# Certain conformal-like infinitesimal symmetries and the curvature of a compact Riemannian manifold.

1 Introduction

Relating the existence of geometrically relevant vector fields on a Riemannian manifold with its curvature properties is a classical topic in Differential Geometry. Recall for instance the well-known Bochner's technique ( and references therein). In this note, we introduce a new class of geometrically relevant vector fields on Riemannian manifolds and obtain an integral inequality under the assumption of the existence of such vector fields. The new notion has been inspired by the behavior of the unit vector field tangent to the parallels of a surface of revolution in the 3-dimensional Euclidean space. A flow of such a vector field provides a linear isometry on its orthogonal complement at any point but, in general, this vector field fails to be Killing unless the surface is a cylinder (see details in Example 1). This situation can be considered as a particular case of a Riemannian submersion with 1-dimensional fibers. In this case, each (local) flow of the unit vertical vector field gives linear isometries on horizontal spaces of the submersion. But the unit vertical vector field of a such Riemannian submersion is Killing if and only if the fibers are totally geodesies (Example 7). On the other hand, if we consider a conformal (resp. Killing) vector field X on a Riemannian manifold (M,g), i.e. the Lie derivative with respect to X of g satisfies [L.sub.x]g = 2[rho]g (resp. [L.sub.x]g = 0), and we assume that X has no zeroes, then Z = (1/[square root of g(X, X))] X satisfies ([L.sub.x]g)(U,V) = 2[rho]g(U,V) (resp. ([L.sub.x]g)(U,V) = 0), for all U,V [perpendicular to] Z. Therefore, any of the flows [[PHI].sub.t] of the vector field Z induces a linear conformal (resp. isometry) mapping from [Z.sup.[perpendicular to].sub.p] onto [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Thus, it seems natural to introduce the following notion: an orthogonally conformal vector field is a unit vector field Z on a Riemannian manifold, (M,g), which satisfies

([L.sub.x]g)(U,V) = 2[rho]g(U,V), (1)

for all U,V [perpendicular to] Z. In particular, when [rho] = 0, we will call Z orthogonally Killing. This class of vector fields properly includes the normalized vector fields of the nowhere zero conformal vector fields (see Section 2). That is, there exist unit vector fields which satisfy (1) but they cannot be obtained from a nowhere zero conformal vector field. Even more, there exists a family of compact Riemannian manifolds such that each one of its members admits a unit vector field satisfying (1) although no conformal vector field is free of zeroes (Example 6).

The main aim of this note is to relate the existence of such orthogonal conformal vector fields on a compact Riemannian manifold with its curvature properties. So, we will obtain a vanishing result in the spirit of the well-known Bochner's technique as follows.

Theorem 1. Let (M,g) be an n([greater than or equal to] 3)-dimensional compact Riemannian manifold. If (M,g) admits an orthogonally conformal vector field Z, then

[[integral].sub.M] Ric(Z,Z)d[[mu].sub.g] [greater than or equal to] 0,

where Ric denotes the Ricci tensor of (M, g) and d[[mu].sub.g] is the canonical measure induced by the Riemannian metric g. Moreover, the equality holds if and only if [[nabla].sub.u]Z = 0 for any U [perpendicular to] Z, and in such case, Z is orthogonally Killing.

As an application, we obtain two consequences.

Corollary 1. If a compact Einstein Riemannian manifold (M,g), with dim M [greater than or equal to] 3 and Ric = [[lambda].sub.g], admits an orthogonally conformal vector field, then [lambda] [greater than or equal to] 0.

The second one can be regarded as a Wu-type result  for orthogonally conformal vector fields.

Corollary 2. If the Ricci tensor of a n([greater than or equal to] 3)-dimensional compact Riemannian manifold, (M,g), is negative semi-definite everywhere and negative definite at some p [member of] M, then (M,g) admits no orthogonally conformal vector field.

2 Examples

It should be recalled that the existence of a nowhere zero vector field on a manifold M imposes some restrictions on its topology. In fact, it holds if and only if M is noncompact or its Euler-Poincare number vanishes. If dimM = 2, then every unit vector field is orthogonally conformal. Therefore, when dimM = 2, (M, g) admits an orthogonally conformal vector field if, and only if, either M is noncompact or M is compact and its Euler-Poincare characteristic vanishes. We would like to point out that if Z is an orthogonally conformal vector field for a Riemannian metric g, then the vector field (1/[square root of [g.sup.1](Z/Z))] Z is also orthogonally conformal for every Riemannian metric [g.sup.1] conformally related to g.

