ON ISOTROPY IRREDUCIBLE RIEMANNIAN MANIFOLDS BY MCKENZIE WANG AND WOLFGANG ZILLER

A connected Riemannian manifold (M, g) is said to be isotropy irreducible if for each point p ∈ M the isotropy group H_{p}, i.e. all isometries of g fixing p, acts irreducibly on T_{p}M via its isotropy representation. This class of manifolds is of great interest since they have a number of geometric properties which follow immediately from the definition. By Schur’s lemma the metric g is unique up to scaling among all metrics with the same isometry group. By the same argument, the Ricci tensor of g must be proportional to g, i.e. g is an Einstein metric. Furthermore, according to a theorem of Takahashi [Ta], every eigenspace of the Laplace operator of (M, g) with eigenvalue λ ≠ 0 and of dimension k+1 gives rise to an isometric minimal immersion into S^{k}(r) with r^{2} = dim M/λ , by using the eigenfunctions as coordinates (see Li [L] and §6 of this paper for further properties of these minimal immersions). By a theorem of D.Bleecker [Bl], these metrics can also be characterised as being the only metrics which are critical points for every natural functional on the space of metrics of volume 1 on a given manifold.

From the definition it follows easily that the isometry group of g must act transitively on M. Hence (M, g) is also a Riemannian homogeneous space. Conversely, we can define a connected effective homogeneous space G/H to be isotropy irreducible if H is compact and Ad_{H} acts irreducibly on 𝔤/𝔥. Given an isotropy irreducible homogeneous space G/H, there exists a G-invariant metric g, unique up to scaling, such that (M, g) is isotropy irreducible in the first sense. But if we start with a Riemannian manifold (M, g) which is isotropy irreducible, it can give rise to several isotropy irreducible homogeneous spaces G/H since G does not have to be the full isometry group of g. The aim of this paper is to classify the Riemannian manifolds as well as the homogeneous spaces which are isotropy irreducible.

If the identity component H_{0} of H also acts irreducibly on 𝔤/𝔥, then G/H is called a strongly isotropy irreducible homogeneous space, and similarly for strongly isotropy irreducible Riemannian manifolds.

Typeset by AMS-TEX

The first author acknowledges partial support from the Natural Sciences and Engineering Research Council of Canada. The second author acknowledges partial support from the National Science Foundation.

The most important examples of strongly isotropy irreducible homogeneous spaces are the irreducible symmetric spaces, classified by E.Cartan [C 1,2] in 1926. The non-symmetric strongly isotropy irreducible homogeneous spaces were classified independently by O.V.Manturov [Ma 1,2,3] in 1961, by J.A.Wolf [Wo 1] in 1968, and by M.Krämer [K] in 1975, but Wolf in addition studied many of their geometric properties. Both the classification of Manturov and Wolf contain some omissions, but the classification of Krämer is complete. For an a priori proof of this classification for quotients of the classical groups, see [WZ 2].

Generalizing a theorem of Wolf [Wo 1,Theorem 1.1] it was shown in [Be,7.46] that a non-compact isotropy irreducible homogeneous space G/H is either flat or is a symmetric space of non-compact type. Furthermore, one easily shows that if M is an isotropy irreducible Riemannian manifold, then its universal cover is also isotropy irreducible, and with the product metric, M × ⋯ × M is also isotropy irreducible. Hence we will first assume that M is compact, de Rham irreducible and simply connected and we will prove:

Theorem A. Let G/H be a compact, simply connected, effective Riemannian homogeneous space which is de Rham irreducible and let G_{0}, H_{0} be the id-component of G, H. Then G/H = G_{0}/H_{0}, and if G/H is isotropy irreducible but not strongly isotropy irreducible, then G_{0}/H_{0} is listed in Table I and II. Conversely, for every entry in Table I and II there exists in general several isotropy irreducible G/H.

This theorem completely describes the simply connected, compact, isotropy irreducible and de Rham irreducible Riemannian manifolds which are not strongly isotropy irreducible. In most cases one can also easily read off the possibilities for G and H from the other columns in Table I and II . χ_{0} describes the isotropy representation of H_{0} on 𝔤/𝔥 and enables one to determine the action of Ĥ/H_{0} on 𝔤/𝔥, where Ĥ is the full isotropy group of the metric (see §3 for details). The entry (Ĥ/H_{0})_{min} is a subgroup of minimal order (not necessarily unique) which is needed to make G_{0}/H_{0} isotropy irreducible. In a few cases there are some finite group problems involved in trying to determine all possibilities for G, which we do not try to resolve. E.g. for the biinvariant metric on a connected, compact, simple Lie group G one can adjoin any finite subgroup Γ of Aut(G) whose natural representation on 𝔤 is irreducible. Then G ⋉ Γ/Γ is an isotropy irreducible homogeneous space which is not strongly isotropy irreducible, but as a Riemannian manifold it is of course isometric to G. This Riemannian manifold can also be presented as [G × G]/△G, in which case it becomes a strongly isotropy irreducible symmetric space.

If M is simply connected and isotropy irreducible, but not de Rham irreducible, then M is either flat or isometric to a Riemannian product N × ⋯ × N, where N is isotropy irreducible and de Rham irreducible. In the flat case one can write M in many ways as an isotropy irreducible homogeneous space: M = [ℝ^{n} ⋉ H]/H, where H acts irreducibly on ℝ^{n}.

The only entry in Table I and II for which G_{0} is not the full id-component of the isometry group is again a compact simple Lie group with a biinvariant metric.

In §6 we describe how to determine the isotropy irreducible Riemannian manifolds which are not simply connected. This uses the column N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) in Table I and II. There are again some finite group problems, which we will not solve completely.

In §1 we make some general remarks about isotropy irreducible manifolds and describe the most interesting examples we obtain in our classification. In §2 we discuss the general theory of isotropy irreducible manifolds and reduce the classification to the case where G is compact and simple. In that case the negative of the Killing form of G induces the standard homogeneous metric on G/H, which must be Einstein by the above remarks. Such Einstein metrics were classified in [WZ 1]. To select from them the candidates G_{0}/H_{0} which might be isotropy irreducible as Riemannian manifolds, we derive a criterion on the isotropy representation χ_{0} of H_{0} on 𝔤/𝔥 which turns out to be necessary and sufficient:

Theorem B. Let G_{0}/H_{0} be a simply connected, compact, effective Riemannian homogeneous space with H_{0} ≠ 1. Then there exists a Riemannian homogeneous space G/H with G_{0}, H_{0} the id-component of G, H and such that G/H is isotropy irreducible iff there exists a finite group of automorphisms of 𝔥 which permutes transitively the dominant weights of the Ad_{H_{0}} irreducible summands of 𝔤/𝔥.

In particular all irreducible summands of Ad_{H_{0}} are equivalent up to some (possibly outer) automorphism of 𝔥. Included is the possibility that all dominant weights are the same, in which case we can choose the finite group to be trivial. This actually occurs for the Examples 4 and 6 in Table II.

We are able to prove Theorem B directly only when G and H have equal rank. In the unequal rank case it follows from a case by case argument. In these cases we have to determine the full isometry group of G_{0}/H_{0}, in order to see whether an isotropy irreducible G/H exists. This boils down to a problem of extending automorphisms of 𝔥 to 𝔤 and is discussed in §3 . The methods in §3 may be of independent interest since they enable one to determine the full isometry group of the standard Riemannian homogeneous metric in many circumstances, assuming that one knows the id-component of the isometry group already. They extend results of Cartan [C 1,2] for symmetric spaces and of Wolf [Wo 1] for strongly isotropy irreducible spaces. These methods are then applied in §4 and §5 to the equal and unequal rank cases and finish the proof of our classification.

It would be interesting to have a direct proof of Theorem B also in the unequal rank case. It would imply in particular that if G_{0}/H_{0} is a homogeneous space whose isotropy representation is equal to ϕ ⊕ ⋯ ⊕ ϕ for some irreducible representation ϕ, then the universal cover of G_{0}/H_{0} is one of Example 4 or 6 in Table II. Presently this only follows if G_{0} is also simple.

In §6 we determine which subcoverings are still isotropy irreducible and discuss two applications. In the first one we describe the isotropy irreducible homogeneous spaces G^{∗}/H^{∗} with G^{∗} connected, and in the second one we study some of the geometric properties of the minimal isometric immersions as in [L].

We emphasize that, whereas strongly isotropy irreducible is a local condition, isotropy irreducible is in general a global one. In many of our examples the group G/G_{0} can be very large and gives rise to many global symmetry properties of the manifold G_{0}/H_{0}, most of which are reflected in symmetry properties of the isotropy representation Ad_{H_{0}}. We believe that these large global symmetry groups make these examples particularly appealing and may be helpful in other geometric considerations.

It may not be a coincidence that our examples frequently occur in other geometric contexts. E.g. in the classification of homogeneous spaces of positive sectional curvature [Wa], [BB] all the examples are isotropy irreducible as homogeneous spaces (although the metrics in general are not) except for the Aloff-Wallach examples SU(3)/S^{1}. Many of our examples also occur in the classification of homogeneous isoparametric hypersurfaces [TT] and in the classification of 3-symmetric or ksymmetric spaces, see e.g. [WG]. In a forthcoming paper we will also discuss the close relationship between isotropy irreducible homogeneous spaces and primitive subgroups, as defined in [Go], [GR].

We would like to thank Alex Rosa for pointing out the relationship of Example 6 in §1 with finite geometries.

§1. General Remarks and Examples.

Let (M, g) be a Riemannian manifold with isometry group Ĝ and isotropy group Ĥ_{p} for p ∈ M. We say that (M, g) is an isotropy irreducible Riemannian manifold if for every p ∈ M, Ĥ_{p} acts irreducibly on T_{p}M via its isotropy representation. This implies that Ĝ must act transitively on M. Indeed, if p lies in a principal orbit of Ĝ, then Ĥ_{p} leaves the tangent space to the principal orbit invariant, and hence this tangent space is either all of T_{p}M, in which case Ĝ acts transitively, or it is trivial. But the latter case is impossible since then Ĝ is discrete, which contradicts the fact that Ĥ_{p} is non-trivial for each p ∈ M. Hence we can also write M as a homogeneous space Ĝ/Ĥ_{p}.

Conversely, given a connected homogeneous space G/H, where G acts effectively on G/H and H is compact, we say that G/H is an isotropy irreducible homogeneous space if Ad_{H} acts irreducibly on 𝔤/𝔥. There then exists a G-invariant metric g on G/H, which is uniquely determined up to scaling by Schur’s Lemma, and (M = G/H, g) is an isotropy irreducible Riemannian manifold in the above sense. But G does not have to be the full isometry group of (M, g). In fact even the dimension of G could be smaller than the dimension of the full isometry group. Hence an isotropy irreducible homogeneous space gives rise to an isotropy irreducible Riemannian manifold, unique up to scaling, but an isotropy irreducible Riemannian manifold can give rise to several isotropy irreducible homogeneous spaces. Since such descriptions of the same manifold as homogeneous spaces in different ways can sometimes be useful, we try to classify all isotropy irreducible homogeneous spaces in this paper. We will achieve this modulo some finite group problems.

If (M, g) is isotropy irreducible, then its universal Riemannian cover M̃ is also isotropy irreducible, since the isometry group of M̃ contains all the covering transformations, and all isometries of M lift to isometries of M̃. In fact, the isotropy group of p ∈ M is isomorphic to the subgroup of the isotropy group of any p̃ ∈ M̃ above p which normalizes the group of covering transformations, and the isotropy representation of one restricts to the isotropy representation of the other. But not every Riemannian subcover of an isotropy irreducible manifold is necessarily isotropy irreducible.

If we start instead with a Riemannian homogeneous space M = G^{∗}/H^{∗} which is isotropy irreducible but not simply connected and let G̃_{0} be the universal cover of the id-component G_{0}^{∗} of G^{∗}, then the action of G_{0}^{∗} on M lifts to an action of G̃_{0} on M̃ with a possible ineffective kernel N ⊂ Z(G̃_{0}). The isometries in G^{∗} which are not in G_{0}^{∗} also lift to M̃, uniquely modulo the deckgroup. Hence these lifts, the deck group and G̃_{0}/N ⊂ I(M̃, g̃) generate a Lie group G which acts transitively on M̃ with isotropy group H. Then M̃ = G/H is isotropy irreducible as a homogeneous space. Clearly dim G = dim G^{∗}, and G as well as H cannot be connected if G^{∗}/H^{∗} is not strongly isotropy irreducible.

Next, if N is an isotropy irreducible Riemannian manifold, then M = N ×⋯×N (k times) with the product metric is also isotropy irreducible, since the isometry group of M contains the symmetric group S_{k}, which acts by interchanging the factors. Conversely it follows from the de Rham decomposition theorem that if M is isotropy irreducible, simply connected, and a Riemannian product, then M is either flat or all its de Rham factors must be isometric to each other.

Similarly in the homogeneous case. If G/H is isotropy irreducible, then the product of G/H with itself k times becomes an isotropy irreducible homogeneous space, if we adjoin to G × ⋯ × G (k times) the symmetric group S_{k} or a subgroup of it that still acts transitively on the factors. Conversely, if G/H is an isotropy irreducible homogeneous space which is simply connected but de Rham reducible, then it is either flat, or G_{0}/H_{0} is the product of K/L with itself k times. To prove this last statement we need the fact, proved in §2, that the metric on G_{0}/H_{0} is always naturally reductive, if M is not flat. Then the de Rham decomposition theorem for naturally reductive spaces (see [KN, Theorem X.5.2]) implies that G_{0}/H_{0} = K_{1}/L_{1} × ⋯ × K_{k}/L_{k}. But the factors are isometric as Riemannian manifolds and since G/H is isotropy irreducible, there exists an isometry h_{ij} ∈ H which maps the i-th factor into the j-th factor. Moreover, h_{ij} normalizes G_{0} and hence gives an isomorphism of K_{i} with K_{j} which carries L_{i} into L_{j}.

