# Hermitian geometry of Lie algebras with abelian ideals of codimension $2$

Yuqin Guo  Yuqin Guo. School of Mathematical Sciences, Chongqing Normal
University, Chongqing 401331, China<EMAIL_ADDRESS>and  Fangyang Zheng
Fangyang Zheng. School of Mathematical Sciences, Chongqing Normal University,
Chongqing 401331, China<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We examine Hermitian metrics on unimodular Lie algebras which contains a
$J$-invariant abelian ideal of codimension two, and give a classification for
all Bismut Kähler-like and all Bismut torsion-parallel metrics on such Lie
algebras.

###### Key words and phrases:

Hermitian manifold; Chern connection; Bismut connection; balanced metrics;
pluriclosed metrics

###### 2010 Mathematics Subject Classification:

53C55 (primary), 53C05 (secondary)

Zheng is partially supported by National Natural Science Foundations of China
with the grant No.12071050 and 12141101, Chongqing grant cstc2021ycjh-
bgzxm0139, and is supported by the 111 Project D21024.

###### Contents

  1. 1 Introduction and statement of results
  2. 2 Preliminary on Lie-Hermitian manifolds
  3. 3 Almost abelian Lie algebras
  4. 4 Lie algebras with $J$-invariant abelian ideal of codimension $2$

##  1\. Introduction and statement of results

A major difficulty in the study of differential geometry of Hermitian
manifolds is the algebraic complexity of the curvature tensor, which lack of
symmetry due to the presence of torsion. When the universal cover of the
manifold is a Lie group, the situation is much simpler in the sense that every
point looks like every other point. But even in this case the torsion and
curvature of the Hermitian metric already present rich and interesting
algebraic complexity, which could serve as good sample spaces as well as
provide guidance towards the study of the general case. For this reason, it
makes sense to study Hermitian geometry of Lie groups.

Recall that a Lie-Hermitian manifold is a compact Hermitian manifold
$(M^{n},g)$ whose universal cover is $(G,J,g)$, where $G$ is an even-
dimensional (connected and simply-connected) Lie group, equipped with a left-
invariant complex structure $J$ and a left-invariant Riemannian metric $g$
compatible with $J$. The compactness of $M$ forces $G$ to be unimodular.

In the past decades, Lie-Hermitian manifolds were studied from various angles
by A. Gray, S. Salamon, L. Ugarte, A. Fino, L. Vezzoni, F. Podestà, D.
Angella, A. Andrada, and others. There is a vast amount of literature on this
topic, here we will just mention a small sample: [6], [13], [16], [20], [24],
[26], [32], [39], [43], [45], and interested readers could start their
exploration there. For more general discussions on non-Kähler Hermitian
geometry, see for example [1], [5], [29], [41], [42] and the references
therein.

The cases when $G$ is nilpotent or when $\dim G\leq 6$ are relatively well-
understood, while the non-nilpotent and general dimensional cases are less
clear. One exception is perhaps the case of almost abelian groups, where the
Lie algebra ${\mathfrak{g}}$ of $G$ has an abelian ideal of codimension one. A
number of interesting results were obtained about this special type, see for
instance [2], [7], [9], [12], [19], [21], [36], [37], [43] and the references
therein for more details.

Another special types of Lie-Hermitian manifolds are quotients of $2$-step
solvable Lie groups. In the beautiful recent papers [27],[28], Freibert and
Swann studied the Hermitian geometry of such Lie groups, especially on
balanced and pluriclosed (namely, SKT) structures.

The purpose of this article is two fold: on one hand we would like to
demonstrate the convenience of using complex coordinates in the study of
Hermitian structures on Lie algebras, and we would like to illustrate this
point by summarizing some of the existing results about almost abelian Lie
algebras, and give hopefully somewhat simplified computation and argument
there. This will be treated in §3, where we will collect some known results on
almost abelian Lie algebras and also prove a couple of new ones, regarding
Bismut torsion-parallel metrics and astheno-Kähler metrics. The second purpose
is to push this complex coordinate computation to another very special type of
Lie algebras: those which contain an abelian ideal of codimension two. Note
that such a Lie algebra is always solvable of step at most $3$, but in general
it will not be solvable of step $2$.

Let ${\mathfrak{a}}$ be an abelian ideal of codimension $2$ in a Lie algebra
${\mathfrak{g}}$ of real dimension $2n$. If $J$ is an integrable almost
complex structure on ${\mathfrak{g}}$, then the $J$-invariant subspace
${\mathfrak{a}}_{J}:={\mathfrak{a}}\cap J{\mathfrak{a}}$ has codimension
either $2$ or $4$ in ${\mathfrak{g}}$, and it is of codimension $2$ if and
only if $J{\mathfrak{a}}={\mathfrak{a}}$. In this article we will focus on
this simpler case, and mimic the study on almost abelian Lie algebras to
characterize some special types of Hermitian structures on such Lie algebras.
This will be carried out in §4. The main result is Proposition 5 below, which
gives a description of all Bismut torsion-parallel metrics on such Lie
algebras.

We will prove the following statements regarding almost abelian Lie algebras
or Lie algebras with $J$-invariant abelian ideals of codimension $2$. For an
almost abelian Lie algebra ${\mathfrak{g}}$ equipped with a Hermitian
structure $(J,g)$, there always exists a unitary basis
$\\{e_{1},\ldots,e_{n}\\}$ of the complex vector space
${\mathfrak{g}}^{1,0}=\\{x-\sqrt{-1}Jx\,\mid\,x\in{\mathfrak{g}}\\}$ so that
the structure equation of $g$ takes the form

(1)
$\left\\{\begin{split}d\varphi_{1}=-\lambda\,\varphi_{1}\wedge\overline{\varphi}_{1},\hskip
165.02606pt\\\
d\varphi_{i}=-\overline{v}_{i}\,\varphi_{1}\wedge\overline{\varphi}_{1}+\sum_{j=2}^{n}\overline{A_{ij}}\,(\varphi_{1}+\overline{\varphi}_{1})\wedge\varphi_{j},\
\ \ 2\leq i\leq n.\end{split}\right.$

Here $\varphi$ is the unitary coframe dual to $e$, $\lambda\in{\mathbb{R}}$,
$v\in{\mathbb{C}}^{n-1}$, and $A$ is a complex $(n-1)\times(n-1)$ matrix. Such
a frame will be called admissible for the Hermitian almost abelian Lie
algebra.

Recall that a Hermitian metric is said to be Bismut torsion-parallel (BTP in
short) if its Bismut connection has parallel torsion tensor. It is called
Bismut Kähler-like (BKL in short) if the curvature tensor of the Bismut
connection obeys all Kähler symmetries. An equivalent form of a conjecture
proposed by Angella, Otal, Ugarte, Villacampa in [3] states that BKL metrics
are always BTP. This was confirmed by Zhao and the second named author in
[49].

######  Proposition 1.

Let $({\mathfrak{g}},J,g)$ be a Hermitian structure on an almost abelian Lie
algebra. Then $g$ is Bismut torsion-parallel if and only if it is Bismut
Kähler-like if and only if $A+A^{\ast}=0$ and $Av=0$ under any admissible
frame.

Here and from now on $A^{\ast}$ stands for conjugate transpose
$\overline{\,{}^{t}\\!A}$. Note that ${\mathfrak{g}}$ is unimodular if and
only if $\lambda+\mbox{tr}(A)+\overline{\mbox{tr}(A)}=0$. So when
${\mathfrak{g}}$ is unimodular, the skew-Hermitian property of $A$ implies
that $\lambda=0$, and one can choose a unitary frame $e$ so that $A$ is
diagonal.

Recall that a Hermitian manifold $(M^{n},g)$ is said to be astheno-Kähler
[33], if $\partial\overline{\partial}(\omega^{n-2})=0$, where $\omega$ is the
Kähler form of $g$. For almost abelian Lie algebras, we have the following

######  Proposition 2.

Let $({\mathfrak{g}},J,g)$ be a Hermitian structure on a unimodular almost
abelian Lie algebra. Assume that the complex dimension $n\geq 4$. Then $g$ is
astheno-Kähler if and only if $g$ is pluriclosed if and only if
$[A^{\ast},A]=0$ and one of the eigenvalue of $A$ has real part equal to
$-\frac{\lambda}{2}$ while the other eigenvalues have real part equal to $0$.

The characterization for pluriclosed metrics on almost abelian Lie algebras
were obtained by Arroyo and Lafuente [7]. Note that when $n=3$ the astheno-
Kähler and pluriclosed conditions coincide, so the above statement holds
automatically.

If one drops the assumption that ${\mathfrak{g}}$ is unimodular, then from the
proof of Proposition 2 we know that $g$ is astheno-Kähler if and only if
$[A^{\ast},A]=0$ and the real part of the eigenvalues of $A$ satisfy the
following condition:

There exists an integer $0\leq k\leq n-1$ such that $k$ of them equal to
$-\frac{(n-1-k)}{2(n-2)}\lambda$ while the other $n-1-k$ of them equal to
$\frac{(k-1)}{2(n-2)}\lambda$.

In the unimodular case, $k$ is necessarily $1$, so the real part of one
eigenvalue of $A$ equals $-\frac{\lambda}{2}$ while the real part of the other
eigenvalues of $A$ all equal to $0$.

We remark that, the simplicity of the structure equation (1) enables one to
explore Hermitian properties on almost abelian Lie algebras with relative
ease, and the above two propositions were intended to serve as an
illustration.

Next, let ${\mathfrak{g}}$ be a Lie algebra equipped with a Hermitian
structure $(J,g)$, and let ${\mathfrak{a}}\subseteq{\mathfrak{g}}$ be a
$J$-invariant abelian ideal of codimension $2$. Then there always exists
unitary frame $e$ under which the structure equation becomes

(2)
$\left\\{\begin{split}d\varphi_{1}\,=\,-\lambda\varphi_{1}\overline{\varphi}_{1},\hskip
116.65646pt\\\
d\varphi\,=\,-\varphi_{1}\overline{\varphi}_{1}\overline{v}-\varphi_{1}\,^{t}\\!X\varphi+\overline{\varphi}_{1}\overline{Y}\varphi-\varphi_{1}\overline{Z}\,\overline{\varphi}.\end{split}\right.$

Here we wrote $\varphi=\,^{t}\\!(\varphi_{2},\ldots,\varphi_{n})$ as a column
vector, while $\lambda\geq 0$, $v\in{\mathbb{C}}^{n-1}$, and $X$, $Y$, $Z$ are
complex $(n-1)\times(n-1)$ matrices satisfying

(3)
$\left\\{\begin{split}\lambda(X^{\ast}\\!+Y)+[X^{\ast},Y]-Z\overline{Z}\,=\,0,\\\
\lambda Z-(Z\,^{t}\\!X+YZ)\,=\,0.\hskip 34.14322pt\end{split}\right.$

Such a unitary frame $e$ will be called an admissible frame. Note that the
algebraic data involves three matrices satisfying a system of matrix equations
and are tangled up, unlike the almost abelian case. Also, when $\lambda\neq
0$, ${\mathfrak{g}}$ is not $2$-step solvable in general, although it is
always $3$-step solvable.

######  Proposition 3.

Let $({\mathfrak{g}},J,g)$ be a Lie algebra with Hermitian structure and let
${\mathfrak{a}}\subseteq{\mathfrak{g}}$ be a $J$-invariant abelian ideal of
codimension $2$. Then under any admissible frame,

  1. (i)

${\mathfrak{g}}$ is unimodular if and only if
$\,\lambda-\mbox{tr}(X)+\mbox{tr}(Y)=0$,

  2. (ii)

$g$ is balanced if and only if $\,\mbox{tr}(X)=\mbox{tr}(Y)$ and $v=0$,

  3. (iii)

$g$ is Kähler if and only if $v=0$, $\,{}^{t}\\!Z=Z$, and $X=Y$. When
${\mathfrak{g}}$ is unimodular, $g$ is Kähler if and only if $\lambda=0$,
$v=0$, $Z=0$, and $X=Y$ is normal. In this case there is a unitary frame under
which the structure equation becomes

$d\varphi_{1}=0,\ \ \
d\varphi_{i}=(\overline{a}_{i}\overline{\varphi}_{1}-a_{i}\varphi_{1})\varphi_{i},\
\ \ \ 2\leq i\leq n.$

where $a_{2},\ldots,a_{n}$ are arbitrary complex numbers.

  4. (iv)

$g$ is pluriclosed (SKT) if and only if it satisfies (3) and the equation
below:

(4)
$XX^{\ast}-YX^{\ast}+\,^{t}\\!Z\overline{Z}-Z\overline{Z}+Y^{\ast}Y-Y^{\ast}X+\lambda(Y-X)=0.$

From items (ii) and (iv) above, one can write down explicit examples of
balanced metrics or pluriclosed metrics on such ${\mathfrak{g}}$. See §4 for
more details. Similar to the almost abelian case, one could also analyze the
behavior of the Chern and Bismut curvature of $g$. In particular, one has the
following

######  Proposition 4.

Let $({\mathfrak{g}},J,g)$ be a Lie algebra with Hermitian structure and
${\mathfrak{a}}$ be a $J$-invariant abelian ideal of codimension $2$. Then the
Chern curvature satisfies the following

  1. (i)

$g$ is Chern flat if and only if $\lambda=0$, $v=0$, $Z=0$, $[Y,Y^{\ast}]=0$,
and $[Y,X^{\ast}]=0$. In this case one can choose unitary frame so that $Y$ is
diagonal and $X$ is block-diagonal in the form (5) below;

  2. (ii)

The rank of the first Chern Ricci $Ric^{(1)}$ is at most $1$, and it has the
same sign with the Chern scalar curvature $s$. In particular, $Ric^{(1)}=0\
\Longleftrightarrow\ s=0\ \Longleftrightarrow\ \lambda=0\ \mbox{or}\
\lambda=-\mbox{Re}\\{\mbox{tr}(Y)\\}$;

  3. (iii)

If the second Chern Ricci $Ric^{(2)}\geq 0$, then $R=0$;

  4. (iv)

For the (first) Bismut Ricci curvature $Ric^{b}$, its $(2,0)$-part has rank
$2$ or $0$, and its $(1,1)$-part has rank at most $1$ if $v=0$ and has rank at
most $3$ in general. $g$ is Bismut Ricci flat if and only if $Xv=0$,
$Y^{\ast}v+\,^{t}\\!Z\overline{v}=0$, and $s^{b}=0$, where the Bismut scalar
curvature $s^{b}$ is given by (6) below. In particular, $g$ is Bismut Ricci
flat when $v=0$ and $\lambda=0$, or when $v=0$ and
$\lambda=\mbox{Re}\\{\mbox{tr}(X)-2\mbox{tr}(Y)\\}$.

