Critical point equation on almost f-cosymplectic manifolds

1. Introduction

One of the natural ways of finding canonical Riemannian metric, that is, Riemannian metrics with constant curvature in various form on a smooth manifold is to look for metrics which are critical points of a natural functional on the space of all metrics on a given manifold. In this context, it is very interesting to investigate the critical points of total scalar curvature functional S:M→R given by

Let C⊂M be the subset of metrics with constant scalar curvature. If we consider the functional in Eqn (1.1) restricted to C, then it is not difficult to see that the Euler–Lagrangian equation is given by,

We notice that if λ is constant in Eqn (1.2), then λ = 0 and g becomes Einstein. Therefore, we have the following definition:

Besse first conjectured that the solution of the CPE must be Einstein [1]. Since then, we find many articles regarding the solution of the CPE. In [2], Barros and Ribeiro proved that the CPE conjecture is true under the assumption of half conformally flat spaces. Recently, Hwang [3] proved that the CPE conjecture is also true under certain condition on the bounds of the potential function λ. A necessary and sufficient condition for the norm of the gradient of the potential function for a CPE metric to be the Einstein metric was obtained by Neto [4].

It is very interesting to consider the CPE on odd-dimensional Riemannian manifolds. In this direction, Ghosh and Patra considered the K-contact metrics that satisfy the CPE [5], and proved that the CPE conjecture is true for this class of metric. Patra et al. in [6], and De and Mandal in [7] independently considered an almost Kenmotsu manifold with CPE. Recently, present authors in [8], and Blaga and Dey in [9] studied CPE on cosymplectic manifold and three dimensional α-cosymplectic manifold, respectively.

As the generalization of almost Kenmotsu and almost cosymplectic manifolds, the results obtained in [6–9] motivates us to consider almost f-cosymplectic manifolds. In this paper, we classify an almost f-cosymplectic manifold which satisfies CPE.

2. Preliminaries

Let M be a smooth differentiable manifold of dimension 2n + 1 equipped with a triple (ϕ, ξ, η), where ϕ is a (1, 1)-tensor field, ξ is a Reeb vector and η is a one-form such that

An almost contact metric manifold M is said to be almost cosymplectic if dη = 0 and dΦ = 0, where d is the exterior differential operator, and it is said to be cosymplectic if in addition the almost contact structure is normal. An almost α-Kenmotsu manifold is an almost contact metric manifold, in which dη = 0 and dΦ = 2αη ∧ Φ, for a nonzero constant α. More generally, if the constant α is any real number, then almost contact structure is said to be almost α-cosymplectic [10]. Moreover, the authors in [11] generalizes the almost α-cosymplectic manifold by allowing the real number α to any smooth function f, and it is called as an almost f-cosymplectic manifold, which is an almost contact metric manifold M such that dΦ = 2fη ∧ Φ and dη = 0 for a smooth function f satisfying df ∧ η = 0. In addition, a normal almost f-cosymplectic manifold is said to be f-cosymplectic manifold. In particular, M is an almost cosymplectic manifold under the condition f(constant) = 0 and an almost α-Kenmotsu manifold if (α = f ≠ 1).

Besides, we recall that there is an operator h=12 £_ξ ϕ, which is a self-dual operator. We denote by R and Ric the Riemannian curvature tensor and Ricci tensor, respectively. For an almost f-cosymplectic manifold M, the following equations were proved [11]:

3. CPE on normal almost f-cosymplectic manifolds

In this section, we aim to study CPE on normal almost f-cosymplectic manifold. We are aware that if almost contact metric manifold is normal then h = 0. Hence, as a result of Proposition 9 and Proposition 10 of [11] we have the following identities, which are valid on f-cosymplectic manifolds;

Now, we will give some properties, which will be used in the proof of our results.

Proof. Differentiation of Eqn (3.2), and utilization of first term of Eqn (3.1) provides Eqn (3.4). Now differentiating Eqn (3.3) along Z leads to

Taking X = Z = E_i in the above equation and then summing over i shows that

Feeding Eqn (3.7) into Eqn (3.6) and with the help of Eqn (3.4), we obtain

In the following, we will consider an f-cosymplectic manifold M satisfying a CPE and assume that the function f satisfies ξ(f̃)=0.

Proof. Taking scalar product of Eqn (3.8) with ξ and making use of Eqns (3.2) and (3.3), we obtain

Replacing X by ϕX and Y by ξ in above relation, we get

According to Proposition 2.1 of Chen [12], it is know that if (ξf̃)=0, then f̃ is constant. So that, Eqn (3.9) implies

The scalar curvature r of g is constant (as (g, λ) is a solution of the CPE). For a (2n + 1)-dimensional f-cosymplectic manifold, we have ν=−r(λ2n+12n+1), therefore from Eqn (3.10) it appears that

From Eqn (3.11), we have either r=−2n(2n+1)f̃ or ϕDλ = 0.