The flat case is somewhat special and will always be excluded. Notice though that ℝ^{n} can be written as an isotropy irreducible homogeneous space in many ways. If we represent ℝ^{n} as a homogeneous space, then it is of the form [ℝ^{n} ⋉ H]/H for any closed subgroup H ⊂ O(n) and the isotropy representation of H is given by the embedding H ⊂ O(n). Hence [ℝ^{n} ⋉ H]/H is isotropy irreducible iff H is an irreducible subgroup of O(n). There are many such subgroups, especially since H need not be connected. Indeed H can also be a finite group.

Hence both for the homogeneous spaces and the Riemannian manifolds it will suffice to assume first that all manifolds are simply connected, de Rham irreducible and not flat. In §6 we will discuss the non-simply connected case.

Given an isotropy irreducible M = G/H, since H is compact, we can write 𝔤 = 𝔥 ⊕ 𝔪, where 𝔪 is Ad_{H} invariant and can be identified with the tangent space of M at a point. The metric corresponds to an Ad_{H} invariant inner product on 𝔪. As we will see in §2, one easily reduces to the case where G is compact and semisimple. Hence the negative of the Killing form B of 𝔤 induces the standard normal homogeneous metric g_{B} on G/H and from now on we will assume that the metric is of this form. This also means that 𝔪 is chosen so that 𝔥 and 𝔪 are perpendicular with respect to B. In particular, every automorphism of G that leaves H invariant induces an isometry of G/H.

Let H_{0} be the identity component of H and let 𝔪 = 𝔪_{1} ⊕ ⋯ ⊕ 𝔪_{k} be a decomposition of 𝔪 into Ad_{H_{0}} irreducible summands. Since we assume that G/H is not strongly isotropy irreducible, we have k > 1. For G/H to be isotropy irreducible H/H_{0} must act via inner or outer automorphisms on G_{0}, leaving H ∩ G_{0} invariant, and permuting transitively the irreducible factors 𝔪_{i}, at least if all 𝔪_{i} are inequivalent. See (2.3) for a precise statement. Hence the dominant weights of the irreducible factors 𝔪_{i} must in particular be equivalent to each other up to some inner or outer automorphism of H.

We end this section with a number of illustrative examples.

Example 1. Let G be a compact connected simple Lie group with a biinvariant metric. As a symmetric space this is [G × G]/△G and is strongly isotropy irreducible. But there are many other ways to write it as a homogeneous space, most of which are not isotropy irreducible. It will turn out that the only other way to write it as an isotropy irreducible homogeneous space is described as follows.

Let Γ ⊂ G be a finite subgroup, acting on G via the adjoint representation. Then G = [G ⋉ Γ]/Γ with isotropy representation Ad|_{Γ}. Hence, if Ad|_{Γ} is irreducible, [G ⋉ Γ]/Γ is isotropy irreducible, but not strongly isotropy irreducible. We can also use Γ to construct subcoverings of G since the isotropy representation of G/Γ is also Ad|_{Γ}. Hence if Ad|_{Γ} is irreducible, G/Γ is isotropy irreducible, but not strongly isotropy irreducible.

More generally, we can take any subgroup Γ of Aut(G) such that the natural representation on 𝔤 is irreducible. Then [G ⋉ Γ]/Γ becomes isotropy irreducible and if Γ^{∗} ⊆ Γ ∩ Int(G) is a normal subgroup of Γ, then G/Γ^{∗} becomes isotropy irreducible if we adjoin to G the isometries induced by Γ but not in Γ^{∗}.

This example, and the flat metric on ℝ^{n}, are the only examples of isotropy irreducible Riemannian manifolds which can be written in several ways as isotropy irreducible homogeneous spaces such that the id-components of the transitive groups are distinct.

At first sight the condition on Γ appears to be a very restrictive one. However one can find many examples satisfying this condition, including many of the sporadic finite simple groups. We first explain the condition more explicitly for the classical groups.

If π: Γ → SO(n) is a faithful orthogonal representation of Γ, then since the adjoint representation of SO(n) is just the second exterior power of the standard n-dimensional representation, our condition says that ∧^{2}π is irreducible. Clearly we may assume that π is absolutely irreducible. For π: Γ → Sp(n), since the adjoint representation of Sp(n) is the second symmetric power of the standard 2ndimensional representation, the condition becomes S^{2}π is irreducible. Finally, if π: Γ → SU(n), then the condition is that π ⊗ π^{∗} − 1 is irreducible. Here we can assume that π is irreducible and π ≠ π^{∗}. Below we give some of the examples we found satisfying these conditions.

(1) The natural representation of the symmetric group S_{n} on ℝ^{n} is orthogonal and splits into the trivial representation on the diagonal and an irreducible representation on the orthogonal complement. Let π: S_{n} → O(n−1) denote this representation. Then, if n ≥ 4, ∧^{2}π is irreducible, so that O(n−1)/S_{n} is isotropy irreducible. The restriction to the alternating group A_{n} maps into SO(n − 1) and yields the isotropy irreducible space SO(n − 1)/A_{n} (n ≠ 5) , which as a manifold is of course the same as O(n − 1)/S_{n}. More generally, if W is the Weyl group of a compact simple Lie group of rank n, it acts naturally on ℝ^{n} and O(n)/W is isotropy irreducible.

(2) A finite subgroup of SO(3) (respectively SU(2)) satifies the isotropy irreducibility condition iff it is conjugate to the tetrahedral, icosahedral, or octahedral (resp. binary tetrahedral, binary octahedral, or binary icosahedral) group. The subgroups of SO(3) of course follow from (1) since they contain the alternating group and if Γ ⊂ Φ ⊂ G with G/Γ isotropy irreducible, then G/Φ is also isotropy irreducible.

(3) All the Mathieu groups have irreducible orthogonal representations satisfying the isotropy irreducibility condition. For M_{11} the representations have dimension 10 and 11, for M_{12} there are two 11-dimensional representations and for M_{22}, M_{23}, M_{24} there is a representation of dimension 21, 22, 23 respectively. As for unitary representations such that π ⊗π^{∗} −1 is irreducible, there are none for M_{11}, M_{12}, or M_{22}, but one each for M_{23}, M_{24}, both of dimension 45. All this is verified using character tables in [CCNPW].

(4) Among the other sporadic finite simple groups one obtains a unitary representation of dimension 1333 for the Janko group J_{4}, and orthogonal representations for the Baby Monster group B, the Fischer group F i_{22}, the Harada-Norton group HN, the Thompson group T h, the McLaughlin group McL, and the Conway groups Co_{2}, Co_{3} of dimension 4371, 78, 133, 248, 22, 23, and 23 respectively. Similarly one can easily obtain a number orthogonal, unitary, and symplectic representations from [CCNPW] among the finite simple groups of Lie type.

Although a complete enumeration of all such groups is probably complicated, an answer to the following questions would be of interest:

(a) For each compact connected simple Lie group G, does there exist a finite subgroup Γ ⊂ Aut(G) such that Γ acts irreducibly on 𝔤 ?

(b) Is there a characterization of those finite groups Γ which admit a faithful representation π: Γ → Aut(G) for some compact connected simple Lie group G such that Γ acts irreducibly on 𝔤 ?

Example 2. Let H be a compact, simply connected, simple Lie group with center Z and let △H be the diagonal subgroup of G = H ×⋯×H (n times). The isotropy representation of G/△H is equal to (n − 1)Ad_{H}, where 𝔪 = {(X_{1}, …, X_{n}) | X_{i} ∈ 𝔥 and ∑ X_{i} = 0}. Notice though that the decomposition of 𝔪 into Ad_{H} irreducible summands is highly non-unique. The symmetric group S_{n} acts on G = H ×⋯×H as outer automorphisms by permuting the factors. Since this action keeps △H invariant, it induces isometries of the standard normal homogeneous metric on M = G/△H. If we add these isometries to G, then M becomes an isotropy irreducible homogeneous space. Notice that S_{n} does not interchange the Ad_{△H} irreducible factors of any fixed decomposition of 𝔪, and that it acts trivially on the collection of dominant weights since they are all the same. M is simply connected, but not yet effective, ([H × ⋯ × H]/△Z)/(△H/△Z) being the effective version.

Isotropy irreducible subcoverings are obtained by dividing G by a central subgroup such that S_{n} still acts by isometries. E.g. {[(H/A)×⋯×(H/A)]/△Z}/{△(H/A)/△Z} is isotropy irreducible for any A ⊂ Z.

M is a symmetric space if n = 2 (and hence strongly isotropy irreducible), but is not strongly isotropy irreducible if n > 2.

Another description of this example is obtained by observing that as a manifold G/△H is equivariantly diffeomorphic to H × ⋯ × H (n − 1 times), an explicit map being given by

(g_{1}, …, g_{n})△H ∈ G/△H ⟶ (g_{1}g_{n}^{−1}, …, g_{n−1}g_{n}^{−1}) ∈ H × ⋯ × H.

The isotropy irreducible metric on G/△H corresponds under this map to a left invariant metric on H × ⋯ × H, which one easily checks is given at the identity by

|(X_{1}, …, X_{n−1})|^{2} = ∑ |X_{i}|^{2} − \frac{1}{n}|∑ X_{i}|^{2}.

This metric can thus be viewed also as follows. Starting with the metric on H × ⋯ × H (n − 1 times) which is the (n − 1)-fold product of a biinvariant metric on H with itself, we can fibre H × ⋯ × H by left cosets of △H and multiply the metric in direction of these fibres by (n − 1)/n, while leaving the metric unchanged in the direction perpendicular to the fibres.

If one starts with this left invariant metric on H × ⋯ × H (n − 1 times) it is more difficult to see all the isometries that are necessary in order to make it isotropy irreducible. Besides the left translations on H × ⋯ × H one needs the isometries (Ad(g), …, Ad(g)) , g ∈ H, the interchange of any two factors, and the special isometries

(g_{1}, …, g_{n−1}) ⟶ (g_{1}g_{i}^{−1}, …, g_{i−1}g_{i}^{−1}, g_{i}^{−1}, g_{i+1}g_{i}^{−1}, …, g_{n−1}g_{i}^{−1})

for each i = 1, …, n − 1.

Notice that on this manifold there exists another isotropy irreducible metric, namely the (n − 1)-fold product of a biinvariant metric with itself. This metric is not isometric to the above example, since the metric on G/△H is de Rham irreducible, as follows from [KN, Theorem X.5.3] and the fact that the metric on G/△H is given by the standard normal homogeneous metric. Hence this manifold has two non-isometric isotropy irreducible metrics. This phenomenon does not occur among strongly isotropy irreducible manifolds.

Example 3. Let H = G_{2} ⊂ Spin(7) ⊂ Spin(8) = G. The isotropy representation of G/H is ϕ_{7} ⊕ ϕ_{7}, where ϕ_{7} is the seven dimensional representation of G_{2}. G_{2} has no outer automorphisms and one easily sees that the normalizer of G_{2} in Spin(8) is Z(G)H, and hence does not induce any new isometries on G/H. But there are outer automorphisms of Spin(8) which keep G_{2} invariant and hence induce further isometries of G/H. For example, the triality automorphism of Spin(8) fixes G_{2} and one easily checks that it does not keep the decomposition of 𝔪 into irreducible summands invariant. Hence, if we add this isometry to Spin(8), G/H becomes an isotropy irreducible homogeneous space. On the other hand, the order 2 automorphism keeps the decomposition ϕ_{7} ⊕ ϕ_{7} invariant.

As a manifold G/H is diffeomorphic to S^{7} × S^{7} (see e.g. [WZ 2, p.575, Example 4]). But the Riemannian metric is de Rham irreducible, as follows from [KN,Corollary X.5.4]. The product metric of a round sphere metric with itself is of course also isotropy irreducible. Hence this manifold is again an example which carries two non-isometric isotropy irreducible homogeneous metrics. From our classification it seems to follow that Example 2 and 3 are the only manifolds which can carry more than one isotropy irreducible metric. Although this is most likely the case, a rigorous proof would involve showing that none of the examples in Table I and II are diffeomorphic to each other or to a strongly isotropy irreducible space or to products of such, which is not easy to do. As far as we know, even among the strongly isotropy irreducible homogeneous spaces, such a program has not been carried out.

There are several Riemannian homogeneous subcoverings, obtained by dividing Spin(8) by a subgroup of its center, but the only subcovering for which the triality automorphism descends to an isometry is that obtained by dividing by the full center, i.e. (SO(8)/ℤ_{2})/G_{2}, which is diffeomorphic to ℝP^{7} × ℝP^{7}.

This example also arises naturally in the following context. As a homogeneous space the Cayley plane is F_{4}/ Spin(9). Then Spin(8) ⊂ Spin(9) acts on the Cayley plane with principal orbits Spin(8)/G_{2}. As we move through the principal orbits, the induced metric achieves, up to scaling, every Spin(8)-invariant metric on Spin(8)/G_{2}. In particular, the standard normal homogeneous metric is achieved on a specific principal orbit.

Example 4. Let H = SO(k)×⋯×SO(k) (n times) ⊂ SO(nk) = G be given by the usual diagonal embedding. One easily sees that ∑_{i<j}[id ˆ⊗ … ˆ⊗ρ_{k} ˆ⊗ … ˆ⊗ρ_{k} ˆ⊗ … ˆ⊗id] is the isotropy representation of H and that the outer automorphisms of H which interchange the SO(k) factors extend to inner automorphisms of G, which therefore lie in the normalizer of H. Thus conjugation by these elements induce isometries fixing (H) and if we add these to G, we obtain an isotropy irreducible homogeneous space.

There are further isometries which are not needed to make G/H isotropy irreducible, but which can be used to produce subcoverings that are also isotropy irreducible. They are all of the form Ad(diag{±1, …, ±1}).

