#  SKT solvable Lie algebras with codimension two nilradical

Beatrice Brienza  Dipartimento di Matematica “G. Peano”, Università degli
studi di Torino
Via Carlo Alberto 10
10123 Torino, Italy<EMAIL_ADDRESS>and  Anna Fino  Dipartimento di
Matematica “G. Peano”, Università degli studi di Torino
Via Carlo Alberto 10
10123 Torino, Italy
& Department of Mathematics and Statistics, Florida International University
Miami, FL 33199, United States<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In the present paper we study SKT and generalized Kähler structures on
solvable Lie algebras with (not necessarily abelian) codimension two
nilradical. We treat separately the case of $J$-invariant nilradical and non
$J$-invariant nilradical. A classification of such SKT Lie algebras in
dimension six is provided. In particular, we give a general construction to
extend SKT nilpotent Lie algebras to SKT solvable Lie algebras of higher
dimension, and we construct new examples of SKT and generalized Kähler compact
solvmanifolds.

###### Key words and phrases:

Solvable Lie algebra, SKT metric, Generalized Kähler structure

###### 2010 Mathematics Subject Classification:

53C55; 53C05; 53C30; 53C44

##  1\. Introduction

Let $(M,J,g)$ be an Hermitian manifold of complex dimension $n$ with
fundamental form $\omega=g(J\cdot,\cdot)$. A connection $\nabla$ on $TM$ is
said to be _Hermitian_ if $\nabla g=0$ and $\nabla J=0$. In [29], Gauduchon
has introduced an affine line of Hermitian connections, known as _Gauduchon_
or _canonical_ connections, which can be written as

$g(\nabla^{t}_{X}Y,Z)=g(\nabla^{LC}_{X}Y,Z)+\frac{t-1}{4}(d^{c}\omega)(X,Y,Z)+\frac{t+1}{4}(d^{c}\omega)(X,JY,JZ),$

where $d^{c}\omega=-Jd\omega$. We adopt the convention
$Jd\omega(X,Y,Z):=d\omega(JX,JY,JZ)$. When the manifold $(M,J,g)$ is Kähler,
namely, $d^{c}\omega$ is zero, the line of connections collapses to a single
point, which is the Levi-Civita connection $\nabla^{LC}$, whereas, when
$(M,J,g)$ is non-Kähler, the line is non trivial and the connections
$\nabla^{t}$ have non vanishing torsion. For particular values of
$t\in\mathbb{R}$, we recover connections that play a relevant role in complex
(non-Kähler) geometry. For $t=1$, the _Chern connection_
$\nabla^{1}=\nabla^{Ch}$ ([15]), characterized by
$(\nabla^{Ch})^{0,1}=\overline{\partial}$, and for $t=-1$ the _Bismut_ (or
_Strominger_) _connection_ $\nabla^{-1}=\nabla^{B}$ ([9, 40]), characterized
as the only Hermitian connection with totally skew-symmetric torsion. Although
$\nabla^{LC},\nabla^{B},\nabla^{Ch}$ are mutually different connections, any
one of them completely determines the other two, e.g. the Bismut connection
can be defined in terms of the Levi-Civita one as

$g(\nabla^{B}_{X}Y,Z)=g(\nabla^{LC}_{X}Y,Z)-\frac{1}{2}d^{c}\omega(X,Y,Z),$

and its torsion $3$-form $c$, also called the _Bismut torsion_ , is

$c(X,Y,Z)=g(T^{B}(X,Y),Z)=d\omega(JX,JY,JZ)=-d^{c}\omega(X,Y,Z).$

If the torsion $3$-form $c$ is closed, i.e., $dd^{c}\omega=0$ or,
equivalently, $\partial\overline{\partial}\omega=0$, the metric $g$ is said
strong Kähler with torsion (SKT in short) or pluriclosed.
SKT metrics also appear naturally in the setting of _generalized Kähler
geometry_ : according to [30], a _generalized Kähler_ structure on a
$2n$-dimensional manifold $M$ is a pair of commuting generalized complex
structures $(\mathcal{J}_{1},\mathcal{J}_{2})$ such that
$\mathcal{G}=-\mathcal{J}_{1}\mathcal{J}_{2}$ is a positive definite metric on
$TM\oplus T^{*}M$. It turns out that the generalized Kähler condition can be
equivalently described in the language of Hermitian geometry as a bi-Hermitian
structure $(J_{+},J_{-},g)$ on $M$ such that the Bismut torsions $c_{\pm}$ of
$(J_{\pm},g)$ satisfy $c_{+}=-c_{-}$ and $dc_{+}=0$ (see [30] for further
details). In [35], Hitchin has proved that whenever $[J_{+},J_{-}]\neq 0$, the
tensor $\sigma=[J_{+},J_{-}]g^{-1}$ defines a holomorphic Poisson structure.
Generalized Kähler structures $(g,J_{\pm})$ are said _non-split_ if
$[J_{+},J_{-}]\neq 0$ and _split_ otherwise: in the latter case the tensor
$Q=J_{+}J_{-}$ is an involution of $TM$ and induces the splitting
$TM=T_{+}M\oplus T_{-}M$ in terms of its $\pm 1$-eigenbundles.
As one may easily observe, if $(M,J,g)$ is Kähler, then $(M,\pm J,g)$ is
generalized Kähler: generalized Kähler structures which arise in such a way
are said _trivial_. As a consequence, much study has been devoted to the
explicit construction of non-trivial generalized Kähler manifolds, e.g., [1,
4, 5, 12, 31, 16, 21, 20, 25, 11]. In contrast with the case of compact
nilmanifolds, which cannot admit invariant generalized Kähler structures
unless they are tori [14], several families of (non-Kähler) compact
generalized Kähler manifolds are compact solvmanifolds ([21, 20, 25]), namely,
compact quotients of a connected and simply connected solvable Lie group by a
lattice.
By [33], a compact solvmanifold admits a Kähler structure if and only if it is
a finite quotient of a complex torus which has the structure of a complex
torus bundle over a complex torus. No general restrictions are known to the
existence of generalized Kähler structures on compact solvmanifolds: up to now
the only known examples have abelian nilradical, even though it is still an
open question whether this is true in general.
Since the underlying metric of a generalized Kähler structure is in particular
SKT, a more general problem regards the existence of SKT structures on
solvmanifolds $\Gamma\backslash G$. When the complex structure is invariant,
i.e., it descends from a left-invariant complex structure on $G$, exploiting
the symmetrization process ([8]) the problem reduces to investigate the
existence of SKT inner products at the level of the Lie algebra
$\mathfrak{g}=Lie(G)$. Although it is a simplified setting, the solvable case
seems to be harder than the nilpotent case, even in low dimensions. The
existence of a SKT structure on a nilpotent Lie algebra imposes severe
restrictions: Arroyo and Nicolini proved in [7] that the existence of a SKT
metric on a nilpotent Lie algebra implies that the nilpotency step is at most
$2$, as conjectured in [18]. Furthermore, SKT nilpotent Lie algebras of
dimension 6 and 8 have been fully classified in [24] and [18], respectively.
SKT structures on (non-nilpotent) solvable Lie algebras have been instead
studied in several papers [21, 20, 22, 6, 27, 37], however, a full
classification has been obtained only in dimension $4$ in [37]. In dimension
$6$, the second author and Paradiso have classified the SKT almost nilpotent
Lie algebras in [21, 20, 22] and Freibert and Swann have classified in [27,
28] the SKT $2$-step solvable Lie algebras. Furthermore, Hermitian geometry of
solvable Lie algebras with an abelian ideal of codimension $2$ have been
recently investigated in [13, 32].
Another special class of Hermitian structures on complex manifolds is provided
by the _balanced_ structures, namely, Hermitian structures $(g,J)$ whose
fundamental form $\omega$ is co-closed or, equivalently, satisfying
$d\omega^{n-1}=0$. It has been conjectured in [26], that a compact complex
manifold cannot admit both SKT and balanced metrics, unless it admits Kähler
metrics as well. In the locally homogeneous setting, the conjecture holds true
for nilmanifolds ([26]), almost abelian solvmanifolds ([23]), six-dimensional
almost nilpotent solvmanifolds ([22]) and on solvable Lie algebras with an
abelian ideal of codimension $2$ ([13, 32, 28]). Non compact counterexamples
of this conjecture have been instead constructed in [28].
In this work, we mainly focus on Hermitian structures on solvable Lie algebras
$\mathfrak{g}$ with (not necessarily abelian) nilradical $\mathfrak{h}$ of
codimension $2$. Since $\mathfrak{h}$ has even dimension, if
$(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ is Hermitian, we may distinguish
the two cases $J\mathfrak{h}=\mathfrak{h}$ and
$J\mathfrak{h}\neq\mathfrak{h}$, which will be treated separately. Section 2
is devoted to study the first case, i.e., Hermitian Lie algebras
$(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ with codimension $2$
$J$-invariant nilradical $\mathfrak{h}$. In Theorem 2.1 we prove that the
Hermitian structure $(J,\langle\cdot,\cdot\rangle)$ is always Chern Ricci
flat, extending a result given in [32] in the case of $\mathfrak{h}$ being
abelian, and we give necessary conditions for $(J,\langle\cdot,\cdot\rangle)$
to be SKT. In particular we observe that the SKT condition imposes
restrictions on the structure of $\mathfrak{h}$, namely, it has to be $2$-step
nilpotent. As a consequence (see Corollary 2.3) we study the SKT condition
when $\mathfrak{h}$ is abelian. Although in [32] Hermitian structures on Lie
algebras with a $J$-invariant abelian ideal of codimension $2$ are
investigated, we are able to prove that when the ideal coincides with the
nilradical of the Lie algebra, the SKT condition has a more specialized
characterization. Furthermore, we prove that if $\mathfrak{g}$ admits a
generalized Kähler structure $(J_{\pm},\langle\cdot,\cdot\rangle)$ satisfiying
$J_{\pm}\mathfrak{h}=\mathfrak{h}$, then $(J_{\pm},\langle\cdot,\cdot\rangle)$
must be Kähler.
In Section 3 we investigate Hermitian Lie algebras
$(\mathfrak{g},\langle\cdot,\cdot\rangle,J)$ with codimension $2$ nilradical
$\mathfrak{h}$ such that $J\mathfrak{h}\neq\mathfrak{h}$, with a special focus
on the existence of generalized Kähler structures
$(J,I,\langle\cdot,\cdot\rangle)$ such that $J\mathfrak{h}=\mathfrak{h}$ and
$I\mathfrak{h}\neq\mathfrak{h}$. We also construct the first example of
generalized Kähler Lie algebra with non-abelian nilradical: however, the
example is not unimodular. In Section 4, we provide a full classification of
unimodular six-dimensional solvable Lie algebras with codimension $2$
nilradical. Section 5 is devoted to the construction of new examples of SKT
solvable Lie algebras. In particular, we provide a general process to extend
SKT nilpotent Lie algebras of dimension $2n$ to SKT solvable Lie algebras of
dimension $2n+2k$. We apply this construction to the six-dimensional SKT
nilpotent Lie algebras classified in [24] (see also [41]), to obtain new
families of SKT solvable Lie algebras in dimension eight. Finally, in the last
section, we exhibit some results on the existence of generalized Kähler
structures on solvmanifolds with codimension two nilradical and we construct
new examples of compact SKT and generalized Kähler solvmanifolds.

Acknowledgements. The authors are partially supported by Project PRIN 2022
“Geometry and Holomorphic Dynamics” and by GNSAGA (Indam). Anna Fino is also
supported by a grant from the Simons Foundation (#944448). The authors would
like to thank Alejandro Tolcachier for his interest in this paper. The first
author would like to thank Tommaso Sferruzza for useful conversations and
remarks.

##  2\.  Case $J\mathfrak{h}=\mathfrak{h}$

Let $\mathfrak{g}$ be a $2n$-dimensional solvable Lie algebra with a
codimension $2$ nilradical $\mathfrak{h}$, endowed with an almost Hermitian
structure $(J,\langle\cdot,\cdot\rangle)$ such that
$J\mathfrak{h}=\mathfrak{h}$. We can decompose $\mathfrak{g}$ as the
orthogonal sum $\mathfrak{h}\oplus\mathfrak{h}^{\perp}$, where each summand is
$J$-invariant and $\dim_{\mathbb{C}}\mathfrak{h}^{\perp}=1$.
Let $U$ be a unit vector of $\mathfrak{h}^{\perp}$. We have that the Lie
bracket of $\mathfrak{g}$ is given by

$[U,Y]=AY,\ [JU,Y]=BY,\ [Y,W]=\mu(Y,W),\ [U,JU]=V,\quad\forall
Y,W\in\mathfrak{h},$

where $A:=ad_{U}|_{\mathfrak{h}},B:=ad_{JU}|_{\mathfrak{h}}$ are derivations
of $\mathfrak{h}$ and $\mu$ is the Lie bracket $[\cdot,\cdot]_{\mathfrak{h}}$
on $\mathfrak{h}.$ Note that $V\in\mathfrak{h}$, since $\mathfrak{g}^{1}$ is
contained in $\mathfrak{h}$ and $ad_{V}|_{\mathfrak{h}}=[A,B]$.
Furthermore, since the restrictions of $J$ and $\langle\cdot,\cdot\rangle$ to
$\mathfrak{h}^{\perp}$ are completely determined by choosing the orthonormal
basis $\\{U,JU\\}$ of
$(\mathfrak{h}^{\perp},\langle\cdot,\cdot\rangle_{\mathfrak{h}})$, we have
that the almost Hermitian Lie algebra
$(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ is uniquely determined by the
algebraic data
$(U,JU,A,B,V,\mu,J_{\mathfrak{h}},\langle\cdot,\cdot\rangle_{\mathfrak{h}}).$

######  Remark 2.1.

Note that the data $(U,JU,A,B,V,\mu)$ determine a Lie algebra if and only if
$A,B$ are derivations of $\mathfrak{h}$, $[A,B]=ad_{V}|_{\mathfrak{h}}$ and
$\mu$ is a Lie bracket on $\mathfrak{h}$.

######  Theorem 2.1.

Let $(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ be an almost Hermitian
solvable Lie algebra with a $J$-invariant codimension $2$ nilradical
$\mathfrak{h}$ and let $\\{U,JU\\}$ be an orthonormal basis of
$(\mathfrak{h}^{\perp},\langle\cdot,\cdot\rangle_{\mathfrak{h}^{\perp}})$.
Then

  1. (i)

The complex structure $J$ is integrable, if and only if $J_{\mathfrak{h}}$ is
a complex structure on $\mathfrak{h}$ and $A:=ad_{U}|_{\mathfrak{h}}$ and
$B:=ad_{JU}|_{\mathfrak{h}}$ satisfy the following condition

$[J_{\mathfrak{h}},A]J_{\mathfrak{h}}+[J_{\mathfrak{h}},B]=0.$

  2. (ii)

If $J$ is integrable, then the Hermitian Lie algebra
$(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ is Chern Ricci flat.

  3. (iii)

If $J$ is integrable, $\mathfrak{g}$ is unimodular and $[U,JU]=0$, then
$(J,\langle\cdot,\cdot\rangle)$ is balanced if and only if
$(J_{\mathfrak{h}},\langle\cdot,\cdot\rangle_{\mathfrak{h}})$ is balanced.

