A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS.

Transcription

1 A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS. BENJAMIN SCHMIDT, KRISHNAN SHANKAR, AND RALF SPATZIER Abstract. We prove a Schur type theorem for Kählerian manifolds with nearly constant Jacobi operators. We include examples to illustrate that the theorem fails in the Riemannian setting. 1. Introduction Let M be a Riemannian manifold with Levi-Civita connection and curvature tensor R(X, Y )Z = [ X, Y ]Z [X,Y ] Z. Given p M and v S p M, the Jacobi operator is the self-adjoint linear map J v : v v defined by J v (w) = R(w, v)v. The eigenvalues of J v determine the sectional curvatures of planes containing v. Given an integer 0 k < dim(m)/2, we say that a point p M has coisotropy rank at most k if there exists κ(p) R such that rank(j v κ(p) Id) k for each v S p M; when all p M satisfy this property, we say that M has coisotropy rank at most k. Classically, p M is defined to be isotropic when the sectional curvatures of all plane sections σ T p M are equal, or equivalently, when p has coisotropy rank 0. By Schur s theorem [10], manifolds of dimension at least three and coisotropy rank 0 have constant sectional curvatures. Our main theorem is an analogue of Schur s theorem for Kählerian manifolds of coisotropy rank at most one. Theorem A. Let M be a complete, simply-connected Kählerian manifold of real dimension 2n 2. If M has coisotropy rank at most one, then κ : M R is constant. Moreover, (1) if κ > 0, then M is isometric to a symmetric metric on CP n having constant holomorphic curvatures 4κ. (2) if κ < 0, then M is isometric to a symmetric metric on CH n having constant holomorphic curvatures 4κ. (3) if κ = 0, then the open set of non-isotropic points O admits an (n 1)- dimensional parallel distribution that is tangent to a foliation by complete and totally geodesic leaves isometric to C n 1. We remark that in (3), each point p O admits a neighborhood that is locally isometric to a Kählerian product C n 1 Σ. In the case when the Kählerian metric is real analytic, this local product structure extends to a global product structure M = C n 1 Σ by [1, Theorem 8]). We do not know if this remains true for smooth Kählerian metrics (compare with [11, Section 4] and [8, Corollary 2]). Date: September 25, The first named author is partially supported by the NSF grant DMS The second named author is partially supported by the NSF grant?. The third named author is partially supposrted by the NSF grant?. (ADD MORE OF THESE?). 1

2 2 B. SCHMIDT, K. SHANKAR, AND R. SPATZIER 2. Eigenspace distributions on coisotropy rank one unit tangent spheres. In this section we assume that p M is a point of coisotropy rank at most one in a Riemannian manifold of dimension d. For each v S p M, let E v = ker(j v κ(p) Id). Then E v is a subspace of v of codimension at most one. Lemma 2.1. Assume that {v, w} S p M are orthonormal. Then sec(v, w) = κ(p) if and only if v E w. In particular, v E w if and only if w E v. Proof. If v E w, then sec(v, w) = R(v, w, w, v) = g p (J w (v), v) = κ(p). Next assume that sec(v, w) = κ(p). If E w = v then we trivially have v E w. Otherwise there exists an orthonormal J w -eigenbasis e 1,..., e d 1 of v with corresponding eigenvalues λ i = κ(p) for each i = 1,..., d 2 and λ d 1 κ(p). Express v = d 1 i=1 α ie i. A simple calculation yields κ(p) = sec(v, w) = g p (J w (v), v) = κ(p) + α 2 d 1(λ d 1 κ(p)) whence α d 1 = 0, concluding the proof. Convention 2.2. For each v S p M, parallel translation in T p M defines an isomorphism between the subspace v of T p M and the subspace T v (S p M) of T v (T p M). We will use this isomorphism without mention when contextually unambiguous. Definition 2.3. The eigenspace distribution on S p M is the assignment of tangent subspaces S p M v E v T v (S p M). For each v S p M, the tangent subspace E v has dimension d 2 or d 1. Let X p = {v S p M dim(e v ) = d 1}. There are three possibilities: (1) S p M = X p, (2) X p =, or (3) X p and X p S p M. In case (1), p is an isotropic point. In case (2) will will say that the tangent distribution is nonsingular and of codimension one. In case (3) will will say the distribution is singular and of codimension one with singular set X p. We will say the distribution is possibly singular when not wanting to specify whether (1), (2), or (3) holds and possibly singular of codimension one when wanting to specify that (1) does not hold. A possibly singular tangent distribution D on a complete Riemannian manifold S is said to be totally-geodesic if complete geodesics of S that are somewhere tangent to D are everywhere tangent to D. Convention 2.4. Henceforth, unit tangent spheres S p M are equipped with the standard Riemannian metric, denoted by,, induced from the Euclidean metric g p (, ) on T p M. Lemma 2.5. For each p M, the eigenspace distribution E is a (possibly singular) totally-geodesic distribution on S p M. Proof. Let v S p M and w E v. The geodesic c(t) = cos(t)v + sin(t)w in S p M satisfies c(0) = v and ċ(0) = w. Calculate J c(t) (ċ(t)) = sin(t)j w (v)+cos(t)j v (w). By assumption J v (w) = κ(p)w and by Lemma 2.1, J w (v) = κ(p)v. Therefore J c(t) (ċ(t)) = κ(p)( sin(t)v + cos(t)w) = κ(p)ċ(t). Hence ċ(t) E c(t), concluding the proof.

