This paper confirms a conjecture about calculating the index of exceptional symmetric spaces using a novel approach involving slice representations, advancing understanding of their geometric structure.
Contribution
It provides an affirmative proof of a conjecture for exceptional and certain classical symmetric spaces, introducing a new methodology based on slice representations.
Findings
01
Confirmed the conjecture for exceptional symmetric spaces
02
Developed a new method using slice representations
03
Enhanced understanding of totally geodesic submanifolds
Abstract
The index of a Riemannian symmetric space is the minimal codimension of a proper totally geodesic submanifold (Onishchik, 1980). There is a conjecture by the first two authors for how to calculate the index. In this paper we give an affirmative answer to this conjecture for the exceptional Riemannian symmetric spaces and for the classical symmetric spaces Sp(r,R)/U(r). Our methodology is new and based on the idea of using slice representations for studying totally geodesic submanifolds.
Tables2
Table 1. Table 1. The index i ( M ) 𝑖 𝑀 i(M) for irreducible exceptional Riemannian symmetric spaces M = G / K 𝑀 𝐺 𝐾 M=G/K of type III and submanifolds Σ Σ \Sigma of M 𝑀 M with codim ( Σ ) = i ( M ) codim Σ 𝑖 𝑀 \operatorname{codim}(\Sigma)=i(M)
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Full text
The index of exceptional symmetric spaces
Jürgen Berndt
King’s College London, Department of Mathematics, London WC2R 2LS, United Kingdom
The index of a Riemannian symmetric space is the minimal codimension of a proper totally geodesic submanifold (Onishchik [18]). There is a conjecture by the first two authors ([2]) for how to calculate the index. In this paper we give an affirmative answer to this conjecture for the exceptional Riemannian symmetric spaces and for the classical symmetric spaces Spr(R)/Ur. Our methodology is new and based on the idea of using slice representations for studying totally geodesic submanifolds.
2010 Mathematics Subject Classification:
Primary 53C35, 53C40
The second and third author acknowledge financial support from Famaf, UNC and Ciem, CONICET
1. Introduction
Let M be a connected Riemannian manifold and denote by S the set of all connected totally geodesic submanifolds Σ of M with dim(Σ)<dim(M). The index i(M) of M is defined by
[TABLE]
This terminology was introduced by Onishchik in [18].
Riemannian symmetric spaces are among the most distinguished manifolds in Riemannian geometry. These spaces have been studied by numerous mathematicians and many fascinating properties and applications have been discovered. Despite them being so widely studied, it is a remarkable fact that their totally geodesic submanifolds are not yet known in general. Wolf ([20]) classified them in symmetric spaces of rank 1 and Klein ([10],[11],[12],[13]) in symmetric spaces of rank 2.
In previous work ([2],[3],[4]) we developed a systematic approach to the index of Riemannian symmetric spaces. Totally geodesic submanifolds are in one-to-one correspondence to algebraic objects called Lie triple systems. If M=G/K is a Riemannian symmetric space and g=k⊕p is a corresponding Cartan decomposition, then a Lie triple system is a linear subspace V of p such that [[V,V],V]⊆V. A distinguished class of Lie triple systems is formed by those Lie triple systems V for which the orthogonal complement V⊥ of V in p is also a Lie triple system. Such Lie triple systems are called reflective. Algebraically they correspond to certain involutive automorphisms of g, geometrically they correspond to totally geodesic submanifolds Σ for which the geodesic reflection of M in Σ is a well-defined global isometry. These so-called reflective submanifolds have been classified by Leung ([15],[16]) in irreducible Riemannian symmetric spaces. Denote by Sr the set of all connected reflective submanifolds Σ of M with dim(Σ)<dim(M). The reflective index ir(M) of M is defined by
[TABLE]
It is clear that i(M)≤ir(M) and thus ir(M) is an upper bound for i(M). Moreover, from Leung’s work we can calculate ir(M) explicitly for each irreducible Riemannian symmetric space. This was done explicitly in [2], where we formulated the following conjecture:
Conjecture. *For an irreducible Riemannian symmetric space we have i(M)=ir(M) if and only if M=G22/SO4 and M=G2/SO4. *
In our previous work we verified this conjecture for a number of Riemannian symmetric spaces. In [3] we verified the conjecture for all exceptional Riemannian symmetric spaces of types II and IV, that is, for the five exceptional compact Lie groups G2,F4,E6,E7,E8 and their non-compact dual symmetric spaces. The purpose of this paper is to give an affirmative answer to this conjecture for the remaining exceptional Riemannian symmetric spaces, which are of type I (compact) or type III (non-compact). Our main result states:
Theorem 1.1**.**
Let M=G/K be an irreducible exceptional Riemannian symmetric space. Then i(M)=ir(M) if and only if M=G22/SO4 and M=G2/SO4.
Duality between symmetric spaces of compact type and of non-compact type preserves totally geodesic submanifolds. We can therefore restrict to symmetric spaces of non-compact type. In Table 1 we list the exceptional Riemannian symmetric spaces of type III together with their index and a totally geodesic submanifold Σ with codim(Σ)=i(M).
The same methodology that we develop in this paper can be used to determine the index of another series of classical symmetric spaces:
Theorem 1.2**.**
For M=Spr(R)/Ur, r≥3, we have i(M)=ir(M)=2(r−1) and Σ=RH2×Spr−1(R)/Ur−1 is a reflective submanifold of M with codim(Σ)=i(M).
This was proved in [2] for r∈{3,4,5}, but is new for r>5. Taking into account the results from [2], [3] and [4], the conjecture remains open for three series of classical symmetric spaces:
(i)
M=SO2k+2∗/Uk+1 for k≥5. Conjecture: i(M)=2k.
(ii)
M=SU2k+2∗/Spk+1 for k≥3. Conjecture: i(M)=4k.
(iii)
M=Spk,k+l/SpkSpk+l for l≥0 and k≥max{3,l+2}. Conjecture: i(M)=4k.
Our methodology is new and based on the idea of using slice representations for studying totally geodesic submanifolds. In Section 2 we develop basic sufficient criteria for deciding whether a totally geodesic submanifold is reflective. One of these criteria states that a semisimple totally geodesic submanifold of a symmetric space of rank ≥2 is reflective when the kernel of the full slice representation is non-trivial. It is intuitively clear that fixed vectors of the slice representation must play a crucial rule in the theory. We make this precise in Theorem 3.1: a totally geodesic submanifold Σ is maximal and the slice representation has a non-zero fixed vector if and only if Σ is reflective and the complimentary reflective submanifold is semisimple. This relates to the theory of symmetric R-spaces (symmetric real flag manifolds). In Section 3 we give some applications of Theorem 3.1 and in Section 4 we present its proof. In Section 5 we first derive a very useful lower bound for the codimension of a totally geodesic submanifold. Another important ingredient for our theory is Proposition 5.6, which states that a reducible totally geodesic submanifold containing a real hyperbolic space or a complex hyperbolic space as a de Rham factor must either be a product of two real hyperbolic spaces or there exists a reflective submanifold of dimension greater than or equal to the dimension of Σ. This theoretical result will allow us to dismiss many possibilities in our investigations. We eventually apply all these theoretical results to calculate the index of the exceptional symmetric spaces and of Spr(R)/Ur.
2. Preliminaries and basic results
Let M=G/K be a connected, locally irreducible, Riemannian symmetric space of compact type, where G=Io(M) is the identity component of the isometry group I(M) of M. We denote by e the identity of G and by o=eK∈M the base point in M. Let g=k⊕p be the corresponding Cartan decomposition of the Lie algebra g of G. Here, k is the Lie algebra of K and p is the orthogonal complement of k in g with respect to the Cartan-Killing form of g. We denote by ⟨⋅,⋅⟩ the Riemannian metric on M, by ∇ the Riemannian connection on M, and by R the Riemannian curvature tensor of M. By r=rk(M) we denote the rank of M, which by definition is the maximal dimension of a flat totally geodesic submanifold in M. Each X∈g determines a Killing vector field X∗ on M by Xp∗=dtdt=0Exp(tX)(p) for all p∈M, where Exp:g→G is the Lie exponential map. Then X∈k if and only if Xo∗=0. It is well known that [X,Y]∗=−[X∗,Y∗] for all X,Y∈g.
For p∈M we denote by σp∈I(M) the geodesic symmetry of M at p. For v∈ToM we denote by γv:R→M the geodesic in M with γv(0)=o and γ˙v(0)=v. Then ϕtv=σγv(t/2)∘σo∈G induces a 1-parameter group (ϕtv)t∈R of isometries of M. Each ϕtv is a geometric transvection; it translates the geodesic γv by t and dγv(t0)ϕtv coincides with the parallel transport along γv from γv(t0) to γv(t0+t). If Xv∈g with ϕtv=Exp(tXv), then Xv∈p, or equivalently (∇Xv)o=0. Conversely, if X∈p, then Exp(tX) is a geometric transvection defined by the geodesic γv(t)=Exp(tX)(o), v=X∈p≅ToM.
Let γv be a non-trivial closed geodesic in M with γv(0)=o. We may assume, by rescaling v, that γv has (minimal) period 1. Then ϕ1v(o)=o and (ϕ1v)2=ϕ1v∘ϕ1v=ϕ2v=σγv(1)∘σo=σo2=idM is the identity idM of M. The isometry ϕ1v=σγv(1/2)∘σo may be trivial, as is the case when M is a sphere, where γv(1/2) is the antipodal point of o and σγv(1/2)=σo. For constructing a non-trivial isometry ϕ1v on M, we have to pass to a suitable globally symmetric quotient of M and then lift back to M, provided that M is simply connected.
Assume that M is simply connected and irreducible. We define an equivalence relation on M by p∼q if and only if Gp=Gq, where Gp and Gq are the isotropy groups of G at p and q, respectively. Note that the isotropy groups are connected since M is simply connected and G=Io(M) is connected. Denote by Mˉ the quotient space relative to this equivalence relation and by π:M→Mˉ the canonical projection. Since M is irreducible, the isotropy action of Gp on TpM is irreducible and therefore has no fixed non-zero vectors. Hence the action of Gp on M has no fixed points apart from p on a sufficiently small open neighborhood of p in M. It follows that each fibre of the projection π is a discrete subset of M. For all isometries g∈I(M) we have Gg(p)=gGpg−1 and therefore every isometry g∈I(M) maps equivalence classes to equivalence classes. Thus every g∈I(M) descends to an isometry of Mˉ, where we equip Mˉ with the induced Riemannian structure from M via π. We thus can write Mˉ=G/Gπ(o), where G acts almost effectively on Mˉ. This also shows that every geodesic symmetry of M descends to a geodesic symmetry of Mˉ. Thus Mˉ is a connected Riemannian symmetric space of compact type. The isotropy group Gπ(o) is not necessarily connected, but its Lie algebra is equal to k. Note that π:M→Mˉ induces a bijection from the Lie algebra of Killing vector fields on M onto the Lie algebra of Killing vector fields on Mˉ.