In the Chern flat case, since $Y$ is normal and $[X^{\ast},Y]=0$, we can
choose a unitary frame $e$ so that $Y$ is diagonal and $X$ is block-diagonal
in the following way:

(5) $Y=\left[\begin{array}[]{ccc}\lambda_{1}I_{n_{1}}&&\\\ &\ddots&\\\
&&\lambda_{r}I_{n_{r}}\end{array}\right],\ \ \ \ \ \
X=\left[\begin{array}[]{ccc}X_{1}&&\\\ &\ddots&\\\ &&X_{r}\end{array}\right],$

where $n_{1}+\cdots+n_{r}=n-1$ and $\lambda_{1},\ldots,\lambda_{r}$ are
distinct, while each $X_{i}$ is an arbitrary $n_{i}\times n_{i}$ matrix. The
Bismut scalar curvature in (iv) above is given by

(6)
$s^{b}\,=\,-2|v|^{2}-\lambda\left(2\lambda+2\mbox{tr}(Y)+2\overline{\mbox{tr}(Y)}-\mbox{tr}(X)-\overline{\mbox{tr}(X)}\right).$

A Hermitian metric $g$ with zero Bismut (first) Ricci curvature is said to be
Calabi-Yau with torsion, or CYT in short. In [21, Theorem 3.1], Fino and
Paradiso characterized CYT metrics on almost abelian Lie algebras, and item
(iv) above is just analogous to their result.

Also analogous to the almost abelian case, for ${\mathfrak{g}}$ with
$J$-invariant abelian ideal of codimension $2$, one could also ask when will
the metric be Bismut torsion-parallel or Bismut Kähler-like. The following is
the main result of this article:

######  Proposition 5.

Let $({\mathfrak{g}},J,g)$ be a unimodular Lie algebra with Hermitian
structure and ${\mathfrak{a}}$ be a $J$-invariant abelian ideal of codimension
$2$. Then $g$ is Bismut Kähler-like if and only if there exists a unitary
frame under which the structure equation is given by (7) below. $g$ is Bismut
torsion-parallel if and only if there exists a unitary frame under which the
structure equation is given by (7), or (8), or (9).

(7) $\left\\{\begin{split}d\varphi_{1}=0,\hskip 119.50148pt\\\
d\varphi_{2}=-v_{2}\varphi_{1}\overline{\varphi}_{1},\hskip 85.35826pt\\\
d\varphi_{i}=(\overline{a}_{i}\overline{\varphi}_{1}-a_{i}\varphi_{1})\varphi_{i},\
\ \ 3\leq i\leq n;\end{split}\right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

where $v_{2}>0$ and $a_{3},\ldots,a_{n}$ are complex numbers.

(8) $\left\\{\begin{split}d\varphi_{1}=d\varphi_{3}=0,\hskip 89.626pt\\\
d\varphi_{2}=-v_{2}\varphi_{1}\overline{\varphi}_{1}+p\,(\overline{\varphi}_{1}\varphi_{3}-\varphi_{1}\overline{\varphi}_{3}),\hskip
2.84544pt\\\
d\varphi_{i}=(\overline{a}_{i}\overline{\varphi}_{1}-a_{i}\varphi_{1})\varphi_{i},\
\ \ 4\leq i\leq n.\end{split}\right.$

where $v_{2}>0$, $p>0$, and $a_{4},\ldots,a_{n}$ are complex numbers.

(9) $\left\\{\begin{split}d\varphi_{1}=0,\hskip 182.09746pt\\\
d\varphi_{i}=\lambda_{i}\overline{\varphi}_{1}\varphi_{i}-\lambda_{i}\varphi_{1}\sum_{j=2}^{r+1}\overline{W_{ij}}\varphi_{j},\
\ \ \ \ 2\leq i\leq r+1,\hskip 2.84544pt\\\ d\varphi_{i}=0,\ \ \ \ \ \ \ \ \
r+2\leq i\leq 2r+1,\hskip 69.70915pt\\\
d\varphi_{i}=(\overline{a}_{i}\overline{\varphi}_{1}-a_{i}\varphi_{1})\varphi_{i},\
\ \ \ \ 2r+2\leq i\leq n.\hskip 34.14322pt\end{split}\right.$

where $r$ is an integer with $1\leq r\leq\frac{n-2}{2}$, each $a_{i}$ is a
complex number, $S=\mbox{diag}\\{\lambda_{2},\ldots,\lambda_{r+1}\\}>0$ and
$W$ is any unitary symmetric $r\times r$ matrix satisfying $WS=SW$.

Note that (7) and (8) are not balanced, while (9) is balanced. Analogous to
the almost abelian case, one could show that, for any unimodular Lie algebra
with $J$-invariant abelian ideal of codimension $2$, if the metric is Chern
Kähler-like, then it must be Chern flat. So again there is no non-trivial
example of compact Chern Kähler-like manifolds amongst this type of Lie
algebras. Also analogous to the almost abelian case, for unimodular
${\mathfrak{g}}$ with $J$-invariant abelian ideal of codimension $2$, when the
complex dimension $n\geq 4$, then one can write down explicitly the matrix
equation for the metric to be astheno-Kähler, in the form $P=\mu I$ where $P$
is a matrix involving $X$, $Y$, $Z$ and $\mu$ is a constant. The computation
is strictly analogous but slightly more tedious, and we will skip it here.

The article is organized as follows. In §2 we will set up notations and
collect some basic formula for Lie-Hermitian manifolds. In §3 we will recall
some known properties about almost abelian groups and also prove a couple of
new ones: Propositions 1 and 2 stated above. In §4, we will turn our attention
to Lie algebras with $J$-invariant abelian ideals of codimension $2$, and
prove a number of properties including Propositions 3 \- 5 stated above.

##  2\. Preliminary on Lie-Hermitian manifolds

In this section, we will collect some basic formula for Lie-Hermitian
manifolds that will be used later. We will follow the notations in [44].

Let $(M^{n},g)$ be a Lie-Hermitian manifold, with universal cover $(G,J,g)$
where $G$ is an even-dimensional Lie group and $J$ is a left-invariant complex
structure on $G$, while $g$ is a left-invariant Riemannian metric on $G$
compatible with $J$. As we mention before, the compactness of $M$ forces $G$
to be unimodular.

Denote by ${\mathfrak{g}}$ the Lie algebra of $G$, and use the same letter $J$
or $g=\langle,\rangle$ to denote respectively the almost complex structure or
inner product on ${\mathfrak{g}}$ corresponding to that of $G$. As is well
known, the integrability of $J$ is characterized by the property

(10) $[x,y]-[Jx,Jy]+J[Jx,y]+J[x,Jy]=0,\ \ \ \ \forall\ x,y\in{\mathfrak{g}}.$

Denote by ${\mathfrak{g}}^{\mathbb{C}}$ the complexification of
${\mathfrak{g}}$, and by ${\mathfrak{g}}^{1,0}=\\{x-\sqrt{-1}Jx\mid
x\in{\mathfrak{g}}\\}\subseteq{\mathfrak{g}}^{\mathbb{C}}$. The condition (10)
simply says that ${\mathfrak{g}}^{1,0}$ is a complex Lie subalgebra of
${\mathfrak{g}}^{\mathbb{C}}$. Extend $g=\langle,\rangle$ bi-linearly over
${\mathbb{C}}$, and let $e=\\{e_{1},\ldots,e_{n}\\}$ be a unitary basis of
${\mathfrak{g}}^{1,0}$. Following [44], we will use

$C^{j}_{ik}=\langle[e_{i},e_{k}],\overline{e}_{j}\rangle,\ \ \ \ \ \
D^{j}_{ik}=\langle[\overline{e}_{j},e_{k}],e_{i}\rangle$

to denote the structure constants, or equivalently, under the unitary frame
$e$ we have

(11) $[e_{i},e_{j}]=\sum_{k}C^{k}_{ij}e_{k},\ \ \ \ \
[e_{i},\overline{e}_{j}]=\sum_{k}\big{(}\overline{D^{i}_{kj}}e_{k}-D^{j}_{ki}\overline{e}_{k}\big{)}.$

Note that ${\mathfrak{g}}$ is unimodular if and only if $\mbox{tr}(ad_{x})=0$
for any $x\in{\mathfrak{g}}$, which is equivalent to

(12) ${\mathfrak{g}}\ \,\mbox{is unimodular}\ \ \Longleftrightarrow\ \
\sum_{r}\big{(}C^{r}_{ri}+D^{r}_{ri}\big{)}=0,\,\ \forall\ i.$

We should note that two Lie-Hermitian manifolds could be holomorphically
isometric to each other, while their universal covers are not isomorphic as
Lie groups. For instance, there are Lie-Hermitian manifold $M^{n}$ with non-
abelian $G$ whose metric turns out to be Kähler and flat, thus forcing $M^{n}$
to be a quotient of the abelian group ${\mathbb{C}}^{n}$. In general it would
be an interesting question to determine when will two non-isomorphic Lie
groups be holomorphically isometric as Hermitian manifolds, but since our goal
is primarily to find examples of Hermitian manifolds satisfying special
geometric conditions, we will ignore this more subtle issue of group
isomorphism here.

Next, let us denote by $\nabla$ the Chern connection, and by $T$, $R$ its
torsion and curvature tensor. Then one has

$\nabla e_{i}=\sum_{j}\theta_{ij}e_{j},\ \ \ \
\theta_{ij}=\sum_{k}\big{(}\Gamma^{j}_{ik}\varphi_{k}-\overline{\Gamma^{i}_{jk}}\,\overline{\varphi}_{k}\big{)},\
\ \ \ \ \Gamma^{j}_{ik}=D^{j}_{ik},$

where $\\{\varphi_{1},\ldots,\varphi_{n}\\}$ is the coframe dual to $e$. The
torsion tensor $T$ of $\nabla$ has components

(13) $T(e_{i},\overline{e}_{j})=0,\ \ \ \
T(e_{i},e_{j})=\sum_{k}T^{k}_{ij}e_{k},\ \ \ \ \ \
T^{j}_{ik}=-C^{j}_{ik}-D^{j}_{ik}+D^{j}_{ki}.$

The only possibly non-zero components of $R$ are
$R_{i\bar{j}k\bar{\ell}}=\langle
R_{e_{i}\bar{e}_{j}}e_{k},\bar{e}_{\ell}\rangle$, which are equal to

(14) $R_{i\bar{j}k\bar{\ell}}=\sum_{r}\big{(}D^{r}_{ki}\overline{D^{r}_{\ell
j}}-D^{\ell}_{ri}\overline{D^{k}_{rj}}-D^{j}_{ri}\overline{D^{k}_{\ell
r}}-\overline{D^{i}_{rj}}D^{\ell}_{kr}\big{)}.$

Denote by $\omega=\sqrt{-1}\sum_{i}\varphi_{i}\wedge\overline{\varphi}_{i}$
the Kähler form of the metric. Recall that Gauduchon’s torsion $1$-form $\eta$
([30]) is defined by $\partial\omega^{n-1}=-\eta\wedge\omega^{n-1}$. Its
components are

(15) $\eta=\sum_{i}\eta_{i}\varphi_{i},\ \ \ \ \ \
\eta_{i}=\sum_{r}T^{r}_{ri}=\sum_{r}D^{r}_{ir},$

where the last equality holds when ${\mathfrak{g}}$ is unimodular (12). The
structure equation takes the form:

(16)
$d\varphi_{i}=-\frac{1}{2}\sum_{j,k}C^{i}_{jk}\,\varphi_{j}\wedge\varphi_{k}-\sum_{j,k}\overline{D^{j}_{ik}}\,\varphi_{j}\wedge\overline{\varphi}_{k}$

Differentiate the above, we get the first Bianchi identity, which is
equivalent to the Jacobi identity in this case:

(17)
$\displaystyle\sum_{r}\big{(}C^{r}_{ij}C^{\ell}_{rk}+C^{r}_{jk}C^{\ell}_{ri}+C^{r}_{ki}C^{\ell}_{rj}\big{)}\
=\ 0,$ (18)
$\displaystyle\sum_{r}\big{(}C^{r}_{ik}D^{\ell}_{jr}+D^{r}_{ji}D^{\ell}_{rk}-D^{r}_{jk}D^{\ell}_{ri}\big{)}\
=\ 0,$ (19)
$\displaystyle\sum_{r}\big{(}C^{r}_{ik}\overline{D^{r}_{j\ell}}-C^{j}_{rk}\overline{D^{i}_{r\ell}}+C^{j}_{ri}\overline{D^{k}_{r\ell}}-D^{\ell}_{ri}\overline{D^{k}_{jr}}+D^{\ell}_{rk}\overline{D^{i}_{jr}}\big{)}\
=\ 0.$

Next, denote by $\nabla^{b}$ the Bismut connection of $g$, and let
$\Gamma^{bj}_{ik}=\langle\nabla^{b}_{e_{k}}e_{i},\overline{e}_{j}\rangle$ be
the connection coefficients under the frame $e$. We have

$\Gamma^{bj}_{ik}=\Gamma^{j}_{ik}+T^{j}_{ik}=-C^{j}_{ik}+D^{j}_{ki}.$

More generally, for any $t\in{\mathbb{R}}$, denote by
$\nabla^{(t)}=(1-\frac{t}{2})\nabla+\frac{t}{2}\nabla^{b}$ the $t$-Gauduchon
connection ([31]). Its connection coefficients are given by

$\Gamma^{(t)j}_{ik}=\Gamma^{j}_{ik}+\frac{t}{2}T^{j}_{ik}=(1-\frac{t}{2})D^{j}_{ik}+\frac{t}{2}\big{(}-C^{j}_{ik}+D^{j}_{ki}\big{)}.$

The following observation should be well-known to experts, and we give a proof
here for readers’ convenience.

