Geometry of Kenmotsu Manifolds via Q-Curvature Tensor and Schouten–Van Kampen Connection

Abstract

This research paper aims to study the Q-curvature tensor on Kenmotsu manifolds endowed with the Schouten–van Kampen connection. Using the Q-curvature tensor, whose trace is the well-known Z-tensor, we characterized Kenmotsu manifolds by introducing the notion of $\zeta$-$\tilde{Q}$ flat and $\varphi$-$\tilde{Q}$ flat manifolds and novel tensor conditions, such as $\tilde{Q}(\xi ,X)\tilde{Q}=0,$$\tilde{Q}(\xi ,X)\tilde{R}=0,$$\tilde{Q}(\xi ,X)\tilde{C}=0,$$\tilde{Q}(\xi ,X)\tilde{S}=0$, $\tilde{Q}(\xi ,X)\tilde{H}=0$, and $\tilde{Q}(\xi ,X)\tilde{P}=0$, with the Schouten–van Kampen connection. To validate some of our results, we constructed a non-trivial example of Kenmotsu manifolds endowed with the Schouten–van Kampen connection.

1. Introduction

Contact geometry is emerging as a highly effective branch of differential geometry, with numerous applications in geometric optics, thermodynamics, geometric quantization, and the mechanics of dynamical systems with time-dependent Hamiltonians. One of contact geometry’s most recent chapters was initiated in 1972, when Kenmotsu [1] introduced a new class of almost contact Riemannian manifolds, known as Kenmotsu manifolds. As is well known, odd-dimensional spheres permit Sasakian structures, but odd-dimensional hyperbolic spaces do not. They do, however, have Kenmotsu structures. Kenmotsu manifolds are normal, almost-contact Riemannian manifolds. The fundamental properties of the local structure of such manifolds have been extensively studied by many geometers (for details, see [2,3,4,5,6,7,8]).

Mantica [9] introduced the Q tensor notation, which refers to a tensor whose trace corresponds to the Z tensor. Alongside this new tensor, Mantica and Suh [10] also defined a class of manifolds termed pseudo Q-symmetric Riemannian manifolds, which include both pseudo-symmetric and pseudo-concircular symmetric manifolds. In recent years, Yıldırım [11] defined new types known as $\varphi$-Q symmetric and $\varphi$-Q recurrent Kenmotsu manifolds. Meanwhile, Yılmaz [12] investigated Sasakian manifolds under certain conditions admitting the Q tensor, and Yadav examined f-Kenmotsu 3-manifolds that satisfy certain curvature conditions related to the Q-curvature tensor associated with the Schouten–van Kampen connection [13].

The Schouten–van Kampen connection (briefly, SvK-connection) is one of the most natural connections adapted to a pair of complementary distributions on a differential manifold with an affine connection. In 1978, Solov’ev examined hyperdistributions in Riemannian manifolds using this connection [14]. In 2006 [15], Bejancu investigated the foliated manifolds with Schouten–van Kampen connections. Olszak [16] (2013) conducted research on modifying the Schouten–van Kampen connection to a nearly contact metric structure, characterizing several kinds of almost-contact metric manifolds. In recent studies, G. Ghosh [17], T. Mert [18], Vaishali [19], Chakraborty [20], and El Hendi [21] examined the Schouten–van Kampen connection in Sasakian, f-Kenmotsu, and Kenmotsu manifolds. Siddiqi [22] investigated optimal inequalities for submanifolds in trans-Sasakian manifolds or $\left(\alpha ,\beta \right)$-type almost contact manifolds endowed with the Schouten–van Kampen connection.

In this study, we present novel tensor conditions for the Q-curvature tensor on Kenmotsu manifolds with a SvK-connection. We introduced the notions of $\zeta$-$\tilde{Q}$ flat and $\varphi$-$\tilde{Q}$ flat on Kenmotsu manifolds and investigated their results in those with the SvK-connection. By establishing tensor conditions such as $\tilde{Q}(\zeta ,X)\tilde{Q}=0,$$\tilde{Q}(\zeta ,X)\tilde{R}=0,$ $\tilde{Q}(\zeta ,X)\tilde{C}=0,$$\tilde{Q}(\zeta ,X)\tilde{S}=0$, $\tilde{Q}(\zeta ,X)\tilde{H}=0$, and $\tilde{Q}(\zeta ,X)\tilde{P}=0$, with respect to the SvK-connection, we were able to characterize Kenmotsu manifolds using the Q-curvature tensor, whose trace is the well-known Z tensor. The definitions regarding the notations used throughout this paper are provided in Section 2 (Preliminaries). Finally, we constructed a non-trivial example of a Kenmotsu manifold admitting a SvK-connection in order to verify some of our results.

2. Preliminaries

If a smooth Riemannian manifold $(M,g)$ (dimension = $(2m+1)$) admits a global 1-form $\eta$, also known as a contact form, such that $\eta \wedge {\left(d\eta \right)}^m\ne 0$, a $(1,1)$ tensor field $\varphi$, a characteristic vector field $\zeta$, and an indefinite metric g, then it is said to be an almost contact metric manifold. An almost contact metric manifold satisfying the following conditions [23]:

$${\varphi }^2=-I+\eta \otimes \zeta ,\eta \left(\zeta \right)=1,$$

(1)

$$\varphi \zeta =0, \eta \circ \varphi =0,$$

(2)

$$g(\varphi X_1,\varphi X_2)=g(X_1,X_2)-\eta \left(X_1\right)\eta \left(X_2\right),g(X_1,\zeta )=\eta \left(X_1\right).$$

(3)

Also, we note that g is a Riemannian metric compatible with the almost contact structure of $M.$

An almost contact metric manifold is said to be a Kenmotsu manifold [23] if

$$\left({\nabla }_{X_1}\varphi \right)\left(X_2\right)=-g(X_1,\varphi X_2)\zeta -\eta \left(X_2\right)\varphi X_1,$$

(4)

where ∇ denotes the Riemannian connection of g.

