A Note on Cohomology of a Riemannian Manifold

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Int. J. Contemp. ath. Sciences, Vol. 9, 2014, no. 2, 51-56 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ijcms.2014.311131 A Note on Cohomology of a Riemannian anifold Tahsin Ghazal King Saud University, athematics Department Riyadh, Saudi Arabia Copyright c 2014 Tahsin Ghazal. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper we use the vanishing of first cohomology group of a Riemannian manifold (,g) to find a sufficient condition for a closed vector field ξ on to be a concircular vector field. athematics Subject Classification: 53C55 Keywords: concircular transformation, Concircular vector field, Cohomology 1. Introduction Let (,g) beann - dimensional Riemannian manifold. A diffeomorphism ϕ : is said to be a concircular transformation if it maps circle to a circle, and a smooth vector field ξ on (,g) is said to be concircular vector field if its flow consists of concircular transformation (cf. [1], [4]). concircular vector fields have been used in finding characterizations of different Riemannian manifolds (cf.[3]) and are also important in the general theory of relativity. Recall that circular vector fields are closed vector fields; and on a Riemannian manifold (,g) whose first cohomology group H 1 (,R) =0, the closed vector fields are in abundance. Since, concircular vector fields are important in geometry as well as physical sciences, its an interesting question to find sufficient conditions for a closed vector field ξ on a Riemannian manifold to be a concircular vector fields. Recall that on Riemannian manifold (,g), a smooth vector field X is said to be an eigen vector of the Laplacian

52 Tahsin Ghazal operator Δ (also called the rough Laplacian) if there exists a constant λ such that ΔX = λx, λ 0. Note that the position vector field X = xi on the Euclidian space x i (R n,, ) satisfies ΔX =0. Also if Z is a constant vector field on (R n+1,, ), then its tangential projection u on the unit sphere S n satisfies Δu = u, where Z = u + ρn, ρ = Z, N N being the unit normal vector field to S n in (R n+1,, ). In this paper, we consider the question under what conditions a closed vector field ξ on a Riemannian manifold (,g) satisfying H 1 (,R) =0is a concircular vector field?. Note that in the examples discussed above the vector fields X and u are closed vector fields on (R n,, ) and the unit sphere S n respectively and that H 1 (R n, R) =0,H 1 (S n, R) = 0 holds. We answer the above question by proving the following Theorem: Let (,g) be an n dimensional compact Riemannian manifold with H 1 (,R) =0. If ξ is smooth closed vector field on satisfying Δξ = λξ and (div ξ) 2 nλ ξ 2, then ξ is a concircular vector field. 2. Preliminaries Let (,g) be an n dimensional Riemannian manifold. A smooth vector field ξ on is said to be a concircular vector field if X ξ = fx, X X (), (2.1) where f is a smooth function, the Riemannian connection on and X ()is the Lie algebra of smooth vector fields on.suppose that the first cohomology group H 1 (,R) = 0 for the Riemannian manifold (,g). Since each closed 1 - form η on defines a cohomology class in the derham cohomology group H 1 dr () which is isomorphic to H1 (,R) = 0, the closed smooth 1 - form η must be exact. If ξ X () is dual to the smooth closed 1 - form η, that is η (X) =g (X, ξ),x X (), then the conditions that η is closed and exact imply that η = dρ for some function ρ on (,g), that is ξ = ρ, where ρ is the gradient of ρ on (,g). If we define an operator A : X () X (), BY g (A (X),Y)= 1 2 ( ξg)(x, Y ),X,Y X () then A is symmetric that is g (A (X),Y) = g (Y,A(X)) holds and using Koszul s formula we have X ξ = AX, X X (), (2.2)

