Semisimplicity of indefinite extrinsic symmetric spaces and mean curvature
aa r X i v : . [ m a t h . DG ] M a y Semisimplicity of indefinite extrinsic symmetric spaces andmean curvature
Ines KathNovember 26, 2018
Abstract
Improving a result of Eschenburg and Kim we give a criterion for semisimplicityof pseudo-Riemannian extrinsic symmetric spaces in terms of the shape operatorwith respect to the mean curvature vector.
A non-degenerate submanifold M of a pseudo-Euclidean space R p,q is called extrinsicsymmetric if for each x ∈ M the reflection s x : R p,q → R p,q at the affine normal spacemaps M to M . Extrinsic symmetric spaces are exactly those complete submanifolds of R p,q that have a parallel second fundamental form [S]. A large class of examples canbe obtained in the following way. Let g be a Lie algebra and assume that g admits anad( g )-invariant non-degenerate inner product and an orthogonal Cartan decomposition g = k ⊕ p , i.e., [ k , k ] ⊂ k , [ k , p ] ⊂ p and [ p , p ] = k . Define G := h exp ad( X ) | p | X ∈ k i .Suppose that ξ ∈ p satisfies ad( ξ ) = − ad( ξ ). Then the orbit G · ξ is an extrinsicsymmetric space in p ∼ = R p,q . These examples will be called extrinsic symmetric spacesof Ferus type. This notion was introduced in [KE] for the following reason. In thecase where the ambient space is Euclidean Ferus [F1, F2] proved that each extrinsicsymmetric space decomposes into a flat factor and a compact extrinsic symmetric spaceand that, moreover, each compact extrinsic symmetric space M ⊂ R n arises in the waydescribed above from a compact Lie algebra g . This leads to a complete classificationin the case of a Euclidean ambient space.In the pseudo-Riemannian situation we cannot expect to get such a classification forarbitrary ( p, q ) since, in general, a pseudo-Riemannian extrinsic symmetric space is notof Ferus type and even those that are cannot be classified since in ‘most cases’ theLie algebra g is not reductive. However, as a first step one can determine all extrinsicsymmetric spaces that arise by the above described construction from a semisimpleLie algebra g . This was done by Naitoh who gave a complete classification of thesespaces [Nai].In [KE], Theorem B, Eschenburg and Kim gave a sufficient condition for being of Ferustype in terms of the shape operator A and the mean curvature vector h of M . Theyproved that a full and indecomposable extrinsic symmetric space M ⊂ R p,q is of Ferus1ype if A h = 0 . (1)More exactly, for any full extrinsic symmetric space M ⊂ V = R p,q they constructed ametric Lie algebra ( g , h· , ·i ) with an orthogonal Cartan decomposition g = k ⊕ V . Underthe Condition (1) they find an element ξ ∈ V such that M is the G -orbit through ξ ∈ V as explained above.In [K1] we showed that extrinsic symmetric spaces correspond bijectively to so-calledweak extrinsic symmetric triples, which consist of a Lie algebra g together with aninvariant inner product, an involution θ and a derivation D on g satisfying certainconditions. If the extrinsic symmetric space is full then g and θ coincide with the Liealgebra and the Cartan decomposition constructed in [KE]. If, in addition, M ⊂ R p,q is of Ferus type, then D = ad( ξ ).We will show that Condition (1) is very strong. In fact, we will prove: Theorem 1
An indecomposable extrinsic symmetric space satisfies (1) if and only if g is semisimple. Any extrinsic symmetric space that arises from a semisimple Lie algebra g is of Ferustype. Hence Theorem 1 implies that the indecomposable extrinsic symmetric spacessatisfying (1) are exactly those listed in [Nai]. Note that we do not use the assumptionof fullness, which is somewhat inadequate in the indefinite case.I would like to thank Martin Olbrich for several useful discussions. Let M ⊂ R p,q be an n -dimensional extrinsic symmetric space. We denote by α thesecond fundamental form of M and by A its shape operator. Let V be the (possiblydegenerate) smallest affine subspace that