Osserman manifolds and Weyl-Schouten Theorem for rank-one symmetric spaces
aa r X i v : . [ m a t h . DG ] O c t OSSERMAN MANIFOLDS AND THE WEYL-SCHOUTEN THEOREM FORRANK-ONE SYMMETRIC SPACES
Y.NIKOLAYEVSKY
Abstract.
A Riemannian manifold is called Osserman (conformally Osserman, respectively), if theeigenvalues of the Jacobi operator of its curvature tensor (Weyl tensor, respectively) are constant onthe unit tangent sphere at every point. Osserman Conjecture asserts that every Osserman manifoldis either flat or rank-one symmetric. We prove that both the Osserman Conjecture and its conformalversion, the Conformal Osserman Conjecture, are true, modulo a certain assumption on algebraiccurvature tensors in R . As a consequence, we show that a Riemannian manifold having the sameWeyl tensor as a rank-one symmetric space, is conformally equivalent to it. Introduction
The aim of this paper is twofold. Firstly, we consider Osserman and conformally Osserman manifoldsof dimension 16 (which is the only missing dimension in the proof of the Osserman Conjecture). Weshow that both the “genuine” Osserman Conjecture and its conformal version can be reduced to a purelyalgebraic question on algebraic curvature tensors in R . Secondly, we obtain an analogue of the classicalWeyl-Schouten Theorem for rank-one symmetric spaces: a Riemannian manifold of dimension greaterthan four having “the same” Weyl tensor as that of a rank-one symmetric space is locally conformallyequivalent to that space.An algebraic curvature tensor R on a Euclidean space R n is a (3 ,
1) tensor having the same symmetriesas the curvature tensor of a Riemannian manifold. For X ∈ R n , the Jacobi operator R X : R n → R n isdefined by R X Y = R ( X, Y ) X . The Jacobi operator is symmetric and R X X = 0 for all X ∈ R n . Definition 1.
An algebraic curvature tensor R is called Osserman if the eigenvalues of the Jacobioperator R X do not depend on the choice of a unit vector X ∈ R n .One of the algebraic curvature tensors naturally associated to a Riemannian manifold (apart fromthe curvature tensor itself) is the Weyl conformal curvature tensor. Definition 2.
A Riemannian manifold is called (pointwise) Osserman if its curvature tensor at everypoint is Osserman. A Riemannian manifold is called conformally Osserman if its Weyl tensor at everypoint is Osserman.It is well-known (and is easy to check directly) that a Riemannian space locally isometric to aEuclidean space or to a rank-one symmetric space is Osserman. The question of whether the converseis true is known as the Osserman Conjecture [Oss]:
Osserman Conjecture.
Any smooth pointwise Osserman manifold of dimension n = 2 , is either flator locally rank-one symmetric. The study of conformally Osserman manifolds was started in [BG1] and then continued in [BG2,BGNSi, Gil, BGNSt]. Every Osserman manifold is conformally Osserman (which easily follows fromthe formula for the Weyl tensor and the fact that every Osserman manifold is Einstein), as also isevery manifold locally conformally equivalent to an Osserman manifold. This motivates the followingconjecture made in [BGNSi]:
Date : November 3, 2018.2000
Mathematics Subject Classification.
Primary: 53B20.
Key words and phrases.
Osserman manifold, Jacobi operator, Clifford structure.
Conformal Osserman Conjecture.
Any smooth conformally Osserman manifold of dimension n > is either conformally flat or locally conformally equivalent to a rank-one symmetric space. The proof of the Osserman Conjecture for manifolds of dimension not divisible by 4 was given in[Chi], before the conjecture itself was published. The Conformal Osserman Conjecture for manifolds ofdimension n >
6, not divisible by 4, is proved in [BG1], for manifolds with the structure of a warpedproduct, both conjectures are proved in [BGV].At present, both the Osserman Conjecture and the Conformal Osserman Conjecture are proved in allthe cases, with the only exception when n = 16 and one of the eigenvalues of the Jacobi operator hasmultiplicity 7 or 8 [N1, N2, N3, N4, N5]. The main difficulty in this remaining case lies in the followingalgebraic fact: it can be shown that in all the other cases, an Osserman algebraic curvature tensor hasa Clifford structure , so it “looks similar” to the curvature tensor of the complex or the quaternionicprojective space (a Clifford structure arises from an orthogonal representation of a Clifford algebra;see Section 2 for details). However, the curvature tensor of the Cayley projective plane (whose Jacobioperator has eigenvalues with multiplicities exactly 7 and 8) is essentially different. This is the onlyknown Osserman curvature tensor without a Clifford structure, which motivates the following algebraicconjecture.
Conjecture A.
Every Osserman algebraic curvature tensor in R whose Jacobi operator has an eigen-value of multiplicity or of multiplicity either has a Clifford structure or is a linear combination ofthe constant curvature tensor and the curvature tensor of the Cayley projective plane. In the latter case, we will say that an algebraic curvature tensor has a
Cayley structure .Our main result is the following theorem.
Theorem 1.
Assuming Conjecture A, both the Osserman Conjecture and the Conformal OssermanConjecture are true.
As a consequence of Theorem 1, we obtain the following analogue of the Weyl-Schouten Theorem forrank-one symmetric spaces (without assuming Conjecture A):
Theorem 2.
Suppose that for every point x of a smooth Riemannian manifold M n , n > , there existsa linear isometry which maps the Weyl tensor of M n at x on a positive multiple of the Weyl tensor ofa rank-one symmetric space M n . Then M n is locally conformally equivalent to M n . Manifolds satisfying the assumption of Theorem 2 can be viewed as conformal analogues of (a subclassof) curvature homogeneous manifolds [TV, Gil].In dimension four, both theorems are false. For Theorem 2 and the conformal part of Theorem 1, thisfollows from the fact that a four-dimensional Riemannian manifold is conformally Osserman if and onlyif it is either self-dual or anti-self-dual [BG2] and that there exist self-dual K¨ahler metrics on C whichare not locally conformally equivalent to locally-symmetric ones [Der]. For the “genuine” Osserman partof Theorem 1, the counterexample is given by the generalized complex space forms [GSV, Corollary 2.7].However, the Osserman Conjecture is true for four-dimensional globally Osserman manifolds, that is,for those whose Jacobi operator has constant eigenvalues on the whole unit tangent bundle [Chi].The paper is organized as follows. Section 2 gives the algebraic background for the proof of theboth theorems: we consider Osserman algebraic curvature tensors on R , each of which, assumingConjecture A, has either a Clifford structure (discussed in Sections 2.1,2.2), or a Cayley structure(Section 2.3). The proofs of the both theorems are given in Section 3. We first prove the local versionof the conformal part of Theorem 1 using the second Bianchi identity, separately in the Clifford case(Section 3.2) and in the Cayley case (Section 3.3), and then the global version, by showing that the“algebraic type” of the Weyl tensor is the same at all the points of a connected conformally OssermanRiemannian manifold (Section 3.4). The proofs of Theorem 2 and of the “genuine” Osserman part ofTheorem 1 easily follow (see the second and the last paragraphs of Section 3, respectively).The Riemannian manifold M n is assumed to be smooth (of class C ∞ ), although both theoremsremain valid for class C k , with sufficiently large k . SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 3 Osserman algebraic curvature tensors in R and the Clifford structure Both Theorem 1 and Theorem 2 have to be proved only when n = 16 (see Section 1). In this section,we consider all the known Osserman algebraic curvature tensors in R , namely, the algebraic curvaturetensors with a Clifford structure and the algebraic curvature tensors with a Cayley structure.2.1. Clifford structure.
The property of an algebraic curvature tensor R to be Osserman is quitealgebraically restrictive. In the most cases, such a tensor can be obtained by the following construc-tion, suggested in [GSV], which generalizes the curvature tensors of the complex and the quaternionicprojective spaces. Definition 3. A Clifford structure
Cliff( ν ; J , . . . , J ν ; λ , η , . . . , η ν ) on a Euclidean space R n is a set of ν ≥ J i and ν + 1 real numbers λ , η , . . . η ν , with η i = 0.An algebraic curvature tensor R on R n has a Clifford structure Cliff( ν ; J , . . . , J ν ; λ , η , . . . , η ν ) if(1) R ( X, Y ) Z = λ ( h X, Z i Y − h Y, Z i X ) + X νi =1 η i (2 h J i X, Y i J i Z + h J i Z, Y i J i X − h J i Z, X i J i Y ) . When this does not create ambiguity, we abbreviate Cliff( ν ; J , . . . , J ν ; λ , η , . . . , η ν ) to just Cliff( ν ). Remark . As it follows from Definition 3, the operators J i are skew-symmetric, orthogonal and satisfythe equations h J i X, J j X i = δ ij k X k and J i J j + J j J i = − δ ij id, for all i, j = 1 , . . . , ν , and all X ∈ R n .This implies that every algebraic curvature tensor with a Clifford structure is Osserman, as by (1) theJacobi operator has the form R X Y = λ ( k X k Y −h Y, X i X )+ P νi =1 η i h J i X, Y i J i X , so for a unit vector X , the eigenvalues of R X are λ (of multiplicity n − − ν provided ν < n − λ +3 η i , i = 1 , . . . , ν .From the fact that the J i ’s are anticommuting almost Hermitian structures it easily follows that theoperators J i . . . J i m with pairwise nonequal i j ’s, are skew-symmetric, if m ≡ , n = 16, and also in many cases when n = 16, as follows from [N3] (Proposition 1and the second last paragraph of the proof of Theorem 1 and Theorem 2), [N2, Proposition 1] and[N4, Proposition 2.1]. The only known counterexample is an algebraic curvature tensor with a Cayleystructure: R = aR O + bR S , where R S and R O are the curvature tensors of the unit sphere S (1) and ofthe Cayley projective plane O P , respectively, and a = 0.A Clifford structure Cliff( ν ) on the Euclidean space R n turns it the into a Clifford module (we refer to[ABS, Part 1], [Hus, Chapter 11] for standard facts on Clifford algebras and Clifford modules). DenoteCl( ν ) a Clifford algebra on ν generators x , . . . , x ν , an associative unital algebra over R defined by therelations x i x j + x j x i = − δ ij (this condition determines Cl( ν ) uniquely). The map ρ : Cl( ν ) → R n defined on generators by ρ ( x i ) = J i (and ρ (1) = id) is a representation of Cl( ν ) on R n . As all the J i ’sare orthogonal and skew-symmetric, ρ gives rise to an orthogonal multiplication defined as follows. Inthe Euclidean space R ν , fix an orthonormal basis e , . . . , e ν . For every u = P νi =1 u i e i ∈ R ν and every X ∈ R n , define(2) J u X = X νi =1 u i J i X (when u = e i , we abbreviate J e i to J i ). The map J : R ν × R n → R n defined by (2) is an or-thogonal multiplication: k J u X k = k u k k X k (similarly, we can define an orthogonal multiplication J : R ν +1 × R n → R n by J u X = u X + P νi =1 u i J i X , for u = P νi =0 u i e i ∈ R ν +1 , where e , e , . . . , e ν isan orthonormal basis for the Euclidean space R ν +1 ).For X ∈ R n , introduce the subspaces J X = Span( J X, . . . , J ν X ) , I X = Span( X, J X, . . . , J ν X ) . Later we will also use the complexified versions of these subspaces, which we denote J C X and I C X respectively, for X ∈ C n . Y.NIKOLAYEVSKY
Algebraic curvature tensors of dimension 16 with a Clifford structure.
To find all thealgebraic curvature tensors with a Clifford structure in dimension 16, we need to find all the possibleways of turning R into a Cl( ν )-module. A convenient way to describe them is by using the octonions.In general, the proof of Theorem 1 extensively uses computations in the octonion algebra O (inparticular, the standard identities like a ∗ = 2 h a, i − a, h a, b i = h a ∗ , b ∗ i = ( a ∗ b + b ∗ a ) , a ( ab ) = a b, h a, bc i = h b ∗ a, c i = h ac ∗ , b i , ( ab ∗ ) c + ( ac ∗ ) b = 2 h b, c i a, h ab, ac i = h ba, ca i = k a k h b, c i , for any a, b, c ∈ O , and similar ones, see e.g. [BG]) and the fact that O is a division algebra (in particular, anynonzero octonion is invertible: a − = k a k − a ∗ ). We will also use the bioctonions O ⊗ C , the algebraover C with the same multiplication table as that for O . As all the above identities are polynomial, theystill hold for bioctonions, with the complex inner product on C , the underlying linear space of O ⊗ C .However, the bioctonion algebra is not a division algebra (and has zero-divisors: (i1 + e )(i1 − e ) = 0).In the following lemma, (which contains known facts, but will be convenient for us to refer to) we call arepresentation ρ of Cl( ν ) in R n orthogonal , if all the ρ ( x i ) are orthogonal. Representation ρ , ρ are calledequivalent (respectively, orthogonally equivalent), if there exists T ∈ GL( n ) (respectively, T ∈ O( n ))such that ρ ( x ) = T ρ ( x ) T − , for all x ∈ Cl( ν ). For a representation ρ , the representation − ρ is definedon the generators of Cl( ν ) by ( − ρ )( x i ) = − ρ ( x i ) (induced by the automorphism α : x i → − x i of Cl( ν )). Lemma 1.
1. For any representation of a Clifford algebra
Cl( ν ) in R , ν ≤ . There is exactlyone, up to orthogonal equivalence, orthogonal representations ρ of Cl(8) in R . It can be defined asfollows. Identify R and R with O ⊕ ⊥ O and O respectively, via linear isometries. Then the orthogonalmultiplication (2) defined by ρ is given by (3) J p ( a, b ) = ( bp, − ap ∗ ) , for p ∈ R = O , X = ( a, b ) ∈ R = O ⊕ ⊥ O .
2. Any orthogonal representation of a Clifford algebra
Cl( ν ) , ν ≥ , in R is either a restrictionof the representation ρ to Cl( ν ) ⊂ Cl(8) , or, up to orthogonal equivalence, is ± ρ , where ρ is thefollowing reducible orthogonal representation of Cl(7) . Identify R and R with O ⊕ ⊥ O and O ′ = 1 ⊥ respectively, via linear isometries. Then the orthogonal multiplication (2) defined by ρ is given by (4) J p ( a, b ) = ( ap, bp ) , for p ∈ R = O ′ , X = ( a, b ) ∈ R = O ⊕ ⊥ O . Proof.
It is easy to see that if two orthogonal representations are equivalent, then they are orthogonallyequivalent. Indeed, suppose that ρ ( x i ) = J i and ρ ( x i ) = T J i T − , where T ∈ GL( n ). Then, asboth ρ and ρ are orthogonal, we get ( T t T ) J i = J i ( T t T ). As every J i commutes with T t T , it alsocommutes with S = √ T t T , the unique symmetric positive definite matrix such that S = T t T . Butthen T S − ∈ O( n ) and ρ ( x i ) = ( T S − ) J i ( T S − ) − .By [Hus, Table 6.5], N = N = N = Z , N = Z ⊕ Z , where N ν is the free abelian group generatedby irreducible representations of Cl( ν ).1. The fact that ν ≤ R (which follows from the octonion identity ( ap ∗ ) q + ( aq ∗ ) p =2 h p, q i a ). As N = Z , a representation of Cl(8) in R is unique, up to orthogonal equivalence, henceany orthogonal representation is orthogonally equivalent to the one defined by (3).2. The restriction of ρ to any Cl( ν ) ⊂ Cl(8) defines an orthogonal representation of Cl( ν ) in R . As N = N = Z , any orthogonal representation of Cl( ν ) , ν = 5 ,
6, in R is equivalent (hence orthogonallyequivalent) to it.For the algebra Cl(7), the group N = Z ⊕ Z is generated by two inequivalent representations in R .These are ± σ , where on generators, σ ( x i ) is the right multiplication in O by the imaginary octonion x i (note that Q i =1 σ ( x i ) = ± id, with ± replaced by ∓ for − σ ). Then there are exactly three (orthogonally)inequivalent (orthogonal) representations of Cl(7) in R : ± σ and ( − σ ) ⊕ σ . As it follows from (4), ρ = 2 σ . Moreover, neither of ± ρ can be a restriction of ρ to Cl(7), as Q i =1 ρ ( x i ) = ± id, so Q i =1 ρ ( x i ) = ± ρ ( x ), which is skew-symmetric, thus contradicting Remark 1. (cid:3) Note that an algebraic curvature tensor with a Clifford structure does not change, if we change thesigns of (some or all of) the J i ’s, so it does not matter, which of ± ρ is defined by (4).In the proof of Theorem 1 we will use the following Lemma. SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 5
Lemma 2.
1. Suppose that a Clifford structure on R is given by (3) . Let N : R → R be a quadraticform such that for all i = 1 , . . . , the cubic polynomial h N ( X ) , J i X i is divisible by k X k . Then there exista linear operator A : R → R and vectors V, U ∈ R such that N ( X ) = J A ( X ) X + h V, X i X − U k X k .2. Suppose that a Clifford structure on R is given by (4) . Let N = ( N , N ) : R → R be aquadratic form, u be a unit imaginary octonion, and p ∈ R be such that for all X = ( a, b ) ∈ R , a ∗ N ( X ) + b ∗ N ( X ) = h p, X ih a ∗ b, u i . Then there exists m ∈ R such that N ( X ) = k X k ( m − π I X m )= k X k m − h m, X i X − P i =1 h J i X, m i J i X , where π I X m is the orthogonal projection to I X .Proof.