3 A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS (Possibly singular) codimension one totally-geodesic distributions on unit spheres. In this section, we classify (possibly singular) codimension one totally-geodesic distributions on unit spheres S d E d+1. They are in correspondence with nonzero skew-symmetric linear maps A : E d+1 E d+1. This fact is the first source of rigidity in Theorem A. Given a non-zero skew-symmetric linear map A : E d+1 E d+1 and v S d, parallel translation in E d+1 identifies v and T v S d. As A is skew-symmetric and non-zero, the assignment S d v Av T v S d defines a non-zero Killing field on S d. Let D v = span{v, Av} denote the subspace of T v S d orthogonal to Av. Then S d v D v defines a codimension one totally-geodesic distribution on S d with singular set X = {x S d D x = T x S d 1 } = ker(a) S d as a consequence of the following well known lemma. Lemma 3.1. Let X be a Killing field on a complete Riemannian manifold (S, g). If a geodesic c(t) satisfies g(ċ(0), X(c(0)) = 0, then g(ċ(t), X(c(t)) = 0 for all t R. Replacing the skew-symmetric linear map A with a nonzero real multiple ra in the above construction yields the same (possibly singular) codimension one totallygeodesic distribution D on S n. An elegant application of the fundamental theorem of projective geometry shows that all nonsingular totally-geodesic codimension one distributions on unit spheres arise in this fashion [4]. While no a priori continuity assumptionis made, a posteriori the distribution is algebraic. Theorem 3.2 (Hangan and Lutz). Let D be a nonsingular codimension one totallygeodesic distribution on a unit sphere S d E d+1. Then d is odd and there exists a nonsingular projective class [A] of skew-symmetric linear maps such that for each x S d, D x = span{x, Ax}. In Theorem 3.5 below, we extend Theorem 3.2 to allow nonempty singular sets X = {x S d D x = T x S d }. We adopt the following notation. Given a subset U S d, we denote by S(U) = span{u} S d, the smallest totally geodesic subsphere of S d containing U. We begin with two simple lemmas. Lemma 3.3. The singular set X satisfies S(X ) = X. Proof. There is nothing to prove if X is empty. If x X, note that x X since each great circle through x also passes through x. It remains to prove that for linearly independent x 1, x 2 X, the great circle C 1 := S({x 1, x 2 }) X. If x 3 C 1 \ {±x 1, ±x 2 }, then the line L 1 := T x3 C 1 is a subspace of D x3 since x 1 X. It remains to prove that if L 2 is any other line in T x3 S d, then L 2 is also a subspace of D x3. Let C 2 denote the great circle containing x 3 with tangent line L 2. The two great circles C 1 and C 2 both lie in a unique round 2-subsphere S. Let p C 2 \ {±x 3 }. As x 1, x 2 X are linearly independent, the tangent lines at p to the great circles in S joining x 1 to p and x 2 to p are transverse subspaces of D p T p S. Therefore T p S D p. In particular, the tangent line to C 2 at p is a subspace of D p whence the line L 2 is a subspace of D x3 as required. Lemma 3.4. Let n = dim(x ) if X = and n = 1 if X =. Then d n is even. Proof. We induct on the dimension d; the base case d = 1 is trivially verified. Now assume that d > 1. If n = 1, then d is odd by Theorem 3.2, concluding the proof in

4 4 B. SCHMIDT, K. SHANKAR, AND R. SPATZIER this case. Assume now that n 0 and let p X. Let H S d denote the collection of points at spherical distance π 2 from p, a totally-geodesic subsphere of dimension d 1. For each h H, let ˆD h = D h T h H and note that the tangent distribution h ˆD h is a codimension one (possibly singular) totally-geodesic distribution on H with singular set X H, a round subsphere of H of dimension n 1. By induction (d 1) (n 1) = d n is even, concluding the proof. Theorem 3.5. Let D be a codimension one (possibly singular) totally-geodesic distribution on a unit sphere S d E d+1. There exists a projective class [A] of skew-symmetric linear maps such that for each x S d, D x = span{x, Ax}. Proof. Let E 1 = span X, E 2 = E 1, and Y = E 2 S d. Then E d+1 = E 1 E 2 and Y S d is a totally-geodesic subsphere of odd dimension d n 1 by Lemma 3.4. Consider the restriction of D to Y defined by ˆD v = D v T v Y for each v Y. We claim that ˆD is a nonsingular codimension one totally-geodesic distribution on Y. Indeed, as both D and Y are totally-geodesic, ˆD is totally-geodesic. Moreover, if v Y and w E 1 is a unit vector, then since w X and v, w = 0, we have that v D w whence w D v by the totally-geodesic property. Conclude that E 1 D v. As T v S d = v = E 1 T v Y and D v has codimension one in T v S d, ˆDv has codimension one in T v Y, concluding the proof of the claim. Theorem 3.2 gives a non-degenerate skew-symmetric linear map  : E 2 E 2 with the property that ˆD v = span{v, Âv} E 2 for each v Y. Extend  to a non-zero skew-symmetric linear map A : E d+1 E d+1 by Av = Âv 2 for each v = v 1 + v 2 E 1 E 2 = E d+1. We claim that D v = span{v, Av} for each v S d. As ker A = X, the claim is immediate for v X. The argument in the previous paragraph establishes the claim for v Y. We therefore assume that v = v 1 + v 2 is a unit vector with 0 v i E i for i = 1, 2. As v 2 0, dim(d v ) = d 1. As span{v, Av} = (v 1 E 1 ) (span{v 2, Âv 2} E 2 ) R( v 2, v 2 v 1 v 1, v 1 v 2 ) also has dimension d 1, it suffices to prove that each of the three subspaces in the above orthogonal sum are subspaces of D v. Let w 1 (v 1 E 1 ) be a unit vector. Then w 1 X and w 1, v = 0 imply v D w1, whence w 1 D v by the totally-geodesic property. Therefore (v 1 E 1 ) D v. Let w 2 (span{v 2, Âv 2} E 2 ). As w 2 Y, D w2 = span{w 2, Âw 2}. As v, w 2 = 0 and v, Âw 2 = v 1, Âw 2 + v 2, Âw 2 = v 2, Âw 2 = Âv 2, w 2 = 0, we have that v D w2, whence w 2 D v by the totally-geodesic property. Therefore (span{v 2, Âv 2} E 2 ) D v. Finally, the geodesic c(t) = cos(t)(v 1 + v 2 ) + sin(t)( v 2, v 2 v 1 v 1, v 1 v 2 ) in S d satisfies c(0) = v and ċ(0) = v 2, v 2 v 1 v 1, v 1 v 2. Let T = cot 1 ( v 1, v 1 ) and note that c(t ) X, whence ċ(t ) D c(t ). By the totally-geodesic property v 2, v 2 v 1 v 1, v 1 v 2 = ċ(0) D c(0) = D v, concluding the proof. 4. Relating the complex structure and the eigenspace distributions. In the remainder of the paper, we assume that M is a Kählerian manifold with complex structure J : T M T M and complex dimension n = dim C M 2. As

