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Article

Trans-Sasakian 3-Manifolds with Reeb Flow Invariant Ricci Operator

1
Department of Mathematics, College of Science, Henan University of Technology, Zhengzhou 450001, Henan, China
2
Wenjie Wang, School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, Liaoning, China
*
Author to whom correspondence should be addressed.
Mathematics 2018, 6(11), 246; https://doi.org/10.3390/math6110246
Submission received: 19 September 2018 / Revised: 3 November 2018 / Accepted: 6 November 2018 / Published: 9 November 2018
(This article belongs to the Special Issue Differential Geometry)

Abstract

:
Let M be a three-dimensional trans-Sasakian manifold of type ( α , β ) . In this paper, we obtain that the Ricci operator of M is invariant along Reeb flow if and only if M is an α -Sasakian manifold, cosymplectic manifold or a space of constant sectional curvature. Applying this, we give a new characterization of proper trans-Sasakian 3-manifolds.

1. Introduction

A trans-Sasakian manifold is usually denoted by ( M , ϕ , ξ , η , g , α , β ) , where both α and β are smooth functions and ( ϕ , ξ , η , g ) is an almost contact metric structure. M is said to be proper if either α = 0 or β = 0 . When β = 0 , α is a constant if dim M 5 (see [1]) and in this case M becomes an α -Sasakian manifold if α R * or a cosymplectic manifold if α = 0 . This conclusion is not necessarily true for dimension three. However, unlike the above case, when α = 0 , β is not necessarily a constant even if dim M 5 or M is compact for dimension three (see [2]). The set of all trans-Sasakian manifolds of type ( 0 , β ) coincides with that of all f-cosymplectic manifolds (see [3]) or f-Kenmotsu manifolds (see [4,5,6]). A trans-Sasakian manifold of dimension 5 must be proper (see [1]). In the geometry of trans-Sasakian 3-manifolds, there exists a basic interesting problem, that is:
Under what condition is a trans-Sasakian 3-manifold proper?
De [7,8,9,10,11,12], Deshmukh [13,14,15], Wang and Liu [16] and Wang [2,17] answered this question from various points of view. In this paper, we study this question under a new geometric condition. Before stating our main results, we recall some results related with such a condition.
On an almost contact metric manifold ( M , ϕ , ξ , η , g ) , the Ricci operator of M is said to be Reeb flow invariant if it satisfies
L ξ Q = 0 ,
where L , ξ and Q are the Lie derivative, Reeb vector field and the Ricci operator, respectively. Cho in [18] proved that a contact metric 3-manifold satisfies Equation (1) if and only if it is Sasakian or locally isometric to S U ( 2 ) (or S O ( 3 ) ), S L ( 2 , R ) (or O ( 1 , 2 ) ), the group E ( 2 ) of rigid motions of Euclidean 2-plane. Cho in [19] proved that an almost cosymplectic 3-manifold satisfies (1) if and only if it is either cosymplectic or locally isometric to the group E ( 1 , 1 ) of rigid motions of Minkowski 2-space. In addition, Cho and Kimura in [20] proved that an almost Kenmotsu 3-manifold satisfies (1) if and only if it is of constant sectional curvature 1 or a non-unimodular Lie group. Reeb flow invariant Ricci operators were also investigated on the unit tangent sphere bundle of a Riemannian manifold (see [21]), even on real hypersurfaces in complex two-plane Grassmannians (see [22]). In this paper, we obtain a new characterization of proper trans-Sasakian 3-manifolds by employing (1) and proving
Theorem 1.
The Ricci operator of a trans-Sasakian 3-manifold is invariant along Reeb flow if and only if the manifold is an α-Sasakian manifold, cosymplectic manifold or a space of constant sectional curvature.
According to calculations shown in Section 3, we observe that Ricci parallelism with respect to the Levi–Civita connection (i.e., Q = 0 ) is stronger than a Reeb flow invariant Ricci operator. Thus, we have
Remark 1.
Theorem 1 is an extension of Wang and Liu [16] (Theorem 3.12).
Some corollaries induced from Theorem 1 are also given in the last section.

