# Fino-Vezzoni conjecture on Lie algebras with abelian ideals of codimension
two

Kexiang Cao  Kexiang Cao. 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.

In this paper, we confirm the Fino-Vezzoni Conjecture for unimodular Lie
algebras which contain abelian ideals of codimension two, a natural
generalization to the class of almost abelian Lie algebras. This provides new
evidence towards the validity of the conjecture on this very special type of
$3$-step solvmanifolds.

###### Key words and phrases:

Hermitian manifolds; balanced metrics; pluriclosed metrics; Fino-Vezzoni
Conjecture

###### 2010 Mathematics Subject Classification:

53C55 (primary), 53C05 (secondary)

The corresponding author Zheng is partially supported by National Natural
Science Foundations of China with the grant No.12071050 and 12141101,
Chongqing grant cstc2021ycjh-bgzxm0139 and CQYC2021059145, and is supported by
the 111 Project D21024.

###### Contents

  1. 1 Introduction and statement of results
  2. 2 Preliminary on Hermitian Lie algebras
  3. 3 Structure constants
  4. 4 Balanced metrics
  5. 5 Pluriclosed metrics

##  1\. Introduction and statement of results

An important open question in non-Kähler geometry is the following interesting
conjecture raised by Fino and Vezzoni ([17, 18]):

######  Conjecture 1 (Fino-Vezzoni).

Let $M^{n}$ be a compact complex manifold. If $M^{n}$ admits a balanced
Hermitian metric and a pluriclosed Hermitian metric, then it must admit a
Kähler metric.

Recall that a Hermitian metric $g$ on $M^{n}$ is said to be balanced if
$d(\omega^{n-1})=0$, and $g$ is said to be pluriclosed if
$\partial\overline{\partial}\omega=0$. Here $\omega$ is the Kähler form of
$g$. Balanced metrics and pluriclosed metrics are natural extensions of Kähler
metrics, and they are two of the special Hermitian structures that are
extensively studied. See [12, 21, 26, 30, 44] for a sample of discussion on
balanced metrics and see [4, 9, 16, 33, 34] for pluriclosed metrics.

The aforementioned conjecture simply says that these two types of special
Hermitian metrics cannot coexist on any compact complex manifold, unless the
manifold is Kählerian (i.e., admitting a Kähler metric).

A few quick remarks about the conjecture. The first one is that the dimension
$n$ starts with $3$ here, as in $n=2$ the balanced condition coincides with
the Kähler condition, which is $d\omega=0$. Also, the compactness assumption
in the conjecture is necessary, and there are counterexamples in the non-
compact case. Finally, if a Hermitian metric $g$ is both balanced and
pluriclosed, then it must be Kähler by a theorem of Alexandrov and Ivanov [1].
So the main difficulty of the conjecture is caused by the fact that the
balanced metric and the pluriclosed metric in the assumption are allowed to be
different.

Since its inception, the conjecture has drawn attention from many complex
geometers and it has been confirmed in a number of special situations. For
example, Verbitsky in [39] showed that any twistor space of non-Kählerian type
cannot admit any pluriclosed metric, so the Fino-Vezzoni Conjecture holds for
all twistor spaces. In [7], Chiose proved that the conjecture holds for
manifolds in the Fujiki ${\mathcal{C}}$ class, which means compact complex
manifolds that are bimeromorphic to compact Kähler manifolds. Fu, Li, and Yau
in [22] proved that the conjecture holds for a special type of non-Kähler
Calabi-Yau threefolds. Also, Fei [10] confirmed the conjecture for generalized
Calabi-Gray manifolds, and Otiman [29] proved that the conjecture holds for
all Oeljeklaus-Toma manifolds.

In [43], Zhao and the second named author proved that the conjecture holds on
any compact non-Kähler Bismut Kähler-like (BKL) manifolds. Here BKL means that
the curvature tensor of the Bismut connection obeys all Kähler symmetries.
Such a metric is always pluriclosed, and when a compact complex manifold
admits a non-Kähler BKL metric, then it cannot admit any balanced metric.

Next we consider Lie-complex manifolds, which means compact complex manifolds
with universal cover $(G,J)$, where $G$ is a Lie group and $J$ a left-
invariant complex structure on $G$. So Lie-complex manifolds are in the form
$M=G/\Gamma$ where $\Gamma$ is a discrete subgroup in the automorphism group
of $(G,J)$. A commonly occurred special case is when $\Gamma$ is a discrete
subgroup of $G$ (acting as left-multiplications on $G$) with smooth compact
quotient (i.e., uniform lattice). When $G$ is nilpotent or solvable, $M$ is
called a nilpotent Lie-complex manifold or solvable Lie-complex manifold. In
this article, we will simply call it a nilmanifold or solvmanifold.

Nilpotent groups admit uniform lattices if and only if they have rational
structure constants by the result of Malcev. But for a given solvable group,
whether it admits a uniform lattice or not is often a challenging problem in
Lie group theory. See [6, 8, 23] as a glimpse of examples. Also, given an
even-dimensional Lie group, whether or not it admits a left-invariant complex
structure might be complicated as well and it depends on the structure of the
Lie group. Clearly, left-invariant complex structures on $G$ are in one-one
correspondence with the complex structures on its Lie algebra
${\mathfrak{g}}$, namely, linear maps
$J:{\mathfrak{g}}\rightarrow{\mathfrak{g}}$ satisfying $J^{2}=-I$ and the
integrability condition

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

By the work of Fino-Grantcharov [11] and Ugarte [38], when the complex
structure is left-invariant, if the compact quotient $M$ admits a balanced
metric, then through the averaging process it also admits a balanced metric
that is left-invariant on $G$. Similarly, if $M$ admits a pluriclosed metric,
then it also admits a pluriclosed metric whose lift on $G$ is left-invariant.
So in order to verify the Fino-Vezzoni Conjecture for a Lie-complex manifold,
it suffices to consider Hermitian structures $(J,g)$ on the Lie algebra
${\mathfrak{g}}$, where $J$ is a complex structure and $g=\langle,\rangle$ is
an inner product on ${\mathfrak{g}}$ compatible with $J$ (i.e., making $J$
orthogonal).

In [17], Fino and Vezzoni proved that their conjecture holds for all
$6$-dimensional nilpotent groups and all $6$-dimensional solvable groups of
Calabi-Yau type. In [18], they confirmed the conjecture for all $2$-step
nilpotent groups in any (even) dimension. They also conjectured that if a
nilpotent group admits a pluriclosed metric, then it must be of step at most
$2$. This last statement was proved recently by Arroyo and Nicolini in [5]. So
in particular we know that Conjecture 1 holds for all nilmanifolds.

Beyond nilmanifolds, [12] and [31] confirmed the conjecture for all compact
semi-simple Lie groups. Giusti and Podestà [24] proved the conjecture for all
regular complex structures on non-compact real semi-simple Lie groups. In
their recent work [13, 14, 15], Fino and Paradiso confirmed the conjecture in
the almost abelian case (meaning that ${\mathfrak{g}}$ contains an abelian
ideal of codimension $1$) and the almost nilpotent solvable case (meaning that
the nilradical of ${\mathfrak{g}}$ has $1$-dimensional commutator). In [19,
20], Freibert and Swann systematically studied the Hermitian structures on
$2$-step solvable Lie algebras, and they proved that Conjecture 1 holds for a
large subset, namely for those that are ‘pure type’.

As a natural generalization to the almost abelian case, one could consider
those Lie algebras ${\mathfrak{g}}$ which contain an abelian ideal
${\mathfrak{a}}$ of codimension $2$. Suppose $J$ is a complex structure on
${\mathfrak{g}}$. Then ${\mathfrak{a}}_{J}={\mathfrak{a}}\cap J{\mathfrak{a}}$
is an ideal in ${\mathfrak{g}}$ of codimension equal to either $2$ or $4$,
depending on whether $J{\mathfrak{a}}={\mathfrak{a}}$ or not. Of course the
Hermitian geometric study in the first case (i.e., when ${\mathfrak{a}}$ is
$J$-invariant) is much easier.

In [25], Guo and the second named author studied Hermitian structures on
${\mathfrak{g}}$ with $J$-invariant ${\mathfrak{a}}$ and obtained
characterizations for balanced, pluriclosed, and other special types of
Hermitian metrics. Utilizing this, Li and the second named author in [28]
confirmed Conjecture 1 in this special case:

######  Theorem 1 (Li-Z [28]).

Let ${\mathfrak{g}}$ be a unimodular Lie algebra which contains an abelian
ideal ${\mathfrak{a}}$ of codimension $2$, and $J$ a complex structure on
${\mathfrak{g}}$ satisfying $J{\mathfrak{a}}={\mathfrak{a}}$. Then the Fino-
Vezzoni Conjecture holds, namely, if $({\mathfrak{g}},J)$ admits compatible
metrics $g$ and $h$ so that $g$ is balanced and $h$ is pluriclosed, then it
must admit a compatible Kähler metric.

Note that ${\mathfrak{g}}$ is unimodular if $\,\mbox{tr}(\mbox{ad}_{x})=0$ for
all $x\in{\mathfrak{g}}$. This is a necessary condition for $G$ to have
compact quotients. So in the study of Fino-Vezzoni Conjecture we always
restrict ourselves to unimodular Lie algebras, and without this assumption
there are plenty of counterexamples.

It is quite natural to wonder what happens when
$J{\mathfrak{a}}\neq{\mathfrak{a}}$. In this case the Hermitian geometry on
$({\mathfrak{g}},J)$ becomes much more complicated algebraically, and so far
we are unable to obtain clear-cut characterizations parallel to [25]. However,
we are able to get basic descriptions on balanced metrics and pluriclosed
metrics, which enable us to confirm the Fino-Vezzoni Conjecture for such
$({\mathfrak{g}},J)$. The following is the main result of this article:

######  Theorem 2.

Let ${\mathfrak{g}}$ be a unimodular Lie algebra which contains an abelian
ideal ${\mathfrak{a}}$ of codimension $2$, and $J$ a complex structure on
${\mathfrak{g}}$ satisfying $J{\mathfrak{a}}\neq{\mathfrak{a}}$. Then the
Fino-Vezzoni Conjecture holds.

Combining the two theorems above, we get the following:

######  Corollary 3.

The Fino-Vezzoni Conjecture holds for any unimodular Lie algebra which
contains an abelian ideal of codimension $2$.

We would like to point out that, despite the parallelness in the appearance of
statement for the above two theorems, the $J{\mathfrak{a}}={\mathfrak{a}}$
case and the $J{\mathfrak{a}}\neq{\mathfrak{a}}$ case are actually quite
different. In the former case, if a complex structures $J$ on ${\mathfrak{g}}$
admits both a balanced metric $g$ and a pluriclosed metric $h$, then through
natural algebraic modifications one can construct another metric $g_{0}$ which
is Kähler. In the latter case, however, for the most part $J$ does not admit
any Kähler metric at all. So one has to show that $J$ admitting balanced
metrics and $J$ admitting pluriclosed metrics will result in conflicting
features so they cannot coexist on the same $J$.

