17_914_de.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 3, No. 6, 2010, 1137-1140 ISSN 1307-5543 – www.ejpam.com SPECIAL ISSUE ON COMPLEX ANALYSIS: THEORY AND APPLICATIONS DEDICATED TO PROFESSOR HARI M. SRIVASTAVA, ON THE OCCASION OF HIS 70TH BIRTHDAY A Note on Kaehler Manifolds Uday Chand De Department of Pure Mathematics, University of Calcutta, 35, Ballygunje Circular Road, Kolkata 700019, West Bengal, India Abstract. The object of the present paper is to prove that in a Kaehler manifold of dimension n≥ 4, div R = 0 and div C = 0 are equivalent, where ’div’ denotes divergence and R and C denote the curvature tensor and Weyl conformal curvature tensor, respectively. 2000 Mathematics Subject Classifications: 53C25. Key Words and Phrases: Kaehler manifold, divergence, Weyl conformal curvature tensor. 1. Introduction Let M be an n-dimensional Kaehler manifold. Then the Kaehler metric g of M satisfies g(JX , JY ) = g(X , Y ) and ∇J = 0, where J and ∇ denote the complex structure and the covariant differentiation of M , respectively. Let R, S and C denote the curvature tensor, Ricci tensor and Weyl conformal curvature tensor of M , respectively. It is well known that a Kaehler manifold with parallel Ricci tensor is Einstein if M is irreducible. In a Riemannian manifold it can be easily verified from the differential Bianchi identity that div R = 0 holds if and only if (∇X S)(Y, Z) = (∇Y S)(X , Z), where ’div’ denotes divergence. It is well known [1] that if the Ricci tensor S satisfies (∇X S)(Y, Z) = (∇Y S)(X , Z) in a Kaehler manifold, then the Ricci tensor is parallel. In a Riemannian manifold it is also known [1] that the statements (i) div R = 0, (ii) div C = 0 and the scalar curvature is constant are equivalent. In the present paper we prove that in a Kaehler manifold of dimension n ≥ 4, div R= 0 and div C= 0 are equivalent. Email address: u _de�yahoo. om http://www.ejpam.com 1137 c© 2010 EJPAM All rights reserved. U. De / Eur. J. Pure Appl. Math, 3 (2010), 1137-1140 1138 2. Preliminaries In a Riemannian manifold Weyl conformal curvature tensor C is defined by C(X , Y )Z = R(X , Y )Z − 1 n− 2 [g(Y, Z)QX − g(X , Z)QY + S(Y, Z)X − S(X , Z)Y ] + r (n− 1)(n− 2) [g(Y, Z)X − g(X , Z)Y ], (1) where Q is the Ricci operator defined by g(QX , Y ) = S(X , Y ) and r denotes the scalar curva- ture. It is well known [3] that in a Riemannian manifold of dimension n> 3, (divC)(X , Y )Z = n− 3 n− 2 [{(∇X S)(Y, Z)− (∇Y S)(X , Z)} + 1 2(n− 1) {dr(X )g(Y, Z)− dr(Y )g(X , Z)}]. (2) In a Kaehler manifold the following relations hold [5]: g(X , JY ) = −g(JX , Y ), (3) S(X , JY ) = −S(JX , Y ), (4) ∇X JY = J∇X Y. (5) 3. Main Result Theorem 1. Let M be a Kaehler manifold of dimension n≥ 4. Then div R = 0 and div C = 0 are equivalent. To prove the theorem we first state and prove the following: Lemma 1. In a Kaehler manifold (∇ZS)(JX , Y ) = −(∇ZS)(X , JY ) holds. Proof. In a Kaehler manifold the Ricci tensor S satisfies S(JX , Y ) = −S(X , JY.) Now (∇ZS)(JX , Y ) = ∇ZS(JX , Y )− S(∇Z JX , Y )− S(JX ,∇Z Y ) = −∇ZS(X , JY )− S(J∇Z X , Y )− S(X , J∇Z Y ), using (5) = −∇ZS(X , JY ) + S(∇Z X , JY ) + S(X ,∇Z JY ), by (5) = −(∇ZS)(X , JY ). This completes the proof. Lemma 2. In a Kaehler manifold the Ricci tensor S satisfies the condition∑n i=1(∇ei S)(JX , ei) = 1 2 dr(JX ), where {ei} is an orthonormal basis of the tangent space at each point of the manifold. U. De / Eur. J. Pure Appl. Math, 3 (2010), 1137-1140 1139 Proof. From S(X , Y ) = g(QX , Y ) we easily get [2] (∇ZS)(X , Y ) = g((∇ZQ)X , Y ). (6) Replacing X by JX in (6) yields (∇ZS)(JX , Y ) = g((∇ZQ)JX , Y ). (7) Putting Y = Z = ei in (7) and taking summation over i, i = 1,2, . . . , n, we get (∇ei S)(JX , ei) = g((∇ei Q)JX , ei). We know (divQ)(X ) = tr(Z → (∇ZQ)(X )) = ∑ i g((∇ei Q)(X ), ei). But it is known [4] that (div Q)(X) = 1 2 dr(X ). Hence (∇ei S)(JX , ei) = 1 2 dr(JX ), which completes the proof. Proof. [of the main theorem] Suppose div C = 0. Then from (2) we have (∇ZS)(X , Y )− (∇X S)(Z , Y ) = 1 2(n− 1) [dr(Z)g(X , Y )− dr(X )g(Z , Y )]. (8) It is known [5] that in a Kaehler manifold the Ricci tensor S satisfies (∇ZS)(X , Y ) = (∇X S)(Z , Y ) + (∇JY S)(JX , Z). (9) Using (9) in (8) we obtain (∇JY S)(JX , Z) = 1 2(n− 1) [dr(Z)g(X , Y )− dr(X )g(Z , Y )]. (10) Replacing Y by JY in (10) we obtain −(∇Y S)(JX , Z) = 1 2(n− 1) [dr(Z)g(X , JY )− dr(X )g(Z , JY )]. (11) Using (3) and Lemma 1 we get from (11) (∇Y S)(X , J Z) = 1 2(n− 1) [dr(Z)g(X , JY ) + dr(X )g(J Z , Y )]. (12) Taking X = Y = ei in (12) we get 1 2 dr(J Z) = 1 2(n− 1) dr(J Z) which implies dr(J Z) = 0, since n ≥ 4. Hence dr(Z) = 0, that is, r =constant. Using r =constant in (8) we get (∇ZS)(X , Y ) = (∇X S)(Z , Y ). REFERENCES 1140 Therefore div R = 0. This completes the proof. From Theorem 1 and the known result mentioned in the introduction we obtain that if the conformal curvature tensor is divergence free in a Kaehler manifold of dimension ≥ 4, then the Ricci tensor is parallel. Conversely, if the Ricci tensor is parallel, then from (2) it follows that div C = 0. Thus we conclude that in a Kaehler manifold of dimension ≥ 4, the statements (i) div C = 0 and (ii) the Ricci tensor is parallel are equivalent. References [1] A. L. Besse. Einstein Manifolds, Springer-Verlag, 1987. [2] U. C. De and A. A. Shaikh. Differential Geometry of Manifolds, Alpha Science publishers, U. K., 2007. [3] L. P. Eisenhart. Riemannian Geometry, Princeton University Press, 1949. [4] P. Peterson. Riemannian Geometry, Springer, p-33. [5] K. Yano and M. Kon. Structures on manifolds, World Sci., 1984.