######  Lemma 1.

Let ${\mathfrak{g}}$ be a unimodular Lie algebra with a Hermitian structure
$(J,g)$, then $g$ is always Gauduchon, namely,
$\partial\overline{\partial}\omega^{n-1}=0$. Here $\omega$ is the Kähler form
of $g$.

###### Proof.

By the defining equation of $\eta$, one gets

$\partial\overline{\partial}\omega^{n-1}=(\overline{\partial}\eta+\eta\wedge\overline{\eta})\wedge\omega^{n-1}.$

So $g$ is Gauduchon if and only if $\chi=|\eta|^{2}$, where
$|\eta|^{2}=\sum_{i}|\eta_{i}|^{2}$ and $\chi=\sum_{i}\eta_{i,\bar{i}}$ under
any unitary frame. Here and from now on the index after comma stands for
covariant derivatives with respect to the Chern connection $\nabla$. Since
$\Gamma^{j}_{ik}=D^{j}_{ik}$ and ${\mathfrak{g}}$ is unimodular, by (15) we
have

$\chi=\sum_{i}\eta_{i,\bar{i}}=\sum_{i,r}\eta_{r}\overline{\Gamma^{i}_{ri}}=\sum_{r}\eta_{r}\overline{\eta_{r}}=|\eta|^{2}.$

Therefore $g$ is always Gauduchon. ∎

Now let us discuss the geometric meaning of the constants $C$ and $D$. Note
that $J$ makes ${\mathfrak{g}}$ a complex Lie algebra when and only when
$D=0$, so the tensor $D$ in a way measures how far $({\mathfrak{g}},J)$ is
deviated from being a complex Lie algebra. When $D=0$, we know from (14) that
$g$ is Chern flat. Note that the converse of this not true: if the Hermitian
manifold $(G,J,g)$ is Chern flat, then $D$ might not be identically zero.

For the tensor $C$, $C=0$ means $[Jx,Jy]=[x,y]$ for any
$x,y\in{\mathfrak{g}}$. In this case $J$ is called an abelian complex
structure, representing a rather special type of Lie-Hermitian manifolds. The
condition $D=-C$ is equivalent to $g$ being a bi-invariant metric on $G$, and
in this case the Bismut connection $\nabla^{b}$ ([8]) has coefficients
$\Gamma^{b}=0$, hence $g$ is Bismut flat. A beautiful recent result of
Lafuente and Stanfield [34], settling a question raised in [47], states that
for any $t\neq 0,2$ if a Hermitian manifold has flat $t$-Gauduchon connection
$\nabla^{(t)}$, then $g$ must be Kähler. In the special case of Lie-Hermitian
manifolds, their result implies that

######  Proposition 6 ([34]).

For any $t\neq 0,2$, if $({\mathfrak{g}},J,g)$ satisfies $\Gamma^{(t)}=0$,
then $g$ is Kähler, namely, $T=0$. In other words, if $t\neq 0,2$,
$D^{j}_{ik}=-D^{j}_{ki}$, and $C=\frac{2(1-t)}{t}D$, then $C=D=0$ thus
${\mathfrak{g}}$ is abelian.

Note that for Lie-Hermitian manifolds, $T=0$ does not imply ${\mathfrak{g}}$
is abelian in general, as we mentioned before. But in this particular case,
since $T=-C-D+\,^{t}\\!D=-C-2D$ as $\,{}^{t}\\!D=-D$, so when $T=0$ we get
$C=-2D=-2(1-\frac{1}{t})D$, which leads to $D=0$ hence $C=0$. We remark that,
the result of Lafuente and Stanfield is highly non-trivial even in the special
case of Lie-Hermitian manifolds.

Recall that a Hermitian metric $g$ is called pluriclosed (also called strong
Kähler with torsion, or SKT in short in some literature), if its Kähler form
$\omega$ satisfies $\partial\overline{\partial}\omega=0$. For Lie-Hermitian
manifolds, one has

######  Lemma 2.

Let $({\mathfrak{g}},J,g)$ be any Lie algebra with a Hermitian structure. It
holds that

$\sqrt{-1}\partial\overline{\partial}\omega=\sum_{i,j,k,\ell}\sum_{r}\left(-\frac{1}{2}T^{r}_{ik}\overline{C^{r}_{j\ell}}-T^{j}_{ir}\overline{D^{k}_{r\ell}}+T^{j}_{kr}\overline{D^{i}_{r\ell}}\right)\varphi_{i}\varphi_{k}\overline{\varphi}_{j}\overline{\varphi}_{\ell}.$

In particular, $g$ is pluriclosed if and only if

(20)
$\sum_{r}\left(-T^{r}_{ik}\overline{C^{r}_{j\ell}}-T^{j}_{ir}\overline{D^{k}_{r\ell}}+T^{j}_{kr}\overline{D^{i}_{r\ell}}+T^{\ell}_{ir}\overline{D^{k}_{rj}}-T^{\ell}_{kr}\overline{D^{i}_{rj}}\right)=0$

for any $1\leq i<k\leq n$ and any $1\leq j<\ell\leq n$.

###### Proof.

From the structure equation (16), we have

$\displaystyle\sqrt{-1}\partial\omega$ $\displaystyle=$
$\displaystyle-\sum_{i,j,k}T^{j}_{ik}\varphi_{i}\varphi_{k}\overline{\varphi}_{j}$
$\displaystyle\sqrt{-1}\partial\overline{\partial}\omega$ $\displaystyle=$
$\displaystyle-d(\sqrt{-1}\partial\omega)\ =\
\sum_{i,j,k}T^{j}_{ik}\left((d(\varphi_{i}\varphi_{k}))^{2,1}\overline{\varphi}_{j}+\varphi_{i}\varphi_{k}\overline{(d\varphi_{j})^{2,0}}\right)$
$\displaystyle=$
$\displaystyle\sum_{i,j,k}T^{j}_{ik}\left(\sum_{p,q}\big{(}\overline{D^{p}_{iq}}\varphi_{p}\varphi_{k}+\overline{D^{p}_{kq}}\varphi_{i}\varphi_{p}\big{)}\overline{\varphi}_{q}\overline{\varphi}_{j}-\sum_{p,q}\frac{1}{2}\overline{C^{j}_{pq}}\varphi_{i}\varphi_{k}\overline{\varphi}_{p}\overline{\varphi}_{q}\right)$
$\displaystyle=$
$\displaystyle\sum_{i,j,k,\ell}\sum_{r}\left(-\frac{1}{2}T^{r}_{ik}\overline{C^{r}_{j\ell}}-T^{j}_{ir}\overline{D^{k}_{r\ell}}+T^{j}_{kr}\overline{D^{i}_{r\ell}}\right)\varphi_{i}\varphi_{k}\overline{\varphi}_{j}\overline{\varphi}_{\ell}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{i,j,k,\ell}\sum_{r}\left(-T^{r}_{ik}\overline{C^{r}_{j\ell}}-T^{j}_{ir}\overline{D^{k}_{r\ell}}+T^{j}_{kr}\overline{D^{i}_{r\ell}}+T^{\ell}_{ir}\overline{D^{k}_{rj}}-T^{\ell}_{kr}\overline{D^{i}_{rj}}\right)\varphi_{i}\varphi_{k}\overline{\varphi}_{j}\overline{\varphi}_{\ell}.$

This completes the proof of the lemma. ∎

Let us denote by $\zeta_{i}=\sum_{r}D^{r}_{ri}$ under any unitary frame $e$.
When ${\mathfrak{g}}$ is unimodular, we also have
$\zeta_{i}=-\sum_{r}C^{r}_{ri}$. Define $\zeta=\sum_{i}\zeta_{i}\varphi_{i}$,
where $\varphi$ is the coframe dual to $e$, then clearly $\zeta$ is
independent of the choice of $e$, hence is a globally defined left-invariant
$(1,0)$-form on $G$. By the structure equation (16), a straight-forward
computation leads to

######  Lemma 3.

Let $(G,J,g)$ be the universal cover of a Lie-Hermitian manifold. Then one has

$d(\varphi_{1}\varphi_{2}\cdots\varphi_{n})=\overline{\zeta}\wedge\varphi_{1}\varphi_{2}\cdots\varphi_{n}.$

where $\varphi$ is a unitary coframe of left-invariant $(1,0)$-forms. So when
$\zeta=0$, the global $(n,0)$-form $\varphi_{1}\varphi_{2}\cdots\varphi_{n}$
is a nowhere zero holomorphic $n$-form, hence $g$ is Chern Ricci flat.

Conversely, if $G$ admits a global nowhere zero holomorphic $n$-form, then
$\zeta=\partial f$ for some smooth function $f$ on $G$, namely, $\zeta$ is
$\partial$-exact.

By taking trace of the Chern curvature tensor $R$ (see (14)) in three
different ways, one gets the first, second, and third Chern Ricci curvature
tensors:

(21) $\displaystyle Ric^{(1)}_{i\bar{j}}$ $\displaystyle=$
$\displaystyle\sum_{r}R_{i\bar{j}r\bar{r}}\ =\
-\sum_{r}\big{(}\zeta_{r}\overline{D^{i}_{rj}}+\overline{\zeta}_{r}D^{j}_{ri}\big{)},$
(22) $\displaystyle Ric^{(2)}_{i\bar{j}}$ $\displaystyle=$
$\displaystyle\sum_{r}R_{r\bar{r}i\bar{j}}\ =\
\sum_{r,s}\big{(}D^{r}_{is}\overline{D^{r}_{js}}-D^{j}_{rs}\overline{D^{i}_{rs}}\big{)}-\sum_{r}\big{(}\eta_{r}\overline{D^{i}_{jr}}+\overline{\eta}_{r}D^{j}_{ir}\big{)},$
(23) $\displaystyle Ric^{(3)}_{i\bar{j}}$ $\displaystyle=$
$\displaystyle\sum_{r}R_{r\bar{j}i\bar{r}}\ =\
-\sum_{r,s}D^{j}_{rs}\overline{D^{i}_{sr}}-\sum_{r}\eta_{r}\overline{D^{i}_{rj}}.$

From the first equation, we could also see that $\zeta=0$ implies
$Ric^{(1)}=0$. In the second and third line, we assumed that ${\mathfrak{g}}$
is unimodular, so $\eta_{i}=\sum_{r}D^{r}_{ir}$. By taking trace again on
these Ricci tensors, one gets the Chern scalar curvature $s$ and altered
scalar curvature $\hat{s}$:

(24) $\displaystyle s$ $\displaystyle=$
$\displaystyle\sum_{i}Ric^{(1)}_{i\bar{i}}\ =\ \sum_{i}Ric^{(2)}_{i\bar{i}}\
=\
-\sum_{i}\big{(}\zeta_{i}\overline{\eta}_{i}+\overline{\zeta}_{i}\eta_{i}\big{)},$
(25) $\displaystyle\hat{s}$ $\displaystyle=$
$\displaystyle\sum_{i}Ric^{(3)}_{i\bar{i}}\ =\
-\sum_{r,s,t}D^{t}_{rs}\overline{D^{t}_{sr}}-|\eta|^{2}.$

Note that on any Hermitian manifolds, $\hat{s}=s-\chi$ always holds, and
$\hat{s}$ and $\chi$ are both real. For Lie-Hermitian manifolds, we have
$|\eta|^{2}=\chi$, and by (19) one gets

$\sum_{r,s}D^{j}_{rs}\overline{D^{i}_{sr}}=\sum_{r}\big{(}\zeta_{r}\overline{D^{i}_{rj}}+\overline{\zeta}_{r}D^{j}_{ri}\big{)},\
\ \ \ \ \forall\ i,j.$

Thus $\hat{s}=s-\chi$ holds, and
$s=-\sum_{r,s,t}D^{t}_{rs}\overline{D^{t}_{sr}}$. This last equality leads to
the following

######  Lemma 4.

If a Lie-Hermitian manifold $M^{n}$ has symmetric $D$ tensor, namely,
$D^{j}_{ik}=D^{j}_{ki}$ for any $i,j,k$, then
$s=-\sum_{r,s,t}|D^{t}_{rs}|^{2}\leq 0$, and $s<0$ unless $D=0$. Similarly, if
$D$ is skew-symmetric: $D^{j}_{ik}=-D^{j}_{ki}$, then
$s=\sum_{r,s,t}|D^{t}_{rs}|^{2}\geq 0$, and $s>0$ unless $D=0$. Note that when
$s>0$ the Kodaira dimension of $M^{n}$ is necessarily $-\infty$.

Finally, let us close this section by recalling the covariant differentiation
formula for Chern torsion:

(26) $\displaystyle T^{j}_{ik,\ell}$ $\displaystyle=$
$\displaystyle\sum_{r}\left(-T^{j}_{rk}\Gamma^{r}_{i\ell}-T^{j}_{ir}\Gamma^{r}_{k\ell}+T^{r}_{ik}\Gamma^{j}_{r\ell}\right)$
(27) $\displaystyle T^{j}_{ik,\bar{\ell}}$ $\displaystyle=$
$\displaystyle\sum_{r}\left(T^{j}_{rk}\overline{\Gamma^{i}_{r\ell}}+T^{j}_{ir}\overline{\Gamma^{k}_{r\ell}}-T^{r}_{ik}\overline{\Gamma^{r}_{j\ell}}\right)$

for any $1\leq i,j,k,\ell\leq n$, where index after comma stands for covariant
derivatives with respect to the Chern connection $\nabla$. Of course when
$\nabla$ is replaced by another Hermitian connection, then the same
differentiation formula holds when $\Gamma$ is replaced by the connection
coefficients of that connection.

