ,−β2 Aα1 +β2 + N−β1 ,α2 A−β1 +α2 + Nβ1 ,α2 A−β1 −α2 + N−β1 ,−β2 A−β1 −β2 + Nβ1 ,−β2 A−β1 +β2 ), √ √ i [X2 , A12q+1 X2 ] = [Aα2 + A−β2 , [ −1hΛk , Aα2 + A−β2 ]] = 2 −1α2 (hΛk )(hα2 − h−β2 ) −2α2 (hΛk )Nα2 ,−β2 Bα2 −β2 , √ i [X1 , A12q+1 X2 ] = [Aα1 + A−β1 , [ −1hΛl , Aα2 + A−β2 ]] = α2 (hΛl )(Nα1 ,α2 Bα1 +α2 + Nα1 ,−α2 Bα1 −α2 −Nα1 ,−β2 Bα1 −β2 − Nα1 ,β2 Bα1 +β2 + N−β1 ,α2 B−β1 +α2 + N−β1 ,−α2 B−β1 −α2 − N−β1 ,−β2 B−β1 −β2 −N−β1 ,β2 B−β1 +β2 ). It follows that i

i

i

i

(µ1q+1 − µ1q )[X1 , X2 ] ∩ ([X2 , A12q+1 X2 ] + [X1 , A12q+1 X2 ]) = {0}. Since (−β1 ± α2 ) 6= (α1 ± α2 ) 6= (−β1 ± β2 ) and (α1 ± α2 ) 6= (α1 ± β2 ), then we have i i i i that (µ1q+1 − µ1q )[X1 , X2 ] ∈ k1 if and only if µ1q+1 = µ1q . Since (−β1 ± α2 ) 6= (α1 ± α2 ) 6= (−β1 ± β2 ), and (α1 ± α2 ) 6= (α1 ± β2 ), it follows that i

i

[X2 , A12q+1 X2 ] ∈ k,

[X1 , A12q+1 X2 ] ∈ m

This implies that i

i

[X2 , A12q+1 X2 ] ∩ [X1 , A12q+1 X2 ] = 0. i

Therefore we have [X1 , A12q+1 X2 ] ∈ k1 if and only if i

[X1 , A12q+1 X2 ] = 0. It is easy to check that i

[X1 , A12q+1 X2 ] 6= 0, i

which is a contradiction. Hence we have Ajkq+1 = 0. i i i i i i We prove µ1q+1 = µ2q , µ2q+1 = µ1q , µ2q+1 = µ2q by the same method as above. i i i i i i i Hence if Ajkq = 0, we prove that Ajkq+1 = 0 and µ1q+1 = µ1q , µ1q+1 = µ2q , µ2q+1 = i i i µ1q , µ2q+1 = µ2q . i i i If Ajkq+1 = 0, by the same method as above we prove that Ajkq = 0 and µ1q+1 = i i i i i i i µ1q , µ1q+1 = µ2q , µ2q+1 = µ1q , µ2q+1 = µ2q . i i Case (c2). We assume that Ajkq 6= 0 and Ajkq+1 6= 0. We will prove that we get a contradiction. i i i i Since Ajkq 6= 0 and Ajkq+1 6= 0, we assume that Ajkq and Ajkq+1 are the matrix representations of the following maps √ i i ad( −1hΛl ) : njq → nkq , j 6= k ∈ {1, 2} and

√ i i ad( −1hΛk ) : njq+1 → nkq+1 ,

respectively. (i) Assume that α1 = β1 , α2 6= β2 .

j 6= k ∈ {1, 2}

ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO

18

We choose X1 = Aα1 and X2 = Aα2 + A−β2 . Then we have i

i

i

i

(µ1q+1 − µ1q )[X1 , X2 ] = (µ1q+1 − µ1q )(Nα1 ,α2 Aα1 +α2 + N−α1 ,α2 Aα1 −α2 + Nα1 ,−β2 Aα1 −β2 +N−α1 ,−β2 Aα1 +β2 ), √ √ i [X1 , A12q X1 ] = [Aα1 , [ −1hΛl , Aα1 ]] = 2 −1α1 (hΛ1 )hα1 , √ √ i [X2 , A12q+1 X2 ] = [Aα2 + A−β2 , [ −1hΛk , Aα2 + A−β2 ]] = 2 −1α2 (hΛk )(hα2 − h−β2 ) −2α2 (hΛk )Nα2 ,−β2 Bα2 −β2 , √ i [X1 , A12q+1 X2 ] = [Aα1 , [ −1hΛl , Aα2 + A−β2 ]] = α2 (hΛl )(Nα1 ,α2 Bα1 +α2 + Nα1 ,−α2 Bα1 −α2 −Nα1 ,−β2 Bα1 −β2 − Nα1 ,β2 Bα1 +β2 ), √ i [X2 , A12q X1 ] = [Aα2 + A−β2 , [ −1hΛk , Aα1 ]] = α1 (hΛk )(Nα2 ,α1 Bα2 +α1 + Nα2 ,−α1 Bα2 −α1 +N−β2 ,α1 B−β2 +α1 + N−β2 ,−α1 B−β2 −α1 ). It follows that i

i

i

i

i

i

(µ1q+1 −µ1q )[X1 , X2 ] ∩([X1 , A12q X1 ] + [X2 , A12q+1 X2 ] + [X1 , A12q+1 X2 ] + [X2 , A12q X1 ]) = {0}. Since α2 6= β2 , by the same method as in 1) we prove that (α1 ± α2 ) 6= (α1 ± β2 ), i i i i then we have that (µ1q+1 − µ1q )[X1 , X2 ] ∈ k1 if and only if µ1q+1 = µ1q . i i Since [X1 , A12q X1 ] + [X2 , A12q X2 ]) ∈ k and (α1 ± α2 ) 6= (α1 ± β2 ), it follows that i