The topic of this article lies in the area of non-Kähler geometry. The main
focus has been on studying the geometry and topology of various special types
of Hermitian structures, where balanced and pluriclosed being two of the most
extensively studied. We refer the readers to [2, 3, 27, 32, 35, 36, 37, 41,
42, 45, 46] as a sampler of references.

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 consider
Lie algebras with codimension $2$ abelian ideals, and analyze properties
involving the structure constants. In §4, we will turn our attention to
balanced metrics on such Lie algebras. In §5 we study pluriclosed metrics and
establish the proof of Theorem 2 stated above.

##  2\. Preliminary on Hermitian Lie algebras

We start by fixing some terminologies and notations, following [40] and [25].
Let $(M^{n},g)$ be a compact Hermitian manifold with universal cover
$(G,J,g)$, where $G$ is an even-dimensional Lie group, $J$ is a left-invariant
complex structure on $G$, and $g$ is a left-invariant Riemannian metric on $G$
compatible with $J$. As we mention before, the compactness of $M$ forces $G$
to be unimodular.

Let ${\mathfrak{g}}$ be the Lie algebra of $G$, and for convenience we will
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 $G$. The integrability of $J$ is characterized by the property (1).

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 (1)
simply means that ${\mathfrak{g}}^{1,0}$ is a complex Lie subalgebra of
${\mathfrak{g}}^{\mathbb{C}}$. Let us extend $g=\langle,\rangle$ bilinearly
over ${\mathbb{C}}$, and let $e=\\{e_{1},\ldots,e_{n}\\}$ be a unitary basis
of ${\mathfrak{g}}^{1,0}$. Again following [40], 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

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

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

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

Denote by $\nabla$ the Chern connection and by $T$ its torsion tensor. We have

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

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

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

Denote by $\omega=\sqrt{-1}\sum_{i}\varphi_{i}\wedge\overline{\varphi}_{i}$
the Kähler form of the metric $g$. The structure equation takes the form:

(5)
$d\varphi_{i}=-\frac{1}{2}\sum_{j,k}C^{i}_{jk}\,\varphi_{j}\wedge\varphi_{k}-\sum_{j,k}\overline{D^{j}_{ik}}\,\varphi_{j}\wedge\overline{\varphi}_{k}$

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

(6)
$\left\\{\begin{split}\sum_{r}\big{(}C^{r}_{ij}C^{\ell}_{rk}+C^{r}_{jk}C^{\ell}_{ri}+C^{r}_{ki}C^{\ell}_{rj}\big{)}\
=\ 0,\hskip 88.2037pt\\\
\sum_{r}\big{(}C^{r}_{ik}D^{\ell}_{jr}+D^{r}_{ji}D^{\ell}_{rk}-D^{r}_{jk}D^{\ell}_{ri}\big{)}\
=\ 0,\hskip 82.51282pt\\\ \ \
\sum_{r}\big{(}C^{r}_{ik}\overline{D^{r}_{j\ell}}-C^{j}_{rk}\overline{D^{i}_{r\ell}}+C^{j}_{ri}\overline{D^{k}_{r\ell}}-D^{\ell}_{ri}\overline{D^{k}_{jr}}+D^{\ell}_{rk}\overline{D^{i}_{jr}}\big{)}\
=\ 0.\end{split}\right.$

From the structure equation (5), a direct computation leads us to the
following

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

Note that in the last equality for $\eta_{i}$, we used the assumption that
${\mathfrak{g}}$ is unimodular (3). In particular, one has
$\partial\overline{\partial}\omega^{n-1}=0$, namely any unimodular Hermitian
Lie algebra $({\mathfrak{g}},J,g)$ is always Gauduchon. By definition and the
proof of Lemma 2 in [25], we have the following:

######  Lemma 1.

Let ${\mathfrak{g}}$ be a unimodular Lie algebra with a Hermitian structure
$(J,g)$. Then the metric $g$ is balanced if and only if

(7) $\sum_{k}D^{k}_{ik}=0,\ \ \ \ \forall\ 1\leq i\leq n.$

The metric $g$ is pluriclosed if and only if

(8)
$\sum_{r}\left(-T^{r}_{ik}\overline{C^{r}_{j\ell}}-T^{j}_{ir}\overline{D^{k}_{r\ell}}+T^{j}_{kr}\overline{D^{i}_{r\ell}}+T^{\ell}_{ir}\overline{D^{k}_{rj}}-T^{\ell}_{kr}\overline{D^{i}_{rj}}\right)=0$

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

Another algebraic fact that will be used later is the following:

######  Lemma 2.

Suppose $A$ and $B$ are two $n\times n$ matrices satisfying $[A,B]=cA$ for
some constant $c\neq 0$. Then $A$ must be nilpotent.

Since this is a standard fact in linear algebra and is well-known, we will
omit its proof here.

##  3\. Structure constants

Throughout this section, we will assume that ${\mathfrak{g}}$ is a unimodular
Lie algebra of real dimension $2n$ containing an abelian ideal
${\mathfrak{a}}$ of codimension $2$, $J$ a complex structure on
${\mathfrak{g}}$, and $g=\langle,\rangle$ an inner product on ${\mathfrak{g}}$
compatible with $J$.

We will also make a couple of assumptions that will be used throughout the
rest of the article:

(9) $J{\mathfrak{a}}\neq{\mathfrak{a}},\ \ \mbox{and}\ \
{\mathfrak{g}}/{\mathfrak{a}}\ \mbox{is non-abelian}.$

This is because the $J{\mathfrak{a}}={\mathfrak{a}}$ case has been treated in
[25] and [28], while when ${\mathfrak{g}}/{\mathfrak{a}}$ is abelian, the Lie
algebra ${\mathfrak{g}}$ must be $2$-step solvable, where the Hermitian
structures in general and Fino-Vezzoni Conjecture in particular were treated
by Freibert and Swann in [20]. Write

(10) ${\mathfrak{a}}_{J}:={\mathfrak{a}}\cap J{\mathfrak{a}},\ \ \ \ \
{\mathfrak{a}}^{\prime}:={\mathfrak{a}}+[{\mathfrak{g}},{\mathfrak{g}}],\ \ \
\ \ {\mathfrak{b}}={\mathfrak{a}}\cap J{\mathfrak{a}}^{\prime}.$

Under our assumption (9), ${\mathfrak{a}}_{J}$ and ${\mathfrak{a}}^{\prime}$
are of codimension $4$ and $1$ in ${\mathfrak{g}}$, respectively. Using the
integrality condition (1), it is not hard to show that

######  Lemma 3.

Let $({\mathfrak{g}},J,g)$ be a unimodular Hermitian Lie algebra which
contains an abelian ideal of codimension $2$ satisfying (9). Then
${\mathfrak{a}}_{J}$ is an ideal of ${\mathfrak{g}}$ and

${\mathfrak{a}}_{J}\subsetneq{\mathfrak{b}}\subsetneq{\mathfrak{a}}\subsetneq{\mathfrak{a}}^{\prime}\subsetneq{\mathfrak{g}}.$

Furthermore, if $x\in{\mathfrak{b}}\setminus{\mathfrak{a}}_{J}$, then
$Jx\in{\mathfrak{a}}^{\prime}\setminus{\mathfrak{a}}$; while if
$y\in{\mathfrak{a}}\setminus{\mathfrak{b}}$, then
$Jy\notin{\mathfrak{a}}^{\prime}$.

######  Definition (admissible frame).

Let $({\mathfrak{g}},J,g)$ be a unimodular Hermitian Lie algebra which
contains an abelian ideal ${\mathfrak{a}}$ of codimension $2$ satisfying (9).
Suppose $\dim_{\mathbb{R}}{\mathfrak{g}}=2n$. A unitary basis
$\\{e_{1},\ldots,e_{n}\\}$ of ${\mathfrak{g}}^{1,0}$ is called an admissible
frame of ${\mathfrak{g}}$, if

${\mathfrak{a}}_{J}=\mbox{span}\\{e_{j}+\overline{e}_{j},i(e_{j}-\overline{e}_{j});\
3\leq j\leq n\\},\ \ \ \
{\mathfrak{a}}={\mathfrak{a}}_{J}+\mbox{span}\\{x,y\\},$

where $\\{x,y\\}$ is perpendicular to ${\mathfrak{a}}_{J}$ and orthonormal,
with $x\in{\mathfrak{b}}$ and

$e_{1}=\frac{1}{\sqrt{2}}(x-iJx),\ \ \ Y=\frac{1}{\sqrt{2}}(y-iJy)=i\delta
e_{1}+\delta^{\prime}e_{2}.$

Here $\delta=\langle Jx,y\rangle\in(-1,1)$ and
$\delta^{\prime}=\sqrt{1-\delta^{2}}\in(0,1)$.

Note that once the metric $g$ is given on $({\mathfrak{g}},J)$, both $x$ and
$y$ are uniquely determined up to $\pm$ sign, as
$x\in{\mathfrak{b}}\cap{\mathfrak{a}}_{J}^{\perp}$ and
$y\in{\mathfrak{a}}\cap{\mathfrak{b}}^{\perp}$.

######  Lemma 4.

Let $({\mathfrak{g}},J,g)$ be a unimodular Hermitian Lie algebra which
contains an abelian ideal ${\mathfrak{a}}$ of codimension $2$ satisfying (9),
and $e$ an admissible frame. Then the structure constants $C$ and $D$ satisfy

$\displaystyle C^{\ast}_{ij}=D^{\ast}_{\ast i}=C^{\alpha}_{\beta
i}=D^{i}_{\alpha\beta}=0,$ $\displaystyle C^{\ast}_{1i}=\overline{D^{i}_{\ast
1}},\ \ \ C^{\ast}_{2i}=\overline{D^{i}_{\ast 2}}-2t\overline{D^{i}_{\ast
1}},\ \ \ C^{\ast}_{12}=\overline{D^{2}_{\ast 1}}-\overline{D^{1}_{\ast
2}}+2t\overline{D^{1}_{\ast 1}},$

for any $1\leq\alpha,\beta\leq 2$ and any $3\leq i,j\leq n$, any
$1\leq\ast\leq n$. Here $t=\frac{i\delta}{\delta^{\prime}}$.