The space G/H can also be described as the set of all oriented flags V_{1} ⊂ V_{2} ⊂ ⋯ ⊂ V_{n} = ℝ^{nk} with dim V_{i} = ik. These flags can be represented by an element A of SO(nk) such that the first ik columns span V_{i}. If we denote the columns of A by w_{1}, …, w_{nk}, then the isometry which interchanges the i-th and j-th factor in H, but viewed as a right translation on G/H, acts on the flags by interchanging the vectors w_{(i−1)k+1}, …, w_{ik} with the vectors w_{(j−1)k+1}, …, w_{jk}. They form a group S_{n} of isometries which are fixed point free. Additional isometries are obtained by changing the orientation of one or several of the subspaces V_{i}, i = 1, …, n−1. These isometries can be viewed as right translations on G/H by D = diag{±1, …, ±1} where D contains an even number of −1 on the diagonal. This group ℤ_{2}^{n−1}, which acts freely on G/H, gives rise to many isotropy irreducible subcoverings of G/H, including the space of unoriented flags in ℝ^{nk}. One can also divide out by S_{n} or subgroups of S_{n} to obtain isotropy irreducible subcoverings. This subgroup has to be chosen so that it acts transitively on the set of unordered pairs. If nk is even, there exists one additional isometry, given by reflecting a flag in a hyperplane. But the component of the full isometry group containing this isometry does not contain any fixed point free isometries. For more details see §4 and §5.

Similar examples arise by using SU(k) and Sp(k) instead of SO(k).

Example 5. If rank H = rank G, we will see that a necessary and sufficient condition for G/H to be isotropy irreducible as a Riemannian manifold is the existence of a group of outer automorphisms of 𝔥 which acts transitively on the set of dominant weights of the Ad_{H_{0}} irreducible factors of 𝔪. Hence for almost all the examples in Table I it is obvious that G/H is isotropy irreducible, without any computation.

One particularly interesting example is [SU(5) × SU(5)]/△ℤ_{5} ⊂ E_{8}, which is the only maximal subalgebra of maximal rank such that the corresponding homogeneous space is not strongly isotropy irreducible. Hence all maximal subalgebras of maximal rank give rise to isotropy irreducible homogeneous spaces.

Another interesting example is G/T, where G is a compact, simple, centerless Lie group such that all roots of 𝔤 have the same length, and T is a maximal torus in G. Then the Weyl group acts transitively on all roots and hence transitively on the irreducible factors of the isotropy representation of G/T. If we add these isometries to G, G/T becomes an isotropy irreducible space. Further isometries are obtained from the diagram automorphisms of 𝔤.

Example 6. Let G_{0} = E_{7} or E_{8} and H_{0} = [SU(2)]^{7}/ℤ_{2}^{4} or [SU(2)]^{8}/ℤ_{2}^{4}. The isotropy representations are given in Table I. It turns out that in order to make G_{0}/H_{0} isotropy irreducible, we have to add a group of automorphisms of G_{0} which is a subgroup of the group of automorphisms of the projective plane (resp. affine 3- space) over the field 𝔽_{2} of 2 elements. To explain the connection, we consider first the case of E_{7}.

Each copy in the isotropy representation of G_{0}/H_{0} is described by a 4-tuple of integers from 1 to 7, where e.g. (1234) corresponds to the representation ∘^{1} ˆ⊗ ∘^{1} ˆ⊗ ∘^{1} ˆ⊗ ∘^{1} ˆ⊗ ∘ ˆ⊗ ∘ ˆ⊗∘ of [SU(2)]^{7}. We can instead describe this representation by the complementary 3-tuple of numbers (567) and thus obtain 7 such 3-tuples. These 3-tuples correspond to all possible lines on the projective plane over 𝔽_{2} as follows. Label the points in this projective plane by 1 = (1, 0, 0), 2 = (0, 1, 0), 3 = (0, 0, 1), 4 = (1, 1, 1), 5 = (0, 1, 1), 6 = (1, 0, 1), 7 = (1, 1, 0). Then each line goes through precisely 3 points and we can label a line by these points. We again obtain 7 3-tuples, and the 3-tuples in both cases agree. An element of Ĥ/H_{0} corresponds to a permutation of the integers 1 to 7, which preserves the 4-tuples and hence the 3-tuples which describe the isotropy representation. Thus it is described by a transformation of the projective plane which preserves lines, and therefore by an element of Aut(P^{2}(𝔽_{2})) = P GL(3, 2) = GL(3, 2). Hence Ĥ/H_{0} = GL(3, 2), a simple group of order 8·7·3. The 7-Sylow subgroup ℤ_{7} is generated by the matrix

( \begin{matrix}0 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 0 \end{matrix} ) or equivalently by the permutation (1254673). It permutes the lines

transitively and hence, if Γ ⊃ ℤ_{7}, then [G_{0} ⋉ Γ]/[H_{0} ⋉ Γ] is isotropy irreducible.

For the case E_{8} we have to relabel the circles according to the scheme 1 → 0, 2 → 5, 3 → 4, 4 → 1, 5 → 3, 6 → 7, 7 → 2, and 8 → 6. Then the isotropy representation becomes

(0145) ⊕ (2367) ⊕ (0347) ⊕ (1256) ⊕ (0246) ⊕ (1357) ⊕ (0136)⊕
(2457) ⊕ (0127) ⊕ (3456) ⊕ (0235) ⊕ (1467) ⊕ (0567) ⊕ (1234)

We now put these 4-tuples in 1-1 correspondence with the set of 2-planes in 3dimensional affine space over 𝔽_{2}. We label the points of this space with the integers from 0 to 7 as before. Then each 2-plane contains exactly 4 points, and if we describe the 2-plane by this 4-tuple of numbers, we obtain the same set of 4-tuples as above. Hence an element of Ĥ/H_{0} corresponds to a transformation of affine 3-space which takes 2-planes to 2-planes, and therefore an element of T (3, 2) ⋉ GL(3, 2), where T (3, 2) are the translations in 𝔽_{2}^{3}. This group has order 2^{3}·168. If ℤ_{7} denotes the 7-Sylow subgroup of GL(3, 2) as above, then T (3, 2) ⋉ ℤ_{7} already acts transitively on the 2-planes, and one easily sees that there is no subgroup of smaller order which does. Hence, if Γ ⊃ T (3, 2) ⋉ ℤ_{7}, then [G_{0} ⋉ Γ]/[H_{0} ⋉ Γ] is isotropy irreducible.

It would be interesting to obtain a better understanding of the relationship of these examples with the corresponding finite geometries.

§2. General Theory.

The first result is essentially already contained in [Be, 7.46], but we include a proof here for completeness.

(2.1) Theorem. Let M = G/H be an isotropy irreducible Riemannian homogeneous space. Then

(a) If M is non-compact, M is either flat or G/H is a symmetric space of non-compact type. In either case, M is simply connected.

(b) If 𝔤 is not semisimple, then M is flat.

Proof. Let B be the Killing form of 𝔤. Since G/H is effective and H is compact, it easily follows that B|𝔥 is negative definite (see e.g. [Be,7.35 and 7.36]). Hence we can choose an Ad_{H} invariant complement 𝔪 with B(𝔥, 𝔪) = 0. B|𝔪 is a symmetric Ad_{H} invariant bilinear form, and so is the metric g. Hence B|𝔪 = cg for some constant c. If c > 0, then (𝔤, 𝔥) is a symmetric pair of non-compact type and hence so is G/H. If c < 0, then B < 0 and hence 𝔤 is semisimple and compact. If c = 0, then 𝔪 is the nullspace of B and hence an ideal. In particular [𝔪, 𝔪] ⊂ 𝔪 and since [𝔪, 𝔪] is Ad_{H} invariant, isotropy irreducibility implies [𝔪, 𝔪] = 𝔪 or [𝔪, 𝔪] = 0. But if [𝔪, 𝔪] = 𝔪, then 𝔪 would be semisimple, which would contradict B|𝔪 = 0, since 𝔪 is an ideal. Hence 𝔪 is abelian, which implies that any G-invariant metric on G/H is flat. A flat homogeneous metric is isometric to the product of the flat metric on ℝ^{k} and a flat metric on T^{n−k} (see e.g. [Wo 2, 2.7.1] ). Since M is isotropy irreducible and non-compact, M must be isometric to ℝ^{n}. This proves (a) and (b). □

We already saw in §1 that we can assume that M is simply connected and de Rham irreducible. We will do so now.

(2.2) Theorem. Let M = G/H be a simply connected, compact, isotropy irreducible Riemannian homogeneous space which is also de Rham irreducible. If 𝔤 is semisimple but not simple, then G/H is the Example 2 of §1.

Proof. Let G_{0} and H_{0} be the id-component of G, H. Since M is compact and simply connected, G_{0} acts transitively on M and the isotropy group, which must be connected, is equal to H_{0}. We can assume that the metric on M is the standard homogeneous metric and hence [KN,Corollary X.5.3] implies that M is de Rham reducible iff 𝔤 = 𝔤_{1} ⊕ ⋯ ⊕ 𝔤_{k} and 𝔥 = 𝔥_{1} ⊕ ⋯ ⊕ 𝔥_{k} with 𝔥_{i} ⊂ 𝔤_{i} and k ≥ 2.

Let 𝔪 be an Ad_{H} invariant complement of 𝔥. If 𝔤 is semisimple but not simple, let 𝔤_{0} be a non-trivial simple ideal of 𝔤 and let 𝔤_{1} be the ideal Ad_{H}(𝔤_{0}). If 𝔤_{1} is not all of 𝔤, there exists a complementary Ad_{H} invariant ideal 𝔤_{2}. Decompose 𝔥 into ideals such that 𝔥 = 𝔥_{1}⊕𝔥_{2}⊕△𝔥_{3}, where 𝔥_{i}⊕𝔥_{3} are subalgebras of 𝔤_{i} and △𝔥_{3} is embedded diagonally into 𝔥_{3}⊕𝔥_{3} ⊂ 𝔤_{1}⊕𝔤_{2}. Then 𝔪 = 𝔪_{1}⊕𝔪_{2}⊕△𝔪_{3}, where 𝔪_{i} are orthogonal complements of 𝔥_{i} ⊕ 𝔥_{3} in 𝔤_{i} and △𝔪_{3} = {(X, −X) | X ∈ 𝔥_{3}} ⊂ 𝔥_{3} ⊕ 𝔥_{3} ⊂ 𝔤_{1} ⊕ 𝔤_{2}. 𝔥_{3} ≠ 0 since otherwise M would be de Rham reducible. Hence △𝔪_{3} ≠ 0 and since 𝔪_{i} and △𝔪_{3} are Ad_{H} invariant, we have 𝔪_{1} = 𝔪_{2} = 0. Since G/H is effective, this implies 𝔥_{1} = 𝔥_{2} = 0 and hence 𝔤_{1} = 𝔤_{2} = 𝔥_{3} with 𝔥 embedded diagonally. 𝔥 must then be simple, since otherwise M would be de Rham reducible. Hence G/H is as in Example 2.

We can therefore assume that Ad_{H}(𝔤_{0}) = 𝔤. Hence 𝔤 = 𝔤_{1}⊕⋯⊕𝔤_{k}, k ≥ 2, where 𝔤_{i} are isomorphic simple ideals permuted transitively by Ad_{H}. In particular 𝔥 is invariant under the automorphisms which permute the 𝔤_{i}. Let π_{i} be the projection of 𝔤 onto 𝔤_{i}. Then 𝔥^{∗} = π_{1}(𝔥) ⊕ ⋯ ⊕ π_{k}(𝔥) is Ad_{H} invariant and contains 𝔥. Hence either 𝔥^{∗} = 𝔥 or 𝔥^{∗} = 𝔤. If 𝔥^{∗} = 𝔥 (which includes the possibility of 𝔥 = 0), then 𝔥 = 𝔥_{1} ⊕ ⋯ ⊕ 𝔥_{k} with 𝔥_{i} ⊂ 𝔤_{i}, which implies that M is de Rham reducible. Hence 𝔥^{∗} = 𝔤, in other words, π_{i}(𝔥) = 𝔤_{i}. For each simple ideal 𝔥_{1} of 𝔥, π_{i}|_{𝔥_{1}} is either an isomorphism onto 𝔤_{i} or trivial, and if 𝔥_{2} is a second simple ideal of 𝔥, then π_{i}|_{𝔥_{1}} ≠ 0 implies π_{i}|_{𝔥_{2}} = 0 since 𝔥_{1} and 𝔥_{2} commute. Hence, if 𝔥 is not simple, M is again de Rham reducible. Thus 𝔥 is simple and π_{i}|𝔥 is an isomorphism onto 𝔤_{i} for each i. We can therefore assume that 𝔤 = 𝔥 ⊕ ⋯ ⊕ 𝔥 and 𝔥 is embedded diagonally, which is Example 2 in §1. □

We can now assume that G is compact and simple and hence the metric is given by the standard normal homogeneous metric g_{B}, which also must be Einstein. In [WZ 1] we classified all simply connected effective homogeneous spaces such that G is simple, compact, and such that the standard normal homogeneous metric is Einstein. To select among these manifolds the ones that are isotropy irreducible, we prove:

(2.3) Theorem. Let G/H be an isotropy irreducible homogeneous space which is not strongly isotropy irreducible and assume that 𝔥 ≠ 0. Let 𝔤 = 𝔥 ⊕ 𝔪 be an Ad_{H} invariant decomposition, 𝔱 a maximal abelian subalgebra of 𝔥 and Φ a fundamental system of roots of (𝔥, 𝔱). Given any Ad_{H_{0}} irreducible summand 𝔪_{0} of 𝔪, there exist elements h_{1}, …, h_{k} of H such that 𝔪 = Ad_{h_{1}}(𝔪_{0}) ⊕ ⋯ ⊕ Ad_{h_{k}}(𝔪_{0}) is a decomposition of 𝔪 into Ad_{H_{0}} irreducible summands. Furthermore Γ = {h ∈ H|Ad_{h}(𝔱) ⊂ 𝔱 and Ad_{h}^{∗}Φ ⊂ Φ} permutes transitively the dominant weights of the Ad_{H_{0}} irreducible summands of 𝔪 .