5 A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS. 5 J = 0, one has X JY = J X Y. The following curvature symmetries are easily deduced R(X, Y, Z, W ) = R(JX, JY, Z, W ) = R(X, Y, JZ, JW ) = R(JX, JY, JZ, JW ). Let p M be a point of coisotropy rank at most one. By Lemma 2.5 and Theorem 3.5, there exists a projective class of skew-symmetric linear maps [A p ] P L(T p M) such that E v = span{v, A p v} for each v S p M and any A p [A p ]. Let K p = ker(a p ) and M p = Kp. Then T p M = K p M p is an orthogonal direct sum into A p -invariant subspaces of even real dimensions. Lemma 4.1. For each v S p M, E Jpv = J p (E v ). In particular, the orthogonal decomposition T p M = K p M p is J p -invariant. Proof. The lemma is an immediate consequence of Lemma 2.1 and the fact that J p preserves the sectional curvatures of plane sections. Let Âp and Ĵp denote the restrictions of A p and J p to the jointly invariant subspace M p. For a unit vector v M p, A p v = Âpv = span{v, E v }. Consequently, Â p v is an eigenvector of J v with eigenvalue λ(v) := sec(v, Âpv) κ(p). In particular, λ(v) is either the maximal or minimal value of the sectional curvature function when restricted to the submanifold of plane sections σ M p containing the vector v. In particular, either sec(σ) κ(p) for all plane sections σ M p or sec(σ) κ(p) for all plane sections σ M p. Convention 4.2. We adopt the following notational convention. Given r 1, r 2 R, (1) if sec p κ(p) on M p, then r 1 r 2 (respecitvely, r 1 r 2 ) if and only if r 1 < r 2 (respectively, r 1 r 2 ), and (2) if sec p κ(p) on M p, then r 1 r 2 (respectively, r 1 r 2 ) if and only if r 1 > r 2 (respectively, r 1 r 2 ). As Âpv is orthogonal to both v and Â2 pv we have that λ(v) λ(âpv) with equality if and only if v and Â2 pv are linearly dependent. Assume that V = σ 1 σ 2 M p is an orthogonal direct sum of two Âp-invariant planes. There exists real numbers 0 < µ 1 and 0 < µ 2 such that Âpv i = µ i for each unit vector v i σ i. Let λ i = sec(σ i ) for i = 1, 2. Without loss of generality, we may assume that µ 1 µ 2 and that if equality µ 1 = µ 2 holds, then λ 1 λ 2. For a vector v i σ i, let v i = Âpv i /µ i. Note that with this notation, v i = v i. Lemma 4.3. Assume that {u, v, w} V are orthonormal vectors with u, v σ i and w σ j with i j {1, 2}. Then R(u, v, w, u) = 0 and R(u, w, w, u) = κ(p). Proof. As u σ i, an Âp-invariant plane, the orthogonal plane σ j is contained in E u. In particular, w E u, from which the curvature components are easily derived. Lemma 4.4. Let v i σ i, i = 1, 2, be unit-vectors. If v = αv 1 +βv 2 is a unit-vector, then λ(v) = α 2 λ 1 + β 2 λ 2. Proof. We have that λ(v) Âpv 2 = R(v, Âpv, Âpv, v) or equivalently λ(v)(α 2 µ β 2 µ 2 2) = R(αv 1 + βv 2, αµ 1 v 1 + βµ 2 v 2, αµ 1 v 1 + βµ 2 v 2, αv 1 + βv 2 ). Expanding the above, using Lemma 4.3, and simplifying yields

6 6 B. SCHMIDT, K. SHANKAR, AND R. SPATZIER (4.1) λ(v)(α 2 µ β 2 µ 2 2) = α 4 µ 2 1λ 1 + κ(p)α 2 β 2 (µ µ 2 2) + β 4 µ 2 2λ 2 + Φ where (4.2) Φ = 2α 2 β 2 µ 1 µ 2 [R(v 1, v 1, v 2, v 2 ) + R(v 1, v 2, v 1, v 2 )]. The vector w = βµ 2 v 1 αµ 1 v 2 is orthogonal to both v and Âpv so that κ(p) = sec(v, w). Equivalently κ(p)(α 2 µ β 2 µ 2 2) = R(αv 1 + βv 2, βµ 2 v 1 αµ 1 v 2, βµ 2 v 1 αµ 1 v 2, αv 1 + βv 2 ). Expanding the above, using Lemma 4.3, and simplifying yields (4.3) Φ = α 2 β 2 (µ 2 2λ 1 + µ 2 1λ 2 ) + κ(p)α 2 µ 2 1(α 2 1) + κ(p)β 2 µ 2 2(β 2 1). Substituting (4.3) into (4.1) and simplifying using α 2 + β 2 = 1 yields the desired formula for λ(v). Corollary 4.5. If µ 1 < µ 2, then λ 1 λ 2. Proof. In the notation of Lemma 4.4, choose the vector v so that α = β = 2/2. As µ 1 < µ 2, the vectors v = αv 1 + βv 2 and Â2 pv = (αµ 2 1v 1 + βµ 2 2v 2 ) are linearly independent. Therefore λ(v) = sec(v, Âpv) sec(âpv, Â2 pv) = λ(âpv)/ Âpv ). By Lemma 4.4, 1 2 λ λ 2 µ2 1 λ µ µ2 2 λ 1 +µ2 2 µ 2 2, or equivalently, ( 1 1 +µ2 2 2 µ2 1 )λ µ µ2 2 µ ( µ 2 1 +µ2 2 2 )λ 2. If λ 2 λ 1, it follows that 1 2 µ2 1 µ2 µ µ2 2 µ 2 1 +µ2 2 2, a contradiction. Given unit vectors e i σ i, i = 1, 2, consider the following components of the curvature tensor: α = R(e 1, ē 1, e 2, ē 2 ), β = R(ē 1, e 2, e 1, ē 2 ), and γ = R(e 2, e 1, ē 1, ē 2 ). By the Bianchi identity, (4.4) α + β + γ = 0. Lemma 4.6. In the notation above, 0 β = γ, α = 2γ 0, and µ 2 1(λ 2 κ(p)) + µ 2 2(λ 1 κ(p)) = 6µ 1 µ 2 γ. Moreover, (λ 1 κ(p))(λ 2 κ(p)) 9γ 2. Proof. Set v 1 = e 1, v 2 = e 2, and α = β = 2/2 and use (4.2) and (4.3) to deduce (4.5) µ 2 1(λ 2 κ(p)) + µ 2 2(λ 2 κ(p)) = 2µ 1 µ 2 (β α). Set v 1 = e 1, v 2 = ē 2, and α = β = 2/2 and use (4.2) and (4.3) to deduce (4.6) µ 2 1(λ 2 κ(p)) + µ 2 1(λ 2 κ(p)) = 2µ 1 µ 2 (γ α). As 0 < µ i and κ(p) λ i, (4.5) and (4.6) imply that β = γ. By (4.4) α = 2γ which upon substitution into (4.6) yields µ 2 1(λ 2 κ(p)) + µ 2 2(λ 1 κ(p)) = 6µ 1 µ 2 γ from which the remaining inequalities are easily deduced. Lemma 4.7. In the notation above, λ 1 3γ + κ(p) λ 2. Equality holds in either case only if λ 1 = λ 2 = 3γ + κ(p) and µ 1 = µ 2.