###### Proof.

These properties are direct consequences of the fact that ${\mathfrak{a}}$ is
abelian and is an ideal. For instance, from the fact that
$[{\mathfrak{a}}_{J},{\mathfrak{a}}_{J}]=0$ we get
$[e_{i},e_{j}]=[e_{i},\overline{e}_{j}]=0$ for any $3\leq i,j\leq n$, hence by
(2) we have $C^{\ast}_{ij}=D^{i}_{\ast j}=0$. The other properties follows
similarly, and we omit the straight forward calculations here. ∎

Based on the above lemma, under an admissible frame $e$ all the $C$ components
can be expressed in terms of $D$ components, and the only possibly non-zero
components of $D$ are the following:

(11)
$D_{\alpha}=(D^{\,\ast}_{\ast\alpha})=\left[\begin{array}[]{ll}E_{\alpha}&0\\\
V_{\alpha}&Y_{\alpha}\end{array}\right],\ \ \
E_{\alpha}=\left[\begin{array}[]{ll}D^{\,1}_{1\alpha}&D^{\,2}_{1\alpha}\\\
D^{\,1}_{2\alpha}&D^{\,2}_{2\alpha}\end{array}\right],\ \ \
V_{\alpha}=[v^{1}_{\alpha},v^{2}_{\alpha}]\ \ \ \ $

for $\alpha=1$ and $2$, where each $v^{\beta}_{\alpha}$ (for $\beta=1,2$) is a
column vector in ${\mathbb{C}}^{n-2}$, while each
$Y_{\alpha}=(D^{j}_{i\alpha})$ is an $(n-2)\times(n-2)$ matrix. In order to
prove the Fino-Vezzoni Conjecture, we need to find the exact value of the
entries of $E_{1}$ and $E_{2}$. Towards this end, let us write

(12) $\left\\{\begin{split}[Jx,x]=ax+by+{\mathfrak{a}}_{J}\ \,\\\
[Jx,y]=cx+dy+{\mathfrak{a}}_{J}\ \,\\\ \
[Jy,x]=a^{\prime}x+b^{\prime}y+{\mathfrak{a}}_{J}\\\ \
[Jy,y]=c^{\prime}x+d^{\prime}y+{\mathfrak{a}}_{J}\end{split}\right.$

where $a,\ldots,d^{\prime}$ are real constants. By the choice of admissible
frames, $x\in{\mathfrak{b}}={\mathfrak{a}}\cap J{\mathfrak{a}}^{\prime}$, so
$Jx\in J{\mathfrak{b}}\subseteq{\mathfrak{a}}^{\prime}$. Thus $[Jx,Jy]=\sigma
Jx+{\mathfrak{a}}$ for some constant $\sigma$, and this constant cannot be
zero as otherwise ${\mathfrak{g}}/{\mathfrak{a}}$ would be abelian,
contradicting with (9). By the integrability condition (1) and the fact that
$[x,y]=0$, we get

$[Jx,Jy]=J\\{[Jx,y]-[Jy,x]\\}=(c-a^{\prime})Jx+(d-b^{\prime})Jy+{\mathfrak{a}}_{J}=\sigma
Jx+{\mathfrak{a}}.$

Therefore we must have

(13) $[Jx,Jy]=\sigma Jx+{\mathfrak{a}}_{J}\ \ \ \ \mbox{and}\ \ \ \
a^{\prime}=c-\sigma,\ \ b^{\prime}=d.$

On the other hand, by Jacobi identity,

$[[Jx,Jy],z]=[Jx,[Jy,z]]-[Jy,[Jx,z]]$

holds for any $z\in{\mathfrak{g}}$. Applying this to $z=x$ and $z=y$, we get

$[A,B]=-\sigma A,\ \ \ \mbox{where}\ \ A=\left[\begin{array}[]{ll}a&b\\\
c&d\end{array}\right],\ \ B=\left[\begin{array}[]{ll}a^{\prime}&b^{\prime}\\\
c^{\prime}&d^{\prime}\end{array}\right]$

By Lemma 2, we know that the $2\times 2$ matrix $A$ must be nilpotent, so it
has vanishing trace and determinant. Therefore we have

(14) $d=-a,\ \ \ a^{2}+bc=0,\ \ \ a^{\prime}=c-\sigma,\ \ \ b^{\prime}=-a,$

while the equation $[A,B]=-\sigma A$ becomes

(15) $\left\\{\begin{split}\,bc^{\prime}=-a(c+\sigma)\\\ \
bd^{\prime}=a^{2}-2b\sigma\ \,\\\ 2ac^{\prime}+cd^{\prime}=c^{2}\
\end{split}\right.$

For convenience in future discussions, let us introduce the following
terminology:

######  Definition (types of $J$).

Let ${\mathfrak{g}}$ be a unimodular Lie algebra containing a codimension $2$
abelian ideal ${\mathfrak{a}}$ such that ${\mathfrak{g}}/{\mathfrak{a}}$ is
not abelian, and $J$ a complex structure on ${\mathfrak{g}}$ such that
$J{\mathfrak{a}}\neq{\mathfrak{a}}$. We will call $({\mathfrak{g}},J)$ to be
generic if $[J{\mathfrak{b}},{\mathfrak{b}}]\not\subseteq{\mathfrak{b}}$; call
it half-generic if $[J{\mathfrak{b}},{\mathfrak{b}}]\subseteq{\mathfrak{b}}$
and $[J{\mathfrak{b}},{\mathfrak{a}}]\not\subseteq{\mathfrak{a}}_{J}$; and
call it degenerate if
$[J{\mathfrak{b}},{\mathfrak{a}}]\subseteq{\mathfrak{a}}_{J}$.

Clearly, these three cases are mutually disjoint, and any $J$ with
$J{\mathfrak{a}}\neq{\mathfrak{a}}$ has to be one of them. Also, for any given
metric $g$ compatible with $J$, then under any admissible frame we have

$\left\\{\begin{split}J\ \mbox{is generic}\,\Longleftrightarrow\,b\neq
0;\hskip 62.59596pt\\\ J\ \mbox{is half-generic}\,\Longleftrightarrow\,b=0\
\mbox{and}\ c\neq 0;\\\ J\ \mbox{is
degenerate}\,\Longleftrightarrow\,b=c=0.\hskip 31.2982pt\end{split}\right.$

######  Lemma 5.

Let $({\mathfrak{g}},J,g)$ be a unimodular Hermitian Lie algebra which
contains an abelian ideal ${\mathfrak{a}}$ of codimension $2$ satisfying (9),
and $e$ an admissible frame. Then we have

$\left\\{\begin{split}\mbox{When}\ J\ \mbox{is generic}:\ \ \ b\neq 0,\
c=-\frac{a^{2}}{b},\ c^{\prime}=-\frac{a}{b}(c+\sigma),\
d^{\prime}=-c-2\sigma;\ \\\ \mbox{When}\ J\ \mbox{is half-generic}:\ \ c\neq
0,\ a=b=0,\ d^{\prime}=c,\ \mbox{and}\ c^{\prime}\ \mbox{is artbitrary};\\\
\mbox{When}\ J\ \mbox{is degenerate}:\ \ \ a=b=c=0,\ c^{\prime}\ \mbox{and}\
\,d^{\prime}\ \mbox{are artbitrary}\hskip 34.14322pt\end{split}\right.$

Note that we always have $\sigma\neq 0$ and $a^{\prime}=c-\sigma$,
$d=b^{\prime}=-a$ by (13). In each of the above three cases, the data is
determined by three independent real constants: by $b\sigma\neq 0$ and $a$ in
the generic case; by $c\sigma\neq 0$ and $c^{\prime}$ in the half-generic
case; by $\sigma\neq 0$ and $c^{\prime}$, $d^{\prime}$ in the degenerate case.
We have the following:

######  Lemma 6.

Let $({\mathfrak{g}},J,g)$ be a unimodular Hermitian Lie algebra which
contains an abelian ideal ${\mathfrak{a}}$ of codimension $2$ satisfying (9),
and $e$ an admissible frame. Then we have

(16) $C^{1}_{12}=-\frac{\ i\sigma}{\sqrt{2}\delta^{\prime}},\ \ \
C^{2}_{12}=0,$

where $\sigma\neq 0$ is the constant given by $[Jx,Jy]=\sigma
Jx+{\mathfrak{a}}_{J}$ in (13).

###### Proof.

By the choice of admissible frames, we have

$\displaystyle 2\delta^{\prime}[e_{1},e_{2}]$ $\displaystyle=$ $\displaystyle
2[e_{1},Y]\ =\ [x-iJx,y-iJy]$ $\displaystyle=$
$\displaystyle-[Jx,Jy]+i\\{[Jy,x]-[Jx,y]\\}$ $\displaystyle\equiv$
$\displaystyle-\sigma Jx+i(-\sigma x)$ $\displaystyle=$
$\displaystyle-i\sigma(x-iJx)\ =\ -i\sigma\sqrt{2}e_{1}.$

Here and from now on ‘$\equiv$’ means equality modular
${\mathfrak{a}}_{J}\otimes{\mathbb{C}}$. On the other hand, by (2), we know
that $[e_{1},e_{2}]\equiv C^{1}_{12}e_{1}+C^{2}_{12}e_{2}$, hence the above
computation gives us the expression of $C^{1}_{12}$ and $C^{2}_{12}$, and the
proof of Lemma 6 is completed. ∎

Following similar pattern of computation, we have

$\displaystyle[e_{1},\overline{e}_{1}]$ $\displaystyle=$
$\displaystyle-i[Jx,x]\ \equiv\ -iax-iby$ $\displaystyle=$
$\displaystyle-\frac{\,ia}{\sqrt{2}}(e_{1}+\overline{e}_{1})-\frac{\,ib}{\sqrt{2}}(Y+\overline{Y})$
$\displaystyle=$
$\displaystyle-\frac{\,ia}{\sqrt{2}}(e_{1}+\overline{e}_{1})-\frac{\,ib}{\sqrt{2}}(i\delta
e_{1}+\delta^{\prime}e_{2}-i\delta\overline{e}_{1}+\delta^{\prime}\overline{e}_{2})$
$\displaystyle=$ $\displaystyle\frac{b\delta-
ia}{\sqrt{2}}e_{1}-\frac{\,ib\delta^{\prime}}{\sqrt{2}}e_{2}-\frac{b\delta+ia}{\sqrt{2}}\overline{e}_{1}-\frac{\,ib\delta^{\prime}}{\sqrt{2}}\overline{e}_{2}$

Compare this with formula (2) for $[e_{1},\overline{e}_{1}]$, we obtain the
following

(17) $D^{1}_{11}=\frac{b\delta+ia}{\sqrt{2}},\ \ \ \
D^{1}_{21}=\frac{\,ib\delta^{\prime}}{\sqrt{2}}.$

Analogously, we compute

(18) $\displaystyle 2[Y,\overline{e}_{1}]$ $\displaystyle=$
$\displaystyle[y-iJy,x+iJx]\ =\ -[Jx,Jy]-i\\{[Jx,y]+[Jy,x]\\}$
$\displaystyle\equiv$ $\displaystyle-\sigma Jx-i(2c-\sigma)x+i2ay$
$\displaystyle=$ $\displaystyle i\sigma(x+iJx)-i2cx+i2ay$ $\displaystyle=$
$\displaystyle
i\sigma\sqrt{2}\,\overline{e}_{1}-ic\sqrt{2}(e_{1}+\overline{e}_{1})+ia\sqrt{2}(Y+\overline{Y})$
$\displaystyle=$ $\displaystyle
i\sigma\sqrt{2}\,\overline{e}_{1}-ic\sqrt{2}(e_{1}+\overline{e}_{1})+ia\sqrt{2}(i\delta
e_{1}+\delta^{\prime}e_{2}-i\delta\overline{e}_{1}+\delta^{\prime}\overline{e}_{2})$
$\displaystyle=$
$\displaystyle-(ic+a\delta)\sqrt{2}\,e_{1}+ia\delta^{\prime}\sqrt{2}\,e_{2}+\\{a\delta+i(\sigma-c)\\}\sqrt{2}\,\overline{e}_{1}+ia\delta^{\prime}\sqrt{2}\,\overline{e}_{2}$