Proof. We choose a set of representatives {h_{i}} of H/H_{0} such that Ad_{h_{i}} leaves 𝔱 and Φ invariant. Then ∑_{i} Ad_{h_{i}}(𝔪_{0}) ⊂ 𝔪 is Ad_{H} invariant and hence 𝔪 = ∑_{i} Ad_{h_{i}}(𝔪_{0}). After reindexing, there is a subset {h_{1}, …, h_{k}} such that 𝔪 = ⨁_{i} Ad_{h_{i}}(𝔪_{0}). If h ∈ Γ, 𝔪 = Ad_{h}(𝔪) = ⨁_{i} Ad_{h}(Ad_{h_{i}}(𝔪_{0})) is another decomposition of 𝔪 into Ad_{H_{0}} irreducible summands and hence Ad_{h} permutes the dominant weights. For G/H to be isotropy irreducible, Γ must permute these weights transitively. □

Remark. As we saw in Example 2 and 3 in §1, Ad_{h}, h ∈ Γ, need not permute the irreducible summands 𝔪_{i} = Ad_{h_{i}}(𝔪_{0}), although this will necessarily be the case if all 𝔪_{i} are inequivalent as Ad_{H_{0}} representations. The same examples also show that there can be only a single dominant weight among the irreducible summands.

As a consequence of (2.3), all the Ad_{H_{0}} irreducible summands of 𝔪 must be equivalent to each other up to some inner or outer automorphism of 𝔥. Using this criterion, one can easily select from [WZ 1] all the candidates for isotropy irreducible spaces with G simple. The result is given in Table I and II, except that item 4 in Table II should be deleted. We will actually see that all these spaces are isotropy irreducible.

To proceed further, we have to determine the full isometry group of all the candidates. We first do this for the id-component, and then (in the next section) for the full isometry group.

(2.4) Theorem. Let M = G/H be a simply connected isotropy irreducible homogeneous space which is not strongly isotropy irreducible. Then G_{0} is the idcomponent of the isometry group unless

(a) G/H = [ℝ^{n} ⋉ H]/H is flat with isometry group ℝ^{n} ⋉ O(n).

(b) G/H = [K ⋉ Γ]/Γ with K connected, compact, and simply connected and Γ ⊂ Aut(K) a finite subgroup acting irreducibly on 𝔨. Here the idcomponent of the isometry group is (K × K)/△Z(K) acting via left and right translations on K.

Proof. If M is flat, we are in case (a). If M is not flat, the de Rham reducible case immediately reduces to the de Rham irreducible case by the comments in §1, and by (2.1) we can assume that G is semisimple. By [WZ 1, Theorem 5.1] and the note added in proof in [WZ 1], if G_{0} is a simple Lie group, then G_{0} is the id-component of the isometry group, unless we are in the case of a biinvariant metric on a compact simple Lie group, which is case (b).

If G_{0} is semisimple but not simple, we are in the case of Example 2 of §1 as follows from (2.2). For this example G_{0} = [H × ⋯ × H]/△Z(H) and we can argue as follows. If Ĝ_{0} is the id-component of the full isometry group of M, then Ĝ_{0} is semisimple by (2.1). If Ĝ_{0} is simple, it easily follows from [O,Table VII] that Ĝ_{0} = G_{0}. If Ĝ_{0} is not simple, then Ĝ_{0}/Ĥ_{0} is again as in Example 2 of §1 and hence M is diffeomorphic to L × … L for some compact, simple, simply connected Lie group L with Ĝ_{0} = [L × ⋯ × L]/△Z(L). But from H × ⋯ × H = L × ⋯ × L it easily follows, using rational and ℤ_{2} cohomology rings, that both products contain the same number of factors and that H = L. Hence Ĝ_{0} = G_{0}. □

Remark. If G/H is an irreducible symmetric space, it is well known that G_{0} is the id-component of the isometry group. For strongly isotropy irreducible homogeneous spaces, Wolf showed [Wo, Theorem 7.1] that G_{0} is the id-component of the isometry group, unless G/H = Spin(7)/G_{2} = S^{7} with isometry group O(8) or G/H = G_{2}/ SU(3) = S^{6} with isometry group O(7).

§3. Extensions and Restrictions of Automorphisms.

In this section we let M = G_{0}/H_{0} be a simply connected, effective homogeneous space with G_{0} connected, compact, semisimple, and H_{0} compact. On M we consider the standard normal homogeneous metric g_{B} and we assume furthermore that G_{0} is the id-component of the isometry group of g_{B}. Let Ĝ be the full isometry group of g_{B} with isotropy group Ĥ and hence M = Ĝ/Ĥ. The goal of this section is to determine Ĝ/G_{0} or equivalently Ĥ/H_{0}. Every element of Ĥ gives rise to an automorphism of 𝔤 via the adjoint representation AdĜ of Ĝ, whose kernel is CĜ(G_{0}) = centralizer of G_{0} in Ĝ.

Let Aut(𝔤, 𝔥) = {automorphisms of 𝔤 leaving 𝔥 invariant}

and Int(𝔤, 𝔥) = {inner automorphisms of 𝔤 leaving 𝔥 invariant}.

Then we have:

(3.1) Theorem. Under the above assumptions, the restriction of AdĜ to Ĥ induces isomorphisms

(a) Ĥ ≈ Aut(𝔤, 𝔥) which is also equal to the set of automorphisms of G_{0} leaving H_{0} invariant.

(b) Ĥ/H_{0} ≈ Aut(𝔤, 𝔥)/AdĜ(H_{0}) ≈ Aut(𝔤, 𝔥)/H_{0}

(c) N_{G_{0}}(H_{0})/Z(G_{0}) ≈ Int(𝔤, 𝔥)

(d) N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) ≈ Int(𝔤, 𝔥)/AdĜ(H_{0})

Proof. AdĜ clearly maps Ĥ into Aut(𝔤, 𝔥). The kernel is CĜ(G_{0}) ∩ Ĥ. If g ∈ CĜ(G_{0}) ∩ Ĥ, then g is an isometry of Ĝ/Ĥ fixing the coset (Ĥ), whose differential at (Ĥ) is the identity. Thus g = 1, which shows that AdĜ|Ĥ is injective. To show surjectivity, let φ ∈ Aut(𝔤, 𝔥). Then φ induces an automorphism φ̃ of G̃_{0}, the universal cover of G_{0}, which leaves invariant H̃_{0}, the connected subgroup of G̃_{0} with Lie algebra 𝔥. Hence φ̃ induces an isometry of (G̃_{0}/H̃_{0}, g_{B}). Since G_{0}/H_{0} is simply connected, G̃_{0}/H̃_{0} = G_{0}/H_{0} and so φ̃ also induces an automorphism of G_{0} leaving H_{0} invariant and hence an isometry of (G_{0}/H_{0}, g_{B}) fixing (H_{0}). So φ̃ ∈ Ĥ, which proves (a) and (b). Restricting AdĜ to N_{G_{0}}(H_{0}), the kernel is CĜ(G_{0}) ∩ N_{G_{0}}(H_{0}) = Z(G_{0}) and the image clearly lies in Int(𝔤, 𝔥). Surjectivity is clear in this case, which proves (c) and (d). □

We will determine Ĥ/H_{0} and N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) in all our examples, the second group being necessary to determine the Riemannian subcoverings of G_{0}/H_{0} which are isotropy irreducible. We proceed as follows. Let φ be an automorphism of 𝔥. We must determine all extensions (if any) of φ to automorphisms of 𝔤. Clearly, for the existence of extensions we need only consider outer automorphisms of 𝔥.

To determine the number of extensions modulo Ad_{G}(H_{0}), we have to determine all automorphisms φ of 𝔤 which fix 𝔥. If φ is inner, then φ = Ad(g) with g ∈ C_{G_{0}}(H_{0}). Hence we have to determine C_{G_{0}}(H_{0})/Z(G_{0})Z(H_{0}) as well as those outer automorphisms of 𝔤 that fix 𝔥. For the latter, we only need to determine which elements of Aut(𝔤)/ Int(𝔤) have representatives that fix 𝔥.

If rank 𝔥 = rank 𝔤, all these questions are answered by the following result:

(3.2) Theorem. Under the above assumptions and rank 𝔥 = rank 𝔤 the following hold:

(a) Let φ be an automorphism of 𝔥, which, after composing with a suitable inner automorphism of 𝔥, satifies φ(𝔱) ⊂ 𝔱 and φ^{∗}(Φ) ⊂ Φ, where 𝔱 is a maximal abelian subalgebra of 𝔥 and Φ a fundamental system of roots of (𝔥, 𝔱). Then φ extends to an automorphism of 𝔤 iff it permutes the dominant weights of the Ad_{G}|_{H_{0}} irreducible summands of 𝔪.

(b) Z(G_{0}) = 1 and C_{G_{0}}(H_{0}) = Z(H_{0}). Furthermore any two extensions of an automorphism of 𝔥 are equal mod Ad_{G}(Z(H_{0})). In particular, any two extensions are either both inner or both outer.

(c) If 𝔥 = 𝔥_{s} ⊕𝔷, where 𝔥_{s} is semisimple and 𝔷 abelian, then Ĥ/H_{0} is isomorphic to the subgroup Γ of automorphisms of 𝔥 which are sums of automorphisms of 𝔥_{s} and linear isomorphisms of 𝔷 and which permute the dominant weights of the Ad_{G}|_{H_{0}} irreducible summands of 𝔪. Hence G_{0}/H_{0} is isotropy irreducible as a Riemannian manifold iff Γ in addition acts transitively on these weights.

Proof. Let π_{λ_{i}} be the Ad_{G}|_{H_{0}} irreducible summands of 𝔪 with dominant weight λ_{i}. Since rank 𝔥 = rank 𝔤, 𝔱 is a maximal abelian subalgebra of 𝔤 and hence all the weights of π_{λ_{i}} are roots of 𝔤. This implies that all π_{λ_{i}} are inequivalent representations. Assume that φ permutes the λ_{i}, then it permutes the π_{λ_{i}} and their weights, and hence all the roots of 𝔤. So φ|𝔱 admits an extension to an automorphism σ of 𝔤 which leaves 𝔥 invariant. σ|𝔥 ∘ φ^{−1} fixes 𝔱 ⊂ 𝔥 and hence σ|𝔥 ∘ φ^{−1} = Ad_{x} for some x ∈ T. Therefore φ extends to Ad_{x}^{−1} ∘ σ. The converse clearly holds also and hence (a) is proved.

Z(G_{0}) = 1 since G_{0}/H_{0} is effective and Z(G_{0}) is contained in every maximal torus of G_{0} and hence in T ⊂ H_{0}. Similarly C_{G_{0}}(H_{0}) ⊂ ⋂_{T ⊂H_{0}} T = Z(H_{0}) and hence C_{G_{0}}(H_{0}) = Z(H_{0}). If φ and ψ are two automorphisms of 𝔤 with φ|𝔥 = ψ|𝔥, then φ and ψ agree on every maximal abelian subalgebra of 𝔥 and hence φ ∘ ψ^{−1} = Ad_{x} with x ∈ Z(H_{0}). This proves (b), and (c) follows from (a) and (b). □

Therefore in the equal rank case, the fact that every entry in Table I is isotropy irreducible is almost obvious from the representation χ_{0} of Ad_{G}|_{H_{0}} on 𝔪. In the unequal rank case the following observations will help.

(3.3) Theorem. Under the above assumptions and rank 𝔥 < rank 𝔤 the following hold:

(a) If 𝔥 is a maximal subalgebra in 𝔤, then C_{G_{0}}(H_{0}) = Z(G_{0}) and Z(H_{0}) = 1.

(b) Let G_{0} be simple but not of type D_{4}, and assume that H_{0} is semisimple and that the representation of Ad_{G}|_{H_{0}} on 𝔪 contains no trivial representations. If φ is an outer automorphism of 𝔤 that fixes 𝔥, then there exists some integer q such that φ^{q} is an involutive automorphism with fixed point set 𝔨 ⊇ 𝔥. In particular (𝔤, 𝔨) is a symmetric pair.

(c) If in addition to the assumptions in (b), 𝔥 is a maximal subalgebra in 𝔤 and (𝔤, 𝔥) is not a symmetric pair, then Ĥ/H_{0} is isomorphic to the subgroup of diagram automorphisms of 𝔥 that extend to automorphisms of 𝔤.

Proof. Let g ∈ C_{G_{0}}(H_{0}) and let C_{g} be the centralizer of g in G_{0}. Then H_{0} ⊂ C_{g}^{0} and since g is contained in a maximal torus T of G_{0}, we have g ∈ T ⊂ C_{g}^{0}, and rank C_{g}^{0} = rank G_{0}. The maximality of 𝔥 in 𝔤 now implies C_{g}^{0} = G_{0} since C_{g}^{0} = H_{0} would contradict rank 𝔥 < rank 𝔤. Thus g ∈ Z(G_{0}) and so C_{G_{0}}(H_{0}) = Z(G_{0}). In particular Z(H_{0}) = 1 since G_{0}/H_{0} is effective. This proves (a).

To prove (b), we first observe that the assumptions imply that C_{G_{0}}(H_{0}) is finite. Indeed the Lie algebra of C_{G_{0}}(H_{0}) is equal to 𝔷(𝔥)⊕𝔪_{0}, where 𝔪_{0} is the subalgebra of 𝔪 on which Ad_{G}|_{H_{0}} acts trivially. Now φ induces a diagram automorphism of 𝔤 of order 2 since 𝔤 is simple ≠ D_{4}. Then there exists an outer automorphism σ of 𝔤 with order 2 inducing the same diagram automorphism as φ. A theorem of de Siebenthal [Wo 2, 8.6.9], applied to the group Aut(𝔤), implies that we can find x, g ∈ G_{0} such that φ = Ad_{g} ∘ σ ∘ Ad_{x} ∘ Ad_{g}^{−1} and [σ, Ad_{x}] = 0. Then φ^{2} = Ad_{g} ∘ Ad_{x}^{2} ∘ Ad_{g}^{−1} = Ad_{gx^{2}g^{−1}} is inner. Now y = gx^{2}g^{−1} ∈ C_{G_{0}}(H_{0}) because φ fixes 𝔥 and hence y^{q} = 1 for some integer q. But then φ^{2q} = id, and φ^{q} is an involutive automorphism, which implies (b). (c) follows from (a) and (b). □

§4. The Case of Equal Ranks.