Example 1. Let x{v) and z(v) be smooth functions on an open interval] a,b [with x(v) > 0 for every v [member of]]a,b[. Let S be the surface of revolution in the Euclidean space [E.sup.3] with rotation axis z obtained from the above data. Thus, a parametrization for S is given by

[mu](u,v) = (x(v) cosu, x(v) sinu, z(v)).

We consider the unit vector field tangent to parallels, Z [member of] X(S), given by

Z([mu](u,v)) = 1/x(v)[partial derivative][mu]/[partial derivative]u [|.sub.(u,v)].

From the well-known formulas for the Christoffel symbols of a surface of revolution, we deduce that Z is orthogonally Killing. If we suppose that Z is Killing then Z is a geodesic vector field. Therefore all the parallels of the surface of revolution S are geodesies and so x is constant, and S is a cylinder.

Example 2. Let ([T.sup.2],g) be a 2-dimensional flat Riemannian torus and let Z be a unit vector field on [T.sup.2] which is not parallel. We assert that the orthogonally conformal vector field Z is not the normalization of a conformal vector field K. Indeed, if such K exists, then as a consequence of the classical result by H. Wu , which improved a previous one by K. Yano , the conformal vector field K is parallel. Hence Z must be also parallel. This contradicts our choice of Z.

Example 3. Let (I, [dt.sup.2]) be an open interval endowed with its standard metric and let (M, g) be a Riemannian manifold. Given a smooth function f > 0 defined on [??] = I x M, we consider on [??] the twisted metric [??] = [[pi].sup.*.sub.I][dt.sup.2] + [f.sup.2][[pi].sup.*.sub.M]g, where [[pi].sub.I] and [[pi].sub.M] are the corresponding projections onto I and M, respectively. The vector field [[partial derivative].sub.t] on [??] is orthogonally conformal. Indeed, given U, V [perpendicular to] [[partial derivative].sub.t], clearly [[[partial derivative].sub.t], U] = [[[partial derivative].sub.t], V] = 0 and a direct computation shows that the Lie derivative of [??] with respect to [[partial derivative].sub.t] satisfies

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

This computation also holds if (I, [dt.sup.2]) is replaced by the 1-sphere ([S.sup.1], d[[theta].sup.2]). Therefore, for M compact, we obtain orthogonally conformal vector fields on compact Riemannian manifolds.

Remark 1. In Example 3, if the function f only depends on t then [??] is in fact the warped product of [dt.sup.2] and g with warping function f. In this case, it is not difficult to show that [[nabla].sub.X](f[[partial derivative].sub.t]) = f'X for any vector field X on [??], [3, Prop. 7.35]. Therefore, f[[partial derivative].sub.t] is conformal with [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and [[partial derivative].sub.t] is the normalized vector field obtained from f[[partial derivative].sub.t]. Clearly, the function [[partial derivative].sub.t](log(f)) is now constant along each slice {t} x M.

Example 4. In order to show examples of orthogonally conformal vector fields which cannot be obtained from nowhere zero conformal vector fields, we add in Example 3 the condition that the twisting function f satisfies that [[partial derivative].sub.t] (log(f)) is not constant along each slice {t} x M. Suppose now that [[partial derivative].sub.t] is the normalization of a nowhere zero conformal vector field, i.e. for a certain smooth function h > 0 defined on [??], the vector field h[[partial derivative].sub.t] is conformal. Given U [perpendicular to] [[partial derivative].sub.t], direct computations show

U(h) = 0 and [[partial derivative].sub.t](log(h)) = [[partial derivative].sub.t](log(f)).

Therefore,

U([[partial derivative].sub.t](log(f))) = U([[partial derivative].sub.t](log(h))) = [[partial derivative].sub.t] (U(log(h))) = 0,

which implies that [[partial derivative].sub.t] (log(f)) only depends on t, contrary to our assumption.

