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\title{Remarks on Gauduchon K\"{a}hler-like manifolds}
\alttitle{Remarques sur les variétés de Gauduchon de type Kähler}

\author{\firstname{Jieming} \lastname{Yang}}
\address{Department of mathematics, Wenzhou University, Zhejiang, China}
\email{20230120@wzu.edu.cn}

\subjclass{53B35, 53C07, 53C55}

\keywords{\kwd{Gauduchon connections} \kwd{Ricci curvatures} \kwd{K\"{a}hler-like}}
\altkeywords{\kwd{Connexions de Gauduchon} \kwd{courbures de Ricci} \kwd{de type Kähler}}

\begin{abstract}
We introduce conditions weaker than Gauduchon K\"{a}hler-like and use them to derive rigidity results. These results extend those obtained in~\cite{ZZ22,ZZ23} under the stronger Gauduchon K\"{a}hler-like assumption.
\end{abstract}

\begin{altabstract}
Nous introduisons des conditions plus faibles que celles de Gauduchon de type Kähler et les utilisons pour en déduire des résultats de rigidité. Ces résultats étendent ceux obtenus dans~\cite{ZZ22,ZZ23} sous l'hypothèse plus forte de Gauduchon de type Kähler.
\end{altabstract}

\dateposted{2026-03-09}
\begin{document}
%\input{CR-pagedemetas}
%\end{document}
\maketitle

\section{Introduction}

Let $X$ be a complex manifold with $\dim_{\mathbb{C}}X=n\geq2$. Let
\[
\omega=\sqrt{-1}g_{i\bar{j}} \dif z^{i}\wedge \dif \bar{z}^{j}
\]
be a Hermitian metric on $X$.

A metric $\omega$ is called Gauduchon if $\partial\bar{\partial}\omega^{n-1}=0$, balanced if $ \dif \omega^{n-1}=0$, and K\"{a}hler if $ \dif \omega=0$. While there are obstructions to the existence of balanced (or K\"{a}hler) metrics, a classical result due to Gauduchon~\cite{Gauduchon} asserts the following: on any compact complex manifold $X$ with $n\geq2$, there exists a unique (up to a positive constant) Gauduchon metric in the conformal class of every Hermitian metric.

Let $\nabla$ and $\nabla^{B}$ be the Chern and Bismut connections of a Hermitian metric $\omega$. For any $t\in\mathbb{R}$, the corresponding Gauduchon connection $\nabla^{t}$ is defined by~\cite{G}
\[
\nabla^{t}=(1-t)\nabla+t\nabla^{B}.
\]
If $\omega$ is non-K\"{a}hler, the Gauduchon connections $\nabla^{t}$ are all distinct. In particular, the Lichnerowicz connection $\nabla^{\frac{1}{2}}$ coincides with the restriction of the complexified Levi-Civita connection to the holomorphic tangent bundle $T^{1,0}X$. For the geometry of these connections, see~\cite{FZ,LY,WY} and references therein.

The connection $\nabla^{t}$ is called K\"{a}hler-like if its curvature $F$ satisfies all the symmetries of a K\"{a}hler metric's curvature; namely, its $(2,0)$- and $(1,1)$-parts satisfy $F^{(2,0)}=0$ and $F^{(1,1)}$ is symmetric (see~\eqref{11155}). For background on K\"{a}hler-like manifolds, we refer to~\cite{AOUV,LS,YZ,ZZ22,ZZ23} and the references therein.

Let $F^{t,1}_{i\bar{j}}$, $F^{t,2}_{i\bar{j}}$, $F^{t,3}_{i\bar{j}}$ and $F^{t,4}_{i\bar{j}}$ denote the four Ricci curvatures of $F^{(1,1)}$ (see Definition~\ref{11153}). If $\nabla^{t}$ is K\"{a}hler-like, then these four Ricci curvatures coincide.

Our starting point is the following.

\begin{proposition}\label{11154}
There exists a Hermitian metric $\omega$ on the Hopf surface $S^{3}\times S^{1}$, whose curvatures of the Bismut connection $\nabla^{B}$ satisfy
\[
F^{(2,0)}=0,
\qquad
F^{1,1}_{i\bar{j}}=F^{1,2}_{i\bar{j}}=F^{1,3}_{i\bar{j}}=F^{1,4}_{i\bar{j}}.
\]
While $\nabla^{B}$ is not K\"{a}hler-like.
\end{proposition}

Our main results are the following.

\begin{theorem}\label{11151}
Let $(X, \omega)$ be a compact Hermitian manifold with $n\geq3$. If the curvature of the Gauduchon connection $\nabla^{t}$ satisfies
\[
F^{t,1}_{i\bar{j}}=F^{t,3}_{i\bar{j}}
\]
for some $t\in\parens[\big]{(-\infty, 2-\sqrt{3}]\cup[2+\sqrt{3}, +\infty)}\setminus\{0\}$, then $\omega$ is K\"{a}hler. While for $t=0$, $\omega$ is balanced.
\end{theorem}

As an instance of Theorem~\ref{11151} with $t=0$, consider the Iwasawa threefold with its natural metric~$\omega$. It is well known that $\omega$ is balanced and has a flat Chern connection $\nabla$. Furthermore, for the special case of compact complex surfaces, the conclusion of Theorem~\ref{11151} can be strengthened under an additional hypothesis.

\begin{theorem}\label{11152}
Let $(X, \omega)$ be a compact Hermitian surface. If the curvature of the Gauduchon connection $\nabla^{t}$ satisfies
\[
F^{t,1}_{i\bar{j}}=F^{t,2}_{i\bar{j}}=F^{t,3}_{i\bar{j}}
\]
for some $t\neq1$, then $\omega$ is K\"{a}hler.
\end{theorem}

During the proof of Theorem~\ref{11152}, we note that the Lichnerowicz connection $\nabla^{\frac{1}{2}}$ has a special property valid for all dimensions $n$ (see~\eqref{111512}). This leads us to study Gauduchon connections $\nabla^{t}$ satisfying
\begin{equation}\label{12071}
F^{(2,0)}=0,
\qquad
F^{t,1}_{i\bar{j}}=F^{t,2}_{i\bar{j}}=F^{t,3}_{i\bar{j}}.
\end{equation}
If $\nabla^{t}$ is K\"{a}hler-like, then $\nabla^{t}$ satisfies~\eqref{12071}.

\begin{proposition}\label{11252}
Let $(X, \omega)$ be a compact Gauduchon manifold with $n\geq3$. If the curvature of the Lichnerowicz connection $\nabla^{\frac{1}{2}}$ satisfies~\eqref{12071}, then $\omega$ is K\"{a}hler.
\end{proposition}

The compactness condition can be replaced by introducing an additional connection $\nabla^{s}$.

\begin{proposition}\label{12073}
Let $(X, \omega)$ be a Hermitian surface. Let $s$ and $t$ be two real numbers with $s\neq t$. If both the curvatures of Gauduchon connections $\nabla^{s}$ and $\nabla^{t}$ satisfy~\eqref{12071}, then $\omega$ is K\"{a}hler.
\end{proposition}

\begin{proposition}\label{12072}
Let $(X, \omega)$ be a Hermitian manifold with $n\geq3$. Let $s$ and $t$ be two real numbers with $(s-t)\parens[\big]{t-\frac{1}{2}}\neq0$. If both the curvatures of Gauduchon connections $\nabla^{s}$ and $\nabla^{t}$ satisfy~\eqref{12071}, then $\omega$ is K\"{a}hler.
\end{proposition}

Theorems 1~and~2 and Proposition~1 of~\cite{ZZ22} were proved under the assumption that $\nabla^{t}$ is K\"{a}hler-like. Our present results (Theorems~\ref{11151} and~\ref{11152}, and Propositions~\ref{11252}, \ref{12073} and \ref{12072}) extend those in~\cite{ZZ22} to weaker conditions.