(4.1) Theorem. Let G/H be a simply connected, compact isotropy irreducible homogeneous space which is de Rham irreducible, but not strongly isotropy irreducible. If rank H = rank G, then (G_{0}, H_{0}) can be found in Table I and conversely, every entry in Table I is an isotropy irreducible Riemannian manifold with full isotropy group Ĥ.

Proof. As explained earlier, the entries in Table I and χ_{0} are collected from [WZ 1] and in most cases (3.2) easily implies that G_{0}/H_{0} is isotropy irreducible and also explains the entries for Ĥ/H_{0} and (Ĥ/H_{0})_{min}. The global form of G_{0} and H_{0} (which is not given in [WZ 1]), easily follows in most cases from the fact that Z(G_{0}) = 1 and H_{0} = H̃_{0}/Ker(χ_{0}), where H̃_{0} is the universal cover of H_{0}. The only somewhat tricky case is No. 7, which we discuss seperately. In this case the isotropy representation χ_{0} was listed incorrectly in [WZ 1].

For the entry N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = N_{G_{0}}(H_{0})/H_{0} we only have to determine which outer automorphisms of 𝔥 extend to inner automorphisms of 𝔤. Of course, in most of the cases 𝔤 has no outer automorphisms and hence N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = Ĥ/H_{0}.

We now discuss some of the cases separately and use the numbering in Table I.

Example No.2. S(U(k) × ⋯ × U(k))/ℤ_{nk} ⊂ SU(nk)/ℤ_{nk} (n factors)

It is clear that the outer automorphism on the semisimple part determines what linear isomorphism is needed on the center of 𝔥 in order to permute the Ad_{H_{0}} irreducible summands. The group S_{n} acts as outer automorphisms on 𝔥 by permuting the simple factors of 𝔥. The interchange of the i-th and j-th simple factor extends to an inner automorphism of 𝔤, i.e. conjugation by

( \begin{matrix}I \\ & . & . & . \\ & & & & I \\ & & & & & 0 & & & & & & I \\ & & & & & & I \\ & & & & & & & . & . & . \\ & & & & & & & & & & I \\ & & & & & −I & & & & & & 0 \\ & & & & & & & & & & & & I \\ & & & & & & & & & & & & & . & . & . \\ & & & & & & & & & & & & & & & & I \end{matrix} )

where I is the k × k identity matrix. The diagram automorphism of a single factor does not extend to 𝔤, since it does not preserve χ_{0}, but if we take the product of all diagram automorphisms of all simple factors, then it coincides with X → −X^{t}, which is an outer automorphism of 𝔤. This outer automorphism clearly commutes with S_{n} and hence Ĥ/H_{0} = S_{n} × ℤ_{2} and N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = S_{n}.

Example No.3. [SO(k) × ⋯ × SO(k)] /ℤ_{2} ⊂ SO(nk)/ℤ_{2} (n factors, k even)

This is similar to the previous example, except that the outer automorphism Ad(I_{1,k−1}) of each simple factor (where I_{1,k−1} is the matrix diag{−1, 1, …, 1}) does extend to the outer automorphism Ad (diag{I, …, I_{1,k−1}, …, I}) of 𝔤. Thus the product of an even number of such outer automorphisms of 𝔥 extends to an inner automorphism of 𝔤. Hence Ĥ/H_{0} = (ℤ_{2})^{n} ⋉ S_{n} and N_{G_{0}}(H_{0})/H_{0} = (ℤ_{2})^{n−1} ⋉ S_{n}, where S_{n} acts on (ℤ_{2})^{n} by permuting the generators.

In both No.2 and No.3 (and similarly No.4), it is difficult to determine subgroups of Ĥ/H_{0} of minimal order that still make G_{0}/H_{0} isotropy irreducible. This is equivalent to finding subgroups of S_{n} of minimal order which act transitively on the set of unordered pairs and depends strongly on the value of n.

Example No.6. [Spin(8) × U(1) × U(1)] /(ℤ_{4} × ℤ_{2}) ⊂ E_{6}/ℤ_{3} As we will see, the representation χ_{0} of G_{0}/H_{0} was given incorrectly in [WZ 1]. We first determine the global form of H_{0}. On the level of Lie algebras we have

inclusions 𝔰𝔭𝔦𝔫(8) ⊕ 𝔲(1) ⊕ 𝔲(1) ⊂ 𝔰𝔭𝔦𝔫(10) ⊕ 𝔲(1) ⊂ 𝔢_{6}. Since Z(G_{0}) = 1, the global form of G_{0} is E_{6}/ℤ_{3} and since Z(Spin(10)) = ℤ_{4}, Z(E_{6}) must be contained in the 𝔲(1) factor of 𝔰𝔭𝔦𝔫(10) ⊕ 𝔲(1). The isotropy representation of 𝔰𝔭𝔦𝔫(10) ⊕ 𝔲(1) in 𝔢_{6} is [ ∘—∘—∘<∘^{∘ 1} ⊗ 𝔱^{1} ]_{ℝ} which has kernel △ℤ_{4} ⊂ ℤ_{4} × U(1) = Z(Spin(10)) × U(1) and hence [Spin(10) × U(1)]/(ℤ_{3} × △ℤ_{4}) ⊂ E_{6}/ℤ_{3}, which we can also write as [Spin(10) × U(1)]/△ℤ_{4} ⊂ E_{6}/ℤ_{3} since ℤ_{3} is contained in U(1).

Next we claim that the global form of 𝔰𝔬(8) ⊕ 𝔲(1) in 𝔰𝔭𝔦𝔫(10) is [Spin(8) × U(1)]/△ℤ_{2} ⊂ Spin(10). Indeed, if we lift the inclusion SO(8) × SO(2) ↪ SO(10) to a homomorphism Spin(8) × U(1) → Spin(10), it has kernel (−1, −1). Furthermore, the generator of Z(Spin(10)) = ℤ_{4} is equal to (e, i), where e is the kernel of one of the spin representations of Spin(8). Hence

H_{0} = {[Spin(8) × U(1)] /△ℤ_{2} × U(1)} /△ℤ_{4} = [Spin(8) × U(1) × U(1)]/(ℤ_{2} × ℤ_{4})

where ℤ_{2} is generated by (−1, −1, 1) and ℤ_{4} is generated by (e, i, i).

To determine the isotropy representation χ_{0}, we first observe that the isotropy representation of SO(8) × SO(2) ⊂ SO(10) is given by [ ∘^{1}—∘<∘ ∘ ⊗ 𝔱^{1} ]_{ℝ} and since U(1) → SO(2) is the double cover, the isotropy representation of [Spin(8)×U(1)]/△ℤ_{2} ↪ Spin(10) is equal to [∘^{1}—∘<∘ ∘ ⊗ 𝔱^{2} ]_{ℝ}. Furthermore the representation ∘—∘—∘<∘^{∘ 1} of 𝔰𝔭𝔦𝔫(10) restricted to 𝔰𝔭𝔦𝔫(8) ⊕ 𝔲(1) becomes [ ∘—∘<∘^{∘ 1} ⊗ 𝔱^{1} ]_{ℝ} ⊕ [ ∘—∘<∘_{∘1} ⊗ 𝔱^{−1} ]_{ℝ}. Hence χ_{0} is equal to

[∘^{1}—∘<∘ ∘ ⊗ 𝔱^{2} ⊗ 𝔱 ]_{ℝ} ⊕ [∘—∘<∘^{∘ 1} ⊗ 𝔱^{1} ⊗ 𝔱^{1} ]_{ℝ} ⊕ [∘—∘<∘_{∘ 1} ⊗ 𝔱^{−1} ⊗ 𝔱^{1} ]_{ℝ}

In [WZ 1] the 2 over 𝔱 was mistakenly replaced by 1.

We now check that G_{0}/H_{0} is isotropy irreducible. If φ is an outer automorphism of 𝔥 such that φ|_{𝔰𝔭𝔦𝔫(8)} is the triality automorphism, we need a linear isomorphism on 𝔱 ⊕ 𝔱 whose transpose takes (2, 0) to ±(1, 1), (1, 1) to ±(−1, 1), and (−1, 1) to ±(2, 0). One easily checks that the only possibilities are ±\frac{1}{2} ( \begin{matrix}1 & −3 \\ 1 & 1 \end{matrix} ) which takes (2, 0) to ±(1, 1), (1, 1) to ±(−1, 1), and (−1, 1) to ∓(2, 0). These φ_{±} extend to automorphisms φ̃_{±} of 𝔢_{6}. Hence G_{0}/H_{0} is isotropy irreducible as a Riemannian manifold.

Next, if τ is an outer automorphism of 𝔥 such that τ|_{𝔰𝔭𝔦𝔫(8)} is the order 2 diagram automorphism of 𝔰𝔭𝔦𝔫(8), then τ|_{𝔱⊕𝔱} must take (2, 0) to ±(2, 0), (−1, 1) to ±(1, 1),

and (1, 1) to ±(−1, 1). The only possible choices are ± ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ). These τ_{±} again extend to automorphisms τ̃_{±} of 𝔢_{6}.

To determine the group Ĥ/H_{0}, one easily checks the following identities mod Ad_{H_{0}}:

φ̃_{+}^{6} = 1, φ̃_{−}^{3} = 1, τ̃_{±}^{2} = 1, φ̃_{+}^{4} = φ̃_{−}, τ̃_{+}φ̃_{+}^{3} = φ̃_{+}^{3}τ̃_{+} = τ̃_{−}, τ̃_{+}φ̃_{+}^{2}τ̃_{+}^{−1} = φ̃_{+}^{−2}

by verifying them on 𝔰𝔭𝔦𝔫(8) and 𝔱 ⊕ 𝔱 separately. Hence Ĥ/H_{0} is isomorphic to S_{3} × ℤ_{2}, with S_{3} generated by φ̃_{+}^{2} and τ̃_{+}, and ℤ_{2} generated by φ̃_{+}^{3}. Notice that φ̃_{+}^{3} acts as id on 𝔰𝔭𝔦𝔫(8) and as −id on 𝔱 ⊕ 𝔱.

To determine N_{G_{0}}(H_{0})/H_{0}Z(G_{0}), we examine which one of these automorphisms extend to inner automorphisms of 𝔢_{6}. First observe that φ̃_{+}^{2} must be inner since the group of diagram automorphisms of 𝔢_{6} has order 2. Next observe that φ̃_{+}^{3} and τ̃_{+} must be outer, since they act non-trivially on the center of 𝔢_{6}, which is generated by (1, 1, e^{2πi/3}) in [Spin(8) × U(1) × U(1)]/(ℤ_{4} × ℤ_{2}). Hence N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) is isomorphic to S_{3}, sitting diagonally in S_{3} × ℤ_{2}, and is generated by φ̃_{+}^{2} and τ̃_{+}φ̃_{+}^{3}.

Examples 8 and 12 were discussed in §1 and hence this finishes the proof of (4.1). □

Remark. To determine the group structure of Ĝ, we observe that since Z(G_{0}) = 1, every extension of G_{0} is determined by the character Ĝ/G_{0} → Aut(G_{0})/ Int(G_{0}) and since there exists a lift Aut(G_{0})/ Int(G_{0}) → Aut(G_{0}), Ĝ is the semidirect product of G_{0} with Ĝ/G_{0} = Ĥ/H_{0}. In most cases G_{0} has no outer automorphisms and hence Ĝ = G_{0} × (Ĥ/H_{0}). In the remaining 4 cases the discussion of the examples usually contains an explicit lift Ĥ/H_{0} → Aut(G_{0}) which describes the semidirect product structure of Ĝ.

§5. The Case of Unequal Ranks.

(5.1) Theorem. Let G/H be a simply connected, compact isotropy irreducible homogeneous space which is de Rham irreducible, but not strongly isotropy irreducible. If rank H < rank G, then (G_{0}, H_{0}) can be found in Table II and conversely every entry in Table II is an isotropy irreducible Riemannian manifold with full isotropy group Ĥ, with the exception of No.5.

Case No.5 is isotropy irreducible as a Riemannian manifold, but G_{0} is not the id-component of the isometry group. See §1 Example 1 for a discussion of this case.

Proof. As explained earlier, the entries in Table II are obtained from [WZ 1] and (2.2). We will now check for each entry that G_{0}/H_{0} is isotropy irreducible as a Riemannian manifold. We also need to determine the global form of G_{0} and H_{0} as well as the groups Ĥ/H_{0} and N_{G_{0}}(H_{0})/H_{0}Z(G_{0}). We will use the numbering in Table II.

Example No.1. SO(k) × ⋯ × SO(k) ⊂ SO(nk) (n factors, k odd)

The diagram automorphisms which interchange two simple factors extend to inner automorphisms of 𝔤 as in the corresponding equal rank case. But they do not extend uniquely. Indeed C_{G_{0}}(H_{0}) = ℤ_{2}^{n−1} since C_{G_{0}}(H_{0}) ⊂ S(O(k)×⋯×O(k)) by Schur’s lemma. If n is odd, then Z(G_{0}) = 1 and G_{0} has no outer automorphisms. Hence Ĥ/H_{0} = N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = (ℤ_{2})^{n−1} ⋉ S_{n}.