1. For X = ( a, b ) ∈ R , let N ( X ) = ( N ( X ) , N ( X )), where N , N : R → R are quadraticforms. By the assumption and (3), for any q ∈ R , the cubic polynomial h N ( X ) , J q X i = h N , bq i −h N , aq ∗ i is divisible by k X k , hence so is the polynomial vector b ∗ N − N ∗ a . Then there exists a linearoperator L : R → R such that b ∗ N ( X ) − N ( X ) ∗ a = ( k a k + k b k ) L ( X ), which simplifies to(5) b ∗ ( N ( X ) − bL ( X )) = ( L ( X ) a ∗ + N ( X ) ∗ ) a. Let for X = ( a, b ), N ( X ) − bL ( X ) = ξ ( b ) + P ( a, b ) + φ ( a ) , L ( X ) a ∗ + N ( X ) ∗ = ξ ( a ) + Q ( a, b ) + ψ ( b ) , where φ, ξ , ξ , ψ : R → R are quadratic forms and P, Q : R × R → R are bilinear forms. Collectingthe terms of the same degree in a and b in (5) we get ξ = ξ = 0 and(6) b ∗ P ( a, b ) = ψ ( b ) a, b ∗ φ ( a ) = Q ( a, b ) a. Taking b = 1 in the second equation we find φ ( a ) = Q ( a, a and so Q ( a, b ) a = b ∗ ( Q ( a, a ) =2 h b, Q ( a, i a − Q ( a, ∗ ( ba ), which implies ( a ∗ b ∗ ) Q ( a,
1) = a ∗ (2 h b, Q ( a, i − Q ( a, b ) ∗ ).For every b from the unit sphere S ⊂ O ′ , we can now apply [N2, Lemma 6], with Y = a ∗ , e = b ∗ , L ( Y ) = Q ( a,
1) and F ( Y ) = 2 h b, Q ( a, i − Q ( a, b ) ∗ . Then Q ( a,
1) = h c ( b ) , a ∗ i h d ( b ) , a ∗ i b ∗ + ap ( b ),for some maps c, d, p : S → O . As the left-hand side does not depend on b , we get a ( p ( b ) − p ( b )) ∈ Span(1 , b , b ), for all a ∈ O and all b , b ∈ S , so p ( b ) = p , a constant. Taking the real and theimaginary parts of the both sides we obtain that c ( b ) = c is also a constant and d ( b ) = 0. Then Q ( a,
1) = h c ∗ , a i ap , and therefore φ ( a ) = Q ( a, a = h m, a i a − k a k p ∗ , where m = c ∗ + 2 p ∗ . Thenfrom (6), Q ( a, b ) a = b ∗ φ ( a ) = ( h m, a i b ∗ − ( b ∗ p ∗ ) a ∗ ) a , so Q ( a, b ) = h m, a i b ∗ − ( b ∗ p ∗ ) a ∗ . Similarly, fromthe first equation of (6) we get ψ ( b ) = h r, b ∗ i b ∗ − k b k q, P ( a, b ) = h r, b ∗ i a − b ( qa ). Then N ( X ) = bL ( X ) + h r, b ∗ i a − b ( qa ) + h m, a i a − k a k p ∗ ,N ( X ) = − aL ( X ) ∗ + h m, a i b − a ( pb ) + h r, b ∗ i b − k b k q ∗ , which implies N ( X ) = J A ( X ) X + h V, X i X − U k X k , for all X ∈ R , where A : R → R is a linearoperator defined by A ( X ) = L ( X ) + b ∗ p ∗ − qa , and V = ( m, r ∗ ) , U = ( p ∗ , q ∗ ) ∈ R .2. From the assumption, N (( a, N ((0 , b )) = 0, so N (( a, b )) = P ( a, b ) + ξ ( b ) , N (( a, b )) = ξ ( a ) + Q ( a, b ), for some quadratic forms ξ , ξ : R → R and bilinear forms P, Q : R × R → R .Collecting the terms of degree two in a we get a ∗ P ( a, b ) + b ∗ ξ ( a ) = h p , a ih a ∗ b, u i , where p ∈ R , p =( p , p ). Substituting b = 1 we get ξ ( a ) = h p , a ih a ∗ , u i − a ∗ P ( a, a ∗ P ( a, b ) − b ∗ ( a ∗ P ( a, h p , a i ( h a ∗ b, u i − h a ∗ , u i b ∗ ). Multiplying by a from the left and taking a ⊥ p we get k a k P ( a, b ) − a ( b ∗ ( a ∗ P ( a, − h a, b i − h a, i b )( a ∗ P ( a, k a k ( bP ( a, − P ( a, b )).Multiplying by ( h a, b i − h a, i b ) ∗ from the left we obtain that all the components of the polynomialvector kh a, b i − h a, i b k ( a ∗ P ( a, I generated by k a k and h p , a i . For a fixednonzero b ⊥
1, the quadratic form kh a, b i − h a, i b k is not in I (as it is nonzero and vanishes onSpan(1 , b ) ⊥ ). As the ideal I is prime, it follows that all the components of a ∗ P ( a,
1) belong to I .Since P ( a,
1) is linear in a , we obtain a ∗ P ( a,
1) = k a k c + h p , a i La , for some c ∈ O and some linearoperator L on O . Multiplying this by a from the left we obtain that a · La is divisible by k a k , so a · La = k a k c , for some c ∈ O , so La = a ∗ c , therefore a ∗ P ( a,
1) = k a k c + h p , a i a ∗ c which implies P ( a,
1) = ac + h p , a i c . Then a ∗ P ( a, b ) − b ∗ ( a ∗ ( ac + h p , a i c )) = h p , a i ( h a ∗ b, u i − h a ∗ , u i b ∗ ), for all a, b ∈ O , so a ∗ P ( a, b ) − k a k b ∗ c = h p , a i ( h a ∗ b, u i − h a ∗ , u i b ∗ + b ∗ ( a ∗ c )).Assume p = 0. Then multiplying by a from the left we obtain that all the components of thepolynomial vector a ( h a ∗ b, u i − h a ∗ , u i b ∗ + b ∗ ( a ∗ c )) are divisible by k a k . As it is linear in b and Y.NIKOLAYEVSKY quadratic in a , we obtain a ( h a ∗ b, u i − h a ∗ , u i b ∗ + b ∗ ( a ∗ c )) = k a k L b for some linear operator L on O , so h a ∗ b, u i − h a ∗ , u i b ∗ + b ∗ ( a ∗ c ) = a ∗ · L b . Taking a = 1 we get L b = h b, u i + b ∗ c , so h a ∗ b, u i − h a ∗ , u i b ∗ − h b, u i a ∗ = a ∗ ( b ∗ c ) − b ∗ ( a ∗ c ), for all a, b ∈ O . This implies that for all orthogonal a, b ∈ O ′ , a ( bc ) ∈ Span( a, b, bc ∈ Span( a, ab, c ∈ Span( ab, a, b ). Thus c = 0, so h a ∗ b, u i − h a ∗ , u i b ∗ − h b, u i a ∗ = 0, for all a, b ∈ O , which leads to a contradiction (take b = u, a ⊥ , u ).It follows that p = 0, so a ∗ P ( a, b ) − k a k b ∗ c = 0 which implies P ( a, b ) = a ( b ∗ c ), and then ξ ( a ) = h p , a ih a ∗ , u i − a ∗ P ( a,
1) = −k a k c . Similarly, Q ( a, b ) = b ( a ∗ c ) and ξ ( b ) = −k b k c , forsome c ∈ O . It follows that N ( X ) = ( a ( b ∗ c ) − k b k c , b ( a ∗ c ) − k a k c ). Then by (4), N ( X ) = k X k ( m − π I X m ) = k X k m − h m, X i X − P i =1 h J i X, m i J i X , where m = ( − c , − c ) ∈ R . (cid:3) Algebraic curvature tensors with a Cayley structure.
The curvature tensor R O of the Cayleyprojective plane O P of the sectional curvature between 1 and 4, is explicitly given in [BG, Eq. 6.12].Identifying the tangent space T x O P with O ⊕ O via a linear isometry, we have for X = ( x , x ) , Y =( y , y ) , Z = ( z , z ) ∈ T x O P = O ⊕ O : R O ( X, Y ) Z = (4 h x , z i y − h y , z i x − ( z y ) x ∗ + ( z x ) y ∗ − ( x y − y x ) z ∗ , h x , z i y − h y , z i x − x ∗ ( y z ) + y ∗ ( x z ) + z ∗ ( x y − y x )) . Introducing the symmetric operators S i ∈ End( R ) , i = 0 , , . . . ,
8, by S ( x , x ) = ( x , − x ), S i ( x , x ) = ( e i x ∗ , x ∗ e i ) , i = 1 , . . . ,
8, where { e i } is an orthonormal basis for O , we obtain(7) R O ( X, Y ) = 3 X ∧ Y + X i =0 S i X ∧ S i Y, where X ∧ Y is the skew-symmetric operator defined by ( X ∧ Y ) Z = h X, Z i Y − h Y, Z i X .As it follows from the definition, the operators S i are orthogonal and satisfy(8) S i S j + S j S i = 2 δ ij id , ≤ i, j ≤ . For every w in the Euclidean space R , introduce the symmetric operator S w = P i =0 w i S i . As it followsfrom (8), the map S : R × R → R defined by ( w, X ) → S w X is an orthogonal multiplication: k S w X k = k w k · k X k , for all w ∈ R , X ∈ R . We usually abbreviate S e i to S i .The operators S i define the structure of the Clifford Cl + (9)-module on the Euclidean space R as follows. Denote Cl + (9) the Clifford algebra on nine generators x , x , . . . , x , an associative unitalalgebra over R defined by the relations x i x j + x j x i = 2 δ ij . The map σ : Cl + (9) → R defined ongenerators by σ ( x i ) = S i (and σ (1) = id) is a representation of Cl + (9) on R . The Clifford algebraCl + (9) is isomorphic to R (16) ⊕ R (16), where R (16) is the algebra of 16 ×
16 real matrices [ABS, § σ is surjective. In particular, as by (8) the operator Q i =0 S i commutes with all the S i ’s (hence with allthe R (16)) and is orthogonal, we have Q i =0 S i = ± id.As by (8), for every nonzero w ∈ R , S w = k w k id and Tr S w = 0 (multiply (8) by S i and take thetrace), S w has two eigenvalues ±k w k , each of multiplicity 8. Denote E k w k ( S w ) the k w k -eigenspace of S w and π E k w k ( S w ) the orthogonal projection of R to E k w k ( S w ). Then π E k w k ( S w ) = ( k w k − S w + id).Introduce the subspaces L k = Span i < ···
1. For k = 0 , . . . , , L − k = L k . Moreover, ⊕ k =0 L k = End( R ) and Sym( R ) = L ⊕L ⊕L , Skew( R ) = L ⊕L , where Skew( R ) and Sym( R ) are the spaces of the skew-symmetricand the symmetric endomorphisms of R respectively, and all the direct sums are orthogonal. Moreover, { S i . . . S i k , i < · · · < i k } is an orthonormal basis for L k , ≤ k ≤ .2. The operator A is symmetric and does not depend on the choice of an orthonormal basis S i for L . Its eigenspaces are L k , ≤ k ≤ , with the corresponding eigenvalues ( − k (9 − k ) . In particular, Skew( R ) and Sym( R ) are invariant subspaces of A .3. For every w ∈ R , Q ∈ End( R ) , A ( QS w ) = −A ( Q ) S w + 2 S w Q, A ( S w Q ) = − S w A ( Q ) + 2 QS w .For every X, Y ∈ R , A ( X ∧ Y ) = P i =0 S i X ∧ S i Y . SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 7
4. For every X ∈ R , X ∈ Span i =0 ( S i X ) and P i =0 h S i X, X i S i X = k X k X . For every X, Y ∈ R , P i =0 (2 h S i X, Y i S i X + h S i X, X i S i Y ) = k X k Y + 2 h X, Y i X . For a unit vector X ∈ R , R O X Y = Y ,when Y ⊥ Span i =0 ( S i X ) and R O X Y = Y , when Y ∈ Span i =0 ( S i X ) , Y ⊥ X .5. Let N : R × R → R be a bilinear skew-symmetric map such that h N ( X, Y ) , Z i = 0 , for every w ∈ R and for every X, Y, Z ∈ E k w k ( S w ) . Then there exists q ∈ R such that (10) N ( X, Y ) = (
A − id)( X ∧ Y ) q = X i =0 ( h S i X, q i S i Y − h S i Y, q i S i X ) − ( h X, q i Y − h Y, q i X ) . Proof.