We denote by kp⊂g the Lie algebra of the isotropy group Gp of G at p∈M and by kπ(p)⊂g the Lie algebra of the isotropy group Gπ(p) of G at π(p)∈Mˉ. Obviously, we have kp=kπ(p). Assume that kπ(p)=kπ(q) for p,q∈M. Then we have kp=kq. Since the isotropy groups Gp and Gq are connected, this implies Gp=Gq and hence p∼q and π(p)=π(q). In other words, different points in Mˉ have different isotropy algebras.
Lemma 2.1**.**
Let pˉ,qˉ∈Mˉ and denote by σˉpˉ,σˉqˉ their geodesic symmetries in Mˉ. Then σˉpˉ=σˉqˉ if and only if pˉ=qˉ.
Proof.
Let sˉpˉ:G→G be the involutive automorphism of G induced by σˉpˉ, that is, sˉpˉ(g)=σˉpˉgσˉpˉ for all g∈G. The differential desˉpˉ is an involutive automorphism of g and kpˉ is the +1-eigenspace of desˉpˉ. Assume that σˉpˉ=σˉqˉ. Then sˉpˉ=sˉqˉ and therefore kpˉ=kqˉ, which implies pˉ=qˉ since two different points in Mˉ have different isotropy algebras.
∎
Lemma 2.1 tells us that σˉpˉ=σˉqˉ if pˉ=qˉ. This means that any geodesic symmetry on Mˉ cannot have another isolated fixed point apart from the obvious one. In other words, there are no poles on Mˉ. The symmetric space Mˉ is also known in the literature as the adjoint space ([7], p. 327) or bottom space ([17], Section 4.2) of M. Note that there are simply connected irreducible Riemannian symmetric spaces without poles, that is, M=Mˉ.
Let pˉ∈Mˉ and vˉ∈TpˉMˉ so that γvˉ is a closed geodesic in Mˉ with (minimal) period 1. Then qˉ=γvˉ(1/2) is an antipodal point of pˉ in Mˉ and gvˉ=σˉqˉ∘σˉpˉ is a non-trivial involutive isometry of Mˉ with gvˉ(pˉ)=pˉ. Moreover, ℓvˉ=dpˉgvˉ coincides with the parallel transport in Mˉ along the geodesic loop γvˉ∣[0,1].
Since qˉ is an antipodal point of pˉ in Mˉ, we have σˉqˉ∘σˉpˉ=σˉpˉ∘σˉqˉ, σˉqˉ(pˉ)=pˉ=σˉpˉ(pˉ) and σˉpˉ(qˉ)=qˉ=σˉqˉ(qˉ).
Remark 2.3**.**
In the notation of Corollary 2.2, let Fˉ be a maximal flat of Mˉ with pˉ∈Fˉ and vˉ∈TpˉFˉ. Such a flat is unique if vˉ is a principal vector for the isotropy action of (Gpˉ)o. Since Fˉ is a torus, it is globally flat. Then, since Fˉ is totally geodesic in Mˉ, the restriction ℓvˉ∣TpˉFˉ:TpˉFˉ→TpˉFˉ is the identity map.
The quotient space Mˉ is only auxiliary and we will lift the isometry gvˉ∈I(Mˉ) to an isometry gv∈I(M). Let p∈M with Gp=K (that is, p∼o). Put pˉ=π(p)∈Mˉ and choose v∈TpM so that γvˉ is a closed geodesic in Mˉ with (minimal) period 1, where vˉ=dpπ(v). Then the linear involutive isometry ℓv=(dpπ)−1∘ℓvˉ∘dpπ:TpM→TpM preserves the curvature tensor Rp of M at p. By the global Cartan Lemma and since M is simply connected, there exists a non-trivial involutive isometry gv of M with gv(p)=p and dpgv=ℓv. The isometry gv∈I(M) is not necessarily in G=Io(M). Note that the linear isometry ℓv also induces an isometry of the dual symmetric space of M, since the curvature tensor of the dual is just −R.
Note that any flat Fˉ of Mˉ can be written as Fˉ=π(F) with some flat F of M.
Proposition 2.4**.**
Let M=G/K be a simply connected, irreducible, Riemannian symmetric space with G=Io(M), o=eK and g=k⊕p the Cartan decompositon of g at o. Let 0=w∈ToM≅p be arbitrary. Then there exists a non-trivial isometry g of M with the following properties:
(i)
g(o)=o;
(ii)
g2=idM;
(iii)
The orbit K⋅w⊂p is invariant under dog, that is, dog(K⋅w)=K⋅w;
(iv)
Each vector in νw(K⋅w)≅Zp(w) is fixed by dog, where νw(K⋅w) is the normal space of K⋅w at w and Zp(w) is the centralizer of w in p≅ToM;
(v)
Assume that rk(M)≥2, or equivalently, that K⋅w is not a sphere. Then the linear subspace νw(K⋅w) of ToM coincides with the set of fixed vectors of dog if and only if K⋅w is an extrinsically symmetric orbit.
Proof.
It suffices to prove the proposition for the case that M is of compact type.
We have Zp(w)=⋃a∈Awa, where Aw is the set of maximal abelian subspaces a of p with w∈a. The intersection a0=⋂a∈Awa is the abelian part of Zp(w). Moreover, as it is standard to prove, there exists a compact flat F of M with o∈F and ToF=a0. Now write K⋅w=K/Kw, where Kw is the isotropy group of K at w, and let Kwo be the identity component of Kw. It is clear that a0 is invariant under the action of Kwo on p, and therefore F is invariant under the action of Kwo on M.
Since the connected group Kwo fixes o, we obtain that Kwo acts trivially on F. It follows that Kwo acts trivially, via the isotropy representation, on the subspace a0 of Zp(w)=νw(K⋅w). In other words, the connected slice representation of Kw on the normal space νw(K⋅w) fixes a0 pointwise.
Now consider the bottom space Mˉ and the canonical projection π:M→Mˉ. The image Fˉ=π(F) is a compact flat of Mˉ. Choose 0=v∈a0=ToF such that the geodesic γvˉ=π∘γv, vˉ=doπ(v), is closed in Fˉ. Since the initial directions of closed geodesics in Fˉ starting at oˉ=π(o) form a dense set in ToˉFˉ≅ToF=a0, we can choose v arbitrarily close to w. The orbits K⋅w and K⋅v are parallel orbits if v is sufficiently close to w (see [1], Corollary 2.3.7). For such v this then implies νv(K⋅v)=νw(K⋅w), or equivalently, Zp(v)=Zp(w). We normalize v so that γvˉ is a closed geodesic in Mˉ with (minimal) period 1 and construct the isometries gvˉ∈I(Mˉ) and gv∈I(M) as above. Using Remark 2.3, and using the fact that Zp(vˉ)=Zp(v) is the union of all abelian subspaces of p≅ToˉMˉ containing vˉ, we obtain that doˉgvˉ fixes Zp(vˉ)=Zp(v) pointwise and, in particular, doˉgvˉ(vˉ)=vˉ. Then g=gv∈I(M) satisfies the properties (i)–(iv) (for (iii) use that gKg−1=K, since K is the identity component of the full isotropy group of M).
The “only if” part of (v) is just the definition of an extrinsically symmetric space. It remains to prove that, if S=K⋅w is an extrinsically symmetric space, then dog coincides with the extrinsic symmetry (and so it has no fixed vector tangent to the orbit K⋅w). So let us assume that S is extrinsically symmetric. It is well known that in this case the orbit K⋅w must be most singular, that is, nearby orbits in the sphere have greater dimension than K⋅w, or equivalently, dim(a0)=1 and so v is a scalar multiple of w, hence we may assume that v=w (note that this also follows from Theorem 4.2 in [2]). Let
[TABLE]
Note that K~ contains the transvections ρp∘ρq, where ρp,ρq are the extrinsic symmetries at p,q∈K⋅v. Then K~⊇K and K~ is not transitive on the sphere of TpM, since K~⋅v=K⋅v. From Simons Holonomy Theorem ([19]) we then get K~=K (see Remark 8.3.5 in [1]) and it follows that (K,Kv) is a symmetric pair.
Since Kv⋅v={v}, the construction of g=gv shows that kgk−1=g for all k∈Kv. Let V={u∈Tv(K⋅v):dog(u)=u}. Then V extends to a K-invariant parallel distribution D on the orbit S=K⋅v. The orthogonal complement V⊥ of V in TvS is exactly the (−1)-eigenspace
of the linear isometry dog. Let α be the second fundamental form of S.
Then we have
[TABLE]
for all u∈V and z∈V⊥, and hence α(V,V⊥)=0.
This implies α(D,D⊥)=0. If D is non-trivial, Moore’s Lemma tells us that S splits extrinsically (see [1], Lemma 1.7.1). In this case K does not act irreducibly, which is a contradiction. We conclude that dog coincides with the extrinsic symmetry ρv of S.
∎
Remark 2.5**.**
The above construction of the symmetry for extrinsically symmetric spaces is due to Nagano in the context of fibrations of symmetric spaces (see [17] and its bibliography; see also Section 7 in [6]). However, in Nagano’s construction the closed geodesic must be minimizing until it reaches its antipodal point. In the proof of Proposition 2.4, the closed geodesic does not generally have this property. Let M=G/K be a simply connected irreducible symmetric space. If S=K⋅v is a most singular isotropy orbit, which is not extrinsically symmetric, then expo(tv) is a closed geodesic that is not minimizing until it reaches its antipodal point. Namely, if α1,…,αr is a basis of simple roots, then K⋅v=K⋅Hi for some i,
where H1,…,Hr is the dual basis. From Kobayashi and Nagano [14], S is extrinsically symmetric if and only if δi=1, where δ=δ1α1+…+δrαr is the highest root.
From [7], Chapter VII-3, the closed geodesic γHi is not minimizing beyond 2δi1 of its length.
Let M=G/K be a simply connected irreducible symmetric space and Σ be a non-semisimple maximal totally geodesic submanifold of M with o∈M. Then ToΣ=Zp(v), where K⋅v is a most singular orbit. Note that, if K⋅v is a most singular orbit, then Zp(v) is a non-semisimple Lie triple system of ToM≃p whose abelian part is Rv. It was proved in [2] (Theorem 4.2) that Zp(v), which coincides with the normal space νv(K⋅v), is the tangent space to a maximal non-semisimple totally geodesic submanifold of M if and only if K⋅v is an extrinsically symmetric orbit. If K⋅v is extrinsically symmetric, then both νv(K⋅v) and Tv(K⋅v) are Lie triple systems in ToM. In fact, Tv(K⋅v) coincides with the fixed vectors of doσo∘dogv, where σo is the geodesic symmetry of M at o and dogv is the extrinsic symmetry of K⋅v at v (with gv constructed as above).