On the other hand, we have

$\displaystyle\sqrt{2}\,[Y,\overline{e}_{1}]$ $\displaystyle=$ $\displaystyle
i\delta\sqrt{2}\,[e_{1},\overline{e}_{1}]+\delta^{\prime}\sqrt{2}\,[e_{2},\overline{e}_{1}]$
$\displaystyle\equiv$ $\displaystyle i\delta\\{(b\delta-
ia)e_{1}-ib\delta^{\prime}e_{2}-(b\delta+ia)\overline{e}_{1}-ib\delta^{\prime}\overline{e}_{2}\\}+\delta^{\prime}\sqrt{2}\,[e_{2},\overline{e}_{1}]$
$\displaystyle=$
$\displaystyle\\{(a\delta+ib\delta^{2})e_{1}+b\delta\delta^{\prime}e_{2}+(a\delta-
ib\delta^{2})\overline{e}_{1}+b\delta\delta^{\prime}\overline{e}_{2}\\}+\delta^{\prime}\sqrt{2}\,[e_{2},\overline{e}_{1}].$

Combining this with (18), we obtain

$\delta^{\prime}\sqrt{2}\,[e_{2},\overline{e}_{1}]\ \equiv\
-(2a\delta+ic+ib\delta^{2})e_{1}+(-b\delta+ia)\delta^{\prime}e_{2}+i(\sigma-c+b\delta^{2})\overline{e}_{1}+(ia-b\delta)\delta^{\prime}\overline{e}_{2}$

Compare it with formula (2) for $[e_{2},\overline{e}_{1}]$, which is

$[e_{2},\overline{e}_{1}]\ \equiv\
\overline{D^{2}_{11}}e_{1}+\overline{D^{2}_{21}}e_{2}-D^{1}_{12}\overline{e}_{1}-D^{1}_{22}\overline{e}_{2},$

we derive at the following

(19) $\displaystyle D^{2}_{11}\ =\
\frac{1}{\sqrt{2}\delta^{\prime}}(-2a\delta+ic+ib\delta^{2}),\ \ \ D^{2}_{21}\
=\ -\frac{1}{\sqrt{2}}(b\delta+ia),$ (20) $\displaystyle D^{1}_{12}\ =\
\frac{1}{\sqrt{2}\delta^{\prime}}i(c-\sigma-b\delta^{2}),\ \ \ \ \ \
D^{1}_{22}\ =\ \frac{1}{\sqrt{2}}(b\delta-ia).$

Analogously, consider

$\displaystyle[Y,\overline{Y}]$ $\displaystyle=$ $\displaystyle-i[Jy,y]\
\equiv\ -ic^{\prime}x-id^{\prime}y$ $\displaystyle=$
$\displaystyle-\frac{ic^{\prime}}{\sqrt{2}}(e_{1}+\overline{e}_{1})-\frac{id^{\prime}}{\sqrt{2}}\big{(}i\delta
e_{1}+\delta^{\prime}e_{2}-i\delta\overline{e}_{1}+\delta^{\prime}\overline{e}_{2}\big{)}$
$\displaystyle=$ $\displaystyle\frac{d^{\prime}\delta-
ic^{\prime}}{\sqrt{2}}e_{1}-\frac{id^{\prime}\delta^{\prime}}{\sqrt{2}}e_{2}-\frac{d^{\prime}\delta+ic^{\prime}}{\sqrt{2}}\overline{e}_{1}-\frac{id^{\prime}\delta^{\prime}}{\sqrt{2}}\overline{e}_{2},$

On the other hand,

$\displaystyle[Y,\overline{Y}]$ $\displaystyle=$
$\displaystyle-i\delta[Y,\overline{e}_{1}]+\delta^{\prime}[Y,\overline{e}_{2}]\
=\
-i\delta[Y,\overline{e}_{1}]+i\delta\delta^{\prime}[e_{1},\overline{e}_{2}]+\delta^{\prime
2}[e_{2},\overline{e}_{2}]$ $\displaystyle=$
$\displaystyle-i\delta[Y,\overline{e}_{1}]+i\delta[e_{1},\overline{Y-i\delta
e_{1}}]+\delta^{\prime 2}[e_{2},\overline{e}_{2}]$ $\displaystyle=$
$\displaystyle-i\delta\\{[Y,\overline{e}_{1}]+[\overline{Y},e_{1}]\\}-\delta^{2}[e_{1},\overline{e}_{1}]+\delta^{\prime
2}[e_{2},\overline{e}_{2}]$

So by the expressions for $[Y,\overline{e}_{1}]$, $[e_{1},\overline{e}_{1}]$,
and $[Y,\overline{Y}]$, we get

$\displaystyle\delta^{\prime 2}[e_{2},\overline{e}_{2}]$ $\displaystyle=$
$\displaystyle[Y,\overline{Y}]+\delta^{2}[e_{1},\overline{e}_{1}]+i\delta\\{[Y,\overline{e}_{1}]+[\overline{Y},e_{1}]\\}$
$\displaystyle=$
$\displaystyle[Y,\overline{Y}]+\delta^{2}[e_{1},\overline{e}_{1}]+\frac{\sigma\delta}{\sqrt{2}}e_{1}-\frac{\sigma\delta}{\sqrt{2}}\overline{e}_{1}$
$\displaystyle=$
$\displaystyle[Y,\overline{Y}]+\frac{(b\delta^{2}+\sigma)\delta-
ia\delta^{2}}{\sqrt{2}}e_{1}-\frac{(b\delta^{2}+\sigma)\delta+ia\delta^{2}}{\sqrt{2}}\overline{e}_{1}-\frac{ib\delta^{2}\delta^{\prime}}{\sqrt{2}}(e_{2}+\overline{e}_{2})$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\big{(}-i(c^{\prime}+a\delta^{2})+\delta(d^{\prime}+b\delta^{2}+\sigma)\big{)}e_{1}-\frac{1}{\sqrt{2}}i(d^{\prime}+b\delta^{2})\delta^{\prime}e_{2}$
$\displaystyle-\frac{1}{\sqrt{2}}\big{(}i(c^{\prime}+a\delta^{2})+\delta(d^{\prime}+b\delta^{2}+\sigma)\big{)}\overline{e}_{1}-\frac{1}{\sqrt{2}}i(d^{\prime}+b\delta^{2})\delta^{\prime}\overline{e}_{1}.$

Compare this with formula (2) for $[e_{2},\overline{e}_{2}]$, which is

$[e_{2},\overline{e}_{2}]\ \equiv\
\overline{D^{2}_{12}}e_{1}+\overline{D^{2}_{22}}e_{2}-D^{2}_{12}\overline{e}_{1}-D^{2}_{22}\overline{e}_{2},$

we obtain the last two entries of $E_{2}$:

(21) $D^{2}_{12}\ =\ \frac{1}{\sqrt{2}\delta^{\prime
2}}\big{(}i(c^{\prime}+a\delta^{2})+\delta(d^{\prime}+b\delta^{2}+\sigma)\big{)},\
\ \ \ D^{2}_{22}\ =\
\frac{1}{\sqrt{2}\delta^{\prime}}i(d^{\prime}+b\delta^{2}).$

Now (16), (17), (19), (20), (21) give us the key ingredients for our later
discussions. Since $C$ can be expressed in $D$, and $D^{\ast}_{\ast i}=0$ for
all $i\geq 3$, the Jacobi identities (6) become

$[D_{1},D_{2}]=-C^{1}_{12}D_{1}=\frac{i\sigma}{\sqrt{2}\delta^{\prime}}D_{1},$

or equivalently,

(22) $\displaystyle[E_{1},E_{2}]\ =\
\frac{i\sigma}{\sqrt{2}\delta^{\prime}}E_{1},\ \ \ \ \ \ \ \ [Y_{1},Y_{2}]\ =\
\frac{i\sigma}{\sqrt{2}\delta^{\prime}}Y_{1},$ (23) $\displaystyle
V_{1}E_{2}+Y_{1}V_{2}-V_{2}E_{1}-Y_{2}V_{1}\ =\
\frac{i\sigma}{\sqrt{2}\delta^{\prime}}V_{1}.$

In particular, by Lemma 2 we know that both $E_{1}$ and $Y_{1}$ must be
nilpotent matrices.

##  4\. Balanced metrics

Again let $({\mathfrak{g}},J,g)$ be a Hermitian Lie algebra which contains an
abelian ideal ${\mathfrak{a}}$ of codimension $2$ such that (9) is satisfied.
In this section we will analyze the conditions for the metric $g$ to be
balanced.

Recall that the Lie algebra ${\mathfrak{g}}$ is said to be unimodular if
$\mbox{tr}(\mbox{ad}_{x})=0$ holds for all $x\in{\mathfrak{g}}$. This is
independent of the choice of complex structures or metrics, and is a necessary
condition for the corresponding Lie group $G$ to have any compact quotient.
When $(J,g)$ is a Hermitian structure on ${\mathfrak{g}}$, then
${\mathfrak{g}}$ is unimodular when and only when the equality in (3) is
satisfied. In our case, the condition is equivalent by Lemma 4 to the
following:

$\mbox{tr}(E_{1})+\mbox{tr}(Y_{1})-\overline{\mbox{tr}(Y_{1})}=0,\ \ \ \
C^{1}_{12}+\mbox{tr}(E_{2})+\mbox{tr}(Y_{2})-\overline{\mbox{tr}(Y_{2})}+2t\,\overline{\mbox{tr}(Y_{1})}=0.$

Since $E_{1}$ and $Y_{1}$ are both nilpotent, their trace vanish, so we
conclude the following

######  Lemma 7.

Let $({\mathfrak{g}},J,g)$ be a Hermitian Lie algebra containing an abelian
ideal ${\mathfrak{a}}$ of codimension $2$ satisfying (9). Then
${\mathfrak{g}}$ is unimodular if and only if

$\mbox{tr}(Y_{2})-\overline{\mbox{tr}(Y_{2})}=\frac{i}{\sqrt{2}\delta^{\prime}}\big{(}2\sigma-d^{\prime}-c\big{)}.$

Next we want to find out when will the metric $g$ be balanced. By Lemma 1 and
Lemma 4, we know that this happens when

$\sum_{s}D^{s}_{\ast s}=D^{1}_{\ast 1}+D^{2}_{\ast 2}=0$

for any $\ast$. For all $\ast\geq 3$, this means that $v^{1}_{1}+v^{2}_{2}=0$,
while for $\ast=1$ and $2$, it means

$\displaystyle 0\ =\ D^{1}_{11}+D^{2}_{12}\ =\ \frac{1}{\sqrt{2}\delta^{\prime
2}}\big{(}i(c^{\prime}+a)+\delta(d^{\prime}+b+\sigma)\big{)}$ $\displaystyle
0\ =\ D^{1}_{21}+D^{2}_{22}\ =\
\frac{1}{\sqrt{2}\delta^{\prime}}i(d^{\prime}+b).$

Here we used formula (17), (21), and the fact that $\delta^{2}+\delta^{\prime
2}=1$. Since $\sigma\neq 0$, the above means

$c^{\prime}=-a,\ \ \ d^{\prime}=-b,\ \ \ \delta=0.$

Combining this with (14) and (15), we see that $a=c=0$ and $b(b-2\sigma)=0$.
So either $b=0$, and we are in the $J$ degenerate case, or $b=2\sigma$ so we
are in the $J$ generic case. In the latter case $b$ and $\sigma$ happen to
have the same sign. In summary, we have proved the following:

######  Lemma 8.

Let ${\mathfrak{g}}$ be a unimodular Lie algebra containing an abelian ideal
${\mathfrak{a}}$ of codimension $2$, and $J$ a complex structure on
${\mathfrak{g}}$. Assume that (9) holds. If $({\mathfrak{g}},J)$ admits a
balanced Hermitian metric $g$, then either $J$ is degenerate with
$d^{\prime}=0$, or $J$ is generic and with $b$ and $\sigma$ having the same
sign.