2. Trans-Sasakian Manifolds

On a smooth Riemannian manifold ( M , g ) of dimension 2 n + 1 , we assume that ϕ , ξ and η are ( 1 , 1 ) -type, ( 1 , 0 ) -type and ( 0 , 1 ) -type tensor fields, respectively. According to [23], M is called an almost contact metric manifold if
ϕ 2 X = X + η ( X ) ξ , η ( ξ ) = 1 , η ( ϕ X ) = 0 , g ( ϕ X , ϕ Y ) = g ( X , Y ) η ( X ) η ( Y ) , η ( X ) = g ( X , ξ )
for any vector fields X and Y. An almost contact metric manifold is said to be normal if [ ϕ , ϕ ] = 2 d η ξ , where [ ϕ , ϕ ] denotes the Nijenhuis tensor of ϕ .
A normal almost contact metric manifold is called a trans-Sasakian manifold (see [1]) if
( X ϕ ) Y = α ( g ( X , Y ) ξ η ( Y ) X ) + β ( g ( ϕ X , Y ) ξ η ( Y ) ϕ X )
for any vector fields X , Y and two smooth functions α , β . In particular, a three-dimensional almost contact metric manifold is trans-Sasakian if and only if it is normal (see [24,25]).
A normal almost contact metric manifold is called an α-Sasakian manifold if d η = α Φ and d Φ = 0 , where α R * (see [26]). An α -Sasakian manifold reduces to a Sasakian manifold (see [23]) when α = 1 . A normal almost contact metric manifold is called a β-Kenmotsu manifold if it satisfies d η = 0 and d Φ = 2 β η Φ , where β R * (see [26]). A β -Kenmotsu manifold becomes a Kenmotsu manifold when β = 1 . A normal almost contact metric manifold is called a cosymplectic manifold if it satisfies d η = 0 and d Φ = 0 .
Putting Y = ξ into (3) and using (2), we have
X ξ = α ϕ X + β ( X η ( X ) ξ )
for any vector field X. In this paper, all manifolds are assumed to be connected.