  4. (iv)

If $J$ is integrable and the Hermitian Lie algebra
$(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ is SKT, then the nilradical
$\mathfrak{h}$ is at most $2$-step nilpotent and the restrictions
$A_{\mathfrak{z}(\mathfrak{h})},B_{\mathfrak{z}(\mathfrak{h})}$ of $A$ and $B$
to the center $\mathfrak{z}(\mathfrak{h})$ of $\mathfrak{h}$ must satisfy the
following conditions

$A_{\mathfrak{z}(\mathfrak{h})},B_{\mathfrak{z}(\mathfrak{h})}\in\mathfrak{so}(\mathfrak{z}(\mathfrak{h})),\quad[A_{\mathfrak{z}(\mathfrak{h})},J_{\mathfrak{z}(\mathfrak{h})}]=[B_{\mathfrak{z}(\mathfrak{h})},J_{\mathfrak{z}(\mathfrak{h})}]=0.$

###### Proof.

Using that the Nijenhuis tensor $N$ of $J$ satisfies the condition
$N(J\cdot,J\cdot)=-N(\cdot,\cdot)$, one can see that it is enough to check the
vanishing of $N(U,Y)$ and $N(Y,Z)$ for every $Y,Z\in\mathfrak{h}$. By a direct
computation we have that $N(Y,Z)=N_{J_{\mathfrak{h}}}(Y,Z)$ and

$N(U,Y)=[U,Y]+J_{\mathfrak{h}}([JU,Y]+[X,J_{\mathfrak{h}}Y])-[JU,J_{\mathfrak{h}}Y]=(A+J_{\mathfrak{h}}AJ_{\mathfrak{h}}+J_{\mathfrak{h}}B-BJ_{\mathfrak{h}})Y,$

from which _(i)_ follows. To prove _(ii)_ , following [42] (see also [21,
Formula 5.8]), we use that the Ricci form of the Chern Connection $\rho^{Ch}$
is given by $d\eta^{Ch}$, where

$\eta^{Ch}(Y)=\frac{1}{2}\big{(}\tr(ad_{Y}\circ J)-\tr ad_{JY}\big{)},\
\forall Y\in\mathfrak{g}.$

If $Y\in\mathfrak{h}$, then $\eta^{Ch}(Y)=\eta^{Ch}_{\mathfrak{h}}(Y)=0$,
since $\mathfrak{h}$ is nilpotent (a proof can be found in [36, Proposition
2.1]). Furthermore, since

$\begin{split}\eta^{Ch}(U)&=\frac{1}{2}\big{(}\tr(ad_{U}\circ J)-\tr
ad_{JU}\big{)}=\frac{1}{2}\big{(}\tr(A\circ J_{\mathfrak{h}})-\tr B\big{)}\\\
\eta^{Ch}(JU)&=\frac{1}{2}\big{(}\tr(ad_{JU}\circ J)+\tr
ad_{U}\big{)}=\frac{1}{2}\big{(}\tr(B\circ J_{\mathfrak{h}})+\tr A\big{)},\\\
\end{split}$

we obtain

$\eta^{Ch}=\frac{1}{2}\big{(}\tr(A\circ J_{\mathfrak{h}})-\tr
B\big{)}u+\frac{1}{2}\big{(}\tr(B\circ J_{\mathfrak{h}})+\tr A\big{)}Ju,$

where $\\{u,Ju\\}$ is the dual basis of $\\{U,JU\\}$. Then _(ii)_ follows by
differentiating $\eta^{Ch}$, since $du=d(Ju)=0$.
To prove _(iii)_ , we observe that the orthogonal splitting
$\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{h}^{\perp}$, implies that the
fundamental form $\omega$ of $(J,\langle\cdot,\cdot\rangle)$ can be written as
$\omega=\omega_{\mathfrak{h}}+u\wedge Ju$. Its $(n-1)$ power is then given by
$\omega^{n-1}=\omega_{\mathfrak{h}}^{n-1}+(n-1)\omega^{n-2}\wedge u\wedge Ju$.
We first observe that for any $\alpha\in{\bigwedge}^{k}\mathfrak{h}^{*}$,

$d\alpha=u\wedge A^{*}\alpha+Ju\wedge B^{*}\alpha-u\wedge
Ju\wedge\iota_{V}\alpha+d_{\mathfrak{h}}\alpha,$ (1)

where

$C^{*}\gamma=-[\gamma(C\cdot,\dots,\cdot)+\dots+\gamma(\cdot,\dots,C\cdot)],\quad\forall\gamma\in{\bigwedge}^{k}\mathfrak{h}^{*},\,C\in\mathfrak{gl}(\mathfrak{h}),$

and $d_{\mathfrak{h}}$ stands for the exterior differential of the nilpotent
Lie algebra $(\mathfrak{h},\mu)$.
Exploiting that $du=dJu=0$ and that $V=[U,JU]=0$, we get

$d\omega^{n-1}=d\omega_{\mathfrak{h}}^{n-1}+(n-1)d_{\mathfrak{h}}\omega^{n-2}\wedge
u\wedge Ju.$

We claim that $d\omega_{\mathfrak{h}}^{n-1}=0$. Assume by contradiction that
$d\omega_{\mathfrak{h}}^{n-1}=u\wedge
A^{*}\omega_{\mathfrak{h}}^{n-1}+Ju\wedge B^{*}\omega_{\mathfrak{h}}^{n-1}\neq
0$, hence, at least one between $A^{*}\omega_{\mathfrak{h}}^{n-1}$ and
$B^{*}\omega_{\mathfrak{h}}^{n-1}$ is non-zero. If the first holds, then one
may consider $d(\omega_{\mathfrak{h}}^{n-1}\wedge Ju)=u\wedge
A^{*}\omega_{\mathfrak{h}}^{n-1}\wedge Ju\neq 0$. Moreover, this provides a
contradiction: since $\mathfrak{g}$ is unimodular, any $2n-1$ form on
$\mathfrak{g}$ is closed. The other case proceeds in the same way by
considering $d(\omega_{\mathfrak{h}}^{n-1}\wedge u)$. Therefore,
$d\omega^{n-1}=(n-1)d_{\mathfrak{h}}\omega^{n-2}\wedge u\wedge Ju$ is zero if
and only if $d_{\mathfrak{h}}\omega^{n-2}=0$.
The first part of _(iv)_ follows from [7]. Indeed, if the Hermitian Lie
algebra $(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ is SKT, then
$(\mathfrak{h},J_{\mathfrak{h}},\langle\cdot,\cdot\rangle_{\mathfrak{h}})$ is
SKT (see [7, Proposition 3.1]). In particular,
$(\mathfrak{h},J_{\mathfrak{h}},\langle\cdot,\cdot\rangle_{\mathfrak{h}})$ is
at most $2$-step nilpotent ([7, Theorem 4.8]). Since,
$(\mathfrak{h},\langle\cdot,\cdot\rangle_{\mathfrak{h}},J_{\mathfrak{h}})$ is
SKT and at most $2$-step nilpotent we have the orthogonal decomposition
$\mathfrak{h}=\mathfrak{z}(\mathfrak{h})\oplus\mathfrak{z}(\mathfrak{h})^{\perp}$,
where $\mathfrak{z}(\mathfrak{h})$ is the center of $\mathfrak{h}$. Observe
that each summand in the decomposition
$\mathfrak{h}=\mathfrak{z}(\mathfrak{h})\oplus\mathfrak{z}(\mathfrak{h})^{\perp}$
is $J_{\mathfrak{h}}$-invariant by [18, Proposition 3.5].
Hence, $J_{\mathfrak{h}}$ is determined by $J_{\mathfrak{z}(\mathfrak{h})}$
and $J_{\mathfrak{z}(\mathfrak{h})^{\perp}}$. Furthermore, since $A,B\in
Der(\mathfrak{h})$, then they must preserve the center. Indeed, if
$Z\in\mathfrak{z}(\mathfrak{h})$, then

$0=A\big{(}\mu(Z,Y)\big{)}=\mu(AZ,Y)+\mu(Z,AY)=\mu(AZ,Y),\ \forall
Y\in\mathfrak{h},$

from which follows that $AZ\in\mathfrak{z}(\mathfrak{h})$, and analogously for
$B$. With respect to the decomposition
$\mathfrak{h}=\mathfrak{z}(\mathfrak{h})\oplus\mathfrak{z}(\mathfrak{h})^{\perp}$
we have that

$A=\begin{pmatrix}A_{\mathfrak{z}(\mathfrak{h})}&\ast_{A}\\\
0&A_{\mathfrak{z}(\mathfrak{h})^{\perp}}\end{pmatrix}\ \ \text{and}\ \
B=\begin{pmatrix}B_{\mathfrak{z}(\mathfrak{h})}&\ast_{B}\\\
0&B_{\mathfrak{z}(\mathfrak{h})^{\perp}}\end{pmatrix}.$

In particular, $\mathfrak{z}(\mathfrak{h})$ is a $J$-invariant ideal of
$\mathfrak{g}$ and the integrability condition involving $A,B$ and
$J_{\mathfrak{h}}$ reads
$A_{\mathfrak{z}(\mathfrak{h})}+J_{\mathfrak{z}(\mathfrak{h})}A_{\mathfrak{z}(\mathfrak{h})}J_{\mathfrak{z}(\mathfrak{h})}+J_{\mathfrak{z}(\mathfrak{h})}B_{\mathfrak{z}(\mathfrak{h})}-B_{\mathfrak{z}(\mathfrak{h})}J_{\mathfrak{z}(\mathfrak{h})}=0$
on $\mathfrak{z}(\mathfrak{h})$.
Let $c$ be the Bismut torsion $3$-form of $(J,\langle\cdot,\cdot\rangle)$.
Then, for any $Z\in\mathfrak{z}(\mathfrak{h})$ using the formula for the $dc$
in [17] (see also [7, Formula 3]) we get

$\displaystyle dc(JZ,Z,U,JU)=$
$\displaystyle\norm{AJZ}^{2}+\norm{BJZ}^{2}+\norm{AZ}^{2}+\norm{BZ}^{2}$ (2)
$\displaystyle-\langle AJAJZ,Z\rangle-\langle JAJAZ,Z\rangle-\langle
BJBJZ,Z\rangle-\langle JBJBZ,Z\rangle.$

By _(i)_ , this is equivalent to

$\displaystyle dc(JZ,Z,X,JX)=$
$\displaystyle\norm{AJZ}^{2}+\norm{BJZ}^{2}+\norm{AZ}^{2}+\norm{BZ}^{2}$
$\displaystyle+\langle\big{(}2(A^{2}+B^{2}+AJ_{\mathfrak{h}}B-BJ_{\mathfrak{h}}A)-J_{\mathfrak{h}}[A,B]-[A,B]J_{\mathfrak{h}}\big{)}Z,Z\rangle.$

Moreover, since $[A,B]=ad_{V}|_{\mathfrak{h}}$, and $Z$ is in the center of
$\mathfrak{h}$, $[A,B]Z=0$. Analogously, since $\mathfrak{z}(\mathfrak{h})$ is
$J_{\mathfrak{h}}$ invariant, also $[A,B]J_{\mathfrak{h}}Z=0$. Hence,

$\displaystyle dc(JZ,Z,X,JX)=$
$\displaystyle\norm{AJZ}^{2}+\norm{BJZ}^{2}+\norm{AZ}^{2}+\norm{BZ}^{2}$
$\displaystyle+\langle\big{(}2(A^{2}+B^{2}+AJ_{\mathfrak{h}}B-BJ_{\mathfrak{h}}A)\big{)}Z,Z\rangle.$

Hence, the SKT condition yields

$\norm{AJZ}^{2}+\norm{BJZ}^{2}+\norm{AZ}^{2}+\norm{BZ}^{2}+\langle\big{(}2(A^{2}+B^{2}+AJ_{\mathfrak{h}}B-BJ_{\mathfrak{h}}A)\big{)}Z,Z\rangle=0$
(3)

for any $Z\in\mathfrak{z}(\mathfrak{h})$. Let $\\{e_{1},\dots,e_{2r}\\}$ be
any orthonormal basis of $\mathfrak{z}(\mathfrak{h})$. Without loss of
generality, we may assume that $Je_{2j-1}=e_{2j}$ for each $j=1,\dots,r$.
Using (3), we get that $\sum_{j=1}^{2r}dc(Je_{j},e_{j},X,JX)$ is equal to

$\begin{split}&\sum_{j=1}^{2r}\norm{AJe_{j}}^{2}+\norm{BJe_{j}}^{2}+\norm{Ae_{j}}^{2}+\norm{Be_{j}}^{2}+\langle\big{(}2(A^{2}+B^{2}+AJ_{\mathfrak{h}}B-BJ_{\mathfrak{h}}A)\big{)}e_{j},e_{j}\rangle=\\\
&2\big{(}\norm{A_{\mathfrak{z}(\mathfrak{h})}}^{2}+\norm{B_{\mathfrak{z}(\mathfrak{h})}}^{2}+\tr(A_{\mathfrak{z}(\mathfrak{h})}^{2})+\tr(B_{\mathfrak{z}(\mathfrak{h})}^{2})+\tr(A_{\mathfrak{z}(\mathfrak{h})}J_{\mathfrak{z}(\mathfrak{h})}B_{\mathfrak{z}(\mathfrak{h})}-B_{\mathfrak{z}(\mathfrak{h})}J_{\mathfrak{z}(\mathfrak{h})}A_{\mathfrak{z}(\mathfrak{h})})\big{)}.\end{split}$

Furthermore, since
$[A_{\mathfrak{z}(\mathfrak{h})},B_{\mathfrak{z}(\mathfrak{h})}]=[A,B]_{\mathfrak{z}(\mathfrak{h})}=0$,
then
$\tr(A_{\mathfrak{z}(\mathfrak{h})}J_{\mathfrak{z}(\mathfrak{h})}B_{\mathfrak{z}(\mathfrak{h})}-B_{\mathfrak{z}(\mathfrak{h})}J_{\mathfrak{z}(\mathfrak{h})}A_{\mathfrak{z}(\mathfrak{h})})=0$.
Hence, the SKT condition implies that

$\norm{A_{\mathfrak{z}(\mathfrak{h})}}^{2}+\norm{B_{\mathfrak{z}(\mathfrak{h})}}^{2}+\tr(A_{\mathfrak{z}(\mathfrak{h})}^{2})+\tr(B_{\mathfrak{z}(\mathfrak{h})}^{2})=0.$
(4)

We claim
$\norm{A_{\mathfrak{z}(\mathfrak{h})}}^{2}+\tr(A_{\mathfrak{z}(\mathfrak{h})}^{2})\geq
0$ and
$\norm{B_{\mathfrak{z}(\mathfrak{h})}}^{2}+\tr(B_{\mathfrak{z}(\mathfrak{h})}^{2})\geq
0$. Indeed,

$\norm{A_{\mathfrak{z}(\mathfrak{h})}}^{2}+\tr(A_{\mathfrak{z}(\mathfrak{h})}^{2})=2{{\sum}}_{i=1}^{2r}\absolutevalue{a_{ii}}^{2}+{{\sum}}_{i<j}(\absolutevalue{a_{ij}}^{2}+\absolutevalue{a_{ji}}^{2})+{{\sum}}_{i<j}2({a_{ij}}\cdot{a_{ji}})=2{{\sum}}_{i=1}^{2r}\absolutevalue{a_{ii}}^{2}+{{\sum}}_{i<j}(a_{ij}+a_{ji})^{2}.$

The same argument holds for $B_{\mathfrak{z}(\mathfrak{h})}$. More
specifically, (4) implies that if $(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$
is SKT, then $A_{\mathfrak{z}(\mathfrak{h})}$ and
$B_{\mathfrak{z}(\mathfrak{h})}$ are skew-symmetric matrices with respect to
any orthonormal basis of
$(\mathfrak{h},J_{\mathfrak{h}},\langle\cdot,\cdot\rangle_{\mathfrak{h}})$. In
fact, the choice of the orthonormal basis does not affect the previous
computations. This proves that $A_{\mathfrak{z}(\mathfrak{h})}$ and
$B_{\mathfrak{z}(\mathfrak{h})}$ are in $\mathfrak{so}(\mathfrak{h})$.
With respect to $\\{e_{1},\dots,e_{2r}\\}$ we may write
$A_{\mathfrak{z}(\mathfrak{h})}$ and $B_{\mathfrak{z}(\mathfrak{h})}$ using
$2\times 2$ block matrices $A_{ij}$ and $B_{ij}$ as follows

$A_{\mathfrak{z}(\mathfrak{h})}=\begin{pmatrix}A_{1,1}&\dots&A_{1,r}\\\
\vdots&\ddots&\vdots\\\ A_{r,1}&\dots&A_{r,r}\end{pmatrix},\quad
B_{\mathfrak{z}(\mathfrak{h})}=\begin{pmatrix}B_{1,1}&\dots&B_{1,r}\\\
\vdots&\ddots&\vdots\\\ B_{r,1}&\dots&B_{r,r}\end{pmatrix},$

where, for $i=1,\ldots,r$,

$\displaystyle A_{i,i}=$ $\displaystyle\begin{pmatrix}0&a_{2i-1,2i}\\\
-a_{2i-1,2i}&0\\\ \end{pmatrix},\quad B_{i,i}=\begin{pmatrix}0&b_{2i-1,2i}\\\
-b_{2i-1,2i}&0\\\ \end{pmatrix},$

and, for $i<j$,

$\displaystyle
A_{i,j}=\begin{pmatrix}\begin{array}[]{ll}a_{2i-1,2j-1}&a_{2i-1,2j}\\\
a_{2i,2j-1}&a_{2i,2j}\\\ \end{array}\end{pmatrix},\ A_{j,i}=-(^{t}A_{i,j}),\
B_{i,j}=\begin{pmatrix}\begin{array}[]{ll}b_{2i-1,2j-1}&b_{2i-1,2j}\\\
b_{2i,2j-1}&b_{2i,2j}\\\ \end{array}\end{pmatrix},\ B_{j,i}=-(^{t}B_{i,j}).$

Moreover, since we choose $\\{e_{1},\dots,e_{2r}\\}$ satisfying
$Je_{2j-1}=e_{2j}$ for each $j=1,\dots,r$, with respect to such a basis the
complex structure $J_{\mathfrak{z}(\mathfrak{h})}$ can be represented by the
diagonal block matrix $\operatorname{diag}(\Lambda_{1},\dots,\Lambda_{r})$
with

$\Lambda_{i}=\begin{pmatrix}0&-1\\\ 1&0\\\ \end{pmatrix}.$

If one imposes the integrability condition
$A_{\mathfrak{z}(\mathfrak{h})}+J_{\mathfrak{z}(\mathfrak{h})}A_{\mathfrak{z}(\mathfrak{h})}J_{\mathfrak{z}(\mathfrak{h})}+J_{\mathfrak{z}(\mathfrak{h})}B_{\mathfrak{z}(\mathfrak{h})}-B_{\mathfrak{z}(\mathfrak{h})}J_{\mathfrak{z}(\mathfrak{h})}=0$
then the following linear conditions hold:

$\begin{cases}a_{2i-1,2j-1}=a_{2i,2j}+b_{2i,2j-1}+b_{2i-1,2j}\\\
b_{2i,2j}=a_{2i-1,2j}+a_{2i,2j-1}+b_{2i-1,2j-1}.\\\ \end{cases}$ (5)