The following results can be proved by arguments analogous to those for Theorems~\ref{11151}, \ref{11152} and Proposition~\ref{11252}; their proofs are therefore omitted.

\begin{theorem}\label{12121}
Let $(X, \omega)$ be a compact Hermitian manifold with $n\geq3$. If the curvature of the Bismut connection $\nabla^{B}$ satisfies~\eqref{12071}, then $\omega$ is a Gauduchon metric satisfying
\[
\Lambda_{\omega}\partial\bar{\partial}\omega=0,
\qquad
\nabla^{B}\tau=0,
\]
where $\tau=\Lambda_{\omega}\partial\omega$.
\end{theorem}

\goodbreak

\begin{theorem}\label{12122}
Let $(X, \omega)$ be a Hermitian surface. Then the following statements are equivalent:
\begin{enumerate}
\item the Bismut connection $\nabla^{B}$ satisfies~\eqref{12071};
\item $\nabla^{B}T=0$;
\item $\omega$ is Vaisman.
\end{enumerate}
Here $T$ is the torsion of the Chern connection $\nabla$ of $\omega$.
\end{theorem}

Theorem~\ref{12121} can be seen as an analogue, to some extent, of~\cite[Theorem~1]{ZZ23}. The cited theorem characterizes when the Bismut connection $\nabla^{B}$ of a (not necessarily compact) Hermitian manifold $(X, \omega)$ is K\"{a}hler-like by the conditions $\partial\bar{\partial}\omega=0$ and $\nabla^{B}T=0$. (We remark that the condition $\Lambda_{\omega}\partial\bar{\partial}\omega=0$ appears in works such as~\cite{FWW,Yang}.) In a similar vein, Theorem~\ref{12122} corresponds to~\cite[Theorem~2]{ZZ23}.

Section~\ref{section_2} sets up the notation and main definitions, and contains the proof of Proposition~\ref{11154}. Theorems~\ref{11151}, \ref{11152} and Proposition~\ref{11252} are proved in Sections \ref{section_3}~and~\ref{section_4}.

The proofs of Propositions~\ref{12073} and~\ref{12072} are omitted, as they follow directly from the arguments for~\cite[Theorem~2]{ZZ22}.

\section{Preliminaries}\label{section_2}

Let $X$ be a complex manifold with $\dim_{\mathbb{C}}X=n\geq2$. Let
\[
\omega=\sqrt{-1}g_{i\bar{j}} \dif z^{i}\wedge \dif \bar{z}^{j}
\]
be a Hermitian metric on $X$. Denote
\[
\Gamma^{p}_{ik}=g^{p\bar{j}}\partial_{k}g_{i\bar{j}}.
\]
The Chern connection $\nabla$ of $\omega$ is the unique connection which is compatible with the complex structure of $X$ and the Hermitian metric $\omega$. Locally, we have
\[
\nabla_{k}\parens[\Bigg]{\frac{\partial}{\partial z^{i}}} = \Gamma^{p}_{ik}\frac{\partial}{\partial z^{p}},
\qquad
\nabla_{\bar{l}}\parens[\Bigg]{\frac{\partial}{\partial z^{i}}}=0,
\]
where $\nabla_{k}=\nabla_{\frac{\partial}{\partial z^{k}}}$ and $\nabla_{\bar{l}}=\nabla_{\frac{\partial}{\partial\bar{z}^{l}}}$ denote the covariant derivatives of $\nabla$ along the vector fields $\frac{\partial}{\partial z^{k}}$ and $\frac{\partial}{\partial\bar{z}^{l}}$. Hence, for $\theta=\theta_{\bar{l}} \dif \bar{z}^{l}$, we have
\[
\nabla_{k}\theta_{\bar{l}}=\partial_{k}\theta_{\bar{l}},
\qquad
\nabla_{\bar{l}}\bar{\theta}_{\bar{k}}=\partial_{\bar{l}}\bar{\theta}_{\bar{k}}.
\]
These terms may be used interchangeably.

The Chern connection $\nabla$ has torsion tensor $T$ whose components are
\[
T_{ki\bar{j}}=\partial_{k}g_{i\bar{j}}-\partial_{i}g_{k\bar{j}}.
\]
Then, we have $T_{ki\bar{j}}=-T_{ik\bar{j}}$. Denote
\[
T^{p}_{ik} \coloneqq g^{p\bar{j}}T_{ik\bar{j}}=\Gamma^{p}_{ki}-\Gamma^{p}_{ik}
\]
and
\[
T_{i} \coloneqq g^{k\bar{l}}T_{ik\bar{l}}.
\]
Then the torsion $(1,0)$-form $\tau$ of $\nabla$ is defined by
\[
\tau \coloneqq \Lambda\partial\omega=T_{i} \dif z^{i}.
\]
Since each Hermitian metric $\omega$ satisfies
\begin{equation}\label{06041}
\partial\omega^{n-1}=\tau\wedge\omega^{n-1},
\end{equation}
it is direct to verify that $\omega$ is balanced if and only if
\[
\tau=0,
\]
and that $\omega$ is Gauduchon if and only if
\[
\partial^{*}\tau=0.
\]
Moreover, for a Hermitian metric $\omega$, we calculate
\begin{equation}\label{11161}
-\partial^{*}\tau=g^{k\bar{l}}\nabla_{\bar{l}}T_{k}+\abs{\tau}^{2},
\end{equation}
where
\[
g^{k\bar{l}}\nabla_{\bar{l}}T_{k}=g^{k\bar{l}}\nabla_{k}\bar{T}_{l},
\qquad
\abs{\tau}^{2}=g^{k\bar{l}}T_{k}\bar{T}_{l}.
\]
Integration by parts will be used frequently. Let $(X, \omega)$ be a compact Hermitian manifold. For $\theta=\theta_{\bar{l}} \dif \bar{z}^{l}$, we obtain from~\eqref{06041} that
\begin{equation}\label{05271}
\begin{split}
0 & = n\int_{X}\sqrt{-1} \dif \parens[\big]{(\theta-\bar{\theta})\wedge\omega^{n-1}}\\
& = n\int_{X}\sqrt{-1}(\partial\theta-\bar{\partial}\bar{\theta}+\tau\wedge\theta+\bar{\theta}\wedge\bar{\tau})\wedge\omega^{n-1}\\
& = \int_{X}2\Real\parens[\big]{g^{k\bar{l}}(\nabla_{k}\theta_{\bar{l}}+T_{k}\theta_{\bar{l}})}\omega^{n}.
\end{split}
\end{equation}
Hereafter we may omit the volume form $\frac{\omega^{n}}{n!}$.