We recall that $J$ is degenerate when and only when $b=c=0$, and $J$ is
generic when and only when $b\neq 0$. We claim that in the $J$ generic case,
the sign of the non-zero constants $\sigma b$ depends only on $J$, while when
$J$ is degenerate, $d^{\prime}=0$ is a condition independent of the choice of
metrics. In other words, we have the following two lemmas:

######  Lemma 9.

Let ${\mathfrak{g}}$ be a Lie algebra containing an abelian ideal
${\mathfrak{a}}$ of codimension $2$, and $J$ a complex structure on
${\mathfrak{g}}$. Assume that (9) holds. Suppose $J$ is generic and $g$ is any
Hermitian metric on $({\mathfrak{g}},J)$, then the sign of $\sigma b$ depends
only on $J$, and is independent of the choice of $g$.

###### Proof.

Under the metric $g$, let us take an orthonormal pair $\\{x,y\\}$ in
${\mathfrak{a}}\cap{\mathfrak{a}}_{J}^{\perp}$ so that $x\in{\mathfrak{b}}$.
Then the equations (modular the ideal ${\mathfrak{a}}_{J}$)

$[Jx,Jy]=\sigma Jx+{\mathfrak{a}}_{J},\ \ \ \ \ \
[Jx,x]=ax+by+{\mathfrak{a}}_{J}$

give us the constants $\sigma$ and $b$. Now suppose $\tilde{g}$ is another
Hermitian metric on $({\mathfrak{g}},J)$, and we take an
$\tilde{g}$-orthonormal pair $\\{\tilde{x},\tilde{y}\\}$ in ${\mathfrak{a}}$
so that $\tilde{x}\in{\mathfrak{b}}$ and both $\tilde{x}$ and $\tilde{y}$ are
perpendicular to ${\mathfrak{a}}_{J}$ with respect to $\tilde{g}$. We have

$[J\tilde{x},J\tilde{y}]=\tilde{\sigma}J\tilde{x}+{\mathfrak{a}}_{J},\ \ \ \ \
\
[J\tilde{x},\tilde{x}]=\tilde{a}\tilde{x}+\tilde{b}\tilde{y}+{\mathfrak{a}}_{J}$

Since both $x$ and $\tilde{x}$ are in ${\mathfrak{b}}$, we have

$\tilde{x}=\lambda_{1}x+u,\ \ \ \tilde{y}=\mu x+\lambda_{2}y+v$

for some constants $\lambda_{1}$, $\lambda_{2}$, $\mu$ with
$\lambda_{1}\lambda_{2}\neq 0$ and some $u$, $v\in{\mathfrak{a}}_{J}$. Plug
into the above defining equations for $\sigma$ and $b$, we get

$\tilde{\sigma}=\lambda_{2}\sigma,\ \ \ \
\tilde{b}=\frac{\lambda_{1}^{2}}{\lambda_{2}}b.$

From this we get $\tilde{\sigma}\tilde{b}=\lambda_{1}^{2}\sigma b$. This shows
that for all Hermitian metrics on $({\mathfrak{g}},J)$, the product $\sigma b$
have the same sign, which is either positive or negative, depending only on
$J$ and not on the metric. ∎

######  Lemma 10.

Let ${\mathfrak{g}}$ be a Lie algebra containing an abelian ideal
${\mathfrak{a}}$ of codimension $2$, and $J$ a complex structure on
${\mathfrak{g}}$. Assume that (9) holds. Suppose $J$ is degenerate and $g$ is
a Hermitian metric on $({\mathfrak{g}},J)$ satisfying $d^{\prime}=0$, then for
any other Hermitian metric $\tilde{g}$ on $({\mathfrak{g}},J)$, the
corresponding constant $\tilde{d^{\prime}}=0$ as well.

###### Proof.

By definition, in the degenerate case we have

$[Jx,x]\equiv[Jx,y]\equiv 0,\ \ \ [Jy,x]\equiv-\sigma x,\ \ \ [Jy,y]\equiv
c^{\prime}x+d^{\prime}y$

where ‘$\equiv$’ means equality modular ${\mathfrak{a}}_{J}$. Therefore
$d^{\prime}=0$ when and only when
$[{\mathfrak{g}},{\mathfrak{a}}]\subseteq{\mathfrak{b}}$. This condition is
clearly independent of the choice of the metric. This completes the proof of
the lemma. ∎

##  5\. Pluriclosed metrics

Let $({\mathfrak{g}},J,g)$ be a Hermitian Lie algebra which contains an
abelian ideal ${\mathfrak{a}}$ of codimension $2$ such that (9) is satisfied.
In this section we will analyze the conditions for the metric $g$ to be
pluriclosed. By Lemma 1, the metric $g$ is pluriclosed if and only if equation
(8) is satisfied. Let $k=\alpha$ and $\ell=\beta$ be in $\\{1,2\\}$, $i$ and
$j$ be in $\\{3,\ldots,n\\}$, then by Lemma 4, equation (8) becomes

$\sum_{r=1}^{n}\\{-T^{r}_{i\alpha}\overline{C^{r}_{j\beta}}-T^{j}_{ir}\overline{D^{\alpha}_{r\beta}}+T^{j}_{\alpha
r}\overline{D^{i}_{r\beta}}\\}=0.$

Since $C^{\sigma}_{j\beta}=D^{i}_{\sigma\beta}=0$ for any $\sigma=1,2$, we
know that for the first and third terms on the left hand side of the above
equation, $r$ sums from $3$ to $n$. Similarly, by
$T^{a}_{bc}=-C^{a}_{bc}-D^{a}_{bc}+D^{a}_{cb}$, we have $T^{\ast}_{ij}=0$ for
any $i,j\geq 3$, thus for the second term in the above equation $r$ runs from
$1$ to $2$. Therefore, the above equation becomes:

$\sum_{r=3}^{n}\\{(C^{r}_{i\alpha}+D^{r}_{i\alpha})\overline{C^{r}_{j\beta}}+(-C^{j}_{\alpha
r}+D^{j}_{r\alpha})\overline{D^{i}_{r\beta}}\\}+\sum_{\sigma=1}^{2}(C^{j}_{i\sigma}+D^{j}_{i\sigma})\overline{D^{\alpha}_{\sigma\beta}}=0,$

Again by Lemma 4, we know that

$C^{\ast}_{\alpha i}=\overline{D^{i}_{\ast\alpha}}-2t\delta_{\alpha
2}\overline{D^{i}_{\ast 1}},$

Plug into the previous line we get

$\displaystyle\sum_{r=3}^{n}\\{\big{(}-\overline{D^{i}_{r\alpha}}+2t\delta_{\alpha
2}\overline{D^{i}_{r1}}+D^{r}_{i\alpha}\big{)}\big{(}-D^{j}_{r\beta}+2\bar{t}\delta_{\beta
2}D^{j}_{r1}\big{)}+\big{(}-\overline{D^{r}_{j\alpha}}+2t\delta_{\alpha
2}\overline{D^{r}_{j1}}+D^{j}_{r\alpha}\big{)}\overline{D^{i}_{r\beta}}\,\\}+$
$\displaystyle+\
\,\big{(}-\overline{D^{i}_{j1}}+D^{j}_{i1}\big{)}\,\overline{D^{\alpha}_{1\beta}}+\big{(}-\overline{D^{i}_{j2}}+2t\overline{D^{i}_{j1}}+D^{j}_{i2}\big{)}\,\overline{D^{\alpha}_{2\beta}}\
\ =\ \ 0.$

Since $\bar{t}=-t$, the matrix form of the above equation becomes

$\displaystyle\big{(}Y_{\alpha}-Y_{\alpha}^{\ast}+2t\delta_{\alpha
2}Y_{1}^{\ast}\big{)}\big{(}-Y_{\beta}-2t\delta_{\beta
2}Y_{1}\big{)}+Y^{\ast}_{\beta}\big{(}Y_{\alpha}-Y_{\alpha}^{\ast}+2t\delta_{\alpha
2}Y_{1}^{\ast}\big{)}+$ (24) $\displaystyle+\
\,\overline{D^{\alpha}_{1\beta}}\,\big{(}Y_{1}-Y_{1}^{\ast}\big{)}+\overline{D^{\alpha}_{2\beta}}\,\big{(}Y_{2}-Y_{2}^{\ast}+2tY_{1}^{\ast}\big{)}\,\
\ =\ \ 0,$

for any $1\leq\alpha,\beta\leq 2$. Here $Y_{\alpha}=(D^{j}_{i\alpha})$ is the
$(n-2)\times(n-2)$ matrix which is the lower right corner of $D_{\alpha}$,
$t=\frac{i\delta}{\delta^{\prime}}$, and $A^{\ast}$ stands for the conjugate
transpose of $A$. The above equation is a necessary condition for the metric
$g$ to be pluriclosed.

######  Lemma 11.

Let ${\mathfrak{g}}$ be a unimodular Lie algebra containing an abelian ideal
${\mathfrak{a}}$ of codimension $2$, and $J$ a complex structure on
${\mathfrak{g}}$. Assume that (9) holds. Suppose $J$ is generic and $g$ is a
pluriclosed metric on $({\mathfrak{g}},J)$. Then $\sigma b<0$. In particular,
in this case $({\mathfrak{g}},J)$ cannot admit any balanced metric.