i

i

i

([X1 , A12q X1 ] + [X2 , A12q+1 X2 ]) ∩ ([X1 , A12q+1 X2 ] + [X2 , A12q X1 ]) = {0}. Then we have i

i

i

i

i

i

([X1 , A12q+1 X2 ] + [X2 , A12q X1 ]) ∈ k1 if and only if [X1 , A12q+1 X2 ] + [X2 , A12q X1 ] = 0. It is easy to check that ([X1 , A12q+1 X2 ] + [X2 , A12q X1 ]) = 0 if and only if α1 (hΛk ) = α2 (hΛl ) = 0. But α1 (hΛk ) 6= 0 and α2 (hΛl ) 6= 0, which i i is a contradiction. Hence we have that at least one of Ajkq , Ajkq+1 is equal zero. By i i i i Case c1 we have that Ajkq = Ajkq+1 = 0 and µ1q+1 = µ1q . Similarly, we prove that i i i i i i µ1q+1 = µ2q , µ2q+1 = µ1q , µ2q+1 = µ2q . (ii) Assume that α1 6= β1 , α2 = β2 . We choose X1 = Aα1 + A−β1 and X2 = Aα2 . By the same method as above we get i i a contradiction. Hence we have that at least one of Ajkq , Ajkq+1 is equal zero. By Case i i i i i i i i i i c1 we have that Ajkq = Ajkq+1 = 0 and µ1q+1 = µ1q , µ1q+1 = µ2q , µ2q+1 = µ1q , µ2q+1 = µ2q . (iii) if α1 6= β1 , α2 6= β2 .

RIEMANNIAN M -SPACES WITH HOMOGENEOUS GEODESICS

19

We choose X1 = Aα1 + A−β1 and X2 = Aα2 + A−β2 . Then we have i

i

i

i

(µ1q+1 − µ1q )[X1 , X2 ] = (µ1q+1 − µ1q )(Nα1 ,α2 Aα1 +α2 + N−α1 ,α2 Aα1 −α2 + Nα1 ,−β2 Aα1 −β2 +N−α1 ,−β2 Aα1 +β2 + N−β1 ,α2 A−β1 +α2 + Nβ1 ,α2 A−β1 −α2 + N−β1 ,−β2 A−β1 −β2 + Nβ1 ,−β2 A−β1 +β2 ), √ √ i [X1 , A12q X1 ] = [Aα1 + A−β1 , [ −1hΛl , Aα1 + A−β1 ]] = 2 −1α1 (hΛ1 )(hα1 − h−β1 ) −2α1 (hΛl )Nα1 ,−β1 Bα1 −β1 , √ √ i [X2 , A12q+1 X2 ] = [Aα2 + A−β2 , [ −1hΛk , Aα2 + A−β2 ]] = 2 −1α2 (hΛk )(hα2 − h−β2 ) −2α2 (hΛk )Nα2 ,−β2 Bα2 −β2 , √ i [X1 , A12q+1 X2 ] = [Aα1 + A−β1 , [ −1hΛl , Aα2 + A−β2 ]] = α2 (hΛl )(Nα1 ,α2 Bα1 +α2 + Nα1 ,−α2 Bα1 −α2 −Nα1 ,−β2 Bα1 −β2 − Nα1 ,β2 Bα1 +β2 + N−β1 ,α2 B−β1 +α2 + N−β1 ,−α2 B−β1 −α2 − N−β1 ,−β2 B−β1 −β2 −N−β1 ,β2 B−β1 +β2 ), √ i [X2 , A12q X1 ] = [Aα2 + A−β2 , [ −1hΛk , Aα1 + A−β1 ]] = α1 (hΛk )(Nα2 ,α1 Bα2 +α1 + Nα2 ,−α1 Bα2 −α1 −Nα2 ,−β1 Bα2 −β1 − Nα2 ,β1 Bα2 +β1 + N−β2 ,α1 B−β2 +α1 + N−β2 ,−α1 B−β2 −α1 − N−β2 ,−β1 B−β2 −β1 −N−β2 ,β1 B−β2 +β1 ).

It follows that i

i

i

i

i

i

(µ1q+1 −µ1q )[X1 , X2 ] ∩([X1 , A12q X1 ] + [X2 , A12q+1 X2 ] + [X1 , A12q+1 X2 ] + [X2 , A12q X1 ]) = {0}. Since α1 6= β1 and α2 6= β2 , it follows that (−β1 ± α2 ) 6= (α1 ± α2 ) 6= (−β1 ± β2 ) i i and (α1 ± α2 ) 6= (α1 ± β2 ), then we have that (µ1q+1 − µ1q )[X1 , X2 ] ∈ k1 if and only if i i µ1q+1 = µ1q . i i Since [X1 , A12q X1 ] + [X2 , A12q X2 ]) ∈ k and (−β1 ± α2 ) 6= (α1 ± α2 ) 6= (−β1 ± β2 ), (α1 ± α2 ) 6= (α1 ± β2 ), it follows that i

i

i

i

([X1 , A12q X1 ] + [X2 , A12q+1 X2 ]) ∩ ([X1 , A12q+1 X2 ] + [X2 , A12q X1 ]) = {0}.

