# On Hermitian manifolds with Bismut-Strominger parallel torsion

Quanting Zhao  Quanting Zhao. School of Mathematics and Statistics, and Hubei
Key Laboratory of Mathematical Sciences, Central China Normal University, P.O.
Box 71010, Wuhan 430079, P. R. China.
zhaoquanting<EMAIL_ADDRESS>and  Fangyang Zheng
Fangyang Zheng. School of Mathematical Sciences, Chongqing Normal University,
Chongqing 401331, China<EMAIL_ADDRESS>
###### Abstract.

In this article, we study Hermitian manifolds whose Bismut-Strominger
connection has parallel torsion tensor, which will be called Bismut torsion
parallel manifolds, or BTP manifolds for short. We obtain a necessary and
sufficient condition characterizing this class in terms of the curvature
tensor. In particular, Bismut flat or Bismut Kähler-like manifolds are BTP
manifolds, known by our earlier results. We also obtain structural results for
non-balanced BTP manifolds, and classification theorems for non-balanced BTP
threefolds and balanced BTP threefolds.

###### Key words and phrases:

Bismut torsion parallel, Kähler-like, Bismut connection, admissible frame,
degenerate torsion, Chern flat, Fano manifold, Vaisman manifold

###### 2010 Mathematics Subject Classification:

53C55 (Primary), 53C05 (Secondary)

Zhao is partially supported by NSFC with the grant No.11801205 and 12171180.
Zheng is partially supported by NSFC grant 12071050 and 12141101, a Chongqing
grant cstc2021ycjh-bgzxm0139, and the 111 Project D21024.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Torsion and curvature of Bismut connection
  4. 4 Manifolds with Bismut parallel torsion
  5. 5 Non-balanced BTP manifolds
  6. 6 Balanced BTP threefolds
  7. 7 The Fano case
  8. 8 The Wallach threefold
  9. 9 Balanced BTP threefolds of middle type
    1. 9.1 The structural theorem
    2. 9.2 Examples of Lie-Hermitian threefolds
    3. 9.3 Generalization to higher dimensions

##  1\. Introduction

Given a smooth manifold equipped with a connection, the parallelness of
torsion often forms strong geometric restrictions. For instance, by the result
of Kamber and Tondeur in [18], if a manifold admits a complete connection that
is flat and having parallel torsion, then its universal cover must be a Lie
group (equipped with a flat left invariant connection). The converse is
certainly true. Also, by the classic result of Ambrose and Singer [1], if a
complete Riemannian manifold admits a metric connection with parallel torsion
and curvature, then the universal cover is a homogeneous Riemannian manifold
(and vice versa). The complex version of this is also true, proved by Sekigawa
in [25], namely, if a complete Hermitian manifold admits a Hermitian
connection (that is, a connection compatible with both the metric and the
almost complex structure) which has parallel torsion and curvature, then the
universal cover is a homogeneous Hermitian manifold (and vice versa).

On a given Hermitian manifold, Bismut connection (also known as Strominger
connection, see [7, 26]) is the unique Hermitian connection with totally skew-
symmetric torsion. It serves as a bridge between Levi-Civita connection, which
is the unique metric connection that is torsion free, and Chern connection,
the unique metric connection compatible with the complex structure. When the
metric is Kähler, these three canonical connections coincide, but when the
metric is not Kähler, they are mutually distinct, leaving us with three
different kinds of geometry.

In this article, we will focus on Bismut connection, and we want to know when
Bismut connection will have parallel torsion. For convenience, we will call
the class of Hermitian manifolds whose Bismut connection has parallel torsion
simply as Bismut torsion parallel manifolds, or BTP manifolds in short.

Let $(M^{n},g)$ be a Hermitian manifold. Denote by $\nabla^{b}$ the Bismut
connection and by $T^{b}$, $R^{b}$ its torsion and curvature tensor,
respectively. Let us introduce the notation

$Q_{X\overline{Y}Z\overline{W}}=R^{b}_{X\overline{Y}Z\overline{W}}-R^{b}_{Z\overline{Y}X\overline{W}},\
\ \
\mathrm{Ric}(Q)_{X\overline{Y}}=\sum_{i=1}^{n}\big{(}R^{b}_{X\overline{Y}e_{i}\overline{e_{i}}}-R^{b}_{e_{i}\overline{Y}X\overline{e_{i}}}\big{)}$

where $X$, $Y$, $Z$, $W$ are type $(1,0)$ tangent vectors on $M$ and
$\\{e_{1},\ldots,e_{n}\\}$ is any local unitary frame. Denote by $\omega$ the
Kähler form of $g$. Recall that Gauduchon’s torsion $1$-form is the global
$(1,0)$-form $\eta$ on $M$ defined by
$\partial(\omega^{n-1})=-2\eta\wedge\omega^{n-1}$. Let $\chi$ be the type
$(1,0)$ vector field on $M$ dual to $\eta$, namely, $\langle
Y,\overline{\chi}\rangle=\eta(Y)$ for any $Y$ where $g=\langle,\rangle$ is the
(Riemannian) metric, extended bi-linearly over ${\mathbb{C}}$.

With these notations specified, we can now state the first theorem of this
article, which says that the BTP condition $\nabla^{b}T^{b}=0$ is equivalent
to some conditions involving only $R^{b}$:

######  Theorem 1.1.

Let $(M^{n},g)$ be a Hermitian manifold. The BTP condition $\nabla^{b}T^{b}=0$
is equivalent to the combination of the following

(1.1) $\displaystyle R^{b}_{XYZ\overline{W}}=0$ (1.2) $\displaystyle
R^{b}_{X\overline{Y}Z\overline{W}}=R^{b}_{Z\overline{W}X\overline{Y}}$ (1.3)
$\displaystyle\nabla^{b}\mathrm{Ric}(Q)=0$ (1.4)
$\displaystyle\mathrm{Ric}(Q)_{\chi\overline{Y}}=0$

for any type $(1,0)$ tangent vectors $X$, $Y$, $Z$, $W$.

Recall that the Bismut Kähler-like (BKL) condition, also known as Strominger
Kähler-like (SKL) condition (cf. [3, 12, 13, 33, 34]), for a Hermitian
manifold is defined as

$R^{b}_{XYZ\overline{W}}=0,\ \ \ \
R^{b}_{X\overline{Y}Z\overline{W}}=R^{b}_{Z\overline{Y}X\overline{W}}$

for any type $(1,0)$ tangent vectors $X$, $Y$, $Z$, $W$. The main result of
[34] says that

$\mbox{{\em BKL}}\ \Longleftrightarrow\ \mbox{{\em BTP}}+\mbox{pluriclosed}.$

The main technical part of the proof in [34] is to show that BKL implies BTP.
Note that by our Theorem 1.1 above, this implication is immediate, since the
BKL condition simply means (1.1) and $Q=0$. In this case, all four conditions
in Theorem 1.1 are obviously satisfied, hence BTP. So Theorem 1.1 can be
regarded as a generalization of the main result of [34].

We note that in Theorem 1.1, the conditions (1.3) and (1.4) are automatically
satisfied if $\mathrm{Ric}(Q)=0$. Also, under the BTP assumption, one always
has $\nabla^{b}Q=0$. But there are examples of non-Kähler BKL manifolds with
$\nabla^{b}R^{b}\neq 0$. In other words, for BTP manifolds, the Bismut
connection $\nabla^{b}$ is not always an Ambrose-Singer connection.

Recall that a Hermitian manifold is said to be Gauduchon if
$\partial\overline{\partial}(\omega^{n-1})=0$, and is said to be balanced if
its Gauduchon’s torsion $1$-form $\eta$ vanishes, or equivalently, if
$d(\omega^{n-1})=0$. Also, the metric is said to be strongly Gauduchon in the
sense of Popovici [22] if there exists a global $(n,n-2)$-form $\Omega$ such
that $\partial\omega^{n-1}=\overline{\partial}\Omega$. Clearly,

$\mbox{balanced}\ \ \Longrightarrow\ \ \mbox{strongly\ Gauduchon}\ \
\Longrightarrow\ \ \mbox{Gauduchon}.$

As a consequence to Theorem 1.1, we have

######  Corollary 1.2.

Let $(M^{n},g)$ be a Hermitian manifold satisfying $\nabla^{b}T^{b}=0$. Then
$g$ is Gauduchon, i.e., $\partial\overline{\partial}(\omega^{n-1})=0$. If in
addition $M^{n}$ is compact and $g$ is not balanced, then $g$ is not strongly
Gauduchon and the Dolbeault group $H^{0,1}_{\overline{\partial}}(M)\neq 0$. In
particular, any compact non-balanced BTP manifold is not a
$\partial\overline{\partial}$-manifold and $M$ does not admit any Kähler
metric.

It has been shown in [34, Theorem 7 and Proposition 4] that compact non-Kähler
BKL manifolds, or equivalently non-balanced BKL manifolds, does not admit any
strongly Gauduchon metric and the Dolbeault group
$H^{0,1}_{\overline{\partial}}(M)$ is non-trivial. We speculate that the same
should be true for compact non-balanced BTP manifolds as well:

######  Question 1.3.

Let $(M^{n},g)$ be a compact non-balanced Hermitian manifold satisfying
$\nabla^{b}T^{b}=0$. Then $M$ does not admit any strongly Gauduchon metric.

By [34, Theorem 2], the following equivalence holds for $n=2$:

$\mbox{{\em BKL}}\quad\Longleftrightarrow\ \quad\mbox{{\em
BTP}}\quad\Longleftrightarrow\quad\mathrm{Vaisman}$

where Vaisman means a locally conformally Kähler manifold whose Lee form
(which is $-2(\eta+\overline{\eta})$) is parallel under the Levi-Civita
connection. Compact Vaisman surfaces are fully classified by Belgun [4] and a
structure theorem for Vaisman manifolds is proved by Ornea and Verbitsky [21].

In dimension $n\geq 3$, locally conformally Kähler manifolds can no longer be
(non-Kähler) BKL, shown in [33], but still can be BTP. For instance, it is
easy to check that for any $n$ the classic Hopf manifold

$M^{n}=({\mathbb{C}}^{n}\setminus\\{0\\})/\langle f\rangle,\ \ \ f(z)=\lambda
z,\ \ \ |\lambda|>1,\ \ \ \
\omega=\sqrt{-1}\frac{\partial\overline{\partial}|z|^{2}}{|z|^{2}},$

is BTP. More generally, all Vaisman manifolds are BTP. In fact, a locally
conformally Kähler manifold is BTP if and only if it is Vaisman, a result due
to Andrada and Villacampa [2, Theorem 3.6]. Here we give a slightly more
general statement:

######  Proposition 1.4.

A locally conformally Kähler manifold $(M^{n},g)$ is BTP if and only if it is
Vaisman. Furthermore, if two conformal Hermitian manifolds $(M^{n},g)$ and
$(M^{n},e^{2u}g)$ are both BTP, with $du\neq 0$ in an open dense subset of
$M$, then both manifolds are locally conformally Kähler (thus Vaisman).

In other words, for each conformal class of Hermitian metrics, there is at
most one BTP metric (up to constant multiples) unless the metric is locally
conformally Kähler. When the manifold is compact, this is clearly true since
BTP metrics are Gauduchon.

Next we consider BTP manifolds amongst the class of complex nilmanifolds with
nilpotent $J$. Complex nilmanifolds form an important class of Hermitian
manifolds, and are often used to test and illustrate theory in non-Kähler
geometry. See [11], [15], [20], [23], [27], [30], [19] as a sample of
examples. Following the discussions of [35], we have

######  Proposition 1.5.

Let $(M^{n},g)$ be a complex nilmanifold, namely, a compact Hermitian manifold
whose universal cover is a nilpotent Lie group $G$ equipped with a left-
invariant complex structure $J$ and a compatible left-invariant metric $g$.
Assume that $J$ is nilpotent in the sense of [10]. Then $g$ is BTP if and only
if there exists a unitary left-invariant coframe $\varphi$ on $G$ and an
integer $1\leq r\leq n$ such that

$\left\\{\begin{array}[]{ll}d\varphi_{i}=0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \forall\ 1\leq i\leq r;\\\ d\varphi_{\alpha}=\sum_{i=1}^{r}Y_{\alpha
i}\,\varphi_{i}\wedge\overline{\varphi}_{i},\ \ \forall\ r<\alpha\leq
n.\end{array}\right.$

Note that the metric $g$ will be balanced when and only when
$\sum_{i=1}^{r}Y_{\alpha i}=0$ for each $\alpha$.

In particular, $G$ (when not abelian) is a 2-step nilpotent group and $J$ is
abelian. Let us take the $n=3$ case as an example. In this case, either $r=3$,
where $G$ is abelian and $g$ is Kähler and flat, or $r=2$ and there exists a
unitary coframe $\varphi$ such that

$d\varphi_{1}=d\varphi_{2}=0,\ \
d\varphi_{3}=a\varphi_{1}\wedge\overline{\varphi}_{1}+b\varphi_{2}\wedge\overline{\varphi}_{2},$

where $a$, $b$ are constants. By a unitary change if necessary, we may assume
that $a>0$. Under such a coframe, the metric $g$ is balanced if and only if
$b=-a$, while $g$ is BKL if and only if $b\in\sqrt{-1}{\mathbb{R}}$ is pure
imaginary. For any other choice of $b$ values, $g$ is a non-BKL, non-balanced
BTP metric. We will denote the following example by $N^{3}$,

(1.5) $d\varphi_{1}=d\varphi_{2}=0,\ \ \ \ \
d\varphi_{3}=\varphi_{1}\wedge\overline{\varphi}_{1}-\varphi_{2}\wedge\overline{\varphi}_{2},$

which will appear in the study of Lie-Hermitian threefolds that are balanced
BTP, in §9.2.

The examples above show that when $n\geq 3$, there are more BTP manifolds than
BKL manifolds, and there are more non-balanced BTP manifolds than balanced
ones. For complex nilmanifolds with $J$ not nilpotent, when $n=3$ one can show
by a lengthy computation that they cannot be BTP, but for $n\geq 4$, we do not
know if they can be BTP or not. More generally, it would be a very interesting
question on how to classify all BTP Lie-Hermitian manifolds (see for example
[30] for a discussion on Lie-Hermitian manifolds).

For $n\geq 3$, one can naturally divide the discussion of (non-Kähler) BTP
manifolds into two cases: non-balanced ones and balanced ones. All (non-
Kähler) BKL manifolds and all (non-Kähler) Vaisman manifolds belong to the
first subset, where the $\nabla^{b}$-parallel vector field $\chi$ (dual to
$\eta$) will play an important role. In §5 we will discuss some general
properties and structural results of such spaces following the ideas of [33].

######  Proposition 1.6.

Let $(M^{n},g)$ be a non-balanced BTP manifold with $n\geq 3$. Then the
following holds:

  1. $(1)$

Locally there always exist admissible frames, and the holonomy group of the
Bismut connection is contained in $U(n-1)$.

  2. $(2)$

The metric $g$ satisfies the LP condition if and only if it has degenerate
torsion. In particular, for $n=3$, non-balanced BTP threefolds coincide with
GCE threefolds of Belgun.

For the special case when $(M^{n},g)$ is Vaisman, the conclusion (1) is due to
Andrada and Villacampa [2]. Recall that a local unitary frame $e$ on a non-
balanced BTP manifold is said to be admissible, if $\chi=\lambda e_{n}$ and
$\phi^{j}_{i}=\lambda a_{i}\delta_{ij}$ for any $i$, $j$, where $\lambda>0$
and $a_{i}$ are globally defined constants, as in Definition 5.1. Here the
tensor $\phi$ is defined by
$\phi_{i}^{j}=\sum_{k}\overline{\eta}_{k}T_{ik}^{j}$, where $\eta_{k}$ and
$T_{ik}^{j}$ are components of $\eta$ and the Chern torsion $T$ under local
unitary frame. The metric $g$ is said to have degenerate torsion if under any
admissible frame $e$ it holds $T^{\ast}_{ik}=0$ for any $i<k<n$, as given in
Definition 5.4. The Lee potential (LP) condition of Belgun [6] means that a
Hermitian manifold satisfying the condition

$\partial\eta=0,\quad\partial\omega=c\,\eta\wedge\partial\overline{\eta},$

where $c\neq 0$ is a constant, and a Hermitian manifold is generalized Calabi-
Eckmann (GCE ) if it is both BTP and LP.

The main result of §5 is the following classification theorem of non-balanced
BTP threefolds.

######  Theorem 1.7.

Let $(M^{3},g)$ be a non-balanced BTP threefold. If we denote by
$\\{\lambda_{1},\lambda_{2},0\\}$ the three eigenvalues of the tensor $\phi$,
it yields that

  1. $(1)$

when $\lambda_{1}\overline{\lambda_{2}}+\overline{\lambda_{1}}\lambda_{2}\neq
0$ and
$\lambda_{1}\overline{\lambda_{2}}-\overline{\lambda_{1}}\lambda_{2}\neq 0$,
$(M^{3},g)$ is locally twisted Sasakian product $N_{1}\times_{\kappa}N_{2}$,

  2. $(2)$

when $\lambda_{1}\overline{\lambda_{2}}+\overline{\lambda_{1}}\lambda_{2}=0$,
$(M^{3},g)$ is BKL,

  3. $(3)$

when $\lambda_{1}\overline{\lambda_{2}}+\overline{\lambda_{1}}\lambda_{2}\neq
0$ and
$\lambda_{1}\overline{\lambda_{2}}-\overline{\lambda_{1}}\lambda_{2}=0$,
$(M^{3},g)$ is generalized Vaisman. In particular, $(M^{3},g)$ is Vaisman when
$\lambda_{1}=\lambda_{2}$.

The twisted Sasakian product here is actually the modified Hermitian structure
on the Sasakian product, introduced by Belgun [6], which we will see in
Definition 5.7 and 5.8. A Hermitian manifold $(M^{n},g)$ is said to be
generalized Vaisman if it is BTP and the Lee form, defined as
$-2(\eta+\overline{\eta})$ in this paper, is $d$-closed. It follows from
Proposition 1.4 that a Vaisman manifold is necessarily generalized Vaisman.

The balanced BTP manifolds, on the other hand, form a rather restrictive but
non-empty subset. In the following, we will discuss the classification problem
for such manifolds in dimension 3, starting from a technical lemma obtained in
[36] for special frames on balanced threefolds. Recall that the $B$ tensor of
a Hermitian manifold $(M^{n},g)$ is defined by
$B_{i\overline{j}}=\sum_{r,s}T^{j}_{rs}\overline{T^{i}_{rs}}$ under any local
unitary frame, where $T^{j}_{ik}$ are components of the Chern torsion under
the frame.

######  Proposition 1.8.

Let $(M^{3},g)$ be a non-Kähler balanced BTP threefold. Then, for any $p\in
M$, there exists a local unitary frame near $p$ such that

$B=\begin{bmatrix}c&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}\\!\\!,\
\begin{bmatrix}c&0&0\\\ 0&c&0\\\ 0&0&0\end{bmatrix}\ \text{or}\
\begin{bmatrix}c&0&0\\\ 0&c&0\\\ 0&0&c\end{bmatrix}\\!\\!,$

for some constant $c>0$.

Such a local unitary frame will be called _a special frame_ in §6. We will
carry out the classification of non-Kähler balanced BTP threefolds according
to the rank of the tensor $B$.

######  Definition 1.9.

A balanced BTP threefold $(M^{3},g)$ is said to be of middle type, if
$\mathrm{rank}\,B=2$, or equivalently, for any $p\in M$, under a special frame
near $p$

$B=\begin{bmatrix}c&0&0\\\ 0&c&0\\\ 0&0&0\end{bmatrix}$

for some constant $c>0$.

We would like to illustrate two non-middle type examples. First let us
consider the complex Lie group $SO(3,{\mathbb{C}})$ consisting of all $3\times
3$ complex matrices $A$ such that $\det(A)=1$ and $\,{}^{t}\\!AA=I$. Let
$\varphi$ be a left-invariant coframe satisfying

(1.6) $d\varphi_{1}=\varphi_{2}\wedge\varphi_{3},\ \ \
d\varphi_{2}=\varphi_{3}\wedge\varphi_{1},\ \ \
d\varphi_{3}=\varphi_{1}\wedge\varphi_{2}.$

Let $g$ be the Hermitian metric with $\varphi$ being a unitary coframe. Then
it is easy to check that $g$ is balanced BTP and its $B$ tensor takes the form

$B=\frac{1}{2}\\!\\!\begin{bmatrix}1&0&0\\\ 0&1&0\\\
0&0&1\end{bmatrix}\\!\\!.$

So any compact quotient of $SO(3,{\mathbb{C}})$ is a non-Kähler, balanced BTP
threefold of non-middle type. Note that this example has trivial canonical
line bundle and is Chern flat. We remark that for this Hermitian threefold,
the holonomy group of $\nabla^{b}$ is reduced to $SO(3)\subseteq U(3)$, but
not contained in $U(2)\times 1$ like in the non-balanced BTP case.

There is another example of balanced BTP threefold which is not of the middle
type: the Wallach threefold $(X,g)$. As a complex manifold, $X$ is the flag
threefold $X={\mathbb{P}}(T_{{\mathbb{P}}^{2}})$. Let $\tilde{\omega}$ be the
Kähler-Einstein metric on $X$ with
$\mbox{Ric}(\tilde{\omega})=2\tilde{\omega}$. Then the Kähler form of $g$ is
given by $\omega=\tilde{\omega}-\sigma$, where $\sigma$ is the $(1,1)$-form on
$X$ given by a holomorphic section $\xi\in
H^{0}(X,\Omega_{X}\\!\otimes\\!L)\cong{\mathbb{C}}$ with
$\parallel\\!\xi\\!\parallel^{2}_{\tilde{\omega}}=\frac{1}{2}$. Here
$\Omega_{X}$ is the holomorphic cotangent bundle, and $L$ is the ample line
bundle such that $L^{\otimes 2}$ is the anti-canonical line bundle of $X$.

We will see in §8 that $(X,g)$ is non-Kähler, balanced BTP. We will also show
that its Levi-Civita (Riemannian) connection has constant Ricci curvature $6$
and has non-negative sectional curvature. Its $B$ tensor takes the form

$B=\frac{1}{2}\\!\\!\begin{bmatrix}1&0&0\\\ 0&0&0\\\
0&0&0\end{bmatrix}\\!\\!,$

so $(X,g)$ is not of middle type. This example is Fano, namely, its anti-
canonical line bundle is ample. The Chern connection of $g$ has nonnegative
bisectional curvature and positive holomorphic sectional curvature.

As observed by Wallach in [28], all the homogeneous metrics with positive
sectional curvature on $X$ are Hermitian. This set of metrics (after scaling)
forms a peculiar plane region (see Figure 1 in [9]), and our metric $g$ lies
in the boundary of this region.

We have the following classification theorem of balanced BTP threefolds, which
is the second main theorem of this article. The proofs of the three cases
below will be given in §6, §7 and §9.1.