The curvature $R_{\omega}$ of $\nabla$ is
\[
\nabla\circ\nabla\parens[\Bigg]{\frac{\partial}{\partial z^{^{i}}}}=R^{p}_{ik\bar{l}} \dif z^{k}\wedge \dif \bar{z}^{l}\otimes\frac{\partial}{\partial z^{p}},
\]
where (e.g., \cite[Chapter~1]{Kobayashi})
\begin{equation}\label{12152}
R^{p}_{ik\bar{l}}=g^{p\bar{j}}(-\partial_{\bar{l}}\partial_{k}g_{i\bar{j}}+g^{m\bar{n}}\partial_{\bar{l}}g_{m\bar{j}}\partial_{k}g_{i\bar{n}}).
\end{equation}
Denote
\[
R_{i\bar{j}k\bar{l}}=g_{p\bar{j}}R^{p}_{ik\bar{l}}=-\partial_{\bar{l}}\partial_{k}g_{i\bar{j}}+g^{m\bar{n}}\partial_{\bar{l}}g_{m\bar{j}}\partial_{k}g_{i\bar{n}}.
\]
Then we have
\begin{equation}\label{12151}
R_{i\bar{j}k\bar{l}}=\bar{R}_{j\bar{i}l\bar{k}}.
\end{equation}
The components of the four Ricci curvatures are
\[
R_{i\bar{j}}=g^{k\bar{l}}R_{k\bar{l}i\bar{j}},
\qquad K_{i\bar{j}}=g^{k\bar{l}}R_{i\bar{j}k\bar{l}}, 
\qquad R^{3}_{i\bar{j}}=g^{k\bar{l}}R_{i\bar{l}k\bar{j}},
\qquad R^{4}_{i\bar{j}}=g^{k\bar{l}}R_{k\bar{j}i\bar{l}}.
\]
Then we obtain from~\eqref{12151} that
\[
R_{i\bar{j}}=\bar{R}_{j\bar{i}},
\qquad K_{i\bar{j}}=\bar{K}_{j\bar{i}},
\qquad R^{3}_{i\bar{j}}=\bar{R}^{4}_{j\bar{i}}.
\]
The Bianchi identities are well-known (e.g., \cite[(2.6)]{SW}):
\begin{equation}\label{11022}
\begin{aligned}
	R_{i\bar{j}k\bar{l}}-R_{k\bar{j}i\bar{l}} & = \nabla_{\bar{l}}T_{ik\bar{j}}, \qquad R_{i\bar{j}k\bar{l}}-R_{i\bar{l}k\bar{j}}=\nabla_{k}\bar{T}_{jl\bar{i}},
\\	R_{i\bar{j}k\bar{l}}-R_{k\bar{l}i\bar{j}} & = \nabla_{\bar{l}}T_{ik\bar{j}}+\nabla_{i}\bar{T}_{jl\bar{k}}=\nabla_{\bar{j}}T_{ik\bar{l}}+\nabla_{k}\bar{T}_{jl\bar{i}}.
\end{aligned}
\end{equation}
In the local expression~\eqref{12152}, we follow the notations in~\cite{Kobayashi}. So the first two indices and the last two indices of the curvature tensor in~\cite[(2.4)]{SW} have been swapped. We obtain from~\eqref{11022} that
\begin{equation}\label{11281}
\begin{aligned}
	& R^{3}_{i\bar{j}} = R_{i\bar{j}}+\nabla_{\bar{j}}T_{i},
\\	& K_{i\bar{j}}-R_{i\bar{j}} = \frac{1}{2}(\nabla_{\bar{j}}T_{i}+\nabla_{i}\bar{T}_{j})-\frac{1}{2}g^{k\bar{l}}(\nabla_{\bar{l}}T_{ki\bar{j}}+\nabla_{k}\bar{T}_{lj\bar{i}}).
\end{aligned}
\end{equation}
The Bismut connection $\nabla^{B}$ of $\omega$ is defined by
\[
\nabla^{B}_{k}\parens[\Bigg]{\frac{\partial}{\partial z^{i}}}=\Gamma^{p}_{ki}\frac{\partial}{\partial z^{p}},
\qquad
\nabla^{B}_{\bar{l}}\parens[\Bigg]{\frac{\partial}{\partial z^{i}}}=g^{p\bar{q}}\bar{T}_{lq\bar{i}}\frac{\partial}{\partial z^{p}}.
\]
Then for every $t\in\mathbb{R}$, the Gauduchon connection $\nabla^{t}$ is defined by~\cite{G}
\begin{equation}\label{11251}
\nabla_{k}\parens[\Bigg]{\frac{\partial}{\partial z^{i}}}=\parens[\big]{(1-t)\Gamma^{p}_{ik}+t\Gamma^{p}_{ki}}\frac{\partial}{\partial z^{p}},
\qquad
\nabla_{\bar{l}}\parens[\Bigg]{\frac{\partial}{\partial z^{i}}}=tg^{p\bar{q}}\bar{T}_{lq\bar{i}}\frac{\partial}{\partial z^{p}}.
\end{equation}
The curvature of the Gauduchon connection $\nabla^{t}$ is
\[
\nabla^{t}\circ\nabla^{t}\parens[\Bigg]{\frac{\partial}{\partial z^{^{i}}}}=\parens[\Bigg]{\frac{1}{2}F^{p}_{imk} \dif z^{m}\wedge \dif z^{k} +F^{p}_{ik\bar{l}} \dif z^{k}\wedge \dif \bar{z}^{l} +\frac{1}{2}F^{p}_{i\bar{n}\bar{l}} \dif \bar{z}^{n}\wedge \dif \bar{z}^{l}}\otimes\frac{\partial}{\partial z^{^{p}}},
\]
where (e.g., \cite[Proposition~4.2]{FZ})
\[
F^{p}_{imk} =tg^{p\bar{j}}\parens[\big]{\nabla_{i}T_{mk\bar{j}}+(1-t)(T_{kr\bar{j}}T^{r}_{mi}+T_{mr\bar{j}}T^{r}_{ik})} =-g^{p\bar{j}}\bar{F}^{q}_{j\bar{m}\bar{k}}g_{i\bar{q}}
\]
and
\[
F^{t}_{i\bar{j}k\bar{l}}=g_{p\bar{j}}F^{p}_{ik\bar{l}} =R_{i\bar{j}k\bar{l}}+t(\nabla_{\bar{l}}T_{ki\bar{j}}+\nabla_{k}\bar{T}_{lj\bar{i}}) +t^{2}g^{p\bar{q}}(T_{ki\bar{q}}\bar{T}_{lj\bar{p}}-T_{pk\bar{j}}\bar{T}_{ql\bar{i}}).
\]
By~\eqref{12151}, we have
\begin{equation}\label{12156}
F^{t}_{i\bar{j}k\bar{l}}=\bar{F}^{t}_{j\bar{i}l\bar{k}}.
\end{equation}
Hence, $\nabla^{t}$ is K\"{a}hler-like if and only if (e.g., \cite[Section~2]{ZZ22})
\begin{equation}\label{11155}
F^{p}_{imk}=0,
\qquad
F^{t}_{i\bar{j}k\bar{l}}=F^{t}_{k\bar{j}i\bar{l}}.
\end{equation}
Indeed, we obtain from~\eqref{12156} and the second identity that in~\eqref{11155} that $F^{t}_{k\bar{l}i\bar{j}}=F^{t}_{i\bar{j}k\bar{l}}$.

Denote
\[
Q^{1}_{i\bar{j}}=g^{k\bar{l}}g^{p\bar{q}}T_{ik\bar{q}}\bar{T}_{jl\bar{p}},
\qquad Q^{2}_{i\bar{j}}=g^{k\bar{l}}g^{p\bar{q}}T_{pk\bar{j}}\bar{T}_{ql\bar{i}}
\]
and
\[
U_{i\bar{j}}=g^{p\bar{q}}\bar{T}_{q}T_{pi\bar{j}}, 
\qquad V_{i\bar{j}}=g^{k\bar{l}}\nabla_{\bar{l}}T_{ki\bar{j}}.
\]
Then we have
\[
Q^{1}_{i\bar{j}}=\bar{Q}^{1}_{j\bar{i}},
\qquad Q^{2}_{i\bar{j}}=\bar{Q}^{2}_{j\bar{i}}.
\]

\begin{definition}\label{11153}
Let $(X, \omega)$ be a Hermitian manifold. The four Ricci curvatures of the Gauduchon connection $\nabla^{t}$ are
\begin{align*}
	& F^{t,1}_{i\bar{j}}=g^{k\bar{l}}F^{t}_{k\bar{l}i\bar{j}}=R_{i\bar{j}}+t(\nabla_{\bar{j}}T_{i}+\nabla_{i}\bar{T}_{j}),
\\	& F^{t,2}_{i\bar{j}}=g^{k\bar{l}}F^{t}_{i\bar{j}k\bar{l}}=K_{i\bar{j}}+t(V_{i\bar{j}}+\bar{V}_{j\bar{i}})+t^{2}(Q^{1}_{i\bar{j}}-Q^{2}_{i\bar{j}}),
\\	& F^{t,3}_{i\bar{j}}=g^{k\bar{l}}F^{t}_{i\bar{l}k\bar{j}}=R^{3}_{i\bar{j}}-t(\nabla_{\bar{j}}T_{i}+\bar{V}_{j\bar{i}})-t^{2}(Q^{1}_{i\bar{j}}+\bar{U}_{j\bar{i}}),
\\	& F^{t,4}_{i\bar{j}}=g^{k\bar{l}}F^{t}_{k\bar{j}i\bar{l}}=R^{4}_{i\bar{j}}-t(\nabla_{i}\bar{T}_{j}+V_{i\bar{j}})-t^{2}(Q^{1}_{i\bar{j}}+U_{i\bar{j}}).
\end{align*}
\end{definition}

Then we have (by~\eqref{12156})
\[
F^{t,1}_{i\bar{j}}=\bar{F}^{t,1}_{j\bar{i}},
\qquad F^{t,2}_{i\bar{j}}=\bar{F}^{t,2}_{j\bar{i}}, 
\qquad F^{t,3}_{i\bar{j}}=\bar{F}^{t,4}_{j\bar{i}}.
\]

\begin{remark}
Let $(X, \omega)$ be a compact Hermitian manifold.

If $V_{i\bar{j}}=0$, then
\[
g^{i\bar{j}}\nabla_{\bar{j}}T_{i}=g^{i\bar{j}}V_{i\bar{j}}=0.
\]
Integration by parts (see~\eqref{05271}) yields
\[
0=-\int_{X}\partial^{*}\tau=\int_{X}\parens[\big]{g^{i\bar{j}}\nabla_{\bar{j}}T_{i}+\abs{\tau}^{2}}=\int_{X}\abs{\tau}^{2},
\]
which implies $\omega$ is balanced.