##  3\. Almost abelian Lie algebras

In this section let us specialize to Lie-Hermitian manifolds where
${\mathfrak{g}}$ is an almost abelian Lie algebra, namely, when
${\mathfrak{g}}$ is non-abelian but contains an abelian ideal of codimension
one.

An equivalent definition for this is a non-abelian Lie algebra
${\mathfrak{g}}$ which contains an abelian subalgebra of codimension one. In
this case, ${\mathfrak{g}}$ always contains an abelian ideal of codimension
one by [10, Prop. 3.1].

Let ${\mathfrak{g}}$ be an almost abelian Lie algebra, with
${\mathfrak{a}}\subseteq{\mathfrak{g}}$ an abelian ideal of codimension one.
Assume that $(J,g)$ is an Hermitian structure on ${\mathfrak{g}}$, with
$\dim{\mathfrak{g}}=2n$. Since ${\mathfrak{a}}_{J}:={\mathfrak{a}}\cap
J{\mathfrak{a}}$ is the intersection of two hyperplanes and is $J$-invariant,
it is necessarily of codimension $2$ in ${\mathfrak{g}}$. Thus we can choose a
unitary basis $\\{e_{1},\ldots,e_{n}\\}$ of ${\mathfrak{g}}^{1,0}$ so that
${\mathfrak{a}}$ is spanned by

${\mathfrak{a}}=\mbox{span}_{\mathbb{R}}\\{\sqrt{-1}(e_{1}-\overline{e}_{1});\
(e_{i}+\overline{e}_{i}),\,\sqrt{-1}(e_{i}-\overline{e}_{i});\ 2\leq i\leq
n\\}.$

We will call such a unitary basis an admissible frame, and we will choose $e$
to be admissible from now on. Since ${\mathfrak{a}}$ is abelian, we get

$[e_{i},e_{j}]=[e_{i},\overline{e}_{j}]=[e_{1}-\overline{e}_{1},e_{i}]=0,\ \ \
\ \forall\ 2\leq i,j\leq n.$

From this and (11) we deduce that

$C^{\ast}_{ij}=D^{j}_{\ast i}=D^{1}_{\ast
i}=C^{\ast}_{1i}+\overline{D^{i}_{\ast 1}}=0,\ \ \ \ \forall\ 2\leq i,j\leq
n.$

Since ${\mathfrak{a}}$ is an ideal,
$[e_{1}+\overline{e}_{1},{\mathfrak{a}}]\subseteq{\mathfrak{a}}$, which leads
to $C^{1}_{1i}=0$ and $D^{1}_{11}=\overline{D^{1}_{11}}$. Putting these
together, we know that the only possibly non-trivial components of $C$ and $D$
are

(28) $D^{1}_{11}=\lambda\in{\mathbb{R}},\ \ \
D^{1}_{i1}=v_{i}\in{\mathbb{C}},\ \ \ D^{j}_{i1}=A_{ij},\ \ \
C^{j}_{1i}=-\overline{A_{ji}},\ \ \ \ \ \ 2\leq i,j\leq n.$

So the only algebraic data for the structure constants are
$\lambda\in{\mathbb{R}}$, a complex column vector $v\in{\mathbb{C}}^{n-1}$,
and a complex $(n-1)\times(n-1)$ matrix $A=(A_{ij})_{2\leq i,j\leq n}$. In
terms of the dual coframe $\varphi$, the structure equation (16) now takes the
following form:

######  Lemma 5.

Let ${\mathfrak{g}}$ be an almost abelian Lie algebra with a Hermitian
structure $(J,g)$. Then there exists a unitary coframe $\varphi$ under which
the structure equation is

$\left\\{\begin{split}d\varphi_{1}=-\lambda\,\varphi_{1}\wedge\overline{\varphi}_{1},\hskip
165.02606pt\\\
d\varphi_{i}=-\overline{v}_{i}\,\varphi_{1}\wedge\overline{\varphi}_{1}+\sum_{j=2}^{n}\overline{A_{ij}}\,(\varphi_{1}+\overline{\varphi}_{1})\wedge\varphi_{j},\
\ \ 2\leq i\leq n.\end{split}\right.$

This is well-known to experts studying almost abelian Lie algebras and is
stated as (1) in §1. Clearly, $d^{2}\varphi_{1}=d^{2}\varphi_{i}=0$, so (17) -
(19) do not impose further restriction on $\lambda$, $v$, or $A$. These three
terms are not all zero, as ${\mathfrak{g}}$ is assumed to be non-abelian.

The following properties about Hermitian structures on almost abelian Lie
algebras are discovered by various people and we summarize into a proposition,
and include a proof here for readers’ convenience. Item (iii) was discovered
by Lauret and Will [36], item (iv) was studied extensively by Fino and
Paradiso [19], [21], and item (v) was proved by Arroyo and Lafuente [7].

######  Proposition 7.

Let ${\mathfrak{g}}$ be an almost abelian Lie algebra with Hermitian structure
$(J,g)$. Then there exists unitary coframe with structure equation (1) and the
following holds:

  1. (i)

${\mathfrak{g}}$ is nilpotent $\ \Longleftrightarrow\ $ $\lambda=0$ and
$A^{n-1}=0$;

  2. (ii)

${\mathfrak{g}}$ is unimodular $\ \Longleftrightarrow\ $
$\lambda+\mbox{tr}(A)+\overline{\mbox{tr}(A)}=0$;

  3. (iii)

$g$ is Kähler $\ \Longleftrightarrow\ $ $v=0$ and $A+A^{\ast}=0$.

  4. (iv)

$g$ is balanced $\ \Longleftrightarrow\ $ $v=0$ and
$\mbox{tr}(A)+\overline{\mbox{tr}(A)}=0$.

  5. (v)

$g$ is pluriclosed $\ \Longleftrightarrow\ $
$\lambda(A+A^{\ast})+(A+A^{\ast})^{2}+[A^{\ast},A]=0$ $\ \Longleftrightarrow\
$ $[A,A^{\ast}]=0$ and the real part of each eigenvalue of $A$ is either $0$
or $-\frac{\lambda}{2}$.

###### Proof.

Let $e$ be an admissible frame. The components of $C$ and $D$ are given by the
real constant $\lambda$, complex column vector $v$, and complex square matrix
$A$, with structure equation as (1). Since $ad_{e_{1}+\overline{e}_{1}}$
corresponds to the matrix

$\widetilde{A}:=\left[\begin{array}[]{ll}\lambda&0\\\ v&A\end{array}\right],$

we know that ${\mathfrak{g}}$ will be nilpotent if and only if $\lambda=0$ and
$A$ is nilpotent (i.e., $A^{n-1}=0$). For any $i>1$, we have
$\sum_{r=1}^{n}(C^{r}_{ri}+D^{r}_{ri})=0$, while

$\sum_{r=1}^{n}(C^{r}_{r1}+D^{r}_{r1})=\lambda+\mbox{tr}(A)+\overline{\mbox{tr}(A)},$

so ${\mathfrak{g}}$ is unimodular if and only if
$\lambda+\mbox{tr}(A)+\overline{\mbox{tr}(A)}=0$. By (13), the Chern torsion
components are given by

(29) $T^{\ast}_{ij}=0,\ \ \ T^{1}_{1i}=v_{i},\ \ \
T^{j}_{1i}=A_{ij}+\overline{A_{ji}},\ \ \ \ \ 2\leq i,j\leq n.$

Thus the metric is Kähler if and only if $v=0$ and $A^{\ast}=-A$. As
$\eta_{a}=\sum_{r}T^{r}_{ra}$, we have $\eta_{i}=v_{i}$ for $i>1$ and
$\eta_{1}=-(\mbox{tr}(A)+\overline{\mbox{tr}(A)})$. So the metric $g$ is
balanced if and only if $v=0$ and $\mbox{tr}(A)+\overline{\mbox{tr}(A)}=0$.
For (v), the metric $g$ is pluriclosed if and only if equation (20) holds for
any $1\leq i<k\leq n$ and any $1\leq j<\ell\leq n$. Since $D^{\ast}_{\ast
p}=T^{\ast}_{pq}=D^{p}_{11}=0$ for any $p,q>1$, one just needs to check the
$j=1$ and $i=1$ case of (20), which is

$-\sum_{r>1}T^{r}_{1k}\overline{C^{r}_{1\ell}}+\sum_{r>1}T^{\ell}_{1r}\overline{D^{k}_{r1}}-T^{\ell}_{k1}\overline{D^{1}_{11}}=\sum_{r>1}\\{T^{r}_{1k}D^{\ell}_{r1}+\overline{D^{k}_{r1}}T^{\ell}_{1r}\\}+\lambda
T^{\ell}_{1k}=0,\ \ \ \forall\ k,\ell>1.$

In other words, the metric $g$ is pluriclosed if and only if

(30) $(A+A^{\ast})A+A^{\ast}(A+A^{\ast})+\lambda(A+A^{\ast})=0.$

This is discovered by Arroyo and Lafuente in [7] and they also showed that it
is equivalent to:

(31) $[A,A^{\ast}]=0,\ \ \ \mbox{and the real part of each eigenvalue of}\ A\
\mbox{is either}\ 0\ \mbox{or}\ -\frac{\lambda}{2}.$

To see this equivalence, write $X=A+A^{\ast}$, then (30) says that
$[A,X]=\lambda X+X^{2}$. Since $X$ is Hermitian, one can take a unitary basis
change of $\\{e_{2},\ldots,e_{n}\\}$ to make it into a real diagonal matrix.
Group the same diagonal values into blocks, then the equation implies that $A$
is block-diagonal, and within each block, $X$ is a constant multiple (say
$\lambda_{i}$) of the identity. By taking trace in this block, we know that
the constant $\lambda_{i}$ must be either $0$ or $-\lambda$, and $[A,X]=0$.
Thus $[A,A^{\ast}]=0$ which means $A$ is normal hence diagonalizable by
unitary similarity. Note that $\lambda_{i}$ is twice the real part of an
eigenvalue of $A$, so the assertion (31) holds. Thus (30) implies (31). The
converse is certainly true as well. This establishes the equivalence of (v). ∎

From the above, we see that unimodular plus balanced means that $\lambda=0$,
$v=0$, while $A$ can be any complex matrix with trace in $i{\mathbb{R}}$. By
Schur’s triangulation theorem, $A$ is unitary similar to an upper triangular
matrix. Analogously, unimodular plus pluriclosed means that the real part of
only one eigenvalue of $A$ is $-\frac{\lambda}{2}$ while the others are zero,
so after unitary similarity, $A$ is equal to a diagonal matrix with diagonal
entries $\\{ib_{2},\ldots,ib_{n\\!-\\!1},ib_{n}\\!-\\!\frac{\lambda}{2}\\}$,
with all $b_{j}\in{\mathbb{R}}$, while $\lambda\in{\mathbb{R}}$ and
$v\in{\mathbb{C}}^{n\\!-\\!1}$ are arbitrary.

In [37], Paradiso characterized all Hermitian almost abelian Lie algebra that
are locally conformally balanced, and in [15], Djebbar, Ferreira, Fino, and
Youcef give characterization of all locally conformally pluriclosed ones.

Note that since we are primarily interested in compact examples, we need the
Lie group $G$ to admit a co-compact lattice which might be difficult to find
for general solvable groups. But in the almost ableian case there is a nice
sufficient condition given by Bock [9, Prop 2.1] (see also §4 of [15] and [2],
[12]).

Next let us look at the Chern curvature of an almost abelian Lie algebra. By
(14) and (28), we deduce that $R_{i\bar{\ast}\ast\bar{\ast}}=0$ and

$R_{1\bar{1}1\bar{1}}=-2\lambda^{2}-|v|^{2},\ \ \
(R_{1\bar{1}i\bar{1}})=-A^{\ast}v,\ \ \
(R_{1\bar{1}i\bar{j}})=vv^{\ast}+[A,A^{\ast}]-\lambda(A+A^{\ast}),$

for any $i,j>1$. By (25) and (28), we get

$\hat{s}=-\lambda^{2}-|\eta|^{2}=-\lambda^{2}-|v|^{2}-(\mbox{tr}(A)+\overline{\mbox{tr}(A)})^{2}.$

From these formula we conclude that

######  Lemma 6.

Let ${\mathfrak{g}}$ be an almost abelian Lie algebra with a Hermitian
structure $(J,g)$. Then it is Chern flat if and only if $\lambda=0$, $v=0$,
and $[A,A^{\ast}]=0$. It has altered scalar curvature $\hat{s}=0$ if and only
if $\lambda=0$, $v=0$ and $\mbox{tr}(A)+\overline{\mbox{tr}(A)}=0$. In the
latter case ${\mathfrak{g}}$ is necessarily unimodular.

Using the definition of three Ricci curvatures, a direct calculation leads to
the following:

######  Lemma 7.

Assume that ${\mathfrak{g}}$ is as in Lemma 6 and is unimodular, namely,
$\lambda+\mbox{tr}(A)+\overline{\mbox{tr}(A)}=0$. Then the following holds:

  * •

$s=-\lambda^{2}\leq 0,\ \mbox{and}\ s=0\ \Longleftrightarrow\ Ric^{(1)}=0\
\Longleftrightarrow\ \lambda=0$;

  * •

$\hat{s}=-2\lambda^{2}-|v|^{2}\leq 0,\ \mbox{and}\ \hat{s}=0\
\Longleftrightarrow\ Ric^{(3)}=0\ \Longleftrightarrow\ \lambda=0,\,v=0$;

  * •

$Ric^{(2)}=0\ \Longleftrightarrow\ R=0\ \Longleftrightarrow\
\lambda=0,\,v=0,\,[A,A^{\ast}]=0$.