7 A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS. 7 Proof. If 3γ + κ(p) λ 1, then 9γ 2 (λ 1 κ(p)) 2 (λ 1 κ(p))(λ 2 κ(p)) 9γ 2 by Lemma 4.6, implying that λ 1 = λ 2 = 3γ + κ(p) (and µ 1 = µ 2 by Corollary 4.5). Lemma 4.6 and the derivation of Berger s curvature inequality [2, 7] imply 2γ = α = R(ē 1, e 1, e 2, ē 2 ) = 1 6 [sec(ē 1 + ē 2, e 1 + e 2 ) + sec(e 1 + ē 2, ē 1 e 2 )] [sec(ē 1 ē 2, e 1 + e 2 ) + sec(e 1 ē 2, ē 1 + e 2 )] 1 6 [sec(ē 1 ē 2, e 1 + e 2 ) + sec(e 1 ē 2, ē 1 e 2 )] 1 6 [sec(ē 1 + ē 2, e 1 e 2 ) + sec(e 1 + ē 2, ē 1 + e 2 )]. If σ V is a plane section and v σ is a unit vector, then sec(σ) λ(v) λ 2 where the last inequality is a consequence of Lemma 4.4. Hence κ(p) sec λ 2 on V. These inequalities and the above formula for 2γ yields the inequality 3γ +κ(p) λ 2 where equality holds only if sec(ē 1 + ē 2, e 1 + e 2 ) = λ 2. Hence, equality holds only if λ 2 = sec(ē 1 + ē 2, e 1 + e 2 ) λ(e 1 + e 2 ) = 1 2 (λ 1 + λ 2 ) λ 2, or equivalently if λ 1 = λ 2 (and µ 1 = µ 2 by Corollary 4.5). Lemma 4.8. Either ÂpĴp = ĴpÂp or ÂpĴp = ĴpÂp. Proof. Let v M p be a unit vector. By Lemma 4.1, Ĵ p (E v ) = EĴp(v). As Ĵp acts orthogonally, it follows that ĴpÂpv and ÂpĴpv are both perpendicular to the subspace span{ĵpv, EĴpv } M p of codimension one in M p. Therefore there exists a nonzero constant c(v) such that ÂpĴpv = c(v)ĵpâpv. As both ÂpĴp and ĴpÂp are non-degenerate, the constant c(v) is independent of v. Taking the determinant yields c dim R(M p) = 1, whence c = ±1. Proposition 4.9. A p J p = J p A p. Proof. As the restrictions of A p J p and J p A p to K p are both zero, it suffices to prove ÂpĴp = ĴpÂp. Let σ 1 M p be a plane section with Âp(σ 1 ) = σ 1. If Ĵ p (σ 1 ) = σ 1, then the restrictions of  p and Ĵp to σ 1 differ by a scalar, hence commute, concluding the proof in this case by Lemma 4.8. Therefore, we may assume that ÂpĴp = ĴpÂp and that σ 1 is not invariant under Ĵp. We obtain a contradiction in what follows. Let {e 1, e 2 } be an orthonormal basis of σ 1. There exists a nonzero constant µ such that Âpe 1 = µe 2 and Âpe 2 = µe 1. After rescaling Âp, we may assume that µ = 1. In particular, letting  p denote the adjoint of Âp, we have that  p = Âp on the subspace σ 1. As Ĵp is orthogonal, {e 3 = Ĵpe 1, e 4 = Ĵpe 2 } is an orthonormal basis of σ 2 := Ĵ p (σ 1 ). We claim that the vectors {e 1, e 2, e 3, e 4 } form an orthonormal 4-frame. As g p (e 1, e 3 ) = g p (e 1, Ĵpe 1 ) = 0 = g p (e 2, Ĵpe 2 ) = g p (e 2, e 4 ), it remains to verify the equalities g p (e 1, e 4 ) = 0 = g p (e 2, e 3 ). Calculate g p (e 1, e 4 ) = g p (e 1, Ĵpe 2 ) = g p (e 1, ĴpÂpe 1 ) = g p (e 1, ÂpĴpe 1 ) = g p (  pe 1, Ĵpe 1 ) = g p (Âpe 1, Ĵpe 1 ) = g p (e 2, Ĵpe 1 ) = g p (Ĵpe 2, e 1 ) = g p (e 4, e 1 )