contains M . Then M ⊂ V is a full extrinsicsymmetric space. Obviously, α coincides with the second fundamental form of M as asubmanifold of V , which we also denote by α . Furthermore, the shape operator ¯ A of M ⊂ V satisfies A η = ¯ A η for each η ∈ T ⊥ M ∩ V .The vector h = 1 n tr α ∈ V is called mean curvature vector. In other words, if x ∈ M and e , . . . , e n is an orthonor-mal basis of T x M , then h ( x ) = 1 n n X j =1 κ j α ( e j , e j ) , where κ j = h e j , e j i = ± M ⊂ R p,q is a so-called weak extrinsic symmetric triple. An extrinsic symmetrictriple ( g , Φ , h· , ·i ) consists of a metric Lie algebra ( g , h· , ·i ) and a pair Φ = ( D, θ ),where θ is an isometric involution on g and D is an antisymmetric derivation on g such that Dθ = − θD and D = − D . Besides θ there is a further involution on g ,2amely τ = exp πD . We denote by g + and g − the eigenspaces of θ , by g + and g − theeigenspaces of τ and by g ++ = g + ∩ g + , etc. the intersections of these eigenspaces. Werequire that g ++ = [ g − + , g − + ]. A weak extrinsic symmetric triple is defined in the sameway but the inner product is allowed to be degenerate on g + − . If ( g , Φ , h· , ·i ) is a weakextrinsic symmetric triple, then its metric radical R := g ∩ g ⊥ ⊂ g + − is contained in thecentre of g and g /R is an extrinsic symmetric triple. A weak extrinsic symmetric tripleis called full if g + − = [ g − + , g −− ].There is a one-to-one correspondence between full extrinsic symmetric spaces (in apossibly degenerate ambient space) and full weak extrinsic symmetric triples. Letus briefly recall this correspondence, for details see [K1]. As vector spaces V = g − .Moreover, if φ is the map from g + to the Lie algebra iso ( g − ) = so ( g − ) ⋉ g − of Iso( g − )defined by φ ( X ) = ((ad X ) | g − , − D ( X )) , X ∈ g + , then M is the orbit of x = 0 under the action of the group h exp( φ ( X )) | X ∈ g + i ⊂ Iso( g − ) . In particular, T ⊥ x M = g + − , T x M = g −− . With this identification the second fundamentalform and the shape operator are given by α ( u, v ) = [ Du, v ] , A η u = − [ Du, η ] (2)for all u, v ∈ g −− and η ∈ g + − . The Riemannian curvature tensor and the normalcurvature satisfy R M ( u, v ) w = − [[ u, v ] , w ] , R ⊥ ( u, v ) η = − [[ u, v ] , η ]for u, v, w ∈ g −− and η ∈ g + − . Remark 1
1. If M ⊂ V = R p,q is full and of Ferus type, then g = g + ⊕ g − coincideswith g = k ⊕ V as a metric Lie algebra with orthogonal Cartan decomposition.Moreover D equals ad( ξ ). The extrinsic symmetric spaces in both models areessentially the same, they just differ by a translation by ξ .2. The extrinsic symmetric triple g /R is associated with the projection of the im-mersion of M into V to an immersion of M into V / ( V ∩ V ⊥ ) described in [KE],Section 2.3. The derivation D belonging to a weak extrinsic symmetric triple is inner, i.e., D = ad( ξ ) for some ξ ∈ g − + , if and only if the associated extrinsic symmetricspace is of Ferus type. In particular, if g is semisimple, then the associatedextrinsic symmetric space is of Ferus type.An extrinsic symmetric space M ⊂ R p,q is called indecomposable if the embedding M ⊂ R p,q does not decompose as a non-trivial direct product of embeddings M i ⊂ R p i ,q i , i = 1 ,
2, with M = M × M , p = p + p , q = q + q . Here we want to assumethat M ⊂ R p,q is indecomposable. Then ( g , h· , ·i , Φ) is indecomposable, i.e., it is not3he direct sum of non-trivial weak extrinsic symmetric triples. We have two kinds ofindecomposable weak extrinsic symmetric triples, namely those that are semisimpleand those that do not have simple ideals. We will show that the semisimple ones satify A h = 0 and that for those without simple ideals A h = 0 holds. This will prove thetheorem. Proposition 1 If g is semisimple, then h = λξ , where λ = 0 is a complex number if g is a complex Lie algebra (considered as a real one) or a real number if g is not complex.In particular, A h = − λ · id . Proof.