3. Reeb Flow Invariant Ricci Operator on Trans-Sasakian 3-Manifolds

In this section, we give a proof of our main result Theorem 1. First, we introduce the following two important lemmas (see [12]) which are useful for our proof.
Lemma 1.
On a trans-Sasakian 3-manifold of type ( α , β ) we have
ξ ( α ) + 2 α β = 0 .
Lemma 2.
On a trans-Sasakian 3-manifold of type ( α , β ) , the Ricci operator is given by
Q = r 2 + ξ ( β ) α 2 + β 2 id r 2 + ξ ( β ) 3 α 2 + 3 β 2 η ξ + η ( ϕ ( α ) β ) + g ( ϕ ( α ) β , · ) ξ ,
where by f we mean the gradient of a function f.
We also need the following lemma (see [17])
Lemma 3.
On a trans-Sasakian 3-manifold of type ( α , β ) , the following three conditions are equivalent:
(1) 
The Reeb vector field is minimal or harmonic.
(2) 
The following equation holds: ϕ α β + ξ ( β ) ξ = 0 (⇔ α + ϕ β + 2 α β ξ = 0 ).
(3) 
The Reeb vector field is an eigenvector field of the Ricci operator.
Lemma 4.
The Ricci operator on a cosymplectic 3-manifold is invariant along the Reeb flow.
The above lemma can be seen in [19]
Lemma 5.
The Ricci operator on an α-Sasakian 3-manifold is invariant along the Reeb flow.
Proof. 
According to Lemma 2 and the definition of an α -Sasakian 3-manifold, the Ricci operator is given by
Q X = r 2 α 2 X r 2 3 α 2 η ( X ) ξ ,
for any vector field X and certain nonzero constant α . Moreover, according to [16] (Corollary 3.10), we observe that the scalar curvature r is invariant along the Reeb vector field ξ , i.e., ξ ( r ) = 0 . In fact, such an equation can be deduced directly by using the formula div Q = 1 2 r and (7). Applying ξ ( r ) = 0 , it follows directly from (7) that L ξ Q = 0 . □
Proof of Theorem 1.
Let M be a trans-Sasakian 3-manifold and e be a unit vector field orthogonal to ξ . Then, { ξ , e , ϕ e } forms a local orthonormal basis on the tangent space for each point of M. The Levi–Civita connection ∇ on M can be written as the following (see [12])
ξ ξ = 0 , ξ e = λ ϕ e , ξ ϕ e = λ e , e ξ = β e α ϕ e , e e = β ξ + γ ϕ e , e ϕ e = α ξ γ e , ϕ e ξ = α e + β ϕ e , ϕ e e = α ξ δ ϕ e , ϕ e = β ξ + δ e ,
where λ , γ and δ are smooth functions on some open subset of the manifold. We assume that the Ricci operator is invariant along the Reeb flow. From (1) and (4), we have
0 = ( L ξ Q ) X = ( ξ Q ) X + α ϕ Q X α Q ϕ X + β η ( Q X ) ξ β η ( X ) Q ξ
for any vector field X.
By using the local basis { ξ , e , ϕ e } and Lemma 2, the Ricci operator can be rewritten as the following:
Q ξ = ϕ α β + ( 2 α 2 2 β 2 ξ ( β ) ) ξ , Q e = r 2 + ξ ( β ) α 2 + β 2 e ( ϕ e ( α ) + e ( β ) ) ξ , Q ϕ e = r 2 + ξ ( β ) α 2 + β 2 ϕ e + ( e ( α ) ϕ e ( β ) ) ξ .
Replacing X in (9) by ξ , we obtain
ξ ( ϕ α β ) + ξ ( 2 α 2 2 β 2 ξ ( β ) ) ξ + α ( α + ξ ( α ) ξ ϕ β ) + 2 β ( α 2 β 2 ξ ( β ) ) ξ β ( ϕ α β ) β ( 2 α 2 2 β 2 ξ ( β ) ) ξ = 0 .
Taking the inner product of the above equation with ξ , e and ϕ e , respectively, we obtain
ξ ( ξ ( β ) ) + 2 β ξ ( β ) + 4 α 2 β = 0 , α e ( α ) β ϕ e ( α ) β e ( β ) α ϕ e ( β ) = 0 , β e ( α ) + α ϕ e ( α ) + α e ( β ) β ϕ e ( β ) = 0 ,
where we have employed Lemma 1. The addition of the second term of (12) multiplied by α to the third term of (12) multiplied by β gives
( α 2 + β 2 ) ( e ( α ) ϕ e ( β ) ) = 0 .
Following (13), we consider the following several cases.
Case i: α 2 + β 2 = 0 , or equivalently, α = β = 0 . In this case, the manifold becomes a cosymplectic 3-manifold. The proof for this case is completed because of Lemma 4.
Case ii: α 2 + β 2 0 . It follows immediately from (13) that e ( α ) ϕ e ( β ) = 0 , or equivalently, g ( α + ϕ β , e ) = 0 . Because e is assumed to be an arbitrary vector field, it follows that α + ϕ β = η ( α + ϕ β ) ξ , i.e.,
