Totally geodesic submanifolds of a trans-Sasakian manifold

We consider invariant submanifolds of a trans-Sasakian manifold and obtain the conditions under which the submanifolds are totally geodesic. We also study invariant submanifolds of a trans-Sasakian manifold satisfying Z(X ,Y ).h = 0, where Z is the concircular curvature tensor.


INTRODUCTION
Invariant submanifolds of a contact manifold have been a major area of research for a long time since the concept was borrowed from complex geometry.It helps us to understand several important topics of applied mathematics; for example, in studying non-linear autonomous systems the idea of invariant submanifolds plays an important role [9].A submanifold of a contact manifold is said to be totally geodesic if every geodesic in that submanifold is also geodesic in the ambient manifold.In 1985, Oubina [14] introduced a new class of almost contact manifolds, namely, trans-Sasakian manifold of type (α, β ), which can be considered as a generalization of Sasakian, Kenmotsu, and cosymplectic manifolds.Trans-Sasakian structures of type (0, 0), (0, β ), and (α, 0) are cosymplectic [2], β -Kenmotsu [10], and α-Sasakian [10], respectively.Kon [12] proved that invariant submanifolds of a Sasakian manifold are totally geodesic if the second fundamental form of the immersion is covariantly constant.On the other hand, any submanifold M of a Kenmotsu manifold is totally geodesic if and only if the second fundamental form of the immersion is covariantly constant, provided ξ ∈ T M [11].Recently, Sular and Özgür [16] proved some equivalent conditions regarding the submanifolds of a Kenmotsu manifold to be totally geodesic.Several studies ( [5,17]) have been done on invariant submanifolds of trans-Sasakian manifolds.Recently, Sarkar and Sen [15] proved some equivalent conditions of an invariant submanifold of trans-Sasakian manifolds to be totally geodesic.In the present paper we rectify proofs of most of the major theorems of [15] and [17], show some theorems of [15] as corollary of our present results, and also introduce some new equivalent conditions for an invariant submanifold of a trans-Sasakian manifold to be totally geodesic.

PRELIMINARIES
Let M be a connected almost contact metric manifold with an almost contact metric structure (φ , ξ , η, g), that is, φ is a (1, 1)-tensor field, ξ is a vector field, η is a one-form, and g is the compatible Riemannian for all X,Y ∈ T M ( [2,18]).The fundamental two-form Φ of the manifold is defined by for X,Y ∈ T M.
An almost contact metric structure (φ , ξ , η, g) on a connected manifold M is called a trans-Sasakian structure [14] if (M × R, J, G) belongs to the class W 4 [8], where J is the almost complex structure on M × R defined by for all vector fields X on M and smooth functions f on M × R, and G is the product metric on M × R.This may be expressed by the condition [3] for smooth functions α and β on M.Here we say that the trans-Sasakian structure is of type (α, β ).From the formula (2.5) it follows that ∇X ξ In a (2n + 1)-dimensional trans-Sasakian manifold we also have the following: where S is the Ricci tensor of type (0, 2) and R is the curvature tensor of type (1,3).Let M be a submanifold of a contact manifold M. We denote by ∇ and ∇ the Levi-Civita connections of M and M, respectively, and by T ⊥ (M) the normal bundle of M. Then for vector fields X,Y ∈ T M, the second fundamental form h is given by the formula where ∇ ⊥ denotes the normal connection of M. The second fundamental form h and A N are related by The submanifold M is totally geodesic if and only if h = 0.An immersion is said to be parallel and semi-parallel [6] if for all X,Y ∈ T M we get ∇.h = 0 and R(X,Y ).h = 0, respectively.
It is said to be pseudo-parallel [7] if for all X,Y ∈ T M we get where f denotes a real function on M and Q(E, T ) is defined by where X ∧ E Y is defined by If f = 0, the immersion is semi-parallel.
Similarly, an immersion is said to be 2-pseudo-parallel if for all X,Y ∈ T M we get R(X,Y ).∇h = f Q(g, ∇h), and Ricci generalized pseudo-parallel [13] The second fundamental form h satisfying where A is a nonzero one-form, is said to be recurrent.It is said to be 2-recurrent if h satisfies where B is a nonzero two-form.
Proposition 2.2.[5] Let M be an invariant submanifold of a trans-Sasakian manifold M. Then we have for any vector fields X and Y on M.
For a Riemannian manifold, the concircular curvature tensor Z is defined by for vectors X,Y,V ∈ T M, where τ is the scalar curvature of M. We also have (2.21) In the next section we consider the submanifold M to be tangent to ξ .Proof.The second fundamental form h is said to be recurrent if