###### Proof.

Let us write $Y_{\alpha}=H_{\alpha}+S_{\alpha}$ where $H_{\alpha}$ is
Hermitian symmetric and $S_{\alpha}$ is skew-Hermitian symmetric. Take
$\alpha=\beta=1$ in equation (24), we get

(25)
$[H_{1},S_{1}]-2S^{2}_{1}+\overline{D^{1}_{11}}\,S_{1}+\overline{D^{1}_{21}}\,(S_{2}+tH_{1}-tS_{1})\
=\ 0.$

Since ${\mathfrak{g}}$ is unimodular and $b\neq 0$, by Lemma 7 and (14), (15),
we have

$\mbox{tr}(S_{2})=\frac{1}{2}\frac{i4\sigma}{\sqrt{2}\delta^{\prime}}=\frac{i\sqrt{2}\sigma}{\delta^{\prime}}.$

Since $Y_{1}$ is nilpotent hence traceless, by taking trace in (25) we get

$2\mbox{tr}(S_{1}^{2})=\overline{D^{1}_{21}}\,\mbox{tr}(S_{2})=\frac{-ib\delta^{\prime}}{\sqrt{2}}\frac{i\sqrt{2}\sigma}{\delta^{\prime}}=\sigma
b.$

Thus $\sigma b=-2\mbox{tr}(S_{1}S_{1}^{\ast})\leq 0$, hence $<0$. This
completes the proof of the lemma. ∎

Next let us turn our attention to the degenerate case. We prove the following:

######  Lemma 12.

Let ${\mathfrak{g}}$ be a unimodular Lie algebra containing an abelian ideal
${\mathfrak{a}}$ of codimension $2$, and $J$ a complex structure on
${\mathfrak{g}}$. Assume that (9) holds. Suppose $J$ is degenerate and $g$ is
a pluriclosed metric on $({\mathfrak{g}},J)$. Then one must have
$d^{\prime}\neq 0$. In particular, in this case $({\mathfrak{g}},J)$ cannot
admit any balanced metric.

###### Proof.

Let us assume the contrary that $d^{\prime}=0$. We want to derive at a
contradiction. Under this assumption, the only possibly non-zero entries of
$E_{1}$ and $E_{2}$ are

$D^{1}_{12}=-\frac{i\sigma}{\sqrt{2}\delta^{\prime}},\ \ \ \ \ \ \
D^{2}_{12}=\frac{ic^{\prime}+\delta\sigma}{\sqrt{2}\delta^{\prime 2}}.$

Hence equation (25) become

$[H_{1},S_{1}]-2S^{2}_{1}=0.$

Taking trace in the above equation, we get
$0=-2\mbox{tr}(S_{1}^{2})=2\mbox{tr}(S_{1}S_{1}^{\ast})$. Hence $S_{1}=0$, so
$Y_{1}=H_{1}$ is Hermitian and can be diagonalized. Alternatively, since
$S_{1}$ is skew-Hermitian, by taking a unitary change of
$\\{e_{3},\ldots,e_{n}\\}$ if necessary, we may assume that $S_{1}$ is
diagonal, say

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

where $\lambda_{1},\ldots,\lambda_{r}$ are distinct real constants, and
$n_{1}+\cdots+n_{r}=n-2$. Now $[H_{1},S_{1}]=2S_{1}^{2}$ is diagonal so
$H_{1}$ must be block-diagonal, and by unitary change of basis in each block,
we may assume that both $S_{1}$ and $H_{1}$ are diagonal, so $Y_{1}$ is
diagonal.

On the other hand, by the condition
$[Y_{1},Y_{2}]=\frac{i\sigma}{\sqrt{2}\delta^{\prime}}Y_{1}$ and $\sigma\neq
0$, we know by Lemma 2 that $Y_{1}$ must be nilpotent. Therefore we conclude
that $Y_{1}=0$.

Now the only non-trivial case of equation (24) occur when $\alpha=\beta=2$ and
in this case the equation becomes

$[H_{2},S_{2}]-2S^{2}_{2}=0.$

Taking trace, we get $0=2\mbox{tr}(S_{2}^{2})=-2\mbox{tr}(S_{2}S_{2}^{\ast})$,
hence $S_{2}=0$. On the other hand, since $c=d^{\prime}=0$, by Lemma 7 we get

$\mbox{tr}(S_{2})=\frac{i\sigma}{\sqrt{2}\delta^{\prime}}\neq 0,$

which is a contradiction. This shows that the constant $d^{\prime}$ cannot be
zero, namely, the existence of a pluriclosed metric on $({\mathfrak{g}},J)$
with $J$ degenerate would force
$[{\mathfrak{g}},{\mathfrak{a}}]\not\subseteq{\mathfrak{b}}$. This completes
the proof of the lemma. ∎

Now we are ready to prove Theorem 2.

###### Proof of Theorem 2.

Given a unimodular Lie algebra ${\mathfrak{g}}$ which contains an abelian
ideal ${\mathfrak{a}}$ of codimension $2$, assume that $J$ is a complex
structure on ${\mathfrak{g}}$ with $J{\mathfrak{a}}\neq{\mathfrak{a}}$. If
${\mathfrak{g}}/{\mathfrak{a}}$ is not abelian, then $J$ must be one and
exactly one of the three cases: generic, half-generic, or degenerate. By Lemma
8, the presence of a balanced metric on $({\mathfrak{g}},J)$ would rule out
the half-generic case, and in the generic case $\sigma b>0$ while in the
degenerate case $d^{\prime}=0$. On the other hand, by Lemma 11 and Lemma 12,
we know that in either of these cases the presence of a pluriclosed metric
would lead to a contradiction. So for any given $J$ on ${\mathfrak{g}}$, it
cannot admit both a balanced metric and a pluriclosed metric simultaneously.
That is, the Fino-Vezzoni Conjecture holds for compact complex manifolds whose
universal cover is a Lie group $G$ equipped with a left-invariant complex
structure, with its Lie algebra containing an abelian ideal of codimension $2$
(note that the $J{\mathfrak{a}}={\mathfrak{a}}$ has been proved in [28]
before).

In this article we did not discuss the situation when
${\mathfrak{g}}/{\mathfrak{a}}$ is abelian. The case is easier and can be
dealt with analogously. We omit it because in this case ${\mathfrak{g}}$ must
be $2$-step solvable, and for general $2$-step solvable Lie algebras, Freibert
and Swann [20] give systematic treatment of Hermitian structures on such
${\mathfrak{g}}$ and in particular established the Fino-Vezzoni Conjecture. ∎

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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# Resonant leptogenesis in minimal inverse seesaw ISS(2,2) model

Bikash Thapa and Ng. K. Francis bikash2<EMAIL_ADDRESS>Department of Physics, Tezpur University, Tezpur - 784028, India

Abstract
We investigate the parameter space of the minimal inverse seesaw ISS(2,2)
model for successful leptogenesis. The framework of ISS(2, 2) is realized by
augmenting the Standard Model with two right-handed and two Standard Model
singlet neutrinos. The decay of the heavy sterile states which is essentially
an admixture of the right-handed and SM singlet neutrino states produces the
baryon asymmetry of the universe. In this predictive model of leptogenesis, we
study resonant leptogenesis where the mass splitting between the heavy sterile
states is naturally achieved. We review the possibility of generating the
observed baryon asymmetry of the universe via leptogenesis where the CP
violation comes solely from the low-energy CP phases. In addition, we study
the effect of texture zero in the Dirac mass matrix on the parameter space of
the model for successful resonant leptogenesis.

PACS numbers: 12.60.-i, 14.60.Pq, 14.60.St

##  1 Introduction

In spite of the success of the Standard Model (SM) of particle physics, it
fails to explain the observed tiny neutrino masses and the observed baryon
asymmetry of the Universe (BAU). The remedy to such a limitation is the
extension of the SM with extra sterile singlets like the seesaw models [1, 2,
3, 4, 5]. The simplest extension of the SM is the type-I seesaw mechanism,
where the observed BAU can be explained via thermal leptogenesis [6]. In the
type-I seesaw mechanism, the SM is extended by adding singlet right-handed
(RH) neutrinos and their out-of-equilibrium, lepton number violating decays
produce the lepton asymmetries, which are then processed in baryon asymmetries
via sphaleron interactions [7].

In the standard type-I seesaw with hierarchical RH neutrino mass spectrum,
successful leptogenesis introduces a lower bound on the mass scales of the RH
neutrino of about $10^{9}$ GeV [8]. However, the mass scale for successful
leptogenesis may be lowered in a scenario where the RH neutrinos have a quasi-
degenerate mass spectrum.

Another theoretically motivated extension of SM with low-scale sterile
neutrinos is the inverse seesaw (ISS) mechanism, where two types of sterile
neutrinos: RH neutrinos and SM gauge singlets, are introduced to the SM [9,
10, 11]. In this mechanism, the light neutrino mass is doubly suppressed
allowing the Yukawa coupling to be much larger at the TeV scale of sterile
neutrino mass and the quasi-degeneracy in the sterile neutrinos can be
naturally realized due to the presence of a small lepton-number violating mass
parameter $\mu$ [12, 13, 14, 15, 16, 9, 10]. As the mass parameter $\mu$
allows the sterile states to acquire a quasi-degenerate mass scale, resonantly
enhanced leptogenesis can be naturally realized in such a model without fine-
tuning. The ISS mechanism is a low-scale model making it testable in future
experiments. Since the mass scale of the sterile neutrinos relevant to
leptogenesis is low, one needs to consider a fully-flavored leptogenesis
regime. In a low-scale leptogenesis, the low-energy CP-violating phases can
related to the high-energy CP-violation necessary to produce the BAU.

Motivated by the fact that the measurements of the low-energy CP-phases are
not as precise as the neutrino mixing angles in the neutrino oscillation
experiments, we aim to study the effect of these phases on leptogenesis. We
consider a minimal form of ISS mechanism that can accommodate the neutrino
oscillation experimental data i.e., SM extended with two RH neutrinos and two
sterile gauge singlets [11]. We then proceed to find the parameter space of
the defined model such that the BAU is generated effectively.

This paper is organized as follows. In section 2, we describe the framework of
the ISS(2, 2) model and define the three scenarios on which we study the
implications of leptogenesis. In section 3, we review how resonant
leptogenesis can be naturally achieved within the ISS(2, 2) framework. It also
includes the results of our numerical analysis beginning with section 3.1
where we show the results of the parameter scan for the scenario where the $R$
matrix has complex entries. Section 3.2 presents the results of the analysis
for the case where $R$ has a special form with real entries and the parameter
space for the third scenario with texture zeros in Dirac mass matrix is
presented in section 3.3. We finally summarize our conclusions in section 4.