Let Σ be a connected, complete, totally geodesic submanifold of M with o∈Σ. Then Σ is called reflective if the normal space νoΣ is a Lie triple system as well. Note that Σ is reflective if and only if Σ is the connected component containing o of the
fixed point set of an involutive isometry τ∈I(M) with τ(o)=o (see, for instance, [2]).
In the proof of [2], Theorem 4.2, it is shown that Zp(v) is properly contained in a (proper) Lie triple system of ToM if K⋅v is not extrinsically symmetric. The following corollary of Proposition 2.4 gives an alternative proof of this result with additional information: Zp(v) is properly contained in a (proper) reflective Lie triple system of ToM.
Corollary 2.6**.**
Let M=G/K be a simply connected irreducible Riemannian symmetric space and 0=w∈ToM.
Assume that K⋅w is not extrinsically symmetric. Then Zp(w) is properly contained in a reflective (proper) Lie triple system of ToM.
Proof.
Let g be as in Proposition 2.4. Since g is of order 2, the set of fixed vectors of dog is a reflective Lie triple system containing Zp(w).
∎
We finish this section by proving that if a connected, complete, totally geodesic submanifold Σ of a symmetric space M contains a reflective submanifold of M, then Σ must be reflective as well. For this aim we first generalize Proposition 3.4 of [2] to include the case that the kernel of the slice representation is finite.
Let Σ=GΣ/KΣ be a semisimple totally geodesic submanifold of a (simply connected) symmetric space M=G/K with o∈Σ, where
[TABLE]
KΣ=(GΣ)o is the isotropy group of GΣ at o and I(M) is the full isometry group of M (whose identity component coincides with G). The group GΣ is, in general, neither connected nor effective on Σ. Note that GΣ contains the glide transformations of Σ, that is, the closed subgroup of GΣ with Lie algebra [ToΣ,ToΣ]⊕ToΣ.
The full slice representation of Σ at o is the representation ρ:KΣ→O(νoΣ) given by ρ(k)=dok∣νoΣ, where νoΣ=(ToΣ)⊥ is the normal space of Σ at o.
Remark 2.7**.**
If M=G/K and M^=G^/K are dual symmetric spaces, then g=k+p⊂gC and g^=k+ip⊂gC are the corresponding Cartan decompositions at o=eK∈M and o^=eK∈M^. The isotropy representation of G/K at o on ToM≅p is canonically equivalent to the isotropy representation of G^/K at o^ on To^M^≅ip. If V⊂p is a Lie triple system in p, then iV⊂ip is a Lie triple system in ip, and vice versa. Thus we have a natural bijection between the Lie triple systems on p and ip, and therefore between connected, complete, totally geodesic submanifolds in M and M^.
Let Σ be a (connected, complete) totally geodesic submanifold of M with o∈Σ and let Σ^ be the (connected, complete) totally geodesic submanifold of M^ with To^Σ^=iToΣ. Then KΣ=KΣ^ (via the respective isotropy representations). In fact, if k∈KΣ, then −i(dok)i is a linear isometry of ip=To^M^ leaving iToΣ invariant and preserving the curvature tensor R^o^ of M^ at o^, since R^o^(ix,iy)iz=−iRo(x,y)z for all x,y,z∈ToM. So, by the global Cartan Lemma, −i(dok)i is the differential of an isometry of M^ fixing o^. In particular, Σ is reflective if and only if Σ^ is reflective.
Proposition 2.8**.**
Let M=G/K be an irreducible Riemannian symmetric space with rk(M)≥2, where (G,K) is an effective Riemannian symmetric pair. Let Σ be a semisimple totally geodesic submanifold of M with o∈Σ. Assume that the kernel of the full slice representation of Σ at o is non-trivial. Then Σ is reflective.
Proof.
The proof is similar to the one of Proposition 3.4 in [2]. By Remark 2.7 we may assume that M is of non-compact type and that Σ is complete (and therefore diffeomorphic to ToΣ).
Let KΣ={k∈I(M)o:k(Σ)=Σ} and ρ:KΣ→O(νoΣ) be the full slice representation. The kernel H of ρ is a normal subgroup of KΣ (note that H may be finite). The subspace
[TABLE]
is a Lie triple system in ToΣ. Moreover, since H is a normal subgroup of KΣ, the subspace V⊆ToΣ is invariant under KΣ and, in particular, under the isotropy group K′=(G′)o of the group G′ of glide transformations of Σ. Thus V extends to a G′-invariant, and hence parallel, distribution on Σ. It follows that Σ is the Riemannian product Σ=Σ1×Σ2 of two totally geodesic submanifolds Σ1 and Σ2 with ToΣ1=V and ToΣ2=V⊥∩ToΣ.
The subspace V⊕νoΣ is a Lie triple system in ToM, since it consists of the fixed vectors of the action of H on ToM. This implies that Σ2 is reflective. We write Σ2=G′′/K′′, where G′′⊆G is the group of glide transformations of Σ2 (G′′ acts almost effectively on Σ2) and K′′=(G′′)o. Since Σ=Σ1×Σ2, the group K′′ acts trivially on ToΣ1=V. Consequently V is a subset of the set W of fixed vectors of the action of K′′ on ToΣ3, where Σ3 is the totally geodesic submanifold of M associated with the Lie triple system V⊕νoΣ. By Lemma 3.2 (ii) in [2], the subspace W is the tangent space of a totally geodesic flat submanifold Σ0⊆Σ3. Since TpΣ1=V⊆W, we conclude that Σ1 is flat, which is a contradiction since Σ is semisimple unless V={0}. It follows that Σ=Σ2 is reflective.
∎
Corollary 2.9**.**
Let M=G/K be an irreducible simply connected Riemannian symmetric space with rk(M)≥2. Let Σ1,Σ2 be connected, complete, totally geodesic submanifolds of M with Σ1⊆Σ2. If Σ1 is reflective, then Σ2 is reflective.
Proof.
We can assume that M is of non-compact type and o∈Σ1.
If Σ1 is non-semisimple, it follows from Proposition 4.2 in [4] that ToΣ1 is the normal space of a symmetric isotropy orbit of K. Then, by Theorem 1.2 of [2], Σ1 is maximal and hence Σ2=Σ1 is reflective.
If Σ1 is semisimple, we consider two cases:
Case 1: Σ2 is semisimple. Let τ be the geodesic reflection of M in the reflective submanifolds Σ1 and σ be the geodesic symmetry of M in o. Then h=σ∘τ is involutive and the eigenspaces of doh are ToΣ1 (associated with the eigenvalue −1) and νoΣ1 (associated with the eigenvalue 1). We decompose ToΣ2 into ToΣ2=ToΣ1⊕V with V⊆νoΣ1. Then doh(ToΣ2)=ToΣ2 and thus h∈KΣ2. Moreover, the non-trivial isometry h belongs to the kernel of the full slice representation of Σ2 at o. It follows from Proposition 2.8 that Σ2 is reflective.
Case 2: Σ2 is not semisimple. Write Σ2=Σ0×Σs (Riemannian product), where Σ0 is the Euclidean factor of Σ2 and Σs is the semisimple factor of Σ2. Obviously, we have Σ1⊆Σs. From Case 1 we conclude that Σs is reflective.
However, the semisimple part of a non-semisimple totally geodesic submanifold is never reflective, due to Corollary 3.3 in [2]. So Case 2 cannot occur.
∎
3. Fixed vectors of the slice representation
Let M=G/K be an irreducible, simply connected, Riemannian symmetric space with rk(M)≥2, where G=Io(M), K=Go and o∈M. The corresponding Cartan decomposition at o is g=k⊕p and the tangent space ToM is identified with p in the usual way.
Let Σ be a complete totally geodesic submanifold of M with o∈Σ and put ToΣ≅p′⊆p. Let G′⊆G be the subgroup of G consisting of the glide transformations of Σ and let K′=(G′)o. The Lie algebras of K′ and G′ are k′=[p′,p′]⊆k and g′=k′⊕p′⊆k⊕p=g respectively. Then Σ=G′/K′ and (G′,K′) is an almost effective symmetric pair. Note that normal vectors of Σ that are fixed by the slice representation correspond to G-invariant normal vector fields that are parallel along Σ.
We now state one of our main results; the proof will be given in the next section.
Theorem 3.1**.**
Let M=G/K be an irreducible, simply connected, Riemannian symmetric space with rk(M)≥2, where G=Io(M), K=Go and o∈M. Let Σ=G′/K′ be a (proper) totally geodesic submanifold of M with o∈Σ and dim(Σ)≥21dim(M), where G′ is the subgroup of G consisting of the glide transformations of Σ and K′=(G′)o. Let ρ be the slice representation of (K′)o on SO(νoΣ). Then the following statements are equivalent:
(i)
Σ* is maximal and there exists a non-zero vector in νoΣ that is fixed by the slice representation ρ.*
(ii)
Σ* is reflective and the complementary reflective submanifold is non-semisimple.*
(iii)
ToΣ* coincides, as a linear subspace, with the tangent space Tv(K⋅v) of a symmetric isotropy orbit.*
Note that the assumption in (i) for Σ to be maximal is necessary. Consider for example the symmetric space Spr/Ur and its maximal totally geodesic submanifold Sp1/U1×Spr−1/Ur−1=RH2×Spr−1/Ur−1. If we now consider Σ=Spr−1/Ur−1, then the slice representation of Ur−1 fixes any vector in ToRH2.
It follows from the classification of symmetric R-spaces (see [14]) that a non-semisimple extrinsically symmetric isotropy orbit (briefly, symmetric isotropy orbit) K⋅v⊂ToM has always half the dimension of ToM and that the Euclidean local factor of K⋅v is 1-dimensional. Moreover, Tv(K⋅v) and νv(K⋅v) are K-equivalent Lie triple systems.
In the next proposition we give a conceptual proof of these observations.
Proposition 3.2**.**
Let M=G/K be an irreducible, simply connected, Riemannian symmetric space with rk(M)≥2 and let K⋅v be a non-semisimple symmetric isotropy orbit, where 0=v∈ToM. Then there exists k∈K such that dok(νv(K⋅v))=Tv(K⋅v) (as linear subspaces of ToM). In particular, dim(νv(K⋅v))=dim(Tv(K⋅v))=21dim(M).
Proof.