Example 5. Let us consider the usual Hopf fibration [pi] : [S.sup.2n+1] [right arrow] [C.sup.pn] from the odd dimensional unit sphere [S.sup.2n+1] = {([z.sub.1], ..., [z.sub.n+1]) [member of] [C.sup.n+1]: [[summation].sup.n+1.sub.j=1][[absolute value of [Z.sub.j]].sup.2] = 1} onto the complex projective space [C.sup.pn] endowed with its Fubini-Study metric [g.sub.FS] of constant holomorphic sectional curvature 4. For each u [member of] [C.sup.[infinity]]([S.sup.2n+1]) we set [g.sub.u] = [e.sup.u] [[pi].sup.*] [g.sub.FS] + [omega][cross product][omega], where [omega] to is the 1-form naturally obtained from the usual connection on this fibre bundle. Each [g.sub.u] is a Riemannian metric on [S.sup.2n+1]. The Hopf vector field Z [member of] X([S.sup.2n+1]), given by z [??] iz, satisfies [g.sub.u](Z,Z) = 1 and

([L.sub.z][g.sub.u])(U,V) = Z(u)[g.sub.u](U,V), U,V [perpendicular to] Z.

Therefore Z is orthogonally conformal on every ([S.sup.2n+1], [g.sub.u]). It can be deduced that Z is not conformal whenever Z(u)[not equal to] 0. Note that in this family of examples the distribution [Z.sup.[perpendicular to] is not integrable.

Example 6. Even more, we will construct a compact Riemannian manifold which admits an orthogonally conformal vector field but does not admit a nowhere zero conformal one. We consider an n([greater than or equal to] 3)-dimensional compact Riemannian manifold (M,g) whose Ricci tensor is negative definite (recall that on any n([greater than or equal to] 3)dimensional compact manifold M, there always exists a Riemannian metric g with everywhere negative definite Ricci tensor .) Next, we construct the compact manifold [??] = [S.sup.1] x M endowed with a twisted metric [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and suppose that [[partial derivative].sub.[theta]](log(f)) is not constant along each slice {[e.sup.i[theta]]} x M. As Example 4 shows, [[partial derivative].sub.[theta]] is orthogonally conformal and cannot be obtained by normalizing a nowhere zero conformal vector field. Let [??] [member of] X([??]) be a conformal vector field with [L.sub.[??]][??] = 2[rho][??]. Put K = [??] - [??]([??],[[partial derivative].sub.[theta]]) [[partial derivative].sub.[theta]], then we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and

([L.sub.K][??])(U,V) = 2[sigma](U,V),

for any U, V [perpendicular to] [[partial derivative].sub.[theta]], where [sigma] = [rho] - [??]([??], [[partial derivative].sub.[theta]]) [[partial derivative].sub.[theta]](log(f)). Therefore, for every [theta] we deduce that the vector field [K.sub.[theta]] [member of] X(M) defined by [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is conformal. From our assumption on the Ricci tensor of (M,g) and as a consequence of the classical result by H. Wu  we obtain that [K.sub.[theta]] = 0 for every [theta]. Therefore K = 0, which means that [??] must be proportional to [[partial derivative].sub.[theta]]. In case [??] has no zero, then [[partial derivative].sub.[theta]] should be its normalization.

Example 7. Let [pi] : (M,g) [right arrow] (B,g') be a Riemannian submersion with 1-dimensional fibers. The fundamental tensor fields of [pi] (O'Neill tensors) are given by

[T.sub.E]F = h([[nabla].sub.v(E)] v (F)) + v ([[nabla].sub.v(E)]h(F))

[A.sub.E]F = v([[nabla].sub.h(E)]h(F)) + h([[nabla].sub.h(E)]v(F]),

where E,F [member of] X(M) and v (resp. h) denotes the vertical (resp. horizontal) projection . A unit vertical vector field W [member of] X(M) is orthogonally Killing. In fact, let X, Y [member of] X(M) be horizontal vector fields then,

([L.sub.w]g)(X,Y) = -g([A.sub.x]Y, W) - g([A.sub.Y]X, W) = 0,

since the fundamental tensor A of [pi] satisfies [A.sub.x]Y = -[A.sub.y]X, . Note that VV is a Killing vector field if and only if the fundamental tensor T vanishes identically, that is, the fibers are totally geodesies.