##  2 Model Framework

We have considered the extension of SM by introducing two RH neutrinos and two
SM gauge singlet fermions which results in a minimal form of ISS and is
denoted by ISS(2,2). The Lagrangian of the model which is invariant under the
SM gauge symmetry is,

$-\mathcal{L}_{\nu}=Y_{\nu}\bar{l}_{L}\tilde{H}N_{R}+M_{R}\bar{\left(N_{R}\right)^{c}}\left(S_{L}\right)^{c}+\frac{1}{2}\mu\bar{S_{L}}\left(S_{L}\right)^{c}+h.c.,$
(1)

where $\bar{l}_{L}$ is the SM lepton doublet, $\tilde{H}=i\sigma_{2}H^{*}$
with $H$ being the SM Higgs doublet and $\sigma$ denotes the $2^{nd}$ Pauli
matrix. The extension of the SM includes the RH neutrinos $N_{R}$ and the SM
gauge singlets $S_{L}$. As the Higgs doublet $H$ acquires vacuum expectation
value ($vev$) and the gauge symmetry is broken i.e., $SU(2)_{L}\otimes
U(1)_{Y}\rightarrow U(1)_{EM}$, we obtain the light neutrino mass matrix,

$M_{\nu}=\begin{pmatrix}0&m_{D}&0\\\ m_{D}^{T}&0&M_{R}\\\
0&M_{R}^{T}&\mu\end{pmatrix},$ (2)

where $m_{D}=Y_{\nu}/\sqrt{2}$ represents the Dirac mass matrix, $M_{R}$ is a
complex $2\times 2$ mass matrix and $\mu$ is a complex, symmetric $2\times 2$
matrix. With $\mu\ll m_{D}\ll M_{R}$, diagonalization of equation 2 gives,

$m_{\nu}=m_{D}\left(M_{R}^{T}\right)^{-1}\mu\left(M_{R}\right)^{-1}m_{D}^{T}.$
(3)

As can be seen in equation 3, the light neutrinos are suppressed by the
smallness of $\left(m_{D}M_{R}^{-1}\right)^{2}$ as well as the parameter
$\mu$. In the one-generation case, the masses of the light neutrinos can be
reproduced for the following scale of the different mass states [10],

$\left(\frac{m_{\nu}}{0.1~{}\textrm{eV}}\right)=\left(\frac{m_{D}}{100~{}\textrm{GeV}}\right)^{2}\left(\frac{\mu}{1~{}\textrm{keV}}\right)\left(\frac{M_{R}}{10^{4}~{}\textrm{GeV}}\right)^{-2},$
(4)

The neutrino mass matrix of equation 3 can be approximately diagonalized by
the unitary matrix $U_{\textrm{PMNS}}$ as follows,

$U^{\dagger}_{\textrm{PMNS}}m_{\nu}U^{*}_{\textrm{PMNS}}=\textrm{diag}(m_{1},m_{2},m_{3})=m_{d}$
(5)

where the unitary matrix in the standard parametrization is represented by,

$U_{\textrm{PMNS}}=\begin{pmatrix}c_{12}c_{13}&s_{12}c_{13}&s_{13}e^{-i\delta}\\\
-s_{12}c_{23}-c_{12}s_{23}s_{13}e^{i\delta}&c_{12}c_{23}-s_{12}s_{23}s_{13}e^{i\delta}&s_{23}c_{13}\\\
s_{12}s_{23}-c_{12}c_{23}s_{13}e^{i\delta}&-c_{12}s_{23}-s_{12}c_{23}s_{13}e^{i\delta}&c_{23}c_{13}\end{pmatrix}\times\begin{pmatrix}1&0&0\\\
0&e^{i\alpha_{21}/2}&0\\\ 0&0&e^{i\alpha_{31}/2}\end{pmatrix}$ (6)

A salient feature of the ISS(2,2) model is that one of the three light
neutrinos is massless and thus one of the Majorana CP phases becomes
unphysical. To be more specific, for the case of normal hierarchy (NH) we have
$m_{1}=0<m_{2}<m_{3}$ with the single Majorana phase redefined as
$\sigma=(\alpha_{21}-\alpha_{31})/2$ and $m_{3}=0<m_{1}<m_{2}$ for inverted
hierarchy (IH) with the single Majorana phase given as $\sigma=\alpha_{21}/2$.

Using equations 3 and 5, we can derive the Casas-Ibarra [17] type
parametrization of the Dirac mass matrix for the inverse seesaw model as [18],

$\left(m_{d}^{-1/2}U_{\textrm{PMNS}}^{\dagger}m_{D}\left(M_{R}^{T}\right)^{-1}\mu^{1/2}\right)\cdot\left(\mu^{1/2}M_{R}^{-1}m_{D}^{T}U_{\textrm{PMNS}}^{*}m_{d}^{-1/2}\right)=\mathbb{I},$
(7)

which leads to,

$m_{D}=U_{\textrm{PMNS}}~{}m_{d}^{1/2}~{}R~{}\mu^{-1/2}~{}M_{R}^{T}$ (8)

where R is a complex $3\times 2$ matrix given by,

$\displaystyle R$ $\displaystyle=\begin{pmatrix}0&0\\\
\cos{\zeta}&-\sin{\zeta}\\\
\sin{\zeta}&\cos{\zeta}\end{pmatrix}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\textrm{for
NH}$ (9) $\displaystyle R$
$\displaystyle=\begin{pmatrix}\cos{\zeta}&-\sin{\zeta}\\\
\sin{\zeta}&\cos{\zeta}\\\
0&0\end{pmatrix}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\textrm{for IH}$ (10)

with $z=Re(\zeta)+i~{}Im(\zeta)$, in general being a complex parameter. In the
next section of the paper, we analyze the parameter space for successful
resonant leptogenesis for three different scenarios in the ISS(2, 2) model,
namely, ($i$) $R$ matrix with complex $\zeta$: here the source of CP-violation
comes from the matrix $R$ and the CP-phases present in the PMNS matrix, ($ii$)
$R$ matrix with $Re(\zeta)=\pi/4$ and $Im(\zeta)=0$ : in such a special case
CP-violation necessary for leptogenesis comes from the Dirac and Majorana CP-
phases, and ($iii$) texture zeros in $m_{D}$: in this scenario we consider one
of the elements of $m_{D}$ to be zero such that we can write the parameter
$\zeta$ in terms of the neutrino mass, mixing angles and the CP-phases of the
$U_{\textrm{PMNS}}$ matrix. Thus, just as in scenario ($ii$), the CP-phases
present in the PMNS matrix provide the necessary CP-violation.

##  3 Resonant leptogenesis

In order to study leptogenesis in the ISS(2,2) model, we work in the basis
where the sterile neutrino mass sub-matrix is real and diagonal. The lower
$2\times 2$ block of the mass matrix presented in the equation 2 is given as

$M_{SN}=\begin{pmatrix}0&M_{R}\\\ M_{R}^{T}&\mu\end{pmatrix}$ (11)

We perform a block diagonalization on the above matrix such that we are
working in the basis where the sterile neutrinos are in their mass basis. This
transforms the mass matrix given in the equation 2 by rotating the Yukawa
couplings to the SM leptons and can be written as

$M_{\nu}\rightarrow M_{\nu}\simeq\begin{pmatrix}0&m_{D}^{{}^{\prime}}&0\\\
\left(m_{D}^{{}^{\prime}}\right)^{T}&M_{R}-\frac{1}{2}\mu&0\\\
0&0&M_{R}+\frac{1}{2}\mu\end{pmatrix},$ (12)

where $m_{D}^{{}^{\prime}}$ is the rotated Yukawa coupling matrix. It is clear
from the equation 12 that for small values of $\mu$ the mass spectrum of the
heavy sterile neutrinos becomes degenerate and the scenario of resonant
leptogenesis can be naturally achieved. In resonant leptogenesis, the CP-
violating, out-of-equilibrium decay of the degenerate sterile neutrinos
produces the observed BAU. A non-zero lepton asymmetry is generated from the
CP-violation obtained from the interference of the tree-level decay of the
sterile neutrinos with the one-loop level. In the case of degenerate sterile
neutrinos, the self-energy correction is resonantly enhanced and the flavor-
dependent CP asymmetry is defined as [19, 20, 21]

$\varepsilon_{i}^{\alpha}=\frac{\Gamma\left(N_{i\alpha}\rightarrow
l_{\alpha}\Phi\right)-\Gamma\left(N_{i\alpha}\rightarrow
l_{\alpha}^{c}\Phi^{\dagger}\right)}{\sum_{\alpha}\left[\Gamma\left(N_{i\alpha}\rightarrow
l_{\alpha}\Phi\right)+\Gamma\left(N_{i\alpha}\rightarrow
l_{\alpha}^{c}\Phi^{\dagger}\right)\right]}$ (13)

The CP-asymmetry parameters are given by

$\varepsilon^{\alpha}_{i}=\sum_{i\neq
j}\frac{\textrm{Im}\left[h_{i\alpha}^{\dagger}h_{\alpha
j}\left(h^{\dagger}h\right)_{ij}\right]+\frac{M_{i}}{M_{j}}\textrm{Im}\left[h_{i\alpha}^{\dagger}h_{\alpha
j}\left(h^{\dagger}h\right)_{ji}\right]}{\left(h^{\dagger}h\right)_{ii}\left(h^{\dagger}h\right)_{jj}}\cdot\frac{\left(M_{i}^{2}-M_{j}^{2}\right)\cdot
M_{i}\Gamma_{j}}{\left(M_{i}^{2}-M_{j}^{2}\right)^{2}+M_{i}^{2}\Gamma_{j}^{2}}.$
(14)

where, $h=\frac{\sqrt{2}}{v}m_{D}^{{}^{\prime}}$, $\Gamma_{i}$ is the decay
width of the heavy sterile neutrino state $N_{i}$, and $M_{i}$ is the mass
eigenvalue of $N_{i}$. A lepton asymmetry is generated by utilizing the above
CP-asymmetry parameter by solving the Boltzmann equation. The Boltzmann
equation under consideration is a coupled differential equation describing the
time evolution of the density of heavy sterile neutrinos, $n_{N_{i}}$, and the
lepton number density $n_{N_{\alpha\alpha}}$ (with $\alpha=e,\mu,\tau$) [22],

$\displaystyle\frac{dn_{N_{i}}}{dz}$
$\displaystyle=-D_{i}\left(n_{N_{i}}-n_{N_{i}}^{eq}\right)$
$\displaystyle\frac{n_{N_{\alpha\alpha}}}{dz}$
$\displaystyle=\sum_{i=1}^{2}\varepsilon_{i}^{\alpha}D_{i}\left(n_{N_{i}}-n_{N_{i}}^{eq}\right)-W_{ID}~{}n_{N_{\alpha\alpha}}$
(15)

where

$\displaystyle W_{ID}$ $\displaystyle=\frac{1}{4}Kz^{3}\mathcal{K}_{1}(z),$
(16) $\displaystyle D_{i}$
$\displaystyle=\frac{z}{H(z=1)}\cdot\frac{\Gamma_{i}}{n_{N_{i}}^{eq}},$ (17)

denotes the washout due to inverse decay and the decay term, respectively,
with $\mathcal{K}_{1}(z)$ being the modified Bessel function of the first
kind, $K=\tilde{m}/m_{*}$ is known as the decay parameter, $n_{N_{i}}^{eq}$ is
the equilibrium number density of $N_{i}$ and is defined as

$n^{eq}_{N_{i}}=\frac{3}{8}z^{2}\mathcal{K}_{2}(z)$ (18)

The parameter $H$ is the Hubble parameter, $\tilde{m}$ is the effective
neutrino mass, and $m_{*}$ is the equilibrium neutrino mass.