For dual symmetric spaces the isotropy representations coincide. Hence we may assume that M is of non-compact type. From Lemma 4.1 in [4] we know that Tv(K⋅v) and νv(K⋅v) are complementary Lie triple systems in ToM and the abelian part of νv(K⋅v) coincides with Rv. Let Σ be the connected, complete, totally geodesic submanifold of M with ToΣ=νv(K⋅v). Let K′ be the connected subgroup of K with Lie algebra [νv(K⋅v),νv(K⋅v)]. Since Rv is the abelian part of νv(K⋅v), we have K′⋅v={v} and therefore K′⊆Kv, where Kv is the isotropy group of K at v. Moreover, since K′ is connected, we also have K′⊆(Kv)o. Let 0=w∈Tv(K⋅v) be tangent to the local Euclidean factor of K⋅v at v. Then (Kv)o fixes w, since (K,Kv) is a symmetric pair, and hence K′ fixes w as well. This means that w is a fixed vector of the image of the slice representation of K′ on the normal space νoΣ=Tv(K⋅v). From Lemma 3.2(ii) in [2] we see that the abelian part of Tv(K⋅v) contains Rw. Thus the Lie triple system Tv(K⋅v) in non-semisimple and so, by Proposition 4.2 in [4],
[TABLE]
and Rw is the abelian part of Tv(K⋅v). Since w is an arbitrary non-zero vector tangent to the local Euclidean factor of K⋅v at v, we conclude that the local Euclidean factor of K⋅v is 1-dimensional.
Since K′⊂(Kv)o and (Kv)o leaves invariant the Lie triple system νv(K⋅v), the set of fixed vectors of (Kv)o in νv(K⋅v) is Rv, the abelian part of this Lie triple system. The set of fixed vectors of (Kv)o in Tv(K⋅v) is Rw, the tangent space to the local Euclidean factor of this symmetric isotropy orbit. Altogether we see that V=Rv⊕Rw is the set of fixed vectors of (Kv)o in ToM. Clearly, V is a 2-dimensional Lie triple system. However, V is not abelian because w is not in the centralizer Zp(v)=νv(K⋅v) of v in p≅ToM. It follows that V is the tangent space at o of a totally geodesic real hyperbolic plane RH2 in M. Let Kˉ≅SO2 be the connected subgroup of K with Lie algebra kˉ=[V,V]≅so2. We can assume that ∥v∥=∥w∥. Since Kˉ acts transitively on the spheres in ToRH2, there exists k∈Kˉ with dok(v)=w. Using (3.1), this implies
[TABLE]
which finishes the proof.
∎
Assume that K⋅v is a symmetric isotropy orbit in p≅ToM. It was shown in [4], Lemma 4.1, that Tv(K⋅v) and νv(K⋅v) are complementary Lie triple systems and the abelian part of νv(K⋅v) coincides with Rv. We denote by Σ and Σ⊥ the connected, complete, totally geodesic submanifolds of M with ToΣ=Tv(K⋅v) and ToΣ⊥=νv(K⋅v) respectively. We write Σ=G′/K′, where G′ is the subgroup of G consisting of the glide transformations of Σ and K′=(G′)o.
Lemma 3.3**.**
The identity components of K′ and Kv coincide, that is, (K′)o=(Kv)o. Consequently, if M is of compact type, the symmetric spaces Σ and K⋅v are locally equivalent up to homothety in any local de Rham factor.
Proof.
Let GΣ={g∈I(M):g(Σ)=Σ} and KΣ be the isotropy group of GΣ at o. Then KΣ leaves ToΣ⊥ invariant and hence its identity component (KΣ)o fixes v since the abelian part of νv(K⋅v) coincides with Rv. Thus we have (KΣ)o⊆(Kv)o. On the other hand, Kv leaves ToΣ=Tv(K⋅v) invariant, which implies Kv⊆KΣ and, in particular, (Kv)o⊆(KΣ)o. Altogether this gives (Kv)o=(KΣ)o.
Assume that (K′)o is a proper subgroup of (KΣ)o. Since K′ is an ideal of KΣ and KΣ is compact, there exists a non-trivial connected normal subgroup H of (KΣ)o acting trivially on ToΣ=Tv(K⋅v). Then H is a non-trivial subgroup of Kv acting trivially on Tv(K⋅v). Therefore H acts trivially on the isotropy orbit K⋅v in ToM and also in its affine span, say v+V. Since K acts linearly on ToM, we have K⋅V⊆V, and therefore V=ToM since K acts irreducibly on ToM. So H must be trivial, which is a contradiction, and we conclude that
(K′)o=(Kv)o.
∎
Let M=G/K be an irreducible, simply connected, Riemannian symmetric space and K⋅v be a symmetric isotropy orbit. Then the following statements are equivalent:
(i)
K⋅v* is a non-semisimple symmetric isotropy orbit in ToM.*
(ii)
Tv(K⋅v)* is a non-semisimple Lie triple system in p.*
Since K⋅v is a symmetric isotropy orbit, the normal space νv(K⋅v) is a non-semisimple Lie triple system. If (iii) holds, then Tv(K⋅v)=k(νv(K⋅v)) is also a non-semisimple Lie triple system. Thus (ii) holds.
Finally, assume that (ii) holds. The tangent space Tv(K⋅v) and the normal space νv(K⋅v) of the symmetric isotropy orbit K⋅v are complementary Lie triple systems. By Lemma 3.3 we have (K′)o=(Kv)o. Then the symmetric isotropy orbit K⋅v is non-semisimple since (Kv)o=(K′)o fixes a non-zero vector of the Lie triple system Tv(K⋅v), since by assumption the Lie triple system Tv(K⋅v) is non-semisimple.
∎
Let K⋅v be a symmetric isotropy orbit in ToM. Then the tangent space Tv(K⋅v) and the normal space νv(K⋅v) are complementary reflective Lie triple systems in p≅ToM. By Theorem 1.2 in [2], νv(K⋅v) is a maximal Lie triple system in ToM (the proof of the maximality involves delicate geometric arguments). We will now prove that Tv(K⋅v) is also a maximal Lie triple system in ToM.
Proposition 3.5**.**
Let M=G/K be an irreducible, simply connected, Riemannian symmetric space with rk(M)≥2 and let K⋅v⊂ToM be a symmetric isotropy orbit. Then Tv(K⋅v) is a maximal Lie triple system in ToM.
Proof.
We can assume that M is of non-compact type. Let Σ=G′/K′ be the connected, complete, totally geodesic submanifold of M with ToΣ=Tv(K⋅v), where G′ is the subgroup of G consisting of the glide transformations of Σ. For the corresponding Lie algebras we have g′=[ToΣ,ToΣ]⊕ToΣ and k′=[ToΣ,ToΣ]. Let Σ⊥=G′′/K′′ be the connected, complete, totally geodesic submanifold of M with ToΣ⊥=νv(K⋅v), where G′′ is the subgroup of G consisting of the glide transformations of Σ⊥.
Now consider the slice representation ρ:K′→SO(νv(K⋅v)) and the isotropy representation χ:K′′→SO(νv(K⋅v)). By Lemma 3.2(i) in [2], ρ(K′) is a normal subgroup of χ(K′′). Since M is of non-compact type, Σ is simply connected and therefore K′ is connected. From Lemma 3.3 we then have K′=(Kv)o. Moreover, since Rv is the abelian part of νv(K⋅v), we also have (Kv)o=(KΣ⊥)o, and thus K′=(KΣ⊥)o. The image under the isotropy representation of (KΣ⊥)o on ToΣν=νv(K⋅v) coincides with that of K′′, and therefore ρ(K′)=χ(K′′).
Assume that Σ^ is a proper totally geodesic submanifold of M that properly contains Σ. Then we can write ToΣ^=ToΣ⊕V with {0}=V⊂νv(K⋅v). Note that k′=[ToΣ,ToΣ]⊆[ToΣ^,ToΣ^] and therefore K′ leaves V invariant. Since ρ(K′)=χ(K′′), also K′′ leaves V invariant. This implies that Σ⊥ is a Riemannian product Σ⊥=Σ1×Σ2 with ToΣ1=V. Note that Σ2 and Σ^ are complementary reflective totally geodesic submanifolds and that the isotropy group K2 at o of the glide transformations of Σ2 acts trivially on V=ToΣ1. Thus the image of the slice representation of K2 fixes any element of V. It follows from Lemma 3.2(ii) in [2] that Σ1 is flat.
Since the abelian part of a reflective Lie triple system, in this case ToΣ⊥, is 1-dimensional, this implies ToΣ1=Rv. Hence Σ2 is the semisimple part of the reflective submanifold Σ⊥. Corollary 3.3 in [2] tells us that Σ2 is not reflective, which is a contradiction. It follows that Σ is maximal and hence Tv(K⋅v) is a maximal Lie triple system in ToM.
∎
Lemma 3.6**.**
Let M=G/K be a Riemannian symmetric space of rank 1. Assume that there exists a totally geodesic submanifold
Σ=G′/K′ with dim(Σ)≥2 such that the connected slice representation ρ:(K′)o→SO(νoΣ) is trivial. Then M has constant curvature.
Proof.
From the assumption we obtain that νoΣ is a reflective Lie triple system, which implies that Σ is a reflective submanifold. Assume that M has non-constant curvature. Using duality, we can assume that M is a hyperbolic space over C, H or O.
If M is a complex hyperbolic space CHn=SU1,n/S(U1Un), then Σ is a real hyperbolic space RHn=SO1,no/SOn or a complex hyperbolic space CHk=SU1,k/S(U1Uk) for some k∈{0,…,n−1}. In the first case the slice representation of the isotropy group is equivalent to the standard representation of SOn on Rn, in the second case it is equivalent to the representation of S(U1Uk)≅Uk on Cn−k, where U1 acts canonically and Uk acts trivially.
If M is a quaternionic hyperbolic space HHn=Sp1,n/Sp1Spn, then Σ is a complex hyperbolic space CHn=SU1,n/S(U1Un) or a quaternionic hyperbolic space HHk=Sp1,k/Sp1Spk for some k∈{0,…,n−1}. In the first case the slice representation of the isotropy group is equivalent to the standard representation of S(U1Un)≅Un on Cn, in the second case it is equivalent to the representation of Sp1Spk on Hn−k, where Sp1 acts canonically and Spk acts trivially.
If M is a Cayley hyperbolic plane OH2=F4−20/Spin9, then Σ is a quaternionic hyperbolic plane HH2=Sp1,2/Sp1Sp2 or a Cayley hyperbolic line OH1=Spin1,8/Spin8≅RH8. In the first case the slice representation of the isotropy group is equivalent to the standard representation of Sp1Sp2 on H2, in the second case it is equivalent to one of the two inequivalent spin representations of Spin8 on R8
None of these slice representations is trivial and it follows that M has constant curvature.
∎
From the Slice Lemma 3.1 in [2] and Lemma 3.6 we obtain the following interesting result that generalizes Iwahori’s result in [9].
Corollary 3.7**.**
Let M be an irreducible Riemannian symmetric space and assume that there exists a non-flat proper totally geodesic submanifold Σ of M with trivial connected slice representation. Then M has constant curvature.
