\documentclass[12pt]{umj_eng3} \usepackage[T1,T2A]{fontenc} \usepackage[english]{babel} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsfonts, dsfont, graphicx, srcltx} \newtheorem{lemma}{Lemma}[section] \newtheorem{theorem}{Theorem}[section] \newtheorem{definition}{Definition}[section] \newtheorem{remark}{Remark}[section] \newtheorem{corollary}{Corollary}[section] %\def\udcs{517.958} %Здесь автор определяет УДК своей работы \def\currentyear{2026} \def\currentvolume{18} \def\currentissue{2} \lastpage{99} \urle{umj.ufaras.ru/index.php/umj/article/view/2026-18-2-89} \renewcommand{\baselinestretch}{1.03} \firstpage{89} \renewcommand{\leq}{\leqslant} \renewcommand{\geq}{\geqslant} %%%%% авторские макросы %%%%% \def\tht{\theta} \def\Om{\Omega} \def\om{\omega} \def\e{\varepsilon} \def\g{\gamma} \def\G{\Gamma} \def\l{\lambda} \def\p{\partial} \def\D{\Delta} \def\bs{\backslash} \def\k{\varkappa} \def\E{\mbox{\rm e}} \def\a{\alpha} \def\b{\beta} \def\si{\sigma} \def\Si{\Sigma} \def\d{\delta} \def\L{\Lambda} \def\z{\zeta} \def\vs{\varsigma} \def\r{\rho} \def\vp{\varphi} \def\vw{\varpi} \def\vr{\varrho} \def\vt{\vartheta} \def\h{\widehat} \def\Odr{O} \def\H{W_2} \def\Ho{\mathring{W}_2} \def\Hloc{W_{2,loc}} \def\Hper{W_{2,per}} \def\Hoper{\mathring{W}_{2,per}} \def\Hinf{W_\infty} \def\mD{{D}} \def\mR{{R}} \def\Th{\Theta} \def\I{\mathrm{I}} \def\iu{\mathrm{i}} \def\di{\,d} \def\cH{\mathcal{H}} \def\cS{\mathcal{S}} \def\la{\langle} \def\ra{\rangle} \def\cB{\mathcal{B}} \def\cP{\mathcal{P}} \def\cI{\mathcal{I}} \def\cL{\mathcal{L}} \def\cA{\mathcal{A}} \def\cM{\mathcal{M}} \def\cQ{\mathcal{Q}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\RE}{Re} \DeclareMathOperator{\IM}{Im} \DeclareMathOperator{\spec}{\sigma} \DeclareMathOperator{\discspec}{\sigma_{disc}} \DeclareMathOperator{\essspec}{\sigma_{ess}} \DeclareMathOperator{\dist}{dist} \DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator*{\essinf}{ess\,inf} \DeclareMathOperator{\rot}{rot} \DeclareMathOperator{\dvr}{div} \DeclareMathOperator*{\esssup}{ess\,sup} \DeclareMathOperator{\const}{const} \DeclareMathOperator{\Dom}{\mathfrak{D}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \numberwithin{equation}{section} \begin{document} %УДК \udcs \thispagestyle{empty} \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} \vskip20pt \begin{thebibliography}{999} \bibitem{4} V.F. 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