Hence, for $t=0$, we obtain from~\eqref{11281} and Definition~\ref{11153} that
\[
F^{0,1}_{i\bar{j}}=F^{0,2}_{i\bar{j}}=F^{0,3}_{i\bar{j}}=F^{0,4}_{i\bar{j}}=R_{i\bar{j}}.
\]
Moreover, if $\omega$ is of pointwise constant (Chern) holomorphic sectional curvature $u\in\mathcal{A}^{0}_{\mathbb{R}}(X)\setminus\{0\}$, then $\omega$ is K\"{a}hler and $u$ is a (non-zero) constant. The proof is similar to~\cite[Theorem~1.1]{T}, so we omit it.
\end{remark}

We close this section by a proof of Proposition~\ref{11154}.

Suppose $n\geq2$ and $(\delta_{1}, \dots, \delta_{n})\in\mathbb{C}^{n}\setminus\{0\}$ with $\abs{\delta_{1}}=\dots=\abs{\delta_{n}}\neq1$. Let $S^{2n-1}\times S^{1}=\parens[\big]{\mathbb{C}^{n}\setminus\{0\}}\big/\mathpunct{\sim}$ be the Hopf manifold, where
\[
(z^{1}, \dots, z^{n})\sim(\delta_{1}z^{1}, \dots, \delta_{n}z^{n}).
\]
For $\alpha>0$ and $\beta>-\alpha$, we consider the natural metric $\omega=\sqrt{-1}g_{i\bar{j}} \dif z^{i}\wedge \dif \bar{z}^{j}$ on $S^{2n-1}\times S^{1}$, where
\[
g_{i\bar{j}}=\alpha\frac{\delta_{ij}}{\abs{z}^{2}}+\beta\frac{\bar{z}^{i}z^{j}}{\abs{z}^{4}}
\]
for $\abs{z}^{2}=\sum_{l=1}^{n}\abs{z^{l}}^{2}$. Denote $\gamma=\frac{\beta}{\alpha}>-1$.

\begin{proof}[Proof of Proposition~\ref{11154}]
Direct calculation yields that the $(1,1)$-part of the curvature of $\nabla^{t}$ of $\omega$ is (e.g., \cite[Section~3]{Barbaro} for $t=1$)
\begin{multline*}
F^{t}_{i\bar{j}k\bar{l}} = \parens[\Big]{\parens[\big]{(1+\gamma)(2-t)t-\gamma}\delta_{il}\delta_{kj} +\parens[\big]{1-2(1+\gamma)t}\delta_{ij}\delta_{kl}}\frac{\alpha}{\abs{z}^{4}}
\\	\begin{aligned}
		& + \parens[\Big]{\parens[\big]{(1+\gamma t)(\gamma-t)-(\gamma t)^{2}}\parens{\bar{z}^{i}z^{l}\delta_{kj}+\bar{z}^{k}z^{j}\delta_{il}}
	\\	& + \parens[\big]{\parens[\big]{(1+\gamma)t-\gamma}^{2}+2\gamma}\bar{z}^{i}z^{j}\delta_{kl}+\parens[\big]{2(1+\gamma)t-1}\bar{z}^{k}z^{l}\delta_{ij}}\frac{\alpha}{\abs{z}^{6}}
	\\	& + \alpha\gamma\parens[\big]{(1+\gamma)t^{2}-(3+\gamma)}\frac{\bar{z}^{i}z^{j}\bar{z}^{k}z^{l}}{\abs{z}^{8}},
	\end{aligned}
\end{multline*}
which implies
\[
F^{t,1}_{i\bar{j}}=\parens[\Bigg]{\delta_{ij}-\frac{\bar{z}^{i}z^{j}}{\abs{z}^{2}}}\frac{n-2(n-1)(1+\gamma)t}{\abs{z}^{2}}
\]
and
\begin{multline*}
F^{t,2}_{i\bar{j}} = \parens[\big]{n-\parens[\big]{t^{2}+2(n-2)t+1}(1+\gamma)}\frac{\delta_{ij}}{\abs{z}^{2}}
+ \parens[\Big]{\parens[\big]{2n-1+(n-1)\gamma}\gamma-2(1+\gamma)\parens[\big]{1+(n-1)\gamma}t
\\	+ \parens[\big]{(n-1)\gamma+n}(1+\gamma)t^{2}}\frac{\bar{z}^{i}z^{j}}{\abs{z}^{4}}
\end{multline*}
and
\[
\begin{split}
F^{t,3}_{i\bar{j}}
	& = \parens[\big]{-n(1+\gamma)t^{2}+(2n-3)(1+\gamma)t+1-(n-1)\gamma}\frac{\delta_{ij}}{\abs{z}^{2}}
\\	& \quad + \parens[\Big]{(n-1)\gamma-1+(1+\gamma)\parens[\big]{(n-1)\gamma-(n-2)}t
+ \parens[\big]{1-(n-1)\gamma}(1+\gamma)t^{2}}\frac{\bar{z}^{i}z^{j}}{\abs{z}^{4}}
\\	& = F^{t,4}_{i\bar{j}}.
\end{split}
\]
A necessary condition for $F^{t,1}_{i\bar{j}}=F^{t,2}_{i\bar{j}}$ is
\[
n-2(n-1)(1+\gamma)t=n-\parens[\big]{t^{2}+2(n-2)t+1}(1+\gamma),
\]
i.e.,
\[
(1+\gamma)(t-1)^{2}=0.
\]
The only choice is $t=1$ since $1+\gamma>0$.

Then we have
\[
F^{1,1}_{i\bar{j}}=F^{1,2}_{i\bar{j}}=\parens[\Bigg]{\delta_{ij}-\frac{\bar{z}^{i}z^{j}}{\abs{z}^{2}}}\frac{n-2(n-1)(1+\gamma)}{\abs{z}^{2}}
\]
and
\[
F^{1,3}_{i\bar{j}}=F^{1,4}_{i\bar{j}}=\parens[\Bigg]{\delta_{ij}-\frac{\bar{z}^{i}z^{j}}{\abs{z}^{2}}}\frac{n-2(1+\gamma)}{\abs{z}^{2}}.
\]
Hence $F^{1,1}_{i\bar{j}}=F^{1,3}_{i\bar{j}}$ if and only if
\[
(n-2)(1+\gamma)=0,
\]
i.e., $n=2$ since $1+\gamma>0$.

Direct calculation yields
\[
\Gamma^{p}_{ik}=\parens[\Bigg]{\gamma\bar{z}^{i}\delta_{pk}-\bar{z}^{k}\delta_{pi}-\gamma\frac{\bar{z}^{i}\bar{z}^{k}z^{p}}{\abs{z}^{2}}}\frac{1}{\abs{z}^{2}}
\qquad \text{and} \qquad
T_{ki\bar{j}}=\parens{\bar{z}^{i}\delta_{kj}-\bar{z}^{k}\delta_{ij}}\frac{\alpha(1+\gamma)}{\abs{z}^{4}},
\]
which implies
\[
\nabla_{p}T_{ki\bar{j}}=\partial_{p}T_{ki\bar{j}}-T_{mi\bar{j}}\Gamma^{m}_{kp}-T_{km\bar{j}}\Gamma^{m}_{ip}=0.
\]
Then the $(2,0)$-part of the curvature of $\nabla^{t}$ of $\omega$ is $F^{p}_{imk}=\frac{1}{2}g^{p\bar{j}}\nabla_{i}T_{mk\bar{j}}=0$.