Proposition 3.8**.**
Let M=G/K be a simply connected Riemannian symmetric space of non-positive curvature and Σ=G′/K′ be a semisimple totally geodesic submanifold of M with o∈Σ, where G′⊆G are the glide transformations of Σ. Let Σ=Σ1×…×Σl and M=M0×…×Mg be the de Rham decompositions of Σ and M respectively, where the Euclidean factor M0 may be trivial. Assume that the slice representation ρ:K′→SO(νoΣ) is trivial. Then l≤g and, up to a permutation of the indices, Σi⊆Mi for 1≤i≤l. Moreover, if Σi is strictly contained in Mi, then Mi and Σi have constant curvature.
Proof.
Let i≥1 and consider the irreducible de Rham factor Mi of M. Let 0=z∈ToMi and write z=v+w with v∈ToΣ and w∈νoΣ. If k′∈K′, then k′v+w=k′v+k′w=k′z∈ToMi and hence k′z−z=k′v−v∈ToΣ. The space
[TABLE]
coincides with the affine subspace of ToΣ generated by the orbit K′⋅v. This affine subspace must contain 0∈ToΣ, because otherwise, the vector 0=u∈v+\mboxspan{k′v−v:k′∈K′} of minimal distance to [math] would be fixed by K′, which is a contradiction since Σ is semisimple and so K′ has no fixed non-zero vectors. Then −v∈\mboxspan{k′v−v:k′∈K′} and hence v∈\mboxspan{k′v−v:k′∈K′}⊆ToΣ. But k′v−v=k′z−z′∈ToMi for all k′∈K′, and therefore v∈ToMi. This shows that
[TABLE]
Assume that ToMi∩ToΣ={0}. A priori, ToMi∩ToΣ is not necessarily irreducible. However, since ToMi∩ToΣ is K′-invariant, it is the tangent space of a Riemannian factor Σi′ of Σ. By assumption, the slice representation of Σi′, considered as a totally geodesic submanifold of the de Rham factor Mi of M, is trivial. Observe that Mi is not the Euclidean factor, because otherwise Σi′ would be flat and so Σ would be non-semisimple. It follows that Σi′ is irreducible and thus a de Rham factor Σi. It then follows from Corollary 3.7 that Mi, and hence also Σi, has constant curvature.
∎
If in Proposition 3.8 the submanifold Σ is not semisimple, then the analogous conclusion holds by adding a flat in the factors of M which are not factors of Σ.
4. Extrinsic isometries of maximal totally geodesic submanifolds
Let Σ=G′/K′ be a connected, complete, totally geodesic submanifold of M with o∈Σ, where G′⊂G is the group of glide transformations of Σ and K′=(G′)o. The subgroup GΣ={g∈I(M):g(Σ)=Σ} of I(M) is in general neither connected nor effective on Σ. We always have σo∈GΣ, where σo is the geodesic symmetry of M at o. Note that G′ is a normal subgroup of GΣ and that K′ is a normal subgroup of KΣ=(GΣ)o.
Without loss of generality we may assume that M is of non-compact type. Then Σ is simply connected and hence K′ must be connected.
Remark 4.1**.**
Let X∈g and X∗ be the corresponding Killing vector field on M=G/K (see Section 2). For p∈Σ we denote by XpΣ the orthogonal projection of Xp∗ onto TpΣ. Then XΣ is a Killing vector field on Σ. It is well known, and standard to show, that there exists Z∈g′=[ToΣ,ToΣ]⊕ToΣ⊆g such that ZΣ=(Z∗)∣Σ=XΣ. The key fact of the argument is to show that a Killing vector field induced by G projects constantly along any flat totally geodesic submanifold.
Lemma 4.2**.**
If Σ is maximal and dim(k′)<dim(kΣ) (or equivalently, dim(g′)<dim(gΣ)), then Σ is a reflective submanifold of M.
Proof.
Let 0=X∈kΣ∖k′. By Remark 4.1, there exists Z∈k′ such that ZΣ=XΣ. By adding −Z, we may assume that XΣ=0=X∗∣Σ, where the last equality holds because X∈k′ and so the restriction of X∗ to Σ is always tangent to Σ.
Then Exp(tX) is a non-trival one-parameter group of isometries of M acting trivially on Σ (in particular, fixes o and leaves Σ invariant) and ϕt=doExp(tX) is a one-parameter group of linear isometries of ToM with ϕt∣ToΣ=idToΣ for all t∈R. Since Σ is a maximal totally geodesic submanifold of M, we have ToΣ={u∈ToM:ϕt(u)=u\mboxforallt∈R}. It follows that the dimension of νoΣ is even, say equal to 2d for some 0<d∈Z. We can find an orthonormal basis e1,f1,…,ed,fd of νoΣ and a1≥a2≥⋯≥ad>0 such that
[TABLE]
for all i∈{1,…,d}.
Assume that a1>ad and put t0=a11. Then V={u∈ToM:ϕto(u)=u} contains ToΣ⊕Re1⊕Rf1 and is perpendicular to Red⊕Rfd. Let Σ′ be the connected, complete, totally geodesic submanifold of M with ToΣ′=V. Then Σ′ is a proper submanifold of M and properly contains Σ, which is a contradiction to the maximality of Σ. It follows that there exists 0<a∈R so that a1=…=ad=a. Then ϕ2a1 is the orthogonal reflection of ToM in the subspace ToΣ. The corresponding isometry Exp(2a1X) is the geodesic reflection of M in Σ. Consequently, Σ is a reflective submanifold of M.
∎
We will now proceed with the proof of Theorem 3.1. The equivalence of (ii) and (iii) is an immediate consequence of Proposition 4.2 in [4]. If (iii) holds, then Σ is maximal by Proposition 3.5. Moreover, from Proposition 4.2 in [4], the abelian part of the Lie triple system νv(K⋅v)=Zp(v) is 1-dimensional and so it must coincide with Rv. Then ρ((K′)o) must leave Rv invariant. Since ρ((K′)o)⊂SO(νv(K⋅v)), the group ρ((K′)o) must fix v. This proves (i). Note that all these implications do not use the assumption that dim(Σ)≥21dim(M).
It remains to prove that (i) implies (ii). We only need to show that Σ is reflective. In fact, if ρ((K′)o) fixes v, then the complementary reflective submanifold is non-semisimple according to Lemma 3.2(ii) in [2]. The proof that Σ is reflective if (i) is satisfied will be done in several steps.
Case 1: Σ is non-semisimple.
Since Σ is maximal, it follows from Theorem 1.2 in [2] that ToΣ is the normal space of a symmetric isotropy orbit. Then, by Lemma 4.1 of [4], Σ is reflective. Moreover, from Lemma 3.2(ii) in [2] we see that νoΣ is non-semisimple and so, by Corollary 3.4, k(νoΣ)=ToΣ for some k∈K.
Case 2: Σ is simple.
By duality, we can assume that M is of compact type. We will use again the Riemannian symmetric space Mˉ=G/Gπ(o)=M/∼ that we encountered in Section 2, where p∼q if Gp=Gq and π:M→Mˉ is the canonical projection. We put oˉ=π(o) and identify ToM with ToˉMˉ by means of the isomorphism doπ:ToM→ToˉMˉ. Then we have (Goˉ)o=K. In this way, the identity component of the isotropy group at oˉ of the glide transformations of Σˉ=π(Σ) is canonically identified with the identity component of K′. Note that Σˉ is a maximal simple totally geodesic submanifold of Mˉ since Σ is a maximal simple totally geodesic submanifold of M.
We define V={ξ∈ToˉMˉ:ξ\mboxisfixedbyK′}. From our assumption (i) we have V={0}. Let Σˉ′ be the connected, complete, compact, totally geodesic submanifold of Mˉ with ToˉΣˉ′=V. Since Σˉ is simple, (K′)o acts irreducibly on ToˉΣˉ, which implies that V is perpendicular to ToˉΣˉ.
Since Σˉ′ is a compact Riemannian symmetric space, there exists a non-trivial closed geodesic γvˉ(t) of (minimal) period 1 for some vˉ∈V. Let gvˉ∈I(Mˉ) be as in Corollary 2.2. Since vˉ is fixed by (K′)o, gvˉ commutes with (K′)o and hence doˉgvˉ(ToˉΣˉ) is a (K′)o-invariant subspace on which (K′)o acts irreducibly. Then doˉgvˉ(ToˉΣˉ) is, as well as ToˉΣˉ, perpendicular to vˉ. Since dim(doˉgvˉ(ToˉΣˉ))=dim(ToˉΣˉ)≥21dim(ToˉMˉ) by assumption, these two subspaces intersect in a non-trivial (K′)o-invariant subspace. Since (K′)o acts irreducibly on ToˉΣˉ, this implies doˉgvˉ(ToˉΣˉ)=ToˉΣˉ, or equivalently, gvˉ(Σˉ)=Σˉ.
Let E±1⊂ToˉMˉ be the eigenspaces corresponding to the eigenvalues ±1 of doˉgvˉ. Note that E1 (resp. E−1) is the set of fixed vectors of doˉgvˉ (resp. of doˉ(σoˉ∘gvˉ)). Since gvˉ commutes with (K′)o, both eigenspaces are (K′)o-invariant. Let Σ±1 be the connected, complete, totally geodesic submanifold of Mˉ with ToˉE±1=E±1. By construction Σ1 and Σ−1 are reflective submanifolds of Mˉ with ToˉΣ±1=νoˉΣ∓1. We have ToˉMˉ=E1⊕E−1 and ToˉΣˉ=Eˉ1⊕Eˉ−1, where Eˉ±1=E±1∩ToˉΣˉ. Since (K′)o acts irreducibly on ToˉΣˉ we have either Eˉ1={0} or Eˉ−1={0}. Assume that Eˉ−1={0}. Then Eˉ1=ToˉΣˉ is a proper subset of E1, since vˉ∈E1 and vˉ is perpendicular to ToˉΣˉ. Thus Σˉ is a proper totally geodesic submanifold of Σ1, which contradicts the maximality of Σˉ. It follows that Eˉ1={0} and thus ToˉΣˉ=Eˉ−1. Since Σˉ is maximal, we get Σˉ=Σ−1. This shows that Σˉ, and hence also Σ, are reflective submanifolds.
Case 3: Σ is semisimple but not simple.
By duality, we can assume that M is of non-compact type. Then Σ=Σ1×…×Σa, where a≥2 and Σi is a simple totally geodesic submanifold of M. We put di=dim(Σi) and arrange the factors so that d1≤…≤da. Let Gi⊂G be the group of glide transformations of Σi and Ki=(Gi)o. Then the group G′ of glide transformation of Σ is G′=G1×…×Ga and we have K′=(G′)o=K1×…×Ka. We put Vi=ToΣi. By assumption, there exists 0=v∈νoΣ which is fixed under the slice representation of K′. Let R be the Riemannian curvature tensor of M at o.