We compute the components $dc(e_{2j-1},Je_{2j-1},X,JX)$, using (2). With a
straightforward computation one gets

$\begin{split}dc(e_{2j-1},Je_{2j-1},X,JX)=&-2{{\sum}}_{k=1}^{j-1}(b_{2k,2j-1}+b_{2k-1,2j})^{2}+(a_{2k-1,2j}+a_{2k,2j-1})^{2}\\\
&-2{{\sum}}_{k=j+1}^{r-1}(b_{2j,2k-1}+b_{2j-1,2k})^{2}+(a_{2j-1,2k}+a_{2j,2k-1})^{2}.\end{split}$
(6)

Repeating the same argument for each $j=1,\dots,r$, the vanishing of (6) leads
to the identities $b_{2i,2j-1}=-b_{2i-1,2j}$ and $a_{2i,2j-1}=-a_{2i-1,2j}.$
Moreover, plugging these identities in (5) we get

$\begin{cases}a_{2i,2j}=a_{2i-1,2j-1}\\\ b_{2i,2j}=b_{2i-1,2j-1}.\\\
\end{cases}$

Hence we have that for any $i<j$, the matrices $A_{ij}$ and $B_{ij}$ are of
the kind

$A_{ij}=\begin{pmatrix}a_{2i-1,2j-1}&a_{2i-1,2j}\\\
-a_{2i-1,2j}&a_{2i-1,2j-1}\\\ \end{pmatrix},\
B_{ij}=\begin{pmatrix}b_{2i-1,2j-1}&b_{2i-1,2j}\\\
-b_{2i-1,2j}&b_{2i-1,2j-1}\\\ \end{pmatrix},$

and it is straightforward to observe that
$[A_{\mathfrak{z}(\mathfrak{h})},J_{\mathfrak{z}(\mathfrak{h})}]=[B_{\mathfrak{z}(\mathfrak{h})},J_{\mathfrak{z}(\mathfrak{h})}]=0$.
∎

######  Remark 2.2.

If the Hermitian Lie algebra $(\mathfrak{g},\langle\cdot,\cdot\rangle,J)$ is
Kähler, then $\mathfrak{h}$ is abelian. Exploiting (1), since the fundamental
form $\omega$ splits as the sum $\omega_{\mathfrak{h}}+u\wedge Ju$, we get
that if $(\langle\cdot,\cdot\rangle,J)$ is Kähler, then $0=d\omega=u\wedge
A^{*}\omega_{\mathfrak{h}}+Ju\wedge B^{*}\omega_{\mathfrak{h}}-u\wedge
Ju\wedge\iota_{V}\omega_{\mathfrak{h}}+d_{\mathfrak{h}}\omega_{\mathfrak{h}}$,
which implies that $d_{\mathfrak{h}}\omega_{\mathfrak{h}}=0$. Moreover, since
$\mathfrak{h}$ is nilpotent, then it must be abelian.

######  Corollary 2.2.

Let $\mathfrak{g}$ be a unimodular solvable Lie algebra with codimension $2$
nilradical $\mathfrak{h}$. Assume that $\mathfrak{g}$ is endowed with a
complex structure $J$ such that $J\mathfrak{h}=\mathfrak{h}$. If
$\mathfrak{g}$ admits a $J$-Hermitian SKT metric
$\langle\cdot,\cdot\rangle_{1}$ and a $J$-Hermitian balanced metric
$\langle\cdot,\cdot\rangle_{2}$ such that
$[\mathfrak{h}^{\perp_{\langle\cdot,\cdot\rangle_{2}}},\mathfrak{h}^{\perp_{\langle\cdot,\cdot\rangle_{2}}}]=0$,
then $\mathfrak{g}$ admits also a Kähler metric.

###### Proof.

Since $\mathfrak{g}$ is unimodular and
$[\mathfrak{h}^{\perp_{\langle\cdot,\cdot\rangle_{2}}},\mathfrak{h}^{\perp_{\langle\cdot,\cdot\rangle_{2}}}]=0$,
by Theorem 2.1 statement _(iii)_ we have that $(\mathfrak{h},J)$ admits a
balanced metric. Furthermore, by the proof of statement _(iv)_ in Theorem 2.1,
$(\mathfrak{h},J)$ admits also a SKT metric, and so $\mathfrak{h}$ it is
abelian by [26]. Since $\mathfrak{g}$ contains an abelian ideal of codimension
$2$ which is $J$-invariant, then $(\mathfrak{g},J)$ admits also a Kähler
metric by [13, 32]. ∎

As a corollary of Theorem 2.1 we focus now on the case of $\mathfrak{h}$ being
abelian. Observe that in this case $J_{\mathfrak{h}}$ is trivially integrable,
so the integrability of $J$ simply reduces to the condition
$A+J_{\mathfrak{h}}AJ_{\mathfrak{h}}+J_{\mathfrak{h}}B-BJ_{\mathfrak{h}}=0$.
In [32], Hermitian structures on Lie algebras with a $J$-invariant abelian
ideal of codimension $2$ are investigated. The next Corollary shows that when
the ideal coincides with the nilradical of the Lie algebra, the SKT condition
has a more specialized characterization.

######  Corollary 2.3.

Let $(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ be an almost Hermitian
solvable Lie algebra with a $J$-invariant codimension 2 abelian nilradical
$\mathfrak{h}$ and let $\\{U,JU\\}$ be an orthonormal basis of
$(\mathfrak{h}^{\perp},\langle\cdot,\cdot\rangle_{\mathfrak{h}^{\perp}})$.
Then $(J,\langle\cdot,\cdot\rangle)$ is SKT if and only if one of the
following condition holds

  1. (i)

$A:=ad_{U}|_{\mathfrak{h}},B:=ad_{JU}|_{\mathfrak{h}}\in\mathfrak{so}(\mathfrak{h})$
and $[A,J_{\mathfrak{h}}]=[B,J_{\mathfrak{h}}]=0$;

  2. (ii)

there exists an orthonormal basis $\\{e_{1},\dots,e_{2n-2}\\}$ of
$(\mathfrak{h},\langle\cdot,\cdot\rangle_{\mathfrak{h}})$ with respect to
which

$J_{\mathfrak{h}}=\operatorname{diag}(\Lambda_{1},\dots,\Lambda_{n-1}),\
A=\operatorname{diag}(A_{1},\dots,A_{n-1}),\
B=\operatorname{diag}(B_{1},\dots,B_{n-1}),$

where

$\Lambda_{i}=\begin{pmatrix}0&-1\\\ 1&0\\\ \end{pmatrix},\
A_{i}=\begin{pmatrix}0&a_{i}\\\ -a_{i}&0\\\ \end{pmatrix}\ \text{and}\
B_{i}=\begin{pmatrix}0&b_{i}\\\ -b_{i}&0\\\ \end{pmatrix},$ (7)

with $a_{i},b_{i}\in\mathbb{R}$.

Moreover, if the Hermitian structure $(J,\langle\cdot,\cdot\rangle)$ is SKT,
then

  1. (iii)

$(J,\langle\cdot,\cdot\rangle)$ is Kähler if and only if
$[\mathfrak{h}^{\perp},\mathfrak{h}^{\perp}]=0$;

  2. (iv)

if $\mathfrak{z}(\mathfrak{g})\cap\mathfrak{h}=\\{0\\},$ then $\mathfrak{g}$
admits also a Kähler metric.

###### Proof.

One direction of _(i)_ follows by Theorem 2.1. Indeed, if
$(\langle\cdot,\cdot\rangle,J)$ is SKT and $\mathfrak{h}$ is abelian, then
$A,B\in\mathfrak{so}(\mathfrak{h})$ and
$[A,J_{\mathfrak{h}}]=[B,J_{\mathfrak{h}}]=0$. Let us prove the converse. We
have already observed that $\omega=\omega_{\mathfrak{h}}+u\wedge Ju$. Then,

$d\omega=u\wedge A^{*}\omega_{\mathfrak{h}}+Ju\wedge
B^{*}\omega_{\mathfrak{h}}-u\wedge Ju\wedge\iota_{V}\omega_{\mathfrak{h}},$

with $V=[U,JU]$,
$A^{*}\omega_{\mathfrak{h}}(Y,W)=-\langle(J_{\mathfrak{h}}A+A^{t}J_{\mathfrak{h}})Y,W\rangle_{\mathfrak{h}}$,
and
$B^{*}\omega_{\mathfrak{h}}(Y,W)=-\langle(J_{\mathfrak{h}}B+B^{t}J_{\mathfrak{h}})Y,W\rangle_{\mathfrak{h}}$
for any $Y,Z\in\mathfrak{h}$. Since
$[A,J_{\mathfrak{h}}]=[B,J_{\mathfrak{h}}]=0$ and
$A,B\in\mathfrak{so}(\mathfrak{h})$

$J_{\mathfrak{h}}A+A^{t}J_{\mathfrak{h}}=(A+A^{t})J_{\mathfrak{h}}=0,\
J_{\mathfrak{h}}B+B^{t}J_{\mathfrak{h}}=(B+B^{t})J_{\mathfrak{h}}=0,$

from which follows that $d\omega=-u\wedge
Ju\wedge\iota_{V}\omega_{\mathfrak{h}}$. The Bismut torsion $3$-form $c$ is
hence given by

$c=Jd\omega=-u\wedge Ju\wedge J(\iota_{V}\omega_{\mathfrak{h}})=u\wedge
Ju\wedge\iota_{JV}\omega_{\mathfrak{h}},$

and it is clearly closed as $d\mathfrak{h}^{*}\subset\mathfrak{h}^{*}\otimes
x\oplus\mathfrak{h}^{*}\otimes Jx$.
To prove _(ii)_ we use that by _(i)_ the SKT condition is equivalent to
$[A,J_{\mathfrak{h}}]=[B,J_{\mathfrak{h}}]=0$ and
$A,B\in\mathfrak{so}(\mathfrak{h})$. Hence, since $\mathfrak{h}$ is abelian,
$A,B,J_{\mathfrak{h}}$ are three skew-symmetric endomorphisms which commute
pairwise. As a consequence, there exists an orthonormal basis
$\\{e_{1},\dots,e_{2n-2}\\}$ of
$(\mathfrak{h},\langle\cdot,\cdot\rangle_{|\mathfrak{h}})$ such that
$A,B,J_{\mathfrak{h}}$ can be put simultaneously in their diagonal normal
forms

$J_{\mathfrak{h}}=\operatorname{diag}(\Lambda_{1},\dots,\Lambda_{n-1}),\
A=\operatorname{diag}(A_{1},\dots,A_{n-1}),\
B=\operatorname{diag}(B_{1},\dots,B_{n-1}),$

where $\Lambda_{i}$, $A_{i}$ and $B_{i}$ are gives as in (7).
_(iii)_ follows from the fact that, if $(J,\langle\cdot,\cdot\rangle)$ is SKT
then $d\omega=-u\wedge Ju\wedge\iota_{V}\omega_{\mathfrak{h}}$.
To prove _(iv)_ we exploit the condition
$\mathfrak{z}(\mathfrak{g})\cap\mathfrak{h}=\\{0\\}$ to prove that
$B-AJ_{\mathfrak{h}}$ is invertible. Indeed, since
$(J,\langle\cdot,\cdot\rangle,J)$ is SKT, by _(ii)_ we may always find an
orthonormal basis $\\{e_{1},\dots,e_{2n-2}\\}$ of
$(\mathfrak{h},\langle\cdot,\cdot\rangle_{|\mathfrak{h}})$ such that
$A,B,J_{\mathfrak{h}}$ are in their diagonal normal forms. Then,

$B-AJ_{\mathfrak{h}}=\operatorname{diag}(C_{1},\dots,C_{n-1}),\quad{\text{with
}}\quad C_{i}=\begin{pmatrix}a_{i}&-b_{i}\\\ b_{i}&a_{i}\\\ \end{pmatrix}.$

In particular,
$\det(B-AJ_{\mathfrak{h}})=\prod\det(C_{i})=\prod(a_{i}^{2}+b_{i}^{2})\neq 0.$
In fact, if $\det(B-AJ_{\mathfrak{h}})=0$, then there must exists an index
$\overline{i}$ such that $a_{\overline{i}}^{2}+b_{\overline{i}}^{2}=0$.
Moreover, this would imply that $a_{\overline{i}}=b_{\overline{i}}=0$, i.e.,
$e_{2\overline{i}-1},e_{2\overline{i}}\in\mathfrak{z}(\mathfrak{g})\cap\mathfrak{h}=\\{0\\}$,
a contradiction.
Since $B-AJ_{\mathfrak{h}}$ is invertible, there exists a vector
$Y\in\mathfrak{h}$ such that $(B-AJ_{\mathfrak{h}})Y=V$. We consider the new
$J$-Hermitian metric
$\langle\cdot,\cdot\rangle^{\prime}=\langle\cdot,\cdot\rangle_{|\mathfrak{h}}+u^{\prime
2}+Ju^{\prime 2},$ where $u^{\prime}$ and $Ju^{\prime}$ are the duals of
$U^{\prime}=U+Y$ and $JU^{\prime}=JU+JY$. Then,
$A^{\prime}=ad_{u^{\prime}|\mathfrak{h}}=A$,
$B^{\prime}=ad_{Ju^{\prime}|\mathfrak{h}}=B$, implying that the Hermitian
structure $(J,\langle\cdot,\cdot\rangle^{\prime})$ is again SKT (observe that
$\langle\cdot,\cdot\rangle^{\prime}_{|\mathfrak{h}}=\langle\cdot,\cdot\rangle_{|\mathfrak{h}}$).
Moreover, since $[U^{\prime},JU^{\prime}]=0$, the Hermitian structure
$(\langle\cdot,\cdot\rangle^{\prime},J)$ is Kähler by _(iii)_. ∎

######  Remark 2.3.

Note that by Theorem 2.1 the Hermitian Lie algebra
$(\mathfrak{g},\langle\cdot,\cdot\rangle,J)$ is always Chern Ricci-flat, but
in general, when $\mathfrak{h}$ is abelian, it is not Chern flat (for further
details see [32, Proposition 4]).

Regarding the existence of generalized Kähler structures we can prove the
following

######  Theorem 2.4.

Let $\mathfrak{g}$ be a solvable Lie algebra with nilradical $\mathfrak{h}$ of
codimension $2$. The following are equivalent

  1. (i)

$\mathfrak{g}$ admits a generalized Kähler structure
$(J_{\pm},\langle\cdot,\cdot\rangle)$ such that
$J_{\pm}\mathfrak{h}=\mathfrak{h}$;

  2. (ii)

$\mathfrak{g}$ admits a Kähler structure $(J_{+},\langle\cdot,\cdot\rangle)$
such that $J_{+}\mathfrak{h}=\mathfrak{h}$.

###### Proof.

$(ii)\Rightarrow(i).$ It suffices to take $J_{-}=-J_{+}$.
$(i)\Rightarrow(ii).$ Let us fix an orthonormal basis $\\{U,U^{\prime}\\}$ of
$\mathfrak{h}^{\perp}$ with dual basis $\\{u,u^{\prime}\\}$. Without loss of
generality, we may assume such that $U^{\prime}=J_{+}U$ and
$J_{-}U=\varepsilon U^{\prime}$, for $\varepsilon\in\\{-1,+1\\}$. Let
$A=ad_{U}|_{\mathfrak{h}},B=ad_{U^{\prime}}|_{\mathfrak{h}}$ and
$V=[U,U^{\prime}]$.
Since the Bismut torsions $3$-forms $c_{\pm}$ of the Hermitian structures
$(J_{\pm},\langle\cdot,\cdot\rangle)$ satisfy $c_{+}=-c_{-}$, then there exist
$\alpha,\beta\in{\bigwedge}^{2}\mathfrak{h}^{*}$ and
$\gamma\in\mathfrak{h}^{*}$ such that

$c_{\pm}=\pm\alpha\wedge u\pm\beta\wedge u^{\prime}\pm\gamma\wedge u\wedge
u^{\prime}+c_{\mathfrak{h}_{\pm}},$

where $c_{\mathfrak{h}_{\pm}}$ are the torsion $3$-forms of the Hermitian
structures
$({J_{\pm}}_{\mathfrak{h}},\langle\cdot,\cdot\rangle_{\mathfrak{h}})$ on
$\mathfrak{h}$, respectively. Since $c_{+}=-c_{-}$, we clearly have
$c_{\mathfrak{h}+}=-c_{\mathfrak{h}-}$.
Using (1) and the SKT condition $dc_{+}=0$, we get

$\begin{split}dc_{+}&=(d_{\mathfrak{h}}\ \alpha-A^{*}c_{\mathfrak{h}+})\wedge
u+(d_{\mathfrak{h}}\ \beta-B^{*}c_{\mathfrak{h}+})u^{\prime}\\\
&\quad+(-B^{*}\alpha+A^{*}\beta+d_{\mathfrak{h}}\gamma-\iota_{V}c_{\mathfrak{h}+})\wedge
u\wedge u^{\prime}+d_{\mathfrak{h}}c_{\mathfrak{h}+}=0.\end{split}$

It follows that $d_{\mathfrak{h}}c_{\mathfrak{h}+}=0$, i.e.,
$(\mathfrak{h},{J_{\pm}}_{\mathfrak{h}},\langle\cdot,\cdot\rangle_{\mathfrak{h}})$
is a generalized Kähler Lie algebra. Moreover, since $\mathfrak{h}$ is
nilpotent, this forces $\mathfrak{h}$ to be abelian by [14].
For any $W\in\mathfrak{h}$

$c_{+}(W,U,U^{\prime})=-\langle[J_{+}W,J_{+}U],U^{\prime}\rangle-\langle[J_{+}U,J_{+}U^{\prime}],W\rangle-\langle[J_{+}U^{\prime},J_{+}W],U\rangle=-\langle
V,W\rangle$

and, analogously, $c_{-}(W,U,U^{\prime})=-\varepsilon^{2}\langle
V,W\rangle=-\langle V,W\rangle.$ Since $c_{+}=-c_{-}$, we must have $V=0$,
i.e., $[\mathfrak{h}^{\perp},\mathfrak{h}^{\perp}]=0$. Moreover, this implies
that the SKT structure $(J_{+},\langle\cdot,\cdot\rangle)$ on $\mathfrak{g}$
is Kähler by Thorem 2.3. ∎

######  Remark 2.4.

One can show that if a solvable Lie algebra $\mathfrak{g}$ with nilradical
$\mathfrak{h}$ of even-codimension admits a generalized Kähler structure
$(\langle\cdot,\cdot\rangle,J_{\pm})$ such that
$J_{\pm}\mathfrak{h}=\mathfrak{h}$, then $\mathfrak{h}$ is abelian. The proof
proceeds in the same way as before, using a generalization of (1).