1. The fact that L − k = L k follows from Q i =0 S i = ± id. Moreover, all the operators S i . . . S i k , i < · · · < i k , ≤ k ≤
4, are orthonormal. Indeed, from (8), the norm of each of them is 1. The innerproduct of two different ones is times the trace of some S ′ = S j . . . S j p , j < · · · < j p , ≤ p ≤ S ′ is skew-symmetric. Otherwise, p = 1 or p = 4 (by L − k = L k ), and in theboth cases, Tr S ′ = 0, as S ′ is a product of a symmetric and a skew-symmetric operator: S i = S i · ( S i S j ),when p = 1, and S i S j S k S l = S i · ( S j S k S l ), when p = 4 (for arbitrary pairwise nonequal i, j, k, l ). Now,as the S i ’s are symmetric, (8) implies that L ⊕ L ⊕ L ⊂ Sym( R ) and that L ⊕ L ⊂ Skew( R ),with both inclusions being in fact equalities by the dimension count.2. 3. Directly follow from (8, 9) and the fact that S i ( X ∧ Y ) S i = S i X ∧ S i Y .4. For X = ( x , x ) = 0 , x , x ∈ O , define w ∈ R = R ⊕ O by w = k X k − (( k x k − k x k ) e +2 x x ). Then S w X = X by the definition of the S i ’s, so X ∈ Span i =0 ( S i X ). As h S i X, S j X i = δ ij k X k by (8), equation P i =0 h S i X, X i S i X = k X k X follows. The second equation is obtained by polarization.Substituting it to the expression for R O X Y obtained from (7), we get the last statement of the assertion.5. For every Z ∈ R , define K ( Z ) ∈ Skew( R ) by h K ( Z ) X, Y i = h N ( X, Y ) , Z i . As π E k w k ( S w ) = ( k w k − S w + id) we have ( k w k − S w + id) K (( k w k − S w + id) Z )( k w k − S w + id) = 0, for all nonzero w ∈ R , so ( S w K ( S w Z ) S w + k w k ( K ( S w Z ) + S w K ( Z ) + K ( Z ) S w )) + k w k ( S w K ( S w Z ) + K ( S w Z ) S w + S w K ( Z ) S w + k w k K ( Z )) = 0. As k w k is not a rational functions, we obtain(11) S w K ( S w Z ) S w + k w k ( K ( S w Z ) + S w K ( Z ) + K ( Z ) S w ) = 0 , for all w ∈ R and all Z ∈ R . It follows that for every fixed Z ∈ R and for every i, j = 0 , . . . , S w K ( S w Z ) S w S i S j ), viewed as a polynomial of w = ( w , . . . , w ), is divisible by k w k (the w i ’s are the coordinates of w relative to the orthonormal basis { e , e , . . . , e } for R such that S e i = S i ). Then from (8) we obtain that k w k | w i Tr( K ( S w Z ) S j S w ) − w j Tr( K ( S w Z ) S i S w ). For every Z ∈ R , define the polynomials F Z,i ( w ) = Tr( K ( S w Z ) S i S w ). Then k w k | w i F Z,j ( w ) − w j F Z,i ( w ). Let I be the ideal of the polynomial ring K = R [ w , w , . . . , w ] generated by k w k and let π : K → K / I bethe natural projection. Then π ( w i F Z,j − w j F Z,i ) = 0, so the 2 × K / I whose i -throw is ( π ( w i ) , π ( F Z,i )) , i = 0 , , . . . ,
8, has rank at most one. As the ring K / I is a unique factorizationdomain [Nag], there exist x, y, u i ∈ K / I , such that π ( w i ) = xu i , π ( F Z,i ) = yu i . Since the elements π ( w i ) = xu i are coprime, x is invertible, so we can take x = 1, hence π ( F Z,i ) = yπ ( w i ). Lifting thisequation up to K we obtain that for every Z ∈ R , there exist polynomials Y Z ( w ) and g Z,i ( w ) suchthat F Z,i ( w ) = Y Z ( w ) w i + k w k g Z,i ( w ). As F Z,i is a quadratic form, the comparison of terms of thesame degree gives that for every Z , we can choose Y Z ( w ) to be a linear form in w and the g Z,i ( w )’sto be constants. Using the linearity by Z be obtain that for some linear maps T, G : R → R ,Tr( K ( S w Z ) S i S w ) = F Z,i ( w ) = h T Z, w i w i + k w k h GZ, e i i , for all Z ∈ R , w ∈ R , i = 0 , , . . . , S w Z as Z we obtain k w k Tr( K ( Z ) S i S w ) = h T S w Z, w i w i + k w k h GS w Z, e i i . It followsthat k w k | h T S w Z, w i = h Z, S w T t w i , for all Z ∈ R , so S w T t w = k w k p , for some p ∈ R . As S w = k w k id, this implies T t w = S w p , so Tr( K ( Z ) S i S w ) = h Z, p i w i + h GS w Z, e i i . Multiplying bothsides by w i and summing up by i = 0 , , . . . , h Z, p ik w k + h GS w Z, w i (the left-hand sidevanishes, as S w = k w k id and K ( Z ) is skew-symmetric). It follows that k w k p + S w G t w = 0, so G t w = − S w p . Hence Tr( K ( Z ) S i S w ) = h Z, p i w i + h S w Z, G t e i i = h Z, p i w i − h S w Z, S i p i . In particular,for w = e j , j = i , we get by (8): Tr( K ( Z ) S i S j ) = h Z, S i S j p i = − h Z ∧ p, S i S j i . As { S i S j , i < j } isan orthonormal basis for L by assertion 1, we obtain π K ( Z ) = − π ( Z ∧ p ), for all Z ∈ R , where π is the orthogonal projection from Skew( R ) to L . Y.NIKOLAYEVSKY
Now for any K ∈ Skew( R ) , π ( K ) = ( A + 3 id) K and π ( K ) = − ( A − K , (where π i isthe orthogonal projection to L i ) by assertions 1 and 2, and A ( S w KS w ) = S w A ( K ) S w and A ( S w K + KS w ) = − S w A ( K ) − A ( K ) S w + 2 S w K + 2 KS w , by assertion 3, so π ( S w K + KS w ) = S w π ( K ) + π ( K ) S w and π ( S w KS w ) = S w π ( K ) S w . Projecting (11) to L we then obtain S w π ( K ( S w Z )) S w + k w k ( π ( K ( S w Z )) + S w π ( K ( Z )) + π ( K ( Z )) S w ) = 0. As it is shown above, π K ( Z ) = − π ( Z ∧ p ), so − S w π (( S w Z ) ∧ p ) S w + k w k ( − π (( S w Z ) ∧ p )+ S w π ( K ( Z ))+ π ( K ( Z )) S w ) = 0, which (using the factthat S w π ( K ) S w = π ( S w KS w ) and that S w ( X ∧ Y ) S w = ( S w X ) ∧ ( S w Y )) simplifies to S w π ( K ( Z )) + π ( K ( Z )) S w = π ( Z ∧ ( S w p ) + ( S w Z ) ∧ p ). As S w ( X ∧ Y ) + ( X ∧ Y ) S w = ( S w X ) ∧ Y + X ∧ ( S w Y ) and π ( S w K + KS w ) = S w π ( K ) + π ( K ) S w , we obtain S w π ( K ( Z ) − Z ∧ p ) + π ( K ( Z ) − Z ∧ p ) S w = 0.Taking w = e i we get S i π ( K ( Z ) − Z ∧ p ) S i = − π ( K ( Z ) − Z ∧ p ), which implies A ( π ( K ( Z ) − Z ∧ p ))= − π ( K ( Z ) − Z ∧ p ), so π ( K ( Z ) − Z ∧ p ) = 0 by assertion 2.As K ( Z ) = π ( K ( Z )) + π ( K ( Z )) by assertion 1, it follows that K ( Z ) = ( − π + π ) ( Z ∧ p ) = ( − ( A + 3 id) − ( A − Z ∧ p ) = − ( A − id)( Z ∧ p ). Since h K ( Z ) X, Y i = − h K ( Z ) , X ∧ Y i and A is symmetric, h N ( X, Y ) , Z i = h K ( Z ) X, Y i = h ( A − id)( Z ∧ p ) , X ∧ Y i = h ( A − id)( X ∧ Y ) , Z ∧ p i = h ( A − id)( X ∧ Y ) p, Z i , so N ( X, Y ) = (
A − id)( X ∧ Y ) q , with q = p , as required. (cid:3) Conformally Osserman manifolds. Proof of Theorem 1 and Theorem 2
In all the cases, except when n = 16, Theorem 1 is already proved: for the Osserman Conjecture, see[N3, Theorem 2], for the Conformal Osserman Conjecture, see [N5, Theorem 1]. In this section, we willfirst prove Theorem 1 for conformally Osserman manifold of dimension 16 (assuming Conjecture A) andthen deduce from it the proof for “genuine” Osserman manifolds.Theorem 2 is an easy corollary of Theorem 1, as in Theorem 2 we consider the Riemannian manifolds M n , for which Conjecture A is already “satisfied” by the assumption: the Weyl tensor of each of themat every point is proportional either to the Weyl tensor of C P n or H P n (hence has a Clifford structure)or to the Weyl tensor of O P . By Theorem 1, M n is locally conformally equivalent to a rank-onesymmetric space, which is, in fact, M n , as the Weyl tensors of different rank-one symmetric spaces aredifferent (for instance, because their Jacobi operators have different multiplicities of the eigenvalues).We start with a brief informal outline of the proof of the conformal part of Theorem 1. Recall thatthe Weyl tensor of a Riemannian manifold M n is defined by(12) R ( X, Y ) = ˆ ρX ∧ Y − ˆ ρY ∧ X + W ( X, Y ) , where ˆ ρ = n − Ric − scal2( n − n − id, Ric is the Ricci operator, scal is the scalar curvature and X ∧ Y isthe skew-symmetric operator defined by ( X ∧ Y ) Z = h X, Z i Y − h Y, Z i X . According to Conjecture A,the Weyl tensor has either a Clifford or a Cayley structure. First of all, in Section 3.1 we show thatboth these structures can be chosen smooth on an open, dense subset M ′ ⊂ M (see Lemma 4 forthe precise statement), so that on every connected component M α of M ′ , the curvature tensor R of M is given by either (13) or (14) (up to a conformal equivalence), with all the operators and thefunctions involved being locally smooth. Then we establish the local version of the theorem, at everypoint x ∈ M ′ , for each of the two cases separately. This is done by using the differential Bianchi identityand the fact that under a conformal change of the metric, the symmetric tensor field ρ (which is a linearcombination of ˆ ρ from (12) and the identity) is a Codazzi tensor, that is, ( ∇ X ρ ) Y = ( ∇ Y ρ ) X . Usingthe result of [DS], we show that ρ must be a constant multiple of the identity, which implies that everyconnected component M α ⊂ M ′ is locally conformally equivalent to a symmetric Osserman manifold.The proof in the Clifford case is given in Section 3.2 (Lemma 5 and Lemma 6), in the Cayley case, inSection 3.3 (Lemma 7). Then, by the result of [GSV, Lemma 2.3], every M α is either locally conformallyflat or is locally conformal to a rank-one symmetric space. In Section 3.4 we prove the conformal partof Theorem 1 globally, by first showing (using Lemma 8) that M splits into a disjoint union of a closedsubset M , on which the Weyl tensor vanishes, and nonempty open connected subsets M α , each of whichis locally conformal to one of the rank-one symmetric spaces. On every M α , the conformal factor f isa well-defined positive smooth function. Assuming that there exists at least one M α and that M = ∅ we show that there exists a point x ∈ M on the boundary of a geodesic ball B ⊂ M α such that both f ( x ) and ∇ f ( x ) tend to zero when x → x , x ∈ B (Lemma 9). Then the positive function u = f / satisfies an elliptic equation in B , with lim x → x ,x ∈ B u ( x ) = 0, hence by the boundary point theorem,the limiting value of the inner derivative of u at x must be positive. This contradiction implies thateither M = M or M = M α , thus proving the conformal part of Theorem 1. The “genuine Osserman”part of Theorem 1 then follows easily using the result of [Nic].3.1. Smoothness of the Clifford and of the Cayley structures.
Let M be a connected smoothRiemannian manifold whose Weyl tensor at every point is Osserman. Define a function N : M → N as follows: for x ∈ M , N ( x ) is the number of distinct eigenvalues of the operator W X | X ⊥ , where W X is the Jacobi operator associated to the Weyl tensor and X is an arbitrary nonzero vector from T x M .As the Weyl tensor is Osserman, the function N ( x ) is well-defined. Moreover, as the set of symmetricoperators having no more than N distinct eigenvalues is closed in the linear space of symmetric operatorson R , the function N ( x ) is lower semi-continuous (every subset { x : N ( x ) ≤ N } is closed in M ).Let M ′ be the set of points where the function N ( x ) is continuous (that is, locally constant). It is easyto see that M ′ is an open and dense (but possibly disconnected) subset of M . The following lemmashows that, assuming Conjecture A, all the “ingredients” of the curvature tensor are locally smooth onevery connected component of M ′ . Lemma 4.
Let M be a smooth conformally Osserman Riemannian manifold. Let M ′ be the (open,dense) subset of all the points of M at which the number of distinct eigenvalues of the Jacobi operatorassociated to the Weyl tensor of M is locally constant.Assume Conjecture A. Then for every x ∈ M ′ , there exists a neighborhood U = U ( x ) with exactly oneof the following properties.(a) There exists ν ≥ , smooth functions η , . . . , η ν : U → R \ { } , a smooth symmetric linear operatorfield ρ and smooth anticommuting almost Hermitian structures J i , i = 1 , . . . , ν , on U such that forall y ∈ U and all X, Y, Z ∈ T y M , the curvature tensor of M has the form (13) R ( X, Y ) Z = h X, Z i ρY + h ρX, Z i Y − h Y, Z i ρX − h ρY, Z i X + X νi =1 η i (2 h J i X, Y i J i Z + h J i Z, Y i J i X − h J i Z, X i J i Y ) . (b) The Riemannian manifold U is conformally equivalent to a Riemannian manifold whose curvaturetensor has the form (14) R ( X, Y ) = ρX ∧ Y − ρY ∧ X + ε X i =0 S i X ∧ S i Y = ρX ∧ Y − ρY ∧ X + ε A ( X ∧ Y ) , at every point y ∈ U , where ε = ± and ρ, S i , i = 0 , . . . , , are smooth fields of symmetric operatorson U satisfying (8) .Proof. On every connected component M α ⊂ M ′ , the number N = N ( x ) is a constant, so the oper-ator W X | X ⊥ , where X is a unit vector, has exactly N distinct eigenvalues µ , µ , . . . , µ N − , with themultiplicities m , m , . . . , m N − . The functions µ i ’s are smooth on M α , and the m i ’s are constants,by the smoothness of the characteristic polynomial of W X | X ⊥ . We label them in such a way that m = max( m , m , . . . , m N − ).Clearly, P N − i =0 m i = 15 and, as Tr W X = 0, we have P N − i =0 m i µ i = 0. It follows that if N = 1, then W = 0, so M α is conformally flat. Then by (12), the curvature tensor has the form (13), with ν = 0and a smooth ρ .Suppose N >
2. By Conjecture A, W either has a Clifford structure, or a Cayley structure. But in thelatter case, the operator W X | X ⊥ has two distinct eigenvalues (from assertion 4 of Lemma 3). It followsthat W has a Clifford structure Cliff( ν ), at every point of M α ( ν may a priori depend on x ∈ M α ). Byassertion 1 of Lemma 1, ν ≤
8, and by Remark 1, for a unit vector X , the eigenvalues of W X | X ⊥ are λ , of multiplicity 15 − ν , and λ + 3 η i , i = 1 , . . . , ν . All of the η i ’s are nonzero (by Definition 3), someof them can be equal, but not all, as otherwise N = 2, so the multiplicity of every eigenvalue λ + 3 η i is at most ν − ≤ − ν , as ν ≤
8. It follows that the maximal multiplicity is m = 15 − ν ( ≥ ν = 15 − m , which is a constant on M α . Moreover, λ = µ (this is automatically satisfied, unless ν = 8and η = · · · = η = η ; in the latter case we have two eigenvalues of multiplicity 7 and we choose thelabeling of the µ i ’s so that µ = λ ). The functions λ and λ + 3 η i are smooth, as each of them equals one of the µ i ’s. Moreover, for every smooth unit vector field X on M α and every i = 1 , . . . , N −
1, the µ i -eigendistribution of W X | X ⊥ (which must be smooth on M α and must have a constant dimension m i )is Span j : λ +3 η j = µ i ( J j X ), by Remark 1. By assertion 3 of Lemma 3 of [N5], there exists a neighborhood U i ( x ) and smooth anticommuting almost Hermitian structures J ′ j (for the j ’s such that λ + 3 η j = µ i )on U i ( x ) such that Span j : λ +3 η j = µ i ( J j X ) = Span j : λ +3 η j = µ i ( J ′ j X ). Let W ′ be a (unique) algebraiccurvature tensor on U = ∩ N − i =1 U i ( x ) with the Clifford structure Cliff( ν ; J ′ , . . . , J ′ ν ; λ , η , . . . , η ν ). Then ν = 15 − m is constant and all the J ′ i , η i and λ are smooth on U . Moreover, for every unit vector field X on U , the Jacobi operators W ′ X and W X have the same eigenvalues and eigenvectors by construction,hence W ′ X = W X , which implies W ′ = W . Then the curvature tensor on U has the form (13), withthe operator ρ given by ρ = n − Ric +( λ − scal2( n − n − ) id, by (12). As λ is a smooth function, theoperator field ρ is also smooth.Now consider the case N = 2. Again, by Conjecture A, W either has a Clifford structure, or aCayley structure. In the former case, by Remark 1, for a unit vector X at every point x ∈ M α , theeigenvalues of W X | X ⊥ are λ , of multiplicity 15 − ν , and λ + 3 η, η = 0, of multiplicity ν . In the lattercase, there are two eigenvalues, of multiplicities m = 8 and m = 7, respectively (as it follows fromassertion 4 of Lemma 3). In the both cases, m ≥
8. It follows that if m >
8, the Weyl tensor W has aClifford structure Cliff( m ), at every point x ∈ M α . Then we can finish the proof as in the case N > M α , the functions λ = µ and λ + 3 η = µ are smooth and Span m j =1 ( J j X ), the µ -eigendistribution of W X | X ⊥ , is smooth and has a constant dimension m . Assertion 3 of Lemma 3of [N5] applies and we obtain the curvature tensor of the form (13) (with ν = m and all the η i ’s equal)on some neighborhood U = U ( x ) of an arbitrary point x ∈ M α .Suppose now that N = 2, and m = 8 , m = 7. Then at every point x ∈ M α , the Weyl tensoreither has a Clifford structure Cliff(7), with η = · · · = η = η , or a Clifford structure Cliff(8), with η = · · · = η = η , or a Cayley structure. Denote M (7) , M (8) and M ( O ) the corresponding subsets of M α , respectively. These three subsets are mutually disjoint. Indeed, if W = aR O + bR S , a = 0, wouldhave a Clifford structure, then the same would be true for R O = a − W − ba − R S (as by Definition 3,the set of algebraic curvature tensors with a Clifford structure is invariant under scaling and shiftingby a constant curvature tensor). This contradicts the fact that R O has no Clifford structure [N3,Remark 1]. Moreover, if x ∈ M (7) ∩ M (8) , then for any unit vector X ∈ T x M the operator W X | X ⊥ has an eigenspace of dimension 7 spanned by orthonormal vectors J X, . . . , J X (where the J i aredefined by the Clifford structure Cliff(7)), and the orthogonal eigenspace of dimension 8 spanned byorthonormal vectors J ′ X, . . . , J ′ X (where the J ′ j are defined by Cliff(8)). Then the fifteen operators J i , J ′ j are anticommuting almost Hermitian structures on R , which contradicts the fact that ν ≤ M α = M (7) ⊔ M (8) ⊔ M ( O ) . Moreover, each of the three subsets M (7) , M (8) , M ( O ) is relatively closed in M α . Indeed, suppose a sequence { x n } ⊂ M (7) converges to apoint x ∈ M α . Then the functions λ ( x n ) and λ ( x n ) + η ( x n ) are well-defined (as M (7) , M (8) , M ( O ) aredisjoint) and bounded (as λ ( x n ) + ( λ ( x n ) + η ( x n )) = µ ( x n ) + µ ( x n ) ), so we can assume that bothsequences converge by choosing a subsequence; moreover, as all the J i ( x n ) are orthogonal operators,we can choose a subsequence such that all of them converge. Then by continuity, W ( x ) has the form(1), with ν = 7, with the J i ’s being anticommuting almost Hermitian structures, and with η = 0, asotherwise N = 1, which contradicts the fact that N = 2 on M α . Therefore, x ∈ M (7) . The abovearguments work for M (8) almost verbatim (by replacing 7 by 8), and for M ( O ) , with a slight modification(by replacing the orthogonal operators J i ’s by the orthogonal operators S i from (7)). Hence, as M α isconnected, it coincides with exactly one of the sets M (7) , M (8) , M ( O ) .Now, if M α = M (7) or if M α = M (8) , the proof follows from the same arguments as in the case N >
2: we have a Clifford structure for W with a constant ν .Suppose M α = M ( O ) . Then by (7, 12), the Weyl tensor on M α has the form W ( X, Y ) = bX ∧ Y + a P i =0 S i X ∧ S i Y , with a = 0 (actually, 5 b = 3 a = 0, as Tr W X = 0; see (46b)), so by (12), the curvaturetensor at every point x ∈ M α has the form(15) R ( X, Y ) = ρX ∧ Y − ρY ∧ X + f X i =0 S i X ∧ S i Y, SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 11 where ρ is a symmetric operator and f = 0 (as N = 2 on M α ). As S i = id (by (8)) and Tr S i = 0 (seethe proof of assertion 1 of Lemma 3), at every point x ∈ M α , the Ricci operator, the scalar curvature andthe Weyl tensor of M are given by Ric X = 14 ρX + (Tr ρ − f ) X, scal = 30 Tr ρ − f, W ( X, Y ) = f (3 X ∧ Y +5 P i =0 S i X ∧ S i Y , respectively. A direct computation using (8) and the fact that the S i ’s aresymmetric, orthogonal and Tr S i = 0 gives k W k = f (see (46c)). As f = 0, it follows that f is asmooth function (hence ρ is a smooth symmetric operator, as scal and Ric are smooth) on M α . Introducean algebraic curvature tensor P defined by P ( X, Y ) = f − ( R ( X, Y ) − ρX ∧ Y + ρY ∧ X ) = P i =0 S i X ∧ S i Y = A ( X ∧ Y ), where the last equation follows from assertion 3 of Lemma 3. As P is smooth, the field A of the endomorphisms of the bundle Skew( M α ) of skew-symmetric endomorphisms over M α is smooth(the fact that A is an endomorphism of the bundle Skew( M α ) follows from assertion 2 of Lemma 3).Then the eigenbundles of A , L ( M α ) and L ( M α ) (assertion 2 of Lemma 3) are also smooth. As thematrix product is smooth, it follows that the subbundle ( L ( M α )) = Span( K K , K , K ∈ L ( M α )) issmooth. By the definition of L and from (8), ( L ( M α )) = L ( M α ) ⊕L ( M α ) ⊕L ( M α ), with the directsum being orthogonal relative to the (smooth) inner product in End( M α ) (assertion 1 of Lemma 3). Since L ( M α ) is smooth and L ( M α ) is a one-dimensional bundle spanned by the identity operator, the bundle L ( M α ) is smooth. Then the bundle ( L ( M α )) = L s =0 L s ( M α ) = L s =0 L s ( M α ) (the latter equationfollows from assertion 1 of Lemma 3) is also smooth. The direct sum on the right-hand side is orthogonalwith respect to the smooth inner product and the bundles L s ( M α ) , s = 0 , , ,
4, are smooth, as it isshown above. Hence L ( M α ) is a smooth subbundle of End( M α ). It follows that on some neighborhood U ( x ) of an arbitrary point x ∈ M α we can choose nine smooth sections S ′ i , i = 0 , , . . . ,
8, of L ( M α ),which are orthogonal and all have norm 4. By assertion 2 of Lemma 3, the operator A does not change,if we replace S i by S ′ i , so by assertion 3 of Lemma 3, the curvature tensor (15) remains unchanged ifwe replace the operators S i by the smooth operators S ′ i . Therefore we can assume f, ρ and the S i ’s in(15) to be smooth on U . Under a conformal change of metric ˜ g = h h· , ·i , for a positive smooth function h = e φ : U → R , the curvature tensor transforms as ˜ R ( X, Y ) = R ( X, Y ) − ( X ∧ KY + KX ∧ Y ), where K = H ( φ ) − ∇ φ ⊗ ∇ φ + k∇ φ k id and H ( φ ) is the symmetric operator associated to the Hessian of φ (see Lemma 8). As X ˜ ∧ Y = hX ∧ Y we obtain ˜ R ( X, Y ) = ˜ ρX ˜ ∧ Y − ˜ ρY ˜ ∧ X + f h − P i =0 S i X ˜ ∧ S i Y , where˜ ρ = h − ( ρ − K ). Taking h = | f | (and dropping the tildes) we obtain (14), with ε = ± f ). (cid:3) Clifford case.