Lemma 4.3**.**
For all 0=u∈ToΣ we have Ru,v∈/k′.
Proof.
We prove this by contradiction. Assume that there exists 0=u∈ToΣ with Ru,v∈k′. Since K′ fixes v, we have [k′,v]={0} and thus Ru,vv=0. Since M is a Riemannian symmetric space, this implies Ru,v=0 and hence Rdok(u),v=Rdok(u),dok(v)=dok∘Ru,v∘(dok)−1=0 for all k∈K′. Thus, if W is the linear span of the orbit K′⋅u in ToΣ, we have Rw,v=0 for all w∈W. There exists a non-empty subset J of {1,…,a} so that the K′-invariant subspace W can be written as W=⨁j∈JVj. Now consider the centralizer ZToM(W)={z∈p:[W,z]={0}}={z∈ToM:RW,z={0}} of W in ToM≅p. Since M is irreducible, ZToM(W) is a proper subset of ToM. Obviously, we have Rv⊕(⨁j∈/JVj)⊂ZToM(W). It follows that W+ZToM(W) is a proper Lie triple system in ToM containing ToΣ as a proper subset. This contradicts the maximality of Σ.
∎
To finish the proof of Theorem 3.1, we need to show that Σ is reflective. We will prove this by contradiction. Thus, assume that Σ is not reflective. Let 0=x∈V1 and define Bx=Rx,v∈k⊂so(ToM). Then, by Lemma 4.3, we have Bx∈/k′. Since R is K-invariant and K^1=K2×…×Ka acts trivially on V1, we have [Bx,k^1]={0}. Define the one-parameter subgroup hxt of SO(ToM) by hxt=exp(tBx) and denote by Hxt the corresponding one-parameter group in K given by hxt=doHxt. By Lemma 4.3 and Lemma 4.2, and since Σ is maximal, Bx(ToΣ) is not contained in ToΣ. Thus we have hxt(ToΣ)=ToΣ for sufficiently small t=0. Equivalently, Σt=Hxt(Σ)=Σ for sufficiently small t=0.
We now consider the totally geodesic submanifold Σ^1=Σ2×…×Σa of M. The group G^1 of glide transformations of Σ^1 is G^1=G2×…×Ga and the isotropy group at o is K^1=K2×…×Ka. Since [Bx,k^1]={0},
the totally geodesic submanifold Σ^1t=Hxt(Σ^1) has the same glide isotropy group K^1 at o as Σ^10=Σ^1. This implies dok(ToΣ^1t)=ToΣ^1t for all t and all k∈K^1. Consequently, V^1t=ToΣ^1t is a K^1-invariant Lie triple system in ToM for all t.
Lemma 4.4**.**
If t=0 is sufficiently small, then V^1t∩ToΣ={0}.
Proof.
The intersection V^1t∩ToΣ is a K^1-invariant subspace of ToΣ (recall that K^1 acts trivially on V1). Thus there exist a (possibly empty) subset Jt of {2,…,a} and a (possibly trivial) subspace Vt of V1 such that V^1t∩ToΣ=Vt⊕(⨁j∈JtVj). Since V1 is perpendicular to V^10=V^1=ToΣ^1, we have Vt={0} for sufficiently small t.
On the one hand, if the intersection V^1t∩ToΣ is non-trivial, it is the tangent space of a product of factors of Σ, which implies ToΣ⊆(V^1t∩ToΣ)+ZToM(V^1t∩ToΣ). In fact, we have equality here since Σ is maximal.
On the other hand, since V^1t∩ToΣ is K^1-invariant, it is the tangent space of a product of factors of Σ^1t (recall that K^1 is the isotropy group at o of the glide transformations of Σ^1t for all t). Since Σ^1t is a (not necessarily simple) factor of Σt=Hxt(Σ), we conclude that this intersection is the tangent space to a (not necessarily simple) factor of Σt. This implies ToΣt⊆(V^1t∩ToΣ)+ZToM(V^1t∩ToΣ).
Altogether we now see that the Lie triple system (V^1t∩ToΣ)+ZToM(V^1t∩ToΣ) contains ToΣ+ToΣt. Recall that, for sufficiently small t=0, the subspace
ToΣt=hxt(ToΣ) is different from ToΣ and so ToΣ is a proper subspace of ToΣ+ToΣt. It follows that ToΣ is a proper subspace of the Lie triple system (V^1t∩ToΣ)+ZToM(V^1t∩ToΣ). Since Σ is maximal, this implies (V^1t∩ToΣ)+ZToM(V^1t∩ToΣ)=ToM and thus V^1t∩ToΣ={0} for sufficiently small t=0.
∎
Let π:ToM→νoΣ be the orthogonal projection from ToM onto the normal space νoΣ. Since νoΣ is K′-invariant and K^1⊆K′, νoΣ is also K^1-invariant. This implies that π is K^1-equivariant. From Lemma 4.4 we then obtain that π:ToΣ^1t→νoΣ is injective for sufficiently small t=0.
The totally geodesic submanifold Σ^1t of M is isometric to Σ^1 via Hxt and we have HxtK^1(Hxt)−1=K^1=K2×…×Ka. Furthermore, ToΣ^1t=hxt(V2)⊕…⊕hxt(Va) and Ki acts irreducibly on hxt(Vi) and trivially on hxt(Vj) for i,j∈{2,…,a}, i=j. From this and the fact that π is K^1-equivariant it is not hard to see that we have the orthogonal decomposition π(ToΣ^1t)=π(hxt(V2))⊕…⊕π(hxt(Va)) and that K^1 acts irreducibly on π(hxt(Vi)) and trivially on π(hxt(Vj)) for i,j∈{2,⋯,a}, i=j. Moreover, the irreducible representation of K^1 on π(hxt(Vi)) is equivalent to the representation of K^1 on Vi=ToΣi for i≥2.
Since v=0 is fixed by K^1, the subspace π(ToΣ^1t) is perpendicular to Rv. Let W⊂νoΣ be the linear span of all K^1-invariant subspaces of νoΣ on which the representation of K^1 is equivalent to the representation of K^1 on Vi for some i∈{2,…,a}. Then W is perpendicular to Rv and contains the subspace π(ToΣ^1t) of dimension d2+…+da=dim(Σ)−d1. If π(ToΣ^1t) is a proper subspace of W, then dim(W)≥(d2+…+da)+di for some i∈{2,…a}. Since d1≤⋯≤da, we obtain that dim(W)≥dim(Σ) and so the codimension of Σ is at least dim(Σ)+1, which is a contradiction. Hence we have W=π(ToΣ^1t) and K^1 acts on W as an s-representation, equivalent to the isotropy representation of K^1 on ToΣ^1.
Next, K1 commutes with K^1 and leaves νoΣ invariant. From the definition of W we therefore see that K1 leaves W invariant. Moreover, K∣W1={k1∣W:k1∈K1} lies in the centralizer, and so in the normalizer of K^∣W1 in SO(W). Since K∣W1 acts as an s-representation, we obtain K∣W1⊆K^∣W1 from Lemma 5.2.2 in [1].
It is well-known that the dimension of the centre of the isotropy group of an irreducible Riemannian symmetric space is either [math] or 1 (and in the latter case the space is Hermitian symmetric). If the irreducible Riemannian symmetric space Σ1 is not a real hyperbolic plane RH2, then K1 is not abelian and K1 does not act effectively on W. Interchanging Σ1 with Σ^1, one can show with similar arguments that there exists a K1-invariant subspace V~1 of νoΣ on which the representation of K1 is equivalent to the one of K1 on ToΣ1=V1. This subspace satisfies V~1∩(Rv⊕W)={0}. Since dim(W)=dim(Σ)−d1, we conclude that dim(νoΣ)≥dim(Σ)+1, which is a contradiction. We conclude that Σ1 is a real hyperbolic plane RH2 and therefore dim(W)=dim(Σ)−2.
This implies dim(νoΣ)≥dim(Σ)−1, since v is perpendicular to W. Thus there exists w∈νoΣ, possibly w=0 if dim(νoΣ)=dim(Σ)−1, perpendicular to Rv⊕W such that νoΣ=Rw⊕Rv⊕W. The group K1 fixes w. Let k∈K1 be non-trivial. Then there exists k^∈K^1 such that kk^−1 acts trivially on W and hence on
νoΣ. Note that, since k^ acts trivially on Σ1, kk^−1 is a non-trivial element in GΣ and lies in the kernel of the slice representation (recall that the image of an s-representation coincides with its own connected normalizer in the full orthogonal group, see e.g. Lemma 5.2.2 in [1]). Then, by Proposition 2.8, Σ is reflective, which contradicts our assumption that Σ is non-reflective. This finishes the proof of Theorem 3.1.
5. The index of exceptional Riemannian symmetric spaces
Let M=G/K be an n-dimensional irreducible Riemannian symmetric space of non-compact type and denote by r the rank of M. Let Σ=G′/K′ be a connected totally geodesic submanifold of M with codimension d≥1 and denote by rΣ the rank of Σ. We can assume that o∈Σ and G′⊆G is the group of glide transformations of Σ. Then we have dim(G′)−dim(K′)=dim(Σ)=n−d. The dimension of a principal orbit of the isotropy action of K′ on Σ is dim(K′)−dim(K0′)=(n−d)−rΣ, where K0′ is the principal isotropy group of K′ (that is, the isotropy group of K′ at a point in a principal orbit of the K′-action on Σ). Altogether this implies
[TABLE]
Let M∗=G∗/K and Σ∗=G′∗/K′ be the Riemannian symmetric spaces of compact type that are dual to M=G/K and Σ=G′/K′ respectively. Let i(G∗) denote the index of G∗, where the compact Lie group G∗ is considered as a Riemannian symmetric space of compact type. Using [3] we get the inequality
Since r−rΣ and dim(K0′) are non-negative, the previous equation implies
[TABLE]
We introduce some notations. By ℓΣ we denote the dimension of the principal isotropy algebra k0′ of Σ=G′/K′, thus ℓΣ=dim(K0′). We define the integers ΩM and ΛΣM by
[TABLE]
We will frequently use some data about symmetric spaces, which we summarize in Table 2.
Inequalities (5.3) and (5.4) give the following estimates for the codimension of Σ:
Proposition 5.1**.**
Let M=G/K be an irreducible Riemannian symmetric space of non-compact type and Σ be a connected totally geodesic submanifold of M with codim(Σ)≥1. Then
[TABLE]
Using the second inequality in Proposition 5.1 we can confirm the conjecture for some symmetric spaces where it was previously unknown.
Theorem 5.2**.**
For the following symmetric spaces the index coincides with the reflective index:
(i)
For M=E62/SU6Sp1 we have i(M)=12.
(ii)
For M=E77/SU8 we have i(M)=27.