α + ϕ β + 2 α β ξ = 0 ,
or equivalently, ϕ α β + ξ ( β ) ξ = 0 , where we have used Lemma 1. When β = 0 , it follows from (14) that α is a nonzero constant. Thus, the proof can be done by applying Lemma 5. In what follows, we consider the last case.
Case iii: α 2 + β 2 0 and β 0 . In this context, (10) becomes
Q ξ = 2 ( α 2 β 2 ξ ( β ) ) ξ , Q e = r 2 + ξ ( β ) α 2 + β 2 e , Q ϕ e = r 2 + ξ ( β ) α 2 + β 2 ϕ e .
Replacing X by e in (9) and using (8), (15), we acquire
0 = ( L ξ Q ) e = ξ r 2 + ξ ( β ) α 2 + β 2 e .
With the aid of Lemma 1 and the first term of (12), from the previous relation, we have
ξ ( r ) = 0 .
From (15), we calculate the derivative of the Ricci operator as the following:
( ξ Q ) ξ = 0 , ( e Q ) e = e ( A ) e β A ξ + 2 β ( α 2 β 2 ξ ( β ) ) ξ , ( ϕ e Q ) ϕ e = ϕ e ( A ) ϕ e β A ξ + 2 β ( α 2 β 2 ξ ( β ) ) ξ ,
where we have used the first term of (8) and (12) and, for simplicity, we put
A = r 2 + ξ ( β ) α 2 + β 2 .
On a Riemannian manifold, we have div Q = 1 2 r . In this context, it is equivalent to
g ( ( ξ Q ) ξ + ( e Q ) e + ( ϕ e Q ) ϕ e , X ) = 1 2 X ( r )
for any vector field X. Replacing X in (19) by ξ and recalling (16) and the first term of (12), we obtain 2 β ( A 2 α 2 + 2 β 2 + 2 ξ ( β ) ) = 0 , or equivalently,
ξ ( β ) α 2 + β 2 = r 6 ,
where we have used the assumption β 0 and (18). According to (15), it is clear to see that the manifold is Einstein, i.e, Q = r 3 id . Because the manifold is of dimension three, then it must be of constant sectional curvature. □
A Riemannian manifold is said to be locally symmetric if R = 0 and this is equivalent to Q = 0 for dimension three. Wang and Liu in [16] proved that a trans-Sasakian 3-manifold is locally symmetric if and only if it is locally isometric to the sphere space S 3 ( c 2 ) , the hyperbolic space H 3 ( c 2 ) , the Euclidean space R 3 , product space R × S 2 ( c 2 ) or R × H 2 ( c 2 ) , where c is a nonzero constant. According to [16], on a locally symmetric trans-Sasakian 3-manifold, the Reeb vector field is an eigenvector field of the Ricci operator. Thus, following Lemma 3 and relations (9) and (10), we observe that Ricci parallelism is stronger than the Reeb flow invariant Ricci operator. Hence, our main result in this paper extends [16] (Theorem 3.12).
From Theorem 1, we obtain a new characterization of proper trans-Sasakian 3-manifolds.
Theorem 2.
A compact trans-Sasakian 3-manifold with Reeb flow invariant Ricci operator is homothetic to either a Sasakian manifold or a cosymplectic manifold.
Proof. 
As seen in the proof of Theorem 1, a trans-Sasakian 3-manifold with Reeb flow invariant Ricci operator is a α -Sasakian manifold, a cosymplectic manifold or a space of constant sectional curvature. It is well known that an α -Sasakian manifold is homothetic to a Sasakian manifold. Moreover, there do exist compact Sasakian and cosymplectic manifolds. To complete the proof, we need only to prove that Case iii in the proof of Theorem 1 cannot occur.
Let M be a trans-Sasakian 3-manifold satisfying Case iii. According to (14) and Lemma 5, we know that the Reeb vector field is minimal or harmonic. It has been proved in [17] (Lemma 5.1) that when ξ of a compact trans-Sasakian 3-manifold is minimal or harmonic, then α is a constant. Because the manifold is of constant sectional curvature, then the scalar curvature r is also a constant. Therefore, the differentiation of (20) along ξ gives
ξ ( ξ ( β ) ) + 2 β ξ ( β ) = 0 .