Let x ∈ M ′ and let U = U ( x ) be the neighborhood of x defined in assertion (a)of Lemma 4. By the second Bianchi identity, ( ∇ U R )( X, Y ) Y + ( ∇ Y R )( U, X ) Y + ( ∇ X R )( Y, U ) Y = 0.Substituting R from (13) and using the fact that the operators J i ’s and their covariant derivatives areskew-symmetric and the operator ρ and its covariant derivatives are symmetric we get: h X, Y i (( ∇ U ρ ) Y − ( ∇ Y ρ ) U ) + k Y k (( ∇ X ρ ) U − ( ∇ U ρ ) X ) + h U, Y i (( ∇ Y ρ ) X − ( ∇ X ρ ) Y )+ h ( ∇ Y ρ ) U − ( ∇ U ρ ) Y, Y i X + h ( ∇ X ρ ) Y − ( ∇ Y ρ ) X, Y i U + h ( ∇ U ρ ) X − ( ∇ X ρ ) U, Y i Y + X νi =1 X ( η i ) h J i Y, U i − U ( η i ) h J i Y, X i ) J i Y + X νi =1 Y ( η i )(2 h J i U, X i J i Y + h J i Y, X i J i U − h J i Y, U i J i X )+ X νi =1 η i (cid:0) (3 h ( ∇ U J i ) X, Y i + 3 h ( ∇ X J i ) Y, U i + 2 h ( ∇ Y J i ) U, X i ) J i Y + 3 h J i X, Y i ( ∇ U J i ) Y + 3 h J i Y, U i ( ∇ X J i ) Y + 2 h J i U, X i ( ∇ Y J i ) Y + h ( ∇ Y J i ) Y, X i J i U + h J i Y, X i ( ∇ Y J i ) U − h ( ∇ Y J i ) Y, U i J i X − h J i Y, U i ( ∇ Y J i ) X (cid:1) = 0 . (16)Taking the inner product of (16) with X and assuming X, Y and U to be orthogonal we obtain k X k h Q ( Y ) , U i + k Y k h Q ( X ) , U i − h Q ( X ) , Y ih Y, U i − h Q ( Y ) , X ih X, U i + X νi =1 X ( η i ) h J i Y, U i − Y ( η i ) h J i X, U i − U ( η i ) h J i Y, X i ) h J i Y, X i + X νi =1 η i (cid:0) (2 h ( ∇ U J i ) X, Y i + h ( ∇ X J i ) Y, U i + h ( ∇ Y J i ) U, X i ) h J i Y, X i− h J i Y, U ih ( ∇ X J i ) X, Y i − h J i X, U ih ( ∇ Y J i ) Y, X i (cid:1) = 0 , (17) where Q : R → R is the quadratic map defined by(18) h Q ( X ) , U i = h ( ∇ X ρ ) U − ( ∇ U ρ ) X, X i . Note that h Q ( X ) , X i = 0. Lemma 5.
In the assumptions of Lemma 4, let x ∈ M ′ and let U be the neighborhood of x introducedin assertion (a) of Lemma 4. Suppose that ν > . For every point y ∈ U , identify T y M with theEuclidean space R via a linear isometry. Then1. There exist m i , b ij ∈ R , i, j = 1 , . . . , ν , such that for all X ∈ R and all i, j = 1 , . . . , ν , Q ( X ) = 3 X νk =1 h m k , X i J k X, (19a) ( ∇ X J i ) X = η − i ( k X k m i − h m i , X i X ) + X νj =1 h b ij , X i J j X, (19b) b ij + b ji = η − i J j m i + η − j J i m j . (19c)
2. The following equations hold for all
X, Y ∈ R and all i, j = 1 , . . . , ν : ∇ η i = 2 J i m i , (20a) η i b ij + η j b ji = 0 , i = j, (20b) ( ∇ Y ρ ) X − ( ∇ X ρ ) Y = X νi =1 (2 h J i Y, X i m i − h m i , Y i J i X + h m i , X i J i Y ) , (20c) b ij ( η i − η j ) = 0 , (20d) J i m i = η i p, for some p ∈ R . (20e) Proof.
1. Equation (17) is a polynomial equation in 48 real variables, the coordinates of the vectors
X, Y, U . It must still hold if we allow
X, Y, U to be complex and extend the J i ’s, the ∇ J i ’s and h· , ·i bycomplex linearity (bilinearity) to C . The complexified inner product h· , ·i is a nonsingular quadraticform on C (not a Hermitian inner product on C ).From (17), for any two vectors X, Y ∈ C with Y ⊥ I C X , we get(21) k X k Q ( Y ) + k Y k Q ( X ) − h Q ( X ) , Y i Y − h Q ( Y ) , X i X − X νi =1 η i (cid:0) h ( ∇ X J i ) X, Y i J i Y + h ( ∇ Y J i ) Y, X i J i X (cid:1) = 0 . By assertion 2 of Lemma 1, the Clifford structure has one of two possible forms. We prove identities(19) separately for each of them.
Case (a)
The representation of Cl( ν ) is a restriction to Cl( ν ) ⊂ Cl(8) of the representation ρ ofCl(8) given by (3).By complexification, we can assume that J p X is given by equation (3), where X = ( a, b ) ∈ C and a, b, p ∈ O ⊗ C . Denote C = { X = ( a, b ) : k X k = 0 } ⊂ C the isotropic cone, and for X ∈ C ,denote I C X the complex linear span of X, J X, . . . , J X . For X ∈ C , the space I C X is isotropic: theinner product of any two vectors from I C X vanishes. Take Y = J q X ∈ I C X , with some q ∈ O ⊗ C .Then from (21) we obtain: P νi =1 η i h ( ∇ X J i ) X, J q X i J i J q X ∈ I C X . Introduce the ( C -)linear operator M X : O ⊗ C → O ⊗ C by M X ( q ) = P νi =1 η i h ( ∇ X J i ) X, J q X i e i . Then, as I C X is isotropic, we get J M X ( q ) J q X ⊥ I C X , for all X ∈ C and all q ∈ O ⊗ C . Then by (3), for any X = ( a, b ) ∈ C and any p ∈ O ⊗ C we obtain (( − aq ∗ ) M X ( q ) , − ( bq )( M X ( q )) ∗ ⊥ ( bp, − ap ∗ ), so(22) ( M X ( q )( bq ) ∗ ) a = b ∗ (( aq ∗ ) M X ( q )) . The bioctonion equation ( m ( bq ) ∗ ) a − b ∗ (( aq ∗ ) m ) = 0 for m ∈ O ⊗ C , with ( a, b, q ) on the algebraic surface S = C × C ⊂ C , can be viewed as a system of eight linear equations for m ∈ C . Let M ( a, b, q ) bethe matrix of this system. As both m = q and m = b ∗ a are solutions to the system, rk M ( a, b, q ) ≤ a, b, q ) from a nonempty Zariski open subset of S . On the other hand, if a = q = 1, b ⊥ , k b k = −
1, the equation has the form mb ∗ = b ∗ m , which implies that m ∈ Span C (1 , b ), sork M ( a, b, q ) ≥ a, b, q ) from a nonempty Zariski open subset of S . It follows that fora nonempty Zariski open subset of S , rk M ( a, b, q ) = 6 and the solution set is Span C ( q, b ∗ a ). Therefore SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 13 from (22), M X ( q ) ∈ Span C ( q, b ∗ a ), for all ( a, b, q ) from a nonempty Zariski open subset of the S , hencerk( q, M X q, b ∗ a ) ≤
3, for all X = ( a, b ) ∈ C , q ∈ C . It follows that the linear operators from C toSkew( C ) defined by q → M X q ∧ ( b ∗ a ) and q → q ∧ ( b ∗ a ) are linearly dependent, for every X = ( a, b ) ∈ C ,so M X q ∧ ( b ∗ a ) = c X q ∧ ( b ∗ a ), for c X ∈ C . Then ( M X q − c X q ) ∧ ( b ∗ a ) = 0, so M X ( q ) = c X q + α X ( q ) b ∗ a ,for all X = ( a, b ) ∈ C such that b ∗ a = 0, where α X is a linear form on C . By the definition of M X ( q ) , c X q + α X ( q ) b ∗ a = P νi =1 η i h ( ∇ X J i ) X, J q X i e i . Substituting q = e j , j ≤ ν , and taking the innerproduct with e k , k ≤ ν , we obtain η − k ( c X δ jk + α X ( e j ) h b ∗ a, e k i ) = h ( ∇ X J k ) X, J j X i . As the right-handside is antisymmetric in j and k , we get 2 η − k c X δ jk + α X ( e j ) · η − k h b ∗ a, e k i + α X ( e k ) · η − j h b ∗ a, e j i = 0.As the rank of the ν × ν -matrix A X defined by ( A X ) jk = α X ( e j ) · η − k h b ∗ a, e k i + α X ( e k ) · η − j h b ∗ a, e j i is at most two and as ν >
4, we obtain that c X = 0, for all X = ( a, b ) ∈ C such that b ∗ a = 0, hence A X = 0. Now, if ν = 8, it follows from A X = 0 , b ∗ a = 0, that α X = 0, so M X = 0. If ν <
8, then, as c X = 0, we have M X ( q ) = α X ( q ) b ∗ a = P νi =1 η i h ( ∇ X J i ) X, J q X i e i , so α X ( q ) h b ∗ a, e s i = 0, for all s > ν .Choosing X = ( a, b ) ∈ C such that all the components of b ∗ a are nonzero, we again obtain that α X = 0,so M X = 0. It follows that M X = 0 for all X from a nonempty open subset of C , hence for all X ∈ C .From the definition of M X ( q ), it follows that for all X ∈ C and all i = 1 , . . . , ν, j = 1 , . . . , h ( ∇ X J i ) X, J j X i = 0, so the polynomials h ( ∇ X J i ) X, J j X i are divisible by k X k over C , and henceover R . Then by assertion 1 of Lemma 2, ( ∇ X J i ) X = P j =1 h b ij , X i J j X + h V i , X i X − k X k U i , for all i = 1 , . . . , ν , where b ij , V i , U i are some vectors from R (note that the summation on the right-handside is up to 8, not up to ν , as in (19b)). As h ( ∇ X J i ) X, X i = 0, we have V i = U i , so(23) ( ∇ X J i ) X = X j =1 h b ij , X i J j X + η − i ( k X k m i − h m i , X i X ) , for some m i ∈ R . Substituting this into (21), complexifying the resulting equation and taking X ∈ C , Y = J q X , with some q ∈ O ⊗ C we get h Q ( X ) , J q X i J q X − h Q ( J q X ) , X i X = 0. As X and J q X are linearly independent for a nonempty Zariski open subset of ( X, q ) ∈ S = C × C , the polynomial h Q ( X ) , J q X i vanishes on S , so for all q ∈ O ⊗ C , the polynomial h Q ( X ) , J q X i is divisible by k X k .Then h Q ( X ) , J i X i is divisible by k X k , for all i = 1 , . . . ,
8, which by assertion 1 of Lemma 2, impliesthat h Q ( X ) , Y i = k X k h Y, U i , for some fixed U ∈ R , where Y ⊥ I X .It then follows from (21) and (23) that for all X, Y ∈ R , with Y ⊥ I X ,(24) k X k T ( Y ) + k Y k T ( X ) = 0 , where the quadratic form T : R → R is defined by T ( X ) = Q ( X ) − h X, U i X − P νi =1 h m i , X i J i X .We want to show that T = 0. Suppose that for some E ∈ R , the quadratic form t ( X ) = h T ( X ) , E i isnonzero. Then from (24), k X k t ( Y ) + k Y k t ( X ) = 0, for all X, Y ∈ R , Y ⊥ I X . If X = ( a, Y =( b, a ⊥ b , then Y ⊥ I X by (3), so k a k t (( b, k b k t (( a, t (( a, a ∈ O . Similarly, t ((0 , b )) = 0, for all b ∈ O . It follows that t (( a, b )) = h La, b i for some L ∈ End( O ).From (3), any X = ( a, b ) , a, b = 0, and any Y = ( k b k − bu, k a k − au ∗ ), with u ⊥ b ∗ a , satisfy Y ⊥ I X .Then from k X k t ( Y ) + k Y k t ( X ) = 0 and t (( a, b )) = h La, b i we obtain k u k h La, b i = h L ( bu ) , au ∗ i , forall u ⊥ b ∗ a (the condition a, b = 0 can be dropped). It follows that hk u k L t b − L ( bu ) · u, a i = 0 for all a ⊥ bu (where L t is the operator transposed to L ), so k u k L t b − L ( bu ) · u k bu , for all b, u ∈ O . Takingthe inner product with bu we get k u k L t b − L ( bu ) · u = 0. Taking u = 1 we get L t = L , so Lb · u ∗ = L ( bu ),for all b, u ∈ O . Substituting b = 1 we obtain Lu = pu ∗ , where p = L
1, so ( pb ∗ ) u ∗ = p ( bu ) ∗ . If p = 0,this equation, with b = p , implies p ∗ u ∗ = u ∗ p ∗ , for all u ∈ O . This contradiction shows that p = 0,hence L = 0, hence t = 0.Therefore, the quadratic form T ( X ) = Q ( X ) − h X, U i X − P νi =1 h m i , X i J i X vanishes, which implies Q ( X ) = h X, U i X + 3 P νi =1 h m i , X i J i X . Substituting this and (23) into (21), with Y ⊥ I X , we get U = 0, which proves (19a).Now, if ν = 8, then (19b) follows from (23). If ν <
8, choose X = 0 , Y = J s X, s > ν . Substitutinginto (21) and using (23) and (19a) we get P νi =1 η i ( h b is , X i J i J s X + h b is , J s X i J i X ) = 0. Taking X = ( a, X = (0 , a ) and using (3) we get h b is , ( a, i = h b is , (0 , a ) i = 0, so b is = 0, for all 1 ≤ i ≤ ν < s ≤ h ( ∇ X J i ) X, J j X i is antisymmetric in i and j . Case (b) ν = 7 and the representation of Cl(7) is given by (4).Let e i , i = 1 , . . . ,
7, be a fixed orthonormal basis in O ′ = 1 ⊥ (or in its complexification), for instance,the one with the multiplication table as in [Bes, Section 3.64].As in Case (a), by complexification, we can assume that J p X is given by equation (4), where X =( a, b ) ∈ C and a, b, p ∈ O ⊗ C , p ⊥
1. We extend J p X to O ⊗ C by complex linearity by defining J a · X = aX , for a ∈ C . Denote C the isotropic cone in C , and for X ∈ C , denote I C X the complexification of I X . Take X ∈ C , q ∈ O ⊗ C . Then Y = J q X ∈ I C X , so by (21), P i =1 η i h ( ∇ X J i ) X, J q X i J i J q X ∈ I C X .Introduce the operator M X ∈ End( O ⊗ C ) by(25) M X ( q ) = X i =1 η i h ( ∇ X J i ) X, J q X i e i . As I C X is isotropic, J M X ( q ) J q X ⊥ J p X , for all q, p ∈ O ⊗ C . Then by (4), for any X ∈ C , q ∈ O ⊗ C ,( aq ) ∗ ( aM X ( q )) + ( bq ) ∗ ( bM X ( q )) = 0 . Consider this bioctonion equation as a linear system for M ∈ End( O ⊗ C ), with X = ( a, b ) ∈ C . A directcomputation shows that M ( q ) = q, a ∗ ( bq ) and M ( q ) = h v, q i , h w, q i a ∗ b , with arbitrary v, w ∈ O ⊗ C ,are the solutions. When a = e , b = i e , these solutions span a subspace of dimension 18 of End( O ⊗ C ),so for all X = ( a, b ) from a nonempty Zariski open subset C ⊂ C , the dimension of the solution space isat least 18. On the other hand, a direct computation, with a = e , b = i e shows that every solution isa linear combination of those above. It follows that the corank of the matrix of the linear system (whoseentries are polynomials in the coordinates of X = ( a, b ) ∈ C ) equals 18 for all the points X = ( a, b ) froma nonempty Zariski open subset C ⊂ C . Then for every X = ( a, b ) ∈ C = C ∩ C , the operator M isbe a linear combination of the four listed above, that is, M X ( q ) = h v X , q i h w X , q i a ∗ b + α X q + β X a ∗ ( bq ) , for all X = ( a, b ) ∈ C , where v, w : C → O ⊗ C , α, β : C → C . From (25), M X (1) = 0 , M X ( q ) ⊥
1, so(26) M X ( q ) = Im( h w X , q i a ∗ b − h w X , i a ∗ ( bq ) + α X q ) , where Im is the operator of taking the imaginary part of a bioctonion: Im( q ) = q − h q, i
1. Define thesymmetric operator D on O ⊗ C by D , De i = η − i e i . From (25) it follows that h M X q, Dq i = 0,for all q ∈ O ⊗ C , so for all X = ( a, b ) ∈ C ,(27) h w X , q ih a ∗ b, Dq i = h w X , ih a ∗ ( bq ) , Dq i − α X h q, Dq i . Substituting q = b ∗ a and using the fact that Dq ⊥ h w X , b ∗ a i − α X ) h ( b ∗ a ) , D ( b ∗ a ) i = 0. Thealgebraic function h ( b ∗ a ) , D ( b ∗ a ) i is not zero on C (for instance, for a = i e , b = 1), hence on a nonemptyZariski open subset C ⊂ C , we have α X = h w X , b ∗ a i .For x, y ∈ O ⊗ C , define the operators L x and x ⊙ y on O ⊗ C by L x q = xq and ( x ⊙ y ) q = h y, q i x + h x, q i y .As L tx = L x ∗ and L a ∗ L b + L b ∗ L a = 2 h a, b i id, equation (27) can be rewritten as(28) D ( a ∗ b ) ⊙ w X = h w X , i [ D, L a ∗ L b ] − h Im w X , b ∗ a i D. Let S be a symmetric operator commuting with D . Multiplying both sides of (28) by S and takingthe trace we get h SD ( a ∗ b ) , w X i = −h Im w X , b ∗ a i Tr SD . Choosing S in such a way that SD = Im weget h Im w X , b ∗ a i = 0, hence h SD ( a ∗ b ) , w X i = 0, for any symmetric S commuting with D . Taking S diagonal relative to the basis e i we obtain h a ∗ b, e i ih w X , e i i = 0, for all i = 1 , . . . ,