##  3\. Case $J\mathfrak{h}\neq\mathfrak{h}$

Let $\mathfrak{g}$ be a $2n$-dimensional solvable Lie algebra with nilradical
$\mathfrak{h}$ of codimension $2$ endowed with an almost Hermitian structure
$(J,\langle\cdot,\cdot\rangle)$ such that $J\mathfrak{h}\neq\mathfrak{h}$. As
a consequence $\mathfrak{g}=\mathfrak{h}+J\mathfrak{h}$. Setting
$\mathfrak{h}_{J}:=\mathfrak{h}\cap J\mathfrak{h}$, we have the orthogonal
decomposition $\mathfrak{g}=\mathfrak{h}_{J}\oplus(\mathfrak{h}_{J})^{\perp},$
where each summand is $J$-invariant. Since $\mathfrak{h}$ has codimension $2$,
$\dim(\mathfrak{h}_{J})=2n-4$. Observe that when $2n=4$, the decomposition
above is trivial. Moreover, since SKT structures on $4$-dimensional solvable
Lie algebras have been fully descripted in [37], we may restrict to consider
$2n>4$. Now, we focus on the $4$-dimensional $J$-invariant space
$(\mathfrak{h}_{J})^{\perp}=\mathfrak{k}\oplus\mathfrak{h}^{\perp}$, where
$\mathfrak{k}$ is the orthogonal complement of $\mathfrak{h}_{J}$ in
$\mathfrak{h}$. Let $\\{e_{2n-1},e_{2n}\\}$ be any orthonormal basis of
$\mathfrak{h}^{\perp}$. Then,

$Je_{2n-1}=J_{34}e_{2n}+h_{2n-2},\quad Je_{2n}=-J_{34}e_{2n-1}+h_{2n-3}$

where $J_{34}\in\mathbb{R}$ and $h_{2n-2},h_{2n-3}$ is a pair of non-zero
orthogonal vectors of $\mathfrak{k}$ such that $J^{2}e_{2n-1}=-e_{2n-1}$ and
$J^{2}e_{2n}=-e_{2n}$. Hence,
$\\{e_{2n-3}:=\frac{h_{2n-3}}{\norm{h_{2n-3}}},e_{2n-2}:=\frac{h_{2n-2}}{\norm{h_{2n-2}}},e_{2n-1},e_{2n}\\}$
is an orthonormal basis of the $J$-invariant subspace
$(\mathfrak{h}_{J})^{\perp}$.
With respect to the decomposition
$\mathfrak{g}=\mathfrak{h}_{J}\oplus(\mathfrak{h}_{J})^{\perp}$, the almost
complex structure $J$ splits as

$J:=\begin{pmatrix}J_{\mathfrak{h}_{J}}&0\\\
0&J_{(\mathfrak{h}_{J})^{\perp}}\end{pmatrix},$

and $J^{2}=-Id$ is equivalent to $J_{\mathfrak{h}_{J}}^{2}=-Id_{2n-4}$ and
$J_{(\mathfrak{h}_{J})^{\perp}}^{2}=-Id_{4}$. With respect to the orthonormal
basis $\\{e_{2n-3},e_{2n-2},e_{2n-1},e_{2n}\\}$ of
$(\mathfrak{h}_{J})^{\perp}$, the restricted almost complex structure
$J_{(\mathfrak{h}_{J})^{\perp}}$ is represented by the skew-symmetric matrix

$J_{(\mathfrak{h}_{J})^{\perp}}=\begin{pmatrix}0&J_{12}&0&J_{14}\\\
-J_{12}&0&J_{23}&0\\\ 0&-J_{23}&0&J_{34}\\\ -J_{14}&0&-J_{34}&0\\\
\end{pmatrix},$

where the entries $J_{ij}$ satisfy the conditions

$\begin{cases}J_{12}^{2}+J_{14}^{2}=1,\quad J_{12}^{2}+J_{23}^{2}=1,\quad
J_{23}^{2}+J_{34}^{2}=1,\quad J_{14}^{2}+J_{34}^{2}=1,\\\
-J_{14}J_{34}+J_{12}J_{23}=0,\quad-J_{14}J_{12}+J_{23}J_{34}=0.\end{cases}$
(8)

Observe that $J_{14}\neq 0$ and $J_{23}\neq 0$, as
$J\mathfrak{h}\neq\mathfrak{h}$. Exploiting the conditions (8), one obtains
two different but equivalent almost complex structures on
$(\mathfrak{h}_{J})^{\perp}$ corresponding to either $J_{12}=-J_{34}$ and
$J_{14}=-J_{23}$ or $J_{12}=J_{34}$ and $J_{14}=J_{23}$. We will consider only
the first case, as they are equivalent up to a change of basis of
$(\mathfrak{h}_{J})^{\perp}$.

######  Remark 3.1.

To summarize, we may always endow $(\mathfrak{h}_{J})^{\perp}$ with an
orthonormal basis $\\{e_{2n-3},e_{2n-2},e_{2n-1},e_{2n}\\}$ such that
$\\{e_{2n-3},e_{2n-2}\\}$ and $\\{e_{2n-1},e_{2n}\\}$ are unitary basis of
$\mathfrak{k}$ and $\mathfrak{h}^{\perp}$ respectively, and the restricted
almost complex structure $J_{(\mathfrak{h}_{J})^{\perp}}$ can be written with
respect to such a basis as

$J_{(\mathfrak{h}_{J})^{\perp}}=\begin{pmatrix}0&J_{12}&0&J_{14}\\\
-J_{12}&0&-J_{14}&0\\\ 0&J_{14}&0&-J_{12}\\\ -J_{14}&0&J_{12}&0\\\
\end{pmatrix},$ (9)

with $J_{12}^{2}+J_{14}^{2}=1$ and $J_{14}\neq 0$.

We restrict now to the case of $\mathfrak{h}$ being abelian, where we can
investigate the existence of generalized Kähler structures.

######  Proposition 3.1.

Let $(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ be an Hermitian solvable Lie
algebra with codimension $2$ abelian nilradical $\mathfrak{h}$ such that
$J\mathfrak{h}\neq\mathfrak{h}$. Then $\mathfrak{h}_{J}:=\mathfrak{h}\cap
J\mathfrak{h}$ is an ideal of $\mathfrak{g}$.
Moreover, if $\\{e_{2n-3},e_{2n-2},e_{2n-1},e_{2n}\\}$ is an orthonormal basis
of $(\mathfrak{h}_{J})^{\perp}$ as in Remark 3.1, we further have that

$[A_{\mathfrak{h}_{J}},J_{\mathfrak{h}_{J}}]=[B_{\mathfrak{h}_{J}},J_{\mathfrak{h}_{J}}]=0,$

where $A_{\mathfrak{h}_{J}},B_{\mathfrak{h}_{J}},J_{\mathfrak{h}_{J}}$ denote
the restriction of
$A:=ad_{e_{2n-1}}|_{\mathfrak{h}},B:=ad_{e_{2n}}|_{\mathfrak{h}}$ and $J$ to
$\mathfrak{h}_{J}$, respectively.

###### Proof.

Let $\\{e_{2n-3},e_{2n-2},e_{2n-1},e_{2n}\\}$ be an orthonormal basis of
$(\mathfrak{h}_{J})^{\perp}$ as in Remark 3.1. To prove that
$\mathfrak{h}_{J}$ is an ideal, since $\mathfrak{h}$ is abelian and
$\mathfrak{g}^{1}\subset\mathfrak{h}$, we only need to check that for any
$X\in\mathfrak{h}_{J}$, $[X,e_{2n}],[X,e_{2n-1}]\in J\mathfrak{h}$. By the
integrability of $J$

$\begin{split}N_{J}(X,e_{2n-2})=&J[X,Je_{2n-2}]-[JX,Je_{2n-2}]\\\
=&J[X,J_{12}e_{2n-3}+J_{14}e_{2n-1}]-[JX,J_{12}e_{2n-3}+J_{14}e_{2n-1}]\\\
=&J_{14}(J[X,e_{2n-1}]-[JX,e_{2n-1}])=-J_{14}(JAX-AJX)=0.\end{split}$

Hence, $[X,e_{2n-1}]=J[-JX,e_{2n-1}]\in J\mathfrak{h}$. Analogously,
$[X,e_{2n}]=J[-JX,e_{2n}]\in J\mathfrak{h}$. Furthermore, we also get
$[A_{\mathfrak{h}_{J}},J_{\mathfrak{h}_{J}}]=[B_{\mathfrak{h}_{J}},J_{\mathfrak{h}_{J}}]=0$.
∎

######  Proposition 3.2.

Let $\mathfrak{g}$ be a solvable Lie algebra with abelian nilradical
$\mathfrak{h}$ of codimension $2$. Assume that $\mathfrak{g}$ is endowed with
a SKT structure $(J,\langle\cdot,\cdot\rangle)$ such that
$J\mathfrak{h}=\mathfrak{h}$. If there exists another complex structure $I$
compatible with $\langle\cdot,\cdot\rangle$ and such that
$I\mathfrak{h}\neq\mathfrak{h}$, then $(J,\langle\cdot,\cdot\rangle)$ is
Kähler.
In particular $\mathfrak{g}$ does not admit any non Kähler generalized Kähler
structure $(I,J,\langle\cdot,\cdot\rangle)$ such that
$I\mathfrak{h}\neq\mathfrak{h}$ and $J\mathfrak{h}=\mathfrak{h}$.

###### Proof.

Let $\\{e_{2n-3},e_{2n-2},e_{2n-1},e_{2n}\\}$ be an orthonormal basis of
$(\mathfrak{h}_{I})^{\perp}=(\mathfrak{h}\cap I\mathfrak{h})^{\perp}$ as in
Remark 3.1).
If we consider any orthonormal basis $\\{e_{1},\dots,e_{2n-4}\\}$ of
$\mathfrak{h}_{I}$, then
$\mathcal{B}=\\{e_{1},\dots,e_{2n-4},e_{2n-3},e_{2n-2}\\}$ is an orthonormal
basis of $\mathfrak{h}$. Furthermore, $\mathfrak{h}_{I}$ is an ideal of
$\mathfrak{g}$ by Proposition 3.1.
Since $(\langle\cdot,\cdot\rangle,J)$ is SKT and $\mathfrak{h}$ is abelian, we
have that $A=ad_{e_{2n-1}}|_{\mathfrak{h}},B=ad_{e_{2n}}|_{\mathfrak{h}}$ are
in $\mathfrak{so}(\mathfrak{h})$ by Corollary 2.3. Hence, with respect to
$\mathcal{B}$

$A=\begin{pmatrix}\begin{array}[]{c|c}A_{\mathfrak{h}_{I}}&0\\\ \hline\cr
0&\begin{matrix}0&c_{12}\\\ -c_{12}&0\end{matrix}\end{array}\end{pmatrix},\ \
B=\begin{pmatrix}\begin{array}[]{c|c}B_{\mathfrak{h}_{I}}&0\\\ \hline\cr
0&\begin{matrix}0&d_{12}\\\
-d_{12}&0\end{matrix}\end{array}\end{pmatrix},\quad{\text{where}}\,\,A_{\mathfrak{h}_{I}},B_{\mathfrak{h}_{I}}\in\mathfrak{so}(\mathfrak{h}_{I}).$

Regarding the Nijenhuis tensor $N_{I}$ as a $(0,3)$-tensor with the aid of the
inner product $\langle\cdot,\cdot\rangle$, namely $N_{I}(X,Y,Z)=\langle
N_{I}(X,Y),Z\rangle$, we get

$N_{I}(e_{2n-2},e_{2n-3},e_{2n-1})=-I_{14}c_{12},\ \
N_{I}(e_{2n-2},e_{2n-3},e_{2n})=-I_{14}d_{12}.$

By the integrability of $I$ and by the fact that $I_{14}\neq 0$ (see Remark
3.1), we have $c_{12}=d_{12}=0$.
Hence, if one computes

$[e_{2n-1},e_{2n}]=-(I[Ie_{2n-1},e_{2n}]+I[e_{2n-1},Ie_{2n}])+[Ie_{2n-1},Ie_{2n}]=I_{12}^{2}[e_{2n-1},e_{2n}],$

one gets $0=(1-I_{12}^{2})[e_{2n-1},e_{2n}]=I_{14}^{2}[e_{2n-1},e_{2n}]$,
i.e., $[\mathfrak{h}^{\perp},\mathfrak{h}^{\perp}]=0$. The result follows by
Corollary 2.3. ∎

######  Remark 3.2.

We have seen in the previous section that if $\mathfrak{g}$ is a solvable Lie
algebra with codimension $2$ nilradical $\mathfrak{h}$ that admits a
generalized Kähler structure $(I_{\pm},\langle\cdot,\cdot\rangle)$ such that
$I_{\pm}\mathfrak{h}=\mathfrak{h}$, then $\mathfrak{h}$ is abelian. It is in
general not true when $I_{\pm}\mathfrak{h}\neq\mathfrak{h}$. Indeed, let us
consider the Lie algebra
$\mathfrak{s}\cong(\mathfrak{h}_{3}\oplus\mathbb{R}^{3})\rtimes\mathbb{R}^{2}=(e^{23}+e^{17},\frac{1}{2}e^{27},\frac{1}{2}e^{37},-e^{48},\frac{1}{2}e^{58}-e^{67},\frac{1}{2}e^{68}+e^{57},0,0),$
where $\\{e_{1},\dots,e_{6}\\}$ is a basis of
$\mathfrak{h}_{3}\oplus\mathbb{R}^{3}$ , $\\{e_{7},e_{8}\\}$ is a basis of
$\mathbb{R}^{2}$ and by $e^{ij}$ we denote $e^{i}\wedge e^{j}$. By the
structure equations, we have that the nilradical $\mathfrak{h}$ is spanned by
$\\{e_{1},\dots,e_{6}\\}$ and it is easy to observe that
$\mathfrak{h}=\mathfrak{h}_{3}\oplus\mathbb{R}^{3}$. Let us define the bi-
Hermitian structure $(\langle\cdot,\cdot\rangle,I_{\pm})$ with
$I_{\pm}\mathfrak{h}\neq\mathfrak{h}$ as

$I_{\pm}e_{1}=e_{7},\ I_{\pm}e_{2}=e_{3},\ I_{\pm}e_{5}=\pm
e_{6},I_{\pm}e_{4}=e_{8},\quad\langle\cdot,\cdot\rangle={\sum}_{i=1}^{8}(e^{i})^{2}.\
$

The corresponding fundamental forms $\omega_{\pm}=e^{1}\wedge
e^{7}+e^{2}\wedge e^{3}\pm e^{5}\wedge e^{6}+e^{4}\wedge e^{8}$ satisfy

$\begin{split}d^{c}_{\pm}\omega_{\pm}=\pm\ e^{4}\wedge e^{5}\wedge
e^{6},\end{split}$

and $e^{4}\wedge e^{5}\wedge e^{6}$ is a closed $3$-form, i.e.,
$(I_{\pm},\langle\cdot,\cdot\rangle)$ is a generalized Kähler structure.
Moreover, we observe that since the Lie algebra is not unimodular, the
corresponding connected and simply connected Lie group does not admit
lattices.

##  4\. Classification in dimension six

In this section we provide a classification of six-dimensional unimodular
solvable Lie algebras with nilradical of codimension $2$ that admit a SKT
structure.