######  Theorem 1.10.

Let $(M^{3},g)$ be a compact non-Kähler, balanced BTP threefold. After scaling
the metric by a constant multiple, it holds that,

  1. $(1)$

when $\mbox{rank}\,B=1$, $M$ is holomorphically isometric to the Wallach
threefold.

  2. $(2)$

when $\mbox{rank}\,B=2$, namely the middle type, then there is an unbranched
cover $\pi:\hat{M}\rightarrow M$ of degree at most $2$ on which there is a bi-
Hermitian structure $(\hat{M},g,J,I)$, where $(\hat{M},g,J)$ is balanced BTP
while $(\hat{M},g,I)$ is Vaisman, with $IJ=JI$ and the Bismut connections of
the two Hermitian structures $(\hat{M},g,J)$ and $(\hat{M},g,I)$ coincide. In
particular, $\hat{M}$ is a smooth fiber bundle over $S^{1}$ with fiber being a
Sasakian 5-manifold. Furthermore, a compact non-Kähler, balanced BTP threefold
of middle type admits no pluriclosed metric.

  3. $(3)$

when $\mbox{rank}\,B=3$, $M$ is holomorphically isometric to a compact
quotient of $SO(3,{\mathbb{C}})$ with global unitary coframe satisfying (1.6).

In §9.2, we will study balanced BTP threefolds of middle type in the category
of Lie-Hermitian manifolds and prove the following theorem

######  Theorem 1.11.

Let $(M^{3},g)$ be a non-Kähler, balanced BTP Lie-Hermitian threefold of
middle type, then after a scaling of the metric by a constant multiple, it is
one of the members in $A_{a,b}$, $B_{u,v,w}^{\pm}$, $C_{u,v}$ or
$D_{u,\rho}^{\pm}$.

Here the constant parameters $a,b\in{\mathbb{R}}$, $u,v,w,\rho\in{\mathbb{C}}$
with $|\rho|=1$, and for $B^{\varepsilon}_{u,v,w}$ the parameters also
satisfies $w-\overline{w}=\varepsilon i(|u-v|^{2}-|u+v|^{2})$, where
$\varepsilon=\pm 1$. See §9.2 for the explicit descriptions of these Lie-
Hermitian threefolds.

Finally, we remark that there are examples of complete non-compact Hermitian
surfaces with zero Chern curvature but non-parallel Chern torsion, so the
curvature characterization of torsion parallelness for Bismut connection does
not hold for Chern connection or other Hermitian connections in general. Of
course it would be a much harder problem if one adds the compactness
assumption on the manifolds, then things get more delicate and we plan to
investigate it in future projects.

The paper is organized as follows. In §2, we will set up notations and also
collect some known results that will be used later. In §3 we will discuss the
properties of BTP manifolds, and in §4, we will prove Theorem 1.1 stated
above. In §5, we will discuss the properties of non-balanced BTP manifolds,
following ideas of [33] where non-Kähler BKL manifolds were investigated, and
will prove Proposition 1.6. In §6, we will discuss the types of the $B$ tensor
of a balanced BTP threefold and prove Proposition 1.8, where we will also show
that the $\mathrm{rank}\,B=3$ case of Theorem 1.10 leads to the compact
quotients of $SO(3,{\mathbb{C}})$. In §7, we will prove that the
$\mathrm{rank}\,B=1$ case of Theorem 1.10 leads to a Fano threefold of even
index, so as a complex manifold it is either ${\mathbb{P}}^{3}$ or a del Pezzo
threefold. All cases will be ruled out by topological or algebro-geometric
argument, except the flag threefold case: the Wallach threefold $(X,g)$. In
§8, we will carry out the detailed computation which shows that the Wallach
threefold is indeed balanced BTP, with nonnegative bisectional curvature and
positive holomorphic sectional curvature. We will also show that its Levi-
Civita connection has non-negative sectional curvature and constant Ricci
curvature $6$. In §9, we will prove the case of $\mbox{rank}\,B=2$ of Theorem
1.10 and investigate balanced BTP Lie-Hermitian threefolds of middle type, as
well as discussing some generalization in higher dimensions.

##  2\. Preliminaries

Let $(M^{n},g)$ be a Hermitian manifold of complex dimension $n$. We will set
up notations following [31, 32, 34, 37, 38]. Let $g=\langle,\rangle$ be the
metric, extended bi-linearly over ${\mathbb{C}}$. The bundle of complex
tangent vector fields of type $(1,0)$, namely, complex vector fields of the
form $v-\sqrt{-1}Jv$, where $v$ is a real vector field on $M$, is denoted by
$T^{1,0}M$. Let $\\{e_{1},\ldots,e_{n}\\}$ be a local frame of $T^{1,0}M$ in a
neighborhood in $M$. Write $e=\ ^{t}\\!(e_{1},\ldots,e_{n})$ as a column
vector. Denote by $\varphi=\ ^{t}\\!(\varphi_{1},\ldots,\varphi_{n})$ the
column vector of local $(1,0)$-forms which is the coframe dual to $e$.

Denote by $\nabla$, $\nabla^{c}$, $\nabla^{b}$ the Levi-Civita, Chern, and
Bismut connection, respectively. Denote by $T^{c}=T$, $R^{c}$ the torsion and
curvature of $\nabla^{c}$, and by $T^{b}$, $R^{b}$ the torsion and curvature
of $\nabla^{b}$. Under the frame $e$, denote the components of $T^{c}$ as

(2.1) $T^{c}(e_{i},\overline{e}_{j})=0,\ \ \
T^{c}(e_{i},e_{j})=2\sum_{k=1}^{n}T^{k}_{ij}e_{k}.$

For Chern connection $\nabla^{c}$, let us denote by $\theta$, $\Theta$ the
matrices of connection and curvature, respectively, and by $\tau$ the column
vector of the torsion $2$-forms, all under the local frame $e$. Then the
structure equations and Bianchi identities are

(2.2) $\displaystyle d\varphi$ $\displaystyle=$ $\displaystyle-\
^{t}\\!\theta\wedge\varphi+\tau,$ (2.3) $\displaystyle d\theta$
$\displaystyle=$ $\displaystyle\theta\wedge\theta+\Theta.$ (2.4)
$\displaystyle d\tau$ $\displaystyle=$ $\displaystyle-\
^{t}\\!\theta\wedge\tau+\ ^{t}\\!\Theta\wedge\varphi,$ (2.5) $\displaystyle
d\Theta$ $\displaystyle=$
$\displaystyle\theta\wedge\Theta-\Theta\wedge\theta.$

The entries of $\Theta$ are all $(1,1)$ forms, while the entries of the column
vector $\tau$ are all $(2,0)$ forms, under any frame $e$. Similar symbols such
as $\theta^{b},\Theta^{b}$ and $\tau^{b}$ are applied to Bismut connection.
The components of $\tau$ are just $T_{ij}^{k}$ defined in (2.1):

$\tau_{k}=\sum_{i,j=1}^{n}T_{ij}^{k}\varphi_{i}\wedge\varphi_{j}\ =\sum_{1\leq
i<j\leq n}2\ T_{ij}^{k}\varphi_{i}\wedge\varphi_{j},$

where $T_{ij}^{k}=-T_{ji}^{k}$. Under the frame $e$, express the Levi-Civita
(Riemannian) connection $\nabla$ as

$\nabla e=\theta_{1}e+\overline{\theta_{2}}\overline{e},\ \ \
\nabla\overline{e}=\theta_{2}e+\overline{\theta_{1}}\overline{e},$

thus the matrices of connection and curvature for $\nabla$ become:

$\hat{\theta}=\left[\begin{array}[]{ll}\theta_{1}&\overline{\theta_{2}}\\\
\theta_{2}&\overline{\theta_{1}}\end{array}\right],\ \ \
\hat{\Theta}=\left[\begin{array}[]{ll}\Theta_{1}&\overline{\Theta}_{2}\\\
\Theta_{2}&\overline{\Theta}_{1}\end{array}\right],$

where

$\displaystyle\Theta_{1}$ $\displaystyle=$ $\displaystyle
d\theta_{1}-\theta_{1}\wedge\theta_{1}-\overline{\theta_{2}}\wedge\theta_{2},$
$\displaystyle\Theta_{2}$ $\displaystyle=$ $\displaystyle
d\theta_{2}-\theta_{2}\wedge\theta_{1}-\overline{\theta_{1}}\wedge\theta_{2},$
$\displaystyle d\varphi$ $\displaystyle=$ $\displaystyle-\
^{t}\\!\theta_{1}\wedge\varphi-\ ^{t}\\!\theta_{2}\wedge\overline{\varphi}.$

When $e$ is unitary, both $\theta_{2}$ and $\Theta_{2}$ are skew-symmetric,
while $\theta$, $\theta_{1}$, $\theta^{b}$, or $\Theta$, $\Theta_{1}$,
$\Theta^{b}$ are all skew-Hermitian. Consider the $(2,1)$ tensor
$\gamma=\frac{1}{2}(\nabla^{b}-\nabla^{c})$ introduced in [31]. Its
representation under the frame $e$ is the matrix of $1$-forms, which by abuse
of notation we will also denote by $\gamma$, is given by

$\gamma=\theta_{1}-\theta.$

Denote the decomposition of $\gamma$ into $(1,0)$ and $(0,1)$ parts by
$\gamma=\gamma^{\prime}+\gamma^{\prime\prime}$. As in [29, Lemma 2], it yields
that

$\theta^{b}=\theta+2\gamma=\theta_{1}+\gamma.$

As observed in [31], when $e$ is unitary, $\gamma$ and $\theta_{2}$ take the
following simple forms

(2.6) $(\theta_{2})_{ij}=\sum_{k=1}^{n}\overline{T^{k}_{ij}}\varphi_{k},\ \ \
\
\gamma_{ij}=\sum_{k=1}^{n}(T_{ik}^{j}\varphi_{k}-\overline{T^{i}_{jk}}\overline{\varphi}_{k}),$

while for general frames the above formula will have the matrix
$(g_{i\overline{j}})=(\langle e_{i},\overline{e}_{j}\rangle)$ and its inverse
involved, which is less convenient. This is why we often choose unitary frames
to work with. By (2.6), we get the expression of the components of the torsion
$T^{b}$ under any unitary frame $e$:

(2.7) $T^{b}(e_{i},e_{j})=-2\sum_{k=1}^{n}T^{k}_{ij}e_{k},\ \ \
T^{b}(e_{i},\overline{e}_{j})=2\sum_{k=1}^{n}\big{(}T^{j}_{ik}\overline{e}_{k}-\overline{T^{i}_{jk}}e_{k}\big{)}.$

Note that in the second equation above we assumed that the frame $e$ is
unitary, otherwise the matrix $g$ needs to appear in the formula. From this,
we deduce that

(2.8) $\nabla^{b}T^{b}=0\ \Longleftrightarrow\nabla^{b}T^{c}=0.\ $

This is observed in [34, the proof of Theorem 3]. More generally, if we denote
by $\nabla^{t}=(1-t)\nabla^{c}+t\nabla^{b}$ the $t$-Gauduchon connection,
where $t$ is any real constant, then its torsion tensor is
$T^{t}(x,y)=T^{c}(x,y)+2t(\gamma_{x}y-\gamma_{y}x)$, whose components under
$e$ are again given by $T^{k}_{ij}$, so it is easy to see that for any fixed
$t\in{\mathbb{R}}$

(2.9) $\nabla^{t}T^{t_{1}}=0\ \Longleftrightarrow\nabla^{t}T^{t_{2}}=0\ \ \ \
\forall\ \ t_{1},t_{2}\in\mathbb{R}.\ $

So for a fixed $t$-Gauduchon connection $\nabla^{t}$, if the torsion of one
Gauduchon connection is parallel with respect to $\nabla^{t}$, then the
torsion of any other Gauduchon connection will be parallel under $\nabla^{t}$
as well.

Next let us discuss the curvature. As usual, the curvature tensor $R^{D}$ of a
linear connection $D$ on a Riemannian manifold $M$ is defined by

$R^{D}(x,y,z,w)=\langle R^{D}_{xy}z,\ w\rangle=\langle
D_{x}D_{y}z-D_{y}D_{x}z-D_{[x,y]}z,\ w\rangle,$

where $x,y,z,w$ are tangent vectors in $M$. We will also write it as
$R^{D}_{xyzw}$ for brevity. It is always skew-symmetric with respect to the
first two positions, and is also skew-symmetric with respect to its last two
positions if the connection is metric, namely, if $Dg=0$. The first and second
Bianchi identities are respectively

$\displaystyle{\mathfrak{S}}\\{R^{D}_{xy}z-(D_{x}T^{D})(y,z)-T^{D}(T^{D}(x,y),z)\\}=0$
$\displaystyle{\mathfrak{S}}\\{(D_{x}R^{D})_{yz}+R^{D}_{T^{D}(x,y)\,z}\\}=0$

where ${\mathfrak{S}}$ means the sum over all cyclic permutation of $x,y,z$.
When the manifold $M$ is complex and the connection $D$ is Hermitian, namely,
satisfies $Dg=0$ and $DJ=0$, where $J$ is the almost complex structure, then
$R^{D}$ satisfies

$R^{D}(x,y,Jz,Jw)=R^{D}(x,y,z,w)$

for any tangent vectors $x,y,z,w$, or equivalently,

(2.10) $R^{D}(x,y,Z,W)=R^{D}(x,y,\overline{Z},\overline{W})=0$

for any type $(1,0)$ tangent vectors $Z$, $W$. From now on, we will use $X$,
$Y$, $Z$, $W$ to denote type $(1,0)$ vectors. For any Hermitian connection
$D$, with the skew-symmetry of first two or last two positions in mind, the
above equation (2.10) says that the only possibly non-zero components of
$R^{D}$ are $R^{D}_{XYZ\overline{W}}$, $R^{D}_{X\overline{Y}Z\overline{W}}$
and their conjugations.

Let us now apply the Bianchi identities to the Chern connection $\nabla$, and
use the fact that $T^{c}(e_{i},\overline{e}_{j})=0$ and
$R^{c}_{e_{i}e_{j}}x=0$ for any $x$ where $e$ is any local frame of type
$(1,0)$ vectors, we get the following

(2.11) $\displaystyle
T^{\ell}_{ij;k}+T^{\ell}_{jk;i}+T^{\ell}_{ki;j}\,=\,2\sum_{r=1}^{n}\big{(}T^{r}_{ij}T^{\ell}_{kr}+T^{r}_{jk}T^{\ell}_{ir}+T^{r}_{ki}T^{\ell}_{jr}\big{)}$
(2.12) $\displaystyle
R^{c}_{k\overline{j}i\overline{\ell}}-R^{c}_{k\overline{j}i\overline{\ell}}\,=\,2T^{\ell}_{ik;\overline{j}}$
(2.13) $\displaystyle
R^{c}_{i\overline{j}k\overline{\ell};\,m}-R^{c}_{m\overline{j}k\overline{\ell};\,i}\,=\,2\sum_{r=1}^{n}T^{r}_{im}R^{c}_{r\overline{j}k\overline{\ell}}$

for any indices $1\leq i,j,k,\ell,m\leq n$. Here the indices after semicolon
stand for covariant derivatives with respect to the Chern connection
$\nabla^{c}$. Note that in formula (2.12) we assumed that the frame $e$ is
unitary, otherwise the right hand side needs to be multiplied on the right by
$g$, the matrix of the metric.

##  3\. Torsion and curvature of Bismut connection

Since we will primarily be interested in Bismut connection $\nabla^{b}$, we
would like to have formula involving the curvature and covariant
differentiation of $\nabla^{b}$. Using the Bianchi identities for general
metric connections, we could get formula for $\nabla^{b}$ which are similar to
the ones for Chern connection mentioned in the previous section. But since we
also need relations between $R^{c}$ and $R^{b}$ later, let us deduce them from
the structure equations. Under a $(1,0)$-frame $e$, the components of the
Chern, Bismut and Riemannian curvature tensors are given by

$R^{c}_{i\overline{j}k\overline{\ell}}=\sum_{p=1}^{n}\Theta_{kp}(e_{i},\overline{e}_{j})g_{p\overline{\ell}},\
\ \ \ \
R^{b}_{abk\overline{\ell}}=\sum_{p=1}^{n}\Theta^{b}_{kp}(e_{a},e_{b})g_{p\overline{\ell}},\
\ \ \ \ R_{abcd}=\sum_{f=1}^{2n}\hat{\Theta}_{cf}(e_{a},e_{b})g_{fd},$

where $i,j,k,\ell,p$ range from $1$ to $n$, while $a,b,c,d,f$ range from $1$
to $2n$ with $e_{n+i}=\overline{e}_{i}$, and $g_{ab}=g(e_{a},e_{b})$. It
follows that, for any Hermitian connection $D$,
$R^{D}_{abij}=R^{D}_{ab\overline{i}\overline{j}}=0$ by the discussion above.

For convenience, we will assume from now on that our local $(1,0)$-frame $e$
is unitary for the Hermitian metric $g$. We have

######  Lemma 3.1.

Let $(M^{n},g)$ be a Hermitian manifold. It follows that

(3.1) $\displaystyle R^{b}_{ijk\bar{\ell}}\ =\
2\big{(}T^{\ell}_{kj,i}-T^{\ell}_{ki,j}\big{)}+4\sum_{r}\big{(}T^{r}_{ij}T^{\ell}_{rk}+T^{r}_{jk}T^{\ell}_{ri}+T^{r}_{ki}T^{\ell}_{rj}\big{)},$
(3.2) $\displaystyle R^{b}_{i\bar{j}k\bar{\ell}}-R^{c}_{i\bar{j}k\bar{\ell}}\
=\
2\big{(}T^{\ell}_{ik,\bar{j}}+\overline{T^{k}_{j\ell,\bar{i}}}\big{)}-4\sum_{r}\big{(}T^{\ell}_{kr}\overline{T^{i}_{jr}}+T^{j}_{ir}\overline{T^{k}_{\ell
r}}+T^{r}_{ik}\overline{T^{r}_{j\ell}}-T^{\ell}_{ir}\overline{T^{k}_{jr}}\big{)},$
(3.3) $\displaystyle T^{k}_{ij,\ell}+T^{k}_{j\ell,i}+T^{k}_{\ell i,j}\ =\
4\sum_{r}\big{(}T^{r}_{\ell
i}T^{k}_{rj}+T^{r}_{j\ell}T^{k}_{ri}+T^{r}_{ij}T^{k}_{r\ell}\big{)},$ (3.4)
$\displaystyle
T^{j}_{ik,\bar{\ell}}+\overline{T^{i}_{j\ell,\bar{k}}}-\overline{T^{k}_{j\ell,\bar{i}}}\
\,=\
\,-\,2\sum_{r}\big{(}T^{r}_{ik}\overline{T^{r}_{j\ell}}+T^{j}_{ir}\overline{T^{k}_{\ell
r}}-T^{j}_{kr}\overline{T^{i}_{\ell
r}}-T^{\ell}_{ir}\overline{T^{k}_{jr}}+T^{\ell}_{kr}\overline{T^{i}_{jr}}\big{)}+$
$\displaystyle\hskip 113.81102pt+\
\frac{1}{2}\,\big{(}R^{s}_{i\bar{\ell}k\bar{j}}-R^{s}_{k\bar{\ell}i\bar{j}}\big{)},$
(3.5) $\displaystyle
R^{b}_{ijk\bar{\ell},p}+R^{b}_{pik\bar{\ell},j}+R^{b}_{jpk\bar{\ell},i}\ =\
-\,2\sum_{r}\big{(}R^{b}_{irk\bar{\ell}}T_{jp}^{r}+R^{b}_{jrk\bar{\ell}}T_{pi}^{r}+R^{b}_{prk\bar{\ell}}T_{ij}^{r}\big{)},$
(3.6) $\displaystyle
R^{b}_{ipk\bar{\ell},\bar{q}}-R^{b}_{i\bar{q}k\bar{\ell},p}+R^{b}_{p\bar{q}k\bar{\ell},i}\
=\
2\sum_{r}\\{R^{b}_{r\bar{q}k\bar{\ell}}T^{r}_{ip}-R^{b}_{p\bar{r}k\bar{\ell}}T^{q}_{ir}+R^{b}_{i\bar{r}k\bar{\ell}}T^{q}_{pr}+$
$\displaystyle\hskip 136.5733pt+\
R^{b}_{prk\bar{\ell}}\overline{T^{i}_{qr}}-R^{b}_{irk\bar{\ell}}\overline{T^{p}_{qr}}\\},$

for any $i,j,k,\ell,p,q$, where indices after comma mean covariant derivatives
with respect to $\nabla^{b}$.

###### Proof.

Fix a point $p\in M$. We will choose our local unitary frame $e$ near $p$ such
that $\theta^{b}(p)=0$, thus $\theta^{c}=-2\gamma$ at $p$. From the structure
equation $\,\Theta^{b}=d\theta^{b}-\theta^{b}\wedge\theta^{b}$, it yields that

$\Theta^{b}=\Theta+2(d\gamma+2\gamma\wedge\gamma)$

at $p$. By comparison for the types of the differential forms we get

$\displaystyle\Theta^{b}_{2,0}$ $\displaystyle=$ $\displaystyle
2(\partial\gamma^{\prime}+2\gamma^{\prime}\wedge\gamma^{\prime}),$
$\displaystyle\Theta^{b}_{1,1}$ $\displaystyle=$
$\displaystyle\Theta+2(\overline{\partial}\gamma^{\prime}-\partial\overline{{}^{t}\\!\gamma^{\prime}}-2\gamma^{\prime}\wedge\overline{{}^{t}\\!\gamma^{\prime}}-2\overline{{}^{t}\\!\gamma^{\prime}}\wedge\gamma^{\prime}),$

hence (3.1) and (3.2) are established. From the first Bianchi identity
$\,d\tau=-^{t}\\!\theta\wedge\tau+^{t}\\!\\!\Theta\wedge\varphi$, it yields at
$p$ that

$\displaystyle\partial\tau$ $\displaystyle=$ $\displaystyle
2^{t}\\!\gamma^{\prime}\wedge\tau,$ $\displaystyle\overline{\partial}\tau$
$\displaystyle=$
$\displaystyle-2\overline{\gamma^{\prime}}\wedge\tau+^{t}\\!\\!\Theta\wedge\varphi,$
$\displaystyle=$
$\displaystyle-2\overline{\gamma^{\prime}}\wedge\tau+^{t}\\!\\!\Theta^{b}_{1,1}\wedge\varphi-2^{t}\\!(\overline{\partial}\gamma^{\prime}-\partial\overline{{}^{t}\\!\gamma^{\prime}}-2\gamma^{\prime}\wedge\overline{{}^{t}\\!\gamma^{\prime}}-2\overline{{}^{t}\\!\gamma^{\prime}}\wedge\gamma^{\prime})\wedge\varphi,$

which imply (3.3) and (3.4). By the second Bianchi identity,
$d\Theta^{b}=\theta^{b}\wedge\Theta^{b}-\Theta^{b}\wedge\theta^{b}$, we have
$d\Theta^{b}=0$ at $p$ since $\theta^{b}=0$ at $p$. This leads to

$\partial\Theta^{b}_{2,0}=0,\quad\overline{\partial}\Theta^{b}_{2,0}+\partial\Theta^{b}_{1,1}=0$

which yields (3.5) and (3.6). ∎

######  Lemma 3.2.

Let $(M,g)$ be a Hermitian manifold. It follows that

(3.7) $\displaystyle T^{\ell}_{ij,k}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}(R^{b}_{jki\bar{\ell}}+R^{b}_{kij\bar{\ell}}),$
(3.8)
$\displaystyle\sum_{r}(T^{r}_{ij}T^{\ell}_{rk}+T^{r}_{jk}T^{\ell}_{ri}+T^{r}_{ki}T^{\ell}_{rj})$
$\displaystyle=$
$\displaystyle-\frac{1}{4}(R^{b}_{ijk\bar{\ell}}+R^{b}_{jki\bar{\ell}}+R^{b}_{kij\bar{\ell}}),$
$\displaystyle T^{j}_{ik,\bar{\ell}}$ $\displaystyle=$
$\displaystyle-\frac{2}{3}\sum_{r}(T^{r}_{ik}\overline{T^{r}_{j\ell}}+T^{j}_{ir}\overline{T^{k}_{\ell
r}}-T^{j}_{kr}\overline{T^{i}_{\ell
r}}-T^{\ell}_{ir}\overline{T^{k}_{jr}}+T^{\ell}_{kr}\overline{T^{i}_{jr}})$
$\displaystyle+\frac{1}{3}(R^{b}_{i\bar{\ell}k\bar{j}}-R^{b}_{k\bar{\ell}i\bar{j}})+\frac{1}{6}(R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}),$
(3.10) $\displaystyle\eta_{i,j}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\sum_{r}(R^{b}_{ijr\bar{r}}+R^{b}_{jri\bar{r}}),$
$\displaystyle\eta_{i,\bar{j}}$ $\displaystyle=$
$\displaystyle-\frac{2}{3}(\sum_{r}\eta_{r}\overline{T^{i}_{jr}}+\overline{\eta}_{r}T^{j}_{ir}-\sum_{t,r}T^{j}_{tr}\overline{T^{i}_{tr}})$
$\displaystyle+\frac{1}{3}\sum_{r}(R^{b}_{r\bar{j}i\bar{r}}-R^{b}_{i\bar{j}r\bar{r}})+\frac{1}{6}\sum_{r}(R^{b}_{r\bar{r}i\bar{j}}-R^{b}_{i\bar{r}r\bar{j}}).$

for any $i,j,k,\ell$, where indices after comma mean the covariant derivative
with respect to $\nabla^{b}$.