7. As for a nonemptyZariski open subset of X = ( a, b ) ∈ C , all the numbers h a ∗ b, e i i are nonzero, we get Im w X = 0, that is, w X = γ X
1, for all X from a nonempty Zariski open subset of C . Then from the above, α X = γ X h a, b i and from (26), M X ( q ) = γ X ( h , q i a ∗ b − a ∗ ( bq ) + h a, b i q − h a, b ih , q i + h b ∗ a, q i ) , for all X = ( a, b ) from a nonempty Zariski open subset of C . It then follows from (25) that for all i = 1 , . . . , h ( ∇ X J i ) X, J q X i = η − i h M X ( q ) , e i i = η − i γ X ( h , q ih a ∗ b, e i i − h a ∗ ( bq ) , e i i + h a, b ih q, e i i ).Introduce the quadratic forms Φ i , Ψ i : C → C by ( ∇ X J i ) X = (Φ i ( X ) , Ψ i ( X )), for i = 1 , . . . , a ∗ Φ i ( X ) + b ∗ Ψ i ( X ) = η − i γ X ( h a ∗ b, e i i − b ∗ ( ae i ) + h a, b i e i ).It follows that for every fixed i = 1 , . . . ,
7, the polynomial vectors a ∗ Φ i ( X ) + b ∗ Ψ i ( X ) and T i ( X ) = h a ∗ b, e i i− b ∗ ( ae i )+ h a, b i e i are linearly dependent for all X = ( a, b ) from a nonempty Zariski open subset SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 15 of C , that is, for all X ∈ C . Note that h a ∗ Φ i ( X )+ b ∗ Ψ i ( X ) , i = h T i ( X ) , i = 0 (the first equation followsfrom h ( ∇ X J i ) X, X i = 0). Then the rank of the 7 × N ( X ) = ( a ∗ Φ i ( X ) + b ∗ Ψ i ( X ) | T i ( X ))(whose j -th row, j = 1 , . . . ,
7, is h ( a ∗ Φ i ( X ) + b ∗ Ψ i ( X ) , e j i | h T i ( X ) , e j i )) is at most one, for all X ∈ C .As R = C [ X ] / (cid:0) k X k (cid:1) , the coordinate ring of C , is a unique factorization domain [Nag], there exist u , u ∈ R and v in the free module R such that π ( N ( X )) = ( u v | u v ), where π : C [ X ] → R isthe natural projection. Let U , V j ∈ C [ X ] be the polynomials of the lowest degree in the cosets π − u and π − v j respectively. Lifting the equation u v j = π ( h T i ( X ) , e j i ) = π ( h ae i , be j i ), for j = i , to C [ X ]we get h ae i , be j i = U ( X ) V j ( X ) + k X k Ξ j ( X ), for some Ξ j ∈ C [ X ]. Then U and V j are nonzero (as h ae i , be j i is not divisible by k X k = k a k + k b k ). Moreover, as the polynomial on the left-hand sideis of degree two in X , and k X k is prime in C [ X ], the polynomials U and V j are homogeneous, withdeg U + deg V j = 2 and Ξ j are constants.Suppose deg U = 2. Then the V j ’s are nonzero constants for all j = i . It follows that for somenontrivial linear combination e ′ of the e j , j = i , h ae i , be ′ i = 0, for all a, b ∈ O , a contradiction.Suppose deg U = 1. Then deg V j = 1, for all j = i . Taking a = 0 we get Ξ i = 0 (as the rank ofthe quadratic form U ((0 , b )) V j ((0 , b )) in b is at most two), so for some nonzero linear forms U and V j , h ae i , be j i = U ( X ) V j ( X ). Taking a = 0 or b = 0 we obtain that U ( X ) = h l, a i , V j ( X ) = h t j , b i (or viceversa), for some nonzero l, t j ∈ O , j = i . Then taking b = a we arrive at a contradiction.It follows that deg U = 0, and without loss of generality, we can take U = 1. Then π ( N ( X )) =( u v | v ), for some u ∈ R , v ∈ R , so, lifting to C [ X ], we obtain a ∗ Φ i ( X ) + b ∗ Ψ i ( X ) = U ( X ) T i ( X ) + k X k Ξ( X ), for some U ( X ) ∈ C [ X ] and Ξ( X ) ∈ C [ X ] , with h Ξ( X ) , i = 0. As the left-hand side is avector whose components are homogeneous cubic polynomials in X , the components of T i are quadraticforms of X , and k X k is prime in C [ X ], U ( X ) is a linear form and Ξ is a linear operator. As bothsides are real for X ∈ R , for every i = 1 , . . . ,
7, there exist p i ∈ R and Ξ i : R → R = O ′ such that a ∗ Φ i ( X )+ b ∗ Ψ i ( X ) = h p i , X i T i ( X )+ k X k Ξ i X = h p i , X i ( h a ∗ b, e i i− b ∗ ( ae i )+ h a, b i e i )+ k X k Ξ i X , for all X = ( a, b ) ∈ R . Let ˆΦ i ( X ) = Φ i ( X ) − a Ξ i X − h p i , X i be i , ˆΨ i ( X ) = Ψ i ( X ) − b Ξ i X + h p i , X i ae i . Then a ∗ ˆΦ i ( X ) + b ∗ ˆΨ i ( X ) = h p i , X ih a ∗ b, e i i . From assertion 2 of Lemma 2, with N ( X ) = ˆΦ i ( X ) , N ( X ) =ˆΨ i ( X ), p = p i , u = e i , it follows that ( ∇ X J i ) X = (Φ i ( X ) , Ψ i ( X )) = ( a Ξ i X + h p i , X i be i , b Ξ i X − h p i , X i ae i ) + k X k m i − h m i , X i X − P j =1 h J j X, m i i J j X , for some m i ∈ R . As h ( ∇ X J i ) X, X i = 0and h Ξ i X, i = 0 we get h p i , X ih e i , b ∗ a i = 0. Since the polynomial h e i , b ∗ a i is not divisible by k X k , weobtain p i = 0. Then by (4), ( a Ξ i X, b Ξ i X ) = J Ξ i X X = P j =1 h Ξ i X, e j i J j X which implies (19b), with b ij = J j m i + Ξ ti e j and with m i replaced by η − i m i . Equation (19c) follows from (19b) and the fact that h ( ∇ X J i ) X, J j X i is antisymmetric in i and j .We next prove (19a). Substituting (19b) into (21) we obtain(29) k X k T ( Y ) + k Y k T ( X ) − h T ( X ) , Y i Y − h T ( Y ) , X i X = 0 , for all X ⊥ I Y, where the quadratic form T : R → R is defined by(30) T ( X ) = Q ( X ) − X i =1 h m i , X i J i X. For a nonzero U ∈ R , let X, Y ⊥ I U be linearly independent. Then the eight-dimensional spaces I X and I Y are both orthogonal to U , so their intersection is nontrivial. But if J u X = J v Y , then J h u,v i u −k u k v X = J k v k u Y , so dim( I X ∩I Y ) ≥
2. It follows that I X, I Y ⊥ V , for some nonzero V ⊥ U .Substituting V for Y into (29) and taking the inner product with U we obtain k X k − h T ( X ) , U i = −k V k − h T ( V ) , U i . Similarly, substituting V for X into (29) and taking the inner product with U we get k Y k − h T ( Y ) , U i = −k V k − h T ( V ) , U i , so k X k − h T ( X ) , U i = k Y k − h T ( Y ) , U i , for all nonzero X, Y ⊥ I U . It follows that for some function f : R → R , which is homogeneous of degree one,(31) h T ( X ) , U i = k X k f ( U ) , for all X ⊥ I U. Taking the inner product of (29) with Z ⊥ I Y we obtain h T ( Y ) , k X k Z −h X, Z i X i + k Y k h T ( X ) , Z i = 0which by (31) implies h T ( X ) , Z i = f ( h X, Z i X ) − k X k Z ), for all X, Z with dim( I X ∪ I Z ) < I X and I Z having a nontrivial intersection. In particular, taking Z = J i X we obtain h T ( X ) , J i X i = −k X k f ( J i X ). Replacing X by J i X we get h T ( J i X ) , X i = −k X k f ( X ), for all X ∈ R . For an arbitrary nonzero X ∈ R , let U i , i = 1 , . . . ,
8, be an orthonormal basis for ( I X ) ⊥ . Denoting Tr T the vector in R whose components are the traces of the corresponding components of T and using the fact that T ( X ) ⊥ X (which follows from (30) and the fact that Q ( X ) ⊥ X ) we get h Tr T, X i = k X k − P i =1 h T ( J i X ) , X i + P i =1 h T ( U i ) , X i = − f ( X )+8 f ( X ) = f ( X ), by (31). Therefore f is a linear form, f ( X ) = h l, X i , for some l ∈ R . Then T ( X ) = k X k − P i =1 h T ( X ) , J i X i J i X + P i =1 h T ( X ) , U i i U i = − P i =1 h l, J i X i J i X + k X k P i =1 h l, U i i U i = k X k ( π ( I X ) ⊥ l − π I X l ) + h l, X i X = k X k ( l − π I X l ) + h l, X i X . Substituting this into (29) and using (31) we get l = π I X l + π I Y l , forall Y ⊥ I X . Let l = ( l , l ). As it follows from (4), for X = ( a, b ) , π I X l = k X k − ( k a k l + a ( b ∗ l ) ,b ( a ∗ l ) + k b k l ), and if a, b = 0, the vector Y = ( k b k aq, −k a k bq ) satisfies Y ⊥ I X , for all q ∈ O .Then the equation l = π I X l + π I Y l implies k q k a ( b ∗ l ) = ( aq )(( bq ) ∗ l ) , k q k b ( a ∗ l ) = ( bq )(( aq ) ∗ l ), forarbitrary a, b, q ∈ O . The first of them implies l = 0 (to see that, take b = q ∗ , then the octonions a, q, l associate, for every a, q , so l ∈ R ; if l = 0, then the octonions a, q, ( bq ) ∗ associate, for every a, q, b , acontradiction), the second one can be obtained from the first one by interchanging a and b , so it implies l = 0. Thus l = 0, so T ( X ) = 0, which is equivalent to (19a) by (30).2. Substitute X = J k Y, U ⊥ X, Y into (17) and consider the first term in the second summation.As h J i Y, X i = k Y k δ ik , that term equals 3 η k (2 h ( ∇ U J k ) X, Y i + h ( ∇ X J k ) Y, U i + h ( ∇ Y J k ) U, X i ) k Y k .As J k is orthogonal and skew-symmetric, h ( ∇ U J k ) X, Y i = h ( ∇ U J k ) J k Y, Y i = −h J k ( ∇ U J k ) Y, Y i = h ( ∇ U J k ) Y, J k Y i = 0. Next, h ( ∇ Y J k ) U, X i = −h ( ∇ Y J k ) J k Y, U i = h J k ( ∇ Y J k ) Y, U i = h ( η − k k Y k J k m k + P νj =1 h b kj , Y i J k J j Y, U i by (19b). Again by (19b), as Y = − J k X , h ( ∇ X J k ) Y, U i = h J k ( ∇ X J k ) X, U i = h J k ( η − k ( k X k m k − h m k , X i X ) + P νj =1 h b kj , X i J j X ) , U i = h η − k k Y k J k m k + P j = k h b kj , J k Y i J j Y −h b kk , J k Y i J k Y, U i . Substituting this into (17) and using (19a, 19b) we obtain after simplification:(32) k Y k ( h J k m k , U i − U ( η k )) + X νj =1 h η k b kj + η j b jk , h J j Y, U i J k Y + h J k J j Y, U i Y i = 0 . Choose i = j such that k = i, j and take the eigenvectors of the symmetric orthogonal operator J j J i J k as U . For each such U, J i J k U = ± J j U , so dim( J J k U + J U ) < ν ≤
16, hence there exists a nonzero Y ⊥ J U + J J k U , which implies U ⊥ J Y + J J k Y . Substituting such U and Y into (32) we obtain h U, J k m k − ∇ η k i = 0. As the eigenvectors of J j J i J k span R , equation (20a) follows.Substituting (20a) into (32) we obtain P νj =1 h η k b kj + η j b jk , h J j Y, U i J k Y + h J k J j Y, U i Y i = 0, whichimplies P j = k ( h η k b kj + η j b jk , J k Y i J j Y + h η k b kj + η j b jk , Y i J k J j Y ) = 0. Equation (20b) now follows fromassertion 2 of [N5, Lemma 3].By (18) and (19a), h ( ∇ X ρ ) U − ( ∇ U ρ ) X, X i = 3 P νi =1 h m i , X ih J i X, U i , for all X, U ∈ R . Po-larizing this equation and using the fact that the covariant derivative of ρ is symmetric we obtain h ( ∇ X ρ ) U, Y i + h ( ∇ Y ρ ) U, X i− h ( ∇ U ρ ) Y, X i = 3 P νi =1 ( h m i , Y ih J i X, U i + h m i , X ih J i Y, U i ). Subtractingthe same equation, with Y and U interchanged, we get h ( ∇ Y ρ ) U − ( ∇ U ρ ) Y, X i = P νi =1 (2 h m i , X ih J i Y, U i + h m i , Y ih J i X, U i − h m i , U ih J i X, Y i ), which implies (20c).To prove (20d), substitute X ⊥ I Y, U = J k Y into (16). Using (19, 20c) we obtain after simplification:3( ∇ X J k ) Y − ( ∇ Y J k ) X = − η − k h m k , Y i X + X νi =1 η − k h η i b ik + 2 δ ik J k m k , Y i J i X mod ( I Y ) . Subtracting three times the polarized equation (19b) (with i = k ) and solving for ( ∇ Y J k ) X we get(33) ( ∇ Y J k ) X = X νi =1 14 η − k h η k b ki − η i b ik − δ ik J k m k , Y i J i X mod ( I Y ) , for all X ⊥ I Y . Choose s = k and define the subset S ks ⊂ R ⊕ R by S ks = { ( X, Y ) :
X, Y = 0, X, J k X, J s X ⊥ J Y } . It is easy to see that ( X, Y ) ∈ S ks ⇔ ( Y, X ) ∈ S ks and that replacing J Y by I Y in the definition of S ks gives the same set S ks . Moreover, the set { X : ( X, Y ) ∈ S ks } (and hencethe set { Y : ( X, Y ) ∈ S ks } ) spans R . If ν <
8, this follows from [N1, Lemma 3.2 (4)]. If ν = 8, theClifford structure is given by (3). Take X = ( a, b ), with k a k = k b k = 1, and Y = ( bu, au ∗ ) for somenonzero u ∈ O . Then the condition X ⊥ J Y is satisfied and the condition J k X ⊥ J Y is equivalent to h ( ae ∗ k ) q, bu i + h ( be k ) q ∗ , au ∗ i = 0, for all q ∈ O , that is, to ( ae ∗ k ) ∗ ( bu ) + ( au ∗ ) ∗ ( be k ) = 0. As ( ae ∗ k ) ∗ ( bu ) +( au ∗ ) ∗ ( be k ) = 2 h ae ∗ k , b i − b ∗ (( ae ∗ k ) u ) + 2 h au ∗ , b i − b ∗ (( au ∗ ) e k ) = 2 h ae ∗ k , b i − h e k , u i b ∗ a + 2 h au ∗ , b i , thelatter condition is satisfied, if we choose a, b and u in such a way that b ∗ a, u and e k are orthogonal.Similar arguments for e s show that for every X = ( a, b ), with k a k = k b k = 1 and b ∗ a ⊥ e k , e s , there SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 17 exists Y = 0 such that ( X, Y ) ∈ S ks . In particular, taking X = ( ± be i , b ), with a fixed e i ⊥ e k , e s andarbitrary unit b ∈ O we obtain that the set { X : ( X, Y ) ∈ S ks } spans R .Now, for ( X, Y ) ∈ S ks , take the inner product of (33) with J s X . Since h ( ∇ Y J k ) X, J s X i is antisym-metric in k and s , we get h (3 − η k η − s ) b ks + (3 − η s η − k ) b sk , Y i = 0, for a set of the Y ’s spanning R .So (3 − η k η − s ) b ks + (3 − η s η − k ) b sk = 0, for all k = s . This and (20b) imply (20d).Now from (20b, 20d) it follows that b ij + b ji = 0 for all i = j , so by (19c), η − i J j m i = − η − j J i m j .Acting by J i J j we obtain that the vector η − i J i m i is the same, for all i = 1 , . . . , ν , which proves (20e). (cid:3) Lemma 6.