######  Theorem 4.1.

A unimodular six-dimensional solvable Lie algebra $\mathfrak{g}$ with
codimension $2$ nilradical $\mathfrak{h}$ admits a SKT structure
$(J,\langle\cdot,\cdot\rangle)$ if and only if $\mathfrak{g}$ is isomorphic to
one of the following Lie algebras

$\displaystyle\begin{aligned}
&\tau_{3,0}\times\tau_{3,0}=(-f^{25},f^{15},-f^{46},f^{36},0,0),\\\
&\mathfrak{g}_{5.35}^{-2,0}\oplus\mathbb{R}=(2f^{15},-f^{25}-f^{36},-f^{35}+f^{26},0,0,0),\end{aligned}$

hence, in particular, $\mathfrak{h}$ must be abelian. An explicit example of
SKT structure is given respectively by

$\displaystyle Jf_{1}=f_{2},\ Jf_{3}=f_{4},\ Jf_{5}=f_{6},\
\langle\cdot,\cdot\rangle=\sum_{i=1}^{6}(f^{i})^{2}$ $\displaystyle
Jf_{1}=f_{5},\ Jf_{2}=f_{3},\ Jf_{4}=f_{6},\
\langle\cdot,\cdot\rangle=\sum_{i=1}^{6}(f^{i})^{2}.$

###### Proof.

We discuss separately the cases $J\mathfrak{h}=\mathfrak{h}$ and
$J\mathfrak{h}\neq\mathfrak{h}$.
Assume that $\mathfrak{g}$ is endowed with a SKT structure
$(J,\langle\cdot,\cdot\rangle)$ such that $J\mathfrak{h}=\mathfrak{h}$. By
Theorem 2.1, $\mathfrak{h}$ is a $4$-dimensional nilpotent SKT Lie algebra, so
we have that either $\mathfrak{h}=\mathbb{R}^{4}$ or
$\mathfrak{h}=\mathfrak{h}_{3}\oplus\mathbb{R}$. If $\mathfrak{h}$ is abelian,
then one can easily prove that $\mathfrak{z}(\mathfrak{g})\cap\mathfrak{h}=0$
(otherwise we get a contradiction with the fact that $\mathfrak{h}$ has
codimension $2$) and so by the proof of statement _(iv)_ in Corollary 2.3, we
get that $\mathfrak{h}=[\mathfrak{g},\mathfrak{g}]$. The classification
follows by [27, 28] and the Lie algebra $\mathfrak{g}$ has to be isomorphic to
$\tau_{3,0}\times\tau_{3,0}$.
Now we deal to the case $\mathfrak{h}=\mathfrak{h}_{3}\oplus\mathbb{R}$,
proving that this case cannot occur. Assume by contradiction that
$\mathfrak{g}$ admits a SKT structure $(\langle\cdot,\cdot\rangle,J)$ such
that $J\mathfrak{h}=\mathfrak{h}$, then

$\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{h}^{\perp}=\mathfrak{z}(\mathfrak{h})\oplus\mathfrak{z}(\mathfrak{h})\oplus\mathfrak{h}^{\perp}.$

Let $e_{1}$ be a generator of
$\mathfrak{h}^{1}=[\mathfrak{h},\mathfrak{h}]\subset\mathfrak{z}(\mathfrak{h})$
which, up to rescaling, we may assume to be unitary. Then an orthonormal basis
of $\mathfrak{g}$ is provided by
$\mathcal{B}=\\{e_{1},e_{2}=Je_{1},e_{3},e_{4}=Je_{3},e_{5},e_{6}=Je_{5}\\}$,
where $\\{e_{1},e_{2}\\}$ is a basis of $\mathfrak{z}(\mathfrak{h})$,
$\\{e_{1},e_{2}\\}$ is a basis of $\mathfrak{z}(\mathfrak{h})^{\perp}$ and
$\\{e_{5},e_{6}\\}$ is a basis of $\mathfrak{h}^{\perp}$.
The Lie algebra $\mathfrak{g}$ is completely determined by the data

$[e_{5},X]=AX,\ [e_{6},X]=BX,\ \forall X\in\mathfrak{h}\ [e_{3},e_{4}]=\eta
e_{1},\ \ [e_{5},e_{6}]=V,$

where $\eta\in\mathbb{R}\setminus\\{0\\}$,
$A=ad_{e_{5}}|_{\mathfrak{h}},B=ad_{e_{6}}|_{\mathfrak{h}}$ are derivations of
$\mathfrak{h}$ satisfying $[A,B]=ad_{V}|_{\mathfrak{h}}$. We have already
observed in the first section that we may decompose $A$ and $B$ as

$A=\begin{pmatrix}A_{\mathfrak{z}(\mathfrak{h})}&\ast_{A}\\\
0&A_{\mathfrak{z}(\mathfrak{h})^{\perp}}\end{pmatrix}\ \ \text{and}\ \
B=\begin{pmatrix}B_{\mathfrak{z}(\mathfrak{n})}&\ast_{B}\\\
0&B_{\mathfrak{z}(\mathfrak{h})^{\perp}}\end{pmatrix},$

with
$A_{\mathfrak{z}(\mathfrak{h})},B_{\mathfrak{z}(\mathfrak{h})}\in\mathfrak{so}(\mathfrak{z}(\mathfrak{h}))$
being such that
$[A_{\mathfrak{z}(\mathfrak{h})},J_{\mathfrak{z}(\mathfrak{h})}]=[B_{\mathfrak{z}(\mathfrak{h})},J_{\mathfrak{z}(\mathfrak{h})}]=0$
by Theorem 2.1.
Moreover, since $A$ and $B$ are derivations of $\mathfrak{h}$,
$A\mathfrak{h}^{1}\subset\mathfrak{h}^{1}$ and
$B\mathfrak{h}^{1}\subset\mathfrak{h}^{1}$, and so
$A_{\mathfrak{z}(\mathfrak{h})}=B_{\mathfrak{z}(\mathfrak{h})}=0$. In
particular, this forces
$\tr(A_{\mathfrak{z}(\mathfrak{h})^{\perp}})=\tr(B_{\mathfrak{z}(\mathfrak{h})^{\perp}})=0$.
Let

$A_{\mathfrak{z}(\mathfrak{h})^{\perp}}=\begin{pmatrix}a_{33}&a_{34}\\\
a_{43}&-a_{33}\end{pmatrix}\ \ \text{and}\ \
B_{\mathfrak{z}(\mathfrak{h})^{\perp}}=\begin{pmatrix}b_{33}&b_{34}\\\
b_{43}&-b_{33}\end{pmatrix}.$

Exploiting that
$A_{\mathfrak{z}(\mathfrak{h})}=B_{\mathfrak{z}(\mathfrak{h})}=0$, the
integrability condition
$[J_{\mathfrak{h}},A]J_{\mathfrak{h}}+[J_{\mathfrak{h}},B]=0$ and the relation
$[A,B]=ad_{V}|_{\mathfrak{h}}$ yield

$\begin{split}&[J_{\mathfrak{h}},A_{\mathfrak{z}(\mathfrak{h})^{\perp}}]J_{\mathfrak{h}}+[J_{\mathfrak{h}},B_{\mathfrak{z}(\mathfrak{h})^{\perp}}]=0\iff
2a_{33}=b_{34}+b_{43},\ \ 2b_{33}=-(a_{34}+a_{43}),\\\
&[A_{\mathfrak{z}(\mathfrak{h})^{\perp}},B_{\mathfrak{z}(\mathfrak{h})^{\perp}}]=0\iff
2a_{33}(b_{34}+b_{43})-2b_{33}(a_{34}+a_{43})=0.\\\ \end{split}$

It is immediate to observe that the two conditions together imply that
$A_{\mathfrak{z}(\mathfrak{h})^{\perp}}$ and
$B_{\mathfrak{z}(\mathfrak{h})^{\perp}}$ are skew. If any of $a_{34}$ or
$b_{34}$ is zero, then the corresponding matrix would be nilpotent, which is a
contradiction with the hypothesis that $\mathfrak{h}$ has codimension $2$.
Hence we must have $a_{34},b_{34}\neq 0$. Moreover, if we consider the
subspace generated by
$\\{e_{1},\dots,e_{4},e_{5}-\frac{a_{34}}{b_{34}}e_{6}\\}$, then it is a
nilpotent ideal which strictly contains $\mathfrak{h}$, providing the
contradiction.
Now, we consider the case of $J\mathfrak{h}\neq\mathfrak{h}$. In this case,
the four-dimensional nilradical is isomorphic to one of
$\mathbb{R}^{4},\mathfrak{h}_{3}\oplus\mathbb{R}$ and
$\mathfrak{h}_{4}=(-24,-34,0,0)$. Moreover, the only six-dimensional solvable
Lie algebra with nilradical $\mathfrak{h}_{4}$ is
$\mathfrak{g}_{6.28}=(0,0,46-13-2.25,56-24,15,-16+26)$ (see [38]), which is
not unimodular.
Hence, as in the previous case, we may restrict to consider either
$\mathfrak{h}=\mathfrak{h}_{3}\oplus\mathbb{R}$ or
$\mathfrak{h}=\mathbb{R}^{4}$. We start by considering the first case, which
is more involved.
Firstly, we prove that if $\mathfrak{g}$ admits a complex structure $J$ such
that $J\mathfrak{h}\neq\mathfrak{h}$, then
$\dim(J(\mathfrak{z}(\mathfrak{h}))\cap\mathfrak{h})=1$. Since the center
$\mathfrak{z}(\mathfrak{h})$ has dimension $2$, it suffices to check that if
$\dim(J(\mathfrak{z}(\mathfrak{h}))\cap\mathfrak{h})=0,2$ we get a
contradiction.
We start by considering
$\dim(J(\mathfrak{z}(\mathfrak{h}))\cap\mathfrak{h})=0$. Let $Z_{1}$ be a
generator of $\mathfrak{h}^{1}=[\mathfrak{h},\mathfrak{h}]$, and let $Z_{2}$
be such that $\mathfrak{z}(\mathfrak{h})=\text{span}_{\mathbb{R}}\langle
Z_{1},Z_{2}\rangle$. Since $\mathfrak{g}=\mathfrak{h}+J\mathfrak{h}$, we may
fix a basis $\\{Z_{1},Z_{2},X,JX,JZ_{1},JZ_{2}\\}$ of $\mathfrak{g}$, with
$X\in\mathfrak{h}_{J}=\mathfrak{h}\cap J\mathfrak{h}$.
Observe that since $ad_{JZ_{1}}|_{\mathfrak{h}},ad_{JZ_{2}}|_{\mathfrak{h}}$
are derivations of $\mathfrak{h}$, they preserve both $\mathfrak{h}^{1}$ and
$\mathfrak{z}(\mathfrak{h})$. Exploiting this fact, the integrability of $J$
and the inclusion $\mathfrak{g}^{1}\subset\mathfrak{h}$, we get the following
structure equations

$\begin{split}&[X,JX]=\eta Z_{1},\ \ [Z_{1},JZ_{1}]=a_{11}Z_{1},\ \
[Z_{1},JZ_{2}]=a_{12}Z_{1},\ \ [Z_{2},JZ_{1}]=a_{12}Z_{1},\\\
&[Z_{2},JZ_{2}]=b_{12}Z_{1}+b_{22}Z_{2},\ \ [X,JZ_{1}]=a_{33}X+a_{43}JX,\ \
[X,JZ_{2}]=b_{33}X+b_{43}JX,\\\ &[JX,JZ_{1}]=-a_{43}X+a_{33}JX,\ \
[JX,JZ_{2}]=-b_{43}X+b_{33}JX,\ \ [JZ_{1},JZ_{2}]=0.\\\ \end{split}$

Computing

$ad_{JZ_{1}}[X,JX]=[ad_{JZ_{1}}X,JX]+[X,ad_{JZ_{1}}JX]$

and

$ad_{JZ_{2}}[X,JX]=[ad_{JZ_{2}}X,JX]+[X,ad_{JZ_{2}}JX]$

we further have $a_{11}=2a_{33}$ and $a_{12}=2b_{33}$.
In particular, by the unimodularity condition, this leads to $a_{11}=a_{33}=0$
and $b_{22}=-4b_{33}$.
Moreover, $[ad_{JZ_{1}},ad_{JZ_{2}}]=ad_{[JZ_{1},JZ_{2}]}=0$ if and only of
$b_{33}=0$.
To sum up, the Lie algebra $\mathfrak{g}$ is determined by the data

$\begin{split}ad_{JZ_{1}}|_{\mathfrak{h}}=\begin{pmatrix}0&0&0&0\\\ 0&0&0&0\\\
0&0&0&a_{43}\\\ 0&0&-a_{43}&0\\\ \end{pmatrix},\ \
ad_{JZ_{2}}|_{\mathfrak{h}}=\begin{pmatrix}0&-b_{12}&0&0\\\ 0&0&0&0\\\
0&0&0&b_{43}\\\ 0&0&-b_{43}&0\\\ \end{pmatrix},\ \ [X,JX]=\eta Z_{1}\neq
0.\end{split}$

Observe that we must have $a_{43}\neq 0$. As $a_{43}\neq 0$, we may consider
the subspace generated by
$\\{Z_{1},Z_{2},X,JX,JZ_{2}-\frac{b_{43}}{a_{43}}JZ_{1}\\}$, that is a
nilpotent ideal which strictly contains $\mathfrak{h}$. The contradiction
follows.
Let us now consider the case of
$\dim(J(\mathfrak{z}(\mathfrak{h}))\cap\mathfrak{h})=2$. In this case
$\mathfrak{h}_{J}=\mathfrak{z}(\mathfrak{h})$, so we may fix a basis
$\\{Z,JZ\\}$ of $\mathfrak{z}(\mathfrak{h})$, with $Z$ being a generator of
$\mathfrak{h}^{1}$. Since $\mathfrak{g}=\mathfrak{h}+J\mathfrak{h}$, we may
complete $Z,JZ$ to a basis $\\{Z,JZ,X,Y,JX,JY\\}$ of $\mathfrak{g}$, where
$\\{Z,JZ,X,Y\\}$ is a basis of $\mathfrak{h}$. Analogously to the previous
case, we obtain the following structure equations

$\begin{split}&[X,Y]=\eta Z,\ \ [Z,JX]=a_{11}Z,\ \ [Z,JY]=b_{11}Z,\ \
[JZ,JX]=a_{11}JZ,\\\ &[JZ,JY]=b_{11}JZ,\ \
[X,JX]=a_{13}Z+a_{23}JZ+a_{33}X+a_{43}Y,\\\
&[X,JY]=b_{13}Z+b_{23}JZ+a_{34}X+a_{44}Y,\ \
[Y,JX]=a_{14}Z+a_{24}JZ+a_{34}X+a_{44}Y,\\\
&[Y,JY]=b_{14}Z+b_{24}JZ+b_{34}X+b_{44}Y,\ \
[JX,JY]=(\eta+(a_{24}-b_{23}))Z+(b_{13}-a_{14})JZ.\\\ \end{split}$

Exploiting

$ad_{JX}[X,Y]=[ad_{JX}X,Y]+[X,ad_{JX}Y]\quad{\text{and}}\quad
ad_{JY}[X,Y]=[ad_{JY}X,Y]+[X,ad_{JY}Y]$

we further have $a_{11}=a_{33}+a_{44}$ and $b_{11}=a_{34}+b_{44}$. This
identities combined with the unimodularity of $\mathfrak{g}$ lead to
$a_{11}=0$, $a_{33}=-a_{44}$ , $b_{11}=0$ and $a_{34}=-b_{44}$.
We hence get that

$\begin{split}&ad_{JX}|_{\mathfrak{h}}=\begin{pmatrix}0&0&-a_{13}&-a_{14}\\\
0&0&-a_{23}&-a_{24}\\\ 0&0&-a_{33}&-a_{34}\\\ 0&0&-a_{43}&a_{33}\\\
\end{pmatrix},\ \
ad_{JY}|_{\mathfrak{h}}=\begin{pmatrix}0&0&-b_{13}&-b_{14}\\\
0&0&-b_{23}&-b_{24}\\\ 0&0&-a_{34}&-b_{34}\\\ 0&0&a_{33}&a_{34}\\\
\end{pmatrix},\\\ &[X,Y]=\eta Z,\
[JX,JY]=(\eta+(a_{24}-b_{23}))Z+(b_{13}-a_{14})JZ\in\mathfrak{z}(\mathfrak{h})=\mathfrak{z}(\mathfrak{g}).\end{split}$

Since $[JX,JY]\in\mathfrak{z}(\mathfrak{h})=\mathfrak{z}(\mathfrak{g})$, as
before $[ad_{JX},ad_{JY}]=ad_{[JX,JY]}=0$. With the relations coming from this
identity, if one compute the spectrum of $ad_{JX}$ one obtains that $ad_{JX}$
has only the eigenvalue $0$, so it is nilpotent. Hence, the subspace generated
by $\\{Z,JZ,X,Y,JX\\}$ is a nilpotent ideal which strictly contains
$\mathfrak{h}$, giving a contradiction.
Hence, we may restrict to consider the case of
$\dim(J(\mathfrak{z}(\mathfrak{h}))\cap\mathfrak{h})=1$ and we prove that if
$\dim(J(\mathfrak{z}(\mathfrak{h}))\cap\mathfrak{h})=1$, then $\mathfrak{g}$
cannot admit any SKT structure $(\langle\cdot,\cdot\rangle,J)$ such that
$J\mathfrak{h}\neq\mathfrak{h}$.
Let $Z_{1},Z_{2}$ be a basis of $\mathfrak{z}(\mathfrak{h})$ such that
$JZ_{1}\in\mathfrak{h}$ and $JZ_{2}\notin\mathfrak{h}$. Since
$\mathfrak{g}=\mathfrak{h}+J\mathfrak{h}$, there always exists a vector
$X\in\mathfrak{h}$ such that $\\{Z_{1},JZ_{1},Z_{2},X,JZ_{2},JX\\}$ is a basis
of $\mathfrak{g}$. We proceed as before. Indeed, using that
$ad_{JZ_{2}},ad_{JX}$ preserve $\mathfrak{z}(\mathfrak{h})$, the integrability
of $J$, and the inclusion $\mathfrak{g}^{1}\subset\mathfrak{h}$, we get