Adding the above equation to the first term of (12) implies that α = 0 because of β 0 . Using this in (14), we have β = ξ ( β ) ξ . The following proof follows directly from [2]. For sake of completeness, we present the detailed proof.
Applying β = ξ ( β ) ξ and (7), we obtain
X β = X ( ξ ( β ) ) ξ + ξ ( β ) ( β X β η ( X ) ξ ) = 0
for any vector field X. Contracting X in the previous relation and using (21), we obtain Δ β = ξ ( ξ ( β ) ) + 2 β ξ ( β ) = 0 . Because the manifold is assumed to be compact, the application of the divergence theorem gives that β is a non-zero constant. Next, we show that this is impossible. In fact, the application of (4) gives that div ξ = 2 β . Since the manifold is assumed to be compact, it follows that β = 0 , a contradiction. This completes the proof. □
Theorem 2 can also be written as follows.
Theorem 3.
A compact trans-Sasakian 3-manifold with Reeb flow invariant Ricci operator is proper.
The curvature tensor R of a trans-Sasakian 3-manifold is given by (see [10,27])
R ( X , Y ) Z = B ( g ( Y , Z ) X g ( X , Z ) Y ) C g ( Y , Z ) η ( X ) ξ + g ( Y , Z ) ( η ( X ) ( ϕ α β ) g ( β ϕ α , X ) ξ ) + C g ( X , Z ) η ( Y ) ξ g ( X , Z ) ( η ( Y ) ( ϕ α β ) g ( β ϕ α , Y ) ξ ) ( g ( β ϕ α , Z ) η ( Y ) + g ( β ϕ α , Y ) η ( Z ) ) X C η ( Y ) η ( Z ) X + ( g ( β ϕ α , Z ) η ( X ) + g ( β ϕ α , X ) η ( Z ) ) X + C η ( X ) η ( Z ) Y
for any vector fields X , Y , Z , where, for simplicity, we set
B = r 2 + 2 ξ ( β ) 2 α 2 + 2 β 2 , C = r 2 + ξ ( β ) 3 α 2 + 3 β 2 .
Substituting (14) and (20) into (22), with the aid of (23), we get
R ( X , Y ) Z = r 6 ( g ( Y , Z ) X g ( X , Z ) Y )
for any vector fields X , Y , Z . This implies that, on a trans-Sasakian 3-manifold satisfying Case iii in the proof of Theorem 1, we do not know whether α = 0 or not. In view of this, we introduce an interesting question:
Problem 1.
Is there a non-proper and non-compact trans-Sasakian 3-manifold of constant sectional curvature?
Remark 2.
According to De and Sarkar [10] (Theorem 5.1), we observe that a compact trans-Sasakian 3-manifold of constant sectional curvature is either α-Sasakian or β-Kenmotsu.
Remark 3.
Given a trans-Sasakian 3-manifold, following proof of Theorem 1, we still do not know whether β is a constant or not even when α = 0 and the manifold is compact (see [2]).

Author Contributions

X.L. introduced the problem. Y.Z. investigated the problem. W.W. wrote the paper.

Acknowledgments

This paper was supported by the research foundation of Henan University of Technology. The authors would like to thank the reviewers for their useful comments and suggestions.

Conflicts of Interest

The authors declare no conflict of interest.

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Zhao, Y.; Wang, W.; Liu, X. Trans-Sasakian 3-Manifolds with Reeb Flow Invariant Ricci Operator. Mathematics 2018, 6, 246. https://doi.org/10.3390/math6110246

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Zhao Y, Wang W, Liu X. Trans-Sasakian 3-Manifolds with Reeb Flow Invariant Ricci Operator. Mathematics. 2018; 6(11):246. https://doi.org/10.3390/math6110246

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Zhao, Yan, Wenjie Wang, and Ximin Liu. 2018. "Trans-Sasakian 3-Manifolds with Reeb Flow Invariant Ricci Operator" Mathematics 6, no. 11: 246. https://doi.org/10.3390/math6110246

APA Style

Zhao, Y., Wang, W., & Liu, X. (2018). Trans-Sasakian 3-Manifolds with Reeb Flow Invariant Ricci Operator. Mathematics, 6(11), 246. https://doi.org/10.3390/math6110246

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