It is direct to check the Bismut connection $\nabla^{B}$ of $\omega$ is not K\"{a}hler-like.
\end{proof}

\section{Proofs of Theorems~\ref{11151} and~\ref{11152}}\label{section_3}

Let $(X, \omega)$ be a compact Hermitian manifold with $n\geq2$.

\begin{proof}[Proof of Theorem~\ref{11151}]
Since $F^{t,3}_{i\bar{j}}=\bar{F}^{t,4}_{j\bar{i}}$ for every Hermitian metric, we obtain from $F^{t,1}_{i\bar{j}}=F^{t,3}_{i\bar{j}}$ that
\[
F^{t,3}_{i\bar{j}}=F^{t,4}_{i\bar{j}}.
\]
By~\eqref{11281} and Definition~\ref{11153}, we get
\[
F^{t,3}_{i\bar{j}}=R_{i\bar{j}}+\frac{1-t}{2}(\nabla_{\bar{j}}T_{i}+\nabla_{i}\bar{T}_{j})-\frac{t}{2}(V_{i\bar{j}}+\bar{V}_{j\bar{i}})-\frac{t^{2}}{2}(U_{i\bar{j}}+\bar{U}_{j\bar{i}})-t^{2}Q^{1}_{i\bar{j}},
\]
which implies
\begin{equation}\label{11157}
(1-3t)(\nabla_{\bar{j}}T_{i}+\nabla_{i}\bar{T}_{j})=t(V_{i\bar{j}}+\bar{V}_{j\bar{i}})+t^{2}(U_{i\bar{j}}+\bar{U}_{j\bar{i}})+2t^{2}Q^{1}_{i\bar{j}}.
\end{equation}
Since
\begin{align*}
	& g^{i\bar{j}}V_{i\bar{j}}=g^{i\bar{j}}\nabla_{\bar{j}}T_{i}=g^{i\bar{j}}\nabla_{i}\bar{T}_{j}=g^{i\bar{j}}\bar{V}_{j\bar{i}},
\\	& g^{i\bar{j}}U_{i\bar{j}}=\abs{\tau}^{2}=g^{i\bar{j}}\bar{U}_{j\bar{i}},
\\	& g^{i\bar{j}}Q^{1}_{i\bar{j}}=g^{i\bar{j}}Q^{2}_{i\bar{j}}=g^{i\bar{j}}g^{k\bar{l}}g^{p\bar{q}}T_{pk\bar{j}}\bar{T}_{ql\bar{i}}=\abs{T}^{2},
\end{align*}
we obtain from~\eqref{11157} that
\[
(1-4t)g^{i\bar{j}}\nabla_{\bar{j}}T_{i}=t^{2}\parens[\big]{\abs{\tau}^{2}+\abs{T}^{2}}.
\]
Integration by parts yields (see~\eqref{11161} and~\eqref{05271})
\begin{equation}\label{11158}
0=(t^{2}-4t+1)\int_{X}\abs{\tau}^{2}+t^{2}\int_{X}\abs{T}^{2}.
\end{equation}
Let $n\geq3$. If $t=0$, the identity~\eqref{11158} is
\[
0=\int_{X}\abs{\tau}^{2},
\]
which implies $\omega$ is balanced. If $t\in\parens[\big]{(-\infty, 2-\sqrt{3}]\cup[2+\sqrt{3}, +\infty)}\setminus\{0\}$, the right-hand side of~\eqref{11158} is non-negative, which implies $\omega$ is K\"{a}hler.
\end{proof}

\begin{remark}
In Theorem~\ref{11151}, if we replace the condition $F^{t,1}_{i\bar{j}}=F^{t,3}_{i\bar{j}}$ by one of the following conditions:
\begin{enumerate}
\item $F^{t,2}_{i\bar{j}}=F^{t,3}_{i\bar{j}}$;
\item $F^{t,1}_{i\bar{j}}+F^{t,2}_{i\bar{j}}=F^{t,3}_{i\bar{j}}+F^{t,4}_{i\bar{j}}$;
\end{enumerate}
we still have the identity~\eqref{11158}, while other combinations of the four Ricci curvatures fail to give interesting identities like~\eqref{11158}.
\end{remark}

To prove Theorem~\ref{11152}, we need the following.

\begin{lemma}\label{11159}
Let $(X, \omega)$ be a compact Hermitian surface. If
\[
F^{\frac{1}{3},1}_{i\bar{j}}=F^{\frac{1}{3},2}_{i\bar{j}}=F^{\frac{1}{3},3}_{i\bar{j}}=F^{\frac{1}{3},4}_{i\bar{j}},
\]
then $\omega$ is K\"{a}hler.
\end{lemma}

With Lemma~\ref{11159} in hand, we can prove Theorem~\ref{11152}.

\begin{proof}[Proof of Theorem~\ref{11152}]
Suppose $F^{t,1}_{i\bar{j}}=F^{t,3}_{i\bar{j}}$. Since (see~\eqref{06041})
\[
\abs{T}^{2}=2\abs{\tau}^{2},
\]
we obtain from~\eqref{11158} that
\[
0=(t-1)(3t-1)\int_{X}\abs{\tau}^{2}.
\]
By Lemma~\ref{11159}, for $t\neq1$, we have $\tau=0$, which implies $\omega$ is K\"{a}hler by~\eqref{06041}.
\end{proof}

We have to prove Lemma~\ref{11159}. We calculate for general $n\geq2$ and $t\in\mathbb{R}$.

Let $(X, \omega)$ be a Hermitian manifold whose Gauduchon connection $\nabla^{t}$ satisfies
\[
F^{t,1}_{i\bar{j}}=F^{t,2}_{i\bar{j}}=F^{t,3}_{i\bar{j}}=F^{t,4}_{i\bar{j}}
\]
for some $t\in\mathbb{R}$.

By the Bianchi identities~\eqref{11022}, we get
\[
\nabla_{\bar{j}}T_{ik\bar{l}}+\nabla_{\bar{l}}T_{ki\bar{j}}=\nabla_{i}\bar{T}_{jl\bar{k}}+\nabla_{k}\bar{T}_{lj\bar{i}},
\]
which implies
\begin{equation}\label{11156}
\nabla_{\bar{j}}T_{i}+V_{i\bar{j}}=\nabla_{i}\bar{T}_{j}+\bar{V}_{j\bar{i}}.
\end{equation}
Since $F^{t,1}_{i\bar{j}}=F^{t,2}_{i\bar{j}}$, we obtain from~\eqref{11281} and Definition~\ref{11153} that
\begin{equation}\label{111510}
(2t-1)(\nabla_{\bar{j}}T_{i}+\nabla_{i}\bar{T}_{j})=(2t-1)(V_{i\bar{j}}+\bar{V}_{j\bar{i}})+2t^{2}(Q^{1}_{i\bar{j}}-Q^{2}_{i\bar{j}}).
\end{equation}
Combining~\eqref{11157} with~\eqref{111510} yields
\[
V_{i\bar{j}}+\bar{V}_{j\bar{i}}=(3-8t)(\nabla_{\bar{j}}T_{i}+\nabla_{i}\bar{T}_{j})-2t^{2}(U_{i\bar{j}}+\bar{U}_{j\bar{i}}+Q^{1}_{i\bar{j}}+Q^{2}_{i\bar{j}})
\]
Inserting the above identity into~\eqref{111510} yields
\begin{equation}\label{111511}
(2t-1)(4t-1)(\nabla_{\bar{j}}T_{i}+\nabla_{i}\bar{T}_{j})=t^{2}(1-2t)(U_{i\bar{j}}+\bar{U}_{j\bar{i}})+2t^{2}(1-t)Q^{1}_{i\bar{j}}-2t^{3}Q^{2}_{i\bar{j}}.
\end{equation}
Since $F^{t,3}_{i\bar{j}}=F^{t,4}_{i\bar{j}}$, we obtain from Definition~\ref{11153} that
\[
(t-1)(\nabla_{\bar{j}}T_{i}-\nabla_{i}\bar{T}_{j})=t(V_{i\bar{j}}-\bar{V}_{j\bar{i}})+t^{2}(U_{i\bar{j}}-\bar{U}_{j\bar{i}}).
\]
Inserting~\eqref{11156} into the above identity yields
\begin{equation}\label{11255}
(2t-1)(\nabla_{\bar{j}}T_{i}-\nabla_{i}\bar{T}_{j})=t^{2}(U_{i\bar{j}}-\bar{U}_{j\bar{i}}).
\end{equation}
If $t=\frac{1}{4}$, then $\omega$ is already K\"{a}hler by~\eqref{11158}. Thus for $t\neq\frac{1}{4}$, we obtain from~\eqref{111511} and~\eqref{11255} that
\begin{equation}\label{111512}
(2t-1)(4t-1)\nabla_{\bar{j}}T_{i}=t^{3}U_{i\bar{j}}+t^{2}(1-3t)\bar{U}_{j\bar{i}}+t^{2}(1-t)Q^{1}_{i\bar{j}}-t^{3}Q^{2}_{i\bar{j}}.
\end{equation}
Take $n=2$. Then we have (see~\eqref{06041})
\[
U_{i\bar{j}}=\abs{\tau}^{2}g_{i\bar{j}}-T_{i}T_{\bar{j}},
\qquad V_{i\bar{j}}=g^{k\bar{l}}\nabla_{\bar{l}}T_{k}g_{i\bar{j}}-\nabla_{\bar{j}}T_{i}
\]
and
\[
Q^{1}_{i\bar{j}}=\abs{\tau}^{2}g_{i\bar{j}},
\qquad Q^{2}_{i\bar{j}}=2\abs{\tau}^{2}g_{i\bar{j}}-2T_{i}T_{\bar{j}}.
\]
Inserting these into~\eqref{111512} for $t=\frac{1}{3}$ yields
\begin{equation}\label{111513}
-3\nabla_{\bar{j}}T_{i}=\abs{\tau}^{2}g_{i\bar{j}}+T_{i}T_{\bar{j}},
\end{equation}
which implies
\[
-g^{i\bar{j}}\nabla_{\bar{j}}T_{i}=\abs{\tau}^{2}.
\]
We obtain from~\eqref{11161} that $\omega$ is Gauduchon.
\begin{proof}[Proof of Lemma~\ref{11159}]
Let $(X, \omega)$ be a compact Hermitian surface.

By~\eqref{06041}, we calculate
\[
[\nabla_{\bar{j}}, \nabla_{\bar{l}}]T_{i}=\nabla_{\bar{q}}T_{i}\bar{T}^{q}_{lj}=\bar{T}_{l}\nabla_{\bar{j}}T_{i}-\bar{T}_{j}\nabla_{\bar{l}}T_{i}.
\]
which implies
\[
-3\cdot2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\nabla_{\bar{j}}\nabla_{\bar{l}}T_{i}) 
\begin{multlined}[t][.6\displaywidth]
= -3\cdot2\Real\parens[\big]{g^{k\bar{l}}T_{k}\nabla_{\bar{l}}(g^{i\bar{j}}\nabla_{\bar{j}}T_{i})}-6\abs{\tau}^{2}g^{k\bar{l}}\nabla_{\bar{l}}T_{k}
\\	+ 3\cdot2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\bar{T}_{j}\nabla_{\bar{l}}T_{i}).
\end{multlined}
\]
Since $\omega$ is Gauduchon, integration by parts (see~\eqref{05271}) for the right-hand side yields
\begin{equation}\label{11282}
-3\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\nabla_{\bar{j}}\nabla_{\bar{l}}T_{i})=6\int_{X}\abs{\tau}^{4}+3\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\bar{T}_{j}\nabla_{\bar{l}}T_{i}).
\end{equation}
On the other hand, we obtain from~\eqref{111513} that
\[
-3\cdot2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\nabla_{\bar{j}}\nabla_{\bar{l}}T_{i})
= 2\Real\parens[\big]{g^{k\bar{l}}T_{k}\nabla_{\bar{l}}\abs{\tau}^{2}}+2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}T_{i}\nabla_{\bar{l}}\bar{T}_{j})
+ 2\abs{\tau}^{2}g^{k\bar{l}}\nabla_{\bar{l}}T_{k}.
\]
Integration by parts (see~\eqref{05271}) yields
\[
\begin{split}
-3\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\nabla_{\bar{j}}\nabla_{\bar{l}}T_{i})
	& = 2\int_{X}2\Real\parens[\big]{g^{k\bar{l}}T_{k}\nabla_{\bar{l}}\abs{\tau}^{2}}-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\bar{T}_{j}\nabla_{\bar{l}}T_{i})
	-2\int_{X}\abs{\tau}^{4}
\\	& = -\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\bar{T}_{j}\nabla_{\bar{l}}T_{i})-2\int_{X}\abs{\tau}^{4}.
\end{split}
\]
Combining the above identity with~\eqref{11282} yields
\[
2\int_{X}\abs{\tau}^{4}=-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\bar{T}_{j}\nabla_{\bar{l}}T_{i}).
\]
Inserting~\eqref{111513} into the right-hand side yields
\[
6\int_{X}\abs{\tau}^{4}=-3\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\bar{T}_{j}\nabla_{\bar{l}}T_{i})=4\int_{X}\abs{\tau}^{4}.
\]
We have $\tau=0$, which implies $\omega$ is K\"{a}hler by~\eqref{06041}.
\end{proof}

\section{Proof of Proposition~\ref{11252}}\label{section_4}

Let $(X, \omega)$ be a compact Hermitian manifold with $n\geq3$. We consider the Lichnerowicz connection $\nabla^{\frac{1}{2}}$ of $\omega$ satisfying
\begin{equation}\label{11256}
F^{p}_{imk}=0,
\qquad F^{\frac{1}{2},1}_{i\bar{j}}=F^{\frac{1}{2},2}_{i\bar{j}}=F^{\frac{1}{2},3}_{i\bar{j}}.
\end{equation}
By the calculation in Section~\ref{section_2} and the first identity in~\eqref{11256}, we have
\begin{equation}\label{11253}
\nabla_{k}T_{i}=-\frac{1}{2}(T_{p}T^{p}_{ki}+T^{p}_{km}T^{m}_{ip}).
\end{equation}
With the second identity in~\eqref{11256} in hand, we obtain from~\eqref{111510} that
\[
Q^{1}_{i\bar{j}}=Q^{2}_{i\bar{j}},
\]
and from~\eqref{11255} that
\[
U_{i\bar{j}}=\bar{U}_{j\bar{i}},
\]
and from~\eqref{11157} and~\eqref{11156} that
\begin{equation}\label{11254}
0=2(\nabla_{\bar{j}}T_{i}+V_{i\bar{j}})+U_{i\bar{j}}+Q^{1}_{i\bar{j}}.
\end{equation}

\begin{lemma}\label{11257}
Let $(X, \omega)$ be a compact Hermitian manifold with $n\geq3$. If\/ $\nabla^{\frac{1}{2}}$ satisfies~\eqref{11256}, then
\[
 \dif (\tau+\bar{\tau})=0.
\]
\end{lemma}
\begin{proof}
By~\eqref{11253}, we have
\[
\nabla_{k}T_{i}-\nabla_{i}T_{k}=T_{p}T^{p}_{ik}.
\]
While by the definition of the Chern connection,
\[
\nabla_{k}T_{i}-\nabla_{i}T_{k}=\partial_{k}T_{i}-\partial_{i}T_{k}+T_{p}T^{p}_{ik}.
\]
Then we have
\[
0=\frac{1}{2}(\partial_{k}T_{i}-\partial_{i}T_{k}) \dif z^{k}\wedge \dif z^{i}=\partial\tau.
\]
We only need to show
\[
\bar{\partial}\tau+\partial\bar{\tau}=0.
\]

\begin{step}
Direct calculation yields
\[
[\nabla_{\bar{l}}, \nabla_{\bar{j}}]T_{i}=\nabla_{\bar{q}}T_{i}\bar{T}^{q}_{jl},
\]
which implies
\begin{equation}\label{11259}
2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}T_{k}[\nabla_{\bar{l}}, \nabla_{\bar{j}}]T_{i}}=-2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}U_{k\bar{j}}).
\end{equation}
Integration by parts (see~\eqref{05271}) for the left-hand side in~\eqref{11259} yields (by~\eqref{11161})
\[
\begin{split}
\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\nabla_{\bar{l}}\nabla_{\bar{j}}T_{i}) & = \int_{X}2\Real\parens[\big]{g^{k\bar{l}}T_{k}\nabla_{\bar{l}}(g^{i\bar{j}}\nabla_{\bar{j}}T_{i})}\\
& = 2\int_{X}(\partial^{*}\tau)g^{i\bar{j}}\nabla_{\bar{j}}T_{i}
\end{split}
\]
and
\[
-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{k}\nabla_{\bar{j}}\nabla_{\bar{l}}T_{i})=\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{j}}T_{k}\nabla_{\bar{l}}T_{i})+\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}T_{k}\bar{T}_{j}).
\]
Integration by parts (see~\eqref{05271}) for the right-hand side in~\eqref{11259} yields
\[
\begin{split}
-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}U_{k\bar{j}}) & = \int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{i}\nabla_{\bar{l}}U_{k\bar{j}})+\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}U_{k\bar{j}}\bar{T}_{l}T_{i})\\
& = \int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{j}}\bar{T}_{l}T_{p}T^{p}_{ki})-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}V_{i\bar{l}}T_{k}\bar{T}_{j}),
\end{split}
\]
where the second equality follows from the definition of $U_{i\bar{j}}$ and $V_{i\bar{j}}$ and
\begin{equation}\label{11283}
2\Real(g^{i\bar{j}}g^{k\bar{l}}U_{i\bar{l}}T_{k}\bar{T}_{j})=2\Real(g^{i\bar{j}}g^{k\bar{l}}T_{p}T^{p}_{ki}\bar{T}_{l}\bar{T}_{j})=0,
\end{equation}
since $T^{p}_{ki}=-T^{p}_{ik}$.

Moreover, we obtain from~\eqref{11253} that
\[
\begin{split}
2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{j}}\bar{T}_{l}T_{p}T^{p}_{ki}) & = -\frac{1}{2}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}(\bar{T}_{q}\bar{T}^{q}_{jl}+\bar{T}^{q}_{jn}\bar{T}^{n}_{lq})T_{p}T^{p}_{ki}}\\
& = -\frac{1}{2}2\Real(g^{i\bar{j}}g^{k\bar{l}}\bar{T}_{q}\bar{T}^{q}_{jl}T_{p}T^{p}_{ki})\\
& = \frac{1}{2}2\Real(g^{i\bar{j}}g^{k\bar{l}}Q^{1}_{i\bar{l}}T_{k}\bar{T}_{j})\\
& =g^{i\bar{j}}g^{k\bar{l}}U_{i\bar{l}}U_{k\bar{j}}=\abs{U}^{2},
\end{split}
\]
where we have used the definition of $Q^{1}_{i\bar{j}}$ and
\[
Q^{1}_{i\bar{j}}=Q^{2}_{i\bar{j}},
\qquad
U_{i\bar{j}}=\bar{U}_{j\bar{i}}.
\]
Then we obtain from~\eqref{11259} that
\[
2\int_{X}(\partial^{*}\tau)g^{i\bar{j}}\nabla_{\bar{j}}T_{i}+\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{j}}T_{k}\nabla_{\bar{l}}T_{i}) =\int_{X}\abs{U}^{2}-\int_{X}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}(\nabla_{\bar{l}}T_{i}+V_{i\bar{l}})T_{k}\bar{T}_{j}},
\]
where (by~\eqref{11254} and~\eqref{11283})
\[
\begin{split}
-\int_{X}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}(\nabla_{\bar{l}}T_{i}+V_{i\bar{l}})T_{k}\bar{T}_{j}} & = \frac{1}{2}\int_{X}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}(U_{i\bar{l}}+Q^{1}_{i\bar{l}})T_{k}\bar{T}_{j}}\\
& = \frac{1}{2}\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}Q^{1}_{i\bar{l}}T_{k}\bar{T}_{j})\\
& = \int_{X}\abs{U}^{2}.
\end{split}
\]
Thus we get
\begin{equation}\label{112510}
\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{j}}T_{k}\nabla_{\bar{l}}T_{i}) =2\int_{X}\abs{U}^{2}-2\int_{X}(\partial^{*}\tau)g^{i\bar{j}}\nabla_{\bar{j}}T_{i}.
\end{equation}
Denote $\Delta=\sqrt{-1}\Lambda_{\omega}\partial\bar{\partial}$. We have
\[
2\Delta\abs{\tau}^{2}=2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}(\nabla_{k}T_{i}\bar{T}_{j}+T_{i}\nabla_{k}\bar{T}_{j})}.
\]
Integration by parts (see~\eqref{05271}) yields (by~\eqref{11161})
\[
2\int_{X}(\partial^{*}\tau)\abs{\tau}^{2}=\int_{X}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\bar{T}_{l}(\nabla_{k}T_{i}\bar{T}_{j}+T_{i}\nabla_{k}\bar{T}_{j})},
\]
where we obtain from~\eqref{11253} that
\[
2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{k}T_{i}\bar{T}_{l}\bar{T}_{j}) =-\frac{1}{2}2\Real(g^{i\bar{j}}g^{k\bar{l}}T^{p}_{km}T^{m}_{ip}\bar{T}_{l}\bar{T}_{j})=-\abs{U}^{2}.
\]
Then we get
\begin{equation}\label{11243}
\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{k}\bar{T}_{j}\bar{T}_{l}T_{i})=\int_{X}\abs{U}^{2}+2\int_{X}(\partial^{*}\tau)\abs{\tau}^{2}.
\end{equation}
By~\eqref{11259}, \eqref{112510} and~\eqref{11243}, we get
\begin{equation}\label{11244}
\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}U_{k\bar{j}})=-3\int_{X}\abs{U}^{2}-2\int_{X}(\partial^{*}\tau)\abs{\tau}^{2}.
\end{equation}
\end{step}