This implies that

i

i

i

i

i

i

([X1 , A12q+1 X2 ] + [X2 , A12q X1 ]) ∈ k1 if and only if [X1 , A12q+1 X2 ] + [X2 , A12q X1 ] = 0. It is easy to check that ([X1 , A12q+1 X2 ] + [X2 , A12q X1 ]) = 0 if and only if α1 (hΛk ) = α2 (hΛl ) = 0. But α1 (hΛk ) 6= 0 and α2 (hΛl ) 6= 0, which is a i i contradiction. Hence we have that at least one of Ajkq , Ajkq+1 is equal zero. By Case c1 i i i i i i i i i i we have that Ajkq = Ajkq+1 = 0 and µ1q+1 = µ1q , µ1q+1 = µ2q , µ2q+1 = µ1q , µ2q+1 = µ2q . Therefore, if G-invariant metrics on G/K1 defined by (24) are g.o. metrics, then we get that A|m is a diagonal matrix and all diagonal elements are equal. Proof of Corollary 1.

ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO

20

√ If dim s = 1 we assume that ΠM = {αk } and it follows that s = R −1hΛk . Since there exists j ∈ {1, . . . , s} such that mj is reducible as an Ad(K1 )-module, we set mj = nj1 + nj2 , where nj1 and nj2 are equivalent and irreducible as Ad(K1 )-modules. Therefore, G-invariant metrics on G/K1 are defined by (24). Then Theorem 1 implies that a g.o. metric g for a M-space G/K1 is h·, ·i = µId|RhΛk + λB(·, ·)|m .

(47)

Let V ∈ s and X ∈ m be eigenvectors of the associate operator Λ with different eigenvalues µ, λ, Proposition 5 implies that there exists k ∈ k1 such that √ [V, X] = j j j µ λ [k, V ]+ µ−λ [k, X]. Since mj = n1 ⊕n2 , by Remark 1 we have that n2 = [ −1hΛk , nj1 ] µ−λ √ √ and nj1 = [ −1hΛk , nj2 ]. We choose V = −1hΛk and X ∈ nj1 and assume that µ 6= λ. Then Proposition 5 implies that √ √ µ λ [ −1hΛk , X] = [k, −1hΛk ] + [k, X]. µ−λ µ−λ √ √ √ λ [k, X]. But [ −1hΛk , X] ∈ nj2 Since [k, −1hΛk ] = 0, it follows that [ −1hΛk , X] = µ−λ and [k, X] ∈ nj1 . This is a contradiction. Hence we get λ = µ, and the conclusion follows. 5. Proof of Theorem 2 and Corollary 2 Proof of part 1) of Theorem 2. Let n be the tangent space To (G/K1 ) at o = eK1 . Since m1 and m2 are irreducible as Ad(K1 )-modules, then we have that n = s ⊕ m1 ⊕ m2

(48)

h·, ·i = µId|s + µ1 B(·, ·)|m1 + µ2 B(·, ·)|m2 , (µ, µ1 , µ2 > 0).

(49)

[k + V + X, Λ(V + X)] ∈ k1 .

(50)

[k + V + X1 + X2 , µV + µ1 X1 + µ2 X2 ] ∈ k1 .

(51)

is an Ad(K1 )-irreducible decomposition. Let h·, ·i = B(Λ·, ·) be an Ad(K1 )-invariant scalar product on n, where Λ is the associated operator. Then we have By Corollary 3 (G/K1 , g) is a g.o. space if and only if for every V ∈ s and X ∈ m there exists k ∈ k1 such that Let X = X1 + X2 , Xi ∈ mi (i = 1, 2). Then (50) is equivalent to Since [k + V, V ] = 0, (51) reduces to

([k + V, µ1 X1 ] + [k + V, µ2 X2 ] + [X1 , µV ] + [X2 , µV ] + (µ1 − µ2 )[X2 , X1 ]) ∈ k1 . (52)

Since [m2 , m1 ] ⊆ m1 , it follows that ([k +V, µ1 X1 ]+[X1 , µV ]+(µ1 −µ2 )[X2 , X1 ]) ∈ m1 and ([k + V, µ2 X2 ] + [X2 , µV ]) ∈ m2 , (52) is equivalent to and

[µ1 k + (µ1 − µ)V + (µ1 − µ2 )X2 , X1 ] = 0

([µ2 k + (µ2 − µ)V, X2 ]) = 0, and this completes the proof. Proof of part 2) of Theorem 2.

(53)

(54)

RIEMANNIAN M -SPACES WITH HOMOGENEOUS GEODESICS

21

We assume that [m1 , m1 ] ⊆ k ⊕ m2 in the decomposition (2), so it follows that dim m1 6= 2. Case 1. Assume that m1 is reducible as Ad(K1 )-module and m2 is irreducible as Ad(K1 )-module. Lemma 2 implies that m1 = n11 ⊕ n12 , where n11 and n12 are equivalent and irreducible as Ad(K1 )-modules. Then we have that n ∼ = To (G/K1 ) = s ⊕ n11 ⊕ n12 ⊕ m2 is the Ad(K1 )-irreducible decomposition. Then G-invariant metrics on G/K1 which are Ad(K1 )-invariant are defined by

A|s 0 0 0 A = 0 A|m1 0 0 A|m2

(55)

Since s and m2 are irreducible as Ad(K1 )-modules, it follows that Λ|s = µId|s and Λ|m2 = λ2 Id|m2 , where µ, λ2 > 0. Λ|m1 has the form

A|m1 =

λ11 Id|n11 A21 1 A12 λ2 Id|n12

, λ11 , λ12 > 0,

(56)

where Ajk : n1j → n1k , j 6= k ∈ {1, 2} is Ad(K1 )-equivalent map, and ΛX = Λ|m1 X = λ1j X + Ajk X for any X ∈ n1j ⊂ m1 , j ∈ {1, 2}. We need to show that Ajk = 0, j 6= k ∈ {1, 2}, so assume the contrary to get a contradiction. Let ΠM = {αp }, p ∈ {1, . . . , l}. Assume that Ajk 6= 0 is the matrix representation of the following map √ ad( −1hΛp ) : n1j → n1k , j 6= k ∈ {1, 2}.