###### Proof.

Equations (3.1) and (3.3) imply (3.7) and (3.8), and equation (3.4) implies
(3.2). Also, since $\eta_{k}=\sum_{i}T^{i}_{ik}$, (3.10) and (3.2) are implied
by (3.7) and (3.2), respectively. ∎

######  Definition 3.3 (See also (27) and (34) in [34]).

Let us introduce the following notations

$P_{ik}^{j\ell}:=\sum_{r}T^{r}_{ik}\overline{T^{r}_{j\ell}}+T^{j}_{ir}\overline{T^{k}_{\ell
r}}-T^{j}_{kr}\overline{T^{i}_{\ell
r}}-T^{\ell}_{ir}\overline{T^{k}_{jr}}+T^{\ell}_{kr}\overline{T^{i}_{jr}}.\ $
$A_{k\overline{\ell}}:=\sum_{r,s}T^{r}_{sk}\overline{T^{r}_{s\ell}},\quad\quad\
\
B_{k\overline{\ell}}:=\sum_{r,s}T^{\ell}_{rs}\overline{T^{k}_{rs}},\quad\quad\quad\quad\quad\
$
$C_{ik}:=\sum_{r,s}T^{r}_{si}T^{s}_{rk},\quad\quad\quad\phi^{\ell}_{k}:=\sum_{r}\overline{\eta}_{r}T^{\ell}_{kr},\quad\quad\quad\quad\quad\quad$
$Q_{i\bar{j}k\bar{\ell}}:=R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}},\quad\mathrm{Ric}(Q)_{i\bar{j}}:=\sum_{r}R^{b}_{i\bar{j}r\bar{r}}-R^{b}_{r\bar{j}i\bar{r}}$

under any unitary frame. It is clear that $C$ is symmetric, $A$, $B$ are
Hermitian symmetric, and $P_{ik}^{j\ell}$ satisfies

$P^{j\ell}_{\,ik}=-P^{j\ell}_{\,ki}=-P^{\ell
j}_{\,ik}=\overline{P^{ik}_{j\ell}}.$

######  Proposition 3.4.

Let $(M^{n},g)$ be a Hermitian manifold, denote by $\omega$ the Kähler form of
$g$. Then

$\sqrt{-1}\partial\overline{\partial}\omega=\left\\{\frac{1}{2}(T^{\ell}_{ik,\overline{j}}-T^{j}_{ik,\overline{\ell}})-P^{j\ell}_{ik}\right\\}\varphi_{i}\wedge\varphi_{k}\wedge\overline{\varphi}_{j}\wedge\overline{\varphi}_{\ell}.$

Hence we have the following equivalence

$\displaystyle\partial\overline{\partial}\omega=0$
$\displaystyle\Longleftrightarrow\quad
T^{j}_{ik,\overline{\ell}}-T^{\ell}_{ik,\overline{j}}=-2P^{j\ell}_{ik}$
$\displaystyle\Longleftrightarrow\quad(R^{b}_{i\overline{\ell}k\overline{j}}-R^{b}_{k\overline{\ell}i\overline{j}})-(R^{b}_{i\overline{j}k\ell}-R^{b}_{k\overline{j}i\overline{\ell}})=-4P^{j\ell}_{ik}.$

###### Proof.

By [34, Lemma 1], we have

$\displaystyle\sqrt{-1}\partial\overline{\partial}\omega$
$\displaystyle=^{t}\\!\tau\overline{\tau}+\,^{t}\\!\varphi\Theta\overline{\varphi}$
$\displaystyle=\sum_{i,j,k,\ell,p}\left(T^{p}_{ik}\overline{T^{p}_{j\ell}}-\frac{1}{4}(R^{c}_{i\bar{j}k\bar{\ell}}-R^{c}_{k\bar{j}i\bar{\ell}}-R^{c}_{i\bar{\ell}k\bar{j}}+R^{c}_{k\bar{\ell}i\bar{j}})\right)\varphi_{i}\wedge\varphi_{k}\wedge\overline{\varphi}_{j}\wedge\overline{\varphi}_{\ell}.$

Note that, by [31, Lemma 7], for any $i,j,k,\ell$,

(3.12)
$2T^{j}_{ik;\overline{\ell}}=R^{c}_{k\overline{\ell}i\overline{j}}-R^{c}_{i\overline{\ell}k\overline{j}},$

where the index after the semicolon stands for covariant derivative with
respect to the Chern connection $\nabla^{c}$, and it is easy to obtain, since
$\nabla^{c}-\nabla^{b}=-2\gamma$,

(3.13) $\displaystyle T^{j}_{ik;\overline{\ell}}$ $\displaystyle=$
$\displaystyle\overline{e}_{\ell}T^{j}_{ik}-\sum_{q}\left(T^{j}_{qk}\,g(\nabla^{c}_{\overline{e}_{\ell}}e_{i},\overline{e}_{q})+T^{j}_{iq}\,g(\nabla^{c}_{\overline{e}_{\ell}}e_{k},\overline{e}_{q})+T^{q}_{ik}\,g(\nabla^{c}_{\overline{e}_{\ell}}\overline{e}_{j},e_{q})\right)$
$\displaystyle=$ $\displaystyle
T^{j}_{ik,\overline{\ell}}+2\sum_{q}\left(T^{j}_{qk}\gamma_{iq}(\overline{e}_{\ell})+T^{j}_{iq}\gamma_{kq}(\overline{e}_{\ell})-T^{q}_{ik}\gamma_{qj}(\overline{e}_{\ell})\right)$
$\displaystyle=$ $\displaystyle
T^{j}_{ik,\overline{\ell}}-2\sum_{q}\left(T^{j}_{qk}\overline{T^{i}_{q\ell}}+T^{j}_{iq}\overline{T^{k}_{q\ell}}-T^{q}_{ik}\overline{T^{q}_{j\ell}}\right),$

where the index after the comma means covariant derivative with respect to
$\nabla^{b}$. Then the equalities (3.12) and (3.13) imply

$\sqrt{-1}\partial\overline{\partial}\omega=\left\\{\frac{1}{2}(T^{\ell}_{ik,\overline{j}}-T^{j}_{ik,\overline{\ell}})-P^{j\ell}_{ik}\right\\}\varphi_{i}\wedge\varphi_{k}\wedge\overline{\varphi}_{j}\wedge\overline{\varphi}_{\ell}.$

Therefore the equivalence in the proposition follows from the equality (3.2).
∎

######  Proposition 3.5.

Let $(M^{n},g)$ be a Hermitian manifold, with $\omega$ its Kähler form. Then
it holds that

(3.14) $\displaystyle\nabla^{b}_{1,0}T^{c}=0$
$\displaystyle\Longleftrightarrow$ $\displaystyle
R^{b}_{ijk\bar{\ell}}=0\quad\Longrightarrow\quad\sum_{r}(T^{r}_{\ell
i}T^{k}_{rj}+T^{r}_{j\ell}T^{k}_{ri}+T^{r}_{ij}T^{k}_{r\ell})=0,$ (3.15)
$\displaystyle\nabla^{b}_{0,1}T^{c}=0$ $\displaystyle\Longleftrightarrow$
$\displaystyle
R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}=-4P_{ik}^{j\ell},$

where $\nabla^{b}_{1,0}T^{c}$ and $\nabla^{b}_{0,1}T^{c}$ are respectively the
$(1,0)$\- and $(0,1)$-components of $\nabla^{b}T^{c}$.

###### Proof.

It is clear that the condition $\nabla^{b}_{1,0}T^{c}=0$ (that is,
$T_{ij,k}^{\ell}=0$) implies $\sum_{r}(T^{r}_{\ell
i}T^{k}_{rj}+T^{r}_{j\ell}T^{k}_{ri}+T^{r}_{ij}T^{k}_{r\ell})=0$ by (3.3).
Thus $R^{b}_{ijk\bar{\ell}}=0$ follows from (3.1). Conversely, by (3.7) and
(3.8), $R^{b}_{ijk\bar{\ell}}=0$ implies that $T_{ij,k}^{\ell}=0$ and
$\sum_{r}(T^{r}_{\ell
i}T^{k}_{rj}+T^{r}_{j\ell}T^{k}_{ri}+T^{r}_{ij}T^{k}_{r\ell})=0$. Therefore,
the conclusion (3.14) is established.

In the mean time, by (3.4), we know that the condition
$\nabla^{b}_{0,1}T^{c}=0$ (that is, $T_{ik,\bar{\ell}}^{j}=0$) implies that

$R^{b}_{i\bar{\ell}k\bar{j}}-R^{b}_{k\bar{\ell}i\bar{j}}=4P_{ik}^{j\ell},$

which yields

$R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}=4P_{ik}^{\ell
j}=-4P_{ik}^{j\ell}.$

Conversely, if
$R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}=-4P_{ik}^{j\ell}$,
then by the fact that $P_{ik}^{j\ell}=-P_{ik}^{\ell j}$ we get

$R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}=-(R^{b}_{i\bar{\ell}k\bar{j}}-R^{b}_{k\bar{\ell}i\bar{j}}),$

so by (3.2) we have

$T_{ik,\bar{\ell}}^{j}=-\frac{2}{3}P_{ik}^{j\ell}-\frac{1}{6}(R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}})=-\frac{2}{3}P-\frac{1}{6}(-4P)=0.$

This establishes the equivalence of (3.15). ∎

######  Proposition 3.6.

Let $(M^{n},g)$ be a Hermitian manifold with $\nabla^{b}T^{c}=0$. Then it
holds that

  1. $(1)$

$R^{b}_{ijk\bar{\ell}}=0$,

  2. $(2)$

$R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$,

  3. $(3)$

$\nabla^{b}Q=0$,

  4. $(4)$

$R^{b}_{xy\chi w}=0$, where $x,y,w$ are any tangent vector and $\chi$ is the
associated vector field of Gauduchon’s torsion $1$-form $\eta$,

for any $i,j,k,\ell$. Also, under the assumption $\nabla^{b}T^{c}=0$, we have

(3.16)
$\displaystyle\partial\overline{\partial}\omega=0\quad\Longleftrightarrow\quad
P_{ik}^{j\ell}=0\quad\Longleftrightarrow\quad
R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{j}i\bar{\ell}},$ (3.17)
$\displaystyle\partial\eta=0,\quad\overline{\partial}\eta=-2\sum_{i,j,k}\eta_{k}\overline{T^{i}_{jk}}\varphi_{i}\wedge\overline{\varphi}_{j},$

In particular, the metric $g$ is Gauduchon, namely,
$\partial\overline{\partial}\omega^{n-1}=0$.

###### Proof.

Let $(M^{n},g)$ be a Hermitian manifold satisfying the condition
$\nabla^{b}T^{c}=0$, which means $\nabla^{b}_{1,0}T^{c}=0$ and
$\nabla^{b}_{0,1}T^{c}=0$. By (3.14) and (3.15), we get (1) and (3)
immediately, as $P_{ik}^{j\ell}$ is $\nabla^{b}$-parallel. Since Gauduchon’s
torsion 1-form
$\eta=\sum_{k}\eta_{k}\varphi_{k}=\sum_{i,k}T^{i}_{ik}\varphi_{k}$ is the
trace of the Chern torsion $T^{c}$, its associated vector field
$\chi=\sum_{k}\overline{\eta_{k}}e_{k}$ is clearly $\nabla^{b}$-parallel and
thus

$R^{b}_{xy\chi
w}=g\left((\nabla^{b}_{x}\nabla^{b}_{y}-\nabla^{b}_{y}\nabla^{b}_{x}-\nabla^{b}_{[x,y]})\chi,w\right)=0,$

which yields (4). As to (2), from
$R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}=-4P_{ik}^{j\ell}$, it
yields that

$\displaystyle R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{\ell}i\bar{j}}$
$\displaystyle=R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}+R^{b}_{k\bar{j}i\bar{\ell}}-R^{b}_{k\bar{\ell}i\bar{j}}$
$\displaystyle=-4P_{ik}^{j\ell}+(\overline{R^{s}_{j\bar{k}\ell\bar{i}}-R^{b}_{\ell\bar{k}j\bar{i}}})$
$\displaystyle=-4P_{ik}^{j\ell}-\overline{4P_{j\ell}^{ki}}$
$\displaystyle=-4P_{ik}^{j\ell}-4P_{ki}^{j\ell}$ $\displaystyle=0.$

Under the assumption $\nabla^{b}T^{c}=0$, the equivalence (3.16) follows from
Proposition 3.4 and (3.15). For any given point $p\in M$, let $e$ be a local
unitary frame near $p$ such that $\theta^{b}|_{p}=0$. The structure equation
gives us
$d\varphi=-\,^{t}\\!\theta\wedge\varphi+\tau=2\,^{t}\\!\gamma\wedge\varphi+\tau$,
hence $\partial\varphi=-\tau$ and
$\overline{\partial}\varphi=2\overline{\gamma^{\prime}}\wedge\varphi$ at $p$.
Therefore

$\displaystyle\partial\eta$
$\displaystyle=\partial(\sum_{i}\eta_{i}\varphi_{i})=-\sum_{i,j}\eta_{i,j}\varphi_{i}\wedge\varphi_{j}-\sum_{i,j,k}\eta_{k}T^{k}_{ij}\varphi_{i}\wedge\varphi_{j},$
$\displaystyle\overline{\partial}\eta$
$\displaystyle=\overline{\partial}(\sum_{i}\eta_{i}\varphi_{i})=-\sum_{i,j}\eta_{i,\bar{j}}\varphi_{i}\wedge\overline{\varphi}_{j}-2\sum_{i,j,k}\eta_{k}\overline{T^{i}_{jk}}\varphi_{i}\wedge\overline{\varphi}_{j}.$

Here the indices after comma mean the covariant derivative with respect to
$\nabla^{b}$. It is clear that $\nabla^{b}\eta=0$ and thus
$\eta_{i,j}=0,\eta_{i,\bar{j}}=0$. Also, (3.14) implies that

$\sum_{r}(T^{r}_{\ell
i}T^{k}_{rj}+T^{r}_{j\ell}T^{k}_{ri}+T^{r}_{ij}T^{k}_{r\ell})=0.$

Let $k=\ell$ in the equation above and sum up, which yields that
$\sum_{r}\eta_{r}T^{r}_{ij}=0$. Hence,

$\partial\eta=0,\quad\overline{\partial}\eta=-2\sum_{i,j,k}\eta_{k}\overline{T^{i}_{jk}}\varphi_{i}\wedge\overline{\varphi}_{j}.$

From the definition of $\eta$, it follows that
$\partial\omega^{n-1}=-2\eta\wedge\omega^{n-1}$ and thus

$-\sqrt{-1}\partial\overline{\partial}\omega^{n-1}=-2\sqrt{-1}(\overline{\partial}\eta+2\eta\wedge\overline{\eta})\wedge\omega^{n-1}=2(\sum_{i}\eta_{i,\bar{i}})\frac{\omega^{n}}{n}=0.$

This completes the proof of the proposition. ∎

##  4\. Manifolds with Bismut parallel torsion

In this section, we will prove Theorem 1.1, which characterizes Hermitian
manifolds with Bismut parallel torsion in terms of the behavior of the Bismut
curvature $R^{b}$.

First we will give the following two technical lemmata, the latter has been
already shown in the proof of [34, Theorem 2], but we include the proof here
for readers’ convenience.

######  Lemma 4.1.

Let $(M^{n},g)$ be a Hermitian manifold. Then the following are equivalent

  1. $(1)$

$R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$,

  2. $(2)$

$Q_{i\bar{j}k\bar{\ell}}=-Q_{i\bar{\ell}k\bar{j}}=\overline{Q_{j\bar{i}\ell\bar{k}}}$,

  3. $(3)$

$T^{j}_{ik,\overline{\ell}}=-T^{\ell}_{ik,\overline{j}}=\overline{T^{i}_{j\ell,\overline{k}}}$,

for any $i,j,k,\ell$.

###### Proof.

First let us assume that
$R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$ holds. It follows
that

$\displaystyle Q_{i\bar{j}k\bar{\ell}}$
$\displaystyle=R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}-R^{b}_{i\bar{\ell}k\bar{j}}=-Q_{i\bar{\ell}k\bar{j}},$
$\displaystyle\overline{Q_{i\bar{j}k\bar{\ell}}}$
$\displaystyle=R^{b}_{j\bar{i}\ell\bar{k}}-R^{b}_{j\bar{k}\ell\bar{i}}=R^{b}_{j\bar{i}\ell\bar{k}}-R^{b}_{\ell\bar{i}j\bar{k}}=Q_{j\bar{i}\ell\bar{k}}.$

By equation (3.2), we get

$T^{j}_{ik,\bar{\ell}}=-\frac{2}{3}P_{ik}^{j\ell}-\frac{1}{6}Q_{i\bar{j}k\bar{\ell}},$

which implies that

$T^{j}_{ik,\overline{\ell}}=-T^{\ell}_{ik,\overline{j}}=\overline{T^{i}_{j\ell,\overline{k}}}.$

Conversely, assuming that the equality
$T^{j}_{ik,\overline{\ell}}=-T^{\ell}_{ik,\overline{j}}=\overline{T^{i}_{j\ell,\overline{k}}}$
holds. Then by (3.4) we get

$Q_{i\bar{\ell}k\bar{j}}=R^{b}_{i\bar{\ell}k\bar{j}}-R^{b}_{k\bar{\ell}i\bar{j}}=6T_{ik,\bar{\ell}}^{j}+4P^{j\ell}_{ik},$

which implies that $Q_{i\bar{j}k\bar{\ell}}=-Q_{i\bar{\ell}k\bar{j}}$ and
$\overline{Q_{i\bar{j}k\bar{\ell}}}=Q_{j\bar{i}\ell\bar{k}}$. Also,

$\displaystyle R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{\ell}i\bar{j}}$
$\displaystyle=R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}+R^{b}_{k\bar{j}i\bar{\ell}}-R^{b}_{k\bar{\ell}i\bar{j}}$
$\displaystyle=Q_{i\bar{j}k\bar{\ell}}+\overline{R^{b}_{j\bar{k}\ell\bar{i}}-R^{b}_{\ell\bar{k}j\bar{i}}}$
$\displaystyle=Q_{i\bar{j}k\bar{\ell}}+\overline{Q_{j\bar{k}\ell\bar{i}}}$
$\displaystyle=0.$

This completes the proof of the lemma. ∎

######  Lemma 4.2.

Let $(M^{n},g)$ be a Hermitian manifold. If $R^{b}_{ijk\bar{\ell}}=0$ for any
$i,j,k,\ell$, then the following commutation formulae hold

  1. $(1)$

$\eta_{i,\overline{j}\,\overline{k}}-\eta_{i,\overline{k}\,\overline{j}}=2\sum_{r}\overline{T^{r}_{kj}}\,\eta_{i,\overline{r}}$,

  2. $(2)$

$T^{i}_{j\ell,\overline{k}\,\overline{p}}-T^{i}_{j\ell,\overline{p}\,\overline{k}}=2\sum_{r}\overline{T^{r}_{pk}}T^{i}_{j\ell,\overline{r}}$,

for any $i,j,k,\ell,p$. Again indices after comma stand for covariant
derivatives with respect to $\nabla^{b}$.

###### Proof.

First let us prove the second equality. From the definition of covariant
derivatives,

$T_{j\ell,\bar{k}}^{i}=\overline{e}_{k}T_{j\ell}^{i}+\sum_{r}T^{i}_{\ell
r}\,\langle\nabla^{b}_{\overline{e}_{k}}e_{j},e_{r}\rangle-T^{i}_{jr}\,\langle\nabla^{b}_{\overline{e}_{k}}e_{\ell},e_{r}\rangle+T^{r}_{j\ell}\,\langle\nabla^{b}_{\overline{e}_{k}}e_{r},e_{i}\rangle.$

Fix any given point in $M$, choose a local unitary frame $e$ so that the
matrix $\theta^{b}=0$ at the point, then we have

$T_{j\ell,\bar{k}\,\bar{p}}^{i}=\overline{e}_{p}\overline{e}_{k}T_{j\ell}^{i}+\sum_{r}T^{i}_{\ell
r}\,\langle\nabla^{b}_{\overline{e}_{p}}\nabla^{b}_{\overline{e}_{k}}e_{j},e_{r}\rangle-T^{i}_{jr}\,\langle\nabla^{b}_{\overline{e}_{p}}\nabla^{b}_{\overline{e}_{k}}e_{\ell},e_{r}\rangle+T^{r}_{j\ell}\,\langle\nabla^{b}_{\overline{e}_{p}}\nabla^{b}_{\overline{e}_{k}}e_{r},e_{i}\rangle,$

which implies that

$T^{i}_{j\ell,\overline{k}\,\overline{p}}-T^{i}_{j\ell,\overline{p}\,\overline{k}}=[\overline{e}_{p},\overline{e}_{k}]T_{j\ell}^{i}+\sum_{r}T^{i}_{\ell
r}\,R^{b}_{\bar{p}\bar{k}j\bar{r}}-T^{i}_{jr}\,R^{b}_{\bar{p}\bar{k}\ell\bar{r}}+T^{r}_{j\ell}\,R^{b}_{\bar{p}\bar{k}r\bar{i}}.$

At that point, we have

$[\overline{e}_{p},\overline{e}_{k}]=\nabla_{\overline{e}_{p}}\overline{e}_{k}-\nabla_{\overline{e}_{k}}\overline{e}_{p}=-2\overline{T^{r}_{kp}}\overline{e}_{r},$

where $\nabla$ is the Riemannian connection. So the second equality in the
lemma is established since $R^{b}_{ijk\bar{\ell}}=0$. By letting $i=j$ in the
second equality and sum over, we get the first equality. ∎

Now we are ready to prove the main result of this article, Theorem 1.1.