$\begin{split}&[JZ_{1},X]=\eta_{1}Z_{1}+\eta_{2}Z_{2},\ \
[Z_{1},JZ_{2}]=a_{11}Z_{1},\ \ [Z_{1},JX]=b_{11}Z_{1}-\eta_{2}Z_{2},\\\
&[JZ_{1},JZ_{2}]=a_{11}JZ_{1},\ \ [JZ_{1},JX]=(\eta_{1}+b_{11})JZ_{1},\ \
[Z_{2},JZ_{2}]=a_{13}Z_{1}+a_{33}Z_{2},\\\
&[Z_{2},JX]=b_{13}Z_{1}+a_{34}Z_{2},\ \
[X,JZ_{2}]=a_{14}Z_{1}+a_{24}JZ_{1}+a_{34}Z_{2},\\\
&[X,JX]=b_{14}Z_{1}+b_{24}JZ_{1}+b_{34}Z_{2}+b_{44}X,\ \
[JZ_{2},JX]=a_{24}Z_{1}+(b_{13}-a_{14})JZ_{1}.\\\ \end{split}$

By the unimodularity of $\mathfrak{g}$, $a_{33}=-2a_{11}$ and
$2b_{11}+\eta_{1}+a_{34}+b_{44}=0$. Observe that if $a_{11}=0$ then
$ad_{JZ_{1}}$ is a strictly upper triangular matrix, and hence, nilpotent.
Hence $a_{11}\neq 0$, and, exploiting
$ad_{JZ_{2}}[JZ_{1},X]=[ad_{JZ_{2}}JZ_{1},X]+[JZ_{1},ad_{JZ_{2}}X]$, we get
$\eta_{2}=0$.
Analogously, if one computes
$ad_{JX}[JZ_{1},X]=[ad_{JX}JZ_{1},X]+[JZ_{1},ad_{JX}X]$, then one obtains
$b_{44}=-\eta_{1}$. Plugging this identity in the unimodularity condition
$2b_{11}+\eta_{1}+a_{34}+b_{44}=0$ this also gives $2b_{11}+a_{34}=0$.
To sum up

$\begin{split}&ad_{JZ_{2}}|_{\mathfrak{h}}=\begin{pmatrix}-a_{11}&0&-a_{13}&-a_{14}\\\
0&-a_{11}&0&-a_{24}\\\ 0&0&2a_{11}&2b_{11}\\\ 0&0&0&0\\\ \end{pmatrix},\ \
ad_{JX}|_{\mathfrak{h}}=\begin{pmatrix}-b_{11}&0&-b_{13}&-b_{14}\\\
0&-(\eta_{1}+b_{11})&0&-b_{24}\\\ 0&0&2b_{11}&-b_{34}\\\ 0&0&0&\eta_{1}\\\
\end{pmatrix},\\\ &[JZ_{2},JX]=a_{24}Z_{1}+(b_{13}-a_{14})JZ_{1}.\end{split}$

Consider any $J$-hermitian inner product $\langle\cdot,\cdot\rangle$ and
denote by $c$ the Bismut torsion $3$-form of
$(J,\langle\cdot,\cdot\rangle,J)$. Then,
$dc(Z_{1},JZ_{1},X,JZ_{2})=-2a_{11}\eta_{1}\norm{Z_{1}}^{2}\neq 0.$ Indeed,
this can be zero only if $\eta_{1}=0$. Moreover in this case
$\mathfrak{h}=\mathbb{R}^{4}$, giving a contradiction.
Hence, we have proved that if $\mathfrak{g}$ is endowed with a SKT structure
$(\langle\cdot,\cdot\rangle,J)$ such that $J\mathfrak{h}\neq\mathfrak{h}$,
then $\mathfrak{h}$ must be abelian. We restrict to consider
$\mathfrak{h}=\mathbb{R}^{4}$.
Let us assume that $\mathfrak{g}$ admits a SKT structure
$(\langle\cdot,\cdot\rangle,J)$ such that $J\mathfrak{h}\neq\mathfrak{h}$,
with $\mathfrak{h}=\mathbb{R}^{4}$. We fix an orthonormal basis
$\mathcal{B}=\\{e_{1},\dots,e_{6}\\}$ of $\mathfrak{g}$ such that
$\\{e_{1},e_{2}\\}$ is an orthonormal basis of $\mathfrak{h}_{J}$ such that
$Je_{1}=e_{2}$ and $\\{e_{3},\dots,e_{6}\\}$ is an orthonormal basis of
$(\mathfrak{h}_{J})^{\perp}$ as in Remark 3.1, namely, with respect to
$\mathcal{B}$ the complex structure $J$ can be written as

$J=\begin{pmatrix}0&1&0&0&0&0\\\ -1&0&0&0&0&0\\\ 0&0&0&J_{12}&0&J_{14}\\\
0&0&-J_{12}&0&-J_{14}&0\\\ 0&0&0&J_{14}&0&-J_{12}\\\ 0&0&-J_{14}&0&J_{12}&0\\\
\end{pmatrix}.$

By Proposition 3.1, $\mathfrak{h}_{J}$ is an ideal of $\mathfrak{g}$ and
$[A_{\mathfrak{h}_{J}},J]=[B_{\mathfrak{h}_{J}},J]=0$, where
$A_{\mathfrak{h}_{J}},B_{\mathfrak{h}_{J}}$ denote the restrictions of
$A:=ad_{e_{5}}|_{\mathfrak{h}},B:=ad_{e_{6}}|_{\mathfrak{h}}$ to
$\mathfrak{h}_{J}$. By the integrability of $J$, we have that

$\begin{split}&A=\begin{pmatrix}\begin{array}[]{c|c}\begin{matrix}a_{11}&a_{12}\\\
-a_{12}&a_{11}\end{matrix}&\begin{matrix}\ast\end{matrix}\\\ \hline\cr
0&\begin{matrix}c_{11}&c_{12}\\\
c_{21}&c_{22}\end{matrix}\end{array}\end{pmatrix},\
B=\begin{pmatrix}\begin{array}[]{c|c}\begin{matrix}b_{11}&b_{12}\\\
-b_{12}&b_{11}\end{matrix}&\begin{matrix}\ast\end{matrix}\\\ \hline\cr
0&\begin{matrix}d_{11}&-c_{11}\\\
d_{21}&-c_{21}\end{matrix}\end{array}\end{pmatrix},\\\
&[e_{5},e_{6}]=V_{\mathfrak{h}_{J}}-\frac{J_{12}}{J_{14}}(c_{12}-d_{11})e_{3}-\frac{J_{12}}{J_{14}}(c_{22}-d_{21})e_{4},\end{split}$

with $2a_{11}+c_{11}+c_{22}=0$ and $2b_{11}+d_{11}-c_{21}=0$ by the
unimodularity of $\mathfrak{g}$. In the following we will denote by

$C:=\begin{pmatrix}c_{11}&c_{12}\\\ c_{21}&c_{22}\end{pmatrix}\ \text{and}\
D:=\begin{pmatrix}d_{11}&-c_{11}\\\ d_{21}&-c_{21}\end{pmatrix}.$

Since by the Jacobi identity $[A,B]=0$, then also $[C,D]=0$.
By the SKT condition, we get

$\begin{split}dc(e_{1},e_{2},e_{3},e_{6})=&2J_{14}(-b_{11}c_{21}+a_{11}d_{21})=0,\\\
dc(e_{1},e_{2},e_{3},e_{5})=&2J_{14}(b_{11}c_{22}+a_{11}c_{21})=0,\\\
dc(e_{1},e_{2},e_{4},e_{5})=&-2J_{14}(b_{11}c_{12}+a_{11}c_{11})=0\\\
dc(e_{1},e_{2},e_{4},e_{6})=&2J_{14}(b_{11}c_{11}-a_{11}d_{11})=0\\\
\end{split}$

Since $J_{14}\neq 0$, the coefficients of $A$ and $B$ must obey to

$\begin{split}&b_{11}c_{21}-a_{11}d_{21}=0,\ \ b_{11}c_{22}+a_{11}c_{21}=0,\ \
b_{11}c_{12}+a_{11}c_{11}=0,\ \ b_{11}c_{11}-a_{11}d_{11}=0.\end{split}$ (10)

We distinguish two cases depending on whether $a_{11}$ is zero or not. We will
do in details the first case. In fact, the second one is analogous.
Let $a_{11}=0$. The conditions (10) becomes

$b_{11}c_{21}=0,\ \ b_{11}c_{22}=0,\ \ b_{11}c_{12}=0,\ \ b_{11}c_{11}=0,$

and furthermore $c_{11}=-c_{22}$ in order to have $\tr(A)=0$.
We claim that $b_{11}\neq 0$. Indeed, assume by contradiction that $b_{11}=0$.
Then (10) are satisfied and $d_{11}=c_{21}$ by the unimodularity condition.
Using $[C,D]=0$, one can show that $C$ and $D$ are nilpotent matrices and so
we must have $a_{12},b_{12}\neq 0$. Indeed, if for instance $a_{12}=0$, then
the subspace generated by $\\{e_{1},e_{2},e_{3},e_{4},e_{5}\\}$ would be a
nilpotent ideal which strictly contains $\mathfrak{h}$, which is a
contradiction with the maximality of the nilradical.
If $a_{12},b_{12}\neq 0$, then we may consider
$e^{\prime}_{5}=e_{5}-\frac{a_{12}}{b_{12}}e_{6}$ with

$ad_{e^{\prime}_{5}}|_{\mathfrak{h}}=\begin{pmatrix}\begin{array}[]{c|c}0&\ast\\\
\hline\cr 0&E\end{array}\end{pmatrix},$

where $E=C-\frac{a_{12}}{b_{12}}D$. Moreover, since $C,D$ are nilpotent and
$[C,D]=0$, so is $E$ and, hence, also $ad_{e^{\prime}_{5}}|_{\mathfrak{h}}$.
Therefore, if we consider the ideal
$\\{e_{1},e_{2},e_{3},e_{4},e^{\prime}_{5}\\}$, then again this is a nilpotent
ideal which strictly contains the nilradical, proving the claim.
Since $b_{11}\neq 0$, by (10) we must have that

$\begin{split}&A=\begin{pmatrix}\begin{array}[]{c|c}\begin{matrix}0&a_{12}\\\
-a_{12}&0\end{matrix}&\begin{matrix}\ast\end{matrix}\\\ \hline\cr
0&\begin{matrix}0&0\\\ 0&0\end{matrix}\end{array}\end{pmatrix},\ \
B=\begin{pmatrix}\begin{array}[]{c|c}\begin{matrix}b_{11}&b_{12}\\\
-b_{12}&b_{11}\end{matrix}&\begin{matrix}\ast\end{matrix}\\\ \hline\cr
0&\begin{matrix}-2b_{11}&0\\\ d_{21}&0\end{matrix}\end{array}\end{pmatrix},\\\
&[e_{5},e_{6}]=V_{\mathfrak{h}_{J}}-2\frac{J_{12}}{J_{14}}b_{11}e_{3}+\frac{J_{12}}{J_{14}}d_{21}e_{4},\end{split}$

with $a_{12}\neq 0$ (otherwise $A$ would be nilpotent). In order to kill the
components of $V$ along $e_{3}$ and $e_{4}$, we take
$e^{\prime}_{5}:=e_{5}+\frac{J_{12}}{J_{14}}e_{3}$. In such a way
$[e^{\prime}_{5},e_{6}]\in\mathfrak{h}_{J}$ and $\\{e^{\prime}_{5},e_{6}\\}$
define a new complement of $\mathfrak{h}$ inside $\mathfrak{g}$. Observe that
since $\mathfrak{h}$ is abelian,
$A^{\prime}:=ad_{e^{\prime}_{5}}|_{\mathfrak{h}}=A$.
Let $e^{\prime}_{3}$ and $e^{\prime}_{4}$ be eigenvectors of $B$ associated to
the eigenvalues $-2b_{11}$ and $0$, respectively. Since the eigenspaces
$V_{-2b_{11}}$ and $V_{0}$ are $1$-dimensional and $[A,B]=0$, we must have
that $e^{\prime}_{3}$ and $e^{\prime}_{4}$ are also eigenvectors of $A$ with
eigenvalue $0$. Hence, with respect the new basis
$\\{e_{1},e_{2},e^{\prime}_{3},e^{\prime}_{4},e^{\prime}_{5},e_{6}\\}$ the Lie
algebra $\mathfrak{g}$ is determined by the data

$\begin{split}&A^{\prime}=\begin{pmatrix}\begin{array}[]{c|c}\begin{matrix}0&a_{12}\\\
-a_{12}&0\end{matrix}&0\\\ \hline\cr 0&\begin{matrix}0&0\\\
0&0\end{matrix}\end{array}\end{pmatrix},\ \
B=\begin{pmatrix}\begin{array}[]{c|c}\begin{matrix}b_{11}&b_{12}\\\
-b_{12}&b_{11}\end{matrix}&0\\\ \hline\cr 0&\begin{matrix}-2b_{11}&0\\\
0&0\end{matrix}\end{array}\end{pmatrix},\ \
[e^{\prime}_{5},e_{6}]\in\mathfrak{h}_{J}.\end{split}$

Finally, let $X\in\mathfrak{h}_{J}$ be such that
$A^{\prime}X=-[e^{\prime}_{5},e_{6}]$ (observe that it is possible since
$\text{Im}A^{\prime}=\mathfrak{h}_{J}$). The basis
$\\{e^{\prime}_{1},\dots,e^{\prime}_{5},e^{\prime}_{6}=e_{6}+X\\}$ is such
that $B:=ad_{e^{\prime}_{6}}|_{\mathfrak{h}}=B^{\prime}$ and
$[e^{\prime}_{5},e^{\prime}_{6}]=0$. The isomorphism between $\mathfrak{g}$
and $\mathfrak{g}_{5.35}^{-2,0}\oplus\mathbb{R}$ is immediate. ∎

######  Remark 4.1.

The Lie algebra $\mathfrak{g}_{5.35}^{-2,0}$ firstly appears in [10].

######  Remark 4.2.

The connected and simply connected solvable Lie groups corresponding to
$\tau_{3,0}\times\tau_{3,0}$ and $\mathfrak{g}_{5.35}^{-2,0}\oplus\mathbb{R}$
admit lattices. The former case is trivial. For the latter, an explicit
construction of a lattice is done in the next section (Theorem 6.5).

######  Corollary 4.2.

A unimodular six-dimensional solvable Lie algebra with codimension $2$
nilradical $\mathfrak{h}$ admits a generalized Kähler structure
$(J_{\pm},\langle\cdot,\cdot\rangle)$ if and only if $\mathfrak{g}$ is
isomorphic to one of the following Lie algebras

$\displaystyle\begin{aligned}
&\tau_{3,0}\times\tau_{3,0}=(-f^{25},f^{15},-f^{46},f^{36},0,0),\\\
&\mathfrak{g}_{5.35}^{-2,0}\oplus\mathbb{R}=(2f^{15},-f^{25}-f^{36},-f^{35}+f^{26},0,0,0).\end{aligned}$

An explicit example of generalized Kähler structure is given respectively by

$J_{\pm}f_{1}=\pm f_{2},\ J_{\pm}f_{3}=\pm f_{4},\ J_{\pm}f_{5}=\pm f_{6},\
\langle\cdot,\cdot\rangle=\sum_{i=1}^{6}(f^{i})^{2}$ $J_{\pm}f_{1}=f_{5},\
J_{\pm}f_{2}=\pm f_{3},\ J_{\pm}f_{4}=f_{6},\
\langle\cdot,\cdot\rangle=\sum_{i=1}^{6}(f^{i})^{2}.$

##  5\. Construction of new SKT and generalized Kähler Lie algebras

This section is devoted to construct new examples of SKT and generalized
Kähler solvable Lie algebras. In particular, we exhibit examples of SKT
solvable Lie algebras $(\mathfrak{g},\langle\cdot,\cdot\rangle,J)$ with
$J\mathfrak{h}=\mathfrak{h}$ and $J\mathfrak{h}\neq\mathfrak{h}$ in Examples
5.1, 5.3, and in Examples 5.4, 5.5, respectively.