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Set Rt+ = {ξ1 , ξ2 }. Since dim m1 6= 2, this implies there exist α1 , α2 ∈ R1+ such that α1 + α2 ∈ R2+ , where Ri+ , (i = 1, 2) are defined by (16). Lemma 3 and Proposition 3 implies that there exist β1 , β2 ∈ R1+ such that α1 |a1 = −β1 |a1 , α2 |a1 = −β2 |a1 and α1 (h) = β1 (h), α2 (h) = β2 (h) for any h ∈ s. Since dim m1 6= 2, Proposition 2 and Remark 3 implies that α1 6= β1 and α2 6= β2 . Since G/K1 is a g.o.space, by Corollary 3 it follows that for any X ∈ n there exists a k ∈ k1 such that [k + X, ΛX] ∈ k1 . We choose non zero vectors X1 , X2 ∈ n11 , with [k + X1 + X2 , Λ(X1 + X2 )] ∈ k1 . Then we have [k + X1 + X2 , λ11 X1 + A12 X1 + λ11 X2 + A21 X2 ] = [k, λ11 X1 + λ11 X2 ] + [k, A21 X2 + A12 X1 ] + [X1 , A12 X1 ] + [X1 , A21 X2 ] + [X2 , A12 X1 ] + [X2 , A21 X2 ]. Since n11 and n12 are Ad(K1 )-invariant, it follows that [k, λ11 X1 + λ11 X2 ] ⊆ n11 ⊂ m1 and [k, A21 X2 + A12 X1 ] ⊆ n12 ⊂ m1 . We choose X1 = Aα1 + A−β1 , X2 = Aα2 + A−β2 .

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ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO

Then we have that √ √ [X1 , A12 X1 ] = [Aα1 + A−β1 , [ −1hΛp , Aα1 + A−β1 ] = 2 −1α1 (hΛp )(hα1 − h−β1 ) −2α1 (hΛp )Nα1 ,−β1 Bα1 −β1 , √ √ [X2 , A21 X2 ] = [Aα2 + A−β2 , [ −1hΛp , Aα2 + A−β2 ] = 2 −1α2 (hΛp )(hα2 − h−β2 ) −2α2 (hΛp )Nα2 ,−β2 Bα2 −β2 , √ [X1 , A21 X2 ] = [Aα1 + A−β1 , [ −1hΛp , Aα2 + A−β2 ]] = α2 (hΛp )(Nα1 ,α2 Bα1 +α2 + Nα1 ,−α2 Bα1 −α2 −Nα1 ,−β2 Bα1 −β2 − Nα1 ,β2 Bα1 +β2 + N−β1 ,α2 B−β1 +α2 + N−β1 ,−α2 B−β1 −α2 − N−β1 ,−β2 B−β1 −β2 −N−β1 ,β2 B−β1 +β2 ), √ [X2 , A12 X1 ] = [Aα2 + A−β2 , [ −1hΛp , Aα1 + A−β1 ]] = α1 (hΛp )(Nα2 ,α1 Bα2 +α1 + Nα2 ,−α1 Bα2 −α1 −Nα2 ,−β1 Bα2 −β1 − Nα2 ,β1 Bα2 +β1 + N−β2 ,α1 B−β2 +α1 + N−β2 ,−α1 B−β2 −α1 − N−β2 ,−β1 B−β2 −β1 −N−β2 ,β1 B−β2 +β1 ). Since ([X1 , A12 X1 ] + [X2 , A21 X2 ]) ∈ k, ([X1 , A21 X2 ] + [X2 , A12 X1 ]) ∈ m2 ⊕ k and ([k, λ11 X1 + λ11 X2 ] + [k, A21 X2 + A12 X1 ]) ⊂ m1 , it follows that

([k, λ11 X1 +λ11 X2 ]+[k, A21 X2 +A12 X1 ])∩([X1 , A12 X1 ]+[X2 , A21 X2 ]+[X1 , A21 X2 ]+[X2 , A12 X1 ]) = {0}. Since α1 6= β1 and α2 6= β2 we have (−β1 ± α2 ) 6= (α1 ± α2 ) 6= (−β1 ± β2 ) and (α1 ± α2 ) 6= (α1 ± β2 ), then we have ([X1 , A21 X2 ] + [X2 , A12 X1 ]) ∈ k1 if and only if α1 (hΛp ) = α2 (hΛp ) = 0. But α1 (hΛp ) 6= 0 and α2 (hΛp ) 6= 0, which is a contradiction. Therefore we have Ajk = 0, (j, k ∈ {1, 2}). Now we prove that λ11 = λ21 in (56). Assume that λ11 6= λ21 in (56). Since G/K1 is a g.o.space, by Proposition 5 we have that for any X1 ∈ n11 , X2 ∈ n12 there exists a k ∈ k1 such that [X1 , X2 ] =

λ11 λ12 [k, X ] + [k, X2 ]. 1 λ11 − λ12 λ11 − λ12

Lemma 3 and Proposition 3 implies that there exist α1 , β1 ∈ R1+ such that α1 |a1 = −β1 |a1 and α1 (h) = β1 (h)√for any h ∈ s. We choose X1 = Aα1 +A−β1 , X2 = Bα1 −B−β1 , then we have [X1 , X2 ] = 2 −1(hα1 −h−β1 )−2Nα1 ,−β1 Bα1 −β1 , it follows that [X1 , X2 ] 6= 0 and [X1 , X2 ] ∈ k. Since n11 and n12 are irreducible as Ad(K1 )-modules, it follows that λ11 λ12 1 1 ( λ1 −λ 1 [k, X1 ] + λ1 −λ1 [k, X2 ]) ∈ m1 , this is a contradiction. Therefore we have λ1 = λ2 . 1 2 1 2 Therefore Ad(K1 )-invariant scalar product on n are reduced to h·, ·i = µId|s + µ1 B(·, ·)|m1 + µ2 B(·, ·)|m2 , (µ, µ1 , µ2 > 0).