The following relations hold in a Kenmotsu manifold [1]:

$${\nabla }_{X_1}\zeta =X_1-\eta \left(X_1\right)\zeta ,$$

(5)

$$\left({\nabla }_{X_1}\eta \right)X_2=g(X_1,X_2)-\eta \left(X_1\right)\eta \left(X_2\right),$$

(6)

$$\eta \left(R(X_1,X_2)X_3\right)=g(X_1,X_3)\eta \left(X_2\right)-g(X_2,X_3)\eta \left(X_1\right),$$

(7)

$$R(X_1,X_2)\zeta =\eta \left(X_1\right)X_2-\eta \left(X_2\right)X_1,$$

(8)

$$R(\zeta ,X_1)X_2=\eta \left(X_2\right)X_1-g(X_1,X_2)\zeta ,$$

(9)

$$S(X_1,X_2)=g\left(L{X}_1,X_2\right),$$

(10)

$$S(\varphi X_1,\varphi X_2)=S(X_1,X_2)+2m\eta \left(X_1\right)\eta \left(X_2\right),$$

(11)

$$S(X_1,\zeta )=-2m\eta \left(X_1\right),$$

(12)

for all vector fields $X_1,X_2,X_3$$\in \mathsf{\Gamma }\left(M\right),$ where R is the Riemannian curvature tensor, S is the Ricci tensor, and L is the Ricci operator. An $\eta$-Einstein Kenmotsu manifold (dimension = $(2m+1)$) is defined as such if two smooth functions, a and b, exist that satisfy the following relation for any $X_1$ and $X_2$ in $\mathsf{\Gamma }\left(M\right)$:

$$S(X_1,X_2)=ag(X_1,X_2)+b\eta \left(X_1\right)\eta \left(X_2\right),$$

(13)

where S is the Ricci tensor. If $b=0,$ then the manifold is an Einstein manifold.

Now, throughout the article we associate ∼ with the quantities admitting the SvK-connection. The SvK-connection $\tilde{\nabla }$ associated with the Levi–Civita connection ∇ is as follows [16]:

$${\tilde{\nabla }}_{X_1}{X}_2={\nabla }_{X_1}{X}_2-\eta \left(X_2\right){\nabla }_{X_1}\zeta +\left({\nabla }_{X_1}\eta \right)\left(X_2\right)\zeta$$

(14)

for any vector fields $X_1,X_2$$\in \mathsf{\Gamma }\left(M\right).$ Using (5) and (6), the above equation yields

$${\tilde{\nabla }}_{X_1}{X}_2={\nabla }_{X_1}{X}_2+g(X_1,X_2)\zeta -\eta \left(X_2\right)X_1.$$

(15)

In a Kenmotsu manifold which admits the SvK-connection, the following relations hold [24]:

$${\tilde{\nabla }}_{X_1}\zeta =0,$$

(16)

$$\tilde{R}(X_1,X_2)X_3=R(X_1,X_2)X_3+g(X_2,X_3)X_1-g(X_1,X_3)X_2,$$

(17)

$$\tilde{R}(X_1,X_2)\zeta =\tilde{R}(X_1,\zeta )X_3=\tilde{R}(\zeta ,X_2)X_3=0,$$

(18)

$$\tilde{S}(X_1,X_2)={\displaystyle \sum _{i=1}^{2m+1}}g\left(\tilde{R}\left(e_i,X_1\right)X_2,e_i\right),$$

(19)

$$\tilde{S}(X_1,X_2)=S\left(X_1,X_2\right)+2mg(X_1,X_2),$$

(20)

$$\tilde{S}(\varphi X_1,\varphi X_2)=S(\varphi X_1,\varphi X_2)+2mg(\varphi X_1,\varphi X_2),$$

(21)

$$\tilde{L}{X}_1=L{X}_1+2m{X}_1,$$

(22)

$$\widetilde{scal}=scal+2m(2m+1),$$

(23)

where $\tilde{R},\tilde{S},\tilde{L}$, and $\widetilde{scal}$ denote, respectively, the Riemannian curvature tensor, Ricci tensor, Ricci operator, and scalar curvature with respect to the SvK-connection.

A generalized $(0,2)$ symmetric Z tensor for any vector field $X_1,X_2$ $\in \mathsf{\Gamma }\left(M\right)$ is defined as [25]

$$Z(X_1,X_2)=S(X_1,X_2)+\tau g(X_1,X_2),$$

where g and S are the Riemannian metric and Ricci tensor on $M,$ respectively, and $\tau$ is an arbitrary scalar function.

The $(1,3)$Q-curvature tensor [10], whose trace is the Z–tensor, is expressed as

$$Q(X_1,X_2)X_3=R(X_1,X_2)X_3-{\displaystyle \frac{\tau }{2m}}\left\{g(X_2,X_3)X_1-g(X_1,X_3)X_2\right\},$$

(24)

where $\tau$ is an arbitrary scalar function.

In addition, the concircular curvature tensor $\tilde{C}$, the projective curvature tensor $\tilde{P}$ and the conharmonic curvature tensor $\tilde{H}$ admitting the SvK-connection are as follows [26]:

$$\tilde{C}(X_1,X_2)X_3=\tilde{R}(X_1,X_2)X_3-{\displaystyle \frac{\widetilde{scal}}{2m(2m+1)}}\left\{g(X_2,X_3)X_1-g(X_1,X_3)X_2\right\},$$

(25)

$$\tilde{P}(X_1,X_2)X_3=\tilde{R}(X_1,X_2)X_3-{\displaystyle \frac{1}{2m}}\{\tilde{S}(X_2,X_3)X_1-\tilde{S}(X_1,X_3)X_2\},$$

(26)

$$\begin{array}{cc}\hfill \tilde{H}(X_1,X_2)X_3& =\tilde{R}(X_1,X_2)X_3\hfill \\ \hfill & -{\displaystyle \frac{1}{2m-1}}\{\tilde{S}(X_2,X_3)X_1-\tilde{S}(X_1,X_3)X_2\hfill \\ \hfill & +g(X_2,X_3)L{X}_1-g(X_1,X_3)L{X}_2\},\hfill \end{array}$$

(27)

for any $X_1,X_2,X_3\in \mathsf{\Gamma }\left(M\right),$ where L is the Ricci operator and $\widetilde{scal}$ is the scalar curvature of $M.$

3. $\mathit{Q}$-Curvature Tensor of Kenmotsu Manifolds Admitting the Schouten–Van Kampen Connection

Definition 1.

The Q-curvature tensor admitting the SvK-connection $\tilde{\nabla }$ is defined as

$$\tilde{Q}(X_1,X_2)X_3=\tilde{R}(X_1,X_2)X_3-{\displaystyle \frac{\tau }{2m}}\left\{g(X_2,X_3)X_1-g(X_1,X_3)X_2\right\},$$

(28)

for all vector fields $X_1,X_2,X_3$ on $M.$ If $\tilde{Q}(X_1,X_2)X_3=0$, then the manifold is called a $\tilde{Q}$-$flat$ manifold.

Proposition 1.

Let the Kenmotsu manifold ($M,g)$ admitting the SvK-connection be a $\tilde{Q}$-$flat$ manifold; then, the scalar curvature is $\left(-1+{\displaystyle \frac{\tau }{2m}}\right).$ Furthermore, if $\tau =0$, then the manifold M is isomorphic to the hyperbolic space $H^{2m+1}(-1).$

Proof. 

Using Equation (17) in (28), we obtain

$$R(X_1,X_2)X_3=\left(-1+{\displaystyle \frac{\tau }{2m}}\right)\left\{g(X_2,X_3)X_1-g(X_1,X_3)X_2\right\}.$$

Thus, the scalar curvature is $\left(-1+{\displaystyle \frac{\tau }{2m}}\right).$ It is clear that M is isomorphic to the hyperbolic space $H^{2m+1}(-1),$ when $\tau =0.$ □

Theorem 1.