In the assumptions of Lemma 4, let x ∈ M ′ and let U be the neighborhood of x introducedin assertion (a) of Lemma 4. Then there exists a smooth metric on U conformally equivalent to theoriginal metric whose curvature tensor has the form (13) , with ρ a constant multiple of the identity.Proof. If ν ≤
4, the proof follows from [N5, Lemma 7]. Suppose ν ≥
4. Let f be a smooth function on U and let h· , ·i ′ = e f h· , ·i . Then W ′ = W, J ′ i = J i , η ′ i = e − f η i and, on functions, ∇ ′ = e − f ∇ , where weuse the dash for the objects associated to metric h· , ·i ′ . Moreover, the curvature tensor R ′ still has theform (13), and all the identities of Lemma 5 remain valid.In the cases considered in Lemma 5, the ratios η i /η are constant, as it follows from (20a,20e). Inparticular, taking f = ln | η | we obtain that η ′ is a constant, so all the η ′ i are constant, m ′ i = 0 by(20a), so ( ∇ ′ Y ρ ′ ) U − ( ∇ ′ U ρ ′ ) Y = 0 by (20c). Dropping the dashes, we obtain that, up to a conformalsmooth change of the metric on U , the curvature tensor has the form (13), with ρ satisfying the identity( ∇ Y ρ ) X = ( ∇ X ρ ) Y , for all X, Y , that is, with ρ being a symmetric Codazzi tensor .Then by [DS, Theorem 1], at every point of U , for any three eigenspaces E β , E γ , E α of ρ , with α / ∈ { β, γ } , the curvature tensor satisfies R ( X, Y ) Z = 0, for all X ∈ E β , Y ∈ E γ , Z ∈ E α . It thenfollows from (13) that(34) X νi =1 η i (2 h J i X, Y i J i Z + h J i Z, Y i J i X − h J i Z, X i J i Y ) = 0 , for all X ∈ E β , Y ∈ E γ , Z ∈ E α , α / ∈ { β, γ } . Suppose ρ is not a multiple of the identity. Let E , . . . , E p , p ≥
2, be the eigenspaces of ρ . If p > E ′ = E , E ′ = E ⊕ · · · ⊕ E p . Then by linearity, (34) holds for any X, Y ∈ E ′ α , Z ∈ E ′ β , suchthat { α, β } = { , } . Hence to prove the lemma it suffices to show that (34) leads to a contradiction, inthe assumption p = 2. For the rest of the proof, suppose that p = 2. Denote dim E α = d α .If ν <
8, the claim follows from the proof of [N5, Lemma 7] (see [N5, Remark 4]). Suppose ν = 8,then the Clifford structure is given by (3).Choosing Z ∈ E α , X, Y ∈ E β , α = β , and taking the inner product of (34) with X we obtain P i =1 η i h J i X, Y ih J i X, Z i = 0. It follows that for every X ∈ E α , the subspaces E and E are invariantsubspaces of the symmetric operator R ′ X ∈ End( R ) defined by R ′ X Y = P i =1 η i h J i X, Y i J i X . So R ′ X commutes with the orthogonal projections π β : R → E β , β = 1 ,
2. Then for all α, β = 1 , α and β can be equal), all X ∈ E α and all Y ∈ R , P i =1 η i h J i X, π β Y i J i X = P i =1 η i h J i X, Y i π β J i X . Taking Y = J j X we get that π β J j X ⊂ J X , that is, π β J X ⊂ J X , for all X ∈ E α , α, β = 1 ,
2. As π + π = id,we obtain J X ⊂ π J X ⊕ π J X ⊂ J X , hence(35) J X = π J X ⊕ π J X, for all X ∈ E ∪ E . As all four functions f αβ : E α → Z , α, β = 1 ,
2, defined by f αβ ( X ) = dim π β J X, X ∈ E α , are lowersemi-continuous and f α ( X ) + f α ( X ) = 8 for all nonzero X ∈ E α , there exist constants c αβ , with c α + c α = 8, such that dim π β J X = c αβ , for all α, β = 1 , X ∈ E α .Consider two cases.First assume that there exist no nonzero Y ∈ E α , Z ∈ E β , α = β , such that Y ⊥ J Z . Thenit follows from (35) that J X ⊃ E β , for α = β . Then d , d ≤
8. As d + d = 16, we obtain d = d = 8 and J X = E β , for every X ∈ E α , α = β . Then (34), with X, Y ∈ E α , Z ∈ E β , α = β , gives P i =1 η i ( h Z, J i X i J i Y − h Z, J i Y i J i X ) = 0. Taking the inner product with J j X we get η j h Z, J j Y ik X k = P i =1 η i h Z, J i X ih J i Y, J j X i . Taking Z = J k X ( ∈ E β ) and assuming X ⊥ Y we obtain ( η j + η k ) h X, J k J j Y i = 0, for k = j . Taking X = J k J j Y ( ∈ E α ) we get η j + η k = 0, for k = j , soall the η i ’s are zeros, a contradiction with ν = 8.Otherwise, assume that there exist nonzero Y ∈ E α , Z ∈ E β , α = β , such that Y ⊥ J Z . Substi-tuting such Y and Z into (34), with X ∈ E α , we obtain P i =1 η i (2 h J i Y, X i J i Z + h J i Z, X i J i Y ) = 0.Taking X ∈ E α orthogonal to π α J Y (and then to π α J Z ) we obtain π α J Y = π α J Z . As the condition Y ⊥ J Z is symmetric in Y and Z , we can interchange Y and Z and α and β to get π β J Y = π β J Z ,which by (35) implies that J Y = J Z , for any two nonzero vectors Y ∈ E α , Z ∈ E β , α = β , such that Y ⊥ J Z . Now, if for some nonzero Y ∈ E α there exists Z ∈ E β , α = β , such that Y ⊥ J Z , then by(35), the space π β J Y is a proper subspace of E β , so c αβ < d β , which implies that for every nonzero X ∈ E α and any nonzero Z from the orthogonal complement to π β J X in E β (which is nontrivial), X ⊥ J Z , hence J X = J Z , from the above.Consider an operator P = Q i =1 J i . As the J i ’s are anticommuting almost Hermitian structures, P is symmetric and orthogonal, and Tr P = 0, as P is the product of the symmetric operator Q i =1 J i andthe skew-symmetric one, J . So its eigenvalues are ±
1, with both eigenspaces V ± of dimension 8. Asthe J i ’s anticommute, each of them interchanges the eigenspaces of P : J i V ± = V ∓ .From the above, for every unit vector X ∈ E , there exists a unit vector Z ⊥ X such that J X = J Z .Therefore, by (2) there exists F : R → R such that for all u ∈ R , J F ( u ) X = J u Z . As the right-handside is linear in u , the map F is also linear: F ( u ) = Au , for some A ∈ End( R ). Moreover, as J u X is an orthogonal multiplication and as X and Z are orthonormal, the operator A is orthogonal andskew-symmetric. Without losing generality, we can assume that an orthonormal basis for R is chosenin such a way that J i X = J i +4 Z , for i = 1 , , ,
4, so J j J j +4 J i +4 J i X = X , for all 1 ≤ i = j ≤
4. Aschanging an orthonormal basis for R may only change the sign of P , it follows that X is an eigenvectorof P , say X ∈ V + . As X is an arbitrary unit vector from E , by continuity, E ⊂ V + . Similarly, E ⊂ V − . But as every J i interchanges the V ± ’s, we get J E = E , which contradicts the assumptionthat there exist nonzero Y ∈ E , Z ∈ E with Y ⊥ J Z .Hence the Codazzi tensor ρ is a multiple of the identity. The definition of the Codazzi tensor easilyimplies that ρ is a constant multiple of the identity on U . (cid:3) By Lemma 6, up to a conformal change of the metric, we can assume that on U , the curvaturetensor has the form (13), with ρ a constant multiple of the identity. Then (18) implies that Q = 0,so m i = 0 by (19a), ∇ η i = 0 by (20a) and ( ∇ X J i ) X = P νj =1 h b ij , X i J j X by (19b). Then from(13), ( ∇ X R )( X, Y ) X = 3 P νi =1 η i ( h ( ∇ X J i ) X, Y i J i X + h J i X, Y i ( ∇ X J i ) X ) = 3 P νi,j =1 h η i b ij + η j b ji , X ih J j X, Y i J i X = 0, as η i b ij + η j b ji = 0 (by (20b) for i = j , and by (19c) for i = j , as m i = 0).It is well-known [Bes, Proposition 2.35] that the equation ( ∇ X R )( X, Y ) X = 0 implies ∇ R = 0, sothe metric on U is locally symmetric. As ρ is a multiple of the identity, it follows from (13) that thecurvature tensor is Osserman. Then by [GSV, Lemma 2.3], U is either flat or is locally isometric to arank-one symmetric space.Thus, for every x ∈ M ′ satisfying assertion (a) of Lemma 4, the metric on the neighborhood U = U ( x )is either conformally flat or is conformally equivalent to the metric of a rank-one symmetric space.3.3. Cayley case.
In this section, we consider the case when the Weyl tensor has a Cayley structure.Let x ∈ M ′ and let U = U ( x ) be neighborhood of x defined in assertion (b) of Lemma 4. Up to aconformal change of the metric, the curvature tensor on U is given by (14). Then(36) ( ∇ Z R )( X, Y ) = ( ∇ Z ρ ) X ∧ Y − ( ∇ Z ρ ) Y ∧ X + ε X i =0 (( ∇ Z S i ) X ∧ S i Y + S i X ∧ ( ∇ Z S i ) Y ) , so the second Bianchi identity has the form(37) σ XY Z (cid:0) A ( Y, X ) ∧ Z + ε X i =0 S i Z ∧ B i ( Y, X ) (cid:1) = 0 , where σ XY Z is the sum taken over the cyclic permutations of (
X, Y, Z ), and the skew-symmetric maps
A, B w : R × R → R are defined by(38) A ( Y, X ) = ( ∇ Y ρ ) X − ( ∇ X ρ ) Y, B w ( Y, X ) = ( ∇ Y S w ) X − ( ∇ X S w ) Y, for w ∈ R , X, Y ∈ R , where S w X = P i =0 w i S i is the orthogonal multiplication defined in Section 2.3. SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 19
Lemma 7.
In the assumptions of Lemma 4, let x ∈ M ′ and let U be the neighborhood of x introducedin assertion (b) of Lemma 4. For every point y ∈ U , identify T y M with the Euclidean space R viaa linear isometry. Then1. There exists a linear map N : R → Skew( R ) , X → N X , such that ∇ X S w = [ S w , N X ] for all w ∈ R , X ∈ R .2. For every unit vector w ∈ R , there exists a linear operator L w : w ⊥ → E ( S w ) such that for every X, Y ∈ E ( S w ) , Z ∈ E − ( S w ) , and every u ∈ R , u ⊥ w , (39) π E ( S w ) B w ( Y, X ) = 0 ,π E ( S w ) A ( Y, X ) = 0 ,π E ( S w ) B u ( Y, X ) = h L w u, X i Y − h L w u, Y i X,π E ( S w ) ( A ( Z, Y ) − εB w ( Z, Y )) = εS L tw Y Z, where E ± ( S w ) are the ± -eigenspaces of S w and π E ( S w ) is the orthogonal projection to E ( S w ) .3. There exists a bilinear skew-symmetric map T : R × R → R such that for all X, Y ∈ R , w ∈ R , A ( Y, X ) = 0 , B w ( Y, X ) = X i =0 ( h T ( w, e i ) , X i S i Y − h T ( w, e i ) , Y i S i X ) .