If ( g , Φ , h· , ·i ) is indecomposable and g is semisimple, we have two possibilities.Either g is simple or g is the sum of two isomorphic simple Lie algebras. In the lattercase the two simple summands are orthogonal to each other and θ is an isometrybetween them. Hence, h X, Y i = B g ( µX, Y ), where B g is the Killing form of the (real)Lie algebra g and µ = 0 is a complex number if g is a complex Lie algebra or a realnumber if g is not complex. Let e , . . . , e n be an orthonormal basis of g −− . For any η ∈ g + − we have n h h, η i = n X j =1 h κ j α ( e j , e j ) , η i = n X j =1 h κ j [ De j , e j ] , η i = n X j =1 κ j h [[ ξ, e j ] , e j ] , η i = n X j =1 κ j h [ η, [ ξ, e j ]] , e j i = 12 (cid:16) n X j =1 κ j h [ η, [ ξ, e j ]] , e j i + n X j =1 κ j h D [ η, [ ξ, e j ]] , De j i (cid:17) = 12 (cid:16) n X j =1 κ j h [ η, [ ξ, e j ]] , e j i + n X j =1 κ j h [ η, [ ξ, De j ]] , De j i (cid:17) . The latter term equals (1 / · tr(ad( η ) ◦ ad( ξ )) | g − , which is equal to (1 / · B g ( η, ξ )because of ad( ξ ) | g + = 0. Hence h h, η i = h λη, ξ i for λ = 1 / (2 nµ ) and, consequently, h = λξ . The second assertion now follows from (2) and ad( ξ ) = − id on g − . ✷ Remark 2
More exactly, if the Lie algebra g is complex, then the involution θ can becomplex linear or antilinear. In the first case, in general, h is a complex multiple of ξ .In the second case h must be a real multiple of ξ . Proposition 2 If A h = 0 , then g is semisimple. We will give two proofs of this proposition. The first one combines results from [KE]with the description of extrinsic symmetric spaces by weak symmetric triples. Thesecond one is a short proof, which relies on the method of quadratic extensions.
First proof or Proposition 2.
We will need the following Lemma. It is essentially dueto Kim and Eschenburg, we just generalise it to the case of weak extrinsic symmetrictriples. 4 emma 1
For any u, v ∈ g −− we have B g ( u, v ) = − n h A h u, v i = B g ( Du, Dv ) . Proof.
Recall that the metric radical R := g ∩ g ⊥ ⊂ g + − belongs to the centre of g (cf. [K1]) and that the quotient ¯ g = g /R is an extrinsic symmetric triple. In particular,¯ g = g − + ⊕ g − + ⊕ g + − /R ⊕ g −− as a vector space. Let pr denote the projection from g to ¯ g .Since R is in the centre of g we have B g ( u, v ) = tr(ad( u ) ◦ ad( v )) = B ¯ g ( u, v ). Now wecan apply [KE], Lemma 5.2., which says that the claim is true if g is non-degenerate (for u ∈ g −− the vector t u ∈ g − + used in [KE] coincides with the vector Du in our notation).Consequently, B ¯ g ( u, v ) = − n h A ¯ h u, v i for¯ h := (1 /n ) · n X j =1 κ j [ De j , e j ] ¯ g = (1 /n ) · n X j =1 κ j pr [ De j , e j ] g = pr h, (3)where e , . . . , e n is an orthonormal basis of g −− . This implies B g ( u, v ) = B ¯ g ( u, v ) = − n h [pr h, Du ] ¯ g , v i ¯ g = − n h [ h, Du ] , v i = − n h A h u, v i . This also implies the second equation since every derivation is antisymmetric withrespect to the Killing form and D = − id holds on g − . ✷ Also the next Lemma relies on ideas developed in [KE].