###### Proof of Theorem 1.1:.

By Proposition 3.6, it is clear that $\nabla^{b}T^{c}=0$ implies the four
conditions in Theorem 1.1, so we just need to prove the converse. We will
follow the strategy of the proof of [34, Theorem 2]. Note that the first
condition is equivalent to $\nabla^{b}_{1,0}T^{c}=0$, by (3.14), which implies

$\sum_{r}(T^{r}_{\ell
i}T^{k}_{rj}+T^{r}_{j\ell}T^{k}_{ri}+T^{r}_{ij}T^{k}_{r\ell})=0,$

and the fourth condition means

(4.1)
$\sum_{i,r}\overline{\eta_{i}}(R^{b}_{i\bar{j}r\bar{r}}-R^{b}_{r\bar{j}i\bar{r}})=0,$

for any $i,j,\ell$. In the mean time, $\mathrm{Ric}(Q)$ is Hermitian symmetric
due to $R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$, as

$\mathrm{Ric}(Q)_{i\bar{j}}=\sum_{r}R^{b}_{i\bar{j}r\bar{r}}-R^{b}_{r\bar{j}i\bar{r}}=\overline{\sum_{r}R^{b}_{j\bar{i}r\bar{r}}-R^{b}_{j\bar{r}r\bar{i}}}=\overline{\sum_{r}R^{b}_{j\bar{i}r\bar{r}}-R^{b}_{r\bar{i}j\bar{r}}}=\overline{\mathrm{Ric}(Q)_{j\bar{i}}}.$

Firstly, we will show that $|\eta|^{2}$ is a constant. We have $\eta_{i,j}=0$
since $R^{b}_{ijk\overline{\ell}}=0$. By (3.2) and the symmetry assumption
$R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$, we get

$|\eta|^{2}_{,\bar{j}}=\sum_{i}\eta_{i,\bar{j}}\overline{\eta_{i}}=-\frac{2}{3}(\sum_{i,r}\eta_{r}\overline{T^{i}_{jr}}+\overline{\eta}_{r}T^{j}_{ir}-\sum_{i,t,r}T^{j}_{tr}\overline{T^{i}_{tr}})\overline{\eta_{i}}-\frac{1}{6}\sum_{i,r}(R^{b}_{i\bar{j}r\bar{r}}-R^{b}_{r\bar{j}i\bar{r}})\overline{\eta_{i}}=0$

for any $j$. Here we used (4.1) and the fact $\sum_{r}\eta_{r}T_{ij}^{r}=0$,
which follows from the identity $\sum_{r}(T^{r}_{\ell
i}T^{k}_{rj}+T^{r}_{j\ell}T^{k}_{ri}+T^{r}_{ij}T^{k}_{r\ell})=0$. Similarly
$|\eta|^{2}_{,j}=0$ also holds, thus $|\eta|^{2}$ is a constant.

Secondly, we will show that $\nabla^{b}\eta=0$, which is equivalent to
$\nabla^{b}_{0,1}\eta=0$. As $|\eta|^{2}$ is a constant, it yields, after the
covariant derivative in $\overline{\ell}$ is taken, that

$\sum_{k}\eta_{k,\bar{\ell}}\,\overline{\eta_{k}}=0.$

Take the covariant derivative in $\ell$ and sum up $\ell$, it follows that

(4.2)
$\sum_{k,\ell}\big{(}|\eta_{k,\bar{\ell}}|^{2}+\eta_{k,\bar{\ell}\,\ell}\,\overline{\eta_{k}}\big{)}=0.$

The equality (3.2) and
$R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$ imply that

(4.3)
$\eta_{i,\bar{j}}=-\frac{2}{3}(\overline{\phi_{j}^{i}}+\phi_{i}^{j}-B_{i\bar{j}})-\frac{1}{6}\mathrm{Ric}(Q)_{i\bar{j}},$

thus $\eta_{i,\bar{j}}$ is Hermitian symmetric in $i,j$ since
$\mathrm{Ric}(Q)_{i\bar{j}}$ is so. It yields

$\displaystyle\sum_{k,\ell}|\eta_{k,\bar{\ell}}|^{2}$
$\displaystyle=\frac{4}{9}|\phi+\phi^{*}-B|^{2}+\frac{1}{36}|\mathrm{Ric}(Q)|^{2}+\frac{2}{9}\mathrm{Re}\left(\phi\mathrm{Ric}(Q)+\overline{\phi\mathrm{Ric}(Q)}-B\mathrm{Ric}(Q)\right),$
$\displaystyle=\frac{4}{9}\left(|\phi+\phi^{*}|^{2}+|B|^{2}-4\mathrm{Re}(\phi
B)\right)+\frac{1}{36}|\mathrm{Ric}(Q)|^{2}+\frac{4}{9}\mathrm{Re}\left(\phi\mathrm{Ric}(Q)\right)-\frac{2}{9}\mathrm{Re}\left(B\mathrm{Ric}(Q)\right),$

where

$|\phi+\phi^{*}-B|^{2}=\sum_{k,\ell}(\phi_{k}^{\ell}+\overline{\phi_{\ell}^{k}}-B_{k\bar{\ell}})\overline{(\phi_{k}^{\ell}+\overline{\phi_{\ell}^{k}}-B_{k\bar{\ell}})},\quad|\mathrm{Ric}(Q)|^{2}=\sum_{k,\ell}\mathrm{Ric}(Q)_{k\bar{\ell}}\overline{\mathrm{Ric}(Q)_{k\bar{\ell}}},$
$\phi\mathrm{Ric}(Q)=\sum_{k,\ell}\phi^{k}_{\ell}\mathrm{Ric}(Q)_{k\bar{\ell}},\quad
B\mathrm{Ric}(Q)=\sum_{k,\ell}B_{k\bar{\ell}}\mathrm{Ric}(Q)_{\ell\bar{k}},\quad\phi
B=\sum_{k,\ell}\phi^{k}_{\ell}\mathrm{Ric}(Q)_{k\bar{\ell}}.$

Note that

$|\phi+\phi^{*}|^{2}=\sum_{k,\ell}(\phi^{\ell}_{k}+\overline{\phi^{k}_{\ell}})(\overline{\phi^{\ell}_{k}+\overline{\phi^{k}_{\ell}}})=2|\phi|^{2}+2\mathrm{Re}(\phi\cdot\phi),$

where

$\phi\cdot\phi=\sum_{k,\ell}\phi_{k}^{\ell}\phi_{\ell}^{k}.$

Then it follows from Lemma 4.2, the $\nabla^{b}$-parallelness of
$\mathrm{Ric}(Q)$, (4.3) and Lemma 4.1 that

$\displaystyle\sum_{\ell}\eta_{k,\bar{\ell}\,\ell}$
$\displaystyle=\sum_{\ell}\left(\overline{\eta_{\ell,\bar{k}}}\right)_{,\ell}=\sum_{\ell}\overline{\eta_{\ell,\bar{k}\,\bar{\ell}}}=\sum_{\ell}\big{(}\overline{\eta_{\ell,\bar{\ell}\,\bar{k}}+2\sum_{r}\overline{T^{r}_{\ell
k}}\eta_{\ell,\bar{r}}}\big{)}$
$\displaystyle=\left(\sum_{\ell}\eta_{\ell,\bar{\ell}}\right)_{,k}+2\sum_{\ell,r}T^{r}_{\ell
k}\eta_{r,\bar{\ell}}$
$\displaystyle=\left(\frac{2}{3}(|T|^{2}-2|\eta|^{2})-\frac{1}{6}\sum_{r}\mathrm{Ric}(Q)_{r\bar{r}}\right)_{\\!,k}+2\sum_{\ell,r}T^{r}_{\ell
k}\eta_{r,\bar{\ell}}$
$\displaystyle=\frac{2}{3}\sum_{i,j,r}T^{i}_{jr}\overline{T^{i}_{jr,\bar{k}}}-\frac{4}{3}\sum_{\ell,r}T^{r}_{\ell
k}\left(\phi_{r}^{\ell}+\overline{\phi_{\ell}^{r}}-B_{r\bar{\ell}}+\frac{1}{4}\mathrm{Ric}(Q)_{r\bar{\ell}}\right)$
$\displaystyle=\frac{2}{3}\sum_{i,j,r}T^{i}_{jr}T^{j}_{ik,\bar{r}}-\frac{4}{3}\sum_{\ell,r}T^{r}_{\ell
k}\left(\phi_{r}^{\ell}+\overline{\phi_{\ell}^{r}}-B_{r\bar{\ell}}+\frac{1}{4}\mathrm{Ric}(Q)_{r\bar{\ell}}\right),$

Hence, it yields that, from (3.2),
$R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$, and
$\sum_{r}\eta_{r}T_{ij}^{r}=0$,

$\displaystyle\sum_{k,\ell}\eta_{k,\bar{\ell}\,\ell}\overline{\eta_{k}}$
$\displaystyle=\frac{2}{3}\sum_{i,j,k,r}\overline{\eta_{k}}T^{i}_{jr}T^{j}_{ik,\bar{r}}-\frac{4}{3}\sum_{k,\ell,r}\overline{\eta_{k}}T^{r}_{\ell
k}\left(\phi_{r}^{\ell}+\overline{\phi_{\ell}^{r}}-B_{r\bar{\ell}}+\frac{1}{4}\mathrm{Ric}(Q)_{r\bar{\ell}}\right)$
$\displaystyle=-\frac{4}{9}\sum_{i,j,k,r}\overline{\eta_{k}}T^{i}_{jr}\left(\sum_{q}T^{q}_{ik}\overline{T^{q}_{jr}}+T^{j}_{iq}\overline{T^{k}_{rq}}-T^{j}_{kq}\overline{T^{i}_{rq}}-T^{r}_{iq}\overline{T^{k}_{jq}}+T^{r}_{kq}\overline{T^{i}_{jq}}+\frac{1}{4}(R^{b}_{i\bar{j}k\bar{r}}-R^{b}_{k\bar{j}i\bar{r}})\right)$
$\displaystyle\quad-\frac{4}{3}(\phi\cdot\phi+|\phi|^{2}-\phi
B+\frac{1}{4}\phi\mathrm{Ric}(Q))$ $\displaystyle=-\frac{4}{9}(\phi B-2\phi
A)-\frac{4}{3}(\phi\cdot\phi+|\phi|^{2}-\phi
B+\frac{1}{4}\phi\mathrm{Ric}(Q))-\frac{1}{9}\sum_{i,j,k,r}\overline{\eta_{k}}T^{i}_{jr}Q_{i\bar{j}k\bar{r}}$
$\displaystyle=\frac{8}{9}(\phi B+\phi
A)-\frac{4}{3}(\phi\cdot\phi+|\phi|^{2})-\frac{1}{3}\phi\mathrm{Ric}(Q)-\frac{1}{9}\sum_{i,j,k,r}\overline{\eta_{k}}T^{i}_{jr}Q_{i\bar{j}k\bar{r}},$

where

$\phi A=\sum_{k,\ell}\phi^{k}_{\ell}A_{k\bar{\ell}}.$

Hence, the equality (4.2) implies

$\sum_{k,\ell}|\eta_{k,\bar{\ell}}|^{2}+\mathrm{Re}(\eta_{k,\bar{\ell}\,\ell}\overline{\eta_{k}})=0,$

which yields

(4.4) $\displaystyle|B|^{2}-2\mathrm{Re}(\phi B)+2\mathrm{Re}(\phi
A)-\frac{1}{2}\mathrm{Re}(B\mathrm{Ric}(Q))$
$\displaystyle=\frac{1}{2}|\phi+\phi^{*}|^{2}-\frac{1}{4}\mathrm{Re}(\phi\mathrm{Ric}(Q))$
$\displaystyle-\frac{1}{16}|\mathrm{Ric}(Q)|^{2}+\frac{1}{4}\mathrm{Re}\left(\sum_{i,j,k,r}\overline{\eta_{k}}T^{i}_{jr}Q_{i\bar{j}k\bar{r}}\right).$

As $\sum_{r}\eta_{r}T^{r}_{ij}=0$ has already been shown, it follows that,
after the covariant derivative in $\bar{\ell}$ is taken,

$\sum_{r}\eta_{r,\bar{\ell}}\,T^{r}_{ij}+\eta_{r}\,T^{r}_{ij,\bar{\ell}}=0,$

which yields

$\displaystyle\sum_{r}\left(\phi_{r}^{\ell}+\overline{\phi_{\ell}^{r}}-B_{r\bar{\ell}}+\frac{1}{4}\mathrm{Ric}(Q)_{r\bar{\ell}}\right)T^{r}_{ij}$
$\displaystyle+\sum_{r}\eta_{r}\left(\sum_{q}(T^{q}_{ij}\overline{T^{q}_{r\ell}}+T^{r}_{iq}\overline{T^{j}_{\ell
q}}-T^{r}_{jq}\overline{T^{i}_{\ell
q}}-T^{\ell}_{iq}\overline{T^{j}_{rq}}+T^{\ell}_{jq}\overline{T^{i}_{rq}})+\frac{1}{4}(R^{b}_{i\bar{r}j\bar{\ell}}-R^{b}_{j\bar{r}i\bar{\ell}})\right)=0,$

or equivalently,

(4.5)
$\sum_{r}\left((\phi_{r}^{\ell}-B_{r\bar{\ell}})T^{r}_{ij}+\frac{1}{4}\mathrm{Ric}(Q)_{r\bar{\ell}}T^{r}_{ij}+\overline{\phi^{j}_{r}}T^{\ell}_{ir}-\overline{\phi^{i}_{r}}T^{\ell}_{jr}+\frac{1}{4}\eta_{r}Q_{i\bar{r}j\bar{\ell}}\right)=0.$

Multiply $\overline{\eta_{j}}$ on both sides of (4.5) and sum up $j$, it
yields

$\sum_{r}\left(\phi_{r}^{\ell}\phi_{i}^{r}-\phi^{r}_{i}B_{r\bar{\ell}}+\frac{1}{4}\phi^{r}_{i}\mathrm{Ric}(Q)_{r\bar{\ell}}+\phi^{\ell}_{r}\overline{\phi^{i}_{r}}+\frac{1}{4}\sum_{j}\eta_{r}\overline{\eta_{j}}Q_{i\bar{r}j\bar{\ell}}\right)=0,$

where $\sum_{r}\eta_{r}T^{r}_{ij}=0$ is used here. Let $i=\ell$ and sum up
$i$, it implies from (4.1) that

$\phi B=\phi\cdot\phi+\frac{1}{4}\phi\mathrm{Ric}(Q)+|\phi|^{2},$

and thus

(4.6) $\mathrm{Re}(\phi
B)=\mathrm{Re}(\phi\cdot\phi)+\frac{1}{4}\mathrm{Re}(\phi\mathrm{Ric}(Q))+|\phi|^{2}=\frac{1}{2}|\phi+\phi^{*}|^{2}+\frac{1}{4}\mathrm{Re}(\phi\mathrm{Ric}(Q)).$

Similarly, multiply $\overline{T^{\ell}_{ij}}$ on both sides of (4.5) and sum
up $i,j,\ell$, it yields

$\phi B-|B|^{2}+\frac{1}{4}B\mathrm{Ric}(Q)+2\overline{\phi
A}+\frac{1}{4}\sum_{i,j,r,\ell}\eta_{r}\overline{T^{\ell}_{ij}}Q_{i\bar{r}j\bar{\ell}}=0,$

and thus, by Lemma 4.1,

(4.7) $|B|^{2}-\mathrm{Re}(\phi B)-2\mathrm{Re}(\overline{\phi
A})-\frac{1}{4}\mathrm{Re}(B\mathrm{Ric}(Q))=-\frac{1}{4}\mathrm{Re}\left(\sum_{i,j,r,\ell}\overline{\eta_{r}}T^{\ell}_{ij}Q_{\ell\bar{i}r\bar{j}}\right).$

Note that the equality (4.1) implies that

$\sum_{i}\overline{\eta_{i}}\mathrm{Ric}(Q)_{i\bar{j}}=0,$

for any $j$. Take the covariant derivative in $k$ of the equality above, it
yields

$\sum_{i}\eta_{i,\bar{k}}\mathrm{Ric}(Q)_{j\bar{i}}=0,$

where $\nabla^{b}\mathrm{Ric}(Q)=0$ is used here, and thus

$\sum_{i}\left(\phi_{i}^{k}+\overline{\phi_{k}^{i}}-B_{i\bar{k}}+\frac{1}{4}\mathrm{Ric}(Q)_{i\bar{k}}\right)\mathrm{Ric}(Q)_{j\bar{i}}=0.$

Let $j=k$ and sum up $j$, we get

$\phi\mathrm{Ric}(Q)+\overline{\phi\mathrm{Ric}(Q)}-B\mathrm{Ric}(Q)+\frac{1}{4}|\mathrm{Ric}(Q)|^{2}=0,$

and thus

(4.8)
$\mathrm{Re}(B\mathrm{Ric}(Q))=2\mathrm{Re}(\phi\mathrm{Ric}(Q))+\frac{1}{4}|\mathrm{Ric}(Q)|^{2}.$

In the meantime, we will also use the equality

$\sum_{r}(T^{r}_{\ell
i}T^{k}_{rj}+T^{r}_{j\ell}T^{k}_{ri}+T^{r}_{ij}T^{k}_{r\ell})=0.$

Multiply $\overline{T^{k}_{\ell j}}$ on both sides and sum up $j,k,\ell$, we
get

$\sum_{r,\ell}T^{r}_{\ell
i}A_{r\bar{\ell}}-\sum_{k,r}T^{k}_{ri}B_{k\bar{r}}+\sum_{j,r}T^{r}_{ji}A_{r\bar{j}}=0.$

Hence

(4.9) $2\sum_{r,\ell}T^{r}_{\ell i}A_{r\bar{\ell}}=\sum_{r,\ell}T^{r}_{\ell
i}B_{r\bar{\ell}}.$

Multiply $\overline{\eta_{i}}$ on both sides above and sum up $i$, we get

$2\phi A=\phi B,$

and thus

(4.10) $2\mathrm{Re}(\phi A)=\mathrm{Re}(\phi B).$

Therefore, the equalities (4.4), (4.6), (4.7), (4.8) and (4.10) imply that

$|B|^{2}=|\phi+\phi^{*}|^{2}+\mathrm{Re}(\phi\mathrm{Ric}(Q))+\frac{1}{16}|\mathrm{Ric}(Q)|^{2},\quad\mathrm{Re}(\phi
B)=\frac{1}{2}|\phi+\phi^{*}|^{2}+\frac{1}{4}\mathrm{Re}(\phi\mathrm{Ric}(Q)),$
$\mathrm{Re}(\phi
A)=\frac{1}{4}|\phi+\phi^{*}|^{2}+\frac{1}{8}\mathrm{Re}(\phi\mathrm{Ric}(Q)),\quad\mathrm{Re}(B\mathrm{Ric}(Q))=2\mathrm{Re}(\phi\mathrm{Ric}(Q))+\frac{1}{4}|\mathrm{Ric}(Q)|^{2},$
$\mathrm{Re}\left(\sum_{i,j,k,r}\overline{\eta_{k}}T^{i}_{jr}Q_{i\bar{j}k\bar{r}}\right)=0.$

which yields

$\sum_{k,\ell}|\eta_{k,\bar{\ell}}|^{2}=\frac{4}{9}\left(|\phi+\phi^{*}|^{2}+|B|^{2}-4\mathrm{Re}(\phi
B)\right)+\frac{1}{36}|\mathrm{Ric}(Q)|^{2}+\frac{4}{9}\mathrm{Re}\left(\phi\mathrm{Ric}(Q)\right)-\frac{2}{9}\mathrm{Re}\left(B\mathrm{Ric}(Q)\right)=0,$

hence $\nabla^{b}\eta=0$ is proved.

Thirdly, we will show that $|T|^{2}$, $\phi_{i}^{j}$ and $B_{i\bar{j}}$ are
$\nabla^{b}$-parallel. To this end, the equality (4.3) implies that

$0=\sum_{i}\eta_{i,\bar{i}}=\frac{2}{3}(|T|^{2}-2|\eta|^{2})-\frac{1}{6}\sum_{i}\mathrm{Ric}(Q)_{i\bar{i}},$

and it follows that $|T|^{2}$ is $\nabla^{b}$-parallel, since $|\eta|^{2}$ and
$\mathrm{Ric}(Q)$ are. It is clear that $\phi_{i,k}^{j}=0$ now due to
$\nabla^{b}\eta=0$, while it is easy to see

$\phi_{i,\bar{k}}^{j}=\sum_{r}\overline{\eta_{r}}T^{j}_{ir,\bar{k}}=-\sum_{r}\overline{\eta_{r}}T^{j}_{ri,\bar{k}}=-\sum_{r}\overline{\eta_{r}T^{r}_{jk,\bar{i}}}=0,$

where Lemma 4.1 and $\sum_{r}\eta_{r}T^{r}_{ij}=0$ are used. Hence
$\phi_{i}^{j}$ is $\nabla^{b}$-parallel. Note that the equality (4.3) implies

$0=\eta_{i,\bar{j}}=-\frac{2}{3}\left(\phi_{i}^{j}+\overline{\phi_{j}^{i}}-B_{i\bar{j}}+\frac{1}{4}\mathrm{Ric}(Q)_{i\bar{j}}\right),$

which yields that $B_{i\bar{j}}$ is $\nabla^{b}$-parallel, as $\phi_{i}^{j}$
and $\mathrm{Ric}(Q)$ are.