Let $\tilde{\nabla }$ be the SvK-connection on M. The $\tilde{Q}$-curvature tensor of $\tilde{\nabla }$ satisfies the following first Bianchi identity:

$$\tilde{Q}(X_1,X_2)X_3+\tilde{Q}(X_2,X_3)X_1+\tilde{Q}(X_3,X_1)X_2=0.$$

Proof. 

Utilizing Equations (17) and (28), the result is clear. □

Proposition 2.

The $\tilde{Q}$-curvature tensor of the Kenmotsu manifold $(M,g)$ which admits the SvK-connection satisfies the following relations:

$$\tilde{Q}(X_1,X_2)\zeta ={\displaystyle \frac{-\tau }{2m}}\left\{\eta \left(X_2\right)X_1-\eta \left(X_1\right)X_2\right\}$$

(29)

$$\tilde{Q}(\zeta ,X_1)X_2={\displaystyle \frac{-\tau }{2m}}\left\{g(X_1,X_2)\zeta -\eta \left(X_2\right)X_1\right\}$$

(30)

$$\tilde{Q}(X_1,\zeta )X_2={\displaystyle \frac{-\tau }{2m}}\left\{\eta \left(X_2\right)X_1-g(X_1,X_2)\zeta \right\}$$

(31)

for all vector fields $X_1,X_2.$

Proof. 

Applying (3) and (18) in (28), the proof is clear. □

Definition 2.

A Kenmotsu manifold with respect to the SvK–connection $\tilde{\nabla }$ is defined as ζ-$\tilde{Q}$ flat if $\tilde{Q}(X_1,X_2)\zeta =0.$

Theorem 2.

Let the Kenmotsu manifold ($M,g)$ admitting the SvK-connection be the ζ-$\tilde{Q}$ manifold, if and only if the $\tilde{Q}$-curvature tensor is equal to the Riemannian curvature.

Proof. 

Let the manifold M with respect to the SvK-connection be a $\zeta$-$\tilde{Q}$ flat, that is, $\tilde{Q}(X_1,X_2)\zeta =0$. Utilizing (29) yields

$$\tilde{Q}(X_1,X_2)\zeta ={\displaystyle \frac{-\tau }{2m}}\left\{\eta \left(X_2\right)X_1-\eta \left(X_1\right)X_2\right\}.$$

Since $\tilde{Q}(X_1,X_2)\zeta =0,$ we obtain $\tau =0$, so the $\tilde{Q}$-curvature tensor is equal to the Riemannian curvature tensor.

Now, we assume that $\tau =0.$ Then, from Equation (29), it follows that $\tilde{Q}(X_1,X_2)\zeta =0,$ so the manifold is a $\zeta$-$\tilde{Q}$ flat manifold. □

Theorem 3.

For a ζ-$\tilde{Q}$ flat Kenmotsu manifold with respect to the SvK-connection, the manifold is a special η-Einstein type.

Proof. 

Let M be a $\zeta$-$\tilde{Q}$ flat Kenmotsu manifold. From Equation (29), we have

$${\displaystyle \frac{-\tau }{2m}}\left\{\eta \left(X_2\right)X_1-\eta \left(X_1\right)X_2\right\}=0.$$

Replacing $X_2$ with $\zeta$ in the above equation, we obtain

$${\displaystyle \frac{\tau }{2m}}\left\{X_1-\eta \left(X_1\right)\zeta \right\}=0.$$

(32)

Considering the inner product of (32) with $X_3,$ we find

$${\displaystyle \frac{\tau }{2m}}\left\{g(X_1,X_3)-\eta \left(X_1\right)\eta \left(X_3\right)\right\}=0.$$

Substituting $X_1$ for $L{X}_1$ yields

$${\displaystyle \frac{\tau }{2m}}\left\{S(X_1,X_3)-S(X_1,\zeta )\eta \left(X_3\right)\right\}=0.$$

Utilizing (12), we obtain

$${\displaystyle \frac{\tau }{2m}}\left\{S(X_1,X_3)+2m\eta \left(X_1\right)\eta \left(X_3\right)\right\}=0$$

which gives

$$S(X_1,X_3)=-2m\eta \left(X_1\right)\eta \left(X_3\right).$$

From (13), the theorem is obvious. □

Definition 3.

A Kenmotsu manifold is said to be ϕ-$\tilde{Q}$ flat with respect to the SvK-connection $\tilde{\nabla }$ if

$$g(\tilde{Q}(\varphi X_1,\varphi X_2)\varphi X_3,\varphi X_4)=0,$$

(33)

for any vector fields $X_1,X_2,X_3$ on M.

Theorem 4.

Let the Kenmotsu manifold M with SvK-connection be ϕ-$\tilde{Q}$ flat; then, M is an η-Einstein manifold.

Proof. 

Using (28) in (33), we have

$$g\left(\tilde{R}(\varphi X_1,\varphi X_2)\varphi X_3-{\displaystyle \frac{\tau }{2m}}\left\{g(\varphi X_2,\varphi X_3)\varphi X_1-g(\varphi X_1,\varphi X_3)\varphi X_2\right\},\varphi X_4\right)=0.$$

(34)

Now, let $\left\{u_i\right\}$ be an orthonormal basis of the tangent space at each point of the manifold M for $i=1,2,...,2m+1$. Putting $X_1=X_4=u_i$ in (34) implies

$${\displaystyle \sum _{i=1}^{2m}}\tilde{R}\left(\varphi u_i,\varphi X_2,\varphi X_3,\varphi u_i\right)={\displaystyle \frac{\tau }{2m}}{\displaystyle \sum _{i=1}^{2m}}\left\{g(\varphi X_2,\varphi X_3)g\left(\varphi u_i,\varphi u_i\right)-g(\varphi u_i,\varphi X_3)g\left(\varphi X_2,\varphi u_i\right)\right\}.$$

Then, taking summation over index i and using Equation (20), we obtain

$$\tilde{S}\left(\varphi X_2,\varphi X_3\right)={\displaystyle \frac{\tau }{2m}}\left(2m-1\right)g\left(\varphi X_2,\varphi X_3\right).$$

Then, in light of (11) and (21), we obtain

$$S\left(X_2,X_3\right)=\left\{{\displaystyle \frac{\tau }{2m}}\left(2m-1\right)-2m\right\}g(\varphi X_2,\varphi X_3)-2m\eta \left(X_2\right)\eta \left(X_3\right).$$

By using Equation (3), we obtain

$$S\left(X_2,X_3\right)=\left\{{\displaystyle \frac{\tau }{2m}}\left(2m-1\right)-2m\right\}g(X_2,X_3)-{\displaystyle \frac{\tau }{2m}}\left(2m-1\right)\eta \left(X_2\right)\eta \left(X_3\right).$$

By virtue of (13), the manifold is $\eta$-Einstein. □

Corollary 1.

If the Kenmotsu manifold M equipped with the SvK-connection is ϕ-$\tilde{Q}$ flat, then its scalar curvature is given by

$$scal=\tau (2m-1)-2m(2m+1).$$

Theorem 5.

A Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{Q}=0$; then, either the $\tilde{Q}$-curvature tensor is equal to the Riemannian curvature or the manifold is η-Einstein.

Proof. 

In a Kenmotsu manifold $\left(M,g\right),$ from the condition $\tilde{Q}(\zeta ,X_1)\tilde{Q}=0,$ we achieve

$$\begin{array}{cc}\hfill \left(\tilde{Q}(\zeta ,X_1)\tilde{Q}\right)\left(X_2,X_3\right)\zeta & =\tilde{Q}(\zeta ,X_1)\tilde{Q}(X_2,X_3)\zeta -\tilde{Q}\left(\tilde{Q}(\zeta ,X_1)X_2,X_3\right)\zeta \hfill \\ \hfill & -\tilde{Q}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta -\tilde{Q}(X_2,X_3)\tilde{Q}(\zeta ,X_1)\zeta \hfill \\ \hfill & =0.\hfill \end{array}$$

By considering Equations (3), (18), (29), (30), and (31), we obtain

$${\displaystyle \frac{-\tau }{2m}}\left[\tilde{R}\left(X_2,X_3\right)X_1+{\displaystyle \frac{\tau }{2m}}\left\{\eta \left(X_3\right)g\left(X_1,X_2\right)\zeta -\eta \left(X_2\right)g\left(X_1,X_3\right)\zeta \right\}\right]=0.$$

Therefore, either $\tau =0$, and so the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature, or the following equation holds:

$$\tilde{R}\left(X_2,X_3\right)X_1=-{\displaystyle \frac{\tau }{2m}}\left\{\eta \left(X_3\right)g\left(X_1,X_2\right)\zeta -\eta \left(X_2\right)g\left(X_1,X_3\right)\zeta \right\}.$$

(35)

Using (17) in (35), we obtain

$$\begin{array}{cc}\hfill R\left(X_2,X_3\right)X_1& =g(X_1,X_2)\eta \left(X_3\right)-g(X_1,X_3)\eta \left(X_2\right)\hfill \\ \hfill & -{\displaystyle \frac{\tau }{2m}}\left\{\eta \left(X_3\right)g\left(X_1,X_2\right)\zeta -\eta \left(X_2\right)g\left(X_1,X_3\right)\zeta \right\}.\hfill \end{array}$$

Taking the inner product of the above equation with $X_4,$ we have

$$\begin{array}{cc}\hfill R(X_2,X_3,X_1,X_4)& =g(X_1,X_2)g(X_3,X_4)-g(X_1,X_3)g(X_2,X_4)\hfill \\ \hfill & -{\displaystyle \frac{\tau }{2m}}\left\{\eta \left(X_3\right)g\left(X_1,X_2\right)g(\zeta ,X_4)-\eta \left(X_2\right)g\left(X_1,X_3\right)g(\zeta ,X_4)\right\}.\hfill \end{array}$$

(36)

Setting $X_2=X_3=u_i$ in (36), taking the summation over index i and using equation, we obtain

$$S(X_1,X_3)=\left(-2m+{\displaystyle \frac{\tau }{2m}}\right)g(X_1,X_3)+{\displaystyle \frac{\tau }{m}}\eta \left(X_1\right)\eta \left(X_3\right).$$

(37)

Thus, the manifold is an $\eta$-Einstein manifold. □

Corollary 2.

Let the Kenmotsu manifold ($M,g)$ admitting the SvK-connection satisfy the condition $\tilde{Q}(\zeta ,X_1)\tilde{Q}=0$; then, the scalar curvature is $-2m(2m+1+2\tau ).$

Proof. 

Taking $X_1=X_3=u_i$ in (37) and contracting it, we obtain

$$scal=-2m(2m+1+2\tau ).$$

□

Theorem 6.

Let the Kenmotsu manifold ($M,g)$ admitting the SvK-connection satisfy the condition $\tilde{Q}(\zeta ,X_1)\tilde{R}=0$; then, either the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature or the manifold is Einstein.

Proof. 

In a Kenmotsu manifold $\left(M,g\right),$ from the condition $\tilde{Q}(\zeta ,X_1)\tilde{R}=0,$ we achieve

$$\begin{array}{cc}\hfill \left(\tilde{Q}(\zeta ,X_1)\tilde{R}\right)(X_2,X_3)\zeta & =\tilde{Q}(\zeta ,X_1)\tilde{R}(X_2,X_3)\zeta -\tilde{R}\left(\tilde{Q}(\zeta ,X_1)X_2,X_3\right)\zeta \hfill \\ \hfill & -\tilde{R}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta -\tilde{R}\left(X_2,X_3\right)\tilde{Q}(\zeta ,X_1)\zeta \hfill \\ \hfill & =0.\hfill \end{array}$$

(38)

Making use of (18) in (38), it can be clearly seen that

$$\tilde{Q}(\zeta ,X_1)\tilde{R}(X_2,X_3)\zeta =\tilde{R}\left(\tilde{Q}(\zeta ,X_1)X_2,X_3\right)\zeta =\tilde{R}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta =0.$$

Therefore, it will be sufficient to calculate the $\tilde{R}\left(X_2,X_3\right)\tilde{Q}(\zeta ,X_1)\zeta .$ In that case, making use of (1), (2), (18), and (28), we obtain

$$\left(\tilde{Q}(\zeta ,X_1)\tilde{R}\right)(X_2,X_3)\zeta ={\displaystyle \frac{\tau }{2m}}\tilde{R}(X_2,X_3)X_1=0.$$

Therefore, either $\tau =0$, and so $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature, or the following equation holds:

$$\tilde{R}(X_2,X_3)X_1=0.$$

Using (17),

$$R(X_2,X_3)X_1=g(X_2,X_1)X_3-g(X_3,X_1)X_2,$$

and taking the inner product of the above equation with $X_4,$ we have

$$R(X_2,X_3,X_1,X_4)=g(X_2,X_1)g(X_3,X_4)-g(X_3,X_1)g(X_2,X_4).$$

(39)

Setting $X_2=X_4=u_i$ in (39) and summing up to $i$, $1\le i\le 2m+1,$ we obtain

$$S(X_1,X_3)=-2mg(X_1,X_3).$$

Thus, the manifold is Einstein. □

Corollary 3.

Let the Kenmotsu manifold ($M,g)$ admitting the SvK-connection satisfy the condition $\tilde{Q}(\zeta ,X_1)\tilde{R}=0$; then, the scalar curvature is $-2m(2m+1).$

Corollary 4.

Let the Kenmotsu manifold $(M,g)$ admitting the SvK-connection satisfy the condition $\tilde{Q}(\zeta ,X_1)\tilde{R}=0$; then, the manifold M is isomorphic to the hyperbolic space $H^{2m+1}(-1).$

Corollary 5.

Let the Kenmotsu manifold $(M,g)$ admitting the SvK-connection. Then, $\tilde{R}(\zeta ,X_1)\tilde{Q}=\tilde{R}(\zeta ,X_1)\tilde{R}.$

Theorem 7.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{S}=0,$ then either the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature or the manifold is Einstein.

Proof. 

The condition $\tilde{Q}(\zeta ,X_1)\tilde{S}=0$ implies that

$$\left(\tilde{Q}(\zeta ,X_1)\tilde{S}\right)(X_2,\zeta )=\tilde{S}\left(\tilde{Q}\left(\zeta ,X_1\right)X_2,\zeta \right)+\tilde{S}\left(X_2,\tilde{Q}\left(\zeta ,X_1\right)\zeta \right)=0.$$

By considering Equations (18)–(20), we achieve

$$\tilde{S}\left(X_2,\tilde{Q}\left(\zeta ,X_1\right)\zeta \right)=S(X_2,\tilde{Q}\left(\zeta ,X_1\right)\zeta )+2mg\left(X_2,\tilde{Q}\left(\zeta ,X_1\right)\zeta \right)=0.$$

Substituting (1), (3), (12), (18), and (28) into the above equation, we obtain $\tau =0$, so the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature or

$$S(X_1,X_2)=-2mg(X_1,X_2).$$

Therefore, the manifold is Einstein. □

Corollary 6.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{S}=0,$ then the scalar curvature is $scal=-2m(2m+1).$

Theorem 8.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{C}=0,$ then the manifold M is isomorphic to the hyperbolic space $H^{2m+1}(-1)$ and is Einstein.

Proof. 

In a Kenmotsu manifold $\left(M,g\right),$ from the condition $\tilde{Q}(\zeta ,X_1)\tilde{C}=0,$ we achieve

$$\begin{array}{cc}\hfill \left(\tilde{Q}(\zeta ,X_1)\tilde{C}\right)(X_2,X_3)\zeta & =\tilde{Q}(\zeta ,X_1)\tilde{C}(X_2,X_3)\zeta -\tilde{C}\left(\tilde{Q}(\zeta ,X_1)(X_2,X_3)\right)\zeta \hfill \\ \hfill & -\tilde{C}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta -\tilde{C}\left(X_2,X_3\right)\tilde{Q}\left(\zeta ,X_1\right)\zeta \hfill \\ \hfill & =0.\hfill \end{array}$$

(40)

In (1), (3), (18), (20), (25), and (28), the terms on the right side of the equations are calculated separately. As such, we obtain

$$\begin{array}{c}\tilde{Q}(\zeta ,X_1)\tilde{C}(X_2,X_3)\zeta ={\displaystyle \frac{\tau }{2m}}{\displaystyle \frac{\widetilde{scal}}{2m(2m+1)}}\left[g(X_1,X_2)\eta \left(X_3\right)\zeta -g(X_1,X_3)\eta \left(X_2\right)\zeta \right],\hfill \\ \tilde{C}\left(\tilde{Q}(\zeta ,X_1)(X_2,X_3)\right)\zeta ={\displaystyle \frac{\tau }{2m}}{\displaystyle \frac{\widetilde{scal}}{2m(2m+1)}}\left[\begin{array}{c}g(X_1,X_2)\eta \left(X_3\right)\zeta -g(X_1,X_3)\eta \left(X_2\right)\zeta \\ -\eta \left(X_2\right)\eta \left(X_3\right)X_1+\eta \left(X_1\right)\eta \left(X_2\right)X_3\end{array}\right],\hfill \\ \tilde{C}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta ={\displaystyle \frac{\tau }{2m}}{\displaystyle \frac{\widetilde{scal}}{2m(2m+1)}}\left[\begin{array}{c}g(X_1,X_3)X_2-g(X_1,X_3)\eta \left(X_2\right)\zeta \\ +\eta \left(X_2\right)\eta \left(X_3\right)X_1-\eta \left(X_1\right)\eta \left(X_3\right)X_2\end{array}\right],\hfill \\ \tilde{C}\left(X_2,X_3\right)\tilde{Q}\left(\zeta ,X_1\right)\zeta ={\displaystyle \frac{\tau }{2m}}{\displaystyle \frac{\widetilde{scal}}{2m(2m+1)}}\left[\eta \left(X_1\right)\eta \left(X_3\right)X_2-\eta \left(X_1\right)\eta \left(X_2\right)X_3\right]+{\displaystyle \frac{\tau }{2m}}\tilde{C}(X_2,X_3)X_1.\hfill \end{array}$$

If we substitute these results into Equation (40), then $\tau =0$, so the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature or

$$\tilde{C}(X_2,X_3)X_1=-{\displaystyle \frac{\widetilde{scal}}{2m(2m+1)}}\left[g(X_1,X_3)X_2-g(X_1,X_2)X_3\right].$$

If the results obtained here are equated with (25), we obtain

$$\tilde{R}\left(X_2,X_3\right)X_1=0.$$

Using Equation (17), we have

$$R\left(X_2,X_3\right)X_1=g(X_2,X_1)X_3-g(X_3,X_1)X_2.$$

Therefore, the manifold M is isomorphic to the hyperbolic space $H^{2m+1}(-1).$ Considering the inner product of the last equation with $X_4,$ we obtain

$$R(X_2,X_3,X_1,X_4)=g(X_1,X_2)g(X_3,X_4)-g(X_1,X_3)g(X_2,X_4).$$

(41)

Taking $X_2=X_4=u_i$ in (41) and summing up to $i$, $1\le i\le 2m+1,$ we obtain

$$S(X_1,X_3)=-2mg(X_1,X_3).$$

Thus, the manifold M is Einstein. □

Corollary 7.

Let the Kenmotsu manifold ($M,g)$ admitting the SvK-connection satisfy the condition $\tilde{Q}(\zeta ,X_1)\tilde{C}=0$; then, the manifold M is isomorphic to the hyperbolic space $H^{2m+1}(-1).$

Corollary 8.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{C}=0,$ then the scalar curvature is $scal=-2m(2m+1).$

Theorem 9.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{H}=0,$ then either the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature or the manifold M is η-Einstein.

Proof. 

In a Kenmotsu manifold $\left(M,g\right),$ from the condition $\tilde{Q}(\zeta ,X_1)\tilde{H}=0,$ we achieve

$$\begin{array}{cc}\hfill \left(\tilde{Q}(\zeta ,X_1)\tilde{H}\right)(X_2,X_3)\zeta & =\tilde{Q}(\zeta ,X_1)\tilde{H}(X_2,X_3)\zeta -\tilde{H}\left(\tilde{Q}(\zeta ,X_1)(X_2,X_3)\right)\zeta \hfill \\ \hfill & -\tilde{H}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta -\tilde{H}\left(X_2,X_3\right)\tilde{Q}\left(\zeta ,X_1\right)\zeta \hfill \\ \hfill & =0.\hfill \end{array}$$

(42)

Now, we calculate the terms on the right side of the equation separately. If $X_1,$ $X_2,$ $X_3$ are written instead of $L{X}_1,$ $L{X}_2,$ $L{X}_3$, and using (1), (3), (18), (20), (22), (27), and (28), we obtain

$$\begin{array}{c}\tilde{Q}(\zeta ,X_1)\tilde{H}(X_2,X_3)\zeta ={\displaystyle \frac{\tau \left(2m+1\right)}{2m\left(2m-1\right)}}\left[g\left(X_1,X_2\right)\eta \left(X_3\right)\zeta -g\left(X_1,X_3\right)\eta \left(X_2\right)\zeta \right],\hfill \\ \tilde{H}\left(\tilde{Q}(\zeta ,X_1)(X_2,X_3)\right)\zeta ={\displaystyle \frac{\tau \left(2m+1\right)}{2m\left(2m-1\right)}}\left[g\left(X_1,X_2\right)X_3+\eta \left(X_1\right)\eta \left(X_2\right)X_3-\eta \left(X_2\right)\eta \left(X_3\right)X_1\right],\hfill \\ \tilde{H}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta ={\displaystyle \frac{\tau \left(2m+1\right)}{2m\left(2m-1\right)}}\left[g\left(X_1,X_3\right)X_2+\eta \left(X_2\right)\eta \left(X_3\right)X_1-\eta \left(X_1\right)\eta \left(X_2\right)X_3\right],\hfill \\ \tilde{H}\left(X_2,X_3\right)\tilde{Q}\left(\zeta ,X_1\right)\zeta ={\displaystyle \frac{\tau \left(2m+1\right)}{2m\left(2m-1\right)}}\left[\eta \left(X_1\right)\eta \left(X_3\right)X_2-\eta \left(X_1\right)\eta \left(X_2\right)X_3\right]+{\displaystyle \frac{\tau }{2m}}\tilde{H}\left(X_2,X_3\right)X_1.\hfill \end{array}$$

If we substitute these results into Equation (42), then $\tau =0$, so the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature or

$$\tilde{H}\left(X_2,X_3\right)X_1={\displaystyle \frac{2m+1}{2m-1}}\left[\begin{array}{c}g\left(X_1,X_2\right)\eta \left(X_3\right)\zeta -g\left(X_1,X_3\right)\eta \left(X_2\right)\zeta \\ -g\left(X_1,X_2\right)X_3-g\left(X_1,X_3\right)X_2\end{array}\right].$$

Using (27), we have

$$R(X_2,X_3)X_1={\displaystyle \frac{2m+1}{2m-1}}\left[g(X_1,X_2)\eta \left(X_3\right)\zeta -g\left(X_1,X_3\right)\eta \left(X_2\right)\zeta -2g\left(X_1,X_2\right)X_3\right].$$

Taking the inner product of above equation with $X_4$ and contracting the last equation, we have

$$S\left(X_1,X_3\right)=-{\displaystyle \frac{6m+3}{2m-1}}g(X_1,X_3)+{\displaystyle \frac{2m+1}{2m-1}}\eta \left(X_1\right)\eta \left(X_3\right).$$

Thus, as shown in the last equation, the manifold M is $\eta$-Einstein. □

Corollary 9.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{H}=0,$ then the scalar curvature is $scal=-{\displaystyle \frac{2(2m+1)\left(3m+1\right)}{2m-1}}.$

Theorem 10.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{P}=0,$ then either the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature or the manifold M is Einstein.

Proof. 

In a Kenmotsu manifold $\left(M,g\right),$ from the condition $\tilde{Q}(\zeta ,X_1)\tilde{P}=0,$ we achieve

$$\begin{array}{cc}\hfill \left(\tilde{Q}(\zeta ,X_1)\tilde{P}\right)(X_2,X_3)\zeta & =\tilde{Q}(\zeta ,X_1)\tilde{P}(X_2,X_3)\zeta -\tilde{P}\left(\tilde{Q}(\zeta ,X_1)(X_2,X_3)\right)\zeta \hfill \\ \hfill & -\tilde{P}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta -\tilde{P}\left(X_2,X_3\right)\tilde{Q}\left(\zeta ,X_1\right)\zeta \hfill \\ \hfill & =0.\hfill \end{array}$$

Using (1), (3), (18), (20), (26), and (28) the terms on the right side of which are calculated separately, we obtain

$$\begin{array}{c}\tilde{Q}(\zeta ,X_1)\tilde{P}(X_2,X_3)\zeta =0,\hfill \\ \tilde{P}\left(\tilde{Q}(\zeta ,X_1)(X_2,X_3)\right)\zeta =0,\hfill \\ \tilde{P}\left(X_2,\tilde{Q}(\zeta ,X_1)X_3\right)\zeta =0,\hfill \\ \tilde{P}(X_2,X_3)\tilde{Q}(\zeta ,X_1)\zeta ={\displaystyle \frac{\tau }{2m}}\tilde{P}(X_2,X_3)X_1\hfill \end{array}$$

Substituting these results into Equation (42), $\tau =0$, so the $\tilde{Q}$-curvature tensor coincides with the Riemannian curvature or

$$\tilde{P}(X_2,X_3)X_1=0.$$

Keeping in mind (26), we have

$$\tilde{P}(X_2,X_3)X_1=\tilde{R}(X_2,X_3)X_1-{\displaystyle \frac{1}{2m}}\left[\tilde{S}(X_1,X_3)X_2-\tilde{S}(X_1,X_2)X_3\right]=0.$$

By considering $L{X}_2\to X_2,$ $L{X}_3\to X_3,$ we obtain

$$R(X_2,X_3)X_1={\displaystyle \frac{1}{2m}}\left[g\left(X_1,X_3\right)X_2-g\left(X_1,X_2\right)X_3\right].$$

(43)

Taking the inner product of (43) with $X_4$ and contracting, we achieve

$$S\left(X_1,X_3\right)=g\left(X_1,X_3\right).$$

Thus, the manifold M is Einstein. □

Corollary 10.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{P}=0,$ then the scalar curvature is $scal=2m+1.$

Corollary 11.