4. The tensor ρ is a constant multiple of the identity on U .5. ∇ X S w = − P i =0 h T ( w, e i ) , X i S i .Proof.
1. Denote T i = ∇ X S i . The operators T i are symmetric and satisfy T i S j + T j S i + S j T i + S i T j = 0,for all i, j = 0 , , . . . ,
8, by (8). In particular, the operators S i T i are skew-symmetric and(40) [ S i , S i T i ] = 2 T i , [ S i , S j T j ] = T i + S j T i S j , i = j. Define N : R → Skew( R ) by N X = (7 id + A ) P j =0 S j T j , for X ∈ R . The fact that N X is indeedskew-symmetric follows from assertion 2 of Lemma 3, as S j T j are skew-symmetric. Moreover, for any i = 0 , . . . ,
8, from assertion 3 of Lemma 3, [ S i , A ( Q )] = − S i , Q ] − A ([ S i , Q ]) = ( − −A )([ S i , Q ]), so[ S i , N X ] = [ S i , (7 id + A ) P j =0 S j T j ] = (5 id −A ) P j =0 [ S i , S j T j ]. Then by (8) and (40), [ S i , N X ] = (5 id −A )(11 id + A ) T i = − ( A + 6 A −
55 id) T i . As T i is symmetric and Tr T i = 0 (which followsfrom T i = S i · S i T i and the fact that S i T i is skew-symmetric), we obtain from assertion 1 of Lemma 3 that T i ∈ L ⊕ L . Then by assertion 2 of Lemma 3, ( A − id)( A + 7 id) T i = 0, which implies [ S i , N X ] = T i .2. As by assertion 2 of Lemma 3, the operator A does not depend on the choice of the orthonormalbasis S i for L , it follows from (36) that we lose no generality by assuming w = e ∈ R . By assertion 1, ∇ X S = [ S , N X ], so from (38), B ( Y, X ) = [ S , N Y ] X − [ S , N X ] Y = ( S − id)( N Y X − N X Y ) = − π E − ( S ) ( N Y X − N X Y ), so π E ( S ) B ( Y, X ) = 0. This proves the first identity of (39).To prove the second one, take
X, Y, Z ∈ E ( S ). Note that by (8), S j X ∈ E − ( S ) for all j = 0(and similarly, for Y and Z ). Then projecting (37) to E ( S ) ∧ E ( S ) and using the first identityof (39), we obtain σ XY Z (( π E ( S ) A ( Y, X )) ∧ Z ) = 0. Assuming X, Y, Z linearly independent andacting by the both sides on a vector U ∈ E ( S ) ∩ (Span( X, Y, Z )) ⊥ we obtain h A ( Y, X ) , U i = 0,so π E ( S ) A ( Y, X ) ∈ Span(
X, Y ) (as Z ∈ E ( S ) \ Span(
X, Y ) can be chosen arbitrarily). Then fororthonormal vectors
X, Y ∈ E ( S ) , π E ( S ) A ( Y, X ) = h A ( Y, X ) , X i X + h A ( Y, X ) , Y i Y , so the co-efficient of X ∧ Z in σ XY Z (( π E ( S ) A ( Y, X )) ∧ Z ) = 0 (with orthonormal X, Y, Z ∈ E ( S )) gives h A ( Y, X ) , X i + h A ( Y, Z ) , Z i = 0 (using the fact that A ( Y, X ) is antisymmetric in
X, Y , by (38)). Then h A ( Y, X ) , X i = 0, hence π E ( S ) A ( Y, X ) = 0, for any orthonormal vectors
X, Y ∈ E ( S ).For the remaining two identities, take X, Y ∈ E ( S ) , Z ∈ E − ( S ) in (37) and project the resultingequation to E ( S ) ∧ E ( S ). As by (8), S i X, S i Y ∈ E − ( S ) , S i Z ∈ E ( S ), for all i ≥
1, we get(41) ε X i =1 S i Z ∧ π E ( S ) ( B i ( Y, X ))+ π E ( S ) (( A − εB )( Z, Y )) ∧ X + π E ( S ) (( A − εB )( X, Z )) ∧ Y = 0 . Taking the inner product of (41) with S j Z ∧ S k Z we find that the expression h ε k Z k B j ( Y, X ) −h S j Z, X i ( A − εB )( Z, Y ) −h S j Z, Y i ( A − εB )( X, Z ) , S k Z i is symmetric in j, k ≥
1, for all
X, Y ∈ E ( S ), Z ∈ E − ( S ). Fix j, k ≥ , j = k , and take Z ⊥ Span a = j,k ( S a X, S a Y ). Then S k π E ( S ) B j ( Y, X ) − S j π E ( S ) B k ( Y, X ) ∈ Span a = j,k ( S a X, S a Y ), so (acting by S j S k on the both sides) S j π E ( S ) B j ( Y, X ) + S k π E ( S ) B k ( Y, X ) ∈ Span a = j,k ( S a X, S a Y ). Let S jk ⊂ E ( S ) × E ( S ) be the set of pairs ( X, Y )such that X = 0 , Y / ∈ Span(
X, S j S k X ). Then S jk is open and dense in E ( S ) × E ( S ), andthe vectors S j X, S k X, S j Y, S k Y are linearly independent for ( X, Y ) ∈ S jk . As S j π E ( S ) B j ( Y, X ) + S k π E ( S ) B k ( Y, X ) is skew-symmetric in
X, Y and symmetric in k, j , there exist (rational) functions f jk , f kj : S jk → R such that S j π E ( S ) B j ( Y, X )+ S k π E ( S ) B k ( Y, X ) = f jk ( X, Y ) S j X + f kj ( X, Y ) S k X − f jk ( Y, X ) S j Y − f kj ( Y, X ) S k Y , for every ( X, Y ) ∈ S jk . Taking Y ′ ∈ E ( S ) \ Span(
X, S j S k X, Y, S j S k Y )(so that the vectors S j X, S k X, S j Y, S k Y, S j Y ′ , S k Y ′ are linearly independent) and replacing Y by aY + bY ′ = 0, so that ( X, aY + bY ′ ) ∈ S jk , we obtain from the linearity of the left-hand sidethat f jk and f kj do not depend on the first argument and are linear in the second one. It followsthat for some vectors v jk ∈ E ( S ) , ≤ j = k ≤ , S j π E ( S ) B j ( Y, X ) + S k π E ( S ) B k ( Y, X ) = h v jk , Y i S j X + h v kj , Y i S k X − h v jk , X i S j Y − h v kj , X i S k Y , for all X, Y ∈ E ( S ). Choose i, l such that i, j, k, l are all distinct and add to the above equation the same one with j, k replaces by i, l . Theleft-hand side of the resulting equation is symmetric in all four indices i, j, k, l , hence the right-handside also is. Choosing X, Y ∈ E ( S ) in such a way that the eight vectors S a X, S a Y, a = i, j, k, l , arelinearly independent (to do that, take X = 0 and Y / ∈ Span a = b, { a,b }⊂{ i,j,k,l } ( X, S a S b X )), we obtainthat v jk = v ′ j , for all j, k ≥ , j = k , with some v ′ j ∈ E ( S ). It follows that S j ( π E ( S ) B j ( Y, X ) −h v ′ j , Y i X + h v ′ j , X i Y ) + S k ( π E ( S ) B k ( Y, X ) − h v ′ k , Y i X + h v ′ k , X i Y ) = 0, for all X, Y ∈ E ( S ) and all j, k ≥ , j = k , which implies π E ( S ) B j ( Y, X ) = h v ′ j , Y i X − h v ′ j , X i Y . This proves the third identity of(39), if we define the operator L e by L e e j = − v ′ j and extend it by linearity to e ⊥ .Substituting the third identity of (39), with w = e , to (41) we obtain ( F ( Z ) X ) ∧ Y = ( F ( Z ) Y ) ∧ X ,where the linear operator F : E − ( S ) → End( E ( S )) is defined by F ( Z ) X = ε P i =1 h L e e i , X i S i Z + π E ( S ) (( A − εB )( X, Z )), for Z ∈ E − ( S ) , X ∈ E ( S ). It follows that F = 0, so π E ( S ) (( A − εB )( X, Z )) = − ε P i =1 h e i , L te X i S i Z = − εS L te X Z , which proves the fourth identity of (39).3. For a unit vector w ∈ R , extend the operator L w from w ⊥ to R by linearity putting L w w = 0,and then define L w : R → R , for all w = 0, by L w = L w/ k w k . The identities (39) then hold for all u, w ∈ R , w = 0, if we replace E ( S w ) by E k w k ( S w ) (= E ( S w/ k w k ). Combining the first and the thirdidentities of (39) we obtain that for all u, w ∈ R , w = 0, and all X, Y ∈ E k w k ( S w ), π E k w k ( S w ) B u ( Y, X ) = h L w u, X i Y − h L w u, Y i X. For every u ∈ R , define the quadratic map Q u : R → R by h Q u ( Y ) , X i = h B u ( Y, X ) , Y i . Takingthe inner product of the above equation with Y ∈ E k w k ( S w ) and then integrating by Y over the unitsphere S ( w ) ⊂ E k w k ( S w ) we obtain h R S ( w ) Q u ( Y ) dY, X i = ω h L w u, X i , for all X ∈ E k w k ( S w ), where ω is the volume of S ( w ). Relative to some orthonormal basis { E i } for R , the i -th component of R S ( w ) Q u ( Y ) dY ∈ R is ω Tr( π E k w k ( S w ) Q u,i ), where Q u,i ∈ Sym( R ) is the operator associated tothe quadratic form Y → h Q u ( Y ) , E i i . As π E k w k ( S w ) = (id + k w k − S w ), we obtain R S ( w ) Q u ( Y ) dY = ω ( Cu + k w k − T ( u, w )), where a linear operator C : R → R and a bilinear map T : R × R → R are defined by h Cu, E i i = Tr Q u,i and h T ( u, w ) , E i i = Tr( Q u,i S w ), for 1 ≤ i ≤
16. It follows that h Cu + k w k − T ( u, w ) , X i = h L w u, X i , for all X ∈ E k w k ( S w ), which gives π E k w k ( S w ) ( B u ( Y, X ) − h Cu + k w k − T ( u, w ) , X i Y + h Cu + k w k − T ( u, w ) , Y i X ) = 0 , for all X, Y ∈ E k w k ( S w ). This equation is satisfied, if we substitute B ′ u ( Y, X ) = h Cu, X i Y − h Cu, Y i X + P i =0 ( h T ( u, e i ) , X i S i Y − h T ( u, e i ) , Y i S i X ) for B u ( Y, X ) (this follows from the fact that the sum on theright-hand side of B ′ u ( Y, X ) does not depend on the choice of an orthonormal basis { e i } for R , so wecan take e = k w k − w ; then S i X, S i Y ∈ E −k w k ( S w ) for i = 0, by (8)). Therefore, for every u ∈ R , thebilinear skew-symmetric map B ′′ u : R × R → R defined by B ′′ u = B u − B ′ u satisfies the hypothesisof assertion 5 of Lemma 3. It follows that for some q : R → R , B ′′ u ( Y, X ) = (
A − id)( X ∧ Y ) q ( u ).As the left-hand side is linear in u , the map q is a linear, q ( u ) = C ′ u , so for all X, Y ∈ R , u ∈ R ,(42) B u ( Y, X ) = X i =0 S i ( X ∧ Y )( T ( u, e i ) + S i C ′ u ) + ( X ∧ Y )( C − C ′ ) u. SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 21
From the first identity of (39) it now follows that h T ( u, u )+ k u k Cu, X i = 0, for all u ∈ R , X ∈ E k u k ( S u ).As π E k u k ( S u ) = (id + k u k − S u ), this implies k u k ( T ( u, u ) + S u Cu ) + ( S u T ( u, u ) + k u k Cu ) = 0. Since k u k is not a rational function, we obtain T ( u, u ) + S u Cu = 0, for all u ∈ R , so the bilinear map T ′ : R × R → R defined by T ′ ( u, w ) = T ( u, w ) + S w Cu is skew-symmetric.By the second identity of (39), the map A ( X, Y ) satisfies the hypothesis of assertion 5 of Lemma 3,so there exists q ∈ R such that for all X, Y ∈ R ,(43) A ( Y, X ) = (
A − id)( X ∧ Y ) q = X i =0 ( h S i X, q i S i Y − h S i Y, q i S i X ) − ( h X, q i Y − h Y, q i X ) . Substituting (42, 43) into the fourth equation of (39), with w = e , we obtain P i =1 h S i q − ε ( T ( e , e i ) + S i C ′ e ) , Y i S i Z + h q + ε ( T ( e , e ) + ( C − C ′ ) e ) , Z i Y = εS L t Y Z , for all Y ∈ E ( S ) , Z ∈ E − ( S ).Taking the inner product of the both sides with S k Z, k >
0, we get h S k q − ε ( T ( e , e k )+ S k C ′ e ) , Y ik Z k + h q + ε ( T ( e , e ) + ( C − C ′ ) e ) , Z ih Y, S k Z i = ε h Y, L e k ik Z k . It follows that the second term on theright-hand side viewed as a polynomial of Z ∈ E − ( S ) (with Y ∈ E ( S ) , k > k Z k . As this term is a product of two linear forms in Z , we get h q + ε ( T ( e , e ) + ( C − C ′ ) e ) , Z i = 0.From the fact that T ( u, w ) = T ′ ( u, w ) − S w Cu , where T ′ is skew-symmetric, h q + ε ( C − C ′ ) e , Z i = 0, forall Z ∈ E − ( S ). As e ∈ R is an arbitrary unit vector, it follows that k u k q + ε ( C − C ′ ) u ∈ E k u k ( S u ), forall u ∈ R , so S u ( k u k q + ε ( C − C ′ ) u ) = k u k ( k u k q + ε ( C − C ′ ) u ), which implies k u k ( S u q − ε ( C − C ′ ) u ) = k u k q − εS u ( C − C ′ ) u . Since k u k is not a rational function, we obtain ( C − C ′ ) u = εS u q , for all u ∈ R , so T ( u, w ) + S w C ′ u = T ′ ( u, w ) − S w Cu + S w C ′ u = T ′ ( u, w ) − εS w S u q = T ′′ ( u, w ) − ε h w, u i q ,where T ′′ ( u, w ) = T ′ ( u, w ) − ε ( S w S u − h w, u i id) q is skew-symmetric, as T ′ is skew-symmetric and by(8). Substituting this to (42) we obtain B u ( Y, X ) = P i =0 ( h T ′′ ( u, e i ) − ε h e i , u i q, X i S i Y − h T ′′ ( u, e i ) − ε h e i , u i q, Y i S i X )+ ε h S u q, X i Y − ε h S u q, Y i X . Substituting this expression and (43) into (37) we get aftersimplification: σ XY Z (2 P i =0 h X, q i S i Y ∧ S i Z − h X, q i Y ∧ Z ) = 0, so ( A − id)( σ XY Z ( h X, q i Y ∧ Z )) = 0.From assertion 2 of Lemma 3 it follows that σ XY Z ( h X, q i Y ∧ Z ) = 0, for all X, Y, Z ∈ R , which easilyimplies q = 0. This proves the assertion (with T ′′ denoted by T ).4. As it follows from (38) and assertion 3, ρ is a Codazzi tensor. By [DS, Theorem 1], for any twoeigenspaces E α , E β of ρ , the exterior product E α ∧ E β is an invariant subspace of the operator R on thespace of bivectors. Suppose ρ is not a multiple of the identity. As in the proof of Lemma 6, by linearity,it suffices to show that the following assumption leads to a contradiction: there exist two orthogonalnontrivial invariant subspaces E , E of ρ such that every E α ∧ E β is an invariant subspace of R . By(14), this is equivalent to the fact that every E α ∧ E β is an invariant subspace of the operator A , whichis then equivalent to the fact that(44) A ( X ∧ Y ) Z = X i =0 ( S i X ∧ S i Y ) Z = 0 , for all X, Y ∈ E α , Z ∈ E β , α = β. Denote dim E α = d α > , α = 1 ,
2. By assertion 2 of Lemma 3, the space Skew( R ) is an invariantsubspace of A , and the restriction of A to it is a symmetric operator, with eigenvalues 5 and −
3, whosecorresponding eigenspaces are L and L . As every E α ∧ E β is an invariant subspace of A , we obtainthat E α ∧ E β = ⊕ k =2 , V αβk , where V αβk = π L k ( E α ∧ E β ). The six subspaces V αβk , ≤ α ≤ β ≤ k = 2 ,
3, are mutually orthogonal. Moreover, as Skew( R ) = ⊕ k =2 , L k = ⊕ ≤ α ≤ β ≤ E α ∧ E β , weget L k = ⊕ ≤ α ≤ β ≤ V αβk (with all the direct sums above being orthogonal) and V αβk = π E α ∧ E β L k =( E α ∧ E β ) ∩ L k , for all 1 ≤ α ≤ β ≤ , k = 2 , K ∈ V αα has the form P i,j =0 a ij S i S j , where a ji = − a ij . Moreover, as V αα ⊥ ( E β ∧ R ) , β = α , the kernel of every such K contains E β , β = α . Choosing an orthonormal basis for R ,relative to which the skew-symmetric matrix a ij has a canonical form, we get K = P i =1 b i S i − S i , with b = 0 (unless all the b i ’s are zeros). Then X ∈ Ker K if and only if X = ( P i =2 c i S S S i − S i ) X , where c i = b i b − , i = 2 , ,
4. Consider symmetric orthogonal operators D i = S S S i − S i ∈ L , i = 2 , , D i D j ∈ L , i = j , and D D D ∈ L (in fact, D D D = ± S ). Then by assertion 1 ofLemma 3, Tr D i = Tr D i D j = Tr D D D = 0 , ≤ i < j ≤
4. It follows that each of the symmetricorthogonal operators D i , D i D j , D D D , ≤ i < j ≤
4, has eigenvalues ±
1, both of multiplicity 8.Furthermore, as the D i ’s pairwise commute (which again follows from (8)), we can choose an orthonormalbasis for R relative to which the matrices of the D i ’s are diagonal. The D i ’s satisfy the above condition on the multiplicities of eigenvalues if and only if the space R splits into the orthogonalsum of two-dimensional subspaces W ( ε , ε , ε ) , ε i = ±