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Since (G/K1 , g) is a g.o. space, assume that µ1 6= µ, Proposition 5 implies that for µ µ1 any V ∈ s and X ∈ n11 there exists k ∈ k1 such that [V, X] = µ−µ [k, V ] + µ−µ [k, X]. 1 1 µ1 1 Since [k, V ] = 0, it follows that [V, X] = µ−µ1 [k, X]. Since [V, X] ∈ n2 and [k, X] ∈ n11 , this is a contradiction. Hence we get that µ1 = µ. Hence, if (G/K1 , g) is a g.o. space, then the corresponding Ad(K1 )-invariant scalar product (55) is reduced to h·, ·i = µB(·, ·)|s⊕m1 + µ2 B(·, ·)|m2 .

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Case 2. Assume that m2 is reducible as Ad(K1 )-module and m1 is irreducible as Ad(K1 )-module. By Lemma 2 we obtain that m2 = n21 ⊕ n22 , where n21 and n22 are equivalent and irreducible as Ad(K1 )-modules. It follows that n ∼ = To (G/K1 ) = s ⊕ m1 ⊕ n21 ⊕ n22 is the Ad(K1 )-irreducible decomposition. Then G-invariant metrics on G/K1 which are Ad(K1 )-invariant are defined by A|s 0 0 0 A = 0 A|m1 (60) 0 0 A|m2

Since s and m1 are irreducible as Ad(K1 )-modules, it follows that Λ|s = µId|s and Λ|m1 = λ1 Id|m1 , where µ, λ1 > 0. Λ|m2 has the form 2 λ1 Id|n21 A21 A|m2 = , λ21 , λ22 > 0, (61) A12 λ22 Id|n22

where Ajk : n2j → n2k , j 6= k ∈ {1, 2} is Ad(K1 )-equivalent map, and ΛX = Λ|m2 X = λ2j X + Ajk X for any X ∈ n2j ⊂ m2 , j ∈ {1, 2}. Assume on the contrary that Ajk 6= 0, j 6= k ∈ {1, 2} and ΠM = {αp }, p ∈ {1, . . . , l}. Let Ajk be the matrix representation of the map √ ad( −1hΛp ) : n2j → n2k , j 6= k ∈ {1, 2}. We consider two cases:

(a) Let dim m2 = 2. + By Remark 2 we obtain that there exists α ∈ RM such that n21 = RAα and n22 = RBα . Since dim m2 = 2 it follows that β ± α ∈ / RM for any β ∈ RK . Since G/K1 is a g.o.space, by Corollary 3 it follows that for any X ∈ n there exists a k ∈√k1 such that [k + X, ΛX] ∈ k1 . We choose nonzero vectors X = Y + X1 , where Y = −1hΛp ∈ s and X1 = Aα ∈ n21 . It is [k + Y + X1 , Λ(X1 + Y )] ∈ k1 . Then we have that √ [k + X1 + Y, λ11 X1 + A12 X1 + µY ] = [k, λ11 X1 + ad( −1hΛp )(X1 )] √ √ +(λ21 − µ)[X1 , Y ] + [X1 , ad( −1hΛp )(X1 )] + [Y, ad( −1hΛp )(X1 )].

It follows that

√ √ [k, λ11 X1 + ad( −1hΛp )(X1 )] + (λ21 − µ)[X1 , Y ] + [Y, ad( −1hΛp )(X1 )] ⊂ m2 .

Since s = {h ∈ a | B(h, ΠK ) = 0} / RM for any β ∈ RK , we √ have that hα ∈ s. √ and β±α ∈ Since α(hΛp ) 6= 0 and √ [X1 , ad( −1hΛp )(X1 )] = [Aα , α(hΛp )Bα ] = 2 −1α(hΛp )hα , it follows that [X1 , ad( −1hΛp )(X1 )] ∈ s, which implies that ([k, λ11 X1 + A12 X1 ] + (λ21 − µ)[X1 , Y ] + [Y, A12 X1 ]) ∩ [X1 , A12 X1 ] = {0},

hence we have [X1 , A12 X1 ] = 0. But

√ [X1 , A12 X1 ] = 2 −1α(hΛp )hα 6= 0,

which is a contradiction, so we have Ajk = 0, (j 6= k ∈ {1, 2}). Thus A|m2 is a diagonal matrix. Therefore Ad(K1 )-invariant g.o. metrics on n are reduced to h·, ·i = µId|s + λ1 B(·, ·)|m1 + λ21 B(·, ·)|n21 + λ22 B(·, ·)|n22 , (µ, λ1, λ21 , λ21 > 0).