First suppose that r=−2n(2n+1)f̃, then we have Dν=−(2n+1)f̃Dλ. Plugging X = ξ in Eqn (3.8) and calling back Lemma 3.1, we aimed at obtaining

From Eqn (3.3), we deduce R(X,ξ)Y=f̃{g(X,Y)ξ−η(Y)X}, by virtue of this the foregoing equation reduces to

Making use of Dν=−(2n+1)f̃Dλ in Eqn (3.12) we reach at

Since ∇_ξξ = 0 and (ξλ) = g(ξ, Dλ), taking into account ∇_XDλ = (λ + 1)QX + νX, we deduce ξ(ξλ)=f̃λ. If possible, let (ξλ) = f(λ + 1) in some open set O of M, then we have f̃λ=((ξf)+f2)(λ+1). By virtue of f̃=(ξf)+f2, one can see λ = λ + 1, that is, 1 = 0, which is absurd. Hence QX=−2nf̃X and M is Einstein.

Next we assume r≠−2n(2n+1)f̃, then from Eqn (3.11) we have ϕDλ = 0. Action of ϕ on this equation gives Dλ = (ξλ)ξ. Differentiating this along X, calling back Eqn (3.1) furnishes

On the other hand, from Eqn (1.2) we can easily find that

Comparing aforementioned equation with Eqn (3.13), we get

Taking X = ξ in the above equation and making use of Eqns (3.2) and (2.1), we obtain

Contraction of Eqn (3.13) with respect to X brings into view

Unifying this with Eqn (3.15) implies

Differentiating Eqn (3.17) along ξ, keeping in mind that f̃ and r are constants, we obtain ξ(ξλ)f+(ξλ)(ξf)=f̃(ξλ), and further, it implies

If f ≢ 0, then we can assume f ≠ 0 on some neighborhood O of M. Thus, Eqn (3.18) implies ξ(ξλ) = (ξλ)f on O. Inserting this into Eqn (3.16), we find Δλ = (2n + 1)f(ξλ). Moreover, applying Eqn (3.17) in the previous relation shows that

Taking trace of CPE (1.2), we obatin 2nΔλ = −λr, and this together with Eqn (3.19) gives that r=−2n(2n+1)f̃, which is contradictory to our assumption. Hence f ≢ 0, and so M is cosymplectic. According to Blair's [13] result, we can easily conclude that M is locally the product of a Kähler manifold and an interval or unit circle S1. This finishes the proof.□

In particular, when dimension of M is three, due to Theorem 3.3 we have the following outcome:

It is known that an α-cosymplectic manifold is actually an f-cosymplectic manifold with f constant. By the reason of this, we obtain the following conclusion from Theorem 3.3.

Now we consider CPE with λ=f̃, and obtain the following result.

Proof. One can easily obtain from Eqn (3.9) that

Suppose that (4n+1)λ+2n+r2n ≢ 0. Due to constancy of r, we see that λ is constant. Next, we assume that (4n+1)λ+2n+r2n ≢ 0$ in a neighborhood O of M. Consequently, one can gets ϕDλ = 0. Applying ϕ to this equation implies Dλ = (ξλ)ξ. In this context (3.13) holds, from which we can get

Differentiating this along ξ gives (2λ + 1)(ξλ) = ξ(ξλ)f + (ξλ)(ξf), due to our assumption λ=f̃=(ξf)+f2 which further implies

Suppose that f = 0, then from Eqn (3.20), we have λ(λ+1)+r2n(2n+1)=0, which means that λ is constant. In the following we suppose f ≠ 0, then as a result of Eqns (3.16), (3.20) and (3.21), we find

Substitute this into 2nΔλ = −λr to obtain

Differentiating the aforesaid relation along ξ, remembering r is constant and applying Eqn (3.21), we reach at

If (ξλ) = 0, then we have Dλ = 0, which means λ is constant. Suppose (ξλ) ≠ 0, then we get f2=12n+3. Due to f ≠ 0, which shows (ξf) = 0. This together with λ=f̃=(ξf)+f2 yields λ = f², showing λ constant. In a word, we have proved that λ is always constant in the neighborhood O of M, thus λ = constant in M. Hence, the proof completes from Eqn (1.2).

4. CPE on non-normal almost f-cosymplectic manifolds

Here, we consider a three dimensional almost f-cosymplectic manifold M with pseudo anti‐commuting Ricci tensor, that is,

This notion was introduced by Jeong and Suh [14], and they made use of this condition to classify a real hypersurface of complex two-plane Grassmannians.