(iii)
For M=E88/SO16 we have i(M)=56.
(iv)
For M=Spr(R)/Ur (r≥3) we have i(M)=2(r−1).
Proof.
The index of compact simple Lie groups was calculated in [3] and the reflective index of irreducible Riemannian symmetric spaces M was calculated in [2]. Using these results we obtain:
If M=G/K=E62/SU6Sp1, then i(E6)=26, dim(M)=40, rk(M)=4, dim(K)=38. This gives codim(Σ)≥21(26+40−4−38)=12=ir(M).
If M=G/K=E77/SU8, then i(E7)=54, dim(M)=70, rk(M)=7, dim(K)=63. This gives codim(Σ)≥21(54+70−7−63)=27=ir(M).
If M=G/K=E88/SO16, then i(E8)=112, dim(M)=128, rk(M)=8, dim(K)=120. This gives codim(Σ)≥21(112+128−8−120)=56=ir(M).
If M=G/K=Spr(R)/Ur, then i(Spr)=4(r−1), dim(M)=r(r+1), rk(M)=r, dim(K)=r2. This gives codim(Σ)≥21(4r−4+r2+r−r−r2)=2r−2=ir(M).
∎
This finishes the proof of Theorem 1.2. To finish the proof of Theorem 1.1, we need to investigate further the symmetric spaces E66/Sp4, E7−5/SO12Sp1, E7−25/E6U1 and E8−24/E7Sp1. We will investigate these four spaces individually, but first summarize in the following corollary what the first inequality in Proposition 5.1 tells us in this situation.
Corollary 5.3**.**
For the index i(M) of M and we have:
(i)
If M=E66/Sp4, then 13≤i(M)≤ir(M)=14.
(ii)
If M=E7−5/SO12Sp1, then 23≤i(M)≤ir(M)=24.
(iii)
If M=E7−25/E6U1, then 13≤i(M)≤ir(M)=22.
(iv)
If M=E8−24/E7Sp1, then 42≤i(M)≤ir(M)=48.
Moreover, if i(M)<ir(M) and Σ is a totally geodesic submanifold of M with codim(Σ)=i(M), then
(i)
ΛΣM=0, if M=E66/Sp4.
(ii)
0≤ΛΣM≤1, if M=E7−5/SO12Sp1.
(iii)
0≤ΛΣM≤16, if M=E7−25/E6U1.
(iv)
0≤ΛΣM≤10, if M=E8−24/E7Sp1.
Proof.
If M=G/K=E66/Sp4, then i(E6)=26, dim(M)=42, rk(M)=6, dim(K)=36. This gives codim(Σ)≥21(26+42−6−36)=13<14=ir(M).
If M=G/K=E7−5/SO12Sp1, then i(E7)=54, dim(M)=64, rk(M)=4, dim(K)=69. This gives codim(Σ)≥21(54+64−4−69)=22.5<24=ir(M).
If M=G/K=E7−25/E6U1, then i(E7)=54, dim(M)=54, rk(M)=3, dim(K)=79. This gives codim(Σ)≥21(54+54−3−79)=13<22=ir(M).
If M=G/K=E8−24/E7Sp1, then i(E8)=112, dim(M)=112, rk(M)=4, dim(K)=136. This gives codim(Σ)≥21(112+112−4−136)=42<48=ir(M).
The statements about ΛΣM follow immediately from the inequality in Proposition 5.1 and the above calculations.
∎
We now proceed with individual arguments for the four remaining exceptional symmetric spaces.
Theorem 5.4**.**
For M=E66/Sp4 we have i(M)=ir(M)=14.
Proof.
We already know from [2] that ir(M)=14 and from Corollary 5.3 that i(M)≥13. Let Σ be a maximal totally geodesic submanifold of M and assume that codim(Σ)=13. Using Corollary 5.3 we obtain rk(M)−rk(Σ)+ℓΣ=ΛΣM=0. Consequently, we have rk(Σ)=rk(M)=6 and ℓΣ=0. The maximal totally geodesic submanifolds of maximal rank in E66/Sp4 were classified by Chen and Nagano in [5], and independently by Ikawa and Tasaki in [8]. Up to congruency, there are only three such submanifolds, namely
(i)
Σ=R×SO5,5o/SO5SO5, which is reflective and codim(Σ)=16;
(ii)
Σ=RH2×SL6(R)/SO6, which is reflective and codim(Σ)=20;
(iii)
Σ=SL3(R)/SO3×SL3(R)/SO3×SL3(R)/SO3, which is non-reflective and codim(Σ)=27.
In all three cases we have codim(Σ)>14=ir(M), which is a contradicition. Thus we can conclude that i(M)=ir(M)=14.
∎
For the remaining three exceptional symmetric spaces we will first prove a theoretical result that will allow us to reduce the number of cases that need to be considered. The following lemma is a slight generalization of a result by Iwahori [9] which states that if an irreducible Riemannian symmetric space admits a totally geodesic hypersurface, then it must be a space of constant curvature.
Lemma 5.5**.**
Let M be a Riemannian symmetric space with de Rham decomposition M=M0×M1×…×Mg, where M0 is a (possibly [math]-dimensional) Euclidean space and M1,…,Mg, g≥1, are irreducible Riemannian symmetric spaces. Let Σ be a totally geodesic hypersurface of M. Then Σ=M0×M1×…Mj−1×Σj×Mj+1×…×Mg, where Σj is a totally geodesic hypersurface of Mj and Mj is a space of constant curvature for some j∈{0,…,g}.
Proof.
We can assume that all spaces contain the base point o∈M. The intersection ToΣ∩ToMj is a Lie triple system in ToMj of dimension equal to dim(Mj)−1 or dim(Mj). Iwahori’s result therefore implies that ToMj⊆ToΣ if Mj is not of constant curvature, or equivalently, Mj⊆Σ if Mj is not of constant curvature. We can therefore assume that Mj has constant curvature for all j∈{1,…,g}. Denote by Rj the Riemannian curvature tensor of Mj, by R the Riemannian curvature tensor of M, and by κj the constant sectional curvature of Mj. Let X∈ToΣ∩ToMj be a unit vector and Y=Y0+…+Yg∈ToΣ with Yj∈ToMj and ⟨X,Y⟩=0. Then we have κjYj=Rj(Yj,X)X=R(Y,X)X∈ToΣ∩ToMj and therefore Yj∈ToΣ∩ToMj for all j∈{1,…,g}. This implies ToΣ=(ToΣ∩ToM0)⊕…⊕(ToΣ∩ToMg) and hence ToMj⊆ToΣ for all but one index j∈{0,…,g}. For this j we define Σj to be the totally geodesic hypersurface of Mj corresponding to the Lie triple system ToΣ∩ToMj in ToMj. This implies the assertion.
∎
Proposition 5.6**.**
Let Σ be a reducible maximal totally geodesic submanifold of an irreducible Riemannian symmetric space M of non-compact type. Assume that the de Rham decomposition of Σ contains a real hyperbolic space RHk (k≥2), a complex hyperbolic space CHk (k≥2), the symmetric space SL3(R)/SO3, or the symmetric space SO2,2+ko/SO2SO2+k (k≥1 odd). Then either Σ=RHk1×RHk2 for some k1,k2≥2, or there exists a reflective submanifold Σ′ of M with dim(Σ′)≥dim(Σ).
Proof.
As usual, we write M=G/K with G=Io(M) and K=Go with o∈M. We can assume that o∈Σ and write Σ=G′/K′, where G′⊂G is the group of glide transformations of Σ and K′=(G′)o. We denote the de Rham factor specified in the assertion by N. We can assume that o∈N and write N=G′′/K′′, where G′′⊂G is the group of glide transformations of N and K′′=(G′′)o.
There exists a 2-dimensional reflective submanifold P of N with o∈P such that the geodesic reflection τ in its perpendicular reflective submanifold P⊥ at o is inner, i.e. τ∈K′′⊆K′⊂K. Explicitly,
(i)
If N=RHk, then P=RH2 and P⊥=RHk−2;
(ii)
If N=CHk, then P=CH1 and P⊥=CHk−1;
(iii)
If N=SL3(R)/SO3, then P=RH2 and P⊥=RH2×R;
(iv)
If N=SO2,2+ko/SO2SO2+k, then P=CH1 and
P⊥=SO2,2+(k−1)o/SO2SO2+(k−1).
If Σ is non-semisimple, then Σ is reflective by Corollary 4.4 in [2] and we can choose Σ′=Σ. We assume from now on that Σ is semisimple and write Σ=N×Σˉ with a semisimple totally geodesic submanifold Σˉ of M containing o.
Let T be the closure of the subgroup of K⊆SO(ToM) (via the isotropy representation) that is generated by τ. Note that doτ is the identity on ToP⊥⊕ToΣˉ and minus the identity on ToP. In particular, the cardinality of {g∣ToΣ:g∈T} is equal to 2.
Assume that dim(T)>0. Then the kernel TΣ of the Lie group homomorphism T→SO(ToΣ),g↦g∣ToΣ has positive dimension. Since K′ acts almost effectively on Σ, we cannot have TΣ⊆K′. Thus for the corresponding Lie algebras we have tΣ⊆k′ and tΣ⊆gΣ. It then follows from Lemma 4.2 that Σ is reflective and we can choose Σ′=Σ.
From now on we assume that dim(T)=0, or equivalently, that τ has finite order. By construction, τ has even order of the form 2sq with 0<q∈Z odd and 0<s∈Z. We replace τ by τq. Then τ has order 2s. If s>1, τ2s−1 is involutive and its set of fixed vectors must coincide with ToΣ since Σ is maximal. This implies that Σ is reflective and we can choose Σ′=Σ.
Thus we are left with the case that τ is an involution. If the set of fixed vectors of τ in the normal space νoΣ is trivial, then P⊥×Σˉ⊂Σ is reflective and it follows from Corollary 2.9 that Σ is reflective and we can choose Σ′=Σ.
So let us assume that the subspace V of fixed vectors of τ in νoΣ satisfies dim(V)>0. Since τ is involutive, the totally geodesic submanifold Σ′ of M with o∈Σ′ and ToΣ′=ToP⊥⊕ToΣˉ⊕V is reflective. If dim(V)≥2, then dim(Σ′)≥dim(Σ).
Thus it remains to analyze the case dim(V)=1. Then Σ^=P⊥×Σˉ is a totally geodesic hypersurface of Σ′. Recall that P⊥ is irreducible unless N=SL3(R)/SO(3), where P⊥=RH2×R. Let Σ^=Σ^0×…×Σ^g be the de Rham decomposition of Σ^, where Σ^0 is the (possibly [math]-dimensional) Euclidean factor. We can arrange the indices so that Σ^g=P⊥ unless N=SL3(R)/SO(3), in which case Σ^g=RH2 and the 1-dimensional Euclidean factor in P⊥=RH2×R coincides with Σ^0.