8 8 B. SCHMIDT, K. SHANKAR, AND R. SPATZIER to conclude that g p (e 1, e 4 ) = 0. Finally, g p (e 2, e 3 ) = g p (Ĵpe 2, Ĵpe 3 ) = g p (e 4, e 1 ) = 0 concluding the proof that {e 1, e 2, e 3, e 4 } are orthonormal. Let λ = sec(σ 1 ) and note that λ = R(e 1, e 2, e 2, e 1 ) = R(Ĵpe 1, Ĵpe 2, Ĵpe 2, Ĵpe 1 ) = R(e 3, e 4, e 4, e 3 ) = sec(σ 2 ). As Âp(σ 2 ) = ÂpĴp(σ 1 ) = ĴpÂp(σ 1 ) = Ĵp(σ 1 ) = σ 2, the plane σ 2 is Âp-invariant. Therefore, Lemma 4.4 implies that κ(p) sec p λ on the subspace V. By Berger s curvature estimate ([2, 7]) λ = R(e 1, e 2, e 2, e 1 ) = R(e 1, e 2, Ĵpe 1, Ĵpe 2 ) = R(e 1, e 2, e 3, e 4 ) 2 (λ κ(p)), 3 or equivalently, λ 2κ(p), contradicting κ(p) λ. Lemma Assume that V = σ 1 σ 2 is an orthogonal sum of  p -invariant and Ĵp-invariant plane sections. Let v i σ i be unit vectors and v i = Âp(v i )/µ i. If Ĵ p v 1 = v 1, then Ĵpv 2 = v 2. If Ĵ p v 1 = v 1, then Ĵpv 2 = v 2. In both cases γ = R(v 2, v 1, v 1, v 2 ) = κ(p) 0. Proof. The assumptions imply that there are constants c 1, c 2 { 1, 1} such that J p v i = c i v i for i = 1, 2. We must argue that c 1 = c 2. We have that γ = R(v 2, v 1, v 1, v 2 ) = R(Ĵpv 2, Ĵpv 1, v 1, v 2 ) = R(c 2 v 2, c 1 v 1, v 1, v 2 ) = c 1 c 2 κ(p) where we have used Lemma 4.3 in the last equality. By Lemma 4.6, 0 γ whence κ(p) 0. Now if c 1 c 2, then γ = κ(p). In this case, the first inequality in Lemma 4.7 implies that κ(p) λ 1 3γ + κ(p) = 2κ(p), a contradiction. Corollary If σ M p is a plane section satisfying Âp(σ) = σ, then Ĵp(σ) = σ. Proof. After scaling Âp, we may find an orthonormal basis {e 1, e 2 } of σ satisfying  p e 1 = e 2 and Âpe 2 = e 1. If Ĵp(σ) σ then Ĵp(σ) σ = {0}. Letting e 3 = Ĵpe 1 and e 4 = Ĵpe 2, the vectors {e 1, e 2, e 3, e 4 } span a 4-dimensional subspace of M p. By Proposition 4.9,  p e 3 = e 4 and Âpe 4 = e 3 since and Let v 1 = e1+e4 2  p e 3 = ÂpĴpe 1 = ĴpÂpe 1 = Ĵpe 2 = e 4  p e 4 = ÂpĴpe 2 = ĴpÂpe 2 = Ĵpe 1 = e 3. and v 2 = e1 e4 2 and use the above to calculate v 1 = e2 e3 2 v 2 = e2+e3 2. Verify that σ 1 = span{v 1, v 1 } and σ 2 = span{v 2, v 2 } are orthogonal  p -invaraint and Ĵp-invariant planes satisfying Ĵpv 1 = v 1 and Ĵpv 2 = v 2, contradicting Lemma Corollary If κ(p) = 0, then dim R (M p ) 2. Proof. We argue the contrapositive. If dim R (M p ) 4, then there exists orthogonal  p -invariant planes σ 1, σ 2 M p. By Corollary 4.11 the planes σ i are Ĵp-invariant. By Lemma 4.10, κ(p) 0. Lemma If κ(p) 0, then M p = T p M. and

9 A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS. 9 Proof. We argue the contrapositive, assuming that dim R K p 2. As dim R M 4, Lemma 4.1 implies that there exists J p -invariant and orthogonal 2-planes σ 1 and σ 2 with σ 2 K p. Let e i σ i be unit vectors and let v = e1+e2 2. Consider the 2-plane σ = span{j p e 1, J p e 2 } that is orthogonal to v and for each θ [0, 2π) let w θ = cos(θ)j p e 1 + sin(θ)j p e 2 σ. Calculate sec(v, w θ ) = cos 2 (θ) sec(v, J p e 1 ) + sin 2 (θ) sec(v, J p e 2 ) +2 sin(θ) cos(θ)r(v, J p e 1, J p e 2, v). As J p e 2 K p, we have that v E Jpe 2 whence J p e 2 E v by Lemma 2.1. Conclude that R(v, J p e 1, J p e 2, v) = 0 and that sec(v, w π/4 ) = sec(v, w 3π/4 ). Therefore R(e 1 + e 2, J p e 1 + J p e 2, J p e 1 + J p e 2, e 1 + e 2 ) = R(e 1 + e 2, J p e 1 J p e 2, J p e 1 J p e 2, e 1 + e 2 ). Expanding the above and simplifying implies that (4.7) R(e 1, J p e 1, J p e 2, e 2 ) = R(e 1, J p e 2, J p e 1, e 2 ). Use (4.7) and the Bianchi identity R(e 1, J p e 1, J p e 2, e 2 ) + R(J p e 1, J p e 2, e 1, e 2 ) + R(J p e 2, e 1, J p e 1, e 2 ) = 0 to deduce (4.8) R(J p e 1, J p e 2, e 1, e 2 ) + 2R(J p e 2, e 1, J p e 1, e 2 ) = 0. As e 2 K p, we have that R(J p e 1, J p e 2, e 1, e 2 ) = R(e 1, e 2, e 1, e 2 ) = κ(p). Similarly, as J p (e 2 ) K p, we have that R(J p e 2, e 1, J p e 1, e 2 ) = R(J p e 2, e 1, e 1, J p e 2 ) = κ(p). Substituting the above into (4.8) yields κ(p) = 0. While we have not assumed that κ : M R is continuous, we have the following continuity property. Corollary The set {p M κ(p) 0} is open in M. Proof. If not, then there exist a sequence of points {p i } i N M with κ(p i ) = 0 for each i N that converges to a point p M with κ(p) 0. By Corollary 4.12, there exists a unit vector v i K pi for each i N. Every plane section containing the vector v i has sectional curvature zero. The vectors v i S pi M have a subsequence that converge to a vector v S p M. By continuity of sectional curvatures, every plane section containing the vector v has curvature zero. Therefore, 0 is the sole eigenvalue of J v, contradicting κ(p) 0.