At first, we have the following lemma:

Let U be the open subset where the tensor h ≠ 0 and U′ be the open subset such that h is identically zero. Thus, U∪U′ is open dense in M. There exists a local orthonormal frame field E = {ξ, e, ϕe} such that he = μe and hϕe = −μϕe, where μ is a positive nonvanishing smooth function of M. The following proposition is obtain from Proposition 12 and Proposition 14 of Öztürk et al. [10]:

From this onwards, we assume that a three dimensional almost f-cosymplectic manifold M satisfies the CPE Eqn (1.2), then its scalar curvature r is constant. As a result of Lemma 4.1, it can be seen that a is constant. Chen in [15] obtained the relation (r−2κ−2f̃−2a)ϕX=2ϕh2X. From this we have (−2f̃−a)ϕX=2ϕh2X, that is, 2h2X=(2f̃+a)ϕ2X as hξ = 0. Further, from Eqn (4.1) we find (ξf) = 0, due to df ∧ η = 0 we obtain f is constant. Moreover, we can also find a = −2(f² + μ²) from Eqn (4.1). Thus μ is constant. In view of Lemma 2 of [10], we have the following (also see [15]):

In view of Eqn (4.2), the relation Eqn (2.5) implies

By virtue of Eqn (3.8), we obtain

Substituting the above relation into Eqn (4.4), we obtain

Employing X by ϕX in Eqn (4.6) we reach at

In orthonormal frame field E, the gradient vector field Dλ can be written as

Thus from Eqn (4.7), one can obtain

First we assume f ≠ 0, because of f is constant and we shall divide this discussion into two cases:

Case 1. If (eλ) = 0, then from Eqn (4.10) we can observe (ϕeλ) = 0. This together with (4.8) yields Dλ = (ξλ)ξ. Differentiating this along X, using Eqn (2.2) gives

Employing X = ξ in the above equation and remembering ∇ξDλ=(λ+1)Qξ+(Δλ−r3)ξ, we aimed at obtaining

One can find from Eqn (4.11) and second term of Eqn (2.2) that

By virtue of the foregoing relation, Eqn (4.12) transforms into (ξλ)=r6f−(λ+1)a2f, which further gives

If a24f2+r2−a=0, then we have ra12f2−r3−a24f2+a=0, which further shows that r6(a2f2+1)=0. The former case implies that the scalar curvature of M vanishes identically. In the latter case, we have a = −2f², which together with a = −2(f² + μ²) implies μ = 0, which is not possible. Next suppose a24f2+r2−a≠0, then from Eqn (4.14) it can easily conclude that λ is constant.

Case 2. If (eλ) ≠ 0 on a neighborhood O of M, then from (4.9) and (4.10) we extract

From the preceding equation, we can easily observe that b is constant because of μ and f are constant. It is easy to seen from Eqns (4.9) and (4.10) that (eλ) and (ϕeλ) are constants in O.

By the support of Eqn (4.3), we may easily compute that

Thus Δλ = 2f(ξλ) + ξ(ξλ). So, utilization of ∇ξDλ=(λ+1)Qξ+(Δλ−r3)ξ followed from (3.14), shows that

As followed by Case 1, we can conclude that either r vanishes or λ is constant.

Next we assume f = 0, then from Eqns (4.9) and (4.10) we find b(eλ)(ϕeλ) = 0 as μ > 0, which further implies either (eλ)(ϕeλ) = 0 or b = 0. We shall also discuss this matter into two cases.

Subcase i. If b = 0, then from Eqn (4.10) we find r = 3a. For three dimensional case, it is known that the Riemannian curvature is

From this, we have

This together with Eqn (4.4) yields QX = aX, which means M is Einstein.

Subcase ii. If b ≠ 0 on some neighborhood O of M, then we find (eλ)(ϕeλ) = 0 on O. If possible, let (eλ) = 0 = (ϕeλ), then from Eqn (4.8) we obtain Dλ = (ξλ)ξ. From this it is not hard to see that Eqn (4.12) holds and Eqn (4.13) implies Δλ = ξ(ξλ). Thus Eqn (4.12) transforms into (λ+1)a−r3=0, which means λ is constant on O.

If (eλ) = 0 and (ϕeλ) ≠ 0, then from Eqns (4.3) and (4.8) we compute

Utilization of above relation in ∇ξDλ=(Δλ−r3)ξ+(λ+1)aξ, we find b(ϕeλ) = 0. Because of b ≠ 0, we have (ϕeλ) = 0, which is a contradiction. In a similar manner, we also come to contradiction if we consider (eλ) ≠ 0 and (ϕeλ) = 0.

From the above detailed discussion, we have concluded that M is Einstein or its scalar curvature vanishes, and so we state following result: