Symmetric Tensors in Different Perspectives ^†

Abstract

The main subject of this paper is the theme of differential operators defined for symmetric tensors on a Riemannian manifold and introduced in several new contexts. Some examples for 2-tensors are given and then the grad div operator for symmetric vector forms is defined. A few original operators in ${\mathbb{R}}^n$ related with grad div operator are discussed. Finally, important notions such as the Kenmotsu manifold, with some interesting examples, are also presented.

1. Introduction

This paper involves considerations about some operators for symmetric tensors on Riemannian and Kenmotsu manifolds. Some new operators, related to each other, are created here. These operators also act on Killing and conformal Killing tensors. Many investigations in this area constitute important parts of past (e.g., [1,2,3,4]) and especially of modern (e.g., [5,6,7,8]) mathematics and physics. Such great works as [9,10,11] provide the methodical and historical background for the differential and, particularly, Riemannian geometry. Additionally, e.g., in the paper [5], the historical aspects of the Riemannian manifolds development in the context of Kenmotsu or Sasakian manifolds are very well and clearly presented, covering the evolution of current research. The Killing, conformal Killing, trace-free Killing (Stäckel) and trace-free conformal Killing tensors appear in various scientific problems, e.g., in [12]. What is more, in the area of differential geometry, many other objects have been systematically studied, such as Hyperkähler manifolds, equipped with a Riemannian metric, complex structures and Kähler forms. For example, in the paper [13], the calibrated geometry in Hyperkähler cones $C^{4n+4}$, their 3-Sasakian links $M^{4n+3}$, and the corresponding twistor spaces $Z^{4n+2}$, emphasizing the relationships between submanifold geometries in various spaces have been investigated. It is worth adding that some partial differential equations (e.g., the Korteweg–De Vries equation) can be used to describe the dynamics of water waves under the assumptions of shallow water, unidirectionality, weak nonlinearity and constant depth (see [6]). Many valuable investigations have also been based on the Kenmotsu manifolds theory. Here, we can distinguish interesting papers such as [8,14,15]. For instance, some properties of the lifts of the curvature tensor, the conformal curvature tensor, and the conharmonic curvature tensor of Kenmotsu manifolds that admit a non-symmetric non-metric connection in the tangent bundle were precisely investigated in [14]. Some basic properties and fundamental formulas focused on the concept of almost $\alpha$-Kenmotsu pseudo-Riemannian structure have been studied in [15].

The most important operators for the examinations in this paper are d^s, grad, div (introduced in Section 2) and the new one, defined in Section 3, grad div operator. The grad = $\mathfrak{a}$ d^s − d^s $\mathfrak{a}$ (see (7)) and div = tr d^s− d^s tr (see (10)) are first-order differential operators and their composition led to a second-order differential operator grad div: $C^{\infty }(S^k\otimes T)\to C^{\infty }(S^k\otimes T)$. The grad div operator is a sum of the following operators of the second order: ${\mathfrak{G}}_1$, ${\mathfrak{G}}_2$, ${\mathfrak{G}}_3$, ${\mathfrak{G}}_4$ (Proposition 6) and is formally selfadjoint with respect to the global scalar product (Theorem 1). These results were briefly presented in the Conference Hypercomplex Seminar 2023, 9–15 July 2023 (see [16]).

In Section 4, the symmetric 2-tensors $e_i^{*}\odot e_j^{*}\in C^{\infty }\left(S^2\right)$ for any $i,j=1,\dots ,n$ are investigated in a few examples, where $e_1^{*},\dots ,e_n^{*}$ denote the dual frame for a local frame of sections of the tangent bundle $TM$. Such 2-tensors are studied here sequentially (Examples 1–7) for operators: of substitution, $\mathfrak{a}$, ∇, d^s, grad and tr grad, div and finally for d^s*. These short examples simply show the actions of considered operators and can be very useful in many auxiliary calculations.

Relations between operators ${\mathfrak{G}}_1$, ${\mathfrak{G}}_2$, ${\mathfrak{G}}_3$, ${\mathfrak{G}}_4$, grad div, the polynomial $r^2$ and the 1-form ${\nu }^{*}$ (defined, respectively, by (14) and (15)) are examined in Section 5. The relation between operator P and the polynomial $r^2$ is treated here as the commutator

$$\left(\mathfrak{C}\right)\hspace{1em}P\left(r^2\mathrm{\Xi }\right)-r^{2}P\left(\mathrm{\Xi }\right),$$

of an operator P and the polynomial $r^2$, where $\mathrm{\Xi }=\phi \in C^{\infty }\left(S^k\right)$, or respectively, $\mathrm{\Xi }=\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$ and is investigated for P = grad div (Proposition 11). First, we consider the commutator $\mathfrak{C}$ for $P={\mathfrak{G}}_i$, $i=1,2,3,4$ (Propositions 9, 10) and then, by obtaining Corollaries 3 and 4, we achieve the proof of the main proposition.

Finally, in Section 6, the important notion of the Kenmotsu manifold is briefly introduced (see also [8]) and additionally some new and interesting examples are given. The most important part of this section is Example 11, which presents some innovations for operators: $\mathfrak{a}$, d^s, ∇, grad and div considered on the Kenmotsu manifold given by Example 9. This novel approach to these basic operators indicates that our considerations can be extended on other objects and continued onwards into the new directions.

2. Preliminaries

Let all the objects and morphisms be smooth, i.e., of class $C^{\infty }$.

Consider an oriented manifold M of dimension n equipped with a Riemannian metric g.

For any $p\in M$, denote by

$$\langle ·,·\rangle =g_p(·,·):T_p\times T_p\to \mathbb{R}$$

a scalar product in $T_p$, where $T_p=T_{p}M$ is the tangent space to M at p.

Routinely, denote by $C^{\infty }\left(M\right)$ the ring of smooth functions on M, by $T=TM$ the tangent bundle and by $T^{*}=T^{*}M$ the cotangent bundle. Moreover, denote by ${T^{*}}^m={T^{*}}^{m}M$ the bundle of covariant m-tensors on M and by $S^m=S^{m}M$ its subbundle of m-symmetric tensors (m-forms). What is more, for any bundle E over M, denote by $C^{\infty }\left(E\right)$ the $C^{\infty }\left(M\right)$—module of sections of E.

Let $\nabla :C^{\infty }\left(T\right)\times C^{\infty }\left(T\right)\to C^{\infty }\left(T\right)$ be the Levi–Civita covariant derivative on M and denote by the same symbol ∇ its transmittion to any tensor bundle over M.

Now, in the same context as in [17], we will cover some further necessary facts.

Denote by ⊙ the symmetric product of symmetric tensors (see, e.g., [18]). Let d^s: $C^{\infty }\left(S^k\right)\to C^{\infty }\left(S^{k+1}\right)$ be the operator of the symmetric derivative.

Extend the scalar product $\langle ·,·\rangle$ naturally to the fibers of any tensor bundle on M, especially to $S^k$.

In $S^k$ and $S^k\otimes T$, let us also consider another scalar product

$$\langle ·|·\rangle ={\displaystyle \frac{1}{k!}}\langle ·,·\rangle .$$

(1)

For $X\in T$ denote by ${\iota }_X:{T^{*}}^k\to {T^{*}}^{k-1}$ the operation of substituting X. Next, denote by tr: ${T^{*}}^k\otimes T\to {T^{*}}^{k-1}$ the trace operator acting on vector forms and by $\widetilde{\mathrm{tr}}$: ${T^{*}}^k\to {T^{*}}^{k-2}$ the trace operator defined by the metric g and acting on scalar forms. The same symbols for the restrictions of operators tr and $\widetilde{\mathrm{tr}}$ to any subbundles of the bundles ${T^{*}}^k\otimes T$ and ${T^{*}}^k$ will be used, respectively.

A local expression for the symmetric derivative by the covariant derivative is as follows:

Proposition 1.

Let $e_1,\dots ,e_n$ be a local frame of sections of T and let $e_1^{*},\dots ,e_n^{*}$ be the dual frame; then,

$${\mathrm{d}}^{\mathrm{s}}\phi =\sum _{j=1}^n{e}_j^{*}\odot {\nabla }_{e_j}\phi ,$$

for $\phi \in C^{\infty }\left(S^k\right)$.

Notice also that, for $X,Y\in C^{\infty }\left(T\right)$,

$${\iota }_X{\nabla }_Y={\nabla }_Y{\iota }_X-{\iota }_{{\nabla }_{Y}X}.$$

(2)

Now, extend the symmetric derivative to

$${\mathrm{d}}^{\mathrm{s}}: C^{\infty }(S^k\otimes T)\to C^{\infty }(S^{k+1}\otimes T)$$

by the formula

$${\mathrm{d}}^{\mathrm{s}}(\phi \otimes X)={\mathrm{d}}^{\mathrm{s}}\phi \otimes X+\phi \odot \nabla X,$$

(3)

for $\phi \otimes X\in C^{\infty }(S^k\otimes T)$ where $\nabla X$ is treated as 1-form with values in T. Locally this form can be given by $\nabla X={\sum }_{j=1}^n{e}_j^{*}\otimes {\nabla }_{e_j}X$.

In analogy to Proposition 1 we get the following:

Proposition 2.

Let $e_1,\dots ,e_n$ be a local frame of sections of T and let $e_1^{*},\dots ,e_n^{*}$ be the dual frame; then,

$${\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }=\sum _{j=1}^n{e}_j^{*}\odot {\nabla }_{e_j}\mathrm{\Phi },$$

(4)

for $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$.

It is also worth recalling that a m-symmetric tensor $\phi$ is called a Killing tensor if d^s$\phi =0$. A m-symmetric tensor $\phi$ is called a conformal Killing tensor if d^s$\phi =g\odot \psi$ for some $(m-1)$-symmetric tensor $\psi$.

For any vector bundle E over M and a scalar product $\langle ·,·\rangle$ in E, define the global scalar product $\left(·,·\right)$ in the space of sections of E by

$$\left(·,·\right)={\int }_M\langle ·,·\rangle {\mathrm{\Omega }}_M,$$

(5)

where ${\mathrm{\Omega }}_M$ is the volume form on M defined by the orientation and the metric g (cf. [10]).

Notice that the global scalar product is defined only for such pairs of sections that the integral exists and is finite and this is always the case when, e.g., at least one of the sections is of compact support.

For the bundle $S^k$, we can consider two global scalar products $\left(·,·\right)$ and $\left(·|·\right)$, related by

$$\left(·|·\right)={\displaystyle \frac{1}{k!}}\left(·,·\right).$$

(6)

Remember that with respect to the global scalar product $\left(·|·\right)$:

The operator d^s*: $C^{\infty }\left(S^{k+1}\right)\to C^{\infty }\left(S^k\right)$ formally adjoint to d^s: $C^{\infty }\left(S^k\right)\to C^{\infty }\left(S^{k+1}\right)$ is of the form

$${\mathrm{d}}^{{\mathrm{s}}^{*}}=-\widetilde{\mathrm{tr}}{\nabla }_{|C^{\infty }\left(S^{k+1}\right)}.$$

Similarly, the operator d^s*: $C^{\infty }(S^{k+1}\otimes T)\to C^{\infty }(S^k\otimes T)$ formally adjoint to d^s*: $C^{\infty }(S^k\otimes T)\to C^{\infty }(S^{k+1}\otimes T)$ is of the form

$${\mathrm{d}}^{{\mathrm{s}}^{*}}=-\widetilde{\mathrm{tr}}{\nabla }_{|C^{\infty }(S^{k+1}\otimes T)}.$$

Recall also two important definitions of gradient and divergence. These operators have been studied in many different contexts, e.g., [19].

Definition 1.

The gradient operator grad: $C^{\infty }\left(S^k\right)\to C^{\infty }(S^k\otimes T)$ is defined by

$$\mathrm{grad}=\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}-{\mathrm{d}}^{\mathrm{s}} \mathfrak{a},$$

(7)

where for $k=1,2,\dots$ the operator $\mathfrak{a}:S^k\to S^{k-1}\otimes T$ is given by the equation

$$\mathfrak{a}\phi =\sum _{i=1}^n{\iota }_{e_i}\phi \otimes e_i$$

(8)

and for $k=0$

$$\mathfrak{a}\phi =0.$$

Here, $e_1,\dots ,e_n$ is an orthonormal basis in T.

Proposition 3.

Let $e_1,\dots ,e_n$ be a local orthonormal frame of sections of T; then, locally,

$$\mathrm{grad}\phi =\sum _{i=1}^n{\nabla }_{e_i}\phi \otimes e_i,$$

(9)

for any $\phi \in C^{\infty }\left(S^k\right)$.

Definition 2.

The divergence operator $\mathrm{div}:C^{\infty }(S^k\otimes T)\to C^{\infty }\left(S^k\right)$ is defined by

$$\mathrm{div}=\mathrm{tr}{\mathrm{d}}^{\mathbf{s}}-{\mathrm{d}}^{\mathbf{s}}\mathrm{tr}.$$

(10)

Proposition 4.

For $\phi \in C^{\infty }\left(S^k\right)$ and $X\in C^{\infty }\left(T\right)$, we have

$$\mathrm{div}(\phi \otimes X)={\nabla }_X\phi +\phi \mathrm{div}X,$$

(11)

where in any local orthonormal basis $e_1,\dots ,e_n$, the $\mathrm{div}X$ is defined locally by $\mathrm{div}X={\sum }_{j=1}^n〈e_j,{\nabla }_{e_j}X〉$.

3. The Grad Div Operator

Proposition 5.

Let $\phi \otimes X\in C^{\infty }(S^k\otimes T)$. Then

$$\begin{array}{cc}\hfill \mathrm{tr}{\mathrm{d}}^{\mathrm{s}}(\phi \otimes X)& =\mathrm{div}(\phi \otimes X)+{\mathrm{d}}^{\mathrm{s}}{\iota }_X\phi \hfill \\ \hfill & =\mathrm{div}(\phi \otimes X)+{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}(\phi \otimes X).\hfill \end{array}$$

(12)

Proof. 

Let first prove (12). With Proposition 2, we get

$$\begin{array}{cc}\hfill & \mathrm{tr}{\mathrm{d}}^{\mathrm{s}}(\phi \otimes X)=\mathrm{tr}\left(\sum _{j=1}^n{e}_j^{*}\odot {\nabla }_{e_j}(\phi \otimes X)\right)\hfill \\ \hfill & =\sum _{j=1}^n\mathrm{tr}(e_j^{*}\odot ({\nabla }_{e_j}\phi \otimes X+\phi \otimes {\nabla }_{e_j}X))\hfill \\ \hfill & =\sum _{j=1}^n{\iota }_X(e_j^{*}\odot {\nabla }_{e_j}\phi )+\sum _{j=1}^n{\iota }_{{\nabla }_{e_j}X}(e_j^{*}\odot \phi )\hfill \\ \hfill & =\sum _{j=1}^n(X^j{\nabla }_{e_j}\phi +e_j^{*}\odot {\iota }_X{\nabla }_{e_j}\phi )+\sum _{j=1}^n(\phi \langle e_j,{\nabla }_{e_j}X\rangle +e_j^{*}\odot {\iota }_{{\nabla }_{e_j}X}\phi )\hfill \end{array}$$

Now using (2) and then with Proposition 4 and Proposition 1 we can continue with

$$\begin{array}{cc}\hfill & ={\nabla }_X\phi +\phi \sum _{j=1}^n\langle e_j,{\nabla }_{e_j}X\rangle +\sum _{j=1}^n{e}_j^{*}\odot {\nabla }_{e_j}{\iota }_X\phi \hfill \\ \hfill & =\mathrm{div}(\phi \otimes X)+{\mathrm{d}}^{\mathrm{s}}{\iota }_X\phi =\mathrm{div}(\phi \otimes X)+{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}(\phi \otimes X).\hfill \end{array}$$

□

Let now consider the composition of operators grad and div. This composition led to a second-order differential operator grad div: $C^{\infty }(S^k\otimes T)\to C^{\infty }(S^k\otimes T)$.

Proposition 6.

The second-order differential operator grad div is the following sum of differential operators:

$$\begin{array}{c}\hfill \mathrm{grad} \mathrm{div}={\mathfrak{G}}_1-{\mathfrak{G}}_2-{\mathfrak{G}}_3+{\mathfrak{G}}_4,\end{array}$$

where

$$\begin{array}{c}\hfill {\mathfrak{G}}_1=\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\end{array}$$

$$\begin{array}{c}\hfill {\mathfrak{G}}_2=\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\end{array}$$

$$\begin{array}{c}\hfill {\mathfrak{G}}_3={\mathrm{d}}^{\mathrm{s}} \mathfrak{a}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\end{array}$$

$$\begin{array}{c}\hfill {\mathfrak{G}}_4={\mathrm{d}}^{\mathrm{s}}\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}.\end{array}$$

Proof. 

By Definition 1 of the gradient and Definition 2 of the divergence, we simply get

$$\begin{array}{cc}\hfill & \mathrm{grad}\mathrm{div}=(\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}-{\mathrm{d}}^{\mathrm{s}} \mathfrak{a})(\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}-{\mathrm{d}}^{\mathrm{s}}\mathrm{tr})\hfill \\ \hfill & =\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}-\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}-{\mathrm{d}}^{\mathrm{s}} \mathfrak{a}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}+{\mathrm{d}}^{\mathrm{s}} \mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}.\hfill \end{array}$$

□

Theorem 1.

The differential operator $\mathrm{grad}\mathrm{div}:C^{\infty }(S^k\otimes T)\to C^{\infty }(S^k\otimes T)$ is formally selfadjoint with respect to the global scalar product $\left(·|·\right)$.

Proof. 

This is a direct consequence of the currently known fact that the operators –grad: $C^{\infty }\left(S^k\right)\to C^{\infty }(S^k\otimes T)$ and $\mathrm{div}:C^{\infty }(S^k\otimes T)\to C^{\infty }\left(S^k\right)$ are formally adjoint to each other with respect to the global scalar product $\left(·|·\right)$. □

4. Some Examples for 2-Tensors

Let $e_1,\dots ,e_n$ be a local frame of sections of T and let $e_1^{*},\dots ,e_n^{*}$ be the dual frame. Let $k=2$ and consider now the symmetric 2-tensor $e_i^{*}\odot e_j^{*}\in C^{\infty }\left(S^2\right)$ for any $i,j=1,\dots ,n$. Let $X\in C^{\infty }\left(T\right)$.

Example 1.

We simply get

$$\begin{array}{cc}\hfill & {\iota }_X(e_i^{*}\odot e_j^{*})={\iota }_X{e}_i^{*}\odot e_j^{*}+e_i^{*}\odot {\iota }_X{e}_j^{*}=e_i^{*}\left(X\right)e_j^{*}+e_j^{*}\left(X\right)e_i^{*}\hfill \\ \hfill & =e_i^{*}\left(\sum _{k=1}^n{X}^k{e}_k\right)e_j^{*}+e_j^{*}\left(\sum _{k=1}^n{X}^k{e}_k\right)e_i^{*}=\sum _{k=1}^n{X}^k{e}_i^{*}\left(e_k\right)e_j^{*}+\sum _{k=1}^n{X}^k{e}_j^{*}\left(e_k\right)e_i^{*}\hfill \\ \hfill & =\sum _{k=1}^n{X}^k{\delta }_k^i{e}_j^{*}+\sum _{k=1}^n{X}^k{\delta }_k^j{e}_i^{*}=X^i{e}_j^{*}+X^j{e}_i^{*}.\hfill \end{array}$$

Assume now that $X=e_k$ for some $k=1,\dots ,n$. Then

$${\iota }_{e_k}(e_i^{*}\odot e_j^{*})={\delta }_k^i{e}_j^{*}+{\delta }_k^j{e}_i^{*}.$$

Example 2.

By Example 1 we have

$$\begin{array}{cc}\hfill & \mathfrak{a}(e_i^{*}\odot e_j^{*})=\sum _{k=1}^n{\iota }_{e_k}(e_i^{*}\odot e_j^{*})\otimes e_k=\sum _{k=1}^n({\delta }_k^i{e}_j^{*}+{\delta }_k^j{e}_i^{*})\otimes e_k\hfill \\ \hfill & =e_j^{*}\otimes e_i+e_i^{*}\otimes e_j.\hfill \end{array}$$

Example 3.

Let $k=1,\dots ,n$. Because ${\nabla }_{e_k}$ is a derivation, we simply get

$${\nabla }_{e_k}(e_i^{*}\odot e_j^{*})={\nabla }_{e_k}{e}_i^{*}\odot e_j^{*}+e_i^{*}\odot {\nabla }_{e_k}{e}_j^{*}.$$

(13)

So, for any $r=1,\dots ,n$ we get

$$\begin{array}{cc}\hfill {\iota }_{e_r}{\nabla }_{e_k}(e_i^{*}\odot e_j^{*})& ={\iota }_{e_r}\left({\nabla }_{e_k}{e}_i^{*}\right)e_j^{*}+\left({\iota }_{e_r}{e}_j^{*}\right){\nabla }_{e_k}{e}_i^{*}+\left({\iota }_{e_r}{e}_i^{*}\right){\nabla }_{e_k}{e}_j^{*}\hfill \\ \hfill & +{\iota }_{e_r}\left({\nabla }_{e_k}{e}_j^{*}\right)e_i^{*}\hfill \\ \hfill & ={\iota }_{e_r}\left({\nabla }_{e_k}{e}_i^{*}\right)e_j^{*}+{\delta }_r^j{\nabla }_{e_k}{e}_i^{*}+{\delta }_r^i{\nabla }_{e_k}{e}_j^{*}+{\iota }_{e_r}\left({\nabla }_{e_k}{e}_j^{*}\right)e_i^{*}.\hfill \end{array}$$

Example 4.

Using Proposition 1 and Example 3 (13), we get

$$\begin{array}{cc}\hfill & {\mathrm{d}}^{\mathrm{s}}(e_i^{*}\odot e_j^{*})=\sum _{k=1}^n{e}_k^{*}\odot {\nabla }_{e_k}(e_i^{*}\odot e_j^{*})\hfill \\ \hfill & =\sum _{k=1}^n{e}_k^{*}\odot ({\nabla }_{e_k}{e}_i^{*}\odot e_j^{*}+e_i^{*}\odot {\nabla }_{e_k}{e}_j^{*})\hfill \\ \hfill & =\sum _{k=1}^n(e_j^{*}\odot e_k^{*})\odot {\nabla }_{e_k}{e}_i^{*}+\sum _{k=1}^n(e_i^{*}\odot e_k^{*})\odot {\nabla }_{e_k}{e}_j^{*}.\hfill \end{array}$$

So, using Example 1 for any $r=1,\dots ,n$ we simply get

$$\begin{array}{cc}\hfill & {\iota }_{e_r}{\mathrm{d}}^{\mathrm{s}}(e_i^{*}\odot e_j^{*})=\sum _{k=1}^n{\iota }_{e_r}\left((e_j^{*}\odot e_k^{*})\odot {\nabla }_{e_k}{e}_i^{*}+(e_i^{*}\odot e_k^{*})\odot {\nabla }_{e_k}{e}_j^{*}\right)\hfill \\ \hfill & =\sum _{k=1}^n\left({\iota }_{e_r}(e_j^{*}\odot e_k^{*})\odot {\nabla }_{e_k}{e}_i^{*}+(e_j^{*}\odot e_k^{*})\odot {\iota }_{e_r}{\nabla }_{e_k}{e}_i^{*}\right)\hfill \\ \hfill & +\sum _{k=1}^n\left({\iota }_{e_r}(e_i^{*}\odot e_k^{*})\odot {\nabla }_{e_k}{e}_j^{*}+(e_i^{*}\odot e_k^{*})\odot {\iota }_{e_r}{\nabla }_{e_k}{e}_j^{*}\right)\hfill \\ \hfill & =\sum _{k=1}^n{\delta }_r^j{e}_k^{*}\odot {\nabla }_{e_k}{e}_i^{*}+e_j^{*}\odot {\nabla }_{e_r}{e}_i^{*}+\sum _{k=1}^n(e_j^{*}\odot e_k^{*})\odot {\iota }_{e_r}{\nabla }_{e_k}{e}_i^{*}\hfill \\ \hfill & +\sum _{k=1}^n{\delta }_r^i{e}_k^{*}\odot {\nabla }_{e_k}{e}_j^{*}+e_i^{*}\odot {\nabla }_{e_r}{e}_j^{*}+\sum _{k=1}^n(e_i^{*}\odot e_k^{*})\odot {\iota }_{e_r}{\nabla }_{e_k}{e}_j^{*}.\hfill \end{array}$$

Example 5.

By Proposition 3 and Example 3 (13), we get

$$\begin{array}{cc}\hfill & \mathrm{grad}(e_i^{*}\odot e_j^{*})=\sum _{k=1}^n{\nabla }_{e_k}(e_i^{*}\odot e_j^{*})\otimes e_k\hfill \\ \hfill & =\sum _{k=1}^n\left({\nabla }_{e_k}{e}_i^{*}\odot e_j^{*}+e_i^{*}\odot {\nabla }_{e_k}{e}_j^{*}\right)\otimes e_k.\hfill \end{array}$$

So, we have

$$\begin{array}{cc}\hfill & \mathrm{tr}\mathrm{grad}(e_i^{*}\odot e_j^{*})=\sum _{k=1}^n{\iota }_{e_k}\left({\nabla }_{e_k}{e}_i^{*}\odot e_j^{*}+e_i^{*}\odot {\nabla }_{e_k}{e}_j^{*}\right)\hfill \\ \hfill & =\sum _{k=1}^n\left({\iota }_{e_k}{\nabla }_{e_k}{e}_i^{*}\right)e_j^{*}+\sum _{k=1}^n\left({\nabla }_{e_k}{e}_i^{*}\right){\iota }_{e_k}{e}_j^{*}\hfill \\ \hfill & +\sum _{k=1}^n\left({\iota }_{e_k}{e}_i^{*}\right){\nabla }_{e_k}{e}_j^{*}+\sum _{k=1}^n\left({\iota }_{e_k}{\nabla }_{e_k}{e}_j^{*}\right)e_i^{*}\hfill \\ \hfill & =\sum _{k=1}^n\left({\iota }_{e_k}{\nabla }_{e_k}{e}_i^{*}\right)e_j^{*}+{\nabla }_{e_j}{e}_i^{*}+{\nabla }_{e_i}{e}_j^{*}+\sum _{k=1}^n\left({\iota }_{e_k}{\nabla }_{e_k}{e}_j^{*}\right)e_i^{*}.\hfill \end{array}$$

Example 6.

Let $k=1,\dots ,n$. By Proposition 4 and Example 3 (13), we get

$$\begin{array}{cc}\hfill & \mathrm{div}((e_i^{*}\odot e_j^{*})\otimes e_k)={\nabla }_{e_k}(e_i^{*}\odot e_j^{*})+(e_i^{*}\odot e_j^{*})\mathrm{div}{e}_k\hfill \\ \hfill & ={\nabla }_{e_k}{e}_i^{*}\odot e_j^{*}+e_i^{*}\odot {\nabla }_{e_k}{e}_j^{*}+(e_i^{*}\odot e_j^{*})\mathrm{div}{e}_k.\hfill \end{array}$$

Now, using Example 1 for any $r=1,\dots ,n$, we have

$$\begin{array}{cc}\hfill & {\iota }_{e_r}\mathrm{div}((e_i^{*}\odot e_j^{*})\otimes e_k)=\left({\iota }_{e_r}{\nabla }_{e_k}{e}_i^{*}\right)e_j^{*}+{\delta }_r^j{\nabla }_{e_k}{e}_i^{*}\hfill \\ \hfill & +{\delta }_r^i{\nabla }_{e_k}{e}_j^{*}+\left({\iota }_{e_r}{\nabla }_{e_k}{e}_j^{*}\right)e_i^{*}+({\delta }_r^i{e}_j^{*}+{\delta }_r^j{e}_i^{*})\mathrm{div}{e}_k.\hfill \end{array}$$

Example 7.

By the shape of ${{\mathrm{d}}^{\mathrm{s}}}^{*}$ operator and Example 3 (13), we get

$$\begin{array}{cc}\hfill & {{\mathrm{d}}^{\mathrm{s}}}^{*}(e_i^{*}\odot e_j^{*})=-\widetilde{\mathrm{tr}}\nabla (e_i^{*}\odot e_j^{*})=-\sum _{k=1}^n{\iota }_{e_k}{\iota }_{e_k}\nabla (e_i^{*}\odot e_j^{*})\hfill \\ \hfill & =-\sum _{k=1}^n{\iota }_{e_k}{\nabla }_{e_k}(e_i^{*}\odot e_j^{*})=-\sum _{k=1}^n{\iota }_{e_k}({\nabla }_{e_k}{e}_i^{*}\odot e_j^{*}+e_i^{*}\odot {\nabla }_{e_k}{e}_j^{*})\hfill \\ \hfill & =-\sum _{k=1}^n({\iota }_{e_k}{\nabla }_{e_k}{e}_i^{*}\odot e_j^{*}+{\delta }_k^j{\nabla }_{e_k}{e}_i^{*})-\sum _{k=1}^n({\delta }_k^i{\nabla }_{e_k}{e}_j^{*}+e_i^{*}\odot {\iota }_{e_k}{\nabla }_{e_k}{e}_j^{*})\hfill \\ \hfill & =e_j^{*}\odot {{\mathrm{d}}^{\mathrm{s}}}^{*}\left(e_i^{*}\right)-{\nabla }_{e_j}{e}_i^{*}-{\nabla }_{e_i}{e}_j^{*}+e_i^{*}\odot {{\mathrm{d}}^{\mathrm{s}}}^{*}\left(e_j^{*}\right).\hfill \end{array}$$

5. The Differential Operators in ${\mathbb{R}}^{\mathit{n}}$ in the Context of Grad Div Operator

Assume now that $M={\mathbb{R}}^n$ and let g be the standard, flat metric. Therefore, $e_j={\partial }_j={\displaystyle \frac{\partial }{\partial x^j}}$, $e_j^{*}=d{x}^j$ for $j\in \{1,\dots ,n\}$ form dual bases of the tangent bundle T and the cotangent bundle $T^{*}$, respectively. Let consider the polynomial $r^2$ and the 1-form ${\nu }^{*}$ given by

$$r^2\left(x\right)=\sum _{i=1}^n{\left(x^i\right)}^2$$

(14)

and

$${\nu }_x^{*}=\sum _{i=1}^n{x}^{i}d{x}^i.$$

(15)

For the purposes of the rest of the paper, let us recall a few more necessary facts.

Definition 3.

The operator of symmetric multiplication by ${\nu }^{*}$ is denoted by ${\epsilon }_{{\nu }^{*}}$ and, as usual, defined by

$${\epsilon }_{{\nu }^{*}}\phi ={\nu }^{*}\odot \phi \hspace{1em}\mathrm{for}\mathrm{any}\phi \in C^{\infty }\left(S^k\right)$$

(16)

and

$${\epsilon }_{{\nu }^{*}}\mathrm{\Phi }={\nu }^{*}\odot \mathrm{\Phi }\hspace{1em}\mathrm{for}\mathrm{any}\mathrm{\Phi }\in C^{\infty }(S^k\otimes T).$$

(17)

Notice that for $\phi \otimes X\in C^{\infty }(S^k\otimes T)$ we have

$${\epsilon }_{{\nu }^{*}}(\phi \otimes X)=({\nu }^{*}\odot \phi )\otimes X.$$

(18)

Proposition 7.

Let $\phi \in C^{\infty }\left(S^k\right)$, $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$ and let g be a Riemannian metric tensor. Then, we have the following:

$${\mathrm{d}}^{\mathrm{s}}{\nu }^{*}=2g,$$

(19)

$${\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\phi =2g\odot \phi +{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\phi ,$$

(20)

$${\mathrm{d}}^{\mathrm{s}}\left(r^2\phi \right)-r^2{\mathrm{d}}^{\mathrm{s}}\phi =2{\epsilon }_{{\nu }^{*}}\phi ,$$

(21)

$$\mathfrak{a}\left(r^2\phi \right)=r^2\mathfrak{a}\phi ,$$

(22)

$${\mathrm{d}}^{\mathrm{s}}\left(r^2\mathrm{\Phi }\right)-r^2{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }=2{\epsilon }_{{\nu }^{*}}\mathrm{\Phi },$$

(23)

$$\mathrm{t}\mathrm{r}\left(r^2\mathrm{\Phi }\right)=r^2\mathrm{t}\mathrm{r}\mathrm{\Phi }.$$

(24)

Proof. 

These all properties had been proved in [17]. □

Notice that proposition presented above is very important for commutator theory investigations, e.g., [17,20].

Now we can present the original contributions of this section.

Proposition 8.

Let $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$. Then

$${\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{t}\mathrm{r} \mathrm{\Phi }-{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{t}\mathrm{r} \mathrm{\Phi }=2g\odot \mathrm{t}\mathrm{r} \mathrm{\Phi }$$

(25)

and

$${\epsilon }_{{\nu }^{*}}\mathrm{t}\mathrm{r} {\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }-{\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{t}\mathrm{r} \mathrm{\Phi }={\epsilon }_{{\nu }^{*}}\mathrm{div}\mathrm{\Phi }-2g\odot \mathrm{t}\mathrm{r} \mathrm{\Phi },$$

(26)

where g is a Riemannian metric tensor.

Proof. 

By Proposition 7 (19), we get

$$\begin{array}{cc}\hfill & {\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi }={\mathrm{d}}^{\mathrm{s}}({\nu }^{*}\odot \mathrm{tr}\mathrm{\Phi })={\mathrm{d}}^{\mathrm{s}}{\nu }^{*}\odot \mathrm{tr}\mathrm{\Phi }+{\nu }^{*}\odot {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }\hfill \\ \hfill & =2g\odot \mathrm{tr}\mathrm{\Phi }+{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }.\hfill \end{array}$$

This gives (25).

Now, by (25) we get

$$\begin{array}{cc}\hfill & {\epsilon }_{{\nu }^{*}}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }={\nu }^{*}\odot \mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }={\nu }^{*}\odot \mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }-{\nu }^{*}\odot {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+{\nu }^{*}\odot {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }\hfill \\ \hfill & ={\epsilon }_{{\nu }^{*}}\mathrm{div}\mathrm{\Phi }+{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }={\epsilon }_{{\nu }^{*}}\mathrm{div}\mathrm{\Phi }+{\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi }-2g\odot \mathrm{tr}\mathrm{\Phi }.\hfill \end{array}$$

This gives (26). □

Corollary 1.

Let $\phi \in C^{\infty }\left(S^k\right)$ be a Killing tensor. Then ${\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\phi$ is a conformal Killing tensor.

Proof. 

It is a direct consequence of (20) in Proposition 7. □

Let us consider the operator:

$$\left(\mathfrak{C}\right)\hspace{1em}P\left(r^2\mathrm{\Xi }\right)-r^{2}P\left(\mathrm{\Xi }\right),$$

as the commutator of an operator P and the polynomial $r^2$, where $\mathrm{\Xi }=\phi \in C^{\infty }\left(S^k\right)$, or respectively, $\mathrm{\Xi }=\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$.

Using Proposition 7 in the context of the commutator $\left(\mathfrak{C}\right)$ of corresponding operators and polynomial $r^2$, we can, respectively, consider the commutator $\left(\mathfrak{C}\right)$ of operators ${\mathfrak{G}}_i$, $i=1,2,3,4$, and polynomial $r^2$.

Proposition 9.

Let $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$. Then

$${\mathfrak{G}}_1\left(r^2\mathrm{\Phi }\right)-r^2{\mathfrak{G}}_1\left(\mathrm{\Phi }\right)=2\mathfrak{a}{\epsilon }_{{\nu }^{*}}\mathrm{t}\mathrm{r} {\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{t}\mathrm{r} {\epsilon }_{{\nu }^{*}}\mathrm{\Phi },$$

(27)

$${\mathfrak{G}}_2\left(r^2\mathrm{\Phi }\right)-r^2{\mathfrak{G}}_2\left(\mathrm{\Phi }\right)=2\mathfrak{a}{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{t}\mathrm{r} \mathrm{\Phi }+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{t}\mathrm{r} \mathrm{\Phi }.$$

(28)

Proof. 

First, we prove (27). By Proposition 7: (21)–(24) we get

$$\begin{array}{cc}\hfill & {\mathfrak{G}}_1\left(r^2\mathrm{\Phi }\right)=\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}(r^2{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2{\epsilon }_{{\nu }^{*}}\mathrm{\Phi })=\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\left(r^2{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }\right)+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi }\hfill \\ \hfill & =\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\left(r^2\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }\right)+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi }\hfill \\ \hfill & =\mathfrak{a}(r^2{\mathrm{d}}^{\mathrm{s}} \mathrm{tr} {\mathrm{d}}^{\mathrm{s}} \mathrm{\Phi }+2{\epsilon }_{{\nu }^{*}}\mathrm{tr} {\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi })+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi }\hfill \\ \hfill & =r^2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr} {\mathrm{d}}^{\mathrm{s}} \mathrm{\Phi }+2\mathfrak{a}{\epsilon }_{{\nu }^{*}}\mathrm{tr} {\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi }\hfill \\ \hfill & =r^2{\mathfrak{G}}_1\left(\mathrm{\Phi }\right)+2\mathfrak{a}{\epsilon }_{{\nu }^{*}}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2\mathfrak{a}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi }.\hfill \end{array}$$

Now, let us prove (28). By Proposition 7: (21), (22) and first (24), we sequentially get

$$\begin{array}{cc}\hfill & {\mathfrak{G}}_2\left(r^2\mathrm{\Phi }\right)=\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}{\mathrm{d}}^{\mathrm{s}}\left(r^2\mathrm{tr}\mathrm{\Phi }\right)=\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}(r^2{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2{\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi })\hfill \\ \hfill & =\mathfrak{a}(r^2{\mathrm{d}}^{\mathrm{s}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2{\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi })\hfill \\ \hfill & =r^2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2\mathfrak{a}{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi }\hfill \\ \hfill & =r^2{\mathfrak{G}}_2\left(\mathrm{\Phi }\right)+2\mathfrak{a}{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi }.\hfill \end{array}$$

□

Corollary 2.

Let $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$. Then

$${\mathfrak{G}}_2\left(r^2\mathrm{\Phi }\right)-r^2{\mathfrak{G}}_2\left(\mathrm{\Phi }\right)=2\mathfrak{a}{\epsilon }_{{\nu }^{*}}({\mathrm{d}}^{\mathrm{s}} \mathrm{t}\mathrm{r}+\mathrm{t}\mathrm{r} {\mathrm{d}}^{\mathrm{s}}-\mathrm{div})\mathrm{\Phi }+2g\odot \mathrm{t}\mathrm{r} \mathrm{\Phi },$$

(29)

where g is a Riemannian metric tensor.

Proof. 

By (28) of Proposition 9 and (26) of Proposition 8 we get

$$\begin{array}{cc}\hfill & {\mathfrak{G}}_2\left(r^2\mathrm{\Phi }\right)-r^2{\mathfrak{G}}_2\left(\mathrm{\Phi }\right)=2\mathfrak{a}{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}{\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi }\hfill \\ \hfill & =2\mathfrak{a}{\epsilon }_{{\nu }^{*}}{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2\mathfrak{a}({\epsilon }_{{\nu }^{*}}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }-{\epsilon }_{{\nu }^{*}}\mathrm{div}\mathrm{\Phi }+2g\odot \mathrm{tr}\mathrm{\Phi })\hfill \\ \hfill & =2\mathfrak{a}{\epsilon }_{{\nu }^{*}}({\mathrm{d}}^{\mathrm{s}}\mathrm{tr}+\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}-\mathrm{div})\mathrm{\Phi }+2g\odot \mathrm{tr}\mathrm{\Phi }.\hfill \end{array}$$

□

Corollary 3.

Let $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$. Then

$$\begin{array}{cc}\hfill & \left({\mathfrak{G}}_1-{\mathfrak{G}}_2\right)\left(r^2\mathrm{\Phi }\right)-r^2\left({\mathfrak{G}}_1-{\mathfrak{G}}_2\right)\left(\mathrm{\Phi }\right)\hfill \\ \hfill & =2\mathfrak{a}{\epsilon }_{{\nu }^{*}}\mathrm{div}\mathrm{\Phi }+2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\left(\mathrm{tr}{\epsilon }_{{\nu }^{*}}-{\epsilon }_{{\nu }^{*}}\mathrm{tr}\right)\mathrm{\Phi }.\hfill \end{array}$$

(30)

Proof. 

Using Proposition 9 (27) and (28) and Definition 2 we simply get the assertion. □

Proposition 10.

Let $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$. Then

$${\mathfrak{G}}_3\left(r^2\mathrm{\Phi }\right)-r^2{\mathfrak{G}}_3\left(\mathrm{\Phi }\right)=2{\epsilon }_{{\nu }^{*}}\mathfrak{a} \mathrm{t}\mathrm{r} {\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2{\mathrm{d}}^{\mathrm{s}}\mathfrak{a} \mathrm{t}\mathrm{r} {\epsilon }_{{\nu }^{*}}\mathrm{\Phi },$$

(31)

$${\mathfrak{G}}_4\left(r^2\mathrm{\Phi }\right)-r^2{\mathfrak{G}}_4\left(\mathrm{\Phi }\right)=2{\epsilon }_{{\nu }^{*}}\mathfrak{a} {\mathrm{d}}^{\mathrm{s}} \mathrm{t}\mathrm{r} \mathrm{\Phi }+2{\mathrm{d}}^{\mathrm{s}} \mathfrak{a} {\epsilon }_{{\nu }^{*}} \mathrm{t}\mathrm{r} \mathrm{\Phi }.$$

(32)

Proof. 

By Proposition 7: (23), (24), (22) and (21) we sequentially get

$$\begin{array}{cc}\hfill & {\mathfrak{G}}_3\left(r^2\mathrm{\Phi }\right)={\mathrm{d}}^{\mathrm{s}} \mathfrak{a} \mathrm{tr}(r^2{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2{\epsilon }_{{\nu }^{*}}\mathrm{\Phi })={\mathrm{d}}^{\mathrm{s}}\mathfrak{a}(r^2\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi })\hfill \\ \hfill & ={\mathrm{d}}^{\mathrm{s}}(r^2\mathfrak{a}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2\mathfrak{a}\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi })\hfill \\ \hfill & =r^2{\mathrm{d}}^{\mathrm{s}}\mathfrak{a}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2{\epsilon }_{{\nu }^{*}}\mathfrak{a}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2{\mathrm{d}}^{\mathrm{s}}\mathfrak{a}\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi }\hfill \\ \hfill & =r^2{\mathfrak{G}}_3\left(\mathrm{\Phi }\right)+2{\epsilon }_{{\nu }^{*}}\mathfrak{a}\mathrm{tr}{\mathrm{d}}^{\mathrm{s}}\mathrm{\Phi }+2{\mathrm{d}}^{\mathrm{s}}\mathfrak{a}\mathrm{tr}{\epsilon }_{{\nu }^{*}}\mathrm{\Phi }.\hfill \end{array}$$

The above calculations give us (31). Next, let us prove (32). By Proposition 7: (24), (21), (22) and (23), we sequentially get

$$\begin{array}{cc}\hfill & {\mathfrak{G}}_4\left(r^2\mathrm{\Phi }\right)={\mathrm{d}}^{\mathrm{s}} \mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\left(r^2\mathrm{tr}\mathrm{\Phi }\right)={\mathrm{d}}^{\mathrm{s}}\mathfrak{a}(r^2{\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2{\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi })\hfill \\ \hfill & ={\mathrm{d}}^{\mathrm{s}}(r^2\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2\mathfrak{a} {\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi })\hfill \\ \hfill & =r^2{\mathrm{d}}^{\mathrm{s}} \mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2{\epsilon }_{{\nu }^{*}}\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2{\mathrm{d}}^{\mathrm{s}} \mathfrak{a} {\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi }\hfill \\ \hfill & =r^2{\mathfrak{G}}_4\left(\mathrm{\Phi }\right)+2{\epsilon }_{{\nu }^{*}}\mathfrak{a} {\mathrm{d}}^{\mathrm{s}}\mathrm{tr}\mathrm{\Phi }+2{\mathrm{d}}^{\mathrm{s}} \mathfrak{a} {\epsilon }_{{\nu }^{*}}\mathrm{tr}\mathrm{\Phi }.\hfill \end{array}$$

□

Corollary 4.

Let $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$. Then

$$\begin{array}{cc}\hfill & \left({\mathfrak{G}}_4-{\mathfrak{G}}_3\right)\left(r^2\mathrm{\Phi }\right)-r^2\left({\mathfrak{G}}_4-{\mathfrak{G}}_3\right)\left(\mathrm{\Phi }\right)\hfill \\ \hfill & =-2{\epsilon }_{{\nu }^{*}}\mathfrak{a}\mathrm{div}\mathrm{\Phi }-2{\mathrm{d}}^{\mathrm{s}}\mathfrak{a}\left(\mathrm{tr}{\epsilon }_{{\nu }^{*}}-{\epsilon }_{{\nu }^{*}}\mathrm{tr}\right)\mathrm{\Phi }.\hfill \end{array}$$

(33)

Proof. 

Using Proposition 10 (31) and (32) and Definition 2, we simply get the assertion. □

Finally, the commutator ($\mathfrak{C}$) of grad div and polynomial $r^2$ is given by the following:

Proposition 11.

Let $\mathrm{\Phi }\in C^{\infty }(S^k\otimes T)$. Then

$$\begin{array}{cc}\hfill & \mathrm{grad}\mathrm{div}\left(r^2\mathrm{\Phi }\right)-r^2\mathrm{grad}\mathrm{div}\left(\mathrm{\Phi }\right)\hfill \\ \hfill & =2(\mathfrak{a}{\epsilon }_{{\nu }^{*}}-{\epsilon }_{{\nu }^{*}}\mathfrak{a})\mathrm{div}\mathrm{\Phi }+2\mathrm{grad}(\mathrm{t}\mathrm{r} {\epsilon }_{{\nu }^{*}}-{\epsilon }_{{\nu }^{*}}\mathrm{tr})\mathrm{\Phi }.\hfill \end{array}$$

(34)

Proof. 

By Proposition 6, Corollaries 3, 4 and Definition 1 we sequentially get

$$\begin{array}{cc}\hfill & \mathrm{grad}\mathrm{div}\left(r^2\mathrm{\Phi }\right)\hfill \\ \hfill & =r^2\left({\mathfrak{G}}_1-{\mathfrak{G}}_2\right)\left(\mathrm{\Phi }\right)+2\mathfrak{a}{\epsilon }_{{\nu }^{*}}\mathrm{div}\mathrm{\Phi }+2\mathfrak{a}{\mathrm{d}}^{\mathrm{s}}\left(\mathrm{tr}{\epsilon }_{{\nu }^{*}}-{\epsilon }_{{\nu }^{*}}\mathrm{tr}\right)\mathrm{\Phi }\hfill \\ \hfill & +r^2\left({\mathfrak{G}}_4-{\mathfrak{G}}_3\right)\left(\mathrm{\Phi }\right)-2{\epsilon }_{{\nu }^{*}}\mathfrak{a}\mathrm{div}\mathrm{\Phi }-2{\mathrm{d}}^{\mathrm{s}}\mathfrak{a}\left(\mathrm{tr}{\epsilon }_{{\nu }^{*}}-{\epsilon }_{{\nu }^{*}}\mathrm{tr}\right)\mathrm{\Phi }\hfill \\ \hfill & =r^2\mathrm{grad}\mathrm{div}\left(\mathrm{\Phi }\right)+2(\mathfrak{a}{\epsilon }_{{\nu }^{*}}-{\epsilon }_{{\nu }^{*}}\mathfrak{a})\mathrm{div}\mathrm{\Phi }+2\mathrm{grad}(\mathrm{tr}{\epsilon }_{{\nu }^{*}}-{\epsilon }_{{\nu }^{*}}\mathrm{tr})\mathrm{\Phi }.\hfill \end{array}$$

□

6. A Kenmotsu Manifold

Definition 4.

Let M be a $(2m+1)$-dimensional manifold. Then, M is said to admit an almost contact metric structure if there exist a $(1,1)$-tensor field φ, a vector field ξ, a 1-form η and a Riemannian metric g such that

$$\phi \xi =0,\hspace{1em}{\phi }^{2}U=-U+\eta \left(U\right)\xi ,\hspace{1em}\eta \left(\xi \right)=1,\hspace{1em}\eta \left(\phi U\right)=0$$

(35)

and

$$g(\phi U,\phi V)=g(U,V)-\eta \left(U\right)\eta \left(V\right),\hspace{1em}g(U,\xi )=\eta \left(U\right),$$

(36)

where U and V are vector fields on M (see, e.g., [8]).

Let us construct the following example:

Example 8.

Let $M=\{(x_1,\dots ,x_{2n+2})\in {\mathbb{R}}^{2n+2}:r^2\left(x\right)=1\}$, where $r^2\left(x\right)={\sum }_{i=1}^{2n+2}{\left(x^i\right)}^2$, and let g be the standard, flat metric in $\overline{M}={\mathbb{R}}^{2n+2}$. Then $e_j={\partial }_j={\displaystyle \frac{\partial }{\partial x^j}}$, $e_j^{*}=d{x}^j$ for $j\in \{1,\dots ,2n+2\}$ are dual bases (of $T\overline{M}$ and $T^{*}\overline{M}$, respectively). Let ${\xi }_x={\nu }_x={\sum }_{i=1}^{2n+2}{x}^i{\partial }_i$ and ${\eta }_x={\nu }_x^{*}={\sum }_{i=1}^{2n+2}{x}^{i}d{x}^i$. Let φ be a tensor field of type $(1,1)$ defined by

$$\phi \left(U\right)=-U+\eta \left(U\right)\xi \hspace{1em}\mathrm{for} \mathrm{chosen} U \mathrm{on} \overline{M}.$$

(37)

Then on M we have:

$$\eta \left(\xi \right)=r^2=1,$$

(38)

$$\eta \left(U\right)=\sum _{j=1}^{2n+2}{x}^{j}d{x}^j\left(U\right)=\sum _{j=1}^{2n+2}d{x}^j\otimes d{x}^j(U,\xi )=g(U,\xi ),$$

(39)

$${\phi }^2\left(U\right)=-\phi \left(U\right)+\eta \left(\phi \left(U\right)\right)\xi =-U+\eta \left(U\right)\xi ,$$

(40)

$$\phi \left(\xi \right)=-\xi +\eta \left(\xi \right)\xi =-\xi +\xi =0,$$

(41)

$$\eta \left(\phi \right(U\left)\right)=\eta (-U+\eta (U\left)\xi \right)=-\eta \left(U\right)+\eta \left(U\right)\eta \left(\xi \right)=0,$$

(42)

$$g(\phi U,\phi V)=g(-U+\eta (U)\xi ,-V+\eta (V\left)\xi \right)=g(U,V)-\eta \left(U\right)\eta \left(V\right).$$

(43)

Therefore, M admits an almost contact metric structure.

Definition 5.

Let M be a submanifold in $\overline{M}$ such that M admits an almost contact metric structure. If

$$\left({\overline{\nabla }}_U\phi \right)V=-g(U,\phi V)\xi -\eta \left(V\right)\phi U,\hspace{1em}{\overline{\nabla }}_U\xi =U-\eta \left(U\right)\xi ,$$

(44)

where $\overline{\nabla }$ is a Levi–Civita connection on $\overline{M}$, then the structure $(M,\phi ,\xi ,\eta ,g)$ is called a Kenmotsu manifold (see, e.g., [8]).

Example 9.

(See [8].) Let $\overline{M}=\{(x,y,z)\in {\mathbb{R}}^3:z\ne 0\}$, where $(x,y,z)$ are the standard coordinates in ${\mathbb{R}}^3$. Then, $\overline{M}$ is a three-dimensional manifold. Consider on $\overline{M}$ the metric g given by:

$$g=\eta \otimes \eta +e^{2z}(dx\otimes dx+dy\otimes dy).$$

(45)

Define now three linearly independent at each point of $\overline{M}$ vector fields $e_1,e_2,e_3$ by:

$$e_1=e^{-z}{\displaystyle \frac{\partial }{\partial x}},\hspace{1em}{e}_2=e^{-z}{\displaystyle \frac{\partial }{\partial y}},\hspace{1em}{e}_3={\displaystyle \frac{\partial }{\partial z}}.$$

(46)

Then for $i,j=\{1,2,3\}$ we have

$$g(e_i,e_j)=\left\{\begin{array}{c}0,\hspace{1em}\mathrm{for}i\ne j,\hfill \\ 1,\hspace{1em}\mathrm{for}i=j.\hfill \end{array}\right.$$

Here, the vector field ξ and the 1-form η are given by:

$$\xi =e_3\hspace{1em}\mathit{\text{and}}\hspace{1em}\eta \left(U\right)=g(U,e_3)\hspace{1em}\mathrm{for} \mathrm{chosen} U \mathrm{on} \overline{M}.$$

(47)

What is more, the following hold:

$$\begin{array}{cc}\hfill & {\overline{\nabla }}_{e_1}{e}_1=-e_3,\hspace{1em}{\overline{\nabla }}_{e_1}{e}_2=0,\hspace{1em}{\overline{\nabla }}_{e_1}{e}_3=e_1,\hfill \\ \hfill & {\overline{\nabla }}_{e_2}{e}_1=0\hspace{1em}{\overline{\nabla }}_{e_2}{e}_2=-e_3,\hspace{1em}{\overline{\nabla }}_{e_2}{e}_3=e_2,\hfill \\ \hfill & {\overline{\nabla }}_{e_3}{e}_1=e_1,\hspace{1em}{\overline{\nabla }}_{e_3}{e}_2=e_2,\hspace{1em}{\overline{\nabla }}_{e_3}{e}_3=0.\hfill \end{array}$$

Finally, taking a tensor field φ of type $(1,1)$ defined by $\phi \left(e_1\right)=0$, $\phi \left(e_2\right)=0$, $\phi \left(e_3\right)=0$, we get that conditions (35) and (36) hold. Hence, this manifold is a Kenmotsu manifold. What is more, φ is the Killing tensor field so the manifold $\overline{M}$ is a Kenmotsu manifold with the Killing tensor field φ.

Example 10.

(See [8].) Let $\overline{M}=\{(x_1,x_2,x_3,x_4,v)\in {\mathbb{R}}^5:v\ne 0\}$, where $(x_1,x_2,x_3,x_4,v)$ are the standard coordinates in ${\mathbb{R}}^5$. Then, $\overline{M}$ is a five-dimensional manifold. Consider on $\overline{M}$ the metric g given by:

$$g=\eta \otimes \eta +e^{2v}\sum _{i=1}^{4}d{x}_i\otimes d{x}_i.$$

(48)

Define five linearly independent vector fields $e_i$, $i\in \{1,\dots ,5\}$ at each point of $\overline{M}$, by:

$$e_1=e^{-v}{\displaystyle \frac{\partial }{\partial x_1}},\hspace{1em}{e}_2=e^{-v}{\displaystyle \frac{\partial }{\partial x_2}},\hspace{1em}{e}_3=e^{-v}{\displaystyle \frac{\partial }{\partial x_3}},\hspace{1em}{e}_4=e^{-v}{\displaystyle \frac{\partial }{\partial x_4}},\hspace{1em}{e}_5={\displaystyle \frac{\partial }{\partial v}}.$$

(49)

Then for $i,j=\{1,\dots ,5\}$ we have

$$g(e_i,e_j)=\left\{\begin{array}{c}0,\hspace{1em}\mathrm{for}i\ne j,\hfill \\ 1,\hspace{1em}\mathrm{for}i=j.\hfill \end{array}\right.$$

Here, the vector field ξ and the 1-form η are given by:

$$\xi =e_5\hspace{1em}\mathrm{and}\hspace{1em}\eta \left(U\right)=g(U,e_5)\hspace{1em}\mathrm{for} \mathrm{chosen} U \mathrm{on} \overline{M}.$$

(50)

Finally, taking a tensor field φ of type $(1,1)$ defined by:

$$\phi \left(e_1\right)=\phi \left(e_2\right)=\phi \left(e_3\right)=\phi \left(e_4\right)=\phi \left(e_5\right)=0,$$

we get that conditions (35) and (36) hold. Hence, this manifold is a Kenmotsu manifold. What is more, φ is the Killing tensor field so the manifold $\overline{M}$ is a Kenmotsu manifold with the Killing tensor field φ.

Now, we are going to present some innovative examples for our basic operators considered on the following Kenmotsu manifold:

Example 11.

Let $\overline{M}$ be a Kenmotsu manifold with the Killing tensor field φ as in Example 9. Let $S^m=S^m\overline{M}$ be a bundle of m-symmetric tensors on $\overline{M}$. Then, for $\psi \in S^m$ of the form $\psi =f{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}$, where $e_1^{*}=e^{z}dx$, $e_2^{*}=e^{z}dy$, $e_3^{*}=dz$, ${\alpha }_1+{\alpha }_2+{\alpha }_3=m$ and for a smooth function f on $\overline{M}$, we have

$$\begin{array}{cc}\hfill & \mathfrak{a}\psi =\sum _{i=1}^3{\iota }_{e_i}\psi \otimes e_i=f\left({\alpha }_1{e_1^{*}}^{{\alpha }_1-1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}\right)\otimes e_1\hfill \\ \hfill & +f\left({\alpha }_2{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2-1}\odot {e_3^{*}}^{{\alpha }_3}\right)\otimes e_2+f\left({\alpha }_3{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3-1}\right)\otimes e_3\hfill \\ \hfill & ={\alpha }_{1}f{e}^{({\alpha }_1+{\alpha }_2-2)z}{dx}^{{\alpha }_1-1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3}\otimes {\displaystyle \frac{\partial }{\partial x}}\hfill \\ \hfill & +{\alpha }_{2}f{e}^{({\alpha }_1+{\alpha }_2-2)z}{dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2-1}\odot {dz}^{{\alpha }_3}\otimes {\displaystyle \frac{\partial }{\partial y}}\hfill \\ \hfill & +{\alpha }_{3}f{e}^{({\alpha }_1+{\alpha }_2)z}{dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3-1}\otimes {\displaystyle \frac{\partial }{\partial z}}.\hfill \end{array}$$

Notice that for any $\overline{\psi }\in C^{\infty }\left(S^m\right)$ we have

$$\begin{array}{cc}\hfill & \mathfrak{a}\overline{\psi }=e^{-2z}\left({\iota }_{{\displaystyle \frac{\partial }{\partial x}}}\overline{\psi }\otimes {\displaystyle \frac{\partial }{\partial x}}+{\iota }_{{\displaystyle \frac{\partial }{\partial y}}}\overline{\psi }\otimes {\displaystyle \frac{\partial }{\partial y}}\right)+{\iota }_{{\displaystyle \frac{\partial }{\partial z}}}\overline{\psi }\otimes {\displaystyle \frac{\partial }{\partial z}},\hfill \end{array}$$

$$\begin{array}{c}\hfill d^s\overline{\psi }=\sum _{j=1}^3{e}_j^{*}\odot {\overline{\nabla }}_{e_j}\overline{\psi }=dx\odot {\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial x}}}\overline{\psi }+dy\odot {\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial y}}}\overline{\psi }+dz\odot {\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial z}}}\overline{\psi };\end{array}$$

After simple calculations, we get the following

$$\begin{array}{cc}\hfill & {\overline{\nabla }}_{e_1}{e}_1^{*}=-e_3^{*},\hspace{1em}{\overline{\nabla }}_{e_1}{e}_2^{*}=0,\hspace{1em}{\overline{\nabla }}_{e_1}{e}_3^{*}=e_1^{*},\hfill \\ \hfill & {\overline{\nabla }}_{e_2}{e}_1^{*}=0\hspace{1em}{\overline{\nabla }}_{e_2}{e}_2^{*}=-e_3^{*},\hspace{1em}{\overline{\nabla }}_{e_2}{e}_3^{*}=e_2^{*},\hfill \\ \hfill & {\overline{\nabla }}_{e_3}{e}_1^{*}=-e_1^{*},\hspace{1em}{\overline{\nabla }}_{e_3}{e}_2^{*}=-e_2^{*},\hspace{1em}{\overline{\nabla }}_{e_3}{e}_3^{*}=0.\hfill \end{array}$$

Therefore, for $\psi =f{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}$ we have

$$\begin{array}{cc}\hfill & {\overline{\nabla }}_{e_1}\psi =\left({\overline{\nabla }}_{e_1}f\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}\hfill \\ \hfill & +f{\alpha }_1{\overline{\nabla }}_{e_1}{e}_1^{*}\odot {e_1^{*}}^{{\alpha }_1-1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}\hfill \\ \hfill & +f{\alpha }_2,{\overline{\nabla }}_{e_1}{e}_2^{*}\odot {e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2-1}\odot {e_3^{*}}^{{\alpha }_3}\hfill \\ \hfill & +f{\alpha }_3,{\overline{\nabla }}_{e_1}{e}_3^{*}\odot {e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3-1}\hfill \\ \hfill & =\left({\overline{\nabla }}_{e_1}f\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}-f{\alpha }_1{e_1^{*}}^{{\alpha }_1-1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3+1}\hfill \\ \hfill & +f{\alpha }_3{e_1^{*}}^{{\alpha }_1+1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3-1};\hfill \end{array}$$

$$\begin{array}{cc}\hfill & {\overline{\nabla }}_{e_2}\psi =\left({\overline{\nabla }}_{e_2}f\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}-f{\alpha }_2{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2-1}\odot {e_3^{*}}^{{\alpha }_3+1}\hfill \\ \hfill & +f{\alpha }_3{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2+1}\odot {e_3^{*}}^{{\alpha }_3-1};\hfill \end{array}$$

$$\begin{array}{cc}\hfill & {\overline{\nabla }}_{e_3}\psi =\left({\overline{\nabla }}_{e_3}f-f({\alpha }_1+{\alpha }_2)\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}.\hfill \end{array}$$

and next

$$\begin{array}{cc}\hfill & d^s\psi =\left({\overline{\nabla }}_{e_1}f\right){e_1^{*}}^{{\alpha }_1+1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}+\left({\overline{\nabla }}_{e_2}f\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2+1}\odot {e_3^{*}}^{{\alpha }_3}\hfill \\ \hfill & +f{\alpha }_3{e_1^{*}}^{{\alpha }_1+2}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3-1}+f{\alpha }_3{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2+2}\odot {e_3^{*}}^{{\alpha }_3-1}\hfill \\ \hfill & +\left({\overline{\nabla }}_{e_3}f-2f({\alpha }_1+{\alpha }_2)\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3+1}.\hfill \end{array}$$