10 10 B. SCHMIDT, K. SHANKAR, AND R. SPATZIER 5. Pointwise Analysis of κ 0. In this section, we assume that M p = T p M, or equivalently that κ(p) 0 by Corollary 4.12 and Lemma At the end of the section, we establish that such points are Einstein and have constant holomorphic curvatures, facts that are easily deduced once it is known that λ(v) is independent of v S p M. Establishing that λ is constant on S p M differs somewhat in the two cases dim R (M) 4 and dim R (M) 6. We start with the latter case, the easier of the two The case when dim R (M) 6. Lemma 5.1. If κ(p) 0, then A p has at most two distinct eigenvalues. Proof. If not, then there exist three orthogonal A p -invariant planes σ i, i = 1, 2, 3 and constants 0 < µ 1 < µ 2 < µ 3 such that Aw i = µ i for each unit vector w i σ i. Let λ i = sec(σ i ). As µ 1 < µ 2, Corollary 4.5 implies that λ 1 λ 2. By Lemmas 4.7 and 4.10, 4κ(p) λ 2. As µ 2 < µ 3, Corollary 4.5 implies that λ 2 λ 3. By Lemmas 4.7 and 4.10, λ 2 4κ(p), a contradiction.. Lemma 5.2. If κ(p) 0, then A p has a single eigenvalue. Proof. If not, then by Lemma 5.1, there exist constants 0 < µ 1 < µ 2 and A p - eigenspaces E 1 and E 2 such that T p M is the orthogonal direct sum T p M = E 1 E 2 and A p v i = µ i for each unit vector v i E i, i = 1, 2. As dim R (M) 6, one of the two eigenspaces E 1 or E 2 has real dimension at least four. Case I: dim R (E 1 ) 4 Choose orthogonal A p -invariant planes σ 1, σ 2 E 1 and σ 3 E 2. Let λ i = sec(σ i ) for each i = 1, 2, 3. As µ 1 < µ 2, Corollary 4.5 implies that λ 1 λ 3 and λ 2 λ 3. By Lemmas 4.7 and 4.10 applied to the four dimensional subspaces σ 1 σ 3 and σ 2 σ 3, λ 1 4κ(p) and λ 2 4κ(p). However, applying Lemmas 4.6 and 4.10 to the four dimensional subspace σ 1 σ 2 yields λ 1 + λ 2 = 8κ(p), a contradiction. Case II: dim R (E 2 ) 4 Choose orthogonal A p -invariant planes σ 1 E 1 and σ 2, σ 3 E 2. Let λ i = sec(σ i ) for each i = 1, 2, 3. As µ 1 < µ 2, Corollary 4.5 implies that λ 1 λ 2 and λ 1 λ 3. By Lemmas 4.7 and 4.10 applied to the four dimensional subspaces σ 1 σ 2 and σ 1 σ 3, 4κ(p) λ 2 and 4κ(p) λ 3. However, applying Lemmas 4.6 and 4.10 to the four dimensional subspace σ 2 σ 3 yields λ 2 +λ 3 = 8κ(p), a contradiction. Proposition 5.3. If κ(p) 0, then λ(v) = 4κ(p) for each v S p M. Proof. Let v T p M be a unit vector. The plane σ 1 = span{v, A p v} is A p - invariant by Lemma 5.2. As dim R (M) 6, there exist orthogonal A p -invariant planes σ 2, σ 3 σ1. Let λ i = sec(σ i ) for i = 1, 2, 3 and note that λ(v) = λ 1. Applying Lemmas 4.6 and 4.10 to the three four-dimensional subspaces σ i σ j, determined by i, j {1, 2, 3} distinct, yields the linear system λ 1 + λ 2 = λ 1 + λ 3 = λ 2 + λ 3 = 8κ(p), whose solution λ 1 = λ 2 = λ 3 = 4κ(p) is unique.

11 A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS The case when dim R (M) = 4. The following lemma is most likely well-known. Lemma 5.4. Let B be an open connected subset of a Riemannian manifold (M, g) admitting a pair of transverse, orthogonal, and totally-geodesic foliations F 1 and F 2. Then B is locally isometric to the product F 1 F 2. Proof. If H = T F i and V = T F 2, then the tangent bundle splits orthogonally T B = H V. By de Rham s splitting theorem, it suffices to prove that the distribution H is parallel on B. Let h, h denote vector fields tangent to H and let v, v denote vector fields tangent to V. As H is integrable, 0 = g([h, h], v), implying g( h h, v) = g( hh, v). As H is totally geodesic, g( h h, v) = g( hh, v). Conclude that (5.1) g( hh, v) = 0. Similarly, the fact that V is integrable and totally-geodesic implies that g( v v, h) = 0 As H and V are orthogonal, this implies (5.2) g( v h, v) = 0. By (5.1) and (5.2), H is parallel on B, concluding the proof. Proposition 5.5. If κ(p) 0, then λ(v) = 4κ(p) for each v T p M. In particular, A p has a single eigenvalue. Proof. The second claim in the proposition is a consequence of the first claim, Lemma 4.4, and Corollary 4.5. If the first claim in the proposition fails, Corollary 4.14 implies that there exists a metric ball B centered at p M with the property that for each b B, there exists a unit vector v S b M such that λ(v) 4κ(b). Define λ 2 : B R by λ 2 (b) = max {λ(v) v S b M} for each b B. For each b B, there exists a plane section σ 2 (b) T b B with sec(σ 2 (b)) = λ 2 (b). Note that for each b B, the plane σ 2 (b) is A b -invariant since otherwise sec(σ 2 (b)) sec(a b (σ 2 (b))), contradicting the definition of λ 2 (b). Let σ 1 (b) = σ 2 (b) for each b B and note that σ 1 (b) is also A b -invariant. Define λ 1 : B R by λ 1 (b) = sec(σ 1 (b)) for each b B. Then λ 1 (b) λ 2 (b) and there exists positive functions µ 1, µ 2 : B R with µ 1 (b) µ 2 (b) such that A b v i = µ i (b) for each unit vector v i E i (b) and b B. After possibly reducing the radius of B, there exist unit vector fields v 1 and v 2 on B tangent to σ 1 and σ 2 respectively. Letting v i = Av i /µ i, we obtain an orthonormal framing {v 1, v 1, v 2, v 2 } of T B with σ i = span{v i, v i } for i = 1, 2. Define γ : B R by γ = R(v 2, v 1, v 1, v 2 ). By Corollary 4.11, the A b -invariant planes σ i (b) are J b -invariant and by Lemma 4.10, γ = κ on B. We claim that (5.3) λ 1 (b) 4κ(b) λ 2 (b) for each b B. To prove the claim, first assume that A b has a single eigenvalue. In this case, Lemma 4.6 implies that λ 1 (b) + λ 2 (b) = 8κ(b). As λ 1 (b) λ 2 (b), either (5.3) holds or λ 1 (b) = λ 2 (b) = 4κ(b). The latter equality is excluded by Lemma 4.4 since b B. Next, assume that A b has two distinct eigenvalues. In this case, Corollary 4.5, implies that λ 1 (b) < λ 2 (b) and Lemma 4.7 implies (5.3). By the second Bianchi identity, B 1 := ( v2 R)(v 1, v 1 ) + ( v1 R)( v 1, v 2 ) + ( v1 R)(v 2, v 1 ) = 0.