A note on cohomology of a Riemannian manifold 53 where we have used dη, = 0, that is the smooth 1 - form η is closed. curvature tensor field R of the Riemannian manifold (,g) is given by The R (X, Y )= X Y Z Y X Z [X,Y ] Z, X, Y, Z X () and the Ricci curvature Ric is given by Ric (X, Y )= n g (R (e i,x) Y,e i ), X,Y X (). where {e 1,e 2,...,e n } is a local orthonormal frame on. The Ricci operator Q is a symmetric operator defined by Q : X () X () Ric (X, Y )=g (QX, Y ), X, Y X (). For a closed vector field ξ X (), using (2.2, ), we get R (X, Y ) ξ =( A)(X, Y ) ( A)(Y,X,) (2.3) where the covariant derivative ( A)(X, Y ) is defined by ( A)(X, Y )= X AY A ( X Y ), X,Y X ()() Using the equation (2.3) we get Ric (X, ξ) = n g (( A)(e i,x) ( A)(X, e i ),e i ), X X (). (2.4) We define a smooth function α on by α = n g (Ae i,e i ), then using symmetry of the operator A in the equation (2.4), we get Ric (X, ξ) = n g (X, ( A)(e i,e i )) X (α), X X (). (2.5) The rough Laplacian operator Δ on a Riemannian manifold (,g) isa self adjoint operator Δ : X () X () with respect to the inner product (, ):X() X () R (X, Y )= g (X, Y ) for compactly supported vector fields X, Y ; defined by

54 Tahsin Ghazal ( ) ΔX = e i e i X ei e i X, X X () (2.6) i and that the operator Δ being elliptic has non - negative eigenvalues, a non - negative number λ satisfying ΔX = λx is called eigenvalue of Δ corresponding to eigen vector X. 3. Proof of the Theorem Let (,g) be an n - dimensional compact Riemannian manifold with first singular homology group H 1 (,R) =0. Let ξ X () be a closed vector field on that satisfies Δξ = λξ (3.1) and (div ξ) 2 nλ ξ 2 (3.2) as required by the theorem, where λ is a constant. If {e 1,e 2,...,e n } is a local orthonormal frame on and α = n g (Ae i,e i ), then equation (2.2) gives We have divξ = α (3.3) div(αξ) =ξ (α)+α 2 which by Stoke s theorem gives (α) = ξ α 2 (3.4) Now using the equation (2.5), we get Q (ξ) = n (A)(e i,e i ) α (3.5) Also, using the definition (2.6), and the equation (2.2) we get Δ(ξ) = n (A)(e i,e i ) (3.6) Thus using the equations (3.1), (3.5) and (3.8), we conclude

A note on cohomology of a Riemannian manifold 55 Q (ξ) = λξ α and taking the inner product with ξ in the above equation, we get Ric (ξ.ξ) = λ ξ 2 ξ (α) (3.7) Note that the smooth 1- form η dual to ξ is closed and as the singular cohomology H 1 (,R) is isomorphic to the derham cohomology group HdR 1 (), we have HdR 1 () = 0 and consequently λ is exact, that is there exists a smooth function ρ : R such that η = dρ. Hence, we get ξ = ρ. We know by (2.2) AX = X ξ = X ρ, X X (), that is A is the Hessian operator of the smooth function ρ. Then by Bochner formula gives (Ric ( ρ, ρ)+ A 2 (Δρ) 2 )=0 (3.8) Note that Δρ = div ( ρ) = div ξ = α, where we used equation (3.3). Thus the equations (3.7, ), (3.8) and ρ = ξ give (λ ξ 2 + ξ (α)+α 2 A 2 )=0, which together with the equation (3.4) gives ( A 2 λ ξ 2 )=0. The above equation could be re -arranged as ( A 2 1 n α2 )+ 1 ( α 2 nλ ξ 2) =0 (3.9) n Not that the Shwarz s inequality for the symmetric operator A states that A 2 1 n (tra)2 = 1 n α2 and the equality holds if and only if A = α I. oreover, the inequality (3.2) n gives α 2 = (div ξ) 2 nλ ξ 2 consequently, both terms in (3.9) are non - negative and we have

56 Tahsin Ghazal A 2 = 1 n α2 and α 2 = nλ ξ 2 (3.10) The equation is the equality in the Shwarz s inequality and hence we have A = α I.Hence equation (2.2) becomes n X ξ = fx, X X (), where f = α n vector field. is a smooth function on. This proves that ξ is a concircular References [1] Kim, I. B., Special concircular vector fields in Riemannian manifolds, Hiroshima ath. J. 12(1982), 77-91. [2] Shandra, I.G., Concircular vector fields on Semi-Riemannian spaces, J. ath. Sci. 142(5)(2007), 2419-2435. [3] Tashiro, Y., Complete Riemannian manifolds and some vector fields, Trans. AS, 117(1965), 251-275. [4] Yano, K. Concircular geometry, Proc. Imp. Acad. Tokyo, 16, 195 200 (1940). Received: November 11, 2013