Next let us consider the Chern Kähler-like condition (cf. [46], [3]), which
means a Hermitian manifold whose Chern curvature tensor obeys the symmetry
$R_{i\bar{j}k\bar{\ell}}=R_{k\bar{j}i\bar{\ell}}$. As is well-known, the first
Bianchi identity implies that

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

for any indices, where index after comma denotes covariant derivatives with
respect to the Chern connection $\nabla$. So the Chern Kähler-like condition
simply means that $\nabla_{\overline{X}}T=0$ for any type $(1,0)$ tangent
vector $X$. Chern flat metrics are certainly Chern Kähler-like, and it would
be very interesting to find examples of compact Chern Kähler-like manifolds
that are not Chern flat. In [50], it was shown that amongst all complex
nilmanifolds, Chern Kähler-like ones are all Chern flat. For almost abelian
ones, the phenomenon continues:

######  Lemma 8.

Let ${\mathfrak{g}}$ be a unimodular almost abelian Lie algebra with a
Hermitian structure $(J,g)$. Then $g$ is Chern Kähler-like if and only if it
is Chern flat.

###### Proof.

Assume that $g$ is Chern Kähler-like. Then we have $T^{\ell}_{ik,\bar{j}}=0$
for any $i,j,k,\ell$. By (27), this means that

$\sum_{r}\left(T^{j}_{rk}\overline{\Gamma^{i}_{r\ell}}+T^{j}_{ir}\overline{\Gamma^{k}_{r\ell}}-T^{r}_{ik}\overline{\Gamma^{r}_{j\ell}}\right)=0.$

Take $i=\ell=1<k$, since $\Gamma=D$, while $T$ and $D$ are given by (29) and
(28), we get from the $j=1$ case that $A^{\ast}v=0$, and from the $j>1$ case
that

$\lambda(A+A^{\ast})+[A^{\ast},A]-v\,v^{\ast}=0.$

Taking trace, we get $-\lambda^{2}-|v|^{2}=0$, hence $\lambda=0$, $v=0$, and
$[A,A^{\ast}]=0$. Thus $R=0$. ∎

Next we mention the result by Fino and Paradiso [21, Theorem 3.1] which
characterize CYT metrics on almost abelian Lie algebras. Recall that a
Hermitian metric is called Calabi-Yau with torsion, or CYT for short, if the
first Ricci curvature of the Bismut connection vanishes. Denote by
$\theta^{b}$, $\Theta^{b}$ the connection and curvature matrix of $\nabla^{b}$
under an admissible frame. Then one has

$\mbox{tr}(\Theta^{b})=d(\mbox{tr}(\theta^{b}))=d\left(\mbox{tr}(\theta)+\eta-\overline{\eta}\right)=d(\alpha-\overline{\alpha}),$

where
$\alpha=(\lambda-\overline{\mbox{tr}(A)})\varphi_{1}+\sum_{r=2}^{n}v_{r}\varphi_{r}$.
By the structure equation (1), and assuming that ${\mathfrak{g}}$ is
unimodular, then one gets

$\mbox{tr}(\Theta^{b})=d(\alpha-\overline{\alpha})=-(2|v|^{2}+3\lambda^{2})\varphi_{1}\overline{\varphi}_{1}+(\varphi_{1}+\overline{\varphi}_{1})\sum_{i,j=2}^{n}\\{\overline{A_{ij}}v_{i}\varphi_{j}-A_{ij}\overline{v}_{i}\overline{\varphi}_{j}\\}.$

Therefore it holds that

######  Proposition 8 ([21]).

For any unimodular almost abelian Lie algebra ${\mathfrak{g}}$ with Hermitian
structure $(J,g)$, it is CYT if and only if $\lambda=0$, $v=0$, and
$\mbox{tr}(A)+\overline{\mbox{tr}(A)}=0$. In particular, it is CYT when and
only when it is balanced.

Next let us prove Proposition 1 stated in the introduction, which describes
all Bismut torsion-parallel or Bismut Kähler-like metrics on almost abelian
Lie algebras.

###### Proof of Proposition 1:.

Let $({\mathfrak{g}},J,g)$ be an almost abelian Lie algebra with Hermitian
structure. Assume that $g$ is Bismut torsion-parallel (BTP), namely,
$\nabla^{b}T=0$. The expression of the covariant derivatives of Chern torsion
$T$ with respect to the Bismut connection $\nabla^{b}$ are given by formula
(26) and (27) with $\Gamma=D$ replaced by $\Gamma^{b}=\Gamma+T=D+T$. So we
know that BTP is equivalent to

(32)
$\displaystyle\sum_{r}\left(-T^{j}_{rk}(D^{r}_{i\ell}+T^{r}_{i\ell})-T^{j}_{ir}(D^{r}_{k\ell}+T^{r}_{k\ell})+T^{r}_{ik}(D^{j}_{r\ell}+T^{j}_{r\ell})\right)=0,$
(33)
$\displaystyle\sum_{r}\left(T^{j}_{rk}(\overline{D^{i}_{r\ell}}+\overline{T^{i}_{r\ell}})+T^{j}_{ir}(\overline{D^{k}_{r\ell}}+\overline{T^{k}_{r\ell}})-T^{r}_{ik}(\overline{D^{r}_{j\ell}}+\overline{T^{r}_{j\ell}})\right)=0$

for any $1\leq i<k\leq n$ and any $j,\ell$. Now by (28), (29), and by taking
$i=j=1<k,\ell$ in (33), we get

$\sum_{r>1}T^{r}_{1k}\overline{T^{r}_{1\ell}}=0,$

which means that $(n-1)\times(n-1)$ matrix $X=(T^{j}_{1i})=A+A^{\ast}$
satisfies $XX^{\ast}=0$, hence $X=0$. That is, the BTP assumption implies
$A+A^{\ast}=0$ for almost abelian Lie algebras. With $A$ being skew-Hermitian,
by (29) we know that the only possibly non-zero torsion components are
$T^{1}_{1i}=v_{i}$. Letting $i=j=\ell=1<k$ in (32), we get $Av=0$.

Conversely, when $A+A^{\ast}=0$ and $Av=0$, the only possibly non-zero torsion
components are $T^{1}_{1i}=v_{i}$, and it is easy to check that both (32) and
(33) are satisfied. Thus for almost ableian Lie algebra, the BTP condition is
equivalent to $A+A^{\ast}=0$ and $Av=0$. This establishes the first statement
in Proposition 1.

Now we check the BKL condition for almost abelian Lie algebra. By [49], we
know that a Hermitian metric $g$ is BKL if and only if it is both BTP and
pluriclosed. Now by Arroyo-Lafuente’s result, (v) of Proposition 7, we know
that an almost abelian Lie algebra will be pluriclosed when and only when $A$
is normal and whose eigenvalues have real part equal to either $0$ or
$-\frac{\lambda}{2}$. But this condition is automatically satisfied when $A$
is skew-Hermitian. So for almost abelian Lie algebras, the BKL condition and
the BTP conditions coincide. Note that the Lie algebra will be unimodular in
this case if and only if $\lambda=0$. This completes the proof of Proposition
1. ∎

On general Hermitian manifolds, BKL and BTP metrics were studied by several
authors in recent years, for more discussions on such special Hermitian
structures, we refer the readers to [3], [4], [22], [23], [34], [46], [48],
[49], [50], [51], [52], [53] and the references therein.

Astheno-Kähler metrics were introduced by Jost-Yau in [33], as a relaxation to
the Kähler condition. It is defined by
$\partial\overline{\partial}(\omega^{n-2})=0$ where $\omega$ is the Kähler
form and $n$ is the complex dimension of the manifold. This special type of
Hermitian metrics has seen a number of recent development, and we refer the
readers to [11], [14], [17], [18], [25], [35], [38], [40] for instance.

###### Proof of Proposition 2:.

Let $({\mathfrak{g}},J,g)$ be a an almost abelian Lie algebra with Hermitian
structure. Then there exists a unitary frame $e$ so that the structure
equation takes the form of (1), namely,
$d\varphi_{1}=-\lambda\varphi_{1}\overline{\varphi}_{1}$,
$\,d\varphi_{i}=-\overline{v}_{i}\varphi_{1}\overline{\varphi}_{1}+\sum_{j}\overline{A_{ij}}(\varphi_{1}+\overline{\varphi}_{1})\varphi_{j}$.
Here and below we set the index range to be $2\leq i,j,k,\ell\leq n$. The
Kähler form $\omega$ is given by
$\omega=\sqrt{-1}\left(\varphi_{1}\overline{\varphi}_{1}+\sum_{i}\varphi_{i}\overline{\varphi}_{i}\right)$.
Let us write $H=A+A^{\ast}$. Since $H$ is Hermitian, by a unitary change of
$e$ we may assume that $H$ is diagonal, say
$H=\mbox{diag}\\{h_{2},\ldots,h_{n}\\}$. We have

$\displaystyle-\sqrt{-1}\partial\omega$ $\displaystyle=$
$\displaystyle-\varphi_{1}\overline{\varphi}_{1}\sum_{i}v_{i}\varphi_{i}+\varphi_{1}\sum_{i,j}H_{ij}\varphi_{i}\overline{\varphi}_{j}\,=\,-\varphi_{1}\overline{\varphi}_{1}\sum_{i}v_{i}\varphi_{i}+\varphi_{1}\sum_{i}h_{i}\varphi_{i}\overline{\varphi}_{i},$
$\displaystyle\partial\omega\wedge\overline{\partial}\omega$ $\displaystyle=$
$\displaystyle-\varphi_{1}\overline{\varphi}_{1}\big{(}\sum_{i}h_{i}\varphi_{i}\overline{\varphi}_{i}\big{)}^{2}\,=\,-\varphi_{1}\overline{\varphi}_{1}\sum_{i,j}h_{i}h_{j}\,\varphi_{i}\overline{\varphi}_{i}\varphi_{j}\overline{\varphi}_{j}\,,$
$\displaystyle\sqrt{-1}\partial\overline{\partial}\omega$ $\displaystyle=$
$\displaystyle-\varphi_{1}\overline{\varphi}_{1}\sum_{i,j}L_{ij}\varphi_{i}\overline{\varphi}_{j},\
\ $

where $L=\lambda H+A^{\ast}H+HA=\lambda H+H^{2}+[A^{\ast},A]$. Now assume that
$n\geq 4$. The astheno-Kähler condition means that

(34)
$\frac{1}{n-2}\partial\overline{\partial}\omega^{n-2}=\partial\overline{\partial}\omega\wedge\omega^{n-3}+(n-3)\partial\omega\wedge\overline{\partial}\omega\wedge\omega^{n-4}=0.$

For any $2\leq i<j\leq n$, if we wedge the above equality with
$\varphi_{i}\overline{\varphi}_{j}$, then the second term on the right hand
side produce zero, while the first term gives us $L_{ij}=0$, $\forall\,i\neq
j$. So $L$, hence $[A^{\ast},A]$, is diagonal. On the other hand, the diagonal
entries

$[A^{\ast},A]_{ii}=[H,A]_{ii}=h_{i}A_{ii}-A_{ii}h_{i}=0,$

so we conclude that $[A^{\ast},A]=0$. Thus $L=\lambda H+H^{2}$. Now if we fix
any $i$ and wedge the equality (34) with $\varphi_{i}\overline{\varphi}_{i}$,
we get

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\sum_{j\neq
i}L_{jj}+\sum_{j\neq k;\,j,k\neq i}h_{j}h_{k}$ $\displaystyle=$
$\displaystyle\mbox{tr}(L)-L_{ii}+(\sum_{j\neq i}h_{j})^{2}-\sum_{j\neq
i}h_{j}^{2}$ $\displaystyle=$
$\displaystyle\mbox{tr}(L)-L_{ii}+(h-h_{i})^{2}-\sum_{j}h_{j}^{2}+h_{i}^{2}$
$\displaystyle=$ $\displaystyle\lambda h+\mbox{tr}(H^{2})-\lambda
h_{i}-h_{i}^{2}+h^{2}-2hh_{i}+h_{i}^{2}-\mbox{tr}(H^{2})+h_{i}^{2}$
$\displaystyle=$ $\displaystyle(\lambda h+h^{2})-(\lambda+2h)h_{i}+h_{i}^{2}$
$\displaystyle=$ $\displaystyle(\lambda+h-h_{i})(h-h_{i}),$

where $h$ stands for $\mbox{tr}(H)$. Therefore each $h_{i}$ is equal to either
$h$ or $\lambda+h$. Suppose there are $k$ entries amongst $h_{2}$ through
$h_{n}$ that equals to $h$, while the other $(n-1-k)$ entries equal to
$\lambda+h$, then we have $h=h_{2}+\cdots+h_{n}=kh+(n-1-k)(\lambda+h)$,
therefore,

(35) $(n-1-k)\lambda+(n-2)h=0.$

To summarize, if $(J,g)$ is a Hermitian structure on an almost abelian Lie
algebra ${\mathfrak{g}}$ and $g$ is astheno-Kähler, then $[A^{\ast},A]=0$, so
$A$ is unitary similar to a diagonal matrix
$\mbox{diag}\\{\lambda_{2},\ldots,\lambda_{n}\\}$ where $\lambda_{i}$ are
eigenvalues of $A$. Furthermore, twice of the real part of $\lambda_{i}$ is
equal to either $\lambda+h$ or $h$, where $h$ is the sum of
$2\mbox{Re}(\lambda_{i})$ for all $2\leq i\leq n$. If we denote by $k$ the
number of these eigenvalues with $2\mbox{Re}(\lambda_{i})=h$, then $k$ and $h$
obeys the equation (35). Equivalently, $A$ can be described in the following
way: there exists a unitary $(n-1)\times(n-1)$ matrix $U$ such that

$UAU^{\ast}=\left[\begin{array}[]{ccc}\lambda_{2}&&\\\ &\ddots&\\\
&&\lambda_{n}\end{array}\right],\ \ \ \ \
\left[\begin{array}[]{ccc}2\mbox{Re}(\lambda_{2})&&\\\ &\ddots&\\\
&&2\mbox{Re}(\lambda_{n})\end{array}\right]=\left[\begin{array}[]{cc}hI_{k}&\\\
&(\lambda+h)I_{n-1-k}\end{array}\right]$

where $k$ is an integer in the range $0\leq k\leq n-1$ and $h$ is determined
by $k$ and $\lambda$ via (35).