If n is even, then Z(G_{0}) = ℤ_{2} and hence C_{G_{0}}(H_{0})/Z(G_{0}) = ℤ_{2}^{n−2}, which implies that N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = ℤ_{2}^{n−2} ⋉ S_{n}. But now G_{0} has, modulo Int(𝔤), exactly one outer automorphism and we have to determine whether a representative of it fixes 𝔥. This is the case since Ad(diag{−I_{k}, I_{k}, …, I_{k}}) is outer and fixes 𝔥. Hence we have again Ĥ/H_{0} = (ℤ_{2})^{n−1} ⋉ S_{n}. Notice that in both cases, n even and n odd, (ℤ_{2})^{n−1} consists of Ad(g) with g = (±I_{k}, …, ±I_{k}) and hence S_{n} acts on (ℤ_{2})^{n−1} = (ℤ_{2})^{n}/ℤ_{2} via permutations on ℤ_{2}^{n}.

Example No.2. 𝔰𝔬(n) ⊕ 𝔰𝔬(n) ⊂ 𝔰𝔬(n^{2})

We first describe the global form of G_{0} and H_{0}. The homomorphism SO(n) × SO(n) → SO(n^{2}) given by the tensor product representation is injective if n is odd and has kernel ℤ_{2} = {±(1, 1)} if n is even. To determine if the quotient is simply connected, we observe that SO(n) × {1} ⊂ SO(n) × SO(n) → SO(n^{2}) is just the diagonal embedding of SO(n) into SO(n^{2}): A → (A, …, A). Since A → (A, 1, …, 1) is always injective on π_{1}, it follows that if n is odd, SO(n^{2})/[SO(n) × SO(n)] is simply connected and effective.

If n is even, say n = 2k, then the fundamental group of each SO(n) factor goes to 0 in SO(n^{2}). But [SO(n) × SO(n)]/ℤ_{2} has a third generator in π_{1}, represented by σ: S^{1} → [SO(n) × SO(n)]/ℤ_{2} where σ(θ) = (R(θ), …, R(θ)) × (R(θ), …, R(θ)), 0 ≤ θ ≤ π, and R(θ) is a 2 × 2 rotation of angle θ. Going into SO(n^{2}) this becomes (R(2θ), …, R(2θ), 1, …, 1) with k^{2} rotations R(2θ). This loop in SO(n^{2}) is therefore non-trivial iff k is odd. So if k is odd, i.e. n ≡ 2 (mod 4), SO(n^{2})/[SO(n) × SO(n)]/ℤ_{2}) is simply connected. To make it effective one divides out by Z(SO(n^{2})) = ℤ_{2} to obtain G_{0} = SO(n^{2})/ℤ_{2} and H_{0} = (SO(n^{2})/ℤ_{2}) × (SO(n^{2})/ℤ_{2}).

If k is even, i.e. n ≡ 0 (mod 4), this space has π_{1} = ℤ_{2}. To obtain the simply connected space, consider the lift i^{∗}: [SO(n) × SO(n)]/ℤ_{2} → Spin(n^{2}) of the inclusion i: [SO(n) × SO(n)]/ℤ_{2} → SO(n^{2}). The lift i^{∗} is injective since i is. Furthermore (−I, I) ∈ [SO(n) × SO(n)]/ℤ_{2} maps to the center of Spin(n^{2}) under i^{∗} since i((−I, I)) = −I. Therefore Spin(n^{2})/i^{∗}([SO(n) × SO(n)]/ℤ_{2}) is simply connected, but not effective. The effective version is (Spin(n^{2})/ℤ_{2})/[(SO(n)/ℤ_{2}) × (SO(n)/ℤ_{2})]. Note that Spin(n^{2})/ℤ_{2} is not equal to SO(n^{2}).

We proceed to determine Ĥ/H_{0}. We first observe that any automorphism of 𝔥 which extends to 𝔤 does so uniquely by (3.3)(c). Next consider the outer automorphism φ of 𝔥 which interchanges the simple factors. Let σ: ℝ^{n} ⊗ ℝ^{n} → ℝ^{n} ⊗ ℝ^{n} be given by σ(u⊗v) = v ⊗u. Then σ ∈ O(ℝ^{n^{2}}) and one checks easily that Ad_{σ}| _{𝔥} = φ. Since ℝ^{n} ⊗ ℝ^{n} = ∧^{2}ℝ^{n} ⊕ S^{2}ℝ^{n} and since σ|_{∧^{2}ℝ^{n}} = −id and σ|_{S^{2}ℝ^{n}} = id, it follows that the fixed point set of Ad_{σ} is SO(∧^{2}ℝ^{n}) × SO(S^{2}ℝ^{n}). Since an involutive automorphism is inner iff the fixed point group has equal rank, Ad_{σ} is outer iff n ≡ 2 (mod 4). If n is even, we must also consider the order 2 automorphism of each simple factor. One checks that the involution corresponding to the first simple factor is induced by Ad_{τ}, where τ: ℝ^{n} ⊗ ℝ^{n} → ℝ^{n} ⊗ ℝ^{n} is given by τ(u ⊗ v) = I_{1,n−1}(u) ⊗ v. Clearly τ = I_{n,n^{2}−n} and hence Ad_{τ} is inner. Furthermore Ad_{στσ} induces the order 2 automorphism of the second simple factor. Hence Ĥ/H_{0} = N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = ℤ_{2} if n is odd. If n is even Ĥ/H_{0} = D_{8}, the dihedral group of order 8, and N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = D_{8} if n ≡ 0 (mod 4) and = ℤ_{2} × ℤ_{2} if n ≡ 2 (mod 4). In all cases the group (Ĥ/H_{0})_{min} is equal to ℤ_{2} and given by Ad_{σ}.

Example No.3 is similar and is left to the reader.

Example No.4. △(K/Z(K)) ⊂ (K × ⋯ × K)/△Z(K)

If g = (g_{1}, …, g_{n}) ∈ N_{G_{0}}(H_{0}), then g_{1}^{−1}g_{i} ∈ Z(K) for all i and hence g ∈ H_{0}Z(G_{0}), i.e. N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = 1.

Secondly, any outer automorphism A of 𝔥 extends to the outer automorphism (A, …, A) and one easily sees by (3.3) that, modulo Ad_{H_{0}}, any outer automorphism of 𝔤 that fixes 𝔥 is an outer automorphism that permutes the simple factors of 𝔤. Hence Ĥ/H_{0} = S_{n} × D, where D is the group of diagram automorphisms of 𝔨.

Example No.5 was already discussed in §1.

Example No.6. G_{2} ⊂ Spin(8)

Since G_{2} has no outer automorphisms, to determine N_{G_{0}}(H_{0})/H_{0}Z(G_{0}), we only need to compute C_{G_{0}}(H_{0}). But if g ∈ C_{G_{0}}(H_{0}) and g ∉ Z(G_{0}), then g and H_{0} are contained in C_{g}^{0}, and hence in a connected maximal subgroup of maximal rank in Spin(8). But the only such groups have Lie algebras 𝔲(4), 𝔰𝔬(2) ⊕ 𝔰𝔬(6), or 𝔰𝔬(4) ⊕ 𝔰𝔬(4), which do not contain 𝔤_{2}. Thus C_{G_{0}}(H_{0}) = Z(G_{0}) and hence N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = 1.

The triality automorphism of Spin(8) has fixed point set G_{2} and there exists an order 2 outer automorphism of Spin(8) which has fixed point set Spin(7) ⊃G_{2}. Hence Ĥ/H_{0} = S_{3} and (Ĥ/H_{0})_{min} = ℤ_{3}.

Example No. 7. 𝔰𝔬(8) ⊂ 𝔰𝔲(8) ⊂ 𝔢_{7}

The understanding of this case rests upon a description of E_{7} due to E.Cartan. Let < , > denote the negative of the Killing form of 𝔥 = 𝔰𝔬(8) as well as its extension to a hermitian form on 𝔥_{ℂ}. On the 56-dimensional space 𝔥_{ℂ} ⊕ 𝔥_{ℂ} there is a skewsymmetric form ω as well as the hermitian form < , > ⊥ < , > which we again denote by < , >. Then Sp(𝔥_{ℂ}⊕ 𝔥_{ℂ}, ω) ∩ SU(𝔥_{ℂ}⊕ 𝔥_{ℂ}, < , >) is a compact connected Lie group isomorphic to Sp(28). In the following we will refer to this Lie group as Sp(28). As usual we describe the elements of Sp(28) as matrices of the form ( \begin{matrix}A & B \\ −B̄ & Ā \end{matrix} ) with AA^{∗} + BB^{∗} = I and AB^{t} = BA^{t}. In particular the subgroup

SU(28) ⊂ Sp(28) has the form ( \begin{matrix}A & 0 \\ 0 & Ā \end{matrix} ).

On 𝔥_{ℂ} ⊕ 𝔥_{ℂ} we have the following homogeneous polynomial of degree 4:

J(X, Y) = Pf(X)+Pf(Y)−\frac{1}{4} tr(XY XY)+\frac{1}{16} (tr(XY))^{2} , X, Y ∈ 𝔥_{ℂ} = 𝔰𝔬(8, ℂ)

where Pf(X) denotes the Pfaffian. E.Cartan has shown that E_{7} is the largest connected subgroup of Sp(28) that leaves J invariant (see [C 2, 143–144] and [Fr 1,2] ). We first observe that the invariance group of J is actually connected:

(5.2) Lemma. E_{7} is the largest subgroup of Sp(28) leaving J invariant.

Proof. Let L be the invariance group of J. Then it follows from the result of Cartan that E_{7} ⊂ L ⊂ Sp(28). But Sp(28)/E_{7} is strongly isotropy irreducible (see [Wo 1]) and hence 𝔢_{7} is a maximal subalgebra in 𝔰𝔭(28). This implies L ⊂ N_{Sp(28)}(E_{7}). Since E_{7} has no outer automorphisms N_{Sp(28)}(E_{7})/E_{7} = C_{Sp(28)}(E_{7})/Z(E_{7}) and the argument in (3.3)(a) implies C_{Sp(28)}(E_{7}) = Z(Sp(28)) = Z(E_{7}) (Notice that Sp(28)/E_{7} is not effective). Hence L = E_{7}. □

With this description of E_{7}, SO(8)/ℤ_{2} has a natural embedding into it given by the composite

SO(8)/ℤ_{2} →^{Ad} SO(𝔥, < , >) ⊂ SU(𝔥_{ℂ}, < , >) = SU(28) ⊂ Sp(28)

because A ∈ SO(8) is mapped to the transformation of 𝔥_{ℂ} ⊕ 𝔥_{ℂ} given by (X, Y) → (AXA^{t}, AY A^{t}), which clearly leaves J invariant. Thus E_{7}/(SO(8)/ℤ_{2}) is simply connected and effective.

We can also easily describe SU(8)/ℤ_{2} ⊂ E_{7} with SO(8)/ℤ_{2} ⊂ SU(8)/ℤ_{2}. Indeed SU(8)/ℤ_{2} ⊂ SU(𝔥_{ℂ}, < , >) is given by A → (X → AXA^{t}) and hence A ∈ SU(8)/ℤ_{2} acts on 𝔥_{ℂ} ⊕ 𝔥_{ℂ} via the embedding SU(𝔥_{ℂ}) ⊂ Sp(28) as (X, Y) → (AXA^{t}, ĀY Ā^{t}) and this clearly leaves J invariant and contains SO(8)/ℤ_{2}.

Since E_{7} has no outer automorphisms, Ĥ/H_{0} = N_{G_{0}}(H_{0})/H_{0}Z(G_{0}). An order 2 outer automorphism of SO(8)/ℤ_{2} is given by Ad(I_{7,1}), which extends to the inner automorphism Ad(diag{ξ, …, ξ, −ξ}), ξ = e^{πi/8}, of SU(8)/ℤ_{2} and hence to an inner automorphism of E_{7}.

To see that the triality automorphism τ of 𝔥 extends to an inner automorphism of 𝔢_{7} will be more complicated and is discussed in the remainder of this section.

First we observe that τ extends to an inner automorphism of 𝔰𝔬(𝔥). In fact τ ∈ SO(𝔥) since τ^{3} = 1 implies det τ = 1. The embedding of 𝔥 into 𝔰𝔬(𝔥) is given by X → ad_{X} and since τ ∘ ad_{X} ∘ τ^{−1} = ad_{τ(X)} it follows that Ad(τ), τ ∈ SO(𝔥),

induces τ on 𝔥. τ ∈ SO(𝔥) goes into T = ( \begin{matrix}τ & 0 \\ 0 & τ \end{matrix} ) ∈ Sp(28) and hence the automorphism τ of 𝔥 extends to the inner automorphism Ad_{T} of 𝔰𝔭(28). We will see that T ∉ E_{7}, but that it will be in E_{7} if we modify T with an element of C_{Sp(28)}(SO(8)/ℤ_{2}). In fact τ has an extension to an inner automorphism of 𝔢_{7} iff T Λ ∈ E_{7} for some Λ ∈ C_{Sp(28)}(SO(8)/ℤ_{2}). Furthermore any two extensions of τ differ by an element of C_{E_{7}}(SO(8)/ℤ_{2}).

(5.3) Lemma.

(a) C_{Sp(28)}(SO(8)/ℤ_{2}) = \{( \begin{matrix}λI & μI \\ −μ̄I & λ̄I \end{matrix} ) | |λ|^{2} + |μ|^{2} = 1, λ, μ ∈ ℂ\} ≈ Sp(1).

(b) C_{E_{7}}(SO(8)/ℤ_{2}) is the quaternion subgroup of Sp(1) of order 8. The center of E_{7} corresponds to the center of the quaternion subgroup and hence C_{E_{7}}(SO(8)/ℤ_{2})/Z(E_{7}) = ℤ_{2} × ℤ_{2}.

Proof. If ( \begin{matrix}A & B \\ −B̄ & Ā \end{matrix} ) commutes with ( \begin{matrix}Ad(C) & 0 \\ 0 & Ad(C) \end{matrix} ) then A and B commute with Ad(C) and hence A = λI and B = μI, which implies (a).