\begin{step}
Motivated by~\cite[Lemma~3.7]{LS}, we calculate
\begin{multline}\label{112511}
2\Real \parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\parens[\big]{\bar{T}_{j}[\nabla^{\frac{1}{2}}_{\bar{l}}, \nabla^{\frac{1}{2}}_{k}]T_{i}-\bar{T}_{l}[\nabla^{\frac{1}{2}}_{\bar{j}}, \nabla^{\frac{1}{2}}_{k}]T_{i}}}
\\	= 2\Real(g^{i\bar{j}}g^{k\bar{l}}\bar{T}_{l}\nabla^{\frac{1}{2}}_{k}\nabla^{\frac{1}{2}}_{\bar{j}}T_{i}) -2\Real(g^{i\bar{j}}g^{k\bar{l}}\bar{T}_{j}\nabla^{\frac{1}{2}}_{k}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}),
\end{multline}
where we have used (by~\eqref{11251} and~\eqref{11253})
\begin{equation}\label{11246}
\nabla^{\frac{1}{2}}_{k}T_{i}=\nabla_{k}T_{i}+\frac{1}{2}T_{p}T^{p}_{ki}=-\frac{1}{2}T^{p}_{km}T^{m}_{ip}=\nabla^{\frac{1}{2}}_{i}T_{k}.
\end{equation}
Since
\[
2\Real(g^{i\bar{j}}g^{k\bar{l}}\bar{T}_{l}\nabla^{\frac{1}{2}}_{k}\nabla^{\frac{1}{2}}_{\bar{j}}T_{i}) =2\Real\parens[\big]{g^{k\bar{l}}\bar{T}_{l}\nabla_{k}(g^{i\bar{j}}\nabla^{\frac{1}{2}}_{\bar{j}}T_{i})}
\]
and
\begin{equation}\label{11245}
\nabla^{\frac{1}{2}}_{\bar{j}}T_{i}=\nabla_{\bar{j}}T_{i}-\frac{1}{2}g^{p\bar{q}}T_{p}\bar{T}_{jq\bar{i}} =\nabla_{\bar{j}}T_{i}+\frac{1}{2}\bar{U}_{j\bar{i}}=\nabla_{\bar{j}}T_{i}+\frac{1}{2}U_{i\bar{j}},
\end{equation}
integration by parts yields (see~\eqref{11161} and~\eqref{05271})
\begin{equation}\label{11241}
\begin{split}
\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\bar{T}_{l}\nabla^{\frac{1}{2}}_{k}\nabla^{\frac{1}{2}}_{\bar{j}}T_{i}) & = 2\int_{X}(\partial^{*}\tau)g^{i\bar{j}}\parens[\Bigg]{\nabla_{\bar{j}}T_{i}+\frac{1}{2}U_{i\bar{j}}}
\\	& = 2\int_{X}(\partial^{*}\tau)g^{i\bar{j}}\nabla_{\bar{j}}T_{i}+\int_{X}(\partial^{*}\tau)\abs{\tau}^{2}.
\end{split}
\end{equation}
By~\eqref{11251}, we get
\[
\nabla^{\frac{1}{2}}_{k}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}=\partial_{k}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}-\nabla^{\frac{1}{2}}_{\bar{l}}T_{p}\Gamma^{p}_{ik} -\frac{1}{2}g^{p\bar{q}}(\nabla^{\frac{1}{2}}_{\bar{l}}T_{p}T_{ik\bar{q}}+\nabla^{\frac{1}{2}}_{\bar{q}}T_{i}T_{kp\bar{l}}).
\]
Then we have
\begin{multline}\label{11242}
-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\bar{T}_{j}\nabla^{\frac{1}{2}}_{k}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i})
= \frac{1}{2}\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}U_{k\bar{j}})+\frac{1}{2}\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}T_{k}\bar{T}_{j})
\\	+ \int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}\nabla_{k}\bar{T}_{j}).
\end{multline}
We obtain from~\eqref{11283}, \eqref{11243}, \eqref{11244} and~\eqref{11245} that
\begin{align*}
	& \frac{1}{2}\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}U_{k\bar{j}})
	\begin{aligned}[t]
		& = \frac{1}{2}\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}U_{k\bar{j}})+\frac{1}{2}\int_{X}\abs{U}^{2}
	\\	& = -\int_{X}\abs{U}^{2}-\int_{X}(\partial^{*}\tau)\abs{\tau}^{2},
	\end{aligned}
\\	&	\frac{1}{2}\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}T_{k}\bar{T}_{j})
	\begin{aligned}[t]
		& = \frac{1}{2}\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}T_{k}\bar{T}_{j})
	\\	& = \frac{1}{2}\int_{X}\abs{U}^{2}+\int_{X}(\partial^{*}\tau)\abs{\tau}^{2},
	\end{aligned}
\\	& \int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}\nabla_{k}\bar{T}_{j})
	\begin{aligned}[t]
		& = 2\int_{X}\abs{\bar{\nabla}\tau}^{2}+\frac{1}{2}\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}U_{i\bar{l}}\nabla_{k}\bar{T}_{j})
	\\	& = 2\int_{X}\abs{\bar{\nabla}\tau}^{2}-\frac{3}{2}\int_{X}\abs{U}^{2}-\int_{X}(\partial^{*}\tau)\abs{\tau}^{2},
	\end{aligned}
\end{align*}
where
\[
\abs{\bar{\nabla}\tau}^{2}=g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}\nabla_{k}\bar{T}_{j}.
\]
Inserting the three identities into~\eqref{11242} yields
\[
-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\bar{T}_{j}\nabla^{\frac{1}{2}}_{k}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}) =2\int_{X}\abs{\bar{\nabla}\tau}^{2}-2\int_{X}\abs{U}^{2}-\int_{X}(\partial^{*}\tau)\abs{\tau}^{2}.
\]
Combining the above identity with~\eqref{112510}, \eqref{112511} and~\eqref{11241}, we get
\begin{equation}\label{11247}
\begin{split}
\int_{X}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\parens[\big]{\bar{T}_{j}[\nabla^{\frac{1}{2}}_{\bar{l}}, \nabla^{\frac{1}{2}}_{k}]T_{i}-\bar{T}_{l}[\nabla^{\frac{1}{2}}_{\bar{j}}, \nabla^{\frac{1}{2}}_{k}]T_{i}}} & = 2\int_{X}\abs{\bar{\nabla}\tau}^{2}-2\int_{X}\abs{U}^{2}+2\int_{X}(\partial^{*}\tau)g^{i\bar{j}}\nabla_{\bar{j}}T_{i}
\\
& = 2\int_{X}\abs{\bar{\nabla}\tau}^{2}-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{j}}T_{k}\nabla_{\bar{l}}T_{i}).
\end{split}
\end{equation}
We deal with the left-hand side of~\eqref{11247}. By definition (see~\eqref{11251}), we get (e.g., \cite[Corollary~2.5]{LS})
\[
[\nabla^{t}_{\bar{l}}, \nabla^{t}_{k}]T_{i}=T_{p}F^{p}_{ik\bar{l}}+tg^{p\bar{q}}(\nabla^{t}_{\bar{q}}T_{i}T_{kp\bar{l}}-\nabla^{t}_{p}T_{i}\bar{T}_{lq\bar{k}}).
\]
For $t=\frac{1}{2}$, we get
\begin{multline*}
2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\parens[\big]{\bar{T}_{j}[\nabla^{\frac{1}{2}}_{\bar{l}}, \nabla^{\frac{1}{2}}_{k}]T_{i}-\bar{T}_{l}[\nabla^{\frac{1}{2}}_{\bar{j}}, \nabla^{\frac{1}{2}}_{k}]T_{i}}}
\\	\begin{aligned}
		& = 2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}(F^{\frac{1}{2},2}_{i\bar{l}}-F^{\frac{1}{2},4}_{i\bar{l}})T_{k}\bar{T}_{j}}+\frac{1}{2}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{k}T_{i}\bar{T}_{q}\bar{T}^{q}_{jl})
	\\	& \quad + \frac{1}{2}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{k}T_{i}\bar{T}_{l}\bar{T}_{j})-\frac{1}{2}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}(U_{k\bar{j}}+T_{k}\bar{T}_{j})},
	\end{aligned}
\end{multline*}
where the first and second terms on the right-hand side vanish due to~\eqref{11256} and~\eqref{11246}, respectively. For the third term, we obtain from~\eqref{11246} again that
\[%\label{11263}
\frac{1}{2}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{k}T_{i}\bar{T}_{l}\bar{T}_{j})=-\frac{1}{2}\abs{U}^{2}.
\]
For the last term, we obtain from~\eqref{11283}, \eqref{11243}, \eqref{11244} and~\eqref{11245} that
\[
\begin{split}
-\frac{1}{2}\int_{X}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}(U_{k\bar{j}}+T_{k}\bar{T}_{j})} & = -\frac{1}{2}\int_{X}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}(U_{k\bar{j}}+T_{k}\bar{T}_{j})}-\frac{1}{2}\int_{X}\abs{U}^{2}\\
& = \frac{1}{2}\int_{X}\abs{U}^{2}.
\end{split}
\]
Summing up yields
\[
\int_{X}2\Real\parens[\big]{g^{i\bar{j}}g^{k\bar{l}}\parens[\big]{\bar{T}_{j}[\nabla^{\frac{1}{2}}_{\bar{l}}, \nabla^{\frac{1}{2}}_{k}]T_{i}-\bar{T}_{l}[\nabla^{\frac{1}{2}}_{\bar{j}}, \nabla^{\frac{1}{2}}_{k}]T_{i}}}=0.
\]
Inserting it into~\eqref{11247} yields
\begin{equation}\label{11284}
0=2\int_{X}\abs{\bar{\nabla}\tau}^{2}-\int_{X}2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{j}}T_{k}\nabla_{\bar{l}}T_{i}).
\end{equation}
We use the completed square
\[
0\leq g^{i\bar{j}}g^{k\bar{l}}(\nabla_{\bar{l}}T_{i}-\nabla_{i}\bar{T}_{l})(\nabla_{k}\bar{T}_{j}-\nabla_{\bar{j}}T_{k}) =2\abs{\bar{\nabla}\tau}^{2}-2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{j}}T_{k}\nabla_{\bar{l}}T_{i}),
\]
where the equality holds if and only if
\[
\nabla_{\bar{l}}T_{i}=\nabla_{i}\bar{T}_{l}.
\]
Then we have $\bar{\partial}\tau+\partial\bar{\tau}=0$.

The conclusion follows from $\partial\tau=0$ and $\bar{\partial}\tau+\partial\bar{\tau}=0$. \qedhere
\end{step}
\let\qed\relax
\end{proof}

\begin{proof}[Proof of Proposition~\ref{11252}]
Let $(X, \omega)$ be a compact Hermitian manifold with $n\geq3$.

We obtain from~\eqref{11245} that
\[
2\abs{\bar{\nabla}^{\frac{1}{2}}\tau}^{2}=2g^{i\bar{j}}g^{k\bar{l}}\nabla^{\frac{1}{2}}_{\bar{l}}T_{i}\nabla^{\frac{1}{2}}_{k}\bar{T}_{j}=2\abs{\bar{\nabla}\tau}^{2}+2\Real(g^{i\bar{j}}g^{k\bar{l}}\nabla_{\bar{l}}T_{i}U_{k\bar{j}})+\frac{1}{2}\abs{U}^{2}.
\]
By~\eqref{11161}, \eqref{112510}, \eqref{11244} and~\eqref{11284}, we have
\[
\int_{X}(\partial^{*}\tau)^{2}=\int_{X}\abs{\bar{\nabla}^{\frac{1}{2}}\tau}^{2}+\frac{1}{4}\int_{X}\abs{U}^{2}.
\]
Suppose $\omega$ is Gauduchon. By~\eqref{11161}, we have
\[
0=\int_{X}\abs{\bar{\nabla}^{\frac{1}{2}}\tau}^{2}+\frac{1}{4}\int_{X}\abs{U}^{2},
\]
which implies
\[
U_{i\bar{j}}=0.
\]
Then we have
\[
\abs{\tau}^{2}=g^{i\bar{j}}U_{i\bar{j}}=0.
\]
Taking $t=\frac{1}{2}$ in~\eqref{11158} yields
\[
0=3\int_{X}\abs{\tau}^{2}=\int_{X}\abs{T}^{2},
\]
which implies $T=0$, i.e., $\omega$ is K\"{a}hler.
\end{proof}

\section*{Acknowledgments}

The author would like to thank Professor Jixiang Fu. Thanks also go to the referees.

\section*{Declaration of interests}

The author does not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and has declared no affiliations other than their research organizations.

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