Note that $h=\mbox{tr}(A)+\overline{\mbox{tr}(A)}$, so ${\mathfrak{g}}$ is
unimodular if and only if $\lambda+h=0$. In this case (35) becomes
$(1-k)\lambda=0$, thus for the real part of eigenvalues of $A$, one of them
equal to $-\frac{\lambda}{2}$ while the other $n-2$ of them are equal to $0$.
By part (ii) and (v) of Proposition 7, we know that this is exactly the
condition for $g$ to be pluriclosed. This completes the proof of Proposition
2. ∎

##  4\. Lie algebras with $J$-invariant abelian ideal of codimension $2$

Now let us consider Lie algebras with codimension $2$ abelian ideals. Let
${\mathfrak{g}}$ be a Lie algebra and ${\mathfrak{a}}\subseteq{\mathfrak{g}}$
be an abelian ideal of codimension $2$. Note that such ${\mathfrak{g}}$ will
always be solvable of step at most $3$, but in general it will not be of
$2$-step solvable.

Suppose $(J,g)$ is a Hermitian structure on ${\mathfrak{g}}$. Then the
codimension of ${\mathfrak{a}}_{J}:={\mathfrak{a}}\cap{\mathfrak{a}}$ is
either $2$ or $4$, and it is $2$ if and only if
$J{\mathfrak{a}}={\mathfrak{a}}$. In this article we will focus on this case,
and leave the other case to a future project.

Analogous to the almost ableian case, we may take a unitary basis
$\\{e_{1},\ldots,e_{n}\\}$ of ${\mathfrak{g}}^{1,0}$ so that

${\mathfrak{a}}=\mbox{span}_{\mathbb{R}}\\{e_{i}+\overline{e}_{i},\,\sqrt{-1}(e_{i}-\overline{e}_{i});\
2\leq i\leq n\\}.$

We will again call such a basis an admissible frame. Since ${\mathfrak{a}}$ is
abelian and is an ideal, we have

$C^{\ast}_{ij}=D^{j}_{\ast i}=C^{1}_{\ast\ast}=D^{i}_{1\ast}=D^{\ast}_{1i}=0,\
\ \ \ \ \forall\ 2\leq i,j\leq n.$

So the only possibly non-zero components of $C$ and $D$ are

(36) $C^{j}_{1i}=X_{ij},\ \ \ D^{1}_{11}=\lambda,\ \ \ D^{j}_{i1}=Y_{ij},\ \ \
D^{1}_{ij}=Z_{ij},\ \ \ D^{1}_{i1}=v_{i},\ \ \ \ \ 2\leq i,j\leq n,$

where $\lambda\geq 0$, $v\in{\mathbb{C}}^{n-1}$ is a column vector and $X$,
$Y$, $Z$ are $(n-1)\times(n-1)$ complex matrices. Here and from now on we have
rotated the angle of $e_{1}$ to assume that $D^{1}_{11}\geq 0$. Note that when
$\\{e_{2},\ldots,e_{n}\\}$ is changed to
$\\{\tilde{e}_{2},\ldots,\tilde{e}_{n}\\}$ by a unitary matrix $U$, then $v$
is changed to $Uv$, $X$ and $Y$ are changed to $UXU^{\ast}$ and $UYU^{\ast}$
respectively, while $Z$ is changed to $UZ\,^{t}\\!U$. For convenience, let us
write $\varphi$ for the column vector
${}^{t}\\!(\varphi_{2},\ldots,\varphi_{n})$, then the structure equation (16)
becomes (2) stated in §1:

$\left\\{\begin{split}d\varphi_{1}\,=\,-\lambda\varphi_{1}\overline{\varphi}_{1},\hskip
116.65646pt\\\
d\varphi\,=\,-\varphi_{1}\overline{\varphi}_{1}\overline{v}-\varphi_{1}\,^{t}\\!X\varphi+\overline{\varphi}_{1}\overline{Y}\varphi-\varphi_{1}\overline{Z}\,\overline{\varphi}.\end{split}\right.$

By the Bianchi equation (17) - (19), or equivalently, by $d^{2}\varphi=0$, we
get the system of matrix equations (3) stated in §1:

$\left\\{\begin{split}\lambda(X^{\ast}\\!+Y)+[X^{\ast},Y]-Z\overline{Z}\,=\,0,\\\
\lambda Z-(Z\,^{t}\\!X+YZ)\,=\,0.\hskip 34.14322pt\end{split}\right.$

By (13), we know that the possibly non-zero components of the Chern torsion
are:

(37) $T^{1}_{1i}=v_{i},\ \ \ \ T^{1}_{ij}=Z_{ji}-Z_{ij},\ \ \ \
T^{j}_{1i}=Y_{ij}-X_{ij},\ \ \ \ \ 2\leq i,j\leq n.$

###### Proof of Proposition 3:.

Let ${\mathfrak{g}}$ be a Lie algebra with an abelian ideal ${\mathfrak{a}}$
of codimension $2$ and with a Hermitian structure $(J,g)$, and assume that
$J{\mathfrak{a}}={\mathfrak{a}}$. By our discussion above, there exists a
unitary frame under which the structure equation is given by (2), where
$\lambda\geq 0$, $v\in{\mathbb{C}}^{n-1}$ is a column vector and $X$, $Y$, $Z$
are complex $(n-1)\times(n-1)$ matrices satisfying (3). By (12), we know that
${\mathfrak{g}}$ is unimodular if and only if
$\sum_{r=1}^{n}(C^{r}_{ri}+D^{r}_{ri})=0$ for each $1\leq i\leq n$. By (36),
we see that this number is $0$ when $i>1$ and is
$\lambda+\mbox{tr}(Y)-\mbox{tr}(X)$ when $i=1$. So (i) holds. Similarly,
Gauduchon’s torsion $1$-form $\eta$ has components
$\eta_{i}=\sum_{r=1}^{n}T^{r}_{ri}$. By (37), we get
$\eta_{1}=\mbox{tr}(X)-\mbox{tr}(Y)$ and $\eta_{i}=v_{i}$ for $i>1$, so (ii)
holds. Also by (37), we know that $T=0$ if and only if $v=0$ plus $X=Y$ and
$\,{}^{t}\\!Z=Z$. Now assuming that ${\mathfrak{g}}$ is unimodular. Then by
(i) we get $\lambda=0$, hence the first equation of (3) reads
$[X^{\ast},X]=Z\overline{Z}=ZZ^{\ast}$. Taking trace on both sides, we get
$\mbox{tr}(ZZ^{\ast})=0$ thus $Z=0$ and $X$ is normal. So by a unitary change
of $\\{e_{2},\ldots,e_{n}\\}$ if necessary, we may assume that $X$ is
diagonal. This leads to the structure equation stated in case (iii) of the
proposition.

For part (iv), by Lemma 2 we know that $g$ will be pluriclosed if and only if
equation (20) is satisfied. When $i$ or $j$ is $1$, the equation (20)
automatically holds. Take $i=j=1<k,\ell$ in (20), by (36) and (37) we get

$\sum_{r>1}\left((X_{kr}-Y_{kr})\overline{X_{\ell
r}}+(Z_{rk}-Z_{kr})\overline{Z_{r\ell}}+(Y_{r\ell}-X_{r\ell})\overline{Y_{rk}}+\lambda(Y_{k\ell}-X_{k\ell})\right)=0.$

That is,

$XX^{\ast}-YX^{\ast}+\,^{t}\\!Z\overline{Z}-Z\overline{Z}+Y^{\ast}Y-Y^{\ast}X+\lambda(Y-X)=0,$

namely (4) holds. So $g$ is pluriclosed if and only if equations (3) and (4)
hold. This proves part (iv) thus we have completed the proof of Proposition 3.
∎

Note that if we take the trace of (4), we would get

$|\\!|X-Y|\\!|^{2}+|\\!|Z|\\!|^{2}=\mbox{tr}(Z\overline{Z})+\lambda(\mbox{tr}(X)-\mbox{tr}(Y))=\lambda(\mbox{tr}(X)+\overline{\mbox{tr}(X)}),$

where the last equality is from taking trace on the first equation of (3).
When $\lambda=0$ or when $\mbox{tr}(X)+\overline{\mbox{tr}(X)}=0$, we would
have $Z=0$ and $X=Y$. If ${\mathfrak{g}}$ is unimodular, then $\lambda=0$ and
$[X,X^{\ast}]=0$, so $X$ is unitary similar to a diagonal matrix. That is,
when ${\mathfrak{g}}$ is unimodular and $\lambda=0$, pluriclosed metrics admit
unitary frame under which the structure equation becomes:

$d\varphi_{1}=0,\ \ \
d\varphi_{i}=-\overline{v}_{i}\varphi_{1}\overline{\varphi}_{1}+(\overline{a}_{i}\overline{\varphi}_{1}-a_{i}\varphi_{1})\varphi_{i},\
\ \ \ i>1.$

Note that when $v\neq 0$, the above metric is neither balanced nor Chern flat.
Also, there are plenty of solutions to equations (3) and (4) with $\lambda>0$.
For instance, for any positive constant $\lambda>0$, if we let

$Z=0,\ \ \ \ X=-Y=\frac{\lambda}{2}\,\mbox{diag}\\{1,0,\ldots,0\\},$

and let $v$ be arbitrary, then ${\mathfrak{g}}$ would be unimodular and
equations (3) and (4) are satisfied, hence $g$ is pluriclosed.

###### Proof of Proposition 4:.

Let $({\mathfrak{g}},J,g)$ be a Hermitian structure on a Lie algebra which
contains a $J$-invariant abelian ideal of codimension $2$. First let us
examine the behavior of the Chern curvature tensor. By (14), (36), and (21) -
(23), we have

(40) $\displaystyle Ric^{(1)}$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}s,&0\\\ 0,&0\end{array}\right],\ \ \ \
\ \ \mbox{where}\ \ \
s=-\lambda(2\lambda+\mbox{tr}(Y)+\overline{\mbox{tr}(Y)}),$ (43)
$\displaystyle Ric^{(2)}$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}-(|v|^{2}+|\\!|Z|\\!|^{2}+2\lambda^{2}),&-(\,^{t}\\!vZ^{\ast}+v^{\ast}Y)\\\
-(Z\bar{v}+Y^{\ast}v),&vv^{\ast}+ZZ^{\ast}+[Y,Y^{\ast}]-\lambda(Y+Y^{\ast})\end{array}\right],$
(46) $\displaystyle Ric^{(3)}$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}\hat{s},&-\,^{t}\\!v\overline{Z}\\\
-Y^{\ast}v,&0\end{array}\right],\ \ \ \ \ \mbox{where}\ \ \
\hat{s}=-|v|^{2}-\lambda(2\lambda+\mbox{tr}(Y)+\overline{\mbox{tr}(X)}).$

In the last line we used the fact that
$\mbox{tr}(Z\overline{Z})=\lambda(\mbox{tr}(Y)+\overline{\mbox{tr}(X)})$ by
taking trace on the first equation of (3). Now if $R=0$, then by
$Ric^{(2)}_{1\bar{1}}=0$, we get $\lambda=0$, $v=0$ and $Z=0$. Thus the only
possibly non-zero components of $D$ are $D^{j}_{i1}=Y_{ij}$ for $i,j>1$. By
(14), for $2\leq k,\ell\leq n$, we have

$\left(R_{1\bar{1}k\bar{\ell}}\right)=YY^{\ast}-Y^{\ast}Y.$

So $Y$ is normal and can be diagonalizable via unitary similarity. On the
other hand, when $\lambda=0$ and $Z=0$, the first equation of (3) says that
$[X^{\ast},Y]=0$, so if we write

$Y=\left[\begin{array}[]{ccc}\lambda_{1}I_{n_{1}}&&\\\ &\ddots&\\\
&&\lambda_{r}I_{n_{r}}\end{array}\right],$

where $n_{1}+\cdots+n_{r}=n-1$ and $\lambda_{1},\ldots,\lambda_{r}$ are
distinct, then $X$ would be block-diagonal

$X=\left[\begin{array}[]{ccc}X_{1}&&\\\ &\ddots&\\\ &&X_{r}\end{array}\right]$

where $X_{i}$ is any $n_{i}\times n_{i}$ complex matrix, $1\leq i\leq r$.
Therefore if $R=0$, then $\lambda=0$, $v=0$, $Z=0$, $Y$ is normal, and
$[X^{\ast},Y]=0$. By a unitary change of the frame, $Y$ can be made diagonal
while $X$ is block-diagonal as above. The converse is certainly true, namely,
in this case the only possibly non-zero components of $D$ are
$D^{i}_{i1}=Y_{ii}$, and by (14) one easily checks that $R=0$. This
establishes (i). Part (ii) is immediate from the above formula for the first
Chern Ricci $Ric^{(1)}$ and Chern scalar curvature $s$. For part (iii), assume
that the second Chern Ricci is non-neagtive. By looking at the component
$Ric^{(2)}_{1\bar{1}}$ in (43), we know that we must have $\lambda=0$, $v=0$,
$Z=0$. The lower right corner of (43) now says that $[Y,Y^{\ast}]\geq 0$. But
the matrix has zero trace, so it must be identically zero, and we end up in
the $R=0$ situation.