Lemma 2
Either A h is invertible or A h = 0 . Proof.
Since M ⊂ V is indecomposable we conclude from Moore’s Lemma that A h has just one conjugate pair λ, ¯ λ of eigenvalues. If λ = 0, then A h is invertible. Nowsuppose λ = 0.Since h is parallel 0 = R ⊥ ( u, v ) h = − [[ u, v ] , h ] for all u, v ∈ g −− . Since g ++ = [ g − + , g − + ] =[ g −− , g −− ] we obtain [ g ++ , h ] = 0, which also implies [ g + − , h ] = 0 by [ g + − , h ] ⊂ g ++ and h [ g + − , h ] , g ++ i = h g + − , [ g ++ , h ] i = 0. Take u, v ∈ g −− and put η := [ Du, v ]. Let e , . . . , e n bean orthonormal basis of g −− . Since [ h, g + ] = 0 we get B g ( η, h ) = n X j =1 κ j (cid:16) h ad( η ) ◦ ad( h )( e j ) , e j i + h ad( η ) ◦ ad( h )( De j ) , De j i (cid:17) = 2 n X j =1 κ j h [ η, [ h, De j ]] , De j i = 2 n X j =1 κ j h [ h, De j ] , [ η, De j ] i = 2 n X j =1 κ j h A h e j , A η e j i = 2 tr A h A η . Since h is parallel the Ricci equation gives 0 = h R ⊥ ( u, v ) h, η i = −h [ A h , A η ] u, v i for all u, v ∈ g −− . Hence [ A h , A η ] = 0, which implies that A h A η is nilpotent, thus tr A h A η = 0.Consequently, B g ( η, h ) = 0. On the other hand, B g ( η, h ) = B g ([ Du, v ] , h ) = − B g ( v, [ Du, h ]) = B g ( v, A h u ) = c h A h v, A h u i = c h A h u, v i c = 0 by Lemma 1. Hence A h = 0. ✷ By Lemma 2 A h is invertible. Now Lemma 1 implies that the Killing form of g isnon-degenerate on g − . Moreover, [ g − , g − ] ⊃ [ g − + , g − + ] ⊕ [ g − + , g −− ] = g ++ ⊕ g + − = g + , thus[ g − , g − ] = g + . This implies that the Killing form of g is non-degenerate. Hence g issemisimple. ✷ Second proof of Proposition 2.