######  Example 5.1.

By Theorem 4.1, we have that the only six dimensional SKT Lie algebra
$(\mathfrak{g},J,\langle\cdot,\cdot\rangle)$ such that
$J\mathfrak{h}=\mathfrak{h}$ is $\tau_{3,0}\times\tau_{3,0}$. Moreover, as
$\tau_{3,0}\times\tau_{3,0}$ has trivial center, there exists a (possibly
different) $J$-Hermitian structure which is Kähler, by Corollary 2.3 _(iv)_.
In higher dimension, this is no longer true. Indeed, let us consider the eight
dimensional solvable Lie algebra $\mathfrak{g}^{b}_{8},$ with $b\neq 0$,
defined by the structure equations

$\begin{split}&[e_{1},e_{7}]=be_{2},\quad[e_{2},e_{7}]=-be_{1},\quad[e_{3},e_{8}]=be_{4},\quad[e_{4},e_{8}]=-be_{3},\quad[e_{7},e_{8}]=e_{5}+e_{6},\\\
\end{split}$

endowed with the complex structure $J$, given by

$Je_{1}=e_{2},\ \ Je_{3}=e_{4},\ \ Je_{5}=e_{6},\ \ Je_{7}=e_{8}.$

In particular, $\mathfrak{h}=\text{span}_{\mathbb{R}}\langle
e_{1},\dots,e_{6}\rangle$ and $J\mathfrak{h}=\mathfrak{h}$. Let us consider
the inner product $\langle\cdot,\cdot\rangle$ with respect to which the basis
$\\{e_{1},\dots,e_{6}\\}$ is orthonormal. Set $A:=ad_{e_{7}}|_{\mathfrak{h}}$
and $B:=ad_{e_{8}}|_{\mathfrak{h}}$. By Corollary 2.3 the Hermitian structure
$J,(\langle\cdot,\cdot\rangle)$ is SKT and it is (non-flat) Chern Ricci flat
by Theorem 2.1.
We show that $(\mathfrak{g}^{b}_{8},J)$ does not admit any Kähler structure.
Let $\\{Z_{1},\dots,Z_{4}\\}$ be a unitary basis of
$(\mathfrak{g}^{b}_{8})^{(1,0)}$, with dual basis
$\\{\varphi^{1},\dots,\varphi^{4}\\}$. Then

$d\varphi^{1}=-\frac{ib}{2}(\varphi^{14}+\varphi^{1\bar{4}}),\
d\varphi^{2}=-\frac{b}{2}(\varphi^{24}-\varphi^{2\bar{4}}),\ \
d\varphi^{3}=-\frac{(1+i)}{2}\varphi^{4\bar{4}},\ \ d\varphi^{4}=0.$

Let us write the generic fundamental form
$\omega=\frac{i}{2}\sum_{j=1}^{4}F_{j\bar{j}}\varphi^{j\bar{j}}+\frac{1}{2}\sum_{j=1}^{4}(F_{j\bar{k}}\varphi^{j\bar{k}}-\overline{F_{j\bar{k}}}\varphi^{k\bar{j}}),$
with $F_{j\bar{j}}\in\mathbb{R}_{>0}$. Since
$d\omega(Z_{3},Z_{4},\overline{Z}_{4})=\frac{1-i}{2}F_{3\bar{3}},$ we have
that $d\omega\neq 0$.

######  Theorem 5.2.

Let $(\mathfrak{h},\langle\cdot,\cdot\rangle_{\mathfrak{h}},J_{\mathfrak{h}})$
be a SKT nilpotent Lie algebra and let
$(\mathfrak{a},J_{\mathfrak{a}},\langle\cdot,\cdot\rangle_{\mathfrak{a}},J_{\mathfrak{a}})$
an abelian Hermitian Lie algebra of dimension $2k$. Consider a Lie algebras
homomorphism

$\theta:\mathfrak{a}\to Der(\mathfrak{h})$

such that
$\theta(\mathfrak{a})\subset\mathfrak{sp}(\mathfrak{h})\cap\mathfrak{so}(\mathfrak{h})$,
namely $\theta(U)^{t}=-\theta(U)$ and
$J_{\mathfrak{h}}\theta(U)+\theta(U)^{t}J_{\mathfrak{h}}=0$, $\forall
U\in\mathfrak{a}$. Then
$(\mathfrak{g}=\mathfrak{h}\rtimes_{\theta}\mathfrak{a},\langle\cdot,\cdot\rangle,J)$,
with
$\langle\cdot,\cdot\rangle=\langle\cdot,\cdot\rangle_{\mathfrak{h}}+\langle\cdot,\cdot\rangle_{\mathfrak{a}}$
and $J=\begin{pmatrix}J_{\mathfrak{h}}&0\\\ 0&J_{\mathfrak{a}}\\\
\end{pmatrix}$, is SKT.

###### Proof.

Let $\\{U_{1},J_{\mathfrak{a}}U_{1},\dots,U_{k},J_{\mathfrak{a}}U_{k}\\}$ be
an orthonormal basis of the abelian Hermitian Lie algebra
$(\mathfrak{a},J_{\mathfrak{a}},\langle\cdot,\cdot\rangle_{\mathfrak{a}},J_{\mathfrak{a}})$
with dual basis
$\\{u_{1},J_{\mathfrak{a}}u_{1},\dots,u_{k},J_{\mathfrak{a}}u_{k}\\}$. Observe
that since $\mathfrak{g}^{1}\subset\mathfrak{h}$,
$du_{i}=d(J_{\mathfrak{a}}u_{i})=0$. To lighten the notation, let us define
$A_{i}:=\theta(u_{i})$ and $B_{i}:=\theta(J_{\mathfrak{a}}u_{i})$.
Firstly, we prove that $J$ is integrable. Exploiting that
$N(J\cdot,J\cdot)=-N(\cdot,\cdot)$, one can see that it is enough to check the
vanishing of $N(U_{i},Y)$, for any $Y\in\mathfrak{h}$ and $i=1,\dots,k$. We
compute

$\displaystyle N(U_{i},Y)$
$\displaystyle=[U_{i},Y]+J_{\mathfrak{h}}([J_{\mathfrak{a}}U_{i},Y]+[U_{i},J_{\mathfrak{h}}Y])-[J_{\mathfrak{a}}U_{i},J_{\mathfrak{h}}Y]$
(11)
$\displaystyle=(A_{i}+J_{\mathfrak{h}}A_{i}J_{\mathfrak{h}}+J_{\mathfrak{h}}B_{i}-B_{i}J_{\mathfrak{h}})Y=[J_{\mathfrak{h}},A]J_{\mathfrak{h}}+[J_{\mathfrak{h}},B]Y.$

As $A_{i},B_{i}\in\mathfrak{sp}(\mathfrak{h})\cap\mathfrak{so}(\mathfrak{h})$,
we get that
$0=J_{\mathfrak{h}}A_{i}+A_{i}^{t}J_{\mathfrak{h}}=J_{\mathfrak{h}}A_{i}-A_{i}J_{\mathfrak{h}}=[J_{\mathfrak{h}},A_{i}]$
and, analogously, $0=[J_{\mathfrak{h}},B_{i}]$. Hence (11) vanishes, and $J$
is integrable.
Since $\mathfrak{h}$ is a $J$-invariant ideal of $\mathfrak{g}$ of even
codimension, we may extend the formula (1) in this case. Indeed, exploiting
that $\mathfrak{a}$ is abelian, for any
$\alpha\in\bigwedge^{k}\mathfrak{h}^{*}$

$d\alpha=\sum_{i=1}^{k}u_{i}\wedge A_{i}^{*}\alpha+J_{\mathfrak{a}}u_{i}\wedge
B_{i}^{*}\alpha+d_{\mathfrak{h}}\alpha,$ (12)

where $d_{\mathfrak{h}}$ stands for the differential of $\mathfrak{h}$.
Let us consider the fundamental form $\omega$ of
$(J,\langle\cdot,\cdot\rangle)$. Then, by construction,

$\omega=\omega_{\mathfrak{h}}+\sum_{i=1}^{k}u_{i}\wedge
J_{\mathfrak{a}}u_{i}.$

Using (12), we get that
$d\omega=d\omega_{\mathfrak{h}}=\sum_{i=1}^{k}u_{i}\wedge
A_{i}^{*}\omega_{\mathfrak{h}}+J_{\mathfrak{a}}u_{i}\wedge
B_{i}^{*}\omega_{\mathfrak{h}}+d_{\mathfrak{h}}\omega_{\mathfrak{h}}.$ Now, we
know that
$A_{i},B_{i}\in\mathfrak{sp}(\mathfrak{h})\cap\mathfrak{so}(\mathfrak{h})$, so
$\forall Y,Z\in\mathfrak{h}$

$\begin{split}A_{i}^{*}\omega_{\mathfrak{h}}(Y,Z)&=-\omega_{\mathfrak{h}}(A_{i}Y,Z)-\omega_{\mathfrak{h}}(Y,A_{i}Z)=-\langle
J_{\mathfrak{h}}A_{i}Y,Z\rangle-\langle J_{\mathfrak{h}}Y,A_{i}Z\rangle\\\
&=-\langle(J_{\mathfrak{h}}A_{i}+A_{i}^{t}J_{\mathfrak{h}})Y,Z\rangle=0,\end{split}$

and, analogously, $B_{i}^{*}\omega_{\mathfrak{h}}=0$. In particular, we get
$d\omega=d_{\mathfrak{h}}\omega_{\mathfrak{h}}.$ Let
$c=Jd\omega=c_{\mathfrak{h}}$ be the Bismut torsion. Again by (12),
$dc=dc_{\mathfrak{h}}=\sum_{i=1}^{k}u_{i}\wedge
A_{i}^{*}c_{\mathfrak{h}}+J_{\mathfrak{a}}u_{i}\wedge
B_{i}^{*}c_{\mathfrak{h}},$ as $d_{\mathfrak{h}}c_{\mathfrak{h}}$ vanishes
since $(\langle\cdot,\cdot\rangle_{\mathfrak{h}},J_{\mathfrak{h}})$ is SKT. To
conclude, we have to prove that
$A_{i}^{*}c_{\mathfrak{h}}=B_{i}^{*}c_{\mathfrak{h}}=0$. We prove the
statement for the former, since the latter is analogous. Let
$Y,Z,W\in\mathfrak{h}$. Then

$\begin{split}A_{i}^{*}c_{\mathfrak{h}}(Y,Z,W)&=-c_{\mathfrak{h}}(A_{i}Y,Z,W)-c_{\mathfrak{h}}(Y,A_{i}Z,W)-c_{\mathfrak{h}}(Y,Z,A_{i}W)\\\
&=\langle[J_{\mathfrak{h}}A_{i}Y,J_{\mathfrak{h}}Z],W\rangle+\langle[J_{\mathfrak{h}}Z,J_{\mathfrak{h}}W],A_{i}Y\rangle+\langle[J_{\mathfrak{h}}W,J_{\mathfrak{h}}A_{i}Y],Z\rangle\\\
&+\langle[J_{\mathfrak{h}}Y,J_{\mathfrak{h}}A_{i}Z],W\rangle+\langle[J_{\mathfrak{h}}A_{i}Z,J_{\mathfrak{h}}W],Y\rangle+\langle[J_{\mathfrak{h}}W,J_{\mathfrak{h}}Y],A_{i}Z\rangle\\\
&+\langle[J_{\mathfrak{h}}Y,J_{\mathfrak{h}}Z],A_{i}W\rangle+\langle[J_{\mathfrak{h}}Z,J_{\mathfrak{h}}A_{i}W],Y\rangle+\langle[J_{\mathfrak{h}}A_{i}W,J_{\mathfrak{h}}Y],Z\rangle.\end{split}$

Exploiting that $[A_{i},J_{\mathfrak{h}}]=0$ and $A_{i}^{t}=-A_{i}$, one gets

$\begin{split}A_{i}^{*}c_{\mathfrak{h}}(Y,Z,W)&=\langle-
A_{i}[J_{\mathfrak{h}}Y,J_{\mathfrak{h}}Z],W\rangle+\langle[A_{i}J_{\mathfrak{h}}Y,J_{\mathfrak{h}}Z],W\rangle+\langle[J_{\mathfrak{h}}Y,A_{i}J_{\mathfrak{h}}Z],W\rangle\\\
&+\langle-
A_{i}[J_{\mathfrak{h}}Z,J_{\mathfrak{h}}W],Y\rangle+\langle[A_{i}J_{\mathfrak{h}}Z,J_{\mathfrak{h}}W],Y\rangle+\langle[J_{\mathfrak{h}}Z,A_{i}J_{\mathfrak{h}}W],Y\rangle\\\
&+\langle-
A_{i}[J_{\mathfrak{h}}W,J_{\mathfrak{h}}Y],Z\rangle+\langle[A_{i}J_{\mathfrak{h}}W,J_{\mathfrak{h}}Y],Z\rangle+\langle[J_{\mathfrak{h}}W,A_{i}J_{\mathfrak{h}}Y],Z\rangle,\\\
\end{split}$

which vanishes since $A_{i}$ is a derivation of $\mathfrak{h}$. ∎

######  Example 5.3.

Six dimensional (non abelian) nilpotent Lie algebras admitting a SKT structure
are classified in [24] (see also [41, Theorem 3.3]). Let $\mathfrak{h}$ be a
six-dimensional nilpotent Lie algebra defined by the structure equations

$de^{1}=de^{2}=de^{3}=de^{4}=0,\ de^{5}=\rho e^{13}-\rho e^{24}+2\delta
e^{34},\ de^{6}=\rho e^{23}+\rho e^{14}-2e^{12}-2\gamma e^{34},$ (13)

with $\rho\in\\{0,1\\},\ \delta,\gamma\in\mathbb{R}$ and $\rho^{2}-2\gamma=0$.
A SKT structure is given by

$J_{\mathfrak{h}}e_{1}=e_{2},\ J_{\mathfrak{h}}e_{3}=e_{4},\
J_{\mathfrak{h}}e_{5}=e_{6},\
\langle\cdot,\cdot\rangle_{\mathfrak{h}}=\sum_{i=1}^{6}(e^{i})^{2}.$ (14)

Four different non-isomorphic Lie algebras are distinguished

$\begin{split}&\rho=0,\ \gamma=0\implies\
\begin{cases}\mathfrak{h}\cong\mathfrak{h}_{2}=(0,0,0,0,12,34)\ \ \text{for}\
\delta\neq 0,\\\ \mathfrak{h}\cong\mathfrak{h}_{8}=(0,0,0,0,0,12)\ \
\text{for}\ \delta=0,\end{cases}\\\
&\rho=1,\gamma=\frac{1}{2}\implies\begin{cases}\mathfrak{h}\cong\mathfrak{h}_{2}=(0,0,0,0,12,34)\
\ \text{for}\ 4\delta^{2}>3,\\\
\mathfrak{h}\cong\mathfrak{h}_{4}=(0,0,0,0,12,14+23)\ \ \text{for}\
4\delta^{2}=3,\\\ \mathfrak{h}\cong\mathfrak{h}_{5}=(0,0,0,0,13+42,14+23)\ \
\text{for}\ 4\delta^{2}<3.\end{cases}\end{split}$ (15)

By [24, Theorem 3.2] these are the only possible six dimensional nilpotent Lie
algebras admitting a SKT structure, up to isomorphism.
We want to exploit Theorem 5.2 to extend the Lie algebras described above to
$8$-dimensional solvable SKT Lie algebras. Let $\mathfrak{h}$ be a six
dimensional nilpotent Lie algebra with structure equations (13) endowed with
the SKT structure (14). Consider the abelian Lie algebra $\mathbb{R}^{2}$
endowed with the standard Hermitian structure and let $\\{U,U^{\prime}\\}$ be
a unitary basis of $\mathbb{R}^{2}$ satisfying $U^{\prime}=JU$ with dual basis
$\\{u,u^{\prime}\\}$. We define the Lie algebra homomorphism
$\theta:\mathbb{R}^{2}\to Der(\mathfrak{h})$ as

$\theta(U)=\begin{pmatrix}\begin{array}[]{cc|cc|cc}0&a&0&0&0&0\\\
-a&0&0&0&0&0\\\ \hline\cr 0&0&0&-a&0&0\\\ 0&0&a&0&0&0\\\ \hline\cr
0&0&0&0&0&0\\\ 0&0&0&0&0&0\end{array}\end{pmatrix}$

and $\theta(U^{\prime})=0$, with respect to the fixed basis
$\\{e_{1},\dots,e_{6}\\}$ of $\mathfrak{h}$. One can prove that
$\theta(\mathbb{R}^{2})\subset Der(\mathfrak{h})$ by direct computation. Since
$\theta(U)$ (and trivially $\theta(U^{\prime})$) is in
$\mathfrak{sp}(\mathfrak{h})\cap\mathfrak{so}(\mathfrak{h})$ with respect to
the diagonal metric
$\langle\cdot,\cdot\rangle_{\mathfrak{h}}=\sum_{i=1}^{6}(e^{i})^{2}$, Theorem
5.2 applies. Hence, the Lie algebra
$(\mathfrak{g}=\mathfrak{h}\rtimes_{\theta}\mathbb{R}^{2},J,\langle\cdot,\cdot\rangle)$
with $\langle\cdot,\cdot\rangle=\sum_{i=1}^{6}(e^{i})^{2}+u^{2}+u^{\prime 2}$
and $J=\begin{pmatrix}\begin{array}[]{ c|cc}J_{\mathfrak{h}}&&\\\
\hline\cr&0&-1\\\ &1&0\\\ \end{array}\end{pmatrix}$ is SKT.