For part (iv), note that Gauduchon’s torsion $1$-form $\eta$ has components
$\eta_{1}=\mbox{tr}(Y-X)$ and $\eta_{i}=v_{i}$ for $i>1$. So by (36), (2) and
(37) we obtain the following

$\displaystyle d\eta$ $\displaystyle=$ $\displaystyle
d(\sum_{r}\eta_{r}\varphi_{r})\,=\,\eta_{1}d\varphi_{1}-\sum_{r>1}v_{r}\left(\overline{v}_{r}\varphi_{1}\overline{\varphi}_{1}+\sum_{i>1}\\{X_{ir}\varphi_{1}\varphi_{i}+\overline{Y}_{ri}\varphi_{i}\overline{\varphi}_{1}+\overline{Z}_{ri}\varphi_{1}\overline{\varphi}_{i}\\}\right)$
$\displaystyle=$
$\displaystyle-(\lambda\eta_{1}+|v|^{2})\varphi_{1}\overline{\varphi}_{1}-\sum_{i,r>1}v_{r}\left(X_{ir}\varphi_{1}\varphi_{i}+\overline{Y}_{ri}\varphi_{i}\overline{\varphi}_{1}+\overline{Z}_{ri}\varphi_{1}\overline{\varphi}_{i}\right)$

Since the trace of the connection matrices for Chern and Bismut are related by
$\mbox{tr}(\theta^{b})=\mbox{tr}(\theta)+\eta-\overline{\eta}$, so the Bismut
Ricci form is given by

$\displaystyle\mbox{tr}(\Theta^{b})$ $\displaystyle=$
$\displaystyle\mbox{tr}(\Theta)+d\eta-d\overline{\eta}\ \,=\
\,s\varphi_{1}\overline{\varphi}_{1}+d\eta-d\overline{\eta}$ $\displaystyle=$
$\displaystyle-\left(2|v|^{2}+\lambda\\{2\lambda+2\mbox{tr}(Y)+2\overline{\mbox{tr}(Y)}-\mbox{tr}(X)-\overline{\mbox{tr}(X)}\\}\right)\varphi_{1}\overline{\varphi}_{1}+$
$\displaystyle+\sum_{i,r>1}\left(\overline{v}_{r}\overline{X_{ir}}\overline{\varphi}_{1}\overline{\varphi}_{i}-v_{r}X_{ir}\varphi_{1}\varphi_{i}\right)-\sum_{i,r>1}\left(v_{r}\overline{Z_{ri}}+\overline{v}_{r}Y_{ri}\right)\varphi_{1}\overline{\varphi}_{i}$
$\displaystyle-\sum_{i,r>1}\left(v_{r}\overline{Y_{ri}}+\overline{v}_{r}Z_{ri}\right)\varphi_{i}\overline{\varphi}_{1}$

That is,

(47)
$\left\\{\begin{split}(Ric^{b})^{2,0}\,=\,\frac{1}{2}\left[\begin{array}[]{cc}0&{}^{t}\\!v\,^{t}\\!X\\\
-Xv&0\end{array}\right],\hskip 73.97733pt\\\
(Ric^{b})^{1,1}\,=\,\left[\begin{array}[]{cc}s^{b}&-(v^{\ast}Y+\,^{t}\\!v\overline{Z})\\\
-(Y^{\ast}v+\,^{t}\\!Z\overline{v})&0\end{array}\right],\end{split}\right.$

where the Bismut scalar curvature $s^{b}$ is given by (6) as follows:

$s^{b}\,=\,-2|v|^{2}-\lambda(2\lambda+2\mbox{tr}(Y)+2\overline{\mbox{tr}(Y)}-\mbox{tr}(X)-\overline{\mbox{tr}(X)}).$

When ${\mathfrak{g}}$ is unimodular, $\lambda=\mbox{tr}(X)-\mbox{tr}(Y)$, and
$s^{b}=-2|v|^{2}-\lambda(\mbox{tr}(Y)+\overline{\mbox{tr}(Y)})$. By (47), we
know that the rank of the $(2,0)$-part of $Ric^{b}$ is either $2$ or $0$ (when
$Xv=0$), while the $(1,1)$-part of $Ric^{b}$ has rank at most $3$, or at most
$1$ when $Y^{\ast}v+\,^{t}\\!Z\overline{v}=0$. The entire $Ric^{b}=0$ if and
only if $s^{b}=0$ and $Xv=0$, $Y^{\ast}v+\,^{t}\\!Z\overline{v}=0$. This
proves part (iv) and completes the proof of Proposition 4. ∎

The remaining part of the article is devoted to the proof of Proposition 5,
where we want to characterize Bismut torsion-parallel (BTP) and Bismut Kähler-
like (BKL) metrics on Lie algebras with $J$-invariant abelian ideal of
codimension $2$. Since BKL means BTP plus pluriclosed [49], we will just focus
on the BTP condition.

###### Proof of Proposition 5:.

Again let $({\mathfrak{g}},J,g)$ be a Hermitian structure on a unimodular Lie
algebra with a $J$-invariant abelian ideal of codimension $2$. Assume that $g$
is BTP. This means that both (32) and (33) are satisfied, which now takes the
form

(48)
$\left\\{\begin{split}\sum_{r}\left(T^{j}_{rk}\tilde{\Gamma}^{r}_{i\ell}+T^{j}_{ir}\tilde{\Gamma}^{r}_{k\ell}-T^{r}_{ik}\tilde{\Gamma}^{j}_{r\ell}\right)=0,\\\
\sum_{r}\left(T^{j}_{rk}\overline{\tilde{\Gamma}^{i}_{r\ell}}+T^{j}_{ir}\overline{\tilde{\Gamma}^{k}_{r\ell}}-T^{r}_{ik}\overline{\tilde{\Gamma}^{r}_{j\ell}}\right)=0\end{split}\right.$

for any indices $1\leq i<k\leq n$ and any $1\leq j,\,\ell\leq n$, where
$\tilde{\Gamma}=D+T$ are the connection coefficients for the Bismut
connection. For simplicity, let us denote by

$A=\,^{t}\\!Z-Z,\ \ \ \ B=Y-X.$

Then the only possibly non-zero components of $T$ and $\tilde{\Gamma}$ are

$\left\\{\begin{split}T^{1}_{1i}=v_{i},\ \ \ T^{1}_{ij}=A_{ij},\ \ \
T^{j}_{1i}=B_{ij};\hskip 105.2751pt\\\ \tilde{\Gamma}^{1}_{11}=\lambda,\ \ \
\tilde{\Gamma}^{1}_{1i}=v_{i},\ \ \ \tilde{\Gamma}^{1}_{ij}=Z_{ji},\ \ \
\tilde{\Gamma}^{j}_{i1}=X_{ij},\ \ \
\tilde{\Gamma}^{j}_{1i}=B_{ij},\end{split}\right.$

where $2\leq i,j\leq n$. We remark that when the basis
$\\{e_{2},\ldots,e_{n}\\}$ is changed to
$\\{\tilde{e}_{2},\ldots,\tilde{e}_{n}\\}$ by a unitary matrix $U\in U(n-1)$,
the matrix $A$ is changed into $UA\,^{t}\\!U$, just like $Z$, while $B$ is
changed to $UBU^{\ast}$, just like $X$ or $Y$. So strictly speaking we should
denote the entries of $B$ as $B_{i}^{\ j}$ or $B_{i\bar{j}}$, but for the sake
of simplicity we will just write it as $B_{ij}$.

Note that for both $T$ and $\tilde{\Gamma}$, the components are zero unless at
least one of the indices is $1$. If we take $i,j,k,\ell>1$ in (48), then only
the $r=1$ terms survive, and we get

(49) $\displaystyle B_{kj}Z_{\ell i}-B_{ij}Z_{\ell k}-A_{ik}B_{\ell j}$
$\displaystyle=$ $\displaystyle 0,$ (50) $\displaystyle
B_{kj}\overline{B_{\ell i}}-B_{ij}\overline{B_{\ell
k}}-A_{ik}\overline{Z_{\ell j}}$ $\displaystyle=$ $\displaystyle 0,$

where $2\leq i,j,k,\ell\leq n$. Write $t_{B}=\mbox{tr}(B)$ and
$t_{Z}=\mbox{tr}(Z)$. By letting $k=j$ and summing up, or $k=\ell$ and summing
up, or $j=\ell$ and summing up, we get

(51) $\displaystyle\left\\{\begin{split}t_{B}Z-Z\,^{t}\\!B+BA=0\\\
t_{B}\overline{B}-\overline{B}\,^{t}\\!B+\overline{Z}A=0\end{split}\right.$
(52) $\displaystyle\left\\{\begin{split}\,{}^{t}\\!ZB-t_{Z}B-AB=0\\\
B^{\ast}B-\overline{t_{B}}B-A\overline{Z}=0\end{split}\right.$ (53)
$\displaystyle\left\\{\begin{split}BZ-\,^{t}\\!Z\,^{t}\\!B+t_{B}A=0\\\
B\overline{B}-B^{\ast}\,{}^{t}\\!B+\overline{t_{Z}}A=0\end{split}\right.$

Next we consider the cases where at least one of the four indices is $1$. If
$i=j=\ell=1$, then by the fact that
$\tilde{\Gamma}^{i}_{11}=\tilde{\Gamma}^{1}_{i1}=0$ for any $i>1$, (48) yields

(54) $Xv=X^{\ast}v=0.$

Similarly, when exactly one of the indices is $1$, the $\ell=1$ case of (48)
automatically holds, while the $j=1$ and $i=1$ cases lead to

(55) $\displaystyle\left\\{\begin{split}v_{k}Z_{\ell i}-v_{i}Z_{\ell
k}-v_{\ell}A_{ik}=0,\\\ v_{k}\overline{B_{\ell i}}-v_{i}\overline{B_{\ell
k}}-\overline{v_{\ell}}A_{ik}=0,\end{split}\right.\ \ \ \ \forall\ 2\leq
i,k,\ell\leq n;$ (56)
$\displaystyle\left\\{\begin{split}v_{\ell}B_{kj}-v_{k}B_{\ell j}=0,\\\
\overline{v_{\ell}}B_{kj}-v_{k}\overline{Z_{\ell j}}=0,\end{split}\right.\ \ \
\ \ \ \ \ \ \ \ \ \forall\ 2\leq j,k,\ell\leq n.$

Now consider the cases when exactly two of the four indices are $1$, (48)
yields the following

(57) $\displaystyle BA=Z\,^{t}\\!B,\ \ \ \
\overline{Z}A=\overline{B}\,^{t}\\!B;$ (58)
$\displaystyle[B,X]=[B,X^{\ast}]=\lambda B;$ (59) $\displaystyle
XA+A\,^{t}\\!X=X^{\ast}A+A\overline{X}=\lambda A.$

Of course the Lie algebra ${\mathfrak{g}}$ will be unimodular if and only if

(60) $\lambda+t_{B}=0.$

The metric $g$ will be BTP when and only when (3) and (49) – (60) are all
satisfied. When $v\neq 0$, it is easy to find out all the solutions to this
system of matrix equations. By a unitary change of $\\{e_{2},\ldots,e_{n}\\}$
if necessary, we may assume that $v_{2}>0$ and $v_{3}=\cdots=v_{n}=0$. By (55)
and (56), we know all rows except the first of $Z$ are zero, and
$B=\overline{Z}$. The other equations now imply that

$\lambda=0,\ \ X=\left[\begin{array}[]{cc}0&0\\\ 0&X_{1}\end{array}\right],\ \
\ Z=\left[\begin{array}[]{cc}0&\,{}^{t}\\!z\\\ 0&0\end{array}\right],$

and $X_{1}z=X_{1}^{\ast}z=0$, $[X_{1},X_{1}^{\ast}]=0$. So depending on
whether $z=0$ or not, we end up in one of the two cases for BTP metrics with
$v\neq 0$ on unimodular ${\mathfrak{g}}$: there exists unitary frame so that
the structure equation takes the form:

$\displaystyle\left\\{\begin{split}d\varphi_{1}=0,\hskip 119.50148pt\\\
d\varphi_{2}=-v_{2}\varphi_{1}\overline{\varphi}_{1},\hskip 85.35826pt\\\
d\varphi_{i}=(\overline{a}_{i}\overline{\varphi}_{1}-a_{i}\varphi_{1})\varphi_{i},\
\ \ 3\leq i\leq n;\end{split}\right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
or $\displaystyle\left\\{\begin{split}d\varphi_{1}=d\varphi_{3}=0,\hskip
89.626pt\\\
d\varphi_{2}=-v_{2}\varphi_{1}\overline{\varphi}_{1}+p\,(\overline{\varphi}_{1}\varphi_{3}-\varphi_{1}\overline{\varphi}_{3}),\hskip
2.84544pt\\\
d\varphi_{i}=(\overline{a}_{i}\overline{\varphi}_{1}-a_{i}\varphi_{1})\varphi_{i},\
\ \ 4\leq i\leq n.\end{split}\right.$

They are respectively (7) and (8) stated in §1 before. Here $v_{2}>0$, $p>0$,
and $a_{i}$ are arbitrary complex numbers.

Next we consider the $v=0$ case. First we claim that $\lambda$ must zero.
Assume otherwise, then by comparing (51) with (57), we see that $B$ and $Z$
must be zero, which forces $\lambda=-t_{B}=0$ since ${\mathfrak{g}}$ is
unimodular. So $\lambda$ must be zero. Next we claim that $BZ=0$. Since
$t_{B}=0$, by the first line of (53), we know that $BZ$ is symmetric. On the
other hand, by the first equation of (57), we get

$\,{}^{t}(BA)=B\,^{t}\\!Z=B(A+Z)=BA+BZ.$

Hence $BZ$ is skew-symmetric. Therefore, we must have $BZ=0$ and the claim is
proved. So for unimodular ${\mathfrak{g}}$ with BTP metric and with $v=0$, the
equations (3) and (49) – (60) become

(61) $\left\\{\begin{split}\lambda=t_{B}=0,\ \ v=0,\ \ BZ=0,\ \ XA,\
X^{\ast}A,\ BA=B\,^{t}\\!Z\ \ \mbox{are symmetric},\\\ [B,X]=[B,X^{\ast}]=0,\
\ [X^{\ast},X]=Z\overline{Z},\ \ Z\,^{t}\\!X+XZ=0,\hskip 66.86414pt\\\
\overline{Z}A=\overline{B}\,^{t}\\!B,\ \ A\overline{Z}=B^{\ast}B,\ \
ZB=t_{Z}B,\ \ \,^{t}\\!(B\overline{B})-B\overline{B}=\overline{t_{Z}}A.\hskip
42.67912pt\end{split}\right.$

Since
$B^{\ast}B=A\overline{Z}=\,^{t}\\!Z\overline{Z}-Z\overline{Z}=\,^{t}\\!Z\overline{Z}-[X^{\ast},X]$,
by taking trace we get $|\\!|B|\\!|^{2}=|\\!|Z|\\!|^{2}$. Therefore, if $B=0$
or $Z=0$, we get $B=Z=A=0$ and the metric $g$ will be Kähler and flat.
Similarly, if $A=0$, then by $\overline{B}\,^{t}\\!B=\overline{Z}A$ we get
$B=0$ hence $Z=0$. So in the following let us assume that $A,Z,B$ are all non-
trivial.