To prove (b), let g = ( \begin{matrix}λI & μI \\ −μ̄I & λ̄I \end{matrix} ) ∈ C_{E_{7}}(SO(8)/ℤ_{2}). Then g maps 𝔱_{ℂ} ⊕ 𝔱_{ℂ} ⊂ 𝔥_{ℂ} ⊕ 𝔥_{ℂ} to itself and leaves J invariant. Let

X = ( \begin{matrix}0 & x_{1} \\ −x_{1} & 0 \\ & & . & . & . \\ & & & & & 0 & x_{4} \\ & & & & & −x_{4} & 0 \end{matrix} ) and Y = ( \begin{matrix}0 & y_{1} \\ −y_{1} & 0 \\ & & . & . & . \\ & & & & & 0 & y_{4} \\ & & & & & −y_{4} & 0 \end{matrix} )

and notice that

J(X, Y) = x_{1}x_{2}x_{3}x_{4} + y_{1}y_{2}y_{3}y_{4} − \frac{1}{2}(x_{1}^{2}y_{1}^{2} + x_{2}^{2}y_{2}^{2} + x_{3}^{2}y_{3}^{2} + x_{4}^{2}y_{4}^{2})

+ \frac{1}{4}(x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} + x_{4}y_{4})^{2}.

Comparing coefficients of the monomials x_{1}x_{2}x_{3}x_{4} and x_{1}^{2}y_{1}^{2} in J(X, Y) = J(g(X, Y)) = J(λX + μY, −μ̄X + λ̄Y), we obtain λ^{4} + μ̄^{4} = 1 and |λ|^{4} + |μ|^{4} − 4|λ|^{2}|μ|^{2} = 1. Since |λ|^{2} + |μ|^{2} = 1, the only solutions are λ = 0 and μ^{4} = 1, or μ = 0 and λ^{4} = 1. Now one easily checks that all possibilities actually leave J invariant and together they form the quaternion group of order 8. □

To proceed further, we need the following identities involving the triality automorphism:

(5.4) Lemma. Let X, Y ∈ 𝔥_{ℂ}. Then we have:

(a) Pf(τ(X)) = −\frac{1}{2} Pf(X) − \frac{1}{16} tr(X^{4}) + \frac{1}{64} (tr(X^{2}))^{2}

(b) tr(τ(X)^{4}) = 12 Pf(X) + \frac{3}{8}(tr(X^{2}))^{2} − \frac{1}{2} tr(X^{4})

(c) tr(τ(X)τ(Y)τ(X)τ(Y)) − tr(τ(X)^{2}τ(Y)^{2}) = tr(XY XY) − tr(X^{2}Y^{2})

(d) tr(X^{4}) + tr(τ(X)^{4}) + tr([τ^{2}(X)]^{4}) = \frac{3}{4} (tr(X^{2}))^{2}

(e) tr(τ(X)^{2}τ(Y)^{2}) + tr([τ^{2}(X)]^{2}[τ^{2}(Y)]^{2}) = \frac{1}{4} tr(X^{2}) tr(Y^{2}) + \frac{1}{2} tr((XY)^{2}) − tr(XY XY)

Proof. Let X, Y ∈ 𝔥 = 𝔰𝔬(8). (a) and (b) are equations in a single variable X , and since each entry in the equations is Ad_{SO(8)} invariant, it is sufficient to check them on a maximal abelian subalgebra 𝔱 ⊂ 𝔰𝔬(8), which is easily done. Since both equations are polynomials in the entries of X and Y, they also hold for X, Y ∈ 𝔥_{ℂ}.

To prove (c) we use the fact that τ is an automorphism of 𝔥_{ℂ} and that the Killing form is invariant under τ. Hence tr([τ(X), τ(Y)]^{2}) = tr(τ([X, Y])^{2}) = tr([X, Y]^{2}). Writing this out using [X, Y] = XY − Y X, we obtain (c).

To prove (d), we observe that (a) and (b) imply

tr(τ(X)^{4}) + tr(τ^{2}(X)^{4}) = 12 Pf(X) + \frac{3}{8}(tr(X^{2}))^{2} − \frac{1}{2} tr(X^{4})

+ 12 Pf(τ(X)) + \frac{3}{8}(tr(τ(X)^{2}))^{2} − \frac{1}{2} tr(τ(X)^{4})

= − tr(X^{4}) + \frac{3}{4}(tr(X^{2}))^{2}.

Finally, (e) is obtained from (d) by polarization. First note that

tr((X + Y)^{4}) + tr((X − Y)^{4}) = 2 tr(X^{4}) + 2 tr(Y^{4}) + 4 tr(XY XY) + 8 tr(X^{2}Y^{2}).

Using this equation to expand the left hand and right hand side of

tr {(X + Y)^{4} + [τ(X + Y)]^{4} + [τ^{2}(X + Y)]^{4}}
+ tr {(X − Y)^{4} + [τ(X − Y)]^{4} + [τ^{2}(X − Y)]^{4}}
=\frac{3}{4}\{tr((X + Y)^{2})\}^{2} + \frac{3}{4}\{tr((X − Y)^{2})\}^{2}

we obtain (e). □

We now prove that τ extends to an automorphism of 𝔢_{7}. Let Λ = ( \begin{matrix}λI & μI \\ −μ̄I & λ̄I \end{matrix} ) ∈ C_{Sp(28)}(SO(8)/ℤ_{2}). We will try to find λ, μ such that T Λ ∈ E_{7}. This will be satisfied if

J(X, Y) = J(λτ(X) + μτ(Y), −μ̄τ(X) + λ̄τ(Y)).

Since T Λ maps 𝔱_{ℂ} ⊕ 𝔱_{ℂ} to itself, we may compare coefficients of monomials on the left and right hand side for (X, Y) ∈ 𝔱_{ℂ} ⊕ 𝔱_{ℂ}. Doing this for x_{1}^{2}y_{1}^{2}, we get λ^{2}μ^{2}+λ̄^{2}μ̄^{2}+2|λ|^{2}|μ|^{2} = 1, while comparing the coefficients for x_{1}x_{2}x_{3}x_{4} we obtain λ^{4} + μ̄^{4} + 6λ^{2}μ̄^{2} = −2. Since |λ|^{2} + |μ|^{2} = 1 also, we easily see that λ^{4} = −1/4 and either λ = μ̄ or λ = −μ̄. So a necessary condition is Λ = ( \begin{matrix}λI & λ̄I \\ −λI & λ̄I \end{matrix} ) or

Λ = ( \begin{matrix}λI & −λ̄I \\ λI & λ̄I \end{matrix} ) with λ^{4} = −1/4. We may focus on the first possibility as the

two differ by ( \begin{matrix}0 & −I \\ I & 0 \end{matrix} ), which lies in C_{E_{7}}(SO(8)/ℤ_{2}). We will show that with this choice of Λ, T Λ ∈ E_{7}.

Let A = λ̄Y and B = λX. Then

J(λτ(X) + λ̄τ(Y), −λτ(X) + λ̄τ(Y))
= J(τ(A + B), τ(A − B))
= Pf(τ(A + B)) + Pf(τ(A − B))

− \frac{1}{4} tr(τ(A + B)τ(A − B)τ(A + B)τ(A − B)) + \frac{1}{16} [tr(τ(A + B)τ(A − B))]^{2}

= \frac{1}{12} tr(τ^{2}(A + B)^{4}) + \frac{1}{24} tr(τ(A + B)^{4}) − \frac{1}{32} (tr(τ(A + B)^{2}))^{2} +

\frac{1}{12} tr(τ^{2}(A − B)^{4}) + \frac{1}{24} tr(τ(A − B)^{4}) − \frac{1}{32} (tr(τ(A − B)^{2}))^{2}

− \frac{1}{4} tr((A + B)(A − B)(A + B)(A − B)) + \frac{1}{4} tr((A + B)^{2}(A − B)^{2})

− \frac{1}{4} tr(τ(A + B)^{2}τ(A − B)^{2}) + \frac{1}{16} (tr(A^{2} − B^{2}))^{2}

= − \frac{1}{24} tr(τ(A + B)^{4}) − \frac{1}{24} tr(τ(A − B)^{4}) + \frac{1}{16} (tr ((A + B)^{2}))^{2} + \frac{1}{16} (tr ((A − B)^{2}))^{2}

− \frac{1}{12} tr((A + B)^{4}) − \frac{1}{12} tr((A − B)^{4}) − \frac{1}{32} (tr((A + B)^{2}))^{2} − \frac{1}{32} (tr((A − B)^{2}))^{2}

− tr(ABAB) + tr(A^{2}B^{2}) − \frac{1}{4} tr(τ(A + B)^{2}τ(A − B)^{2}) + \frac{1}{16} (tr(A^{2} − B^{2}))^{2}

= − \frac{1}{12}[ tr(τ(A)^{4}) + tr(τ(B)^{4}) + 2 tr(τ(A)τ(B)τ(A)τ(B)) + 4 tr(τ(A)^{2}τ(B)^{2})]

− \frac{1}{6}[ tr(A^{4}) + tr(B^{4}) + 2 tr(ABAB) + 4 tr(A^{2}B^{2})]

+ \frac{1}{16} [(tr(A^{2}))^{2} + (tr(B^{2}))^{2} + 2 tr(A^{2}) tr(B^{2}) + 4 (tr(AB))^{2}]

− tr(ABAB) + tr(A^{2}B^{2}) − \frac{1}{4} tr(τ(A + B)^{2}τ(A − B)^{2}) + \frac{1}{16} (tr(A^{2} − B^{2}))^{2}

= − \frac{1}{3}[ tr(τ(A)^{4}) + tr(τ(B)^{4}) − tr(τ(A)τ(B)τ(A)τ(B)) + tr(τ(A)^{2}τ(B)^{2})]

− \frac{1}{6}[ tr(A^{4}) + tr(B^{4}) + 2 tr(ABAB)] + \frac{1}{3} tr(A^{2}B^{2})

+ \frac{1}{8} [(tr(A^{2}))^{2} + (tr(B^{2}))^{2}] + \frac{1}{4} (tr(AB))^{2} − tr(ABAB)

= − \frac{1}{6}[2 tr(τ(A)^{4}) + tr(A^{4}) + 2 tr(τ(B)^{4}) + tr(B^{4})]

+ \frac{1}{8} (tr(A^{2}))^{2} + \frac{1}{8} (tr(B^{2}))^{2} + \frac{1}{4} (tr(AB))^{2} − tr(ABAB)

= − \frac{1}{6} [24 Pf(A) + \frac{3}{4} (tr(A^{2}))^{2} + 24 Pf(B) + \frac{3}{4} (tr(B^{2}))^{2}]

+ \frac{1}{8} (tr(A^{2}))^{2} + \frac{1}{8} (tr(B^{2}))^{2} + \frac{1}{4} (tr(AB))^{2} − tr(ABAB)

= Pf(X) + Pf(Y) + \frac{1}{16} (tr(XY))^{2} − \frac{1}{4} tr(XY XY)
= J(X, Y).

Hence T Λ ∈ E_{7}.

We have now showed that all outer automorphisms of 𝔥 extend to inner automorphisms of 𝔢_{7}. Each one has four extensions since we showed in (5.4)(b) that C_{G_{0}}(H_{0})/Z(G_{0}) = ℤ_{2} × ℤ_{2}. Hence Ĥ/H_{0} = (ℤ_{2} × ℤ_{2}) ⋉ S_{3}. To determine the group structure, choose the generators α = ± ( \begin{matrix}0 & I \\ −I & 0 \end{matrix} ) and β = ± ( \begin{matrix}0 & iI \\ iI & 0 \end{matrix} ) of C_{G_{0}}(H_{0}). If τ ∈ S_{3} represents the triality automorphism and σ ∈ S_{3} represents the order 2 automorphism, then one easily verifies, by a computation with the corresponding matrices in Sp(28), that

τατ^{−1} = αβ , τβτ^{−1} = α , σασ^{−1} = β , σβσ^{−1} = α.

(Ĥ/H_{0})_{min} is of course ℤ_{3} ⊂ S_{3}.

This finishes this example and hence the proof of (5.1). □

Remark. To determine the group structure of Ĝ, we observe that the discussion of each example contains an explicit map Ĥ/H_{0} → Aut(G_{0}) and hence Ĝ is the semidirect product of G_{0} with Ĥ/H_{0}.

§6. Isotropy Irreducible Subcoverings and Applications.

In this section we describe how the general isotropy irreducible homogeneous space M^{∗} = G^{∗}/H^{∗} can be constructed from Table I and II. As explained in §1, the transitive action of G^{∗} lifts to an effective transitive action of G on the universal cover M of M^{∗} such that G is a covering of G^{∗}, and M becomes isotropy irreducible under G. If M is compact and de Rham irreducible but not strongly isotropy irreducible, then (G_{0}, H_{0}) must appear in Table I or II.

If M is non-compact, it follows from (2.1) that M has no isotropy irreducible subcoverings.

If M^{∗} is compact and flat, then M^{∗} = ℝ^{n}/Γ for some lattice Γ and G_{0}^{∗} = ℝ^{n}, H_{0}^{∗} = 1. The full isometry group of ℝ^{n}/Γ is (ℝ^{n}/Γ) ⋉ N_{O(n)}(Γ) and hence M^{∗} is isotropy irreducible iff N_{O(n)}(Γ) acts irreducibly on ℝ^{n}. We do not try to enumerate such examples.

(6.1) Theorem. Let M^{∗} = G^{∗}/H^{∗} be a compact, non-flat, isotropy irreducible homogeneous space .

(a) If the universal cover of M^{∗} is isometric to a compact simply connected Lie group K with G_{0} = K and H_{0} = 1, then there exists a central subgroup N and a finite subgroup Γ of K containing N and such that M^{∗} = K/Γ = (K/N)/(Γ/N). Furthermore G^{∗} = K · Φ, H^{∗} = Φ for some finite subgroup Φ of Aut(K) such that Φ acts irreducibly on 𝔨, Ad(Γ) ⊆ Φ, and Φ leaves Γ invariant.