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Since (G/K1 , g) is a g.o. space, we assume that λ21 6= µ. Proposition 5 implies that λ21 µ for any V ∈ s and X ∈ n21 there exists k ∈ k1 such that [V, X] = µ−λ 2 [k, V ]+ µ−λ2 [k, X]. λ2

1

1

Since [k, V ] = 0, it follows that [V, X] = µ−λ1 2 [k, X]. Since [V, X] ∈ n22 and [k, X] ∈ n21 , 1 this is a contradiction. Hence we get that λ21 = µ. By the same method we obtain that λ22 = µ. Hence, if (G/K1 , g) is a g.o. space, then the corresponding Ad(K1 )-invariant scalar product (55) is reduced to h·, ·i = µB(·, ·)|s⊕m2 + µ1 B(·, ·)|m1 .

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(b) Assume that dim m2 6= 2.

Since dim m2 6= 2, this implies there exist α1 , α2 ∈ R2+ such that α1 − α2 ∈ R1+ . Lemma 3 and Proposition 3 implies that there exist β1 , β2 ∈ R2+ such that α1 |a1 = −β1 |a1 , α2 |a1 = −β2 |a1 and α1 (h) = β1 (h), α2 (h) = β2 (h) for any h ∈ s. Proposition 2 and Remark 3 implies that α1 6= β1 and α2 6= β2 . Since G/K1 is a g.o.space, by Corollary 3 it follows that for any X ∈ n there exists a k ∈ k1 such that [k + X, ΛX] ∈ k1 . We choose non zero vectors X1 , X2 ∈ n21 , with [k + X1 + X2 , Λ(X1 + X2 )] ∈ k1 . Then we have [k + X1 + X2 , λ21 X1 + A12 X1 + λ21 X2 + A12 X2 ] = [k, λ21 X1 + λ21 X2 ] + [k, A12 X2 + A12 X1 ] + [X1 , A12 X1 ] + [X1 , A12 X2 ] + [X2 , A12 X1 ] + [X2 , A12 X2 ]. Since n21 and n22 are Ad(K1 )-invariant, it follows that [k, λ21 X1 + λ21 X2 ] ⊆ n21 ⊂ m2 and [k, A21 X2 + A12 X1 ] ⊆ n22 ⊂ m2 . We choose X1 = Aα1 + A−β1 , X2 = Aα2 + A−β2 . Then we have that √ √ [X1 , A12 X1 ] = [Aα1 + A−β1 , [ −1hΛp , Aα1 + A−β1 ] = 2 −1α1 (hΛp )(hα1 − h−β1 ) −2α1 (hΛp )Nα1 ,−β1 Bα1 −β1 , √ √ [X2 , A12 X2 ] = [Aα2 + A−β2 , [ −1hΛp , Aα2 + A−β2 ] = 2 −1α2 (hΛp )(hα2 − h−β2 ) −2α2 (hΛp )Nα2 ,−β2 Bα2 −β2 , √ [X1 , A12 X2 ] = [Aα1 + A−β1 , [ −1hΛp , Aα2 + A−β2 ]] = α2 (hΛp )(Nα1 ,−α2 Bα1 −α2 −Nα1 ,−β2 Bα1 −β2 + N−β1 ,α2 B−β1 +α2 − N−β1 ,β2 B−β1 +β2 ), √ [X2 , A12 X1 ] = [Aα2 + A−β2 , [ −1hΛp , Aα1 + A−β1 ]] = α1 (hΛp )(Nα2 ,−α1 Bα2 −α1 −Nα2 ,−β1 Bα2 −β1 + N−β2 ,α1 B−β2 +α1 − N−β2 ,β1 B−β2 +β1 ). Since ([X1 , A12 X1 ] + [X2 , A12 X2 ]) ∈ k, ([X1 , A12 X2 ] + [X2 , A12 X1 ]) ∈ m1 ⊕ k and ([k, λ11 X1 + λ11 X2 ] + [k, A21 X2 + A12 X1 ]) ⊂ m2 , it follows that

([k, λ11 X1 +λ11 X2 ]+[k, A21 X2 +A12 X1 ])∩([X1 , A12 X1 ]+[X2 , A21 X2 ]+[X1 , A21 X2 ]+[X2 , A12 X1 ]) = {0}. Since α1 6= β1 and α2 6= β2 we have (−β1 ± α2 ) 6= (α1 ± α2 ) 6= (−β1 ± β2 ) and (α1 ± α2 ) 6= (α1 ± β2 ), then we have ([X1 , A21 X2 ] + [X2 , A12 X1 ]) ∈ k1 if and only if α1 (hΛp ) = α2 (hΛp ) = 0. But α1 (hΛp ) 6= 0 and α2 (hΛp ) 6= 0, which is a contradiction. Therefore we have Ajk = 0, (j, k ∈ {1, 2}). Now we prove that λ21 = λ21 in (56).

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Assume that λ21 6= λ21 in (56). Since G/K1 is a g.o.space, by Proposition 5 we have that for any X1 ∈ n21 , X2 ∈ n22 there exists a k ∈ k1 such that [X1 , X2 ] =

λ21 λ22 [k, X ] + [k, X2 ]. 1 λ21 − λ22 λ21 − λ22

Lemma 3 and Proposition 3 implies that there exist α1 6= β1 ∈ R2+ such that α1 |a1 = −β1 |a1 and α1 (h) = β1 (h)√for any h ∈ s. We choose X1 = Aα1 +A−β1 , X2 = Bα1 −B−β1 , then we have [X1 , X2 ] = 2 −1(hα1 −h−β1 )−2Nα1 ,−β1 Bα1 −β1 , it follows that [X1 , X2 ] 6= 0 and [X1 , X2 ] ∈ k. Since n21 and n21 are irreducible as Ad(K1 )-modules, it follows that λ21 λ22 2 2 ( λ2 −λ 2 [k, X1 ] + λ2 −λ2 [k, X2 ]) ∈ m2 , this is a contradiction. Therefore we have λ1 = λ2 . 1 2 1 2 Then Ad(K1 )-invariant scalar product on n are reduced to h·, ·i = µId|s + µ1 B(·, ·)|m1 + µ2 B(·, ·)|m2 , (µ, µ1 , µ2 > 0).