1, such that D i | W ( ε ,ε ,ε ) = ε i id W ( ε ,ε ,ε ) .From the above, Ker K is the +1-eigenspace of the operator c D + c D + c D . Its eigenvalues are λ = c ε + c ε + c ε , with the corresponding eigenspaces W λ = ⊕ W ( ε , ε , ε ), where the sum is takenover the set s λ = { ( ε , ε , ε ) : λ = c ε + c ε + c ε } , with dim V λ = 2 s λ . Considering the equations c ε + c ε + c ε = 1 , ε i = ±
1, we see that dim Ker K can be equal to 0 , ,
4, or 8, and in the lattercase (up to relabeling), c = ± , c = c = 0, so K is a nonzero multiple of S S + c S S and Ker K is the c -eigenspace of the symmetric operator S S S S . As Ker K ⊃ E β , dim Ker K ≥ d β . It followsfrom d + d = 16 that either one of the spaces V αα is trivial, or d = d = 8 and Ker K = E β , for allnonzero K ∈ V αα (both for ( α, β ) = (1 ,
2) and ( α, β ) = (2 , V = 0. Then L = V ⊕ V ⊂ E ∧ R which implies that h KX, Y i = 0, for all K ∈ L and all X, Y ∈ E , that is, h S i X, S j Y i = 0,for all i, j = 0 , . . . ,
8. As for a nonzero X , dim Span i =0 ( S i X ) = 9, it follows that d = 1. Then for anonzero Z ∈ E we can choose X ∈ E such that Z ⊥ S i X . Substituting such X, Z and an arbitrary Y ∈ E into (44) we find that Z ⊥ S i Y , for all Y ∈ E . This implies that E is an invariant subspaceof all the operators S i , hence E also is. Then Z is an eigenvector of every S i , which contradicts thefact that the operator S i S j , i = j , is orthogonal and skew-symmetric.Suppose now that d = d = 8 and Ker K = E β , for all nonzero K ∈ V αα (for ( α, β ) = (1 ,
2) and( α, β ) = (2 , K ∈ V . As it is shown above, under an appropriate choice ofan orthonormal basis for R , K = c ( S S + εS S ) (for some ε = ± , c = 0) and E = Ker K isthe ε -eigenspace of S S S S . Then E is the ( − ε )-eigenspace of S S S S . As it follows from (8), S S S S S i = ˆ ε i S i S S S S , where ˆ ε i = − i = 1 , . . . , ε i = 1 for i = 0 , , . . . ,
8. It followsthat for any nonzero X ∈ E , S i X ∈ E , when 1 ≤ i ≤ S i X ∈ E , otherwise. Moreover,as for any nonzero X ∈ E , S i S j X = ± S k S l X , where { i, j, k, l } = { , , , } , the dimension of thespace Span i,j =1 ( S i S j X ) , X ∈ E , is at most four, so there exists a nonzero Y ∈ E orthogonal tothis subspace. Then Span i =1 ( S i X ) ⊥ Span i =1 ( S i Y ). Substituting such X and Y into (44) and taking Z = S X ( ∈ E ) we obtain k X k S Y = 0, which is a contradiction.It follows that ρ is a multiple of the identity, at every point y ∈ U . As ( ∇ X ρ ) Y = ( ∇ Y ρ ) X , ρ is infact a constant multiple of the identity.5. By assertion 1, there exists a linear map N : R → Skew( R ) such that ∇ X S w = [ S w , N X ].Then from (38) and assertion 3, [ S w , N Y ] X − [ S w , N X ] Y = P i =0 ( h T ( w, e i ) , X i S i Y − h T ( w, e i ) , Y i S i X ).As from (8) [ S w , P i,j =0 h T ( e i , e j ) , X i S i S j ] = 4 P i =0 h T ( w, e i ) , X i S i , for a linear map N ′ : R → Skew( R ) defined by N ′ X = N X + P i,j =0 h T ( e i , e j ) , X i S i S j , we obtain that [ S w , N ′ Y ] X = [ S w , N ′ X ] Y ,for all X, Y ∈ R . Taking a unit vector w ∈ R and X ∈ E ε ( S w ) , Y ∈ E − ε ( S w ) , ε = ±
1, we get[ S w , N ′ Y ] X = ( S w − ε id) N ′ Y X = − επ E − ε ( S w ) N ′ Y X , and similarly, [ S w , N ′ X ] Y = 2 επ E ε ( S w ) N ′ X Y , so π E − ε ( S w ) N ′ Y X = π E ε ( S w ) N ′ X Y , hence h N ′ Y X, Z i = 0, for all X ∈ E ε ( S w ) , Y, Z ∈ E − ε ( S w ). As N ′ Y depends linearly on Y and is skew-symmetric, we obtain that h N ′ Y X, Z i = 0, for all X ∈ E ( S w ) , Z ∈ E − ( S w ) and all Y ∈ R . It follows that E ( S w ) and E − ( S w ) are invariant subspaces of N ′ Y , so N ′ Y commutes with S w , for any w ∈ R and any Y ∈ R (which, in fact, implies N ′ = 0). Then ∇ X S w = [ S w , N X ] = [ S w , N ′ X − P i,j =0 h T ( e i , e j ) , X i S i S j ] = − P i =0 h T ( w, e i ) , X i S i , as required. (cid:3) It follows from assertions 4 and 5 of Lemma 7 and (36) that after a conformal change of metric on U , ( ∇ Z R )( X, Y ) = ε P i,j =0 ( −h T ( e i , e j ) , Z i S j X ∧ S i Y − h T ( e i , e j ) , Z i S i X ∧ S j Y ) = 0, as T is skew-symmetric. Hence U is a locally symmetric space. Moreover, as ρ is a constant multiple of the identityby assertion 4 of Lemma 7, the curvature tensor (14) is Osserman, so U is locally isometric to a rank-onesymmetric space by [GSV, Lemma 2.3] (in fact, to the Cayley projective plane or its noncompact dual,as these are the only two rank-one symmetric spaces of dimension 16 the Jacobi operator of whosecurvature tensor has an eigenvalue of multiplicity exactly 8).Thus, for every x ∈ M ′ satisfying assertion (b) of Lemma 4, the metric on the neighborhood U = U ( x )is conformally equivalent to the metric of a rank-one symmetric space. SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 23
Proof of Theorem 1.
Lemma 4 and the results of Sections 3.2 and 3.3 imply the conformal partof Theorem 1 at the generic points. Namely, every x ∈ M ′ (the latter is an open, dense subset of M )has a neighborhood U which is conformally equivalent to a domain either of a Euclidean space, or of arank-one symmetric space, that is, of one of the model spaces(45) R , C P , C H , H P , H H , O P , O H , where we normalize the standard metric ˜ g on each of the non-flat spaces above in such a way that thesectional curvature K σ satisfies | K σ | ∈ [1 , x ∈ M ,and secondly, that the model space to a domain of which U is conformally equivalent is the same, forall x ∈ M . Our proof very closely follows the arguments of [N5] from after Remark 4 to the end ofSection 3. We start with the following technical lemma: Lemma 8.
Let ( N , h· , ·i ) be a smooth Riemannian space locally conformally equivalent to one of the O P , O H , so that ˜ g = f h· , ·i , for a positive smooth function f = e φ : N → R . Then the curvaturetensor R and the Weyl tensor W of ( N , h· , ·i ) satisfy (with ε = 1 for O P and ε = − for O H ): R ( X, Y ) = ( X ∧ KY + KX ∧ Y ) + εf (3 X ∧ Y + P ( X, Y )) , where (46a) P ( X, Y ) = A ( X ∧ Y ) = X i =0 S i X ∧ S i Y, K = H ( φ ) − ∇ φ ⊗ ∇ φ + k∇ φ k id ,W ( X, Y ) = W O ,ε ( X, Y ) = εf ( X ∧ Y + P ( X, Y )) , (46b) k W k = f , (46c) ( ∇ Z W )( X, Y ) = εZf ( X ∧ Y + P ( X, Y ))(46d) + ε ([ P ( X, Y ) , ∇ f ∧ Z ] + P (( ∇ f ∧ Z ) X, Y ) + P ( X, ( ∇ f ∧ Z ) Y )) , where H ( φ ) is the symmetric operator associated to the Hessian of φ , and both ∇ and the norm arecomputed relative to h· , ·i .Proof. By (7), the curvature tensor of O P ( ε = 1) and O H ( ε = −
1) has the form ˜ R ( X, Y ) = ε (3 X ˜ ∧ Y + P i =0 S i X ˜ ∧ S i Y ), where ( X ˜ ∧ Y ) Z = ˜ g ( X, Z ) Y − ˜ g ( Y, Z ) X . Under the conformal change ofmetric ˜ g = f h· , ·i = e φ h· , ·i , the curvature tensor transforms as ˜ R ( X, Y ) = R ( X, Y ) − ( X ∧ KY + KX ∧ Y ).As ˜ g ( X, Y ) = f h X, Y i , X ˜ ∧ Y = f X ∧ Y and the S i ’s still satisfy (8) and are symmetric for h· , ·i , equation(46a) follows.Equation (46b) follows from the definition of the Weyl tensor (12); the norm of W in (46c) can becomputed directly using (8) and the fact that the S i ’s are symmetric, orthogonal and Tr S i = 0.Assertion 5 of Lemma 7 is satisfied for ˜ g , so ˜ ∇ Z S i = P j =0 ω ji ( Z ) S i , where ˜ ∇ is the Levi-Civitaconnection for ˜ g and ω ji ( Z ) = −h T ( e i , e j ) , Z i , ω ji + ω ij = 0. As ˜ ∇ Z X = ∇ Z X + Zφ X + Xφ Z − h
X, Z i φ ,we get ∇ Z S i = P j =0 ω ji ( Z ) S j + [ S i , ∇ φ ∧ Z ], so ( ∇ Z P )( X, Y ) = [ P ( X, Y ) , ∇ φ ∧ Z ]+ P (( ∇ φ ∧ Z ) X, Y )+ P ( X, ( ∇ φ ∧ Z ) Y ), which, together with (46b), proves (46d). (cid:3) From Lemma 4, the results of Sections 3.2 and 3.3 and [N5, Theorem 3], for every point x ∈ M ′ , thereexists a neighborhood U = U ( x ) and a positive smooth function f : U → R such that the Riemannianspace ( U , f h· , ·i ) is isometric to an open subset of one of the model spaces of (45), so at every point x ∈ M ′ , the Weyl tensor W of M either vanishes, or has the form W ν,ε given in [N5, Lemma 8,(36b)], with n = 16 and the corresponding ν , or has the form W O ,ε given in (46b). Here W ν,ε is theWeyl tensor of the corresponding model space M ν,ε = C P , C H , H P , H H , where ε = ± ν = 1( ν = 3) for complex (quaternionic) spaces, respectively. The Jacobioperators associated to the different Weyl tensors W ν,ε of [N5, Eq. (36b)] and W O ,ε of (46b) differ bythe multiplicities and the signs of the eigenvalues, so every point x ∈ M ′ has a neighborhood conformallyequivalent to a domain of exactly one of the model spaces. Moreover, the function f > W = 0, as k W k = C νn f by [N5, Eq. (36c)] and k W k = f , by (46c).By continuity, the Weyl tensor W of M n either has the form W ν,ε , or the form W O ,ε , or vanishes, atevery point x ∈ M n (as M ′ is open and dense in M n , see Lemma 4). Moreover, every point x ∈ M n , at which the Weyl tensor has the form W ν,ε or W O ,ε , has a neighborhood, at which the Weyl tensorhas the same form. Hence M n = M ⊔ F α M α , where M = { x : W ( x ) = 0 } is closed, and every M α is a nonempty open connected subset, with ∂M α ⊂ M , such that the Weyl tensor has the same form W ν,ε or W O ,ε at every point x ∈ M α . In particular, M α ⊂ M ′ , for every α , so that each M α is locallyconformally equivalent to one of the nonflat model spaces (45).To prove the conformal part of Theorem 1, we need to show that either M = M or M = ∅ .Suppose that M = ∅ and that there exists at least one component M α . If M α is locally conformallyequivalent to one of the model spaces M ν,ε , we get a contradiction following the arguments of [N5](from after Lemma 8 to the end of Section 3). Suppose M α is locally conformally equivalent to one of O P , O H . Let y ∈ ∂M α ⊂ M and let B δ ( y ) be a small geodesic ball of M centered at y which isstrictly geodesically convex (any two points from B ( y ) can be connected by a unique geodesic segmentlying in B δ ( y ) and that segment realizes the distance between them). Let x ∈ B δ/ ( y ) ∩ M α and let r = dist( x, M ). Then the geodesic ball B = B r ( x ) lies in M α and is strictly convex. Moreover, ∂B contains a point x ∈ M . Replacing x by the midpoint of the segment [ xx ] and r by r/
2, if necessary,we can assume that all the points of ∂B , except for x , lie in M α .The function f is positive and smooth on B \ { x } (that is, on an open subset containing B \ { x } ,but not containing x ). Lemma 9.
When x → x , x ∈ B , both f and ∇ f have a finite limit. Moreover, lim x → x ,x ∈ B f ( x ) = 0 .Proof. The fact that lim x → x ,x ∈ B f ( x ) = 0 follows from (46c) and the fact that W | x = 0 (as x ∈ M ).As the Riemannian space ( B, f h· , ·i ) is locally isometric to a rank-one symmetric space M O ε (where ε = ± M O + = O P , M O − = O H ) and is simply connected, there exists a smooth isometricimmersion ι : ( B, f h· , ·i ) → M O ε . Since f is smooth on B \ { x } and lim x → x ,x ∈ B f ( x ) = 0, the rangeof ι is a bounded domain in M O ε . Moreover, as lim x → x ,x ∈ B f ( x ) = 0, every sequence of points in B converging to x in the metric h· , ·i is a Cauchy sequence for the metric f h· , ·i ). It follows thatthere exists a limit lim x → x ,x ∈ B ι ( x ) ∈ M O ε . Defining for every x ∈ B the point S | x = Span i =0 ( S i )in the Grassmanian G (9 , Sym( T x M )), we obtain that there exists a limit lim x → x ,x ∈ B S | x =: S | x ∈ G (9 , Sym( T x M )). In particular, if Z is a continuous vector field on B , then there exists a unitcontinuous vector field Y on B such that Y ⊥ Span i =0 ( S i Z ) on B . For such two vector fields, thefunction θ ( Y, Z ) = h P j =1 ( ∇ E j W )( E j , Y ) Y, Z i (where E j is an orthonormal frame on B ) is well-definedand continuous on B . Using (8, 46d) and assertion 4 of Lemma 3, we obtain by a direct computation thatat the points of B , θ ( Y, Z ) = εZf . As θ ( Y, Z ) is continuous on B , there exists a limit lim x → x ,x ∈ B Zf .Since Z is an arbitrary continuous vector field on B , ∇ f has a finite limit when x → x , x ∈ B . (cid:3) As lim x → x ,x ∈ B f ( x ) = 0 and the S i ’s are orthogonal, the second term on the right-hand side of (46a)tends to 0 when x → x in B . Then the tensor field defined by ( X, Y ) → ( X ∧ KY + KX ∧ Y ) hasa finite limit (namely R | x ) when x → x in B . It follows that the symmetric operator K has a finitelimit at x . Computing the trace of K and using the fact that φ = ln f we get △ u = F u, where u = f / , F = 7 Tr K on B . Both functions F and u are smooth on B \ { x } and have a finite limit at x . Moreover,lim x → x ,x ∈ B u ( x ) = 0 by Lemma 9 and u ( x ) > x ∈ B \ { x } . The domain B is a small geodesicball, so it satisfies the inner sphere condition (the radii of curvature of the sphere ∂B are uniformlybounded). By the boundary point theorem [Fra, Section 2.3], the inner directional derivative of u at x (which exists by Lemma 9, if we define u ( x ) = 0 by continuity) is positive. But ∇ u = f / ∇ f in B ,so lim x → x ,x ∈ B ∇ u = 0 by Lemma 9, a contradiction.This proves the conformal part of Theorem 1.The “genuine” Osserman part, the Osserman Conjecture (assuming Conjecture A), now easily follows.Indeed, any Osserman manifold M n , n >
4, is Einstein, hence by (12) its Weyl tensor is Osserman,hence M n is locally conformally equivalent to a rank-one symmetric space or to a flat space, as shownabove. Then by [Nic, Theorem 4.4], ∇ W = 0, so, as M n is Einstein, it is locally symmetric, and theproof follows from [GSV, Lemma 2.3]. SSERMAN MANIFOLDS AND WEYL-SCHOUTEN THEOREM FOR RANK-ONE SYMMETRIC SPACES 25
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