Let g be not semisimple. We have to show that A h = 0.By (2) it suffices to show that (ad h ) = 0. This is equivalent to (ad¯ h ) = 0 in ¯ g = g /R ,where ¯ h is defined as in (3). If g is not semisimple, then also ¯ g is not semisimple. Hencewe may assume that h· , ·i is non-degenerate. Since ( g , h· , ·i , Φ) is indecomposable g doesnot contain simple ideals. Hence g has the structure of a quadratic extension. This isproven in [K2]. We do not want to recall this in detail here. We just collect the factsthat we will need in the following. As a vector space the Lie algebra g can be identifiedwith l ∗ ⊕ a ⊕ l for some vector spaces l and a such that the Lie bracket on g has theproperties [ a , a ] ⊂ l ∗ , [ l ∗ , l ∗ ⊕ a ] = 0 , [ l ∗ , l ] ⊂ l ∗ . (4)Moreover, there are derivations D a : a → a , D l : l → l and involutions θ a : a → a , θ l : l → l such that D = ( − D l ) ∗ ⊕ D a ⊕ D l and θ = θ ∗ l ⊕ θ a ⊕ θ l . We define a −− , l −− and( l ∗ ) −− analogously to g −− . Let a , . . . , a k be an orthonormal basis of a −− and κ j := h a j , a j i .Furthermore, let L , . . . , L l be a basis of l −− and denote by Z , . . . , Z l its dual basis of( l ∗ ) −− . Then Z , . . . , Z l , L , . . . , L l , a , . . . , a k is a basis of g −− = ( l ∗ ) −− ⊕ a −− ⊕ l −− and nh = l X i =1 α ( L i , Z i ) + k X j =1 κ j α ( a j , a j ) = l X i =1 DL i , Z i ] + k X j =1 κ j [ Da j , a j ] . Now (4) shows that h belongs to l ∗ . Using (4) once more, this yields (ad h ) = 0. ✷ We have seen that, for an indecomposable extrinsic symmetric space M ⊂ R p,q , theoperator A h is of one of the following three types:1. A h is invertible. This holds if and only if g is semisimple. More exactly, thefollowing holds. If g and θ are complex, then A h = − λ id, where λ ∈ C , λ = 0.Note that in this case M as well as the ambient space have complex structures,thus this condition makes sense. Semisimple extrinsic symmetric triples withcomplex g and complex θ can be obtained as complexifications of the extrinsicsymmetric triples that are associated with compact extrinsic symmetric spacesin Euclidean spaces. If g is not complex or θ is not complex, then A h = − λ id,where λ ∈ R , λ = 0. A complete list of all semisimple extrinsic symmetric triplescan be found in [Nai].2. A h is 2-step nilpotent. Examples for this case can be found in [K1], Section 7 and[K2]. For instance, all extrinsic symmetric embeddings of Cahen-Wallach spaceshave a 2-step nilpotent operator A h .3. A h = 0. If M ⊂ R p,q is full, this is equivalent to h = 0. Examples for this situationcan also be found in [K2]. E.g., the two-dimensional Lorentzian manifold R , −→ V, ( r, s ) ( r, s , s ) , V ∈ { R , , R , } , h· , ·i V = 2 dx dx ± dx , is full and h vanishes. The sameis true for the three-dimensional Lorentzian manifold R , −→ V, ( r, s, t ) ( s, − rt + r / , t, r, r ) , where V = R , with h· , ·i V = 2 dx dx + 2 dx dx + dx .If the minimal subspace containing M is degenerate, then it can happen that A h = 0 but h = 0. The following example arises as an extension of the abovediscussed embedding R , ֒ → V , V ∈ { R , , R , } . Consider R with the metric h· , ·i = 2 dx dx ± dx + 2 dx dx , which is isometric to the standard space R , in the ‘+’-case and to R , in the ‘ − ’-case. Let us denote this pseudo-Euclideanspace by W . Then an easy computation shows that R , −→ W, ( r, s ) ( r, s , s, rs ± s / , h = (0 , , , , = 0 but A h = 0. References [F1] D. Ferus,
Immersions with Parallel Second Fundamental Form.
Math. Z. (1974), 87 - 92.[F2] D. Ferus,
Symmetric Submanifolds of Euclidean Space.
Math. Ann. (1980),81 - 93.[K1] I. Kath,
Indefinite extrinsic symmetric spaces I.
J. reine angew. M., to appear.[K2] I. Kath,
Indefinite extrinsic symmetric spaces II.
J. reine angew. M., to appear.[Ki] J. R. Kim,
Indefinite Extrinsic Symmetric Spaces.
Dissertation Augsburg(2005).[KE] J. R. Kim, J.-H. Eschenburg,
Indefinite extrinsic symmetric spaces. manuscr.math. (2011), 203 - 214.[Nai] H. Naitoh,
Pseudo-Riemannian symmetric R -spaces. Osaka J. Math. (1984),733 - 764.[S] W. Str¨ubing, Symmetric submanifolds of Riemannian manifolds.
Math. Ann.245