12 12 B. SCHMIDT, K. SHANKAR, AND R. SPATZIER Calculate (5.4) (5.5) (5.6) (5.7) g(b 1 (v 1 ), v 2 ) = (λ 1 κ)g( v2 v 2, v 1 ) + 3κg( v2 v 2, v 1 ) = 0, g(b 1 ( v 2 ), v 1 ) = (λ 1 κ)g( v2 v 2, v 1 ) + 3κg( v2 v 2, v 1 ) = 0, g(b 1 ( v 2 ), v 1 ) = (λ 1 κ)g( v2 v 2, v 1 ) 3κg( v2 v 2, v 1 ) = 0, g(b 1 ( v 1 ), v 2 ) = (λ 1 κ)g( v2 v 2, v 1 ) 3κg( v2 v 2, v 1 ) = 0. Use (5.3), (6.1), and (6.2) to deduce (5.8) g( v2 v 2, v 1 ) = g( v2 v 2, v 1 ) = 0. Use (5.3), (6.3) and (6.4) to deduce (5.9) g( v2 v 2, v 1 ) = g( v2 v 2, v 1 ) = 0. By the second Bianchi identity, Calculate (5.10) (5.11) (5.12) (5.13) B 2 := ( v2 R)(v 1, v 1 ) + ( v1 R)( v 1, v 2 ) + ( v1 R)( v 2, v 1 ) = 0. g(b 2 (v 2 ), v 1 ) = (λ 1 κ)g( v2 v 2, v 1 ) + 3κg( v2 v 2, v 1 ) = 0, g(b 2 ( v 1 ), v 2 ) = (λ 1 κ)g( v2 v 2, v 1 ) + 3κg( v2 v 2, v 1 ) = 0, g(b 2 (v 1 ), v 2 ) = (λ 1 κ)g( v2 v 2, v 1 ) 3κg( v2 v 2, v 1 ) = 0, g(b 2 (v 2 ), v 1 ) = (λ 1 κ)g( v2 v 2, v 1 ) 3κg( v2 v 2, v 1 ) = 0. Use (5.3), (6.7), and (6.8) to deduce (5.14) g( v2 v 2, v 1 ) = g( v2 v 2, v 1 ) = 0. Use (5.3), (5.12), and (5.13) to deduce (5.15) g( v2 v 2, v 1 ) = g( v2 v 2, v 1 ) = 0. The plane field σ 2 is integrable and totally-geodesic by (6.5), (6.6), (5.14), and (5.15). By the second Bianchi identity, Calculate (5.16) (5.17) (5.18) (5.19) B 3 := ( v1 R)(v 2, v 2 ) + ( v2 R)( v 2, v 1 ) + ( v2 R)(v 1, v 2 ) = 0. g(b 3 (v 2 ), v 1 ) = (λ 2 κ)g( v1 v 1, v 2 ) + 3κg( v1 v 1, v 2 ) = 0, g(b 3 ( v 1 ), v 2 ) = (λ 2 κ)g( v1 v 1, v 2 ) + 3κg( v1 v 1, v 2 ) = 0, g(b 3 ( v 2 ), v 1 ) = (λ 2 κ)g( v1 v 1, v 2 ) 3κg( v1 v 1, v 2 ) = 0, g(b 3 ( v 1 ), v 2 ) = (λ 2 κ)g( v1 v 1, v 2 ) 3κg( v1 v 1, v 2 ) = 0. Use (5.3), (5.16), and (5.17) to deduce (5.20) g( v1 v 1, v 2 ) = g( v1 v 1, v 2 ) = 0. Use (5.3), (5.18), and (5.19) to deduce (5.21) g( v1 v 1, v 2 ) = g( v1 v 1, v 2 ) = 0. By the second Bianchi identity, Calculate (5.22) (5.23) B 4 := ( v1 R)(v 2, v 2 ) + ( v2 R)( v 2, v 1 ) + ( v2 R)( v 1, v 2 ) = 0. g(b 4 (v 1 ), v 2 ) = (λ 2 κ)g( v1 v 1, v 2 ) + 3κg( v1 v 1, v 2 ) = 0, g(b 4 ( v 2 ), v 1 ) = (λ 2 κ)g( v1 v 1, v 2 ) + 3κg( v1 v 1, v 2 ) = 0,

13 A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS. 13 (5.24) (5.25) g(b 4 (v 1 ), v 2 ) = (λ 2 κ)g( v1 v 1, v 2 ) 3κg( v1 v 1, v 2 ) = 0, g(b 4 (v 2 ), v 1 ) = (λ 2 κ)g( v1 v 1, v 2 ) 3κg( v1 v 1, v 2 ) = 0. Use (5.3), (5.22), and (5.23) to deduce (5.26) g( v1 v 1, v 2 ) = g( v1 v 1, v 2 ) = 0. Use (5.3), (5.24), and (5.25) to deduce (5.27) g( v1 v 1, v 2 ) = g( v1 v 1, v 2 ) = 0. The plane field σ 1 is integrable and totally-geodesic by (5.20), (5.21), (5.26), and (5.27). As the tangent plane fields σ 1 and σ 2 are orthogonal, integrable, and totallygeodesic, B is locally isometric to a Riemannian product by Lemma 5.4. Consequently, σ 2 ker(j v1 ), a contradiction since κ 0 implies that dim R (ker J v1 ) Einstein points and constant holomorphic curvatures. Corollary 5.6. If κ(p) 0, then a 2-plane σ T p M is holomorphic if and only if A p (σ) = σ. Proof. Let σ T p M be a plane section. If A p (σ) = σ then J p (σ) = σ by Corollary Conversely, assume that J p (σ) = σ and let v σ be a nonzero vector. The two plane σ = span{v, A p v} is A p -invariant by Lemma 5.2 and Proposition 5.5. By Corollary 4.11, σ is J p -invariant. As v lies in a unique holomorphic plane, σ = σ, so that σ is A p -invariant. Proposition 5.7. Assume that n = dim C (M) 2 and that p M satisfies κ(p) 0. Then all holomorphic planes in T p M have sectional curvature 4κ(p). Moreover, for each v S p M, Ric p (v) = 2(n + 1)κ(p). Proof. By Corollary 5.6, holomorphic planes are A p -invariant. They have curvature λ = 4κ(p) by Proposition 5.3 and Proposition 5.5. If v S p M, then Ric p (v) = (2n 2)κ(p) + λ(v) = 2(n + 1)κ(p). 6. Global analysis of κ = 0. In this section, we assume that κ(p) = 0 for every p M. By Corollary 4.12, dim R (M p ) 2 for each p M, yielding the decomposition M = I O where points p I satisfy K p = T p M and points p O satisfy dim R M p = 2. Equivalently, I is the closed subset of isotropic points in M and O is the open subset of nonisotripic points in M. For p O, the orthogonal splitting T p M = K p M p is a pointwise analogue of the local splitting in Theorem A (3). We begin with a classical definition (see [3, 9]). Definition 6.1. For p M, the nullity space at p is the subspace N p = {v T p M R(w, v) = 0 for every w T p M}. Its dimension µ(p) = dim(n p ) is the index of nullity at p. Lemma 6.2. If κ(p) = 0, then N p = K p.