In particular, the following families of $8$-dimensional decomposable Lie
algebras

$\displaystyle\mathfrak{s}^{a,\delta}_{1}=(af^{27},-af^{17},-af^{37},af^{47},2\delta
f^{34},-2f^{12},0,0),$ (16) $\displaystyle\ \quad\qquad\ \text{with}\
\delta\neq 0,$
$\displaystyle\mathfrak{s}^{a}_{2}=(af^{27},-af^{17},-af^{37},af^{47},0,-2f^{12},0,0),$
$\displaystyle\mathfrak{s}^{a,\delta}_{3}=(af^{27},-af^{17},-af^{37},af^{47},f^{13}-f^{24}+2\delta
f^{34},f^{23}+f^{14}-2f^{12}-f^{34},0,0),$ $\displaystyle\ \quad\qquad\
\text{with}\ 4\delta^{2}>3,$
$\displaystyle\mathfrak{s}^{a}_{4}=(af^{27},-af^{17},-af^{37},af^{47},f^{13}-f^{24}+\sqrt{3}f^{34},f^{23}+f^{14}-2f^{12}-f^{34},0,0),$
$\displaystyle\mathfrak{s}^{a,\delta}_{5}=(af^{27},-af^{17},-af^{37},af^{47},f^{13}-f^{24}+2\delta
f^{34},f^{23}+f^{14}-2f^{12}-f^{34},0,0),$ $\displaystyle\ \quad\qquad\
\text{with}\ 4\delta^{2}<3$

are extensions via the homomorphism $\theta$ of the nilpotent Lie algebras
listed in (15) and admit a SKT structure. More precisely,
$\mathfrak{s}^{a,\delta}_{1},\mathfrak{s}^{a,\delta}_{3}$ are extensions of
$\mathfrak{h}_{2}$ and
$\mathfrak{s}^{a}_{2},\mathfrak{s}^{a}_{4},\mathfrak{s}^{a,\delta}_{5}$ are
extensions of $\mathfrak{h}_{8},\mathfrak{h}_{4},\mathfrak{h}_{5}$,
respectively. Observe that the Lie algebras above are nilpotent for $a=0$ and
almost nilpotent otherwise. Examples such that the nilpotent Lie algebra
$\mathfrak{h}$ coincides with the nilradical of the extension can be
constructed in the same way as above by setting $\rho=\gamma=0$ in (13) and

$\theta(U)=\begin{pmatrix}\begin{array}[]{cc|cc|cc}0&a&0&0&0&0\\\
-a&0&0&0&0&0\\\ \hline\cr 0&0&0&0&0&0\\\ 0&0&0&0&0&0\\\ \hline\cr
0&0&0&0&0&0\\\ 0&0&0&0&0&0\end{array}\end{pmatrix},\ \
\theta(U^{\prime})=\begin{pmatrix}\begin{array}[]{cc|cc|cc}0&0&0&0&0&0\\\
0&0&0&0&0&0\\\ \hline\cr 0&0&0&b&0&0\\\ 0&0&-b&0&0&0\\\ \hline\cr
0&0&0&0&0&0\\\ 0&0&0&0&0&0\end{array}\end{pmatrix}$

with $a,b\neq 0$. Therefore, the following families of $8$-dimensional
indecomposable solvable Lie algebras

$\displaystyle\begin{aligned}
&\mathfrak{s}^{a,b,\delta}_{6}=(af^{27},-af^{17},bf^{38},-bf^{48},2\delta
f^{34},-2f^{12},0,0),\quad a,b,\delta\neq 0,\\\
&\mathfrak{s}^{a,b}_{7}=(af^{27},-af^{17},bf^{38},-bf^{48},0,-2f^{12},0,0),\quad
a,b\neq 0,\end{aligned}$

have nilradical $\mathfrak{h}_{2}$ and $\mathfrak{h}_{8}$ respectively, and
admit a SKT structure $(J,\langle\cdot,\cdot\rangle)$, given by

$\langle\cdot,\cdot\rangle=\sum_{i=1}^{8}(f^{i})^{2},\ Jf_{1}=f_{2},\
Jf_{3}=f_{4},\ Jf_{5}=f_{6},\ Jf_{7}=f_{8}.$

such that $J\mathfrak{h}=\mathfrak{h}$. We can prove now that
$(\mathfrak{s}^{a,b,\delta}_{6},J)$ and $(\mathfrak{s}^{a,b}_{7},J)$ do not
admit any balanced metric. The proof proceeds in the same way for both Lie
algebras. Assume that there exists a balanced $J$-Hermitian metric
$\langle\cdot,\cdot\rangle_{2}$. Since $(\mathfrak{h},J)$ admits an SKT
metric, then its center $\mathfrak{z}(\mathfrak{h})$ is $J$-invariant, and so
we have the decomposition of the Lie algebra as
$\mathfrak{z}(\mathfrak{h})^{\perp_{\langle,\rangle_{2}}}\oplus\mathfrak{z}(\mathfrak{h})\oplus\mathfrak{h}^{\perp_{\langle,\rangle_{2}}}$
with respect to $\langle\cdot,\cdot\rangle_{2}$. We fix an unitary basis
$\\{e_{1},\dots,e_{8}\\}$ for the metric $\langle\cdot,\cdot\rangle_{2}$, such
that $\\{e_{1},\dots,e_{4}\\}$ is a unitary basis of
$\mathfrak{z}(\mathfrak{h})^{\perp_{\langle,\rangle_{2}}}$ satisfying
$Je_{2i-1}=e_{2i}$, $\\{e_{5},e_{6}\\}$ is a unitary basis of
$\mathfrak{z}(\mathfrak{h})$ satisfying $Je_{5}=e_{6}$ and $\\{e_{7},e_{8}\\}$
is a unitary basis of $\mathfrak{h}^{\perp_{\langle,\rangle_{2}}}$ satisfying
$Je_{7}=e_{8}$. Since $(J,\langle\cdot,\cdot\rangle_{2})$ is balanced, we have
that the Lee form $\theta$, defined as

$\theta(X)=\frac{1}{2}\langle\sum_{i=1}^{8}[e_{i},Je_{i}],JX\rangle_{2}=\frac{1}{2}\langle\sum_{i=1}^{4}[e_{i},Je_{i}],JX\rangle_{2}+\langle[e_{7},e_{8}],JX\rangle_{2},$

is identically zero. Observe that since $\mathfrak{h}$ is $2$-step nilpotent,
$\sum_{i=1}^{4}[e_{i},Je_{i}]\subset\mathfrak{z}(\mathfrak{h})$. Let $X$ be
any vector of $\mathfrak{z}(\mathfrak{h})^{\perp_{\langle,\rangle_{2}}}$. Then
$\theta(X)=\langle[e_{7},e_{8}],JX\rangle_{2}=0$, impliying that
$[e_{7},e_{8}]\in\mathfrak{z}(\mathfrak{h})$. In terms of the previous basis
$\\{f_{1},\dots,f_{8}\\}$, we may write $e_{7}=\lambda f_{7}+\mu f_{8}+Y$,
with $Y\in\mathfrak{h}$, and $e_{8}=Je_{7}=\lambda f_{8}-\mu f_{7}+JY$.
Exploiting that $[f_{7},f_{8}]=0$, we get

$[e_{7},e_{8}]=[f_{7},\lambda JY+\mu Y]+[f_{8},\mu JY-\lambda
Y]+[Y,JY]\in\big{(}\imaginary(ad_{f_{7}}|_{\mathfrak{h}})\oplus\imaginary(ad_{f_{8}}|_{\mathfrak{h}})\oplus\mathfrak{z}(\mathfrak{h})\big{)}\cap\mathfrak{z}(\mathfrak{h}).$

Hence, the components of $[e_{7},e_{8}]$ along
$\imaginary(ad_{f_{7}}|_{\mathfrak{h}})$ and
$\imaginary(ad_{f_{8}}|_{\mathfrak{h}})$ must vanish. Moreover, a
straightforward computation shows that this forces $\lambda=\mu=0$, giving a
contradiction.

######  Example 5.4.

Now, we give an example of a $8$-dimensional solvable Lie algebra with
codimension $2$ nilradical $\mathfrak{h}\cong\mathfrak{h}_{8}$, endowed with a
SKT structure $(\langle\cdot,\cdot\rangle,J)$ such that
$J\mathfrak{h}\neq\mathfrak{h}$.
Consider the $8$-dimensional solvable Lie algebra $\mathfrak{s}_{8}$ defined
by the structure equations

$\begin{split}&de^{1}=e^{23},\ de^{2}=e^{27},\
de^{3}=-e^{37},de^{4}=e^{57}+e^{48},\\\ &de^{5}=-e^{47}+e^{58},\
de^{6}=-2e^{68},\ de^{7}=de^{8}=0,\end{split}$

with nilradical $\\{e_{1},\dots,e_{6}\\}\cong\mathfrak{h}_{8}$. We define the
Hermitian structure $(\langle\cdot,\cdot\rangle,J)$ as

$\langle\cdot,\cdot\rangle=\sum_{i=1}^{8}(e^{i})^{2},\ Je_{1}=e_{2},\
Je_{3}=e_{7},\ Je_{4}=e_{5},\ Je_{6}=e_{8}.$

Then, $J\mathfrak{h}\neq\mathfrak{h}$ and the associated fundamental form is

$\omega=e^{1}\wedge e^{2}+e^{3}\wedge e^{7}+e^{4}\wedge e^{5}+e^{6}\wedge
e^{8}.$

Since the Bismut torsion $c=Jd\omega=-e^{123}-2e^{456}$ is closed, the
Hermitian structure $(\langle\cdot,\cdot\rangle,J)$ is SKT. In Theorem 2.1, we
have proved that when the nilradical $\mathfrak{h}$ is $J$-invariant, the
Hermitian structure is Chern Ricci flat. It is no longer true when the
nilradical is not $J$-invariant, in fact, if one compute the Chern connection
of the Lie algebra above, then $\eta^{Ch}=e^{3}+e^{6}+e^{7}$ and
$\rho^{Ch}=-e^{37}-2e^{68}\neq 0$.
Let
$\\{\varphi^{1}:=e^{1}+ie^{2},\varphi^{2}:=e^{3}+ie^{7},\varphi^{3}:=e^{4}+ie^{5},\varphi^{4}:=e^{6}+ie^{8}\\}$.
An easy computation shows that
$d\varphi^{2}=\frac{i}{2}\varphi^{2\overline{2}}$. We use $\varphi^{2}$ to
prove that $(\mathfrak{s}_{8},J)$ does not admit any balanced metric. Assume
by contradiction that there exists a balanced metric with associated
fundamental form $\sigma$. Then $\varphi^{2}\wedge\sigma^{3}\neq 0$. The
differential
$d(\varphi^{2}\wedge\sigma^{3})=\frac{i}{2}\varphi^{2\overline{2}}\wedge\sigma^{3}$
is non zero, and so we have a (non-zero) $2n-1$ form on a unimodular Lie
algebra which is not closed. The contradiction follows.

######  Example 5.5.

Consider the solvable Lie algebra $\mathfrak{g}_{b,c,c^{\prime}}^{2n}$,
$b,c,c^{\prime}\in\mathbb{R}\setminus\\{0\\}$, defined for $n\geq 3$ by the
structure equations

$\begin{cases}&de^{1}=-e^{1}\wedge e^{2n-1},\\\ &de^{2}=\frac{1}{2}e^{2}\wedge
e^{2n-1}+be^{3}\wedge e^{2n-1}-ce^{3}\wedge e^{2n},\\\ &de^{3}=-be^{2}\wedge
e^{2n-1}+\frac{1}{2}e^{3}\wedge e^{2n-1}+ce^{2}\wedge e^{2n},\\\
&de^{2l}=c^{\prime}e^{2l+1}\wedge e^{2n},\ \ de^{2l+1}=-c^{\prime}e^{2l}\wedge
e^{2n}\ \ \text{for $l=2,\dots,n-2$},\\\ &de^{2n-2}=0,\\\ &de^{2n-1}=0,\\\
&de^{2n}=0.\end{cases}$

Observe that for $n=3$,
$\mathfrak{g}_{b,c,c^{\prime}}^{6}\cong\mathfrak{g}_{5.35}^{-2,0}\oplus\mathbb{R}$.
If we denote by $\\{e_{1},\dots,e_{2n}\\}$ the dual basis, the codimension
$2$-abelian nilradical $\mathfrak{h}$ is spanned by
$\\{e_{1},\dots,e_{2n-2}\\}$, and
$\mathfrak{g}_{b,c,c^{\prime}}^{2n}\cong(\mathbb{R}^{2n-3}\rtimes\mathbb{R}^{2})\times\mathbb{R}$,
where $\mathbb{R}^{2n-3},\mathbb{R}^{2}$ and $\mathbb{R}$ are spanned by
$\\{e_{1},\dots,e_{2n-3}\\}$, $\\{e_{2n-1},e_{2n}\\}$ and $e_{2n-2}$,
respectively. Let us define the bi-Hermitian structure
$(I_{\pm},\langle\cdot,\cdot\rangle)$ as

$I_{\pm}e_{1}=e_{2n-1},\ I_{\pm}e_{2}=\pm e_{3},\ I_{\pm}e_{2l}=e_{2l+1}\
\text{for $l=2,\dots,n-2$},\ I_{\pm}e_{2n-2}=e_{2n},\
\langle\cdot,\cdot\rangle={{\sum}}_{i=1}^{2n}(e^{i})^{2}.\ $

It is straightforward to note that $I_{\pm}\mathfrak{h}\neq\mathfrak{h}$. The
corresponding fundamental forms $\omega_{\pm}=e^{1}\wedge e^{2n-1}\pm
e^{2}\wedge e^{3}+e^{2n-2}\wedge e^{2n}+{{\sum}}_{l=2}^{2n-1}e^{2l}\wedge
e^{2l+1}$ satisfy $d^{c}_{\pm}\omega_{\pm}=\pm\ e^{1}\wedge e^{2}\wedge
e^{3},$ and by the structure equations $e^{1}\wedge e^{2}\wedge e^{3}$ is a
closed $3$-form, i.e., $(I_{\pm},\langle\cdot,\cdot\rangle)$ is a generalized
Kähler structure of split-type, i.e. $I_{\pm}$ commute. If $n\geq 5$, then
$\mathfrak{g}^{2n}_{b,c,c^{\prime}}$ admits also a non-split generalized
Kähler structure. Let the bi-Hermitian structure
$(I_{\pm},\langle\cdot,\cdot\rangle)$ be defined as

$\begin{split}&\langle\cdot,\cdot\rangle={{\sum}}_{i=1}^{2n}(e^{i})^{2},\\\
&I_{+}e_{1}=e_{2n-1},\ I_{+}e_{2}=+e_{3},\ I_{+}e_{4}=e_{5},\\\
&I_{+}e_{6}=-e_{7},\ I_{+}e_{2l}=e_{2l+1}\ \text{for $l=4,\dots,n-2$},\
I_{+}e_{2n-2}=e_{2n},\\\ &I_{-}e_{1}=e_{2n-1},\ I_{-}e_{2}=-e_{3},\
I_{-}e_{4}=-e_{7},\ I_{-}e_{5}=-e_{6},\\\ &I_{-}e_{2l}=e_{2l+1}\ \text{for
$l=4,\dots,n-2$},\ I_{-}e_{2n-2}=e_{2n}.\\\ \end{split}$

Since $I_{+}I_{-}e_{4}=-I_{+}e_{7}=-e_{6},\ \ \text{and}\ \
I_{-}I_{+}e_{4}=I_{-}e_{5}=e_{6},$ the commutator $[I_{+},I_{-}]\neq 0$.
Let $\omega_{\pm}$ be the fundamental forms associated to
$(I_{\pm},\langle\cdot,\cdot\rangle)$, which are respectively

$\omega_{+}=e^{1}\wedge e^{2n-1}+e^{2}\wedge e^{3}+e^{4}\wedge
e^{5}-e^{6}\wedge e^{7}+e^{2n-2}\wedge
e^{2n}+{{\sum}}_{l=4}^{2n-1}e^{2l}\wedge e^{2l+1},$

and $\omega_{-}=e^{1}\wedge e^{2n-1}+e^{2}\wedge e^{3}-e^{4}\wedge
e^{7}+e^{5}\wedge e^{6}+e^{2n-2}\wedge
e^{2n}+{{\sum}}_{l=4}^{2n-1}e^{2l}\wedge e^{2l+1}.$ Then, as before,
$d^{c}_{\pm}\omega_{\pm}=\pm\ e^{1}\wedge e^{2}\wedge e^{3}$ and
$d(e^{1}\wedge e^{2}\wedge e^{3})=0$, implying that
$(I_{\pm},\langle\cdot,\cdot\rangle)$ is a generalized Kähler structure of
non-split type.

##  6\. Results on compact solvmanifolds

Using the Symmetrization process ([8, 19, 41]), one can prove that if
$M=\Gamma\backslash G$ is a compact solvmanifold endowed with a pair of
invariant complex structures $J_{\pm}$, then $M$ admits a generalized Kähler
structure $(J_{\pm},g)$, if and only if the Lie algebra $\mathfrak{g}$ of $G$
admits a generalized Kähler inner product
$(J_{\pm},\langle\cdot,\cdot\rangle)$. Hence, we may prove the following
results

######  Theorem 6.1.

Let $M=\Gamma\backslash G$ be a $2n$-dimensional solvmanifold and let
$J_{\pm}$ be invariant complex structures on $M$. Assume that the nilradical
$\mathfrak{h}$ of the Lie algebra $\mