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\title[Geometry of  Ricci tensor]{Geometry of  Ricci tensor
of harmonic \\ nearly trans--Sasakian manifolds}


\author{A.R. Rustanov, S.V. Kharitonova}

\address{Aligadzhi Rabadanovich Rustanov,
\newline\hphantom{iii}  Institute of Digital Technologies and Modeling in Construction,
\newline\hphantom{iii}
Moscow State University of Civil Engineering
\newline\hphantom{iii}
(National Research University)
\newline\hphantom{iii} Yaroslavskoye shosse, 26,
\newline\hphantom{iii} 450008, Moscow, Russia}
\email{aligadzhi@yandex.ru}

\address{Svetlana Vladimirovna Kharitonova,
\newline\hphantom{iii} Orenburg State University,
\newline\hphantom{iii} Pobedy av.13, % Адрес (улица, дом, строение и т.п.)
\newline\hphantom{iii} 460000, Orenburg, Russia}%  Адрес (почтовый индекс, город, страна)
\email{hcb@yandex.ru}

\thanks{\sc A.R. Rustanov, S.V. Kharitonova, Geometry of  Ricci tensor of harmonic nearly trans--Sasakian manifolds.}
\thanks{\copyright \  Rustanov A.R.,  Kharitonova S.V. \ 2026}
\thanks{\it Submitted December  31, 2024.}

\maketitle {\small
\begin{quote}
\noindent{\bf Abstract.} In this paper, we study the geometry of the Ricci tensor of a harmonic nearly trans--Sasakian manifold. On the space of  associated $G$-structure we introduce fundamental identities of harmonic nearly trans--Sasakian manifolds. We prove that Ricci--flat harmonic nearly trans--Sasakian manifolds are closely cosymplectic. We obtain conditions, which ensure that harmonic nearly trans--Sasakian manifolds are Einstein and $\eta$--Einstein manifolds. We obtain identities for the Ricci tensor of harmonic nearly trans--Sasakian manifolds. We provide local characterizations   for the following harmonic nearly trans-Sasakian manifolds: Einstein manifolds; manifolds, the  Ricci tensor of which is parallel, $\eta$--parallel, the Codazzi tensor, the Killing tensor, and satisfies the three selected identities.


\medskip
\noindent{\bf Keywords:}  harmonic nearly trans--Sasakian manifolds,  tensor Ricci,
Einstein manifold, closely cosympletic manifold.

\medskip
\noindent{\bf Mathematics Subject Classification:} 53D15
\end{quote}
}

\vskip20pt

\section{Introduction}


In this paper, we continue the study of nearly trans--Sasakian (NTS) manifolds, that is, manifolds equipped with an almost contact metric structure, the linear extension of which belongs to the class $W_1\oplus W_4$ in the Gray~---~Hervella classification.
In   \cite{1}, \cite{2}, \cite{3}, harmonic NTS--manifolds were defined and the local structure of these manifolds was described. To obtain a harmonic NTS--manifold, it suffices to take the Cartesian product of any nearly K\"ahler manifold $M$ with the real line $\mathds{R}$ and make a canonical concircular transformation of the exact cosymplectic structure of  manifold $M\times\mathds{R}$. Each harmonic NTS--manifold is (locally) structured in this way.


The paper is organized as follows. In Section 2 we provide basic definitions and formulate known facts from the geometry of harmonic NTS--manifolds. This is important for understanding the further presentation. The study is made by means of the method of associated $G$-structures. In particular, we provide structural equations, expressions for the components of Ricci tensor, and the scalar curvature of a harmonic NTS--manifold on the space of the associated $G$-structure. Then we give the first and second fundamental identities, and a new third fundamental identity of harmonic NTS--manifolds.
  We obtain new results  for Einstein and $\eta$-Einstein harmonic NTS--manifolds.
We  prove that a harmonic NTS--manifold has a $\Phi$-invariant Ricci tensor.

In Section 3, we calculate the components of   covariant derivative of  the Ricci tensor on the space of  associated $G$-structure, obtain certain identities for the Ricci tensor of a harmonic NTS--manifold, and obtain a local characterization of harmonic NTS--manifolds, the Ricci tensor of which satisfies the obtained identities. We also establish   classification theorems for harmonic NTS--manifolds, the Ricci tensor of which is parallel, $\eta$-parallel, the Codazzi tensor, or the Killing tensor.



\vskip20pt

\section{Preliminaries}

We recall that an almost contact metric (AC) structure on a manifold $M$ is the set $(\xi, \eta, \Phi, g= \left\langle \,\cdot\,,\,\cdot\,\right\rangle)$ of tensor fields on $M$, where $\xi$ is a vector field called the characteristic field, $\eta$ is a differential $1$-form called the contact form, $\Phi$ is an endomorphism of the module of smooth vector fields of the manifold $M$  called the structure endomorphism,
$g= \left\langle \,\cdot\,,\,\cdot\,\right\rangle$ is a Riemann  metric.
Moreover,
\begin{align*}
&1)~\eta(\xi)=1; \qquad  2)~\Phi(\xi)=0; \qquad 3)~\eta\circ\Phi=0;  \qquad 4)~\Phi^2=-id+\xi\otimes\eta; \\
&5)~\left\langle \Phi X,\Phi Y\right\rangle=\left\langle X,Y\right\rangle-\eta(X)\eta(Y); \qquad X,Y\in\mathcal{X}(M).
\end{align*}

A manifold admitting an AC--structure is called an almost contact metric manifold, briefly,  AC--manifold.

\begin{definition}[\cite{1}] \label{d2.1}
An AC--structure is called the nearly trans--Sasakian (NTS) structure if its linear extension belongs to the class $W_1\oplus W_4$ of almost Hermitian structures in the Gray~---~Hervella classification. An AC--manifold equipped with an NTS--structure is called the NTS--manifold.
\end{definition}

\begin{definition} [\cite{1}]
An NTS--structure with a closed contact form is called the eigen NTS--structure.
\end{definition}

\begin{definition} [\cite{1}, \cite{3}]
An eigen NTS--manifold with a harmonic contact form is called harmonic, and the number $\chi=-\frac{1}{2n}\delta\eta$ is its characteristics.
\end{definition}

We consider a harmonic NTS--manifold. The Lie form of such a manifold is closed  \cite{1}, which means that the characteristics of such a manifold is constant,
i.e., $d\chi=0$. Moreover, since $\bar{\chi} =\chi$,  the function $\chi$ is  real.


The complete group of structural equations of a harmonic NTS--manifold reads~\cite{1}
\begin{equation}\label{2.1}
\begin{aligned}
&1)\quad d\theta^a=-\theta_b^a\wedge\theta^b+C^{abc}\theta_b\wedge\theta_c+\chi\delta_b^a \theta^b\wedge\theta; \\
&2)\quad d\theta_a=\theta_a^b\wedge\theta_b+C_{abc} \theta^b\wedge\theta^c+\chi\delta_a^b \theta_b\wedge\theta; \\
&3)\quad d\theta=0;	\\
&4)\quad d\theta_b^a+\theta_c^a\wedge\theta_b^c=(A_{bc}^{ad}-2C^{adh}C_{hbc}) \theta^c\wedge\theta_d,	
\end{aligned}
\end{equation}
where $\{A_{bc}^{ad}\}$ is a family of functions on the space of the associated $G$-structure, which serve as components of the so-called curvature tensor of the associated $Q$-algebra \cite{4}, or the structural tensor of   second kind, and,
\begin{equation}\label{2.3}
\begin{aligned}
&1)~A_{[bc]}^{ad}=0; \qquad   2)~A_{ac}^{[bd]}=0; \qquad  3)~\overline{A_{bc}^{ad}}=A_{ad}^{bc};\\
&4)~C^{[abc]}=C^{abc}; \qquad 5)~C_{[abc]}=C_{abc}.
\end{aligned}
\end{equation}
Moreover,
\begin{equation}\label{2.4}
\begin{aligned}
&1)\quad dC^{abc}+C^{dbc}\theta_d^a+C^{adc}\theta_d^b+C^{abd}\theta_d^c=C^{abcd}\theta_d+\chi C^{abc}\theta;\\
&2)\quad dC_{abc}-C_{dbc}\theta_a^d-C_{adc}\theta_b^d-C_{abd}\theta_c^d=C_{abcd}\theta^d+\chi C_{abc}\theta; \\
&3)\quad d\chi=0,
\end{aligned}
\end{equation}
where $C^{abcd}$, $C_{abcd}$ are appropriate functions on the space of  associated $G$-structure, and,
\begin{equation*}
1)~C^{a[bcd]}=0; \qquad 2)~C_{a[bcd]}=0.
\end{equation*}
Making external differentiation of fourth  equation in (\ref{2.1}), we obtain
\begin{equation}
\label{2.6}
dA_{bc}^{ad}+A_{bc}^{hd}\theta_h^a+A_{bc}^{ah}\theta_h^d-A_{hc}^{ad}\theta_b^h-A_{bh}^{ad}\theta_c^h=A_{bch}^{ad}\theta^h+A_{bc}^{adh}\theta_h+2\chi A_{bc}^{ad}\theta,
\end{equation}
where
\begin{equation}\label{2.7}
\begin{aligned}
&1)\quad A_{b[ch]}^{ad}=A_{bc}^{a[dh]}=0;\\
&2)\quad (A_{bc}^{a[d}-2C^{a[d|h}C_{hbc})C^{c|fg]}=0;\\
&3)\quad (A_{b[c}^{ad}-2C^{adf}C_{fb[c})C_{d|hg]}=0.
\end{aligned}
\end{equation}
Making external differentiation of identity  1) in (\ref{2.4}), we obtain
\begin{equation*}
dC^{abcd}+C^{hbcd}\theta_h^a+C^{abcd}\theta_h^d+C^{abhd}\theta_d^c+C^{abch}\theta_h^d=C^{abcdh}\theta_h+2\chi C^{abcd}\theta,
\end{equation*}
where
\begin{equation}
\label{2.9}
1)~C^{abc[dh]}=0;  \qquad  2)~C^{abcg}C_{gdh}=0.
\end{equation}
We  call identity 2) in  (\ref{2.9}) the first fundamental identity, and  identity 3) in (\ref{2.7}) the second fundamental identity of the harmonic NTS--manifold.

It is convenient to adopt the following notation
\begin{align*}
%\label{2.10}
&C_{bc}^{ad}=C^{adh}C_{hbc}, \qquad C_b^a=C_{bc}^{ac}, \qquad A_b^a=A_{bc}^{ac}, \\
&A_b^{ac}=A_{hb}^{hac}, \qquad A_{bc}^a=A_{hbc}^{ha}.
\end{align*}

\begin{lemma}
For a harmonic NTS--manifold, $C=C_a^a$ is nonnegative. Moreover, $C=0$ if and only if the manifold is either cosymplectic or a Kenmotsu manifold, i.e., a manifold obtained from a cosymplectic manifold by a canonical concircular transformation.
\end{lemma}

\begin{proof}
Since $\overline{C^{abc}}=C_{abc}$, we obtain $C=C^{abc}C_{abc}=\sum_{abc}|C_{abc}|^2\geq 0$, and $C=0$ if and only if $C^{abc}=C_{abc}=0$.
Then, by Theorem~6 in \cite{1}, a harmonic NTS--manifold is either a cosymplectic manifold or a Kenmotsu manifold, i.e., a manifold obtained from a cosymplectic manifold by the canonical concircular transformation \cite{5}. It is easy to see that the converse is also true. The proof is complete.
\end{proof}
According to (\ref{2.4}) and (\ref{2.6}) we have
\begin{equation}\label{2.11}
\begin{aligned}
&1)\quad dC_{bc}^{ad}+C_{bc}^{gd}\theta_g^a+C_{bc}^{ag}\theta_g^d-C_{gc}^{ad}\theta_b^g-C_{bg}^{ad}\theta_c^g=C_{bcg}^{ad}\theta^g+C_{bc}^{adg}\theta_g+2\chi C_{bc}^{ad} \theta;\\
&2)\quad dC_b^a+C_b^h\theta_h^a-C_h^a\theta_b^h=C_{bh}^a\theta^h+C_b^{ah}\theta_h+2\chi C_b^a \theta;\\
&3)\quad dA_b^a+A_b^h\theta_h^a-A_h^a\theta_b^h=A_{bh}^a\theta^h+A_b^{ah}\theta_h+2\chi A_b^a \theta.
\end{aligned}
\end{equation}

We  contract identity  3) in (\ref{2.7}), which is the first fundamental identity of the harmonic NTS--manifold, with respect to the  indices $a$ and $b$
\begin{equation}
\label{2.13}
A_c^dC_{hgd}+A_h^dC_{gcd}+A_g^d C_{chd}-2C_c^d C_{hgd}-2C_h^d C_{gcd}-2C_g^d C_{chd}=0.
\end{equation}
Now we contract the third identity in (\ref{2.7}) with respect to the  indices $a$ and $c$ and rename $b$ to $c$. Then, taking into account the symmetry properties of objects $A$ and $C$ (\ref{2.3}), we obtain
\begin{equation}
\label{2.14}
A_c^d C_{hgd}+2C_c^d C_{hgd}-2C_g^d C_{chd}+2C_h^d C_{cgd}=0.
\end{equation}
We deduct term--by--term  (\ref{2.14}) from (\ref{2.13}), and in view of symmetry properties of $C$, see   (\ref{2.3}), we obtain the identity
\begin{equation}
\label{2.15}
A_{[h}^d C_{g]cd}-2C_c^d C_{hgd}=0.
\end{equation}
We call identity (\ref{2.15}) the third fundamental identity of the harmonic NTS--manifold.

The components of the Ricci tensor of the harmonic NTS--manifold on the space of the associated $G$-structure read \cite{2}
\begin{equation}\label{2.17}
\begin{aligned}
&1)\quad S_{00}=-2n\chi^2;\\
&2)\quad S_{a\hat{b}}=A_{ac}^{bc}-3C^{bcd} C_{dca}-2n\chi^2\delta_a^b;\\
&3)\quad S_{\hat{a}b}=A_{bc}^{ac}-3C^{acd} C_{dcb}-2n\chi^2\delta_b^a.
\end{aligned}
\end{equation}
All other components are zero.


It follows from identity 1) in formula (\ref{2.17}) that a Ricci--flat harmonic NTS--manifold is an closely cosymplectic manifold, and therefore it is  locally equivalent to the product of a nearly K\"ahler manifold and the real line. If the manifold is simply connected, then these equivalences can be chosen to be global.

The scalar curvature of a harmonic NTS--manifold is
\begin{equation}
\label{2.18}
r=2A_{ab}^{ab}-6C^{abc}C_{cba}-2n(n+1)\chi^2.
\end{equation}
It follows from (\ref{2.1}) that each Kenmotsu manifold is a harmonic NTS--manifold of characteristics $\chi=-1$, and each closely cosymplectic manifold is a harmonic NTS--manifold of characteristics $\chi=0$. Kenmotsu manifolds and closely cosymplectic manifolds are the most interesting and well--studied examples of harmonic NTS--manifolds.
One should  add the special Kenmotsu manifolds of the second kind to these examples.

\begin{theorem}
\label{th2.1}
A harmonic NTS--manifold is an Einstein manifold if and only if the identity
\begin{equation}
\label{2.333}
A_b^a=3C_b^a
\end{equation}
holds on  the space of the associated $G$-structure.
\end{theorem}

\begin{proof}
Let a harmonic NTS--manifold be an Einstein manifold with the cosmological constant $\epsilon$. Then the components of Ricci tensor $S$ and the components of  metric tensor $g$ are related by the identity $S_{ij}=\epsilon g_{ij}$, where $\epsilon=\const$. In view of  (\ref{2.17}), these relations on the space of   associated $G$-structure are written as
\begin{equation}\label{2.20}
\begin{aligned}
&1)\quad -2n\chi^2=\epsilon;\\
&2)\quad A_a^b-3C_a^b-2n\chi^2 \delta_a^b=\epsilon\delta_a^b,
\end{aligned}
\end{equation}
that is,  $A_b^a=3C_b^a$. The proof is complete.
\end{proof}

\begin{corollary}
A harmonic Einstein NTS--manifold is a manifold of non--positive scalar curvature.
\end{corollary}

\begin{theorem}
\label{th2.2}
A complete harmonic Einstein NTS--manifold is either a Ricci--flat closely cosymplectic manifold, and hence is holomorphically isometrically covered by the product of a Ricci-flat nearly K\"ahler manifold with the real line, or is compact and has a finite fundamental group.
\end{theorem}

\begin{proof}
For  $\epsilon=0$ it follows from identity 1) in (\ref{2.20}) that  $\chi=0$, i.e.,
the manifold is Ricci--flat, closely cosymplectic, and therefore locally holomorphically isometric to a manifold of the form $N^{2n}\times\mathds{R}$, where $N^{2n}$ is a Ricci--flat nearly K\"ahler manifold \cite{5}.


If $\epsilon<0$, then, according to the classical Myers theorem \cite{6}, in the case of completeness the manifold is compact and has a finite fundamental group. The proof is complete.
\end{proof}

\begin{theorem}
\label{th2.3}
A complete harmonic Einstein NTS--manifold is locally equivalent to the product $N^{2n}\times\mathds{R}$
or canonically concircular to the manifold $N^{2n}\times\mathds{R}$ equipped with a cosymplectic structure, where $N^{2n}$ is a K\"ahler manifold that is an Einstein manifold.
\end{theorem}
\begin{proof}
Let a harmonic NTS--manifold be an Einstein manifold with cosmological constant $\epsilon$. Then, taking into account (\ref{2.333}),   third fundamental identity (\ref{2.15}) can be written as
$$3C_h^d C_{gcd}-3C_g^d C_{hcd}-4C_c^d C_{hgd}=0.$$
Symmetrizing this identity in the indices $h$ and $c$, we obtain
\begin{equation}
\label{2.211}
C_h^d C_{gcd}+C_g^d C_{hcd}=0.
\end{equation}
Since $(C_h^d)$ is a Hermitian matrix, at each point of this harmonic NTS--manifold there exists an $A$-frame \cite{5} in which $C_h^d=C_h\delta_h^d$, where $\{C_c\}$ are the eigenvalues of  this matrix. Then identity (\ref{2.211}) can be written as
$C_h \delta_h^d C_{gcd}+C_g \delta_g^d C_{hcd}=0.$ We contract the resulting identity with the object $C^{cdf}$, then
\begin{align*}
&C_h \delta_h^d C_{gcd}C^{cdf}+C_g\delta_g^d C_{hcd}C^{cdf}=0,\\
&C_h C_g \delta_g^f+C_g C_h \delta_h^f=0, \\
&2C_g C_h=0.
\end{align*}
Hence,  $C_h=0$ and then $C_h^d=0$. Contracting this identity in the indices  $d$ and $h$, we obtain
$$\sum_{abc}|C_{abc}|^2 =C^{abc}C_{abc}=C_c^c=0,$$
and hence, $C_{abc}=0$, and the considered  manifold is either a cosymplectic manifold or a Kenmotsu manifold.

Since a cosymplectic manifold is locally equivalent to the product of a K\"ahler manifold with the real line \cite{5}, and the class of Kenmotsu manifolds coincides with the class of almost contact metric manifolds obtained from cosymplectic manifolds by the canonical concircular transformation of the cosymplectic structure
\cite{5}, we obtain the required statements. The proof is complete.
\end{proof}


We consider an Einstein harmonic NTS--manifold. Then its Ricci tensor on the space of  associated $G$-structure has the   components
\begin{equation}
\label{2.21}
S_{ij}=a g_{ij}+b\delta_i^0 \delta_j^0.
\end{equation}

We are going to obtain expressions for $a$ and $b$. In view of  (\ref{2.17}), relations (\ref{2.21}) can be written as
\begin{equation}\label{2.22}
\begin{aligned}
&1)\quad -2n\chi^2=a+b;\\
&2)\quad A_{ac}^{bc}-3C^{bcd}C_{dca}-2n\chi^2 \delta_a^b=a\delta_a^b.
\end{aligned}
\end{equation}

We contract identity 2)  in (\ref{2.22}) in the indices $a$ and $b$, then in view of  (\ref{2.18}) we obtain
\begin{equation}
\label{2.23}
a=\frac{r}{2n}-(n-1) \chi^2.
\end{equation}

Substituting (\ref{2.23}) into the first identity in formula (\ref{2.22}), we obtain
\begin{equation*}
b=-\frac{r}{2n}-(n+1) \chi^2.
\end{equation*}


\begin{definition}
\label{d2.4}
We call the Ricci tensor of an AC--manifold $\Phi$-invariant if
\begin{equation}
\label{2.26}
\Phi Q=Q\Phi.
\end{equation}
\end{definition}

Writing identity (\ref{2.26}) on the space of  associated $G$-structure, we find that for an AC--manifold with $\Phi$-invariant Ricci tensor the following relations hold
\begin{equation}\label{2.27}
\begin{aligned}
&1)\quad S_{0a}=S_{0\hat{a}}=S_{a0}=S_{\hat{a}0}=0; \\
&2)\quad S_{ab}=S_{\hat{a}\hat{b}}=0.
\end{aligned}
\end{equation}
And vice versa, if  relations (\ref{2.27}) hold on the space of  associated $G$-structure, then the AC--manifold has a $\Phi$-invariant Ricci tensor.  Thus, we have proved the following theorem.


\begin{theorem}
\label{th2.4}
An AC--manifold has a $\Phi$-invariant Ricci tensor if and only if the following equalities hold on the space of  associated $G$-structure
$$
\begin{aligned}
&1)\quad S_{0a}=S_{0\hat{a}}=S_{a0}=S_{\hat{a}0}=0; \\
&2)\quad S_{ab}=S_{\hat{a}\hat{b}}=0.
\end{aligned}
$$
\end{theorem}

We consider a harmonic NTS--manifold. By (\ref{2.17}), definition \ref{d2.4}, and Theorem~\ref{th2.4} we obtain the following assertion.


\begin{theorem}
\label{th2.5}
The Ricci tensor of a harmonic NTS--manifold is $\Phi$--invariant.
\end{theorem}

\vskip20pt

\section{Identities for Ricci tensor}

We recall that the tensor components of   Riemannian connection form for a harmonic NTS--manifold on the space of  associated $G$-structure read \cite{2}
\begin{equation}\label{2.28}
\begin{aligned}
&1)~\theta_b^{\hat{a}}=C^{abc}\theta_c; \qquad  &&2)~\theta_b^{\hat{a}}=C_{abc}\theta^c; \qquad &&3)~\theta_0^a=-\chi\delta_b^a\theta^b; \qquad &&4)~\theta_0^{\hat{a}}=-\chi\delta_a^b\theta_b; \\
&5)~\theta_a^0=\chi\delta_a^b \theta_b; \qquad  &&6)~\theta_{\hat{a}}^0=\chi\delta_b^a \theta^b; \qquad &&7)~\theta_0^0=0; \qquad &&8)~\theta_j^i+\theta_{\hat{i}}^{\hat{j}}=0.
\end{aligned}
\end{equation}
Since the Ricci tensor is a tensor of type (2,0), by the Fundamental Theorem of Tensor Analysis, its components on the space of   principal frame bundle over the considered manifold   satisfy the relations \cite{5}
\begin{equation}
\label{2.29}
dS_{ij}-S_{kj}\theta_i^k-S_{ik}\theta_j^k=S_{ij,k}\theta^k,
\end{equation}
where  $\{S_{ij,k}\}$ a system of smooth functions that serve as components of the tensor  $\nabla S$.

Writing   (\ref{2.29}) on the space on the space of the associated $G$-structure, in view of
(\ref{2.28}), (\ref{2.17}), (\ref{2.4}:3) and (\ref{2.11}) we obtain
\begin{equation}\label{2.30}
\begin{aligned}
&1)~S_{0a,\hat{b}}=S_{a0,\hat{b}}=\chi(A_a^b-3C_a^b ); \qquad &&2)~S_{0\hat{a},b}=S_{\hat{a}0,b}=\chi(A_b^a-3C_b^a ); \\
&3)~S_{ab,c}=-2(A_{(a}^d-3C_{(a}^d) C_{|d|b)c}; \qquad &&4)~S_{a\hat{b},0}=S_{\hat{b}a,0}=2\chi(A_a^b-3C_a^b);\\
&5)~S_{a\hat{b},c}=S_{\hat{b}a,c}=A_{ac}^b-3C_{ac}^b; \qquad &&6)~S_{a\hat{b},\hat{c}}=S_{\hat{b}a,\hat{c}}=A_a^{bc}-3C_a^{bc}; \\
&7)~S_{\widehat{a}\hat{b},\hat{c}}=-2(A_d^{(a}-3C_d^{(a}) C^{|d|b)c},
\end{aligned}
\end{equation}
while other components are zero.

\begin{theorem}
\label{th3.0}
The Ricci tensor of a harmonic NTS--manifold satisfies the identities
\begin{align*}
&1)\quad \nabla_X S(\xi,\xi)=0;\\
&2)\quad \nabla_{\xi} S(\chi,\xi)=0;\\
&3)\quad \nabla_{\Phi X}(S)(\xi,\Phi Y)-\nabla_X(S)(\xi,Y)=0;\\
&4)\quad \nabla_{\xi}(S)(\Phi X,\Phi Y)-\nabla_{\xi}(S)(X,Y)=0;\\
&5)\quad \nabla_{\Phi^2 X}(S)(\Phi^2 Y,\Phi^2 Z)-\nabla_{\Phi^2 X}(S)(\Phi Y,\Phi Z)+\nabla_{\Phi X}(S)(\Phi Y,\Phi^2 Z)+\nabla_{\Phi X}(S)(\Phi^2 Y,\Phi Z)=0.
\end{align*}
\end{theorem}
\begin{proof}
It follows from (\ref{2.30}) that
\begin{equation*}
%\label{3.1}
1)~\nabla_X S(\xi,\xi)=0; \qquad 2)~\nabla_{\xi} S(\chi,\xi)=0.
\end{equation*}
Applying the identity recovery procedure \cite{4}, \cite{5} to the identity $S_{0a,b}=0$, we obtain
\begin{equation*}
\nabla_{\Phi^2 X}(S)(\xi,\Phi^2 Y)-\nabla_{\Phi X}(S)(\xi,\Phi Y)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation*}
This identity can be written as
\begin{equation}
\label{3.2}
\nabla_{\Phi X}(S)(\xi,\Phi Y)-\nabla_X(S)(\xi,Y)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation}
Applying the identity recovery procedure to the identity $S_{ab,0}=0$, we get
\begin{equation*}
\nabla_{\xi} (S)(\Phi^2 X,\Phi^2 Y)-\nabla_{\xi} (S)(\Phi X,\Phi Y)=0, \qquad \forall X,Y\in\mathcal{X}(M),
\end{equation*}
which can be written as
\begin{equation}
\label{3.3}
\nabla_{\xi}(S)(\Phi X,\Phi Y)-\nabla_{\xi}(S)(X,Y)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation}
Applying the identity recovery procedure to the identity $S_{ab,\hat{c}}=0$, we get
\begin{equation}\label{3.4}
\begin{aligned}
\nabla_{\Phi^2 X}(S)(\Phi^2 Y,\Phi^2 Z)&-\nabla_{\Phi^2 X}(S)(\Phi Y,\Phi Z)\\
&+\nabla_{\Phi X}(S)(\Phi Y,\Phi^2 Z)+\nabla_{\Phi X}(S)(\Phi^2 Y,\Phi Z)=0, \quad \forall X,Y\in\mathcal{X}(M).
\end{aligned}
\end{equation}
The proof is complete.
\end{proof}

Main non--zero components of the tensor  $\nabla S$ are defined by the following pairs of expressions
\begin{equation}\label{3.5}
\begin{aligned}
&1)~S_{0a,\hat{b}}=\chi(A_a^b-3C_a^b); \qquad &&2)~S_{a\hat{b},0}=2\chi(A_a^b-3C_a^b );\\
&3)~S_{ab,c}=-2(A_{(a}^d-3C_{(a}^d) C_{|d|b)c}; \qquad &&4)~S_{a\hat{b},c}=A_{ac}^b-3C_{ac}^b
\end{aligned}
\end{equation}
and by their adjoint. The most interesting is the study of geometric meaning of vanishing of these components.


Let $S_{0a,\hat{b}}=\chi(A_a^b-3C_a^b)=0$. Then either $\chi=0$, or $A_a^b-3C_a^b=0$. In the first case, the manifold is closely cosymplectic, in the second case (by Theorem~\ref{th2.4}) it is an Einstein manifold with cosmological constant $\epsilon=-2n\chi^2$.

Applying the identity recovery procedure to the identity \cite{4}, \cite{5} to the identity  $S_{0a,\hat{b}}=0$, we obtain
\begin{equation*}
\nabla_{\Phi^2 X}(S)(\xi,\Phi^2 Y)+\nabla_{\Phi X}(S)(\xi,\Phi Y)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation*}
This identity can be written as
\begin{equation}
\label{3.8}
\nabla_{\Phi X}(S)(\xi,\Phi Y)+\nabla_X(S)(\xi,Y)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation}
Since  (\ref{3.2}) holds satisfied for each harmonic NTS--manifold, identity (\ref{3.8}) is equivalent to
\begin{equation*}
\nabla_{\Phi X}(S)(\xi,\Phi Y)=\nabla_X (S)(\xi,Y)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation*}
Hence, the identity   $S_{0a,\hat{b}}=0$ is equivalent to
\begin{equation}
\label{3.10}
\nabla_X (S)(\xi,Y)=0,\qquad \forall X,Y\in\mathcal{X}(M)
\end{equation}

The above facts are summarized in the following theorem.

\begin{theorem}
\label{th3.1}
A harmonic NTS--manifold, the Ricci tensor of which satisfies   identity (\ref{3.10}) is either a precisely cosymplectic manifold or an Einstein manifold with cosmological constant $\epsilon=-2n\chi^2$.
\end{theorem}

In view of Theorem~\ref{th2.3}, Theorem \ref{th3.1} can be formulated as follows.


\begin{theorem}
\label{th3.2}
A harmonic NTS--manifold, the Ricci tensor of which satisfies   identity (\ref{3.10}) is locally equivalent to the product of a nearly K\"ahler manifold with the real line, or canonically concircular to the product of a K\"ahler manifold with the real line  equipped with a cosymplectic structure.
\end{theorem}


Let $S_{a\hat{b},0}=2\chi(A_a^b-3C_a^b)=0$, that is $S_{a\hat{b},0}=0$.  This identity is equivalent to
\begin{equation}
\label{3.12}
\nabla_{\xi}(S)(\Phi^2 X,\Phi^2 Y)+\nabla_{\xi} (S)(\Phi X,\Phi Y)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation}
By  (\ref{3.3}) and (\ref{3.12}) we obtain\
\begin{equation}
\label{3.13}
\nabla_{\xi}(S)(\Phi X,\Phi Y)=\nabla_{\xi}(S)(X,Y)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation}
It follows from  (\ref{3.5}) that  $2S_{0a,\hat{b}}=S_{a\hat{b},0}$. Thus, a harmonic NTS--manifold, the Ricci tensor of which satisfies identity (\ref{3.10}), also satisfies identity (\ref{3.13}), and vice versa.

\begin{theorem}
\label{th3.3}
The class of harmonic NTS--manifolds, the Ricci tensor of which satisfies the identity (\ref{3.10}), coincides with the class of harmonic NTS--manifolds, the Ricci tensor of which satisfies   identity (\ref{3.13}).
\end{theorem}

Let  $S_{ab,c}=-2(A_{(a}^d-3C_{(a}^d) C_{|d|b)c}=0$. In view of  (\ref{3.4}),  identity $S_{ab,c}=0$ is equivalent to
\begin{equation*}
\nabla_{\Phi^2 X}(S)(\Phi^2 Y,\Phi^2 Z)-\nabla_{\Phi^2 X}(S)(\Phi Y,\Phi Z)=0, \qquad \forall X,Y\in\mathcal{X}(M),
\end{equation*}
which in view of  (\ref{3.3}) is written as
\begin{equation}
\label{3.15}
\nabla_X(S)(\Phi Y,\Phi Z)-\nabla_X(S)(Y,Z)=0, \qquad \forall X,Y\in\mathcal{X}(M).
\end{equation}
Since
\begin{equation}
\label{3.16}
(A_{(a}^d-3C_{(a}^d)C_{|d|b)c}=0.	
\end{equation}


It follows from (\ref{2.15}) and (\ref{3.16}) that
\begin{equation*}
2A_a^d C_{bcd}=3C_a^d C_{bcd}+3C_b^d C_{acd}+4C_c^d C_{abd}.
\end{equation*}
We symmetrize the obtained identity in the indices $b$ and $c$, then by symmetry properties  (\ref{2.3}) of  $C$ we obtain
\begin{equation*}
%\label{3.18}
C_b^d C_{cad}+C_c^d C_{bad}=0.
\end{equation*}
In view of this identity,  identity (\ref{3.16}) becomes
\begin{equation}
\label{3.19}
A_a^d C_{bcd}=4C_c^d C_{abd}.
\end{equation}
We alternate this identity in the  indices $a$ and $b$, then, in view of the symmetry properties of object $C$, we obtain the identity
\begin{equation}
\label{3.20}
A_{[a}^d C_{b]cd}=4C_c^d C_{abd}.
\end{equation}
By  (\ref{2.15}) and (\ref{3.20}) we have
\begin{equation*}
%\label{3.21}
C_c^d C_{abd}=0.
\end{equation*}
Since  $(C_c^d)$ is the Hermitian matrix, at each point of   the manifold, there exists an $A$--frame \cite{5}, in which $C_c^d=C_c \delta_c^d$, where $\{C_c\}$ are the eigenvalues of this matrix. Contracting the identity $C_c C_{abc}=0$ with the object $C^{abd}$, we obtain $(C_c )^2 \delta_c^d=0$, and therefore $C_c=0$.
But then $C_c^d=0$. Contracting this identity over the indices $c$ and $d$, we obtain
\begin{equation*}
\sum_{abc}|C_{abc}|^2 =C^{abc}C_{abc}=C_c^c=0,
\end{equation*}
and hence, $C_{abc}=0$, that is, the manifold is either a cosymplectic manifold or a Kenmotsu manifold \cite{1}.

Similarly, by (\ref{3.19}) one can show that $A_c^c=0$, and hence a harmonic NTS--manifold, the Ricci tensor of which satisfies identity (\ref{3.15}), according to (\ref{2.18}), is a manifold of constant scalar curvature $r=-2n(n+1)\chi^2$  since there are no Kenmotsu manifolds of constant curvature different from $(-1)$, and a Kenmotsu manifold is a space of constant curvature $(-1)$ if and only if it is canonically concircular to a manifold $\mathds{C}^n\times\mathds{R}$ equipped with a cosymplectic structure \cite{5}. It is well--known that a cosymplectic manifold is locally equivalent to the product of a K\"ahler manifold and the real line \cite{5}. The K\"ahler component of a cosymplectic manifold is locally holomorphically isometric to $\mathds{C}^n$, and hence the cosymplectic manifold is locally equivalent to the product $\mathds{C}^n\times\mathds{R}$.

The above facts are summarized in the next theorem.

\begin{theorem}
\label{th3.4}
A harmonic NTS--manifold, the Ricci tensor of which satisfies the identity (\ref{3.10}), is locally equivalent to the product $\mathds{C}^n\times\mathds{R}$ or canonically concircular to the manifold $\mathds{C}^n\times\mathds{R}$ equipped with a cosymplectic structure.
\end{theorem}

\begin{definition}[\cite{7}]
\label{d3.1} The Ricci tensor S of an AC--manifold is called parallel if $\nabla S=0$ and  $\eta$--parallel if $\nabla_X(S)(\Phi Y,\Phi Z)=0$ for all $X,Y,Z\in\mathcal{X}(M)$.
\end{definition}

Let a harmonic NTS--manifold have a parallel Ricci tensor, i.e. $\nabla S=0$. The next statement follows directly from Theorems~\ref{th3.2} and~\ref{th3.4}.

\begin{theorem}
\label{th3.5}
A harmonic NTS--manifold with parallel Ricci tensor is locally equivalent to the product $\mathds{C}^n\times\mathds{R}$ equipped with a cosymplectic structure.
\end{theorem}


Next we consider a harmonic NTS--manifold with $\eta$--parallel Ricci tensor. On the space of associated $G$-structure, the $\eta$--parallel  condition, i.e., the identity
\begin{equation*}
\nabla_X(S)(\Phi Y,\Phi Z)=0, \qquad \forall X,Y,Z\in\mathcal{X}(M)
\end{equation*}
is written as
\begin{equation*}
S_{ij,k}\Phi_r^i \Phi_l^j X^k Y^r Z^l=0,
\end{equation*}
which is equivalent to the relations
\begin{equation}
\begin{aligned}%\label{3.25}
&1)\quad S_{ab,c}=-2(A_{(a}^d-3C_{(a}^d) C_{|d|b)c}; \\
&2)\quad S_{a\hat{b},0}=2\chi(A_a^b-3C_a^b );\\
&3)\quad S_{a\hat{b},c}=S_{\hat{b}a,c}=A_{ac}^b-3C_{ac}^b,
\end{aligned}
\end{equation}
i.e., the Ricci tensor of a harmonic NTS--manifold with $\eta$--parallel Ricci tensor is parallel.
Therefore, for a harmonic NTS--manifold with $\eta$--parallel Ricci tensor, Theorem~\ref{th3.5} holds.


Gray in \cite{8} introduced two classes of Riemannian manifolds defined by the covariant derivative of the Ricci tensor. Class A consists of all Riemannian manifolds, the Ricci tensor of which $S$ is a Killing tensor, i.e.
\begin{equation*}
\nabla_X(S)(Y,Z)+\nabla_Y (S)(X,Z)+\nabla_Z (S)(X,Y)=0; \qquad\forall X,Y,Z\in\mathcal{X}(M).
\end{equation*}
The second class B consists of all Riemannian manifolds, the Ricci tensor of which is the Codazzi tensor, i.e.
\begin{equation*}
\nabla_X (S)(Y,Z)=\nabla_Y (S)(X,Z); \qquad \forall X,Y,Z\in\mathcal{X}(M).
\end{equation*}

\begin{definition} [\cite{9}, \cite{10}]
\label{d3.2} A symmetric $2$-tensor field $T$ is called a Codazzi tensor if $dT=0$, i.e., if T satisfies the Codazzi equation
\begin{equation*}
\nabla_X(T)(Y,Z)=\nabla_Y (T)(X,Z), \qquad \forall X,Y,Z\in\mathcal{X}(M).
\end{equation*}
\end{definition}
Let $M^{2n+1}$ be a harmonic NTS--manifold, the Ricci tensor is the Codazzi tensor. Then the identity holds
\begin{equation}
\label{3.26}
\nabla_X(S)(Y,Z)=\nabla_Y (S)(X,Z); \qquad \forall X,Y,Z\in\mathcal{X}(M).
\end{equation}
Identity  (\ref{3.26}) on the space of  associated $G$-structure is written as
\begin{equation}
\label{3.27}
S_{ij,k}=S_{kj,i}.
\end{equation}
In particular, by  (\ref{3.5}) and (\ref{3.27}) we have
$S_{ab,c}=S_{cb,a}$, that is,
\begin{align*}
&(A_{(a}^d-3C_{(a}^d) C_{|d|b)c}=(A_{(c}^d-3C_{(c}^d)C_{|d|b)a},\\
&A_a^d C_{dbc}+A_b^d C_{dac}-3C_a^d C_{dbc}-3C_b^d C_{dac}=A_c^d C_{dba}+A_b^d C_{dca}-3C_c^d C_{dba}-3C_b^d C_{dca},\\
&A_c^d C_{abd}-A_a^d C_{cbd}+2A_b^d C_{acd}=3C_a^d C_{bcd}+3C_c^d C_{abd}+6C_b^d C_{acd}.
\end{align*}
In view of   the third fundamental identity, the resulting identity can be written as
\begin{align*}
&4C_b^d C_{cad}+2A_b^d C_{acd}=3C_a^d C_{bcd}+3C_c^d C_{abd}+6C_b^d C_acd, \\
&2A_b^d C_{acd}=3C_a^d C_{bcd}+3C_c^d C_{abd}+10C_b^d C_{acd}.
\end{align*}


Alternating the last identity by indices $a$ and $b$, taking into account the third fundamental identity and the properties of the object $C^{abc}$, we obtain
\begin{equation}
\label{3.28}
C_c^d C_{abd}=7C_{[b}^d C_{a]cd}.
\end{equation}
Proceeding as in the proof of Theorems \ref{th3.2} and \ref{th3.4}, by identity (\ref{3.28}) we obtain that $C_{abc}=0$, i.e. the manifold is either a cosymplectic manifold or a Kenmotsu manifold.

By (\ref{3.5}) and (\ref{3.27}) we also have $S_{\hat{b}a,0}=S_{0a,\hat{b}}$, that is,
\begin{align*}
&2\chi(A_a^b-3C_a^b)=\chi(A_a^b-3C_a^b),\\
&\chi(A_a^b-3C_a^b )=0.
\end{align*}

Arguing as in the proof of Theorem~\ref{th3.4}, we obtain the next statement.

\begin{theorem}
\label{th3.6}
A harmonic NTS--manifold, the Ricci tensor of which is the Codazzi tensor, is locally equivalent to the product $\mathds{C}^n\times\mathds{R}$ or canonically concircular to the manifold $\mathds{C}^n\times\mathds{R}$ equipped with a cosymplectic structure.
\end{theorem}

\begin{definition} [\cite{9}, \cite{10}]
\label{d3.3}
A symmetric 2-tensor field $T$ is called the Killing tensor if
\begin{equation*}
%\label{3.30}
\nabla_X (T)(Y,Z)+\nabla_Y (T)(X,Z)+\nabla_Z (T)(X,Y)=0;\qquad \forall X,Y,Z\in\mathcal{X}(M).
\end{equation*}
\end{definition}

Now let $M^{2n+1}$ be a harmonic NTS--manifold, the Ricci tensor of which is the Killing tensor. Then the identity holds
\begin{equation*}
%\label{3.31}
\nabla_X (S)(Y,Z)+\nabla_Y (S)(X,Z)+\nabla_Z (S)(X,Y)=0; \qquad \forall X,Y,Z\in\mathcal{X}(M).
\end{equation*}
On the space of  associated $G$-structure this identity is written as
\begin{equation}
\label{3.32}
S_{ij,k}+S_{jk,i}+S_{ki,j}=0.
\end{equation}
In particular, it follows from (\ref{3.5}) and (\ref{3.32}) that
\begin{align*}
&S_{0a,\hat{b}}+S_{a\hat{b},0}+S_{\hat{b}0,a}=0,\\
&\chi(A_a^b-3C_a^b )+2\chi(A_a^b-3C_a^b )+\chi(A_a^b-3C_a^b )=0,\\
&\chi(A_a^b-3C_a^b )=0.
\end{align*}
Arguing as in the proof of Theorem~\ref{th3.4} and using the third fundamental identity, we obtain the following theorem.

\begin{theorem}
\label{th3.7}
A harmonic NTS--manifold, the Ricci tensor of which is a Killing tensor, is locally equivalent to the product $\mathds{C}^n\times\mathds{R}$ or canonically concircular to the manifold $\mathds{C}^n\times\mathds{R}$ equipped with a cosymplectic structure.
\end{theorem}

\begin{remark}
Theorems~\ref{th3.5}, \ref{th3.6}, \ref{th3.7} are invertible.
\end{remark}


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\end{document}