14 14 B. SCHMIDT, K. SHANKAR, AND R. SPATZIER Proof. The inclusion N p K p. When p I, the equality N p = K p is obvious. We therefore assume that p O and will argue that K p N p. Let v, w, z K p. Use 0 = R(w, v)v, 0 = R(w, z)z, and 0 = R(w, v + z)(v + z) to conclude (6.1) R(w, v)z + R(w, z)v = 0. Similarly, (6.2) R(z, w)v + R(z, v)w = 0. Add (6.1) and (6.2) to obtain (6.3) R(w, v)z + R(z, v)w = 0. Add the Bianchi identity R(w, v)z + R(v, z)w + R(z, w)v = 0 and (6.3) to obtain (6.4) 2R(w, v)z + R(z, w)v = 0. Add (6.1) and (6.4) to obtain 3R(w, v)z = 0. Conclude (6.5) K p ker(r(w, v)) v, w K p. Now assume that v, w K p and that z M p. As M p is A p -invariant and twodimensional, A p (z) M p, whence ker(j z ) = E z = Mp = K p. Then R(w, v)z has no component in K p since if y K p, then g p (R(w, v)z, y) = g p (R(w, v)y, z) = 0 where the last equality is a consequence of (6.5). As dim R M p = 2, it follows that R(w, v)z is a multiple of J p z. Calculate R(w, v, z, J p z) = R(v, z, w, J p z)) R(z, w, v, J p z) = R(v, z, J p w, z) + R(z, w, J p v, z) = g p (J z v, J p w) + g p (J z w, J p v) = 0. where the first equality is the Bianchi identity and the last inequality is a consequence of v, w K p = E z. Therefore M p ker(r(w, v)) which with (6.5) gives (6.6) R(w, v) = 0 v, w K p. Now let v K p and z M p. Then for w K p, R(z, v)w = R(w, v)z + R(z, w)v = R(z, w)v by the Bianchi identity and (6.6). By (6.6), R(z, w)v M p. As w K p = E z, R(z, w, v, z) = g p (J z (w), v) = 0 and R(z, w, v, J p z) = R(z, w, J p v, z) = g p (J z (w), J p v) = 0. Conclude that (6.7) K p ker(r(z, v)) z M p v K p Finally, let v K p and z M p. Then R(z, v)z = 0 since v E z and R(z, v)j p z = R(J p z, v)j p z = J Jpz(v) = 0 since v E Jpz. Therefore M p ker(r(z, v)) which with (6.7) gives (6.8) R(z, v) = 0 z M p v K p Hence K p N p by (6.6) and (6.8).

15 A SCHUR TYPE THEOREM FOR KÄHLERIAN MANIFOLDS Proofs of Main Thoerems We start with the proof of Theorem A. Proof. Let U = {p M κ(p) 0}. By Corollary 4.14 U is an open subset in M. If U =, then κ = 0 on all of M. Otherwise, there is a non-empty connected component C of U. For each p C, Ric p = 2(n + 1)κ(p)g p by Proposition 5.7, whence κ is constant on C by Schur s theorem (REFERENCE) for Ricci curvatures. Conclude that C = M and that κ : M R is constant. If κ 0, then M has constant holomorhic curvatures 4κ by Proposition 5.7, concluding the proof in this case by [5, 6]. If κ = 0, the argument given in [1, Theorem 8] applies verbatim to prove (3). References [1] K. Abe, Applications of a Riccati type differential equation to Riemannian manifolds with totally geodesic distributions. Tôhoku Math. Journ. (2) 25 (1973), [2] M. Berger, Sur quelques variétés riemanniennes suffisamment pincées. Bull. Soc. Math. France 88 (1960), [3] S.-s. Chern and N. Kuiper, Some theorems on the isometric imbedding of compact Riemannian manifold in euclidean space. Ann. of Math. (2) 56 (1952), [4] T. Hangan and R. Lutz, Champs d hyperplans totalement géodésiques sur les sphères. Third Schnepfenried geometry conference, Vol. 1 (Schnepfenried, 1982), , Astérisque, , Soc. Math. France, Paris, [5] N.S. Hawley, Constant holomorphic curvature. Canad. J. Math. 5 (1953), [6] J.-i. Igusa, On the structure of a certain class of Kähler varieties. Amer. J. Math. 76 (1954), [7] H. Karcher, A short proof of Berger s curvature tensor esitimates. Proc. Amer. Math. Soc, 26 (1970), [8] G. Liu, Compact Kähler manifolds with nonpositive bisectional curvature. To appear in GAFA. [9] R. Maltz, The nullity spaces of curvature-like tensors. J. Diff. Geom. 7 (1972), [10] F. Schur, Ueber die Deformation der Räume constanten Riemann schen Krümmungsmaasses. Math. Ann., ), no. 2, [11] F. Zheng, Kodaira dimensions and hyperbolicity of nonpositively curved compact Kähler manifolds. Comment. Math. Helv., 77 (2002), Michigan State University University of Oklahoma Univeristy of Michigan Dept. of Mathematics Dept. of Mathematics Dept. of Mathematics 619 Red Cedar Road 601 Elm Avenue 530 Church Street East Lansing, MI, Norman, OK, Ann Arbor, MI, schmidt@math.msu.edu shankar@math.ou.edu spatzier@umich.edu