Using Proposition 3, for any $\overline{\psi }\in C^{\infty }\left(S^m\right)$, we get

$$\begin{array}{cc}\hfill \mathrm{grad}\overline{\psi }& =\sum _{i=1}^3,{\overline{\nabla }}_{e_i}\overline{\psi }\otimes e_i=e^{-2z}\left({\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial x}}}\overline{\psi }\otimes {\displaystyle \frac{\partial }{\partial x}}+{\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial y}}}\overline{\psi }\otimes {\displaystyle \frac{\partial }{\partial y}}\right)\hfill \\ \hfill & +{\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial z}}}\overline{\psi }\otimes {\displaystyle \frac{\partial }{\partial z}}.\hfill \end{array}$$

Let $\psi =f{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}$. Then

$$\begin{array}{cc}\hfill & \mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\psi =\left({\overline{\nabla }}_{e_1}f\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}\otimes e_1-f{\alpha }_1{e_1^{*}}^{{\alpha }_1-1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3+1}\otimes e_1\hfill \\ \hfill & +f{\alpha }_3{e_1^{*}}^{{\alpha }_1+1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3-1}\otimes e_1+\left({\overline{\nabla }}_{e_2}f\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}\otimes e_2\hfill \\ \hfill & -f{\alpha }_2{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2-1}\odot {e_3^{*}}^{{\alpha }_3+1}\otimes e_2+f{\alpha }_3{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2+1}\odot {e_3^{*}}^{{\alpha }_3-1}\otimes e_2\hfill \\ \hfill & +\left({\overline{\nabla }}_{e_3}f-f({\alpha }_1+{\alpha }_2)\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}\otimes e_3.\hfill \end{array}$$

Let $\psi =f{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}$ and $U=U^1{e}_1+U^2{e}_2+U^3{e}_3$. Then

$$\begin{array}{cc}\hfill & {\overline{\nabla }}_U\psi =U^1{\overline{\nabla }}_{e_1}\psi +U^2{\overline{\nabla }}_{e_2}\psi +U^3{\overline{\nabla }}_{e_3}\psi \hfill \\ \hfill & =\left(U^1{\overline{\nabla }}_{e_1}f+U^2{\overline{\nabla }}_{e_2}f+U^3\left({\overline{\nabla }}_{e_3}f-f({\alpha }_1+{\alpha }_2)\right)\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}\hfill \\ \hfill & -U^{1}f{\alpha }_1{e_1^{*}}^{{\alpha }_1-1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3+1}+U^{1}f{\alpha }_3{e_1^{*}}^{{\alpha }_1+1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3-1}\hfill \\ \hfill & -U^{2}f{\alpha }_2{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2-1}\odot {e_3^{*}}^{{\alpha }_3+1}+U^{2}f{\alpha }_3{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2+1}\odot {e_3^{*}}^{{\alpha }_3-1};\hfill \end{array}$$

$$\begin{array}{cc}\hfill \mathrm{div}U& =\sum _{j=1}^{3}g(e_j,{\overline{\nabla }}_{e_j}U)=\sum _{i,j=1}^{3}g(e_j,{\overline{\nabla }}_{e_j}\left(U^i{e}_i\right))\hfill \\ \hfill & =\sum _{i,j=1}^{3}g(e_j,\left({\overline{\nabla }}_{e_j}{U}^i\right)e_i+U^i{\overline{\nabla }}_{e_j}{e}_i)\hfill \\ \hfill & =\sum _{i=1}^3{\overline{\nabla }}_{e_i}{U}^i+\sum _{i,j=1}^{3}g(e_j,U^i{\overline{\nabla }}_{e_j}{e}_i)=\sum _{i=1}^3{\overline{\nabla }}_{e_i}{U}^i+2U^3;\hfill \end{array}$$

$$\begin{array}{cc}\hfill \psi \mathrm{div} U& =f\left(\sum _{i=1}^3{\overline{\nabla }}_{e_i}{U}^i+2U^3\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}.\hfill \end{array}$$

Now, using Proposition 4, we have

$$\begin{array}{cc}\hfill & \mathrm{div}(\psi \otimes U)={\overline{\nabla }}_U\psi +\psi \mathrm{div}U\hfill \\ \hfill & =\left(\sum _{i=1}^3\left(U^i{\overline{\nabla }}_{e_i}f+f{\overline{\nabla }}_{e_i}{U}^i\right)-U^{3}f({\alpha }_1+{\alpha }_2-2)\right){e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}\hfill \\ \hfill & -U^{1}f{\alpha }_1{e_1^{*}}^{{\alpha }_1-1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3+1}+U^{1}f{\alpha }_3{e_1^{*}}^{{\alpha }_1+1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3-1}\hfill \\ \hfill & -U^{2}f{\alpha }_2{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2-1}\odot {e_3^{*}}^{{\alpha }_3+1}+U^{2}f{\alpha }_3{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2+1}\odot {e_3^{*}}^{{\alpha }_3-1}.\hfill \end{array}$$

Notice also that for $\psi =f{e_1^{*}}^{{\alpha }_1}\odot {e_2^{*}}^{{\alpha }_2}\odot {e_3^{*}}^{{\alpha }_3}$ and $U=U^1{e}_1+U^2{e}_2+U^3{e}_3$ we get e.g., :

$$\begin{array}{cc}\hfill & e^{-z({\alpha }_1+{\alpha }_2)}{\mathrm{d}}^{\mathrm{s}} \psi =({\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial x}}}f){dx}^{{\alpha }_1+1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3}\hfill \\ \hfill & +({\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial y}}}f){dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2+1}\odot {dz}^{{\alpha }_3}\hfill \\ \hfill & +f{\alpha }_3{e}^{2z}\left({dx}^{{\alpha }_1+2}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3-1}+{dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2+2}\odot {dz}^{{\alpha }_3-1}\right)\hfill \\ \hfill & +({\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial z}}}f-2f({\alpha }_1+{\alpha }_2)){dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3+1};\hfill \end{array}$$

$$\begin{array}{cc}\hfill & e^{-z({\alpha }_1+{\alpha }_2)}\mathrm{grad}\psi =e^{-2z}({\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial x}}}f){dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3}\otimes {\displaystyle \frac{\partial }{\partial x}}\hfill \\ \hfill & -f{\alpha }_1{e}^{-2z}{dx}^{{\alpha }_1-1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3+1}\otimes {\displaystyle \frac{\partial }{\partial x}}\hfill \\ \hfill & +f{\alpha }_3{dx}^{{\alpha }_1+1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3-1}\otimes {\displaystyle \frac{\partial }{\partial x}}\hfill \\ \hfill & +e^{-2z}({\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial y}}}f){dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3}\otimes {\displaystyle \frac{\partial }{\partial y}}\hfill \\ \hfill & -f{\alpha }_2{dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2-1}\odot {dz}^{{\alpha }_3+1}\otimes {\displaystyle \frac{\partial }{\partial y}}\hfill \\ \hfill & +f{\alpha }_3{dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2+1}\odot {dz}^{{\alpha }_3-1}\otimes {\displaystyle \frac{\partial }{\partial y}}\hfill \\ \hfill & +\left({\overline{\nabla }}_{{\displaystyle \frac{\partial }{\partial z}}}f-f({\alpha }_1+{\alpha }_2)\right){dx}^{{\alpha }_1}\odot {dy}^{{\alpha }_2}\odot {dz}^{{\alpha }_3}\otimes {\displaystyle \frac{\partial }{\partial z}}.\hfill \end{array}$$

Remark 1.

Under the assumptions of Example 10, one can perform similar calculations as in Example 11.

7. Conclusions

Investigations on Riemannian manifolds and related objects are still very much an evolving part of mathematics. For example, in the paper [7], a novel method for classifying sets of images—so-called Riemannian geodesic discriminant analysis–minimum Riemannian mean distance RGDA-MRMD—has been introduced. The method described therein converts image data into symmetric positive definite SPD matrices, which are then mapped onto simpler, flat spaces. It is worth noting that selecting the Riemannian mean center of the dataset as the tangent point for mapping helps to reduce error and this leads to many important applications. It is likely that our research included in Section 5 can be also used in applied mathematics. The grad div operator studied here in $M={\mathbb{R}}^n$ with the standard, flat metric g, in common with the commutator ($\mathfrak{C}$), especially, might have some interesting applications. The grad div operator is also very important for a new project that plans on investigating this operator in the context of ellipticity—in the sense of injectivity of its symbol (see [21]). Accentuate that our examinations are totally putting in the current direction. For instance, Example 11, innovatively considered in our case, has a trend akin to Example 2 contained in [5]. The theme of the paper [5] is focused on non-integrable distributions with a quarter-symmetric non-metric connection QSNMC in generalized Riemannian manifolds and offers plenty of new opportunities for further study. Potential avenues for our topic are also Lorentzian para-Kenmotsu manifolds explorations, with regard to the generalized symmetric metric connection ${\nabla }^G$ of type $(\alpha ,\beta )$, that are contained in the paper [22]. The subject of Kenmotsu manifolds is nowadays very popular and has garnered the attention of researchers worldwide (see also e.g., [14,15]). This ceaselessly developing area of mathematics allows us to be certain that the investigations in our paper are important and should have many applications in future papers.