Finally, we are ready to show that $\nabla^{b}T^{c}=0$. It follows from (4.9)
that

(4.11) $\sum_{i,\ell,r}\left(T^{r}_{\ell
i}A_{r\bar{\ell}}\right)_{,\bar{i}}=\frac{1}{2}\sum_{i,\ell,r}\left(T^{r}_{\ell
i}B_{r\bar{\ell}}\right)_{,\bar{i}}=\frac{1}{2}\sum_{i,\ell,r}T^{r}_{\ell
i,\bar{i}}B_{r\bar{\ell}}=-\frac{1}{2}\sum_{i,\ell,r}T^{i}_{\ell
i,\bar{r}}B_{r\bar{\ell}}=\frac{1}{2}\sum_{\ell,r}\eta_{\ell,\bar{r}}B_{r\bar{\ell}}=0,$

where Lemma 4.1 is used. Then, since $|T|^{2}$ is $\nabla^{b}$-parallel, it
follows, after the covariant derivative in $\bar{\ell}$ is taken,

$0=|T|^{2}_{,\bar{\ell}}=\sum_{i,j,k}\overline{T^{j}_{ik}}T^{j}_{ik,\bar{\ell}}=0.$

Take the covariant derivative in $\bar{\ell}$ and sum up $\ell$, we get

$\sum_{i,j,k,\ell}|T^{j}_{ik,\bar{\ell}}|^{2}+\overline{T^{j}_{ik}}T^{j}_{ik,\bar{\ell}\,\ell}=0.$

Note that, from Lemma 4.1, Lemma 4.2 and $\nabla^{b}\eta=0$,

$\displaystyle\sum_{i,j,k,\ell}\overline{T^{j}_{ik}}T^{j}_{ik,\bar{\ell}\,\ell}$
$\displaystyle=\sum_{i,j,k,\ell}\overline{T^{j}_{ik}}\left(\overline{T_{j\ell,\bar{k}}^{i}}\right)_{,\ell}=\sum_{i,j,k,\ell}\overline{T^{j}_{ik}T_{j\ell,\bar{k}\,\bar{\ell}}^{i}}=\sum_{i,j,k,\ell}\overline{T^{j}_{ik}(T_{j\ell,\bar{\ell}\,\bar{k}}^{i}+2\sum_{r}\overline{T^{r}_{\ell
k}}T^{i}_{j\ell,\bar{r}})}$
$\displaystyle=-\sum_{i,j,k,\ell}\overline{T^{j}_{ik}T^{\ell}_{j\ell,\bar{i}\,\bar{k}}}+2\sum_{i,j,k,\ell,r}T^{r}_{\ell
k}\overline{T^{j}_{ik}}T^{j}_{ir,\bar{\ell}}=\sum_{i,j,k}\overline{T^{j}_{ik}\eta_{j,\bar{i}\,\bar{k}}}+2\sum_{k,\ell,r}T^{r}_{\ell
k}A_{r\bar{k},\bar{\ell}}$
$\displaystyle=-2\sum_{k,\ell,r}\left(T^{r}_{k\ell}A_{r\bar{k}}\right)_{,\bar{\ell}}+2\sum_{k,\ell,r}T^{r}_{k\ell,\bar{\ell}}A_{r\bar{k}}=-2\sum_{k,\ell,r}\left(T^{r}_{k\ell}A_{r\bar{k}}\right)_{,\bar{\ell}}-2\sum_{k,\ell,r}T_{k\ell,\bar{r}}^{\ell}A_{rk}$
$\displaystyle=-2\sum_{k,\ell,r}\left(T^{r}_{k\ell}A_{r\bar{k}}\right)_{,\bar{\ell}}+2\sum_{k,r}\eta_{k,\bar{r}}A_{r\bar{k}}=-2\sum_{k,\ell,r}\left(T^{r}_{k\ell}A_{r\bar{k}}\right)_{,\bar{\ell}}$
$\displaystyle=0,$

where the last equality is due to (4.11). Therefore, $T^{j}_{ik,\bar{\ell}}=0$
and the proof is completed. ∎

Next we prove Corollary 1.2. The proof is the same as the BKL case in [34],
and we include it here for the sake of completeness.

###### Proof of Corollary 1.2: .

By Proposition 3.6, we know that $\partial\eta=0$. If $g$ is strongly
Gauduchon, that is, there exists a $(n,n-2)$-form $\Omega$ on $M^{n}$ such
that $\partial\omega^{n-1}=\overline{\partial}\Omega$, then by the fact that
$\partial\eta=0$ and $\partial\omega^{n-1}=-2\eta\,\omega^{n-1}$, we get

$\int_{M}2\eta\,\overline{\eta}\,\omega^{n-1}=\int_{M}\overline{\eta}\,\partial\omega^{n-1}=\int_{M}\overline{\eta}\,\overline{\partial}\Omega=\int_{M}\overline{\partial}\overline{\eta}\,\Omega=0.$

Hence $|\eta|^{2}=0$, contradicting to our assumption that $g$ is not
balanced. Similarly, if $\overline{\eta}=\overline{\partial}f$ for some smooth
function $f$ on $M^{n}$, then

$\int_{M}\partial\overline{\eta}\wedge\omega^{n-1}=\int_{M}\partial\overline{\partial}f\wedge\omega^{n-1}=0$

as $g$ is Gauduchon. However, Proposition 3.6 implies
$\partial\overline{\eta}=2\sum_{i,j}\phi^{j}_{i}\varphi_{i}\wedge\overline{\varphi}_{j}$.
This leads to the vanishing of $\eta$, hence $g$ is balanced, which is a
contradiction. This completes the proof of Corollary 1.2. ∎

From the proof of Theorem 1.1 above, we know that on a BTP manifold it holds
that

(4.12) $\phi+\phi^{\ast}=B-\frac{1}{4}\mbox{Ric}(Q).$

The right hand side of (4.12) is not necessarily non-negative, so the proof of
[34, Theorem 6] does not go through, that is why we cannot prove Conjecture
1.3 here for non-balanced BTP manifolds. If $\mbox{Ric}(Q)=0$, or more
generally if $4B\geq\mbox{Ric}(Q)$, then Conjecture 1.3 holds.

Now let us prove Proposition 1.4, where the first part is due to Andrada and
Villacampa [2, Theorem 3.6]. For readers convenience we gave a proof here as
well.

###### Proof of Proposition 1.4:.

Let $(M^{n},g)$ be a locally conformally Kähler manifold. Then under any local
unitary frame $e$, the torsion components are given by

(4.13)
$T_{ik}^{j}=\frac{1}{n-1}\big{(}\eta_{k}\delta_{ij}-\eta_{i}\delta_{kj}\big{)}$

and the torsion $1$-form $\eta$ is equal to $\eta=(n-1)\partial u$ for some
smooth real-valued local function $u$. For any given point $p\in M$, choose
the local unitary frame $e$ so that the matrix $\theta^{b}$ of Bismut
connection vanishes at $p$. Denote by $\varphi$ the local unitary coframe dual
to $e$. By [31], the Levi-Civita connection $\nabla$ has connection matrix

$\nabla e=\theta^{1}e+\overline{\theta^{2}}\,\overline{e},\ \ \
\theta^{1}=\theta^{b}-\gamma,\ \ \
\gamma_{ij}=\sum_{k}\big{(}T^{j}_{ik}\varphi_{k}-\overline{T_{jk}^{i}}\overline{\varphi_{k}}\big{)},\
\ \ \theta^{2}_{ij}=\sum_{k}\overline{T^{k}_{ij}}\varphi_{k}.$

Since $\theta^{b}|_{p}=0$, we have at point $p$ that

$\displaystyle\nabla\varphi_{i}$ $\displaystyle=$
$\displaystyle-\theta^{1}_{ji}\otimes\varphi_{j}-\theta^{2}_{ji}\otimes\overline{\varphi}_{j}\
\,=\
\,\gamma_{ji}\otimes\varphi_{j}-\theta^{2}_{ji}\otimes\overline{\varphi}_{j}$
$\displaystyle=$
$\displaystyle\sum_{j,k}\big{(}T^{i}_{jk}\varphi_{k}\otimes\varphi_{j}-\overline{T^{j}_{ik}}\overline{\varphi}_{k}\otimes\varphi_{j}-\overline{T^{k}_{ji}}\varphi_{k}\otimes\overline{\varphi}_{j}\big{)}$
$\displaystyle=$
$\displaystyle\frac{1}{n-1}\\{\eta\otimes\varphi_{i}-\varphi_{i}\otimes\eta+\varphi_{i}\otimes\overline{\eta}-\overline{\eta}\otimes\varphi_{i}+\overline{\eta}_{i}\sum_{k}\big{(}\overline{\varphi}_{k}\otimes\varphi_{k}-\varphi_{k}\otimes\overline{\varphi}_{k}\big{)}\\}$
$\displaystyle=$
$\displaystyle\frac{2}{n-1}\\{\eta\wedge\varphi_{i}+\varphi_{i}\wedge\overline{\eta}-\overline{\eta}_{i}\sum_{k}\varphi_{k}\wedge\overline{\varphi}_{k}\\}$

Therefore

$\displaystyle\nabla\eta$ $\displaystyle=$
$\displaystyle\nabla\sum_{i}\eta_{i}\varphi_{i}\ \,=\
\,\sum_{i,j}\big{(}\eta_{i,j}\varphi_{j}\otimes\varphi_{i}+\varphi_{i,\overline{j}}\overline{\varphi}_{j}\otimes\varphi_{i}\big{)}+\sum_{i}\eta_{i}\nabla\varphi_{i}$
$\displaystyle=$
$\displaystyle\sum_{i,j}\big{(}\eta_{i,j}\,\varphi_{j}\otimes\varphi_{i}+\eta_{i,\overline{j}}\,\overline{\varphi}_{j}\otimes\varphi_{i}\big{)}+\frac{2}{n-1}\\{\eta\wedge\overline{\eta}-\sum_{i}|\eta_{i}|^{2}\sum_{k}\varphi_{k}\wedge\overline{\varphi}_{k}\\},$

where indices after comma denote covariant derivatives with respect to
$\nabla^{b}$. If $(M^{n},g)$ is BTP, then $\nabla^{b}\eta=0$, hence
$\nabla(\eta+\overline{\eta})=0$ and $g$ is Vaisman. Conversely, when $g$ is
Vaisman, which means the manifold is locally conformally Kähler with
$\nabla(\eta+\overline{\eta})=0$, then from the above calculation we get

$\sum_{i,j}\\{\eta_{i,j}\,\varphi_{j}\otimes\varphi_{i}+\eta_{i,\overline{j}}\,\overline{\varphi}_{j}\otimes\varphi_{i}+\overline{\eta_{i,\overline{j}}}\,\varphi_{j}\otimes\overline{\varphi}_{i}+\overline{\eta_{i,j}}\
\overline{\varphi}_{j}\otimes\overline{\varphi}_{i}\\}=0.$

Hence $\nabla^{b}\eta=0$, and $\nabla^{b}T=0$ by (4.13). So we have proved
that for a locally conformally Kähler manifold, the BTP condition is
equivalent to the Vaisman condition.

Next we consider the uniqueness problem for BTP metrics within a conformal
class. Note that when the manifold is compact, such metrics are clearly unique
(up to constant multiples) within each conformal class, if it exists at all,
as BTP metrics are Gauduchon. So the statement in Proposition 1.4 is really
about non-compact cases.

Suppose $(M^{n},g)$ and $(M^{n},e^{2u}g)$ are two BTP manifolds, where $u$ is
a smooth real-valued function on $M$ with $du\neq 0$ in an open dense subset
$U\subseteq M$. Let $e$ be a local unitary frame for $g$ with dual coframe
$\varphi$. Then $\tilde{e}=e^{-u}e$ and $\tilde{\varphi}=e^{u}\varphi$ are
local unitary frame and dual coframe for $\tilde{g}=e^{2u}g$. We have (cf.
[33, the proof of Theorem 3])

$\displaystyle e^{u}\tilde{T}^{j}_{ik}$ $\displaystyle=$ $\displaystyle
T^{j}_{ik}+u_{i}\delta_{kj}-u_{k}\delta_{ij}$ $\displaystyle
e^{u}\tilde{\eta}_{k}$ $\displaystyle=$ $\displaystyle\eta_{k}-(n-1)u_{k}$
$\displaystyle P_{ik}\ \ $ $\displaystyle=$ $\displaystyle
2u_{i}\varphi_{k}-\partial
u\,\delta_{ik}-2\overline{u_{k}}\,\overline{\varphi}_{i}+\overline{\partial}u\,\delta_{ik}$

where $P=\tilde{\theta}^{b}-\theta^{b}$ and $u_{k}=e_{k}(u)$. Since
$\nabla^{b}T=0$ and $\tilde{\nabla}^{b}\tilde{T}=0$, by the analysis of the
$(1,0)$-part, a straight-forward computation leads to

(4.14)
$u_{i,\ell}\delta_{kj}-u_{k,\ell}\delta_{ij}=2u_{k}T^{j}_{i\ell}-2u_{i}T^{j}_{k\ell}+2u_{i}u_{\ell}\delta_{kj}-2u_{k}u_{\ell}\delta_{ij}-2\delta_{j\ell}\sum_{r}u_{r}T^{r}_{ik}$

for any indices $1\leq i,j,k,\ell\leq n$ under any local unitary frame $e$. As
$du\neq 0$ holds on the open dense subset $U$, we may choose $e$ such that
$\varphi^{n}=\frac{\partial u}{|\partial u|_{g}}$, which implies that
$u_{1}=\cdots=u_{n-1}=0$ and $u_{n}=|\partial u|_{g}=\lambda>0$ are
established where the local unitary frame $e$ intersects with $U$.

Assume $i<k$. If we take $j\notin\\{i,k\\}$ in (4.14), we get

(4.15) $\delta_{kn}T^{j}_{i\ell}=\delta_{j\ell}T^{n}_{ik},\qquad i<k,\ j\neq
i,k.$

Take $k=n$ and $j\neq\ell$, we get $T^{j}_{i\ell}=0$. Take $k<n$ and $j=\ell$,
we get $T^{n}_{ik}=0$. So under this particular unitary frame $e$, we have
$T^{j}_{ik}=0$ whenever all three indices are distinct. Now let us take $n=k$
and $j=\ell$ in (4.15), we get $T^{j}_{ij}=T^{n}_{in}$, for all $i,j<n$ with
$i\neq j$. Therefore $T^{\ell}_{\ell i}$ are equal for $\ell\neq i$, and
$T^{\ell}_{\ell i}=\frac{1}{n-1}\eta_{i}$ whenever $i<n$ and $\ell\neq i$, as
$\sum_{\ell}T^{\ell}_{\ell i}=\eta_{i}$.

Take $k=\ell=n$ and $i=j<n$ in (4.14), which yields that

$-u_{n,n}=2\lambda T^{i}_{in}-2\lambda^{2},\qquad i<n.$

This shows that all $T^{i}_{in}$ for $i<n$ are equal, hence equal to
$\frac{1}{n-1}\eta_{n}$. Combining the above results, we conclude that under
this particular frame $e$, it holds that

(4.16)
$T^{j}_{ik}=\frac{1}{n-1}\big{(}\eta_{k}\delta_{ij}-\eta_{i}\delta_{kj}\big{)},\qquad
1\leq i,j,k\leq n.$

which implies that the above equation holds in the open dense subset
$U\subseteq M$, hence in the entire $M$. From this equation, it follows that

$\phi_{i}^{j}=\sum_{r}\overline{\eta}_{r}T^{j}_{ir}=\frac{1}{n-1}\left(\delta_{ij}\sum_{r}|\eta_{r}|^{2}-\eta_{i}\overline{\eta}_{j}\right).$

In particular, $\phi_{i}^{j}=\overline{\phi^{i}_{j}}$ for any $i,j$. By
Proposition 3.6, we know that the torsion $1$-form $\eta$ for any BTP manifold
satisfies $\partial\eta=0$ and
$\overline{\partial}\eta=-2\sum_{i,j}\overline{\phi^{i}_{j}}\varphi_{i}\wedge\overline{\varphi}_{j}$.
Therefore, it yields that

$\overline{\partial}\eta+\partial\overline{\eta}=2\sum_{i,j}\big{(}\phi_{i}^{j}-\overline{\phi^{i}_{j}})\varphi_{i}\wedge\overline{\varphi}_{j}=0,$

which amount to $d(\eta+\overline{\eta})=0$. It implies that locally there
exists smooth, real-valued function $v$ such that $\eta=(n-1)\partial v$. By
(4.16), the torsion of the metric $e^{2v}g$ will vanish, hence $g$ is locally
conformally Kähler. This completes the proof of Proposition 1.4 ∎

Next let us prove Proposition 1.5.

###### Proof of Proposition 1.5:.

Let $(M^{n},g)$ be a complex nilmanifold with nilpotent $J$, namely, its
universal cover is a nilpotent Lie group $G$ equipped with a left-invariant
complex structure $J$ and a compatible left-invariant metric $g$. Assume that
$J$ is nilpotent in the sense of [10] and $\nabla^{b}T^{c}=0$. By Lemma 1,
Lemma 2 and the discussion right after the proof of Lemma 2 in [35], we know
that $G$ is of step most 2 and $J$ is abelian (namely, $C=0$), and there
exists an integer $1\leq r\leq n$ and a unitary coframe $\varphi$ such that

$d\varphi_{i}=0,\ \ \ \forall\ 1\leq i\leq r;\ \ \ \ \ \
d\varphi_{\alpha}=\sum_{i=1}^{r}Y_{\alpha
i}\varphi_{i}\wedge\overline{\varphi}_{i},\ \ \ \forall\ r<\alpha\leq n.$

Conversely, for such a complex nilmanifold, it is easy to check that
$\nabla^{b}T^{c}=0$. This completes the proof of Proposition 1.5. ∎

##  5\. Non-balanced BTP manifolds

When $n\geq 3$, as illustrated by Proposition 1.5, the set of all non-balanced
BTP manifolds in complex dimension $n$ contains the set of all non-Kähler BKL
manifolds in complex dimension $n$ as a proper subset.

Given any non-balanced BTP manifold $(M^{n},g)$, the type $(1,0)$ vector field
$\chi$ dual to $\eta$ is also $\nabla^{b}$-parallel, hence is nowhere zero.
The following notion of admissible frames introduced by the authors in [33,
Definition 4] to study non-Kähler BKL manifolds can be generalized to non-
balanced BTP manifolds:

######  Definition 5.1.

Let $(M^{n},g)$ be a non-balanced BTP manifold and $\chi$ the vector field
dual to Gauduchon’s torsion $1$-form $\eta$. A local unitary frame $e$ on
$M^{n}$ is said to be admissible, if $\chi=\lambda e_{n}$ for some constant
$\lambda>0$ and under $e$ the matrix $\phi=(\phi_{i}^{j})$ is diagonal.

######  Corollary 5.2 (See also Proposition 3.6 of [6] and Lemma 3 of [33]).

Let $(M^{n},g)$ be a non-balanced BTP manifold. Then around any given point
there exit an admissible frame $e$ such that $\chi=\lambda e_{n}$ and
$\phi_{i}^{j}=\lambda a_{i}\delta_{ij}$, where $\lambda>0$ and $a_{i}$ are
global constants with $a_{n}=0$. Therefore, $T^{1,0}M$ admits a
$\nabla^{b}$-parallel, $g$-orthogonal decomposition

$T^{1,0}M=\bigoplus_{i=0}^{k}H_{i},$

where $H_{0}$ is generated by $\chi$ and other distributions
$\\{H_{i}\\}_{i=1}^{k}$ are the eigenspaces of $\phi$ corresponding to
distinct eigenvalues of $\\{\lambda a_{1},\ldots,\lambda a_{n-1}\\}$.

###### Proof.

Clearly, $|\eta|^{2}$ is a positive constant, which will be denoted by
$\lambda^{2}$. Thus we may choose the unitary coframe $\varphi$ such that
$\eta=\lambda\varphi_{n}$. This means that $\eta_{1}=\cdots=\eta_{n-1}=0$ and
$\eta_{n}=\lambda$, thus $\chi=\lambda e_{n}$. From Proposition 3.6,
$R^{b}_{xy\chi w}=0$ yields $R^{b}_{i\bar{j}n\bar{\ell}}=0$ for any
$i,j,\ell$, and by $R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$
we get

$R^{b}_{n\bar{j}k\bar{\ell}}=R^{b}_{i\bar{n}k\bar{\ell}}=R^{b}_{i\bar{j}k\bar{n}}=0$

for any $i,j,k,\ell$. Moreover, the conclusion (3.14) of Proposition 3.5
implies

(5.1) $\sum_{r}\eta_{r}T_{ij}^{r}=0,$

which yields $T^{n}_{ij}=0$ for any $i$, $j$. Hence $\phi_{i}^{j}=\lambda
T^{j}_{in}$, where $\phi_{i}^{n}=\phi_{n}^{i}=0$ for any $i$.

Since $R^{b}_{i\bar{j}k\bar{\ell}}=R^{b}_{k\bar{\ell}i\bar{j}}$, the equality
(3.2) amounts to

$0=\sum_{r}(T^{r}_{ik}\overline{T^{r}_{j\ell}}+T^{j}_{ir}\overline{T^{k}_{\ell
r}}-T^{j}_{kr}\overline{T^{i}_{\ell
r}}-T^{\ell}_{ir}\overline{T^{k}_{jr}}+T^{\ell}_{kr}\overline{T^{i}_{jr}})+\frac{1}{4}(R^{b}_{i\bar{j}k\bar{\ell}}-R^{b}_{k\bar{j}i\bar{\ell}}).$

Take $k=\ell=n$ above and multiply $\lambda^{2}$ on both sides, we get

$\sum_{r}\phi_{i}^{r}\overline{\phi_{j}^{r}}-\phi_{r}^{j}\overline{\phi_{r}^{i}}=0,\
\ \ \forall\ i,j.$

That is, the matrix $\phi=(\phi_{i}^{j})$ satisfies
$\phi\phi^{\ast}=\phi^{\ast}\phi$ hence is normal. Here $\phi^{\ast}$ stands
for the conjugate transpose of $\phi$. So around each given point, by a
unitary change of $\\{\varphi_{1},\ldots,\varphi_{n-1}\\}$ with $\varphi_{n}$
fixed, we could always make $\phi$ diagonal. This gives us an admissible local
unitary frame.

As $\nabla^{b}T^{c}=0$, the tensor $\phi_{i}^{j}$ is also
$\nabla^{b}$-parallel, which implies the eigenvalues $\\{\lambda
a_{i}\\}_{i=1}^{n}$ of $\phi$ are global constants with $a_{n}=0$. It is also
obvious that the eigenspaces of $\phi$ with respect to all the distinct
eigenvalues lead to a $\nabla^{b}$-parallel and $g$-orthogonal decomposition
of $T^{1,0}M$. ∎

######  Remark 5.3.

It is easy to verify that, for non-balanced BTP manifolds, the distribution
$H_{0}\oplus\overline{H}_{0}$ is a foliation since $[\chi,\overline{\chi}]=0$,
which is known as canonical foliation in the study of Vaisman manifolds.

The following notation of degenerate torsion, introduced in [33, Defintion 5],
could also be applied to non-balanced BTP manifolds.

######  Definition 5.4.

A non-balanced BTP manifold $(M^{n},g)$ is said to have degenerate torsion, if
under any admissible frame $e$, $T^{\ast}_{ik}=0$ for any $i,k<n$.

The Lee potential (LP) condition was introduced by Belgun [6] to study
generalized Calabi-Eckmann (GCE) structures (see also [4, 5]).