Since $BZ=0$, $Z$ must be singular. Denote by $r$ its rank, we have $1\leq
r<n-1$. Choose unitary basis $\\{e_{2},\ldots,e_{n}\\}$ so that the first $r$
rows of $Z$ are linearly independent while the other rows are all zero, and
write $B$ in the same block form:

$Z=\left[\begin{array}[]{cc}x&y\\\ 0&0\end{array}\right],\ \ \
B=\left[\begin{array}[]{cc}0&u\\\ 0&w\end{array}\right]$

where $x$ is an $r\times r$ matrix, and the first column blocks of $B$ vanish
because $BZ=0$. Since $\overline{Z}A=\overline{B}\,^{t}\\!B$, the lower right
corner gives us $\overline{w}\,^{t}\\!w=0$ hence $w=0$. Therefore $B$ has only
the upper right corner thus $B\overline{B}=0$, which leads to
$\overline{t_{Z}}A=0$. Therefore we must have $t_{Z}=0$ as $A$ is assumed to
be non-trivial.

Now we claim that the block $x$ must also be zero. Since $w=0$, by comparing
the upper left corner of $A\overline{Z}=B^{\ast}B$, we get
$(\,^{t}\\!x-x)\overline{x}=0$, so $\mbox{tr}(x\overline{x})=|\\!|x|\\!|^{2}$.
On the other hand, since $Z\overline{Z}=[X^{\ast},X]$, we have
$\mbox{tr}(x\overline{x})=\mbox{tr}(Z\overline{Z})=0$, hence $x=0$.

Note that the $r\times(n-1-r)$ matrix $y$ has rank $r$, so after a unitary
change in the last $(n-1-r)$ elements of $e_{i}$, we may assume that
$y=(z,0)$, where $z$ is an $r\times r$ non-singular matrix. Write $u=(b,c)$
where $b$ is $r\times r$, then by $A\overline{Z}=B^{\ast}B$ we know that we
must have $c=0$, hence the matrices now take the form

$Z=\left[\begin{array}[]{ccc}0&z&0\\\ 0&0&0\\\ 0&0&0\end{array}\right],\ \ \
A=\left[\begin{array}[]{ccc}0&-z&0\\\ \,{}^{t}\\!z&0&0\\\
0&0&0\end{array}\right],\ \ \ B=\left[\begin{array}[]{ccc}0&b&0\\\ 0&0&0\\\
0&0&0\end{array}\right],$

where $z$ and $b$ are $r\times r$ matrices with $z$ non-singular. The
equations in (61) implies that

(62) $\overline{z}\,^{t}\\!z=\overline{b}\,^{t}\\!b,\ \ \
^{t}\\!z\overline{z}=b^{\ast}b,\ \ \ b\,^{t}\\!z=z\,^{t}\\!b,$

and via the non-singularity of $b$ that

$X=\left[\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ 0&0&x\end{array}\right],\ \ \ \
[x^{\ast},x]=0.$

All the solutions to the system of matrix equations (62) are given by the
following elementary lemma, and we include its proof here for readers
convenience:

######  Lemma 9.

Two non-singular square matrices $z$ and $b$ satisfy (62) if and only if

$b=USV^{\ast},\ \ \ z=USW\,^{t}\\!V,$

where $S>0$ is diagonal, $U$, $V$, $W$ are unitary, with $\,{}^{t}\\!W=W$ and
$WS=SW$.

###### Proof of Lemma 9:.

Since $\overline{z}\,^{t}\\!z=\overline{b}\,^{t}\\!b$ is positive definite, it
equals to $US^{2}U^{\ast}$ for some unitary matrix $U$ and a positive definite
diagonal matrix $S$. This means that $V^{\ast}=S^{-1}U^{-1}\overline{b}$ and
$Q=S^{-1}U^{-1}\overline{z}$ are both unitary. Plug
$b=\overline{U}S\,^{t}\\!V$ and $z=\overline{U}S\overline{Q}$ into the other
two equations of (62), we see that the unitary matrix $W=Q\overline{V}$ is
symmetric and commutes with $S^{2}$, hence with $S$. This completes the proof
of the lemma. ∎

Now let us continue with the proof of Proposition 5. If we group $S$ into
blocks with equal diagonal values, then $W$ is block-diagonal, with each block
being symmetric and unitary. Write $S_{ij}=\lambda_{i}\delta_{ij}$, $2\leq
i,j\leq r+1$. Clearly, we can choose the unitary basis $e$ so that

$Z=\left[\begin{array}[]{ccc}0&SW&0\\\ 0&0&0\\\ 0&0&0\end{array}\right],\ \ \
X=\left[\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ 0&0&x\end{array}\right],\ \ \
Y=X+B=\left[\begin{array}[]{ccc}0&S&0\\\ 0&0&0\\\ 0&0&x\end{array}\right]$

where $x$ is diagonal. The structure equation now takes the form

$\left\\{\begin{split}d\varphi_{1}=0,\hskip 182.09746pt\\\
d\varphi_{i}=\lambda_{i}\overline{\varphi}_{1}\varphi_{i}-\lambda_{i}\varphi_{1}\sum_{j=2}^{r+1}\overline{W_{ij}}\varphi_{j},\
\ \ \ \ 2\leq i\leq r+1,\hskip 2.84544pt\\\ d\varphi_{i}=0,\ \ \ \ \ \ \ \ \
r+2\leq i\leq 2r+1,\hskip 69.70915pt\\\
d\varphi_{i}=(\overline{a}_{i}\overline{\varphi}_{1}-a_{i}\varphi_{1})\varphi_{i},\
\ \ \ \ 2r+2\leq i\leq n,\hskip 34.14322pt\end{split}\right.$

which is (9) given in §1 before. Here each $a_{i}$ is complex,
$S=\mbox{diag}\\{\lambda_{2},\ldots,\lambda_{r+1}\\}>0$ and $W$ is any unitary
symmetric $r\times r$ matrix satisfying $WS=SW$. This completes the proof of
Proposition 5. ∎

Note that all three cases in Proposition 5 are non-Kähler and not Chern flat,
with the first two cases being not balanced while the third case being
balanced. Also, since a Hermitian metric is Bismut Kähler-like if and only if
it is both BTP and pluriclosed by [49], so from (4) we see that (7) is
pluriclosed hence is Bismut Kähler-like, while (8) and (9) are not plurilcosed
hence not Bismut Kähler-like.

Acknowledgments. The second named author would like to thank Bo Yang and
Quanting Zhao for their interests and/or helpful discussions.

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# Physics Masterclasses in Africa and the World

Uta Bilow<EMAIL_ADDRESS>Kenneth Cecire<EMAIL_ADDRESS>Institut für
Kern- und Teilchenphysik, Technische Universität Dresden, Dresden, 01062,
Germany  Department of Physics and Astronomy, University of Notre Dame, Notre
Dame, 46556, Indiana, USA

###### Abstract

International Masterclasses (IMC) enable high school students and teachers to
work with particle physicists to analyze authentic data from contemporary
experiments and experience being “physicists for a day”. The IMC program has a
worldwide reach, including several universities and research institutes in
Egypt, Algeria, Morocco, Sao Tomé and Principe, and South Africa. As technical
infrastructure in Africa improves, there is a great opportunity for many more
African institutes to offer IMC on their premises. The authors will discuss
the advantages of IMC to Africa, how institutes may join, and ways to overcome
obstacles.

###### keywords:

The African School of Physics , ASP , The African Conference on Fundamental
and Applied Physics , ACP , International Particle Physics Outreach Group ,
IPPOG , International Masterclasses , IMC , education , outreach , particle
physics

contents=  Submitted to the Proceedings of the African Conference on
Fundamental and Applied Physics Second Edition, ACP2021, March 7–11, 2022 —
Virtual Event

, scale=1,placement=top,opacity=1,color=black,position=3.25in,1.2in

##  1 Introduction

International Masterclasses (IMC) are experimental particle physics education
and outreach events for high school students aged 15 to 19 years. In the IMC
concept, students become “scientists for a day”. They are invited to a
research institute or a university, where they attend introductory talks on
topics such as the Standard Model, detectors, and accelerators to help them
prepare them to make their own measurements. Students analyze authentic data
from an actual experiment to be able to answer a research question or to draw
conclusions from the measurement. Currently, students at different
masterclasses are engaged in measurements from the Large Hadron Collider (LHC)
at Conseil Européen pour la Recherche Nucléaire (CERN), Belle II at Kō Enerugī
Kasokuki Kenkyū (KEK), neutrino experiments at Fermilab, and particle therapy
from Gesellschaft für Schwerionenforschung (GSI). The capstone of the
masterclass is an international videoconference with 3-5 other masterclass
institutes that have made measurements from the same experiment. [1] [2]

What makes this event a masterclass is the interaction of high school students
and their teachers with the particle physicists who host and tutor the event.
Most important, as students analyze data for their measurements, they interact
with physicists to ask questions and clarify the meaning of the data. It is in
this interaction that the physicists impart not only specific information but
also their approach to understanding both the experiments and the results. One
way to express the goal of a masterclass is to help students see data as
scientists see data.

This paper is organized as follows. In Sections 2, we explain details of the
masterclass day. In Section 3, we discuss goals, motivations, and
requirements. In Section 4, we flesh out the opportunities from different
“flavors” of masterclass. We conclude in Section 5 with an invitation to
participate.

##  2 The Moving Parts

The masterclass day varies depending on the institute and time zone but
generally follows a plan similar to this Sample Agenda:

Local Time | Activity
---|---
(prep) | (classroom)
09:00-09:30 | registration and welcome
09:30-10:30 | introduction to particle physics
10:30-11:30 | second talk or tour
11:30-12:00 | introduction to measurement
12:00-13:00 | lunch (with physicists)
13:00-15:00 | data analysis
15:00-16:00 | local combination of results and discussion
16:00-17:00 | international videoconference

The international videoconference has the following key features:

  * 1.

45-60 minute duration

  * 2.

3-5 institutes, reflecting international collaboration (where possible)

  * 3.

Same measurement, different data

  * 4.

2-3 moderators (physicists, graduate students)

  * 5.

moderation centers: CERN, Fermilab, KEK, GSI, TriIniversity Meson Facility
(TRIUMF)

  * 6.

Agenda of videoconference:

    * (a)

welcome

    * (b)

combination and discussion of results

    * (c)

general Q&A

    * (d)

quiz (CERN).

[1] [3]

##  3 Why and What

Why do we facilitate and promote International Masterclasses? Why do
physicists invest their time and efforts in them? IMC is a way to help start
development of the next generation of scientists and engineers by giving young
learners a real experience with cutting-edge science. In the same way, IMC
promotes interest in science among all students. Visiting a university or
laboratory where they work side-by-side with physicists near their homes as
well as international collaborators, high school students have the chance to
participate in the excitement of contemporary physics research. Teachers who
bring students to masterclasses year after year often develop their own
expertise and interest that can enhance their instruction in the classroom.

To host a masterclass, local organizers need to secure space to work with
students and computers with high-speed internet (unless the masterclass is
local only). All masterclass packages are web browser-based (although some
need installation of Java and software, e.g. event display). All masterclass
packages are available for educational purposes without cost. The physicist
leadership of the masterclass should consist of an organizer, at least one
tutor for each ten students, coordination with IMC, some flexibility to deal
with local conditions and unforeseen problems, and a welcoming attitude.

##  4 Opportunities

The masterclasses described so far have been from the IMC “regular season”.
These are the mostly standard full masterclasses that occur each year from
February to April and are coordinated by IMC central. However, there are
additional opportunities that arise each year.

World Wide Data Day (W2D2) is a one-day simplified masterclass event held in
November or December each year. W2D2 measurements are from the ATLAS and CMS
detectors in the LHC and can be completed in about two hours, from
introduction to videoconference end. High school teachers organize W2D2
measurements at their own schools and connect their students to physicists in
the videoconferences, which occur in a day-long “shift”. [4]

IMC also holds special masterclasses for the International Day of Women and
Girls in Science on 11th February each year. These are much like standard
masterclasses but are designed to especially benefit young women with female
scientists as role models and discussions of women in physics. [1]

In addition, any organizers can host masterclasses anytime of the year, in or
out of the IMC schedule. IMC gives these initiatives support upon request and
many institutions, from universities and laboratories to high schools, have
successfully created meaningful particle physics masterclasses opportunities
for students.

IMC is open to alternative designs of masterclasses as needed by local
organizers. There have been masterclasses for university physics students,
masterclasses as outreach activities in conferences, and many more examples.

##  5 Conclusion and Invitation

International Masterclasses have spread to countries all around the world.
Africa is part of IMC but there is great room for mutual growth and enrichment
for high school students and teachers between IMC and physicists in Africa.
Some of these opportunities include masterclasses already described,
adaptations to make the masterclasses more appropriate in the local context,
and masterclasses redesigned for African physics experiments, such as HESS or
the African Light Source.

The authors invite physicists and graduate students in Africa and the world to
think about organizing a masterclass, tutoring in a masterclass, becoming a
videoconference moderator, or even creating a new masterclass measurement. The
authors, as co-Coordinators of International Masterclasses, are eager to help.

## References

  * [1] International Masterclasses, https://www.physicsmasterclasses.org/.
  * [2] Uta Bilow, Kenneth Cecire., Current Status of International Particle Physics Masterclasses, https://doi.org/10.22323/1.390.0937 (2020).
  * [3] LHC Masterclasses Project Map 2022, https://quarknet.org/content/lhc-masterclass-library-project-map (2022).
  * [4] World Wide Data Day, https://quarknet.org/content/world-wide-data-day (2022).