(b) If G_{0} is the id-component of the full isometry group, then there exists a central subgroup N ⊆ Z(G_{0}), compact subgroups P and L satisfying H_{0} × N ⊆ P ⊆ N_{G_{0}}(H_{0}), H_{0} ⊆ P/N ⊆ L ⊆ Ĥ (with Ĥ the full isotropy group), such that Ad_{L} acts irreducibly on 𝔪 and leaves N and P invariant, i.e. P/N is normal in L. Furthermore, G^{∗} = (G_{0}/N) ⋉ (L/{P/N}), H^{∗} = (P/N) ⋉ (L/{P/N}) ≃ L, M^{∗} = G^{∗}/H^{∗} = G_{0}/P = (G_{0}/N)/(P/N) and π_{1}(M^{∗}) ≃ P/H_{0}.

Proof. Let M be the universal cover of M^{∗} as above with deck group D and transitive group action G. Since by construction all isometries in G project to isometries in G^{∗}, D must be normal in G and hence D∩G_{0} = N is a central subgroup of G_{0} with G_{0}^{∗} = G_{0}/N. Let P^{∗} be such that M^{∗} = G_{0}^{∗}/P^{∗} and let π: G_{0} → G_{0}^{∗} = G_{0}/N be the canonical projection. Then M^{∗} = G_{0}/P = (G_{0}/N)/(P/N) with P = π^{−1}(P^{∗}) and hence the id-component of P is H_{0}. Furthermore H_{0} ∩ N = 1 and therefore H_{0} × N ⊂ P ⊂ N_{G_{0}}(H_{0}) and D ≃ π_{1}(M^{∗}) ≃ P/H_{0}. If P/N does not yet act irreducibly on 𝔪, then G^{∗} and hence H^{∗} must contain further isometries.

First, assume that we are in case (a). Then H_{0} = 1 and hence P = Γ is a finite group. The further isometries in H^{∗} become, via the adjoint representation, automorphisms of 𝔨 and hence there exists a finite group Φ with Ad(Γ) ⊂ Φ ⊂ Aut(K) such that Φ acts irreducibly on 𝔨 and Φ leaves Γ invariant.

If we are in case (b), notice that N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) ⊂ Ĥ via right multiplication on G_{0}/H_{0} and hence P/N ⊂ Ĥ. To make G_{0}/P isotropy irreducible we need a compact group L with H_{0} ⊂ P/N ⊂ L ⊂ Ĥ such that Ad_{L} acts irreducibly on 𝔪 and keeps N and P invariant. Then L induces further isometries on G_{0}/P and we obtain an isotropy irreducible form of M^{∗}. □

Notice that by (2.4) the two cases in (6.1) cover all possible isotropy irreducible homogeneous spaces which are not strongly isotropy irreducible.

We now discuss some examples of (6.1)(b). Notice that the possible manifolds that can arise as subcoverings are always of the form G_{0}/P with H_{0} ⊂ P ⊂ N_{G_{0}}(H_{0}).

If rank G_{0} = rank H_{0}, then Z(G_{0}) = 1 and hence G_{0}^{∗} = G_{0}. By examining Table I, we can see that a minimal subgroup needed to make G_{0}/H_{0} isotropy irreducible can always be chosen to be in N_{G_{0}}(H_{0}) and hence we can choose P ⊆ L ⊆ N_{G_{0}}(H_{0}) with P normal in L, and such that Ad_{L} acts irreducibly on 𝔪. One possible choice that always works is P = L = N_{G_{0}}(H_{0}) and hence all equal rank examples have isotropy irreducible subcoverings. In several cases N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = S_{3} with (Ĥ/H_{0})_{min} = ℤ_{3} ⊂ S_{3}. Here the only possible choices are P/H_{0} = ℤ_{3} with L/H_{0} = ℤ_{3} or S_{3} and P/H_{0} = L/H_{0} = S_{3}. But in many other examples the possible isotropy irreducible subcoverings become almost impossible to list, since there are so many of them. E.g. in Example 3 in Table I there are many subgroups of (ℤ_{2})^{n−1} invariant under S_{n} giving rise to spaces of partially oriented flags in ℝ^{nk} (see the discussion in §1, Example 4). There are also many subgroups of S_{n} which act transitively on the set of unordered pairs and hence give rise to further isotropy irreducible subcoverings.

If rank H_{0} < rank G_{0}, (Ĥ/H_{0})_{min} sometimes consists of outer automorphisms of G_{0}. For case (1) in Table II there exist many isotropy irreducible subcoverings as in the equal rank case. Notice that for n even, Z(G_{0}) = ℤ_{2} and hence P ⊂ (ℤ_{2})^{n−1} ⋉ S_{n} for n even and odd, just as in the equal rank case. For case 2(a) there clearly exists only one subcovering which becomes isotropy irreducible, whereas for case 3(a) there exists none. Case 2(c) is interesting since the automorphism in (Ĥ/H_{0})_{min} is outer. There are two isotropy irreducible subcovers corresponding to the whole group N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = ℤ_{2} ×ℤ_{2} or to the subgroup △ℤ_{2} ⊂ ℤ_{2} × ℤ_{2}. Case 2(b) has many isotropy irreducible subcoverings including e.g. the ones corresponding to P = N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = D_{8}, two subroups in D_{8} isomorphic to ℤ_{2}, the product of the previous 3 groups with Z(G_{0}) = ℤ_{2}, and two groups ℤ_{2} embedded diagonally in D_{8} × ℤ_{2}. An interesting example is case (6). Here N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = 1 and Z(G_{0}) = ℤ_{2} × ℤ_{2}, but the only subgroup of Z(G_{0}) which is invariant under the triality automorphism, which generates ℤ_{3} = (Ĥ/H_{0})_{min}, is the whole center. Hence the only isotropy irreducible subcovering is (SO(8)/ℤ_{2})/G_{2} = ℝP^{7} × ℝP^{7}. Finally we discuss case (4). Here we have again N_{G_{0}}(H_{0})/H_{0}Z(G_{0}) = 1 and hence we can only divide out by a central subgroup of G_{0}. An obvious choice is N = (A × ⋯ × A)/△(A) for any subgroup A ⊂ Z(K), in fact this may be the only possibility.

To obtain examples which are not locally de Rham irreducible, we can take for M a product of N k times with itself, where N is an entry in Table I or II. P is then a product of the groups discussed above, but for L we need to choose in addition to the product of the above groups a subgroup of the symmetric group S_{k} acting transitively on the de Rham factors of M.

It is natural to ask the question what the isotropy irreducible homogeneous spaces M^{∗} = G^{∗}/H^{∗} are with G^{∗} connected. We can now easily answer this more restrictive question. Of course, if M^{∗} is simply connected, this can only happen if G^{∗}/H^{∗} is strongly isotropy irreducible. In general we have:

(6.2) Theorem. Let M^{∗} = G^{∗}/H^{∗} be a non-flat isotropy irreducible homogeneous space with G^{∗} connected and with universal cover M = G_{0}/H_{0}. Then M is de Rham irreducible. If H_{0} = 1, there exists a central subgroup N of G_{0} and a finite group Γ ⊂ G_{0} such that N ⊂ Γ and M^{∗} = G_{0}/Γ = (G_{0}/N)/(Γ/N) with Ad|_{Γ} acting irreducibly on 𝔤. In all other cases there exists a central subgroup N of G_{0} and a compact group P with H_{0} × N ⊆ P ⊆ N_{G_{0}}(H_{0}) such that Ad_{P} acts irreducibly on 𝔪 and M^{∗} = G^{∗}/H^{∗} = (G_{0}/N)/(P/N) = G_{0}/P. For every entry in Table I and II, such a group P exists, with the exception of Examples 2(c), 3(a), 4, and 6 in Table II.

Proof. Most of the statements in this theorem follow as in the proof of (6.1). Here we only need in addition L = P/N. That M must be de Rham irreducible follows from the fact that in the de Rham reducible case all isometries which interchange the de Rham factors of M are outer automorphisms of G_{0}. Similarly, the four cases in Table II have no isotropy irreducible subcovers of this form since the isometries needed to make G_{0}/H_{0} isotropy irreducible come from outer automorphisms of G_{0}. □

We finally make some comments about the minimal isometric immersions mentionend in the introduction. In general, if ϕ: M → S^{N}(r) is an isometric minimal immersion, then the coordinate functions ϕ_{1}, …, ϕ_{N+1} are eigenfunctions of the Laplace operator with eigenvalue λ = dim M/r^{2}. Conversely, as was observed by Takahashi [T], if M = G/H is an isotropy irreducible homogeneous space and if ϕ_{1}, …, ϕ_{N+1} is an orthonormal basis of the eigenspace E_{λ} of the Laplace operator with eigenvalue λ > 0, then ϕ = (ϕ_{1}, …, ϕ_{N+1}) is an isometric minimal immersion into S^{N}(r) with r^{2} = dim M/λ. Indeed, the pull back of the metric on S^{N}(r) under ϕ is clearly invariant under G and hence is either 0 or a G-invariant metric. But it cannot be 0 since λ > 0. A different choice of orthonormal basis gives rise to a congruent immersion and we call this class of isometric immersions the standard eigenspace immersion. There are in general many other minimal isometric immersions into S^{N}(r), but they are all given by choosing eigenfunctions in E_{λ}, λ = dim M/r^{2}, as coordinates. If the coordinates form a basis of E_{λ}, then we call such a minimal isometric immersion a full eigenspace immersion, otherwise a partial eigenspace immersion. For example, if there is a subspace of E_{λ} invariant under G, then an orthonormal basis of this subspace gives rise to a partial eigenspace immersion. It was shown in [L] that the space of all isometric minimal immersions of G/H into S^{N}(r), r^{2} = dim M/λ, forms a compact convex body in a finite dimensional vector space with interior points corresponding to the full eigenspace immersions and boundary points corresponding to partial eigenspace immersions.

We now discuss some of the other results in [L], since not all of them are correct. In [L, Lemma 3] it was shown that for a full eigenspace immersion ϕ: M = G/H → S^{N}(r), M^{∗} = ϕ(M) is an embedded submanifold and ϕ: M → M^{∗} is a covering map. This is then applied in [L, Theorem 4] to show that every eigenspace immersion is a covering onto its image, that the homogeneous structure on M^{∗} is the one induced from M, and that the deck group of the covering is contained in the center of G. This is actually false, but partially true for the full eigenspace immersions. A good counterexample is given by K/Γ where K is a connected, compact, simple Lie group and Γ a finite subgroup such that Ad|_{Γ} acts irreducibly on 𝔨. Using a full eigenspace immersion for K/Γ, we obtain an isometric minimal immersion K → K/Γ → S^{N}(r) whose image does not carry the homogeneous structure induced from K. Indeed the id-component of the isometry group is equal to [K × K]/△K for K and equal to K for K/Γ. Also, even for a full eigenspace immersions, the deck group need not be contained in Z(G). The correct statement is:

(6.3) Theorem (P.Li). Let M = G/H be a compact isotropy irreducible homogeneous space and ϕ: M → S^{N}(r) a full eigenspace immersion. Then ϕ(M) = M^{∗} is an embedded submanifold, ϕ: M → M^{∗} a covering map, and the deck group D of the covering is normal in the full isometry group Ĝ of M. Hence D commutes with G_{0} and D ∩ G_{0} ⊂ Z(G_{0}). The isometry group of M^{∗} is Ĝ/D, and hence M^{∗} must be an isotropy irreducible homogeneous space.

Proof. We can regard ϕ = (ϕ_{1}, ⋯, ϕ_{N+1}) as an immersion ϕ: M → ℝ^{N+1}. The isometric action of Ĝ on M induces an orthogonal representation of Ĝ on E_{λ}. By using the basis ϕ_{1}, ⋯, ϕ_{N+1} of E_{λ} to identify E_{λ} with ℝ^{N+1}, this induces a linear action of Ĝ on ℝ^{N+1} and the immersion ϕ: M → ℝ^{N+1} becomes equivariant with respect to this action. Notice though that ϕ: M → S^{N}(r) is equivariant only for the standard eigenspace immersion. In any case, the equivariance implies that the immersion has no double points and since M is compact ϕ: M → M^{∗} is a covering map. The equivariance also implies that the isometries in Ĝ induce isometries on M^{∗} and hence D is normal in Ĝ, which implies that D ∩G_{0} must centralize G_{0}. □

The fact that D is normal in Ĝ is actually a strong property, which is not satisfied by many of our isotropy irreducible subcoverings. E.g. for the space of real flags in ℝ^{nk} as in Example 4 in §1, if we divide out by the symmetric group S_{n}, we obtain an isotropy irreducible subcovering which can not be the image of any full eigenspace immersion of the space of oriented flags. This also implies that if we choose an eigenspace immersion for this quotient and lift it to the space of oriented flags, it must become a partial eigenspace immersion of the space of oriented flags. For the Example 3a in Table II, (6.3) implies that every full eigenspace immersion must be an embedding since there are no isotropy irreducible subcoverings.

Notice that [L, Theorem 5] is false, as the example S^{3}/Γ shows, where Γ is e.g. the binary dodecahedral group (see Example 1 in §1). But it is true that a full eigenspace immersion of S^{n}(1) is either an embedding of S^{n}(1) or an embedding of ℝP^{n}(1). It is an interesting question whether the image of some partial eigenspace immersion of S^{n}(1) can be a lens space or more generally if every manifold of constant positive curvature admits a minimal isometric immersion into S^{N}.

[L, Theorem 7] is again false, since among our examples there are coverings M → L → N with M and N isotropy irreducible, but L not.

In [L, Theorem 8] it is shown that if Z(G) is cyclic, then there are infinitely many eigenfunctions which are not invariant under any subgroup of Z(G). Unfortunately, by the above remarks this does not imply [L, Corollary 9] which claims that infinitely many of the full eigenspace immersions are embeddings. It would be interesting to obtain a criterion which guarantees that some of the full eigenspace immersions are embeddings. We suspect that for each isotropy irreducible homogeneous space there exists at least one of these.

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McKenzie Y. Wang, Department of Mathematics and Statistics, McMaster University, Hamilton, Ontario, Canada, L8S 4K1

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