The baryon asymmetry, $\eta_{B}$ can be estimated by solving equations 3, and
the value of $\eta_{B}$ depends on the initial condition of the heavy sterile
neutrinos. The numerical evaluation of baryon asymmetry performed in this work
involves the decay of sterile neutrinos which is a good approximation, and we
do not consider other effects such as the scattering process, spectator
effects, thermal corrections, etc. In our analysis, we choose $M_{R}$ and
$\mu$ matrices to be diagonal with a degenerate mass spectrum of right-handed
neutrinos, $M_{R_{1}}=M_{R_{2}}=1$ TeV. The elements of the matrix $\mu$
determine the level of degeneracy among the heavy sterile states and we take
$\mu_{i}$ to be free parameters that will be constrained by successful
leptogenesis. Further, we choose a vanishing initial abundance of sterile
neutrinos and scan over the parameters of our model, namely:
$\delta,~{}\sigma,~{}\mu_{1},~{}\mu_{2},~{}Re(\zeta),~{}Im(\zeta)$. We make a
parameter scan by using flat prior and define $1\sigma$, $2\sigma$, and
$3\sigma$ regions of agreement with the observed value of baryon asymmetry by
evaluating the log-likelihood function at a point, $p$. The function is given
as

$log~{}L=-\frac{1}{2}\left(\frac{\eta_{B}^{2}(p)-\eta_{B_{CMB}}^{2}}{\Delta\eta_{B_{CMB}}^{2}}\right)$
(19)

Depending on the scenarios discussed in section 2, we will have different
dimensions of $p$.

###  3.1 Scenario I: $R$ matrix with complex $\zeta$

This is the most general case with
$p=(\delta,~{}\sigma,~{}\mu_{1},~{}\mu_{2},~{}Re(\zeta),~{}Im(\zeta))$, being
the point in the 6-dimensional parameter space of the model over which the
scan is made using the MultiNest package [23]. The results of our analysis are
shown in figure 1. It shows the 2-dimensional projection for leptogenesis with
CP violation coming from the complex matrix $R$ as well as the phases of the
PMNS matrix. The mass splitting between the two sterile states, which is
necessary for successful resonant leptogenesis is quantified by the parameters
$\mu_{1}$, and $\mu_{2}$. Figure 1 also demonstrates the variation of baryon
asymmetry $\eta_{B}$ in the $(\mu_{1}-\mu_{2})$ space. The best-fit values for
the parameters of the model are $\delta=194^{\circ}$, $\sigma=152^{\circ}$,
$\log_{10}(\mu_{1}/\textrm{GeV})=-3.6$,
$\log_{10}(\mu_{2}/\textrm{GeV})=-3.0$, $Re(\zeta)=154^{\circ}$,
$Im(\zeta)=193^{\circ}$ in the NH case. Clearly, the measured value of
$\delta$ coincides with the global fit of experimental data presented in NuFit
5.1 [24] for the NH case. For the IH case, the best-fit values are
$\delta=203^{\circ}$, $\sigma=216^{\circ}$,
$\log_{10}(\mu_{1}/\textrm{GeV})=-2.9$,
$\log_{10}(\mu_{2}/\textrm{GeV})=-2.8$, $Re(\zeta)=175^{\circ}$,
$Im(\zeta)=212^{\circ}$. In this case, the measured best-fit value of $\delta$
lies outside the $1\sigma$ range of the global fit of the neutrino
oscillations experimental data. For the range of $\delta$ between
$[92-296]^{\circ}$ we obtain the $1\sigma$ value of the observed baryon
asymmetry, $\eta_{B}$ in the NH case and $[117-289]^{\circ}$ in the IH case.

Figure 1: The projection for leptogenesis with the contours representing the
$1\sigma$, $2\sigma$, and $3\sigma$ confidence levels. The teal and red colour
denotes the case of NH and IH, respectively.

###  3.2 Scenario II: $R$ is a real matrix with $Re(\zeta)=\pi/4$ and
$Im(\zeta)=0$

For the second case, we choose $Re(\zeta)=\pi/4$ and $Im(\zeta)=0$ making the
$R$ matrix real such that the necessary CP violation comes solely from the CP
phases present in the neutrino mixing matrix. It is clear that the generation
of BAU has a contribution from high as well as low energy parameters, however,
in this case, we analyze the effect of low-energy CP phases on leptogenesis in
the ISS(2,2) model. We first make a parameter scan in a 4-dimensional
parameter space with $p=(\delta,~{}\sigma,~{}\mu_{1},~{}\mu_{2})$. Figure 2
shows the allowed region for the parameters of the model. The contours
represent the region for which the observed value of baryon asymmetry can be
obtained within $1\sigma$, $2\sigma$, and $3\sigma$ confidence intervals. The
best-fit values for the parameters are $\delta=167^{\circ}$,
$\sigma=216^{\circ}$, $\log_{10}(\mu_{1}/\textrm{GeV})=-4.0$,
$\log_{10}(\mu_{2}/\textrm{GeV})=-3.9$ in the NH case. For the IH case, the
best-fit values are $\delta=241^{\circ}$, $\sigma=109^{\circ}$,
$\log_{10}(\mu_{1}/\textrm{GeV})=-4.2$,
$\log_{10}(\mu_{2}/\textrm{GeV})=-3.5$. For the range of $\delta$ between
$[46-288]^{\circ}$ we obtain the $1\sigma$ value of the observed baryon
asymmetry, $\eta_{B}$ in the NH case and $[140-342]^{\circ}$ in the IH case.
We find that the value of $\delta$ calculated within the model that obtains
the best-fit value to the observed BAU lies within the $1\sigma$ region of
NuFit 5.1 data for the case of NH, however, it lies slightly outside the
$1\sigma$ range for the IH case.

Figure 2: The projection for leptogenesis with the contours representing the
$1\sigma$, $2\sigma$, and $3\sigma$ confidence levels. The teal and red colour
denotes the case of NH and IH, respectively.

###  3.3 Scenario III: texture zeros in $m_{D}$

We consider one of the elements of the Yukawa matrix, $m_{D}$ to be zero.
Considering the (1, 1) element of $m_{D}$ to be zero we obtain a simple
expression using Casas-Ibarra type parametrization of equation 8

$\left(U_{\textrm{PMNS}}\right)_{21}\sqrt{m_{2}}\cos\zeta+\left(U_{\textrm{PMNS}}\right)_{23}\sqrt{m_{3}}\sin\zeta=0,$
(20)

in the NH case, and,

$\left(U_{\textrm{PMNS}}\right)_{11}\sqrt{m_{1}}\cos\zeta+\left(U_{\textrm{PMNS}}\right)_{22}\sqrt{m_{2}}\sin\zeta=0,$
(21)

in the IH case.

From the above relations, one can write the parameter $\zeta$ in terms of the
neutrino masses, mixing angles, and the CP phases of the PMNS matrix. We have
a 4-dimensional parameter space with a particular point defined as
$p=(\delta,~{}\sigma,~{}\mu_{1},~{}\mu_{2})$. The results of the exploration
are presented in figure 3. The contours represent the region of the parameter
space for which the observed value of baryon asymmetry lies within $1\sigma$,
$2\sigma$, and $3\sigma$ confidence intervals. For the parameters of the model
the best-fit value are $\delta=196^{\circ}$, $\sigma=192^{\circ}$,
$\log_{10}(\mu_{1}/\textrm{GeV})=-4.1$, $\log_{10}(\mu_{2}/\textrm{GeV})=-3.9$
in the NH case. For the IH case, the best-fit values are $\delta=204^{\circ}$,
$\sigma=182^{\circ}$, $\log_{10}(\mu_{1}/\textrm{GeV})=-3.9$,
$\log_{10}(\mu_{2}/\textrm{GeV})=-3.4$. For the range of $\delta$ between
$[86-306]^{\circ}$ we obtain the $1\sigma$ value of the observed baryon
asymmetry, $\eta_{B}$ in the NH case and $[90-318]^{\circ}$ in the IH case.
From the best-fit values of the Dirac CP phase $\delta$, we see that for both
the NH as well as the IH cases the measured value agrees with the experimental
data up to $1\sigma$ confidence level.

Figure 3: The projection for leptogenesis with the contours representing the
$1\sigma$, $2\sigma$, and $3\sigma$ confidence levels. The teal and red colour
denotes the case of NH and IH, respectively.

##  4 Conclusion

In this paper, we have studied resonant leptogenesis in the framework of the
minimal form of the inverse seesaw model ISS(2, 2). Here, the SM is extended
by adding two right-handed and two SM gauge singlet neutrinos. Considering the
quasi-degenerate, quasi-Dirac sterile neutrino states, we study the scenario
of resonant leptogenesis.

To carry out our analysis, we write the Dirac mass matrix of the ISS(2, 2)
model in the form of Casas-Ibarra type parametrization. We investigate the
viable parameter space for resonant leptogenesis in the ISS(2, 2) model. We
explored the parameter space for three different scenarios. Firstly, we
consider the case where the CP violation necessary for successful leptogenesis
comes from both high-energy parameters (in the form of complex $R$ matrix) and
low-energy CP phases. Secondly, we assess the possibility of CP violation
arising exclusively from the low-energy leptonic sector in the form of CP
phases present in the PMNS matrix. Finally, the third case involves texture
zero in the Dirac mass matrix which allows us to write the elements of the
complex $R$ matrix in terms of low-energy parameters. In other words, we
explore the possibility of successful low-energy resonant leptogenesis with CP
violation coming from the phases present in the PMNS matrix.

We take the best-fit values of the three mixing angles and two mass-squared
differences as the input for the Casas-Ibarra parametrization and allow the
complex parameter, $\zeta$, the elements of the matrix $\mu$ and the two CP
phases present in the PMNS matrix to run freely in a particular region as free
parameters. We numerically solve the coupled Boltzmann equation that describes
the evolution of the Lepton asymmetry and eventually gives the baryon
asymmetry. We make a parameter exploration for the free parameters of the
model using the measured value of baryon asymmetry. From the measured value of
$\delta$, we find that the model agrees with the experimental data of the
Dirac CP phase up to $1\sigma$ confidence level in the NH case for all three
scenarios, however, in the IH case it only agrees with the scenario where we
consider texture zeros in the Dirac mass matrix. Future precision experiments
may give much more stringent results on the Dirac CP phase and probe such a
model.

## Acknowledgement

BT acknowledges the Department of Science and Technology (DST), Government of
India for INSPIRE Fellowship vide Grant No. DST/INSPIRE/2018/IF180588. The
research of NKF is funded by DST-SERB, India under Grant No. EMR/2015/001683.

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