(64)

h·, ·i = µB(·, ·)|s⊕m2 + λ1 B(·, ·)|m1 .

(65)

Since (G/K1 , g) is a g.o. space, assume that µ2 6= µ, Proposition 5 implies that for µ2 µ [k, V ] + µ−µ [k, X]. any V ∈ s and X ∈ n21 there exists k ∈ k1 such that [V, X] = µ−µ 2 2 µ2 2 Since [k, V ] = 0, it follows that [V, X] = µ−µ2 [k, X]. Since [V, X] ∈ n2 and [k, X] ∈ n21 , this is a contradiction. Hence we get that µ2 = µ. Hence, if (G/K1 , g) is a g.o. space, then the corresponding Ad(K1 )-invariant scalar product (60) is reduced to Proof of part 3) of Theorem 2. Set mi = ni1 ⊕ ni2 , i = 1, 2. It follows that Λ|mi has the form i Ai21 λ1 Id|ni1 , λi1 , λi2 > 0, A|mi = Ai12 λi2 Id|ni2

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where Aijk : nij → nik , j 6= k ∈ {1, 2} is an Ad(K1 )-equivalent map. If (G/K1 , g) is a g.o. space, we can prove that A|m1 is a diagonal matrix using the same method as the proof of Case 1 of part 2) of Theorem 2 and λ11 = λ12 = µ, we can also prove that A|m2 is diagonal matrixe using the same method as the proof of Case 2 of part 2) of Theorem 2 and λ21 = λ22 = µ, and this completes the proof. Proof of Corollary 2. Since dim m2 = 2, it follows that m2 is reducible as Ad(K1 )-module by Remark 2. If (G/K1 , g) is a g.o. space, case 2) of Theorem 2 implies that Ad(K1 )-invariant scalar product on n is h·, ·i = µB(·, ·)|s⊕m2 + µ1 B(·, ·)|m1 , (µ, µ1 > 0). (67) By Corollary 3 (G/K1 , g) is a g.o. space if and only if for any V ∈ s and for any X1 ∈ m1 , X2 ∈ m2 , there exists k ∈ k1 such that [k + V + X1 + X2 , µ(V + X2 ) + µ1 X1 ] ∈ k1 .

(68)

[µ1 k + (µ1 − µ)V + (µ1 − µ)X2 , X1 ] ∈ k1 .

(69)

Since dim m2 = 2, it follows that [k, X2 ] = 0. Also, [k + V, V ] = 0, so (68) reduces to Since µ1 6= µ, (69) is equivalent to µ1 [ k + V + X2 , X1 ] ∈ k 1 . µ1 − µ

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ANDREAS ARVANITOYEORGOS, YU WANG∗ AND GUOSONG ZHAO

1 Since [m2 , m1 ] ⊆ m1 , it follows that [ µ1µ−µ k + V + X2 , X1 ] ∈ m1 , then we obtain that µ1 k + V + X2 , X1 ] = 0. (71) [ µ1 − µ

6. Proof of Theorem 3 Proof of part 2) of Theorem 3. Since s = 1 in the decomposition √(2), it follows that dim s = 1. We assume that ΠM = {αp } and it follows that s = R −1hΛp . Since m is reducible as Ad(K1 )-module, by Lemma 2 we have m = n1 + n2 , where n1 and n2 are equivalent and irreducible as Ad(K1 )-modules. Therefore, G-invariant metrics on G/K1 are defined by A|s 0 A= , (72) 0 A|m

where A|s = Λ |s = λId |s , (λ > 0), and A|m has the form µ1 Id|n1 A21 A|m = , µ1 , µ2 > 0. (73) A12 µ2 Id|n2 The block matrices A12 and A21 correspond to Ad(K1 )-equivariant maps φ1 : n1 → n2 and φ2 : n2 → n1 respectively. Moreover, the symmetry of Λ implies that A12 = A21 . Consequently, for any vector Xj ∈ nj ⊂ m, j = 1, 2, it is

ΛXj = Λ|m Xj = µj Xj + Ajk Xj , Ajk : nj → nk , j 6= k ∈ {1, 2}. (74) + Lemma 3 implies that there exist α, β ∈ RM such that α|a1 = −β|a1 and α(h) = β(h) for any h ∈ s, where α, β are the lowest and highest roots respectively. Now we prove the necessity of part 2) of Theorem 3. Assume that Ajk 6= 0, j 6= k ∈ {1, 2}. Let Ajk be the matrix representation of the map √ ad( −1hΛp ) : nj → nk , j 6= k ∈ {1, 2}. 1) Assume that α = β. Since G/K1 is a g.o. space, by Corollary 3 we have that for any X = (X1 + Y ) ∈ n there exists a k ∈ k1 such that [k + X, ΛX] ∈ k1 , where X1 ∈ n1 and Y ∈ s. Since ΛX = λY + µ1 X1 + A12 X1 , then we obtain that [k + X, ΛX] = [k + X1 + Y, λY + µ1 X1 + A12 X1 ] = [k, µ1 X1 + A12 X1 ] + (λ − µ1 )[X, Y ] +[X1 , A12 X1 ] + [Y, A12 X1 ]. Since α = β, by Remark 3 we have dim m =√2, By Remark 2 we have n1 = RAαp and n2 = RBαp . We choose X1 = Aαp and Y = −1hΛp . It follows that ([k, µ1 X1 + A12 X1 ] + (λ − µ1 )[X, Y ] + [Y, A12 X1 ]) ∈ m,

which implies that

[X1 , A12 X1 ] ∈ s,

([k, λ11 X1 + A12 X1 ] + (λ − µ1 )[X1 , Y ] + [Y, A12 X1 ]) ∩ [X1 , A12 X1 ] = {0}.