######  Definition 5.5 ([6]).

A Hermitian manifold $(M^{n},g)$ is LP if the Gauduchon torsion $1$-form
$\eta$ satisfies

$\partial\eta=0,\quad\partial\omega=c\,\eta\wedge\partial\overline{\eta},$

where $c\neq 0$ is a constant. A Hermitian manifold is GCE if it is LP and
BTP. As shown in [6, Proposition 3.2], both the standard Hermitian structure
and the modified ones on the Sasakian product are non-Kähler GCE manifolds.

###### Proof of Proposition 1.6:.

By Corollary 5.2, we know that non-balanced BTP manifolds always admit local
admissible frames. Under such a frame $e$, the vector
$e_{n}=\frac{1}{\lambda}\chi$ is parallel under $\nabla^{b}$, so the
connection matrix $\theta^{b}$ of $\nabla^{b}$ satisfies
$\theta^{b}_{n\ast}=\theta^{b}_{\ast n}=0$. Therefore, the curvature matrix
$\Theta^{b}$ under $e$ will also be block-diagonal:

$\Theta^{b}=\left[\begin{array}[]{cc}\ast&0\\\ 0&0\end{array}\right]$

where $\ast$ is an $(n\\!-\\!1)\times(n\\!-\\!1)$ matrix. As is well-known,
the Lie algebra of the holonomy group $\mbox{Hol}(\nabla^{b})$ of $\nabla^{b}$
is generated by conjugation class of $R^{b}_{xy}=\Theta^{b}(x,y)$ under
$\nabla^{b}$-parallel transports, all of them operates in the
$\nabla^{b}$-parallel block $e_{n}^{\perp}$, so
$\mbox{Hol}(\nabla^{b})\subseteq U(n-1)$. This proves the first part of
Proposition 1.6.

For the second part, under an admissible frame $e$, it follows from
Proposition 3.6 that

$\displaystyle\eta\wedge\partial\overline{\eta}$ $\displaystyle=\
2\lambda^{2}\varphi_{n}\wedge\sum_{i,j}T_{in}^{j}\varphi_{i}\wedge\overline{\varphi_{j}}$
$\displaystyle=\
-2\lambda^{2}\sum_{i<n}T_{in}^{i}\varphi_{i}\wedge\varphi_{n}\wedge\overline{\varphi_{i}},$
$\displaystyle-\sqrt{-1}\partial\omega$
$\displaystyle=^{t}\\!\\!\tau\wedge\overline{\varphi}=\sum_{i,j,k}T_{ik}^{j}\varphi_{i}\wedge\varphi_{k}\wedge\overline{\varphi_{j}}$
$\displaystyle=\ 2\sum_{\begin{subarray}{c}i<k\\\
j<n\end{subarray}}T^{j}_{ik}\varphi_{i}\wedge\varphi_{k}\wedge\overline{\varphi_{j}}.$

It is clear that the condition
$\partial\omega=c\,\eta\wedge\partial\overline{\eta}$ implies $T^{j}_{ik}=0$
for any $i,k<n$. Conversely, the torsion degeneracy condition $T^{j}_{ik}=0$
for any $i,k<n$ yields

$-\sqrt{-1}\partial\omega=2\sum_{\begin{subarray}{c}i<n\\\
j<n\end{subarray}}T^{j}_{in}\varphi_{i}\wedge\varphi_{n}\wedge\overline{\varphi_{j}}=2\sum_{i<n}T^{i}_{in}\varphi_{i}\wedge\varphi_{n}\wedge\overline{\varphi_{i}}$

by the definition of admissible frames, thus
$\partial\omega=c\,\eta\wedge\partial\overline{\eta}$, which shows that the LP
condition is equivalent to the torsion degeneracy condition. To prove the last
statement, namely when $n=3$ the set of non-balanced BTP threefolds is equal
to the set of GCE threefolds, it suffices to show that any non-balanced BTP
threefold $(M^{3},g)$ always has degenerate torsion. Let $e$ be a local
admissible frame on $M^{3}$. By definition, it follows that
$\eta_{1}=\eta_{2}=0$, $\eta_{3}=\lambda>0$, and $T^{1}_{23}=T^{2}_{13}=0$
since $\phi$ is diagonal. The equality (5.1) implies $T^{3}_{ij}=0$ for any
$i,j$. Then it yields that

$T^{1}_{12}=\eta_{2}-T^{3}_{32}=0,\ \ \
T^{2}_{12}=-T^{2}_{21}=-\eta_{1}+T^{3}_{31}=0.$

That is, $T^{\ast}_{12}=0$, so the manifold has degenerate torsion by
definition. This completes the proof of Proposition 1.6. ∎

######  Remark 5.6.

In light of the proposition above, a BKL manifold with degenerate torsion is
exactly a Hermitian manifold which is both BKL and GCE. A complete non-Kähler
BKL manifold with degenerate torsion will split off a Kähler factor, on the
universal cover, of complex codimension $2$ or $3$, as in [33, Theorem 9].

In order to get the classification theorem of non-balanced BTP threefolds, we
recall here the following definitions, including Sasakian manifolds, the
standard Hermitian structure and the modified ones on the Sasakian product.

######  Definition 5.7.

Let $(N^{2m+1},g,\xi)$ be a Sasakian manifold, that is, as defined in [33] for
instance, an odd dimensional Riemannian manifold $(N,g)$ equipped with a
Killing vector field $\xi$ of unit length, such that:

  1. $(1)$

The tensor field $\frac{1}{c}\nabla\xi$ (where $c>0$ is a constant), which
sends a tangent vector $X$ to the tangent vector $\frac{1}{c}\nabla_{X}\xi$,
gives an integrable orthogonal complex structure $I$ on the distribution $H$,
where $H$ is the perpendicular complement of $\xi$ in the tangent bundle $TN$.

  2. $(2)$

Denote by $\phi$ the $1$-form dual to $\xi$, namely, $\phi(X)=g(X,\xi)$ for
any $X$, then $\phi\wedge(d\phi)^{m}$ is nowhere zero. That is, $\phi$ gives a
contact structure on $N$.

######  Definition 5.8.

Let $(N_{1}^{2n_{1}+1},g_{1},\xi_{1})$ and $(N_{2}^{2n_{2}+1},g_{2},\xi_{2})$
be two Sasakian manifolds. As introduced by Belgun [6], on the product
Riemannian manifold $M=N_{1}\times N_{2}$, of even dimension
$2n=2n_{1}+2n_{2}+2$, for
$\kappa\in\\{\kappa\in\mathbb{C}\big{|}\mathrm{Im}(\kappa)\neq 0\\}$, the
natural almost complex structure $J_{\kappa}$ is defined by

$\displaystyle
J_{\kappa}\xi_{1}=\mathrm{Re}(\kappa)\xi_{1}+\mathrm{Im}(\kappa)\xi_{2},$
$\displaystyle J_{\kappa}X_{i}=J_{i}X_{i},\quad\forall\ X_{i}\in H_{i},\
i=1,2.$

Define the associated metric $g_{\kappa}$ as

$\displaystyle
g_{\kappa}(\xi_{1},\xi_{1})=g_{\kappa}(J_{\kappa}\xi_{1},J_{\kappa}\xi_{1})=1,\quad
g_{\kappa}(\xi_{1},J_{\kappa}\xi_{1})=0,$ $\displaystyle
g_{\kappa}(\xi_{i},X)=0,\quad\forall\ X\in H_{1}\oplus H_{2},\ i=1,2.$
$\displaystyle g_{\kappa}(X_{i},Y_{i})=g_{i}(X_{i},Y_{i}),\quad
g_{\kappa}(X_{1},X_{2})=0,\quad\forall\ X_{i},Y_{i}\in H_{i},\ i=1,2.$

It is easy to verify that $g_{\kappa}$ is a $J_{\kappa}$-Hermitian metric,
$J_{\kappa}$ is integrable, and when $\kappa=\sqrt{-1}$, the Hermitian
manifold defined above is the standard Hermitian structure on the Sasakian
product. We will call the Hermitian manifold $(N_{1}\times
N_{2},g_{\kappa},J_{\kappa})$ twisted Sasakian product, and denote it by
$N_{1}\times_{\kappa}N_{2}$.

######  Lemma 5.9.

Let $(N_{1},g_{1},\xi_{1}),(N_{2},g_{2},\xi_{2})$ be two $3$-dimensional
Sasakian manifolds and $\kappa=x+y\sqrt{-1}$ for $x,y\in\mathbb{R},y\neq 0$.
Then there exists a natural unitary frame $(e_{1},e_{2},e_{3})$, with dual
frame $(\varphi_{1},\varphi_{2},\varphi_{3})$, on the twisted Sasakian product
$N_{1}\times_{\kappa}N_{2}$, such that the Riemannian connection $\nabla$
under $e$ takes the form

(5.2) $\begin{cases}\nabla
e_{1}=c_{1}\sqrt{-1}\overline{\varphi}_{1}\,\xi_{1}+\sigma_{1}\,e_{1}\\\
\nabla
e_{2}=c_{2}\sqrt{-1}\overline{\varphi}_{2}\,\xi_{2}+\sigma_{2}\,e_{2}\\\
\nabla
e_{3}=\frac{c_{1}\sqrt{-1}}{\sqrt{2}}(\varphi_{1}\,e_{1}-\overline{\varphi}_{1}\,\overline{e}_{1})+\frac{c_{2}(1-x\sqrt{-1})}{y\sqrt{2}}(\varphi_{2}\,e_{2}-\overline{\varphi}_{2}\,\overline{e}_{2}),\end{cases}$

where $e_{3}=\frac{1}{\sqrt{2}}(\xi_{1}-\sqrt{-1}(x\xi_{1}+y\xi_{2}))$,
$c_{i}$ is the defining constant of the Sasakian manifold $N_{i}$, and
$\sigma_{i}$ is a $1$-form satisfying $\overline{\sigma}_{i}=-\sigma_{i}$, for
$i=1,2$.

###### Proof.

From Definition 5.7, for $i=1,2$, we may assume that
$(\xi_{i},e_{i},\overline{e}_{i})$ be a frame of the Sasakian manifold
$(N_{i},g_{i},\xi_{i})$ such that $e_{i}$ is a unitary frame of the
distribution $\xi_{i}^{\bot}$ with respect to the orthogonal complex structure
$\frac{1}{c_{i}}\nabla\xi_{i}$, which yields the following structure equation

$\nabla\\!\begin{bmatrix}\xi_{i}\\\ e_{i}\\\
\overline{e}_{i}\end{bmatrix}=\begin{bmatrix}0&c_{i}\sqrt{-1}\varphi_{i}&-c_{i}\sqrt{-1}\overline{\varphi}_{i}\\\
c_{i}\sqrt{-1}\overline{\varphi}_{i}&\sigma^{\prime}_{i}&0\\\
-c_{i}\sqrt{-1}\varphi_{i}&0&\overline{\sigma}^{\prime}_{i}\\\
\end{bmatrix}\\!\\!\begin{bmatrix}\xi_{i}\\\ e_{i}\\\ \overline{e}_{i}\\\
\end{bmatrix}\\!\\!,$

where $\sigma^{\prime}_{i}$ is a $1$-form satisfying
$\overline{\sigma}^{\prime}_{i}=-\sigma^{\prime}_{i}$ and the dual frame of
$(\xi_{i},e_{i},\overline{e}_{i})$ is denoted by
$(\phi_{i},\varphi_{i},\overline{\varphi}_{i})$. The dual version of the
equation above is

$\nabla\\!\begin{bmatrix}\phi_{i}\\\ \varphi_{i}\\\
\overline{\varphi}_{i}\end{bmatrix}=\begin{bmatrix}0&-c_{i}\sqrt{-1}\overline{\varphi}_{i}&c_{i}\sqrt{-1}\varphi_{i}\\\
-c_{i}\sqrt{-1}\varphi_{i}&\overline{\sigma}^{\prime}_{i}&0\\\
c_{i}\sqrt{-1}\overline{\varphi}_{i}&0&\sigma^{\prime}_{i}\\\
\end{bmatrix}\\!\\!\begin{bmatrix}\phi_{i}\\\ \varphi_{i}\\\
\overline{\varphi}_{i}\\\ \end{bmatrix}\\!\\!.$

As $\nabla$ is torsion free, it follows that

$d\\!\begin{bmatrix}\phi_{i}\\\ \varphi_{i}\\\
\overline{\varphi}_{i}\end{bmatrix}=\begin{bmatrix}0&-c_{i}\sqrt{-1}\overline{\varphi}_{i}&c_{i}\sqrt{-1}\varphi_{i}\\\
-c_{i}\sqrt{-1}\varphi_{i}&\overline{\sigma}^{\prime}_{i}&0\\\
c_{i}\sqrt{-1}\overline{\varphi}_{i}&0&\sigma^{\prime}_{i}\\\
\end{bmatrix}\\!\\!\begin{bmatrix}\phi_{i}\\\ \varphi_{i}\\\
\overline{\varphi}_{i}\\\ \end{bmatrix}\\!\\!.$

From Definition 5.8, we may define

$\displaystyle
e_{3}=\frac{1}{\sqrt{2}}(\xi_{1}-\sqrt{-1}J_{\kappa}\xi_{1})=\frac{1}{\sqrt{2}}\big{(}(1-x\sqrt{-1})\xi_{1}-y\sqrt{-1}\xi_{2}\big{)},$
$\displaystyle\varphi_{3}=\frac{1}{\sqrt{2}}(\phi_{1}-\frac{x}{y}\phi_{2}+\frac{\sqrt{-1}}{y}\phi_{2})=\frac{1}{\sqrt{2}}(\phi_{1}-\frac{x-\sqrt{-1}}{y}\phi_{2}).$

It is easy to verify that $(e_{1},e_{2},e_{3})$ is a unitary frame on the
twisted Sasakian product $N_{1}\times_{\kappa}N_{2}$ and
$(\varphi_{1},\varphi_{2},\varphi_{3})$ is its dual coframe. Then it yields
that

$\begin{cases}d\varphi_{1}=\frac{c_{1}}{\sqrt{2}}\big{(}(x+\sqrt{-1})\varphi_{3}-(x-\sqrt{-1})\overline{\varphi}_{3}\big{)}\varphi_{1}+\overline{\sigma}^{\prime}_{1}\varphi_{1},\\\
d\varphi_{2}=\frac{c_{2}y}{\sqrt{2}}(\varphi_{3}-\overline{\varphi}_{3})\varphi_{2}+\overline{\sigma}^{\prime}_{2}\varphi_{2},\\\
d\varphi_{3}=\sqrt{2}c_{1}\sqrt{-1}\varphi_{1}\overline{\varphi}_{1}-\frac{\sqrt{2}c_{2}(1+x\sqrt{-1})}{y}\varphi_{2}\overline{\varphi}_{2}.\end{cases}$

In matrix form we have
$d\varphi=\overline{\theta}_{1}\varphi+\theta_{2}\overline{\varphi}$, where
$\theta_{1}$ is skew Hermitian and $\theta_{2}$ is skew symmetric, given by

$\displaystyle\theta_{1}=\begin{bmatrix}\sigma_{1}&0&\frac{c_{1}\sqrt{-1}}{\sqrt{2}}\overline{\varphi}_{1}\\\
0&\sigma_{2}&\frac{-c_{2}(1+x\sqrt{-1})}{y\sqrt{2}}\overline{\varphi}_{2}\\\
\frac{c_{1}\sqrt{-1}}{\sqrt{2}}\varphi_{1}&\frac{c_{2}(1-x\sqrt{-1})}{y\sqrt{2}}\varphi_{2}&0\\\
\end{bmatrix}\\!\\!,\quad\theta_{2}=\begin{bmatrix}0&0&\frac{-c_{1}\sqrt{-1}}{\sqrt{2}}\varphi_{1}\\\
0&0&\frac{c_{2}(1+x\sqrt{-1})}{y\sqrt{2}}\varphi_{2}\\\
\frac{c_{1}\sqrt{-1}}{\sqrt{2}}\varphi_{1}&\frac{-c_{2}(1+x\sqrt{-1})}{\sqrt{2}}\varphi_{2}&0\\\
\end{bmatrix}\\!\\!,$
$\displaystyle\sigma_{1}=\sigma_{1}^{\prime}-\frac{c_{1}x}{\sqrt{2}}(\varphi_{3}-\overline{\varphi}_{3}),\qquad\sigma_{2}=\sigma_{2}^{\prime}-\frac{c_{2}}{\sqrt{2}}\big{(}(y-\frac{1-x\sqrt{-1}}{y})\varphi_{3}-(y-\frac{1+x\sqrt{-1}}{y})\overline{\varphi}_{3}\big{)}.$

Note that $\nabla e=\theta_{1}e+\overline{\theta}_{2}\overline{e}$,
$\xi_{1}=\frac{e_{3}+\overline{e}_{3}}{\sqrt{2}}$ and
$\sqrt{-1}\xi_{2}=\frac{-(1+x\sqrt{-1})}{y\sqrt{2}}e_{3}+\frac{(1-x\sqrt{-1})}{y\sqrt{2}}\overline{e}_{3}$.
Then the equation (5.2) in the lemma is established. ∎

###### Proof of Theorem 1.7:.

It follows, from Proposition 1.6 and its proof, that admissible frames,
denoted by $e$, can be established on the non-balanced BTP threefold, and the
connection matrix $\theta^{b}$ of Bismut connection $\nabla^{b}$ under an
admissible frame $e$ satisfies $\theta^{b}_{3\ast}=\theta^{b}_{\ast 3}=0$.
Then we may write $\lambda_{1}=\lambda a$ and $\lambda_{2}=\lambda b$. As the
admissible frame is applied and the torsion is degenerate when $n=3$ by
Proposition 1.6, it yields that the only possibly nonzero components of the
Chern torsion are

$T_{13}^{1}=a,\quad T^{2}_{23}=b,\quad\mbox{with}\ \ a+b=\lambda.$

In case (1), clearly $a,b\neq 0$ and $a\neq b$. This implies that the
connection matrix $\theta^{b}$ is diagonal under $e$, denoted by

$\theta^{b}=\begin{bmatrix}\psi_{1}&0&0\\\ 0&\psi_{2}&0\\\ 0&0&0\\\
\end{bmatrix}\\!\\!,$

where $\overline{\psi}_{1}=-\psi_{1}$ and $\overline{\psi}_{2}=-\psi_{2}$.
Since the connection matrix of the Riemannian connection $\nabla$ is given by
$\nabla e=\theta_{1}e+\overline{\theta}_{2}\overline{e}$, where
$\theta_{1}=\theta^{b}-\gamma$ and $\gamma,\theta_{2}$ are determined by the
Chern torsion $T$, we have

(5.3) $\begin{cases}\nabla
e_{1}=\overline{\varphi}_{1}\,(a\overline{e}_{3}-\overline{a}e_{3})+\sigma_{1}\,e_{1}\\\
\nabla
e_{2}=\overline{\varphi}_{2}\,(b\overline{e}_{3}-\overline{b}\overline{e}_{3})+\sigma_{2}\,e_{2}\\\
\nabla
e_{3}=a(\varphi_{1}\,e_{1}-\overline{\varphi}_{1}\,\overline{e}_{1})+b(\varphi_{2}\,e_{2}-\overline{\varphi}_{2}\,\overline{e}_{2}),\end{cases}$

where $\varphi$ is the dual coframe of $e$, and
$\sigma_{1}=\psi_{1}-a\varphi_{3}+\overline{a}\overline{\varphi}_{3}$,
$\sigma_{2}=\psi_{2}-b\varphi_{3}+\overline{b}\overline{\varphi}_{3}$. Then we
may define two global real vector fields of unit length by

$\xi_{1}=\frac{-\sqrt{-1}}{\sqrt{2}|a|}(a\overline{e}_{3}-\overline{a}e_{3}),\quad\xi_{2}=\frac{-\sqrt{-1}}{\sqrt{2}|b|}(b\overline{e}_{3}-\overline{b}e_{3}),$

where it is easy to verify from the condition of the case (1) that
$\mbox{span}_{\mathbb{C}}\\{e_{3},\overline{e}_{3}\\}=\mbox{span}_{\mathbb{C}}\\{\xi_{1},\xi_{2}\\}$.
It follows that

$e_{3}=\frac{\sqrt{2}|a|b\sqrt{-1}}{a\overline{b}-\overline{a}b}\xi_{1}-\frac{\sqrt{2}|b|a\sqrt{-1}}{a\overline{b}-\overline{a}b}\xi_{2},\quad
J\xi_{1}=\frac{-(a\overline{b}+\overline{a}b)\sqrt{-1}}{a\overline{b}-\overline{a}b}\xi_{1}+\frac{2|a||b|\sqrt{-1}}{a\overline{b}-\overline{a}b}\xi_{2}.$

Let us define

$\displaystyle
x=\frac{-(a\overline{b}+\overline{a}b)\sqrt{-1}}{a\overline{b}-\overline{a}b},\quad
y=\frac{2|a||b|\sqrt{-1}}{a\overline{b}-\overline{a}b},\quad\kappa=x+y\sqrt{-1},$
$\displaystyle
e^{\prime}_{3}=\frac{1}{\sqrt{2}}(\xi_{1}-\sqrt{-1}J\xi_{1})=\frac{1}{\sqrt{2}}\big{(}(1-x\sqrt{-1})\xi_{1}-y\sqrt{-1}\xi_{2}\big{)}.$

It is easy to see that $x,y$ are real constants and
$e^{\prime}_{3}=\frac{\sqrt{-1}\overline{a}}{|a|}e_{3}$. Then the first two
equations of (5.3) enable us to contruct two 3-dimensional Sasakian manifolds
$N_{1}$ and $N_{2}$ by the following two structure equations respectively,

$\nabla\\!\begin{bmatrix}\xi_{1}\\\ e_{1}\\\
\overline{e}_{1}\end{bmatrix}=\begin{bmatrix}0&\sqrt{2}|a|\sqrt{-1}\varphi_{1}&-\sqrt{2}|a|\sqrt{-1}\overline{\varphi}_{1}\\\
\sqrt{2}|a|\sqrt{-1}\overline{\varphi}_{1}&\sigma^{\prime}_{1}&0\\\
-\sqrt{2}|a|\sqrt{-1}\varphi_{1}&0&\overline{\sigma}^{\prime}_{1}\\\
\end{bmatrix}\\!\\!\begin{bmatrix}\xi_{1}\\\ e_{1}\\\ \overline{e}_{1}\\\
\end{bmatrix}\\!\\!,$ $\nabla\\!\begin{bmatrix}\xi_{2}\\\ e_{2}\\\
\overline{e}_{2}\end{bmatrix}=\begin{bmatrix}0&\sqrt{2}|b|\sqrt{-1}\varphi_{2}&-\sqrt{2}|b|\sqrt{-1}\overline{\varphi}_{2}\\\
\sqrt{2}|b|\sqrt{-1}\overline{\varphi}_{2}&\sigma^{\prime}_{2}&0\\\
-\sqrt{2}|b|\sqrt{-1}\varphi_{2}&0&\overline{\sigma}^{\prime}_{2}\\\
\end{bmatrix}\\!\\!\begin{bmatrix}\xi_{2}\\\ e_{2}\\\ \overline{e}_{2}\\\
\end{bmatrix}\\!\\!,$

where the dual coframe of $(\xi_{i},e_{i},\overline{e}_{i})$ is denoted by
$(\phi_{i},\varphi_{i},\overline{\varphi}_{i})$ for $i=1,2$, and we may also
define

$\displaystyle\varphi^{\prime}_{3}=\frac{1}{\sqrt{2}}(\phi_{1}-\frac{x}{y}\phi_{2}+\frac{\sqrt{-1}}{y}\phi_{2})=\frac{1}{\sqrt{2}}(\phi_{1}-\frac{x-\sqrt{-1}}{y}\phi_{2}),$
$\displaystyle\sigma_{1}^{\prime}=\sigma_{1}+x|a|(\varphi^{\prime}_{3}-\overline{\varphi^{\prime}_{3}}),\quad\sigma^{\prime}_{2}=\sigma_{2}+|b|\big{(}(y-\frac{1-x\sqrt{-1}}{y})\varphi^{\prime}_{3}-(y-\frac{1+x\sqrt{-1}}{y})\overline{\varphi^{\prime}_{3}}\big{)}.$

Here $\sigma_{1},\sigma_{2}$ are coefficient forms from the equation (5.3).
Then it follows from Lemma 5.9 that the structure equation of
$N_{1}\times_{\kappa}N_{2}$ is exactly the one (5.3), after we note that
$c_{1}=\sqrt{2}|a|,c_{2}=\sqrt{2}|b|,e^{\prime}_{3}=\frac{\sqrt{-1}\overline{a}}{|a|}e_{3}$
here. Therefore we have shown the case (1).