INVARIANT SUBMANIFOLDS OF
where A is an everywhere nonzero one-form.
We define a function e on M by e 2 = g(h, h). (3.1) Then we have e(Ye) = e 2 A(Y ).So we obtain Ye = eA(Y ), since f is nonzero.This implies that Therefore we get Since the left-hand side of the above equation is identically zero and e is nonzero on M by our assumption, we obtain dA(X,Y ) = 0, ( that is, the one-form Therefore, for a recurrent second fundamental form we have

R(X,Y
).h = 0 for any vectors X,Y on M.
Theorem 3.1.An invariant submanifold of a non-cosymplectic trans-Sasakian manifold is totally geodesic if and only if its second fundamental form is parallel.
Proof.Since h is parallel, we have Hence for a non-cosymplectic trans-Sasakian manifold h(X,Y ) = 0, for all X,Y ∈ T M. The converse part is trivial.Hence the result.
Remark 3.1.In Theorem 3.1 [15] the authors proved the same result, but they actually proved h(Y, ∇ X ξ ) = 0, and h(Y, ξ ) = 0, ∀X,Y ∈ T M. Since ∇ X ξ is not an arbitrary vector of T M, hence from this we can not conclude that the submanifold is totally geodesic.
Remark 3.2.Again in the proof of Theorem 4.8 [17] the authors assumed φ (h(X,Y )) = 0, ∀X,Y ∈ T M, which is not true in general because this condition directly implies that the submanifold is totally geodesic.
Theorem 3.2.An invariant submanifold of a non-cosymplectic trans-Sasakian manifold is totally geodesic if and only if its second fundamental form is semi-parallel.
Proof.Since h is semi-parallel, we have Applying ϕ to both sides of Eq. (3.8) we obtain So from (3.8) and (3.9) we conclude that Hence as in the previous case, for non-cosymplectic trans-Sasakian manifolds the invariant submanifold is totally geodesic.The converse part follows trivially.Now, by Lemma 3.1 we get that if a second fundamental form is recurrent, then it is semi-parallel.Also, the second fundamental form of a totally geodesic submanifold is trivially recurrent, so from Theorem 3.2 we obtain the following:

. An invariant submanifold of a non-cosymplectic trans-Sasakian manifold is totally geodesic if and only if its second fundamental form is recurrent.
Remark 3.3.In Theorem 3.2 [15] the authors proved the above corollary, but they just showed that h(Y, ∇ X ξ ) = 0, and h(Y, ξ ) = 0, ∀X,Y ∈ T M. Since ∇ X ξ is not an arbitrary vector of T M, we can not conclude from this that the submanifold is totally geodesic.
In [1] Aikawa and Matsuyama proved that if a tensor field T is 2-recurrent, then R(X,Y ).T = 0. Also it can be easily seen that in a totally geodesic submanifold the second fundamental form is 2-recurrent.Therefore by Theorem 3.2 we also obtain the following: Corollary 3.2.An invariant submanifold of a non-cosymplectic trans-Sasakian manifold is totally geodesic if and only if its second fundamental form is 2-recurrent.Remark 3.4.In Theorem 3.4 [15] the authors proved the above corollary, but they considered ∇ X ξ as an arbitrary vector of T M, and actually proved h(Y, ∇ X ξ ) = 0, ∀X,Y ∈ T M, hence the proof of Theorem 3.4 [15] is incorrect.

Theorem 3.3. An invariant submanifold of a trans-Sasakian manifold is totally geodesic if and only if its second fundamental form is
Proof.Since, the second fundamental form is 2-semi-parallel, we have Similarly, So putting Applying φ on both sides of (3.11) we get From (3.11) and (3.12) we conclude that Hence the submanifold is totally geodesic.The converse holds trivially.
Theorem 3.4.An invariant submanifold of a trans-Sasakian manifold is totally geodesic if and only if its second fundamental form is pseudo-parallel, provided Proof.Since the second fundamental form is pseudo-parallel, we have Hence the submanifold is totally geodesic.The converse holds trivially.
Theorem 3.5.An invariant submanifold of a trans-Sasakian manifold is totally geodesic if and only if its second fundamental form is 2-pseudo-parallel.
Proof.Since, the second fundamental form is 2-pseudo-parallel, we have From (2.10) and (2.19) we have and Proof.Since the submanifold is Ricci generalized pseudo-parallel, we have Putting Y = V = ξ and applying (2.19) we obtain Hence the submanifold is totally geodesic.The converse holds trivially.
A TRANS-SASAKIAN MANIFOLD WITH α, β = CONSTANT Lemma 3.1.If a non-flat Riemannian manifold has a recurrent second fundamental form, then it is semiparallel.