If a Kenmotsu manifold $\left(M,g\right)$ admitting the SvK-connection satisfies the condition $\tilde{Q}(\zeta ,X_1)\tilde{P}=0,$ then the constant curvature of M is ${\displaystyle \frac{1}{2m}}.$

3.1. Example

Let $M=\left\{(u,v,w):u,v,w\in \mathbb{R},w\ne 0\right\}$ be a three-dimensional manifold, where $(u,v,w)$ are the standard coordinates in ${\mathbb{R}}^3.$ We choose the vector fields [11]

$$\begin{array}{cc}\hfill E_1& =e^{-w}{\displaystyle \frac{\partial }{\partial u}}+{\displaystyle \frac{\partial }{\partial w}},\hfill \\ \hfill E_2& =e^{-w}{\displaystyle \frac{\partial }{\partial v}},\hfill \\ \hfill E_3& ={\displaystyle \frac{\partial }{\partial w}},\hfill \end{array}$$

which are linearly independent at each point of M. Let g be the Riemannian metric defined by

$$\begin{array}{cccccc}\hfill g_{11}& =e^{2w}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & \hfill g_{21}& =0\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & \hfill g_{31}& =-e^w\hfill \\ \hfill g_{12}& =0\hfill & \hfill g_{22}& =e^{2w}\hfill & \hfill g_{32}& =0\hfill \\ \hfill g_{13}& =-e^w\hfill & \hfill g_{23}& =0\hfill & \hfill g_{33}& =1\hfill \end{array}$$

and given by the matrix representation

$$g=\left[g_{ij}\right]=\left[\begin{array}{ccc}{e}^{2w}& 0& -e^w\\ 0& e^{2w}& 0\\ -e^w& 0& 1\end{array}\right].$$

Let $\eta$ be the 1-form defined by $\eta \left(X\right)=g(X,E_3)$ for any vector field X on M and let $\varphi$ be the $(1,1)$ tensor field defined by

$$\varphi \left(E_1\right)=E_2,\varphi \left(E_2\right)=-E_1,\varphi \left(E_3\right)=0.$$

Then, using linearity of g and $\varphi$, we obtain

$${\varphi }^{2}X=-E+\eta \left(X\right)E_3,\eta \left(E_3\right)=1,g(\varphi X,\varphi Y)=g(X,Y)-\eta \left(X\right)\eta \left(Y\right),$$

for any $X,Y\in \chi \left(M\right).$ It can be easily seen that the $\eta$ is closed and for $E_3=\xi ,$ the manifold is an almost contact metric manifold. On the other hand, checking only non-zero components of fundamental form $\mathsf{\Phi },$ we have

$$\mathsf{\Phi }\left({\displaystyle \frac{\partial }{\partial u}},{\displaystyle \frac{\partial }{\partial v}}\right)=g\left(\varphi {\displaystyle \frac{\partial }{\partial u}},{\displaystyle \frac{\partial }{\partial v}}\right)=g\left({\displaystyle \frac{\partial }{\partial v}},{\displaystyle \frac{\partial }{\partial v}}\right)=e^{2w}.$$

This gives $\mathsf{\Phi }=e^{2w}du\wedge dv,$ and hence

$$d\mathsf{\Phi }=2e^{2w}dw\wedge du\wedge dv=2\eta \wedge \mathsf{\Phi }.$$

So, $M\left(\varphi ,\xi ,\eta ,g\right)$ is an almost Kenmotsu manifold. Moreover, the Nijenhuis torsion tensor of $\varphi$ vanishes, so the manifold is normal. Consequently, the manifold is Kenmotsu.

In addition, we have,

$$\left[E_1,E_2\right]=E_2,\left[E_1,E_3\right]=E_1-E_3,\left[E_2,E_3\right]=E_2.$$

Let ∇ be the Levi–Civita connection with respect to the metric g. Then, taking $E_3=\xi$ and using Kozsul’s formula, we obtain the following:

$$\begin{array}{cccccc}\hfill {\nabla }_{E_1}{E}_1& =-E_3,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & \hfill {\nabla }_{E_2}{E}_1& =-E_2,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & \hfill {\nabla }_{E_3}{E}_1& =E_3,\hfill \\ \hfill {\nabla }_{E_1}{E}_2& =0,\hfill & \hfill {\nabla }_{E_2}{E}_2& =E_1-E_3,\hfill & \hfill {\nabla }_{E_3}{E}_2& =0,\hfill \\ \hfill {\nabla }_{E_1}{E}_3& =E_1,\hfill & \hfill {\nabla }_{E_2}{E}_3& =E_2,\hfill & \hfill {\nabla }_{E_3}{E}_3& =0,\hfill \end{array}$$

From the above results, the non-vanishing components of the curvature tensors are:

$$\begin{array}{cccc}\hfill R(E_1,E_2)E_1& =-E_1+E_3,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & \hfill R(E_2,E_1)E_1& =E_1-E_3,\hfill \\ \hfill R(E_1,E_2)E_2& =-2E_2,\hfill & \hfill R(E_2,E_3)E_3& =-E_2,\hfill \\ \hfill R(E_1,E_3)E_1& =E_1+2E_3,\hfill & \hfill R(E_3,E_1)E_1& =-E_1-2E_3,\hfill \\ \hfill R(E_1,E_3)E_2& =-E_1,\hfill & \hfill R(E_3,E_1)E_3& =E_1+E_3,\hfill \\ \hfill R(E_1,E_3)E_3& =-E_1-E_3,\hfill & \hfill R(E_3,E_2)E_1& =-2E_2,\hfill \\ \hfill R(E_2,E_3)E_1& =2E_2,\hfill & \hfill R(E_3,E_2)E_2& =E_1,\hfill \\ \hfill R(E_2,E_1)E_2& =2E_2,\hfill & \hfill R(E_3,E_2)E_3& =E_2.\hfill \end{array}$$

According to the results obtained above, the non-vanishing components of the Q tensor are as follows:

$$\begin{array}{cc}\hfill Q(E_1,E_2)E_1& =-E_1-{\displaystyle \frac{\tau }{2}}{E}_2+E_3,Q(E_2,E_3)E_1=2E_2,\hfill \\ \hfill Q(E_1,E_2)E_2& =-{\displaystyle \frac{\tau }{2}}{E}_1-2E_2,Q(E_2,E_3)E_3=-(1+{\displaystyle \frac{\tau }{2}})E_2,\hfill \\ \hfill Q(E_1,E_3)E_1& =E_1+(2+{\displaystyle \frac{\tau }{2}})E_3,Q(E_3,E_1)E_1=-E_1-(2+{\displaystyle \frac{\tau }{2}})E_3,\hfill \\ \hfill Q(E_1,E_3)E_2& =-E_1+{\displaystyle \frac{\tau }{2}}{E}_3,Q(E_3,E_1)E_3=(1+{\displaystyle \frac{\tau }{2}})E_1+E_3,\hfill \\ \hfill Q(E_1,E_3)E_3& =-(1+{\displaystyle \frac{\tau }{2}})E_1-E_3,Q(E_3,E_2)E_1=-2E_2,\hfill \\ \hfill Q(E_2,E_1)E_1& =E_1-{\displaystyle \frac{\tau }{2}}{E}_2-E_3,Q(E_3,E_2)E_2=E_1-{\displaystyle \frac{\tau }{2}}{E}_3,\hfill \\ \hfill Q(E_2,E_1)E_2& =(2-{\displaystyle \frac{\tau }{2}})E_2,Q(E_3,E_2)E_3=(1+{\displaystyle \frac{\tau }{2}})E_2.\hfill \end{array}$$

3.2. Example

We consider the three-dimensional manifold $M=\left\{(u,v,w):u,v,w\in \mathbb{R},w\ne 0\right\}$, where $(u,v,w)$ are the usual coordinates in ${\mathbb{R}}^3.$ Let $E_1,E_2$, and $E_3$ be the linearly independent vector fields on ${\mathbb{R}}^3$ given by

$$\begin{array}{cc}\hfill E_1& =e^{-w}{\displaystyle \frac{\partial }{\partial u}},\hfill \\ \hfill E_2& =e^{-w}{\displaystyle \frac{\partial }{\partial v}},\hfill \\ \hfill E_3& ={\displaystyle \frac{\partial }{\partial w}}.\hfill \end{array}$$

Let g be the Riemannian metric, which is defined by

$$g=e^{2w}\left(du\otimes du+dv\otimes dv\right)+dw\otimes dw.$$

Now, let the 1-form $\eta$ and $(1,1)$ tensor field $\varphi$ be defined by

$$\eta \left(X\right)=g(X,E_3),\varphi \left(E_1\right)=E_2,\varphi \left(E_2\right)=-E_1,\varphi \left(E_3\right)=0$$

for all X on M. The linearity property of g and $\varphi$ yields

$${\varphi }^{2}X=-X+\eta \left(X\right)E_3,g(\varphi X,\varphi Y)=g(X,Y)-\eta \left(X\right)\eta \left(Y\right),$$

for any $X,Y\in X\left(M\right).$ It is clear that $\eta$ is a closed form, and for $E_3=\zeta ,$ the manifold qualifies as almost-contact metric. Furthermore, by considering only the non-zero components of the fundamental form $\mathsf{\Phi }$, we arrive at the following expression:

$$\mathsf{\Phi }\left({\displaystyle \frac{\partial }{\partial u}},{\displaystyle \frac{\partial }{\partial v}}\right)=g\left(\varphi {\displaystyle \frac{\partial }{\partial u}},{\displaystyle \frac{\partial }{\partial u}}\right)=g\left({\displaystyle \frac{\partial }{\partial v}},{\displaystyle \frac{\partial }{\partial v}}\right)=e^{2w}.$$

The above equation gives $\mathsf{\Phi }=e^{2w}du\wedge dv.$ Thus, $d\mathsf{\Phi }=2e^{2w}dw\wedge du\wedge dv=2\eta \wedge \mathsf{\Phi }.$ So, $M\left(\varphi ,\zeta ,\eta ,g\right)$ is an almost-Kenmotsu manifold. Moreover, noting that the Nijenhuis torsion tensor of $\varphi$ vanishes, the manifold is thus normal. Consequently, the manifold is Kenmotsu. In addition, we have,

$$\left[E_1,E_2\right]=0, \left[E_1,E_3\right]=E_1,\left[E_2,E_3\right]=E_2.$$

Let ∇ be the Levi–Civita connection with respect to the metric g. For $E_3=\zeta$, the Kozsul’s formula yields

$$\begin{array}{cccc}\hfill {\nabla }_{E_1}{E}_1& ={\displaystyle \frac{1}{2}}{E}_2-E_3,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill & \hfill {\nabla }_{E_2}{E}_2& =-E_3,\hfill \\ \hfill {\nabla }_{E_1}{E}_3& =E_{1,}\hfill & \hfill {\nabla }_{E_2}{E}_3& =E_2.\hfill \\ \hfill & \hfill & \end{array}$$

Based on the above results and employing Equation (17), we determine the non-vanishing components of the Riemannian curvature tensor with respect to the SvK-connection as follows:

$$\begin{array}{cccc}\hfill \tilde{R}(E_1,E_2)E_1& =-{\displaystyle \frac{1}{2}}{E}_3-2E_2,\hspace{1em}\hfill & \hfill \tilde{R}(E_2,E_1)E_1& ={\displaystyle \frac{1}{2}}{E}_3+2E_2,\hfill \\ \hfill \tilde{R}(E_1,E_3)E_1& ={\displaystyle \frac{E_2}{2}},\hfill & \hfill \tilde{R}(E_3,E_1)E_1& =-{\displaystyle \frac{1}{2}}{E}_2-2E_3.\hfill & \hfill & \end{array}$$

According to the results obtained above, and using Equation (28), the non-vanishing components of the $\tilde{Q}$-curvature tensor admitting the SvK-connection are as follows:

$$\begin{array}{cc}\hfill \tilde{Q}(E_1,E_2)E_1& =\left(-2+{\displaystyle \frac{\tau }{2}}\right)E_2-{\displaystyle \frac{1}{2}}{E}_3,\tilde{Q}(E_2,E_1)E_1=\left(2-{\displaystyle \frac{\tau }{2}}\right)E_2+{\displaystyle \frac{1}{2}}{E}_3,\hfill \\ \hfill \tilde{Q}(E_1,E_3)E_1& ={\displaystyle \frac{1}{2}}{E}_2+{\displaystyle \frac{\tau }{2}}{E}_3,\tilde{Q}(E_3,E_1)E_1=-{\displaystyle \frac{1}{2}}{E}_2+\left(2-{\displaystyle \frac{\tau }{2}}\right)E_3.\hfill \end{array}$$

4. Conclusions

In this study, we examined the Q-curvature tensor on Kenmotsu manifolds equipped with the Schouten–van Kampen connection, uncovering significant geometric properties and classifications. By introducing and analyzing new curvature conditions, such as $\zeta$-$\tilde{Q}$ flat and $\varphi$-$\tilde{Q}$ flat properties, we provided a classification for Kenmotsu manifolds under specific geometric constraints. Our findings reveal that the interaction between the Q-curvature tensor and the Schouten–van Kampen connection provides new insights into the structural properties of these manifolds. Notably, some of our results show that certain Kenmotsu manifolds under the given conditions exhibit Einstein and $\eta$-Einstein properties, further enriching the geometric structure of these manifolds. Additionally, through the construction of a concrete example, we illustrated the applicability of our theoretical results. Future research can extend these concepts to broader classes of almost-contact metric manifolds, contributing to a deeper understanding of affine connections in differential geometry.