In case (2), either $a=0$ and $b=\lambda$, or $a=\frac{\lambda}{2}(1+\rho)$,
$b=\frac{\lambda}{2}(1-\rho)$ for some $\rho$ satisfying $|\rho|=1$ and
$\mathrm{Im}(\rho)>0$. It is easy to verify that $P_{ik}^{j\ell}=0$ under any
admissible frame $e$ in this case, thus $(M^{3},g)$ is BKL by Proposition 3.6.

In case (3), both $a$ and $b$ are real nonzero constants with $b=\lambda-a$.
It follows from Proposition 3.6 that $d(\eta+\overline{\eta})=0$ and thus
$(M^{3},g)$ is generalized Vaisman. It is easy to verify that, when
$\lambda_{1}=\lambda_{2}$, i.e. when $a=b=\frac{\lambda}{2}$, it holds that
$d\omega=-(\eta+\overline{\eta})\omega$, which implies that $(M^{3},g)$ is
Vaisman by Proposition 1.4. This completes the proof of Theorem 1.7. ∎

##  6\. Balanced BTP threefolds

Let $(M^{3},g)$ be a compact non-Kähler balanced BTP manifold. We will start
with a technical observation in [36, Proposition 2] which says that any
balanced threefold always admits a particular type of unitary frame under
which the Chern torsion takes a simple form, namely, for any given $p\in
M^{3}$, there exists a unitary frame $e$ near $p$ such that the Chern torsion
components satisfy

(6.1) $T^{1}_{1k}=T^{2}_{2k}=T^{3}_{3k}=0,\ \ \ \forall\ 1\leq k\leq 3.$

Under such a frame, the only possibly non-zero torsion components are

(6.2) $a_{1}=T^{1}_{23},\ \ \ a_{2}=T^{2}_{31},\ \ \ a_{3}=T^{3}_{12}.$

By a rotation and a permutation of $\\{e_{1},e_{2},e_{3}\\}$ if needed, we may
assume that $|a_{1}|\geq|a_{2}|\geq|a_{3}|$.

First we claim that each $|a_{i}|$ is a global constant on $M^{3}$. To see
this, recall that tensor $B$ is defined by
$B_{i\overline{j}}=\sum_{r,s}T^{j}_{rs}\overline{T^{i}_{rs}}$. Under our $e$,
we have

$B=2\begin{bmatrix}|a_{1}|^{2}&0&0\\\ 0&|a_{2}|^{2}&0\\\
0&0&|a_{3}|^{2}\end{bmatrix}\\!\\!.$

By the BTP assumption, we have $\nabla^{b}B=0$, so the eigenvalues of $B$ are
all global constants on $M$, thus we know that each $|a_{i}|$ is a global
constant.

Then one can rotate $e$ to make $a_{i}=|a_{i}|$ for each $i$. To see this,
since $|a_{i}|$ is a constant, we can write $a_{i}=\rho_{i}|a_{i}|$ where
$\rho_{i}$ is a smooth local function with $|\rho_{i}|=1$. We will simply let
$\rho_{i}=1$ if $a_{i}=0$. Replace each $e_{i}$ by
$\tilde{e}_{i}=(\overline{\rho}_{j}\overline{\rho}_{k})^{\frac{1}{2}}e_{i}$,
where $(ijk)$ is a cyclic permutation of $(123)$. Then under the new frame we
will have $a_{i}=|a_{i}|$ for each $i$. An appropriate permutation of
$\\{e_{1},e_{2},e_{3}\\}$ yields

(6.3) $T^{1}_{23}\geq T^{2}_{31}\geq T^{3}_{12}\geq 0.$

We will call a local unitary frame $e$ on $M^{3}$ a special frame if both
(6.1) and (6.3) are satisfied.

Let us fix a special frame $e$ in a neighborhood of $p\in M^{3}$. We have
$a_{1}\geq a_{2}\geq a_{3}\geq 0$, with $a_{1}>0$ since $M^{3}$ is not Kähler.
Denote by $\theta^{b}$ the matrix of the Bismut connection $\nabla^{b}$ under
$e$. Since $\nabla^{b}T=0$, and all $T^{j}_{ik}$ are constants, we have

(6.4)
$0=dT^{j}_{ik}=\sum_{r}\big{(}T^{j}_{rk}\theta^{b}_{ir}+T^{j}_{ir}\theta^{b}_{kr}-T^{r}_{ik}\theta^{b}_{rj}\big{)}$

Since the only possibly non-zero components of $T$ are $a_{1}$, $a_{2}$ and
$a_{3}$, if we take $i$, $j$, $k$ all distinct in (6.4), we get

(6.5)
$\left\\{\begin{array}[]{c}a_{1}(\theta^{b}_{22}+\theta^{b}_{33}-\theta^{b}_{11})=0\\\
a_{2}(\theta^{b}_{11}+\theta^{b}_{33}-\theta^{b}_{22})=0\\\
a_{3}(\theta^{b}_{11}+\theta^{b}_{22}-\theta^{b}_{33})=0\end{array}\right.$

Similarly, by taking $j=i\neq k$ in (6.4), we get

(6.6)
$a_{1}\theta^{b}_{12}+a_{2}\theta^{b}_{21}=a_{1}\theta^{b}_{13}+a_{3}\theta^{b}_{31}=a_{2}\theta^{b}_{23}+a_{3}\theta^{b}_{32}=0.$

Denote by $\varphi$ the unitary coframe dual to the special frame $e$, and by
$\theta$, $\tau$ the matrix of connection and column vector of torsion under
$e$ for the Chern connection $\nabla^{c}$. By [32], we have
$\theta=\theta^{b}-2\gamma$ where

(6.7) $\gamma=\begin{bmatrix}0&-\overline{\psi}_{3}&\psi_{2}\\\
\psi_{3}&0&-\overline{\psi}_{1}\\\
-\overline{\psi}_{2}&\psi_{1}&0\end{bmatrix}\\!\\!,\ \ \ \ \
\tau=\begin{bmatrix}2a_{1}\varphi_{2}\varphi_{3}\\\
2a_{2}\varphi_{3}\varphi_{1}\\\
2a_{3}\varphi_{1}\varphi_{2}\end{bmatrix}\\!\\!,$ (6.8)
$\psi_{1}=a_{2}\varphi_{1}+a_{3}\overline{\varphi}_{1},\ \ \
\psi_{2}=a_{3}\varphi_{2}+a_{1}\overline{\varphi}_{2},\ \ \
\psi_{3}=a_{1}\varphi_{3}+a_{2}\overline{\varphi}_{3}.$

We shall divide the classification into the following three cases:

  1. $(1)$

$a_{1}>a_{2}>a_{3}$;

  2. $(2)$

$a_{1}=a_{2}=a_{3}$;

  3. $(3)$

$a_{1}=a_{2}>a_{3}$ or $a_{1}>a_{2}=a_{3}$,

where the third case contains the middle type and is the main part of the
discussion.

###### Proof of Proposition 1.8.

Case 1: $a_{1}>a_{2}>a_{3}$.

In this case, $B$ has distinct eigenvalues. Since its eigenspaces are all
$\nabla^{b}$-parallel, we know that the matrix $\theta^{b}$ is diagonal.

If $a_{3}>0$, by (6.5) we get $\theta^{b}_{ii}=0$ for each $i$, hence
$\theta^{b}=0$. This means that $M^{3}$ is Bismut flat. Such a manifold cannot
be balanced unless it is Kähler, contradicting to our assumption that $M^{3}$
is balanced and non-Kähler, so we must have $a_{3}=0$, which implies
$\psi_{1}=a_{2}\varphi_{1}$ and $\psi_{2}=a_{1}\overline{\varphi}_{2}$. From
Equation (6.5), we get

$\theta^{b}_{33}=0,\ \ \ \theta^{b}_{11}=\theta^{b}_{22}=\alpha,$

and

$\theta=\theta^{b}-2\gamma=\begin{bmatrix}\alpha&2\overline{\psi}_{3}&-2a_{1}\overline{\varphi}_{2}\\\
-2\psi_{3}&\alpha&2a_{2}\overline{\varphi}_{1}\\\
2a_{1}\varphi_{2}&-2a_{2}\varphi_{1}&0\end{bmatrix}\\!\\!.$

Then the structure equation of Chern connection yields

(6.9)
$d\varphi=-\,^{t}\\!\theta\wedge\varphi+\tau=\begin{bmatrix}-\alpha\varphi_{1}+2\psi_{3}\varphi_{2}\\\
-\alpha\varphi_{2}-2\overline{\psi}_{3}\varphi_{1}\\\
2a_{2}\varphi_{2}\overline{\varphi}_{1}-2a_{1}\varphi_{1}\overline{\varphi}_{2}\end{bmatrix}\\!\\!.$

By the first two equations of (6.9), we get

$\displaystyle d(\varphi_{2}\overline{\varphi}_{1})$ $\displaystyle=$
$\displaystyle d\varphi_{2}\wedge\overline{\varphi}_{1}-\varphi_{2}\wedge
d\overline{\varphi}_{1}$ $\displaystyle=$
$\displaystyle(-\alpha\varphi_{2}-2\overline{\psi}_{3}\varphi_{1})\,\overline{\varphi}_{1}-\varphi_{2}(-\overline{\alpha}\,\overline{\varphi}_{1}+2\overline{\psi}_{3}\overline{\varphi}_{2})$
$\displaystyle=$ $\displaystyle
2\overline{\psi}_{3}\,(\varphi_{2}\,\overline{\varphi}_{2}-\varphi_{1}\,\overline{\varphi}_{1}),$

where $\alpha+\overline{\alpha}=0$ is used. Complex conjugation of the
equality above yields

$d(\varphi_{1}\overline{\varphi}_{2})=-\overline{d(\varphi_{2}\overline{\varphi}_{1})}=2\psi_{3}\,(\varphi_{2}\,\overline{\varphi}_{2}-\varphi_{1}\,\overline{\varphi}_{1}).$

The exterior differentiation of the third equation of (6.9) implies

$0=d^{2}\varphi_{3}=2a_{2}\,d(\varphi_{2}\overline{\varphi}_{1})-2a_{1}\,d(\varphi_{1}\overline{\varphi}_{2})=4(a_{2}\overline{\psi}_{3}-a_{1}\psi_{3})\wedge(\varphi_{2}\,\overline{\varphi}_{2}-\varphi_{1}\,\overline{\varphi}_{1}).$

Note that

$a_{2}\overline{\psi}_{3}-a_{1}\psi_{3}=a_{2}(a_{1}\overline{\varphi}_{3}+a_{2}\varphi_{3})-a_{1}(a_{1}\varphi_{3}+a_{2}\overline{\varphi}_{3})=(a_{2}^{2}-a_{1}^{2})\varphi_{3},$

which yields a contradiction. This shows that the case of distinct
$a_{1},a_{2}$ and $a_{3}$ cannot occur.

Case 2: $a_{1}=a_{2}=a_{3}$.

Let us denote by $a>0$ the common value of those $a_{i}$ in this case.
Equalities (6.5) and (6.6) now imply that $\theta^{b}$ is skew-symmetric.
Since $\psi_{i}=a(\varphi_{i}+\overline{\varphi}_{i})=\overline{\psi}_{i}$ for
each $i$, the equality (6.7) shows that $\gamma$ is also skew-symmetric. So
$\theta=\theta^{b}-2\gamma$ is skew-symmetric. Then it yields that

(6.10) $\theta=\begin{bmatrix}0&x&y\\\ -x&0&z\\\ -y&-z&0\end{bmatrix}\\!\\!,\
\ \ \ \tau=2a\begin{bmatrix}\varphi_{2}\varphi_{3}\\\
\varphi_{3}\varphi_{1}\\\ \varphi_{1}\varphi_{2}\end{bmatrix}\\!\\!,$

where $x$, $y$, $z$ are real $1$-forms. As a result, the Chern curvature
matrix $\Theta=d\theta-\theta\wedge\theta$ is also skew-symmetric. The
structure equation of Chern connection gives us

$d\varphi=-\,^{t}\\!\theta\wedge\varphi+\tau=\begin{bmatrix}x\varphi_{2}+y\varphi_{3}+2a\varphi_{2}\varphi_{3}\\\
-x\varphi_{1}+z\varphi_{3}+2a\varphi_{3}\varphi_{1}\\\
-y\varphi_{1}-z\varphi_{2}+2a\varphi_{1}\varphi_{2}\end{bmatrix}\\!\\!.$

It follows that

$\displaystyle d(\varphi_{2}\varphi_{3})$ $\displaystyle=$
$\displaystyle(x\varphi_{3}-y\varphi_{2})\varphi_{1}$ $\displaystyle
d(\varphi_{3}\varphi_{1})$ $\displaystyle=$
$\displaystyle(x\varphi_{3}+z\varphi_{1})\varphi_{2}$ $\displaystyle
d(\varphi_{2}\varphi_{3})$ $\displaystyle=$
$\displaystyle(-y\varphi_{2}+z\varphi_{1})\varphi_{3}.$

On one hand, if we let $\xi=x\varphi_{3}-y\varphi_{2}+z\varphi_{1}$, then the
above equations simply says $d\tau=2a\,\xi\wedge\varphi$. On the other hand,
by (6.10) we get $\theta\tau=2a\,\xi\wedge\varphi$. So by the first Bianchi
identity

$d\tau=\,^{t}\\!\Theta\varphi-\,^{t}\\!\theta\tau,$

we conclude that $\,{}^{t}\\!\Theta\varphi=0$. This means that the Hermitian
threefold $M^{3}$ is Chern Kähler-like, that is, the Chern curvature tensor
$R^{c}$ obeys the Kähler symmetry
$R^{c}_{i\overline{j}k\overline{\ell}}=R^{c}_{k\overline{j}i\overline{\ell}}$
for any $i,j,k,\ell$.

We claim that $R^{c}=0$. If $\\{i,k\\}\cap\\{j,\ell\\}\neq\emptyset$, for
instance $1$ is contained in the intersection, then by the Kähler symmetry,
$R^{c}_{i\overline{j}k\overline{\ell}}$ can be written as
$R^{c}_{a\overline{b}1\overline{1}}$, which has to vanish since
$\Theta_{11}=0$ as $\Theta$ is skew-symmetric. When
$\\{i,k\\}\cap\\{j,\ell\\}=\emptyset$, what we need to show are the equalities
$R_{i\overline{j}k\overline{j}}=0$ and $R_{i\overline{j}i\overline{j}}=0$
where $i,j,k$ are distinct, as the dimension is $3$. From the skew-symmetric
$\Theta$, it yields that

$R^{c}_{i\overline{j}k\overline{j}}=-R^{c}_{i\overline{j}j\overline{k}}=-R^{c}_{i\overline{k}j\overline{j}}=0,$
$R^{c}_{i\overline{j}i\overline{j}}=-R^{c}_{i\overline{j}j\overline{i}}=-R^{c}_{i\overline{i}j\overline{j}}=0.$

So $M^{3}$ is Chern flat. By [8], we know that the universal cover of $M$ is a
connected, simply-connected complex Lie group $G$, and the lifting metric
$\tilde{g}$ of $g$ is left-invariant and compatible with the complex structure
of $G$. Let $\\{\varepsilon_{1},\varepsilon_{2},\varepsilon_{3}\\}$ be a left-
invariant unitary frame of $G$. Denote by $T^{j}_{ik}$ the components of the
Chern torsion under the frame $\varepsilon$, which are all constants. As $M$
is three-dimensional and balanced, by the proof of [36, Proposition 2], we
know that one can make a constant unitary change of $\varepsilon$ so that
$T^{i}_{ij}=0$ for each $1\leq i\leq 3$. Under the case assumption, the tensor
$B$

$B=\begin{bmatrix}c&0&0\\\ 0&c&0\\\ 0&0&c\end{bmatrix}\\!\\!,$

where $c=2a^{2}$. Similarly, by a suitable constant rotation
$\rho_{i}\varepsilon_{i}$, where $|\rho_{i}|=1$ for each $i$, we may assume
that $T^{1}_{23}=T^{2}_{31}=T^{3}_{12}=a>0$ under the left-invariant unitary
frame $\varepsilon$ of $G$, hence the Lie algebra of $G$ is
${\mathfrak{g}}={\mathbb{C}}\\{X,Y,Z\\}$, satisfying

$[X,Y]=2aZ,\ \ \ [Y,Z]=2aX,\ \ \ [Z,X]=2aY.$

Therefore $G$ is isomorphic to $SO(3,{\mathbb{C}})$. Quotients of
$SO(3,{\mathbb{C}})$ give us the only non-Kähler balanced BTP threefolds that
are Chern flat.

Note that since $\theta^{b}$ is skew-symmetric under a special frame, so is
the curvature matrix $\Theta^{b}$, therefore the holonomy group of
$\nabla^{b}$ is contained in $SO(3)\subseteq U(3)$, but unlike the case of
non-balanced BTP manifolds, it is not contained in $U(n-1)\times 1$ here.

Case 3: $a_{1}=a_{2}>a_{3}$ or $a_{1}>a_{2}=a_{3}$ .

In this case $B$ has two distinct eigenvalues. First we will rule out the
possibility of $a_{3}>0$. Assume that $a_{3}>0$, namely, either
$a_{1}=a_{2}>a_{3}>0$ or $a_{1}>a_{2}=a_{3}>0$, we will derive at a
contradiction.

Since the argument for these two situations are exactly analogous, we will
just focus on the case $a_{1}=a_{2}>a_{3}>0$. Write $a_{1}=a_{2}=a$ for
simplicity. Since the eigenspaces of $B$ are $\nabla^{b}$-parallel, it follows
that $\theta^{b}_{13}=\theta^{b}_{23}=0$. By (6.5) and (6.6), it yields that

$\theta^{b}=\begin{bmatrix}0&\alpha&0\\\ -\alpha&0&0\\\
0&0&0\end{bmatrix}\\!\\!,$

where $\overline{\alpha}=\alpha$. Write $\alpha^{\prime}=\alpha+2\psi_{3}$,
where $\psi_{3}=a(\varphi_{3}+\overline{\varphi}_{3})$ is real, and the
structure equation of Chern connection amounts to

(6.11)
$d\varphi=-\,^{t}\\!(\theta^{b}-2\gamma)\varphi+\tau=\begin{bmatrix}\alpha^{\prime}\varphi_{2}+2a_{3}\varphi_{3}\overline{\varphi}_{2}\\\
-\alpha^{\prime}\varphi_{1}-2a_{3}\varphi_{3}\overline{\varphi}_{1}\\\
-2a_{3}\varphi_{1}\varphi_{2}+2a(\varphi_{2}\overline{\varphi}_{1}-\varphi_{1}\overline{\varphi}_{2})\end{bmatrix}\\!\\!.$

Then we get, from the first two lines,

$\displaystyle d(\varphi_{1}\varphi_{2})$ $\displaystyle=$ $\displaystyle
d\varphi_{1}\wedge\varphi_{2}-\varphi_{1}\wedge d\varphi_{2}\ =\
-2a_{3}\varphi_{3}(\varphi_{1}\overline{\varphi}_{1}+\varphi_{2}\overline{\varphi}_{2}),$
$\displaystyle d(\varphi_{1}\overline{\varphi}_{2})$ $\displaystyle=$
$\displaystyle d\varphi_{1}\wedge\overline{\varphi}_{2}-\varphi_{1}\wedge
d\overline{\varphi}_{2}\ =\
\alpha^{\prime}(\varphi_{2}\overline{\varphi}_{2}-\varphi_{1}\overline{\varphi}_{1}).$

Here the fact $\overline{\alpha^{\prime}}=\alpha^{\prime}$ is used. In
particular,

$d(\varphi_{2}\overline{\varphi}_{1})=-\overline{d(\varphi_{1}\overline{\varphi}_{2})}=\alpha^{\prime}(\varphi_{2}\overline{\varphi}_{2}-\varphi_{1}\overline{\varphi}_{1})=d(\varphi_{1}\overline{\varphi}_{2}).$

Take the exterior differentiation of the third equation of (6.11) and we
obtain

$0=d^{2}\varphi_{3}=-2a_{3}d(\varphi_{1}\varphi_{2})+2a\,d(\varphi_{2}\overline{\varphi}_{1}-\varphi_{1}\overline{\varphi}_{2})=4a_{3}^{2}\varphi_{3}(\varphi_{1}\overline{\varphi}_{1}+\varphi_{2}\overline{\varphi}_{2}),$

which is a contradiction. This shows that the case $a_{3}>0$ cannot occur and
either $a_{1}>0=a_{2}=a_{3}$ or $a_{1}=a_{2}>0=a_{3}$ yields, which implies

$B=\begin{bmatrix}c&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}\ \text{or}\
\begin{bmatrix}c&0&0\\\ 0&c&0\\\ 0&0&0\end{bmatrix}\\!\\!,$

where $c=2a_{1}^{2}$. This completes the proof of Proposition 1.8. ∎

Therefore, we have shown the $\mbox{rank}\,B=3$ case of Theorem 1.10. We will
deal with the case $\mathrm{rank}\,B=1$ in §7, which leads to a Fano threefold
and eventually end up with the Wallach threefold $(X,g)$. It will be computed
in details in §8. The case $\mathrm{rank}\,B=2$ is the middle type, which will
be discussed in §9. This will complete the proof of Theorem 1.10.