3 Proofs of the results

We begin this section by recalling a formula which is our main tool. Let us remark that our approach has been inspired from the recent application of Bochner technique to Lorentzian Geometry ,  and . Let (M,g) be a Riemannian manifold and let X be any vector field on M. If [nabla] denotes the Levi-Civita connection of g, then we have a (1,1)-tensor field [A.sub.x] on M given by [A.sub.x] (v) = -[[nabla].sub.v]X, for any v [member of] [T.sub.p]M, p [member of] M. Note that trace [A.sub.x] = -div(X). If M is compact, then the following integral formula holds true,

[[integral].sub.M]{Ric(X,X) + trace([A.sup.2.sub.X]) - [(trace [A.sub.x]).sup.2]} d[[mu].sub.g] = 0. (2)

Proof. Theorem 1 If Z is an orthogonally conformal vector field, we have

g([A.sub.z]U, V) + g(U, [A.sub.Z]V) = -2[rho]g(U, V), U,V [perpendicular to] Z. (3)

Direct computations show

trace [A.sub.z] = -[rho](n - 1) and trace([A.sup.2.sub.Z]) = 2[[rho].sup.2](n - 1) - [parallel][nabla]Z[[parallel].sup.2] + [[parallel].sub.Z]Z[[parallel].sup.2]. (4)

Observe that [parallel][nabla]Z[[parallel].sup.2] [greater than or equal to] [parallel][[nabla].sub.Z][[parallel].sup.2], and equality holds if, and only if, [[nabla].sub.U]Z = 0 for any U [perpendicular to] Z, and in this case [rho] = 0. Now, making use of (4) in (2) we get

[[integral].sub.M] Ric(Z,Z)d[[mu].sub.g] = (n - 1)(n-3) [[integral].sub.M] [[rho].sup.2] d[[mu].sub.g] + [[integral].sub.M] {[parallel][nabla]Z[[parallel].sup.2] - [parallel][[nabla].sub.Z][[parallel].sup.2]} d[[mu].sub.g], (5)

which concludes the proof.

Remark 2. If a unit vector field Z on a Riemannian manifold (M,g) satisfies [[nabla].sub.U]Z = 0 for any U [perpendicular to] Z (hence Z is orthogonally Killing) then the distribution [Z.sup.[perpendicular to]] is integrable with totally geodesic leaves.

Remark 3. When dimM = 2, trace([A.sup.2.sub.Z]) = [(trace [A.sub.Z]).sup.2] and Ric(Z,Z) = K, the Gaussian curvature of (M,g), for any unit vector field Z on (M,g). Hence, in the compact case, equation (2) reduces to[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], which is the well known Gauss-Bonnet theorem for a Riemannian metric on the torus [T.sup.2].

Proof. Corollary 1 If (M, g) admits an orthogonally conformal vector field Z, from Theorem 1 we get 0 [less than or equal to] [[integral].sub.M] Ric(Z, Z)d[[mu].sub.g] = [lambda]vol(M,g), and therefore [lambda][greater than or equal to] 0.

Proof. Corollary 2 Suppose that (M, g) admits an orthogonally conformal vector field Z. From Theorem 1 and the assumption on the Ricci tensor, we deduce that Ric(Z, Z) = 0 on the whole M. Finally, being the Ricci tensor negative definite at a point p [member of] M, we obtain [Z.sub.p] = 0 which is a contradiction. ?

Acknowledgements The authors are grateful to the referee for their deep reading and suggestions toward the improvement of this article.

References

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 A. Romero, The introduction of Bochner's technique on Lorentzian manifolds, Nonlinear Anal. 47 (2001), 3047-3059.

 A. Romero and M. Sanchez, An integral inequality on compact Lorentz manifolds and its applications, Bull. London Math. Soc. 28 (1996), 509-513.

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Miguel Ortega Francisco J. Palomo Alfonso Romero *

* All authors partially supported by the Spanish MICINN Grant MTM2010-18099 and the Junta de Andalucia Regional Grant P09-FQM-4496 with FEDER funds.

Received by the editors December 2009 - In revised form in March 2010. Communicated by L. Vanhecke.

2000 Mathematics Subject Classification : 53C50,57R25.

Departamento de Geometna y Topologfa,

E-mails: aromero@ugr.es, miortega@ugr.es

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