It follows that [X1 , A12 X1 ] ∈ k1 if and only if [X1 , A12 X1 ] = 0. But √ √ [X1 , A12 X1 ] = [Aαp , [ad( −1hΛp ), Aαp ]] = 2 −1α(hΛp )hαp 6= 0,

which is a contradiction, so we have Ajk = 0, (j 6= k ∈ {1, 2}). Thus A|m is a diagonal matrix.

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Therefore Ad(K1 )-invariant g.o. metrics on n are reduced to h·, ·i = λId|s + µ1 B(·, ·)|n1 + µ2 B(·, ·)|n2 , (λ, µ1 , µ2 > 0).

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Since (G/K1 , g) is a g.o. space, we assume that µ1 6= λ. Proposition 5 implies that µ1 λ [k, V ]+ λ−µ [k, X]. for any V ∈ s and X ∈ n1 there exists k ∈ k1 such that [V, X] = λ−µ 1 1 λ Since [k, V ] = 0, it follows that [V, X] = λ−µ1 [k, X]. Since [V, X] ∈ n2 and [k, X] ∈ n1 , this is a contradiction. Hence we get that µ1 = λ. By the same method we obtain that µ2 = λ. Therefore we have that g.o. metric on G/K1 is the standard metric. 2) Assume that α 6= β. Since α 6= β, by Remark 3 it follows that dim m 6= 2, this implies there exist α1 , α2 ∈ + + RM such that α1 + α2 ∈ RM . Lemma 3 and Proposition 3 implies that there exist + β1 , β2 ∈ RM such that α1 |a1 = −β1 |a1 , α2 |a1 = −β2 |a1 and α1 (h) = β1 (h), α2 (h) = β2 (h) for any h ∈ s. Since dim m 6= 2, Proposition 2 and Remark 3 imply that α1 6= β1 and α2 6= β2 . Since G/K1 is a g.o. space, by Corollary 3 we have that for any X = (X1 + X2 ) ∈ n there exists a k ∈ k1 such that [k + X, ΛX] ∈ k1 , where X1 , X2 ∈ n1 . We choose X1 = Aα1 + A−β1 , X2 = Aα2 + A−β2 . By the same method as the proof of Case 1 of part 2) of Theorem 2 we prove that A12 = A21 = 0 and µ1 = µ2 = λ. Hence we have that g.o. metric on M-space G/K1 is the standard metric. The sufficiency of part 2) of Theorem 3 is obvious. References [1] D.V. Alekseevsky: Flag manifolds, in Sbornik Radova, 11th Jugoslav. Geom. Seminar. Beograd. 6 (14) (1997) 3-35. [2] D.V. Alekseevsky and A. Arvanitoyeorgos: Riemannian flag manifolds with homogeneous geodesics, Trans. Amer. Math. Soc. 359 (2007) 3769–3789. [3] D.V. Alekseevsky and Yu. G. Nikonorov: Compact Riemannian manifolds with homogeneous geodesics, SIGMA: Symmetry Integrability Geom. Methods Appl. 5(93) (2009) 16 pages. [4] D. V. Alekseevsky and A. M. Perelomov: Invariant K¨ ahler-Einstein metrics on compact homogeneous spaces, Funct. Anal. Appl. 20 (1986) 171–182. [5] A. Arvanitoyeorgos and I. Chrysikos: Motion of charged parhicles and homogeneous geodesics in k¨ ahler C-spaces with two isotropy summands, Tokyo J. Math. 32(2) (2009) 487–500. [6] A. Arvanitoyeorgos and N.P. Souris: Geodesics in generalized Wallach spaces, J. Geom. 106 (2015) 583–603. [7] A. Arvanitoyeorgos and N.P. Souris: Two-step Homogeneous Geodesics in Homogeneous Spaces, Taiwanse J. Math. 20(6) (2016) 1313–1333. [8] A. Arvanitoyeorgos: Homogeneous geodesics in the flag manifold SO(2l + 1)/U (l − m) × SO(2m + 1), Linear Alg. Appl. 428 (2008) 1117–1126. [9] A. Arvanitoyeorgos and Y. Wang: Homogeneous geodesics in generalized Wallach spaces, Bull. Belgian Math. Soc. - Simon Stevin 24(2) (2017) 257–270. [10] J. Berndt, O. Kowalski and L. Vanhecke:Geodesies in weakly symmetric spaces, Ann. Global analsis and Geom, 15 (1997) 153–156. [11] G. Calvaruso and R.A. Marinosci: Homogeneous geodesics of three-dimensional unimodular Lie groups, Mediterr. J. Math. 3 (2006) 467–481. ´ Cartan:Sur une classe remarquable d’espaces de Riemann, Bull. Soc. Math. France 54 (1926) [12] E. 214–264. ´ Cartan:Sur une classe remarquable d’espaces de Riemann. II, Bull. Soc. Math. France 55 [13] E. (1927) 114–134. [14] Z.Chen and Yu. Nikonorov:Geodesics orbit Riemannian spaces with two isotropy summands. I, arXiv:1704.01913, 2017. [15] Z. Duˇsek: Explicit geodesic graphs on some H-type groups, Rend. Circ. Mat. Palermo Ser. II, Suppl. 69 (2002) 77–88.

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