Then, by Lemma 5.5, we have Σ′=Σ^0×Σ^1×…Σ^j−1×Σj′×Σ^j+1×…×Σ^g, where Σ^j is a totally geodesic hypersurface of Σj′ and Σj′ is a space of constant curvature for some j∈{0,…,g}.
Assume that g≥3. Then there exists i∈{1,…,g−1} with i=j. The subspace Z=ZToM(ToΣ^i)⊕ToΣ^i of ToM is a Lie triple system in ToM containing both ToΣ and ToΣ′. Moreover, Z is a proper subset of ToM since M is irreducible. By construction, ToΣ′ is not contained in ToΣ and therefore ToΣ is a proper subset of Z, which contradicts the maximality of Σ. Consequently, we have g∈{1,2}. If g=1, then Σˉ is trivial, which contradicts the assumption that Σ=N×Σˉ is reducible. Consequently, we have g=2
and therefore Σ^=Σ^0×Σ^1×Σ^2.
If j∈{0,2}, we consider the subspace ZToM(ToΣ^1)⊕ToΣ^1 of ToM, which again is a Lie triple system in ToM containing both ToΣ and ToΣ′. Then the same argument as in the previous paragraph leads to a contradiction. We therefore must have j=1. Then Σˉ=Σ^1=RHk1 and Σ1′=RHk1+1 for some k1≥2, since Σ^1 is a non-flat totally geodesic hypersurface of the irreducible space Σ1′ of constant negative curvature. It follows that Σ=N×RHk1. Finally, following the same arguments as above with RHk1 instead of N, we conclude that there exists a reflective submanifold Σ′ of M with dim(Σ′)≥dim(Σ), or Σ=RHk1×RHk2 for some k1,k2≥2.
∎
We will use Proposition 5.6 to determine the index of E7−5/SO12Sp1, E7−25/E6U1 and E8−24/E7Sp1. We frequently need the values of ℓΣ when Σ is a hyperbolic space and therefore list them here:
[TABLE]
We will also use frequently the fact that ℓΣ=ℓΣ1+…+ℓΣg if Σ is a Riemannian product Σ=Σ1×…×Σg.
Theorem 5.7**.**
For M=E7−5/SO12Sp1 we have i(M)=ir(M)=24.
Proof.
We already know from [2] that ir(M)=24 and from Corollary 5.3 that i(M)≥23. Let Σ be a maximal totally geodesic submanifold of M and assume that codim(Σ)=23, that is dim(Σ)=41. Using Corollary 5.3 we obtain (rk(M)−rk(Σ))+ℓΣ=ΛΣM≤1. Since rk(M)=4, we have (rk(Σ),ℓΣ)∈{(4,0),(3,0),(4,1)}. Since there exist no irreducible symmetric spaces with rank 3 or 4 and of dimension 41, Σ must be reducible.
Assume that the de Rham decomposition of Σ contains a rank one factor Σ1. Since ℓΣ1≤1, we have Σ1∈{RH2,RH3,CH2}. It then follows from Proposition 5.6 that there exists a reflective submanifold Σ′ of M with dim(Σ′)≥dim(Σ), which is a contradiction.
Finally, assume that rk(Σ)=4 and Σ=Σ1×Σ2 with rk(Σi)=2 and Σi irreducible. Since dim(Σ)=41, one of the two factors must have odd dimension. The only odd-dimensional irreducible symmetric space of noncompact type and rank 2 is SL3(R)/SO3. Using again Proposition 5.6 we see that there exists a reflective submanifold Σ′ of M with dim(Σ′)≥dim(Σ), which is a contradiction.
Consequently Σ cannot exist and it follows that i(M)=24=ir(M).
∎
Theorem 5.8**.**
For M=E7−25/E6U1 we have i(M)=ir(M)=22.
Proof.
We already know from [2] that ir(M)=22. From Corollary 5.3 we know that i(M)≥13. Let Σ be a maximal totally geodesic submanifold of M and assume that codim(Σ)∈{13,…,21}, that is dim(Σ)∈{33,…,41}. From the results in [2] we can assume that Σ is semisimple. Using Corollary 5.3 we obtain (rk(M)−rk(Σ))+ℓΣ=ΛΣM≤16. Since rk(M)=3, we have ℓΣ≤14 if rk(Σ)=1, ℓΣ≤15 if rk(Σ)=2, and ℓΣ≤16 if rk(Σ)=3.
The hyperbolic spaces FHk with ℓFHk≤16 are RHk for k∈{2,3,4,5,6,7} (then ℓRHk∈{0,1,3,6,10,15}), CHk for k∈{2,3,4,5} (then ℓCHk∈{1,4,9,16}), and HHk for k∈{2,3} (then ℓHHk∈{6,13}). Since dim(Σ)≥33 and ℓΣ≤16, we easily see that Σ cannot have rank 1 or be a Riemannian product of rank 1 symmetric spaces.
Assume that rk(Σ)=2. Since Σ cannot be a Riemannian product of two rank 1 symmetric spaces, Σ is irreducible. The irreducible rank 2 symmetric spaces Σ with dim(Σ)∈{33,…,41} are
SO2,qo/SO2SOq (q∈{17,18,19,20}), SU2,q/S(U2Uq) (q∈{9,10}) and Sp2,5/Sp2Sp5. In all cases we have ℓΣ>15 (see Table 2) and therefore no such Σ exists.
Assume that rk(Σ)=3. We already know that Σ cannot be the product of three rank 1 symmetric spaces. Assume that Σ=Σ1×Σ2, where Σ1 has rank 1 and Σ2 has rank 2. Using Proposition 5.6 we see that Σ1 must be HH2 or HH3. Thus dim(Σ1)∈{8,12} and hence 21≤dim(Σ2)≤33 and ℓΣ2≤10. Using Table 2 we see that the only possibility is Σ2=Sp2,3/Sp2Sp3, which satisfies dim(Σ2)=24 and ℓΣ2=9. However, since dim(Σ)≥33, we must have Σ1=HH3, which gives ℓΣ=ℓΣ1+ℓΣ2=13+9=22, which is a contradiction. We conclude that Σ cannot be the Riemannian product of a rank 1 symmetric space and a rank 2 symmetric space. Finally, assume that Σ is irreducible. Using Table 2 we obtain that the irreducible rank 3 symmetric spaces Σ with dim(Σ)∈{33,…,41} and ℓΣ≤16 are Σ=SU3,6/S(U3U6) (then dim(Σ)=36 and ℓΣ=11) and Σ=Sp3,3/Sp3Sp3 (then dim(Σ)=36 and ℓΣ=9). We have ΩM=26. For Σ=SU3,6/S(U3U6) we have ΛMΣ=ℓΣ=11 and hence codim(Σ)=18<18.5=21(ΩM+ΛMΣ), which contradicts Proposition 5.1. Thus Σ=SU3,6/S(U3U6) is not possible. The restricted root system of M=E7−25/E6U1 is of type (C3), the six short roots have multiplicity 8 and the three long roots have multiplicity 1. The restricted root system of Σ=Sp3,3/Sp3Sp3 is also of type (C3), but the six short roots have multiplicity 4 and the three long roots have multiplicity 3. Due to the multiplicities of the long roots, it is not possible to realize the second root system as a subsystem of the first one, which implies that Σ=Sp3,3/Sp3Sp3 is not possible either.
∎
Theorem 5.9**.**
For M=E8−24/E7Sp1 we have i(M)=ir(M)=48.
Proof.
We already know from [2] that ir(M)=48. From Corollary 5.3 we know that i(M)≥42. Let Σ be a maximal totally geodesic submanifold of M and assume that codim(Σ)∈{42,…,47}, that is, dim(Σ)∈{65,…,70}. From the results in [2] we can assume that Σ is semisimple. Using Corollary 5.3 we obtain (rk(M)−rk(Σ))+ℓΣ=ΛΣM≤10. Since rk(M)=4, we have ℓΣ≤7 if rk(Σ)=1, ℓΣ≤8 if rk(Σ)=2, ℓΣ≤9 if rk(Σ)=3 and ℓΣ≤10 if rk(Σ)=4.
The hyperbolic spaces FHk with ℓFHk≤10 are RHk for k∈{2,3,4,5,6} (then ℓRHk∈{0,1,3,6,10}), CHk for k∈{2,3,4} (then ℓCHk∈{1,4,9}), and HH2 (then ℓHH2=6). Since dim(Σ)≥65 and ℓΣ≤10, we easily see that Σ cannot have rank 1 or be a Riemannian product of rank 1 symmetric spaces.
Assume that rk(Σ)=2. Since Σ cannot be a Riemannian product of two rank 1 symmetric spaces, Σ is irreducible. The irreducible rank 2 symmetric spaces Σ with dim(Σ)∈{65,…,70} are
SO2,qo/SO2SOq (q∈{33,34,35}) and SU2,17/S(U2U17). In all cases we have ℓΣ>8 (see Table 2) and therefore no such Σ exists.
Assume that rk(Σ)=3. We already know that Σ cannot be the product of three rank 1 symmetric spaces. Assume that Σ=Σ1×Σ2, where Σ1 has rank 1 and Σ2 has rank 2. We must have ℓΣ1+ℓΣ2=ℓΣ≤9. Using Proposition 5.6 we obtain that Σ1=HH2, which implies dim(Σ2)∈{57,…,62} and ℓΣ2≤3. From Table 2 we see that no such Σ exists. Finally, for the case that Σ is irreducible, we see from Table 2 that there exists no such Σ with dim(Σ)∈{65,…,70} and ℓΣ≤9.
Assume that rk(Σ)=4. We already know that Σ cannot be the product of four rank 1 symmetric spaces. Assume that Σ=Σ1×Σ2×Σ3, where Σ1,Σ2 have rank 1 and Σ3 has rank 2. We must have ℓΣ1+ℓΣ2+ℓΣ3=ℓΣ≤10. It follows that at least one of the two rank 1 symmetric spaces is a real or complex hyperbolic space. Proposition 5.6 then implies that this case cannot occur. Assume that Σ=Σ1×Σ2, where Σ1 has rank 1 and Σ2 has rank 3. Using Proposition 5.6 we obtain that Σ1=HH2, which implies dim(Σ2)∈{57,…,62} and ℓΣ2≤4. From Table 2 we see that no such Σ exists. Assume that Σ is irreducible. Then dim(Σ)∈{65,…,70}, rk(Σ)=4 and ℓΣ≤10, and from Table 2 we see that no such Σ exists. Finally, assume that Σ=Σ1×Σ2, where Σ1 and Σ2 are irreducible and of rank 2. Then one of the two spaces, say Σ1, must satisfy dim(Σ1)≥33, rk(Σ1)=2 and ℓΣ1≤10. From Table 2 we see that no such Σ exists.
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