##  7\. The Fano case

In this section, we deal with the case $\mathrm{rank}\,B=1$. Eventually, we
will show that this leads us to a unique example, the Wallach threefold, up to
a scaling of the metric by a constant.

Let $(M^{3},g)$ be a compact, non-Kähler, balanced BTP threefold with $B$
tensor

$B=\begin{bmatrix}2a_{1}^{2}&0&0\\\ 0&0&0\\\ 0&0&0\end{bmatrix}\\!\\!,$

under a special frame $e$, where the only non-zero component of the Chern
torsion tensor is $a_{1}=T^{1}_{23}>0$. We may assume for simplicity that
$a_{1}=\frac{1}{2}$, after a scaling of the metric $g$ by a suitable constant.

From (6.5) and (6.6) we get $\theta^{b}_{12}=\theta^{b}_{13}=0$ and
$\theta^{b}_{11}=\theta^{b}_{22}+\theta^{b}_{33}$. Then it follows that

$\theta^{b}=\begin{bmatrix}x+y&0&0\\\ 0&x&\alpha\\\
0&-\overline{\alpha}&y\end{bmatrix}\\!\\!,\ \ \ \
\gamma=\frac{1}{2}\begin{bmatrix}0&-\overline{\varphi}_{3}&\overline{\varphi}_{2}\\\
\varphi_{3}&0&0\\\ -\varphi_{2}&0&0\end{bmatrix}\\!\\!,\ \ \
\tau=\begin{bmatrix}\varphi_{2}\varphi_{3}\\\ 0\\\ 0\end{bmatrix}\\!\\!,$

where $\overline{x}=-x$ and $\overline{y}=-y$. From
$\theta=\theta^{b}-2\gamma$ and $d\varphi=-\,^{t}\\!\theta\wedge\varphi+\tau$,
it yields that

$\theta=\begin{bmatrix}x+y&\overline{\varphi}_{3}&-\overline{\varphi}_{2}\\\
-\varphi_{3}&x&\alpha\\\ \varphi_{2}&-\overline{\alpha}&y\end{bmatrix}\\!\\!,\
\ \ \ d\varphi=\begin{bmatrix}-(x+y)\varphi_{1}-\varphi_{2}\varphi_{3}\\\
\varphi_{1}\overline{\varphi}_{3}-x\varphi_{2}+\overline{\alpha}\varphi_{3}\\\
-\varphi_{1}\overline{\varphi}_{2}-\alpha\varphi_{2}-y\varphi_{3}\end{bmatrix}\\!\\!.$

The matrix of the curvature of $\nabla^{b}$ is
$\Theta^{b}=d\theta^{b}-\theta^{b}\wedge\theta^{b}$, whose entries are

(7.1) $\left\\{\begin{array}[]{lll}\Theta^{b}_{22}\ =\
dx+\alpha\overline{\alpha},\ \ \ \ \ \Theta^{b}_{33}\ =\
dy-\alpha\overline{\alpha},\\\ \Theta^{b}_{23}\ =\ d\alpha-x\alpha-\alpha y,\
\ \ \Theta^{b}_{12}\ =\ \Theta^{b}_{13}\ =\ 0,\\\ \Theta^{b}_{11}\ =\ dx+dy\
=\ \Theta^{b}_{22}+\Theta^{b}_{33}.\end{array}\right.$

For convenience, we will use $\varphi_{ij}$ and $\varphi_{i\bar{j}}$ as the
abbreviation of $\varphi_{i}\wedge\varphi_{j}$ and
$\varphi_{i}\wedge\overline{\varphi}_{j}$, respectively. From the exterior
differentiation $d\varphi$ of $\varphi$, the first Bianchi identity of
$\nabla^{b}$ amounts to

(7.2) $\left\\{\begin{array}[]{lll}0=d^{2}\varphi_{1}\ =\
\\{\varphi_{2\overline{2}}+\varphi_{3\overline{3}}-\Theta^{b}_{11}\\}\wedge\varphi_{1},\\\
0=d^{2}\varphi_{2}\ =\
\\{\varphi_{1\overline{1}}-\varphi_{3\overline{3}}-\Theta^{b}_{22}\\}\wedge\varphi_{2}-\Theta^{b}_{32}\wedge\varphi_{3},\\\
0=d^{2}\varphi_{3}\ =\
\\{\varphi_{1\overline{1}}-\varphi_{2\overline{2}}-\Theta^{b}_{33}\\}\wedge\varphi_{3}-\Theta^{b}_{23}\wedge\varphi_{2}\end{array}\right.$

It follows from Theorem 1.1 that $\nabla^{b}T=0$ implies

$R^{b}_{ijk\overline{\ell}}=0\ \ \ \mbox{and}\ \ \
R^{b}_{i\overline{j}k\overline{\ell}}=R^{b}_{k\overline{\ell}i\overline{j}}$

for any $i,j,k,\ell$. So by $\Theta^{b}_{12}=\Theta^{b}_{13}=0$ and
$\Theta^{b}_{11}=\Theta^{b}_{22}+\Theta^{b}_{33}$, we get

$R^{b}_{1\overline{b}i\overline{j}}=R^{b}_{i\overline{j}1\overline{b}}=0,\ \ \
\ \
R^{b}_{1\overline{1}i\overline{j}}=R^{b}_{i\overline{j}1\overline{1}}=R^{b}_{i\overline{j}2\overline{2}}+R^{b}_{i\overline{j}3\overline{3}}=R^{b}_{2\overline{2}i\overline{j}}+R^{b}_{3\overline{3}i\overline{j}}$

for any $i$, $j$ and any $b\in\\{2,3\\}$. Write

$\displaystyle\Theta^{b}_{22}$ $\displaystyle=$ $\displaystyle
A\varphi_{2\overline{2}}+B\varphi_{3\overline{3}}+E\varphi_{2\overline{3}}+\overline{E}\varphi_{3\overline{2}}+(A+B)\varphi_{1\overline{1}}$
$\displaystyle\Theta^{b}_{33}$ $\displaystyle=$ $\displaystyle
B\varphi_{2\overline{2}}+C\varphi_{3\overline{3}}+F\varphi_{2\overline{3}}+\overline{F}\varphi_{3\overline{2}}+(B+C)\varphi_{1\overline{1}}$
$\displaystyle\Theta^{b}_{23}$ $\displaystyle=$ $\displaystyle
E\varphi_{2\overline{2}}+F\varphi_{3\overline{3}}+D\varphi_{2\overline{3}}+G\varphi_{3\overline{2}}+(E+F)\varphi_{1\overline{1}}$
$\displaystyle\Theta^{b}_{11}$ $\displaystyle=$
$\displaystyle(A+B)\varphi_{2\overline{2}}+(B+C)\varphi_{3\overline{3}}+(E+F)\varphi_{2\overline{3}}+(\overline{E}+\overline{F})\varphi_{3\overline{2}}+(A+2B+C)\varphi_{1\overline{1}}$

where $A,B,C,G$ are local real smooth functions. Then the first Bianchi
identity (7.2) indicates

$E+F=0,\ \ \ \ A+B=B+C=G-B=1.$

In particular,

(7.3)
$\Theta^{b}_{11}=2\varphi_{1\overline{1}}+\varphi_{2\overline{2}}+\varphi_{3\overline{3}}$

######  Remark 7.1.

The pattern of the Bismut curvature implies that the holonomy group of
$\nabla^{b}$ in this case would have its Lie algebra contained in the
subalgebra

${\mathfrak{h}}=\left\\{\begin{bmatrix}\mbox{tr}(X)&0\\\ 0&X\end{bmatrix}\mid
X\in{\mathfrak{u}}(2)\right\\}\subseteq{\mathfrak{u}}(3).$

Denote by $\Theta$ the curvature matrix of the Chern connection $\nabla^{c}$
under $e$, and the balanced condition $\eta=0$ implies that

$\mathrm{tr}\,\Theta=\mathrm{tr}\,\Theta^{b}=2\Theta^{b}_{11}.$

Let
$\omega=\sqrt{-1}\big{(}\varphi_{1}\overline{\varphi}_{1}+\varphi_{2}\overline{\varphi}_{2}+\varphi_{3}\overline{\varphi}_{3}\big{)}$
be the Kähler form of the Hermitian metric $g$, and let

(7.4)
$\tilde{\omega}=\sqrt{-1}\big{(}2\varphi_{1}\overline{\varphi}_{1}+\varphi_{2}\overline{\varphi}_{2}+\varphi_{3}\overline{\varphi}_{3}\big{)}.$

be the Kähler form of another Hermitian metric $\tilde{g}$. Clearly
$\tilde{\omega}$ is independent of the choice of special frames hence is
globally defined on $M$. The Chern Ricci form of $g$ is
$\mathrm{Ric}(\omega)=\sqrt{-1}\mathrm{tr}\,\Theta=2\tilde{\omega}$. As
$\mathrm{Ric}(\omega)$ is always closed, we know $\tilde{\omega}$ is Kähler.
Note that $\tilde{\omega}^{3}=2\omega^{3}$ and thus

(7.5) $\mbox{Ric}(\tilde{\omega})=\mbox{Ric}(\omega)=2\tilde{\omega}.$

Therefor $\tilde{\omega}$ is a Kähler-Einstein metric with positive Ricci
curvature and $M^{3}$ is a Fano threefold.

Denote by $E$ and $F$ the $C^{\infty}$ complex vector bundle on $M$ with
fibers $E_{x}={\mathbb{C}}\\{e_{2}(x),e_{3}(x)\\}$ and
$F_{x}={\mathbb{C}}e_{1}(x)$, respectively. They are both globally defined
since $E$ is the eigenspace of $B$ corresponding to the eigenvalue $0$, and
$F$ is the orthogonal complement of $E$ in $T^{1,0}M$.

We claim that $E$ is a holomorphic subbundle of $T^{1,0}M$. To see this, for
any $i\in\\{2,3\\}$ and any $j$, we have

$\langle\nabla^{c}_{\overline{j}}e_{i},\overline{e}_{1}\rangle=\langle\nabla^{b}_{\overline{j}}e_{i}-2\gamma_{\overline{j}}e_{i},\overline{e}_{1}\rangle=\theta^{b}_{i1}(\overline{e}_{j})-2\gamma_{i1}(\overline{e}_{j})=0.$

This means that $\nabla^{c}_{\overline{X}}E\subseteq E$ for any type $(1,0)$
vector field $X$, so $E$ is holomorphic. Note that the distribution $E$ is not
a foliation, while $F$ on the other hand is a foliation, but is not
holomorphic.

Equip $E$ with the restriction metric from $(M^{3},g)$. By the formula of
Chern connection matrix $\theta$ on $M$, the matrices of connection and
curvature of the Hermitian bundle $E$ under the local frame
$\\{e_{2},e_{3}\\}$ are respectively

$\theta^{E}=\begin{bmatrix}x&\alpha\\\
-\overline{\alpha}&y\end{bmatrix}\\!\\!,\ \ \ \ \
\Theta^{E}=d\theta^{E}-\theta^{E}\wedge\theta^{E}=\begin{bmatrix}\Theta^{b}_{22}&\Theta^{b}_{23}\\\
\Theta^{b}_{32}&\Theta^{b}_{33}\end{bmatrix}\\!\\!.$

In particular,
$\sqrt{-1}\mathrm{tr}\Theta^{E}=\sqrt{-1}(\Theta^{b}_{22}+\Theta^{b}_{33})=\frac{\sqrt{-1}}{2}\mathrm{tr}\,\Theta^{b}=\frac{\sqrt{-1}}{2}\,\mathrm{tr}\Theta=\tilde{\omega}$.
This means that

(7.6) $c_{1}(E)=\frac{1}{2}c_{1}(M)=[\tilde{\omega}].$

Denote by $L$ the holomorphic line bundle on $M$ which is the quotient of $E$
in $T^{1,0}M$, namely, the exact sequence follows

(7.7) $0\rightarrow E\rightarrow T^{1,0}M\rightarrow L\rightarrow 0.$

Let $h$ be the first Chern class $c_{1}(L)$ of $L$. The above short exact
sequence implies

$c_{1}(E)+h=c_{1}(M),\ \ \ c_{2}(E)+hc_{1}(E)=c_{2}(M),\ \ \
c_{2}(E)h=c_{3}(M).$

Then the equality (7.6) yields

$c_{1}(E)=h,\quad c_{1}(M)=2h,\quad c_{2}(E)=c_{2}(M)-h^{2},\quad
c_{2}(M)h-h^{3}=c_{3}(M).$

In particular, $L$ is an ample line bundle on $M$, the anti-canonical line
bundle $K_{M}^{-1}=2L$ as holomorphic line bundles are uniquely determined by
their Chern classes on Fano manifolds, and the Chern numbers of $M^{3}$
satisfy

(7.8) $c_{1}(M)c_{2}(M)=2h(c_{2}(E)+h^{2})=2c_{3}(M)+\frac{1}{4}c_{1}^{3}(M).$

Recall that the index of a Fano manifold $X^{n}$ is the largest positive
integer $r$ so that $K^{-1}_{X}=rA$ for an ample line bundle $A$. It is
necessarily less than or equal to $n+1$, where $r=n+1$ if and only if
$X={\mathbb{P}}^{n}$ and $r=n$ if and only if $X={\mathbb{Q}}^{n}$, the smooth
quadric hypersurface in ${\mathbb{P}}^{n+1}$. Fano manifolds satisfying
$r=n-1$ are called del Pezzo manifolds, which are classified by Fujita [14] as
one of the following seven types, according to their degree $d$, which is the
self intersection number $A^{n}$:

  1. $(1)$

$d=1$: $X^{n}_{6}\subset{\mathbb{P}}(1^{n-1},2,3)$, a degree $6$ hypersurface
in the weighted projective space.

  2. $(2)$

$d=2$: $X^{n}_{4}\subset{\mathbb{P}}(1^{n},2)$, a degree $4$ hypersurface in
the weighted projective space.

  3. $(3)$

$d=3$: $X^{n}_{3}\subset{\mathbb{P}}^{n+1}$, a cubic hypersurface.

  4. $(4)$

$d=4$: $X^{n}_{2,2}\subset{\mathbb{P}}^{n+2}$, a complete intersection of two
quadrics.

  5. $(5)$

$d=5$: $Y^{n}$, a linear section of
${\mathbb{G}r}(2,5)\subset{\mathbb{P}}^{9}$.

  6. $(6)$

$d=6$:
${\mathbb{P}}^{1}\\!\times\\!{\mathbb{P}}^{1}\\!\times\\!{\mathbb{P}}^{1}$, or
${\mathbb{P}}^{2}\\!\times\\!{\mathbb{P}}^{2}$, or the flag threefold
${\mathbb{P}}(T_{\\!{\mathbb{P}}^{2}})$.

  7. $(7)$

$d=7$: ${\mathbb{P}}^{3}\\#\overline{{\mathbb{P}}^{3}}$, the blow-up of
${\mathbb{P}}^{3}$ at a point.

For $n=3$, del Pezzo threefolds were classified by Iskovskikh [16] earlier,
and in Table 12.2 of [17] we can find the third betti number $b_{3}$, hence
the Euler number $c_{3}=4-b_{3}$ of del Pezzo threefolds of degree $1\leq
d\leq 5$:

(7.9) $c_{3}(X^{3}_{6})=-38,\ \ \ c_{3}(X^{3}_{4})=-16,\ \ \
c_{3}(X^{3}_{3})=-6,\ \ \ c_{3}(X^{3}_{2,2})=0,\ \ \ c_{3}(Y^{3})=4.$

Let us return to our manifold $M^{3}$, where it holds that $K_{M}^{-1}=2L$ for
the ample $L$, so the index of $M^{3}$ is either $4$ or $2$, which means
$M^{3}$ is biholomorphic to either ${\mathbb{P}}^{3}$ or a del Pezzo
threefold. It is well-known that $c_{1}c_{2}=24$ holds for any Fano threefold,
so the equality (7.8) implies

(7.10) $c_{3}(M)=12-\frac{1}{8}c_{1}^{3}.$

If $M^{3}$ is a del Pezzo threefold of degree $d$, then $c_{1}^{3}=8d$, hence
the equality (7.10) yields $c_{3}=12-d$. This rules out the possibility of
$1\leq d\leq 5$ by (7.9). The case
${\mathbb{P}}^{3}\\#\overline{{\mathbb{P}}^{3}}$ of degree $d=7$ has the Euler
number $c_{3}=6$, which is not equal to $12-7$. Similarly, for the case
${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ of degree
$d=6$, its Euler number $c_{3}=8$ is not equal to $12-6$, which indicates that
neither can be our $M^{3}$. Therefore only two possibilities are left, namely,
$M^{3}$ is either the flag threefold ${\mathbb{P}}(T_{{\mathbb{P}}^{2}})$ or
${\mathbb{P}}^{3}$. Note that in both of these two cases the short exact
sequence (7.7) exists, for which we have to dig further.

Consider the short exact sequence (7.7) of holomorphic vector bundles on our
threefold $M^{3}$, where $E$ has the fiber ${\mathbb{C}}\\{e_{2},e_{3}\\}$
under a special frame $e$. Denote by $\Omega$ the holomorphic cotangent bundle
of $M^{3}$ and write $\Omega(L)=\Omega\otimes L$. Let $\xi\in
H^{0}(M,\Omega(L))$ be the nowhere zero holomorphic section which gives the
map $T^{1,0}M\rightarrow L$ in (7.7). The Kähler-Einstein metric
$\tilde{\omega}$ on $M^{3}$ naturally induces Hermitian metrics on $\Omega$
and $L=-\frac{1}{2}K_{M}$, hence on $\Omega(L)$. We have the following claim

######  Claim 7.2.

The norm $\parallel\\!\xi\\!\parallel^{2}$ is a constant under the Kähler-
Einstein metric $\tilde{\omega}$.

###### Proof.

To see this, let $e$ be a local special frame of $(M^{3},g)$, with dual
coframe $\varphi$. Let $s$ be a local holomorphic frame of $L$. Since $\xi$ is
a $L$-valued $1$-form, we have $\xi=\psi\otimes s$ where $\psi$ is a nowhere-
zero local holomorphic $1$-form. The kernel of the map given by $\xi$ is $E$,
so $\psi=f\varphi_{1}$ for some local smooth function $f$, which is nowhere
zero. By the structure equation, it follows that
$d\varphi_{1}=-(x+y)\varphi_{1}-\varphi_{2}\varphi_{3}$, so the holomorphicity
of $\psi$ gives us

$0=\overline{\partial}\psi=\overline{\partial}f\wedge\varphi_{1}-f(x+y)^{0,1}\wedge\varphi_{1}.$

Hence $(x+y)^{0,1}=\overline{\partial}\log f$. By the fact that
$\overline{x}=-x$ and $\overline{y}=-y$, we get

$x+y=-\partial\log\overline{f}+\overline{\partial}\log f,\ \ \ \
\frac{1}{\sqrt{-1}}\tilde{\omega}=\Theta^{b}_{11}=d(x+y)=\partial\overline{\partial}\log|f|^{2}.$

On the other hand, $L=-\frac{1}{2}K_{M}$ is equipped with the induced metric
from $\tilde{\omega}$, then it yields

$\frac{1}{\sqrt{-1}}\tilde{\omega}=\Theta^{L}=-\partial\overline{\partial}\log\parallel\\!s\\!\parallel^{2}.$

Combine the above two equations and we get

$\partial\overline{\partial}\log(|f|^{2}\parallel\\!s\\!\parallel^{2})=0.$

Since $\\{\sqrt{2}\varphi_{1},\varphi_{2},\varphi_{3}\\}$ is a local unitary
coframe for $\tilde{\omega}$, the norm square of $\varphi_{1}$ under
$\tilde{\omega}$ is $\frac{1}{2}$, hence

$\parallel\\!\xi\\!\parallel^{2}=\frac{1}{2}\,|f|^{2}\parallel\\!s\\!\parallel^{2}.$

It is a global positive function on $M^{3}$, and its log is pluriclosed by the
above equation, hence it must be a constant. This establishes the claim above.
∎

Then we will use Claim 7.2 to rule out the possibility of $M^{3}$ being
${\mathbb{P}}^{3}$. Assume that $M^{3}$ is ${\mathbb{P}}^{3}$. In this case
$L={\mathcal{O}}(2)$. Any nowhere-zero holomorphic section $\xi\in
V:=H^{0}({\mathbb{P}}^{3},\Omega(2))$ gives the so-called null-correlation
bundle (see for example [24]), which is $E(-1)$ in our notation. Let
$\tilde{\omega}$ be the (scaled) Fubini-Study metric of ${\mathbb{P}}^{3}$
with Ricci curvature $2$. It has constant holomorphic sectional curvature $1$.
Let $[Z_{0}\\!:\\!Z_{1}\\!:\\!Z_{2}\\!:\\!Z_{3}]$ be the standard unitary
homogeneous coordinate of ${\mathbb{P}}^{3}$. In the coordinate neighborhood
$U_{0}=\\{Z_{0}\neq 0\\}$, let $z_{i}=\frac{Z_{i}}{Z_{0}}$, $1\leq i\leq 3$
and it follows that

$\sqrt{-1}\,\Theta^{L}=\tilde{\omega}=\frac{1}{2}\mbox{Ric}(\tilde{\omega})=2\sqrt{-1}\,\partial\overline{\partial}\log(1+|z|^{2}),\
\ \ \ \ \ \parallel\\!Z_{0}^{2}\\!\parallel^{2}=\frac{1}{(1+|z|^{2})^{2}}$

where $Z_{0}^{2}$ is a local frame of $L$ in $U_{0}$ and
$|z|^{2}=|z_{1}|^{2}+|z_{2}|^{2}+|z_{3}|^{2}$. Under the coordinate $z$,

$\tilde{g}_{i\overline{j}}=\frac{2}{1+|z|^{2}}\delta_{ij}-\frac{2}{(1+|z|^{2})^{2}}\overline{z}_{i}z_{j},\
\ \ \ \ \
\tilde{g}^{\overline{i}j}=\frac{1}{2}(1+|z|^{2})\big{(}\delta_{ij}+\overline{z}_{i}z_{j}\big{)}.$

As is well-known, $V\cong{\mathbb{C}}^{6}$ has a basis
$\\{\lambda_{ij}\\}_{0\leq i<j\leq 3}$, where
$\lambda_{ij}=Z_{i}dZ_{j}-Z_{j}dZ_{i}$. Suppose that $\xi\in V$ is a nowhere
zero section. It follows that $\xi=\sum a_{ij}\lambda_{ij}$ for some constants
$a_{ij}$. In $U_{0}$,

$\lambda_{0i}=Z_{0}^{2}dz_{i},\ \ \ \
\lambda_{ij}=Z_{0}^{2}(z_{i}dz_{j}-z_{j}dz_{i})$

for any $i,j>0$, hence
$\xi=Z_{0}^{2}(\ell_{1}dz_{1}+\ell_{2}dz_{2}+\ell_{3}dz_{3})$, where

$\displaystyle\ell_{1}$ $\displaystyle=$ $\displaystyle
a_{01}-a_{12}z_{2}-a_{13}z_{3}$ $\displaystyle\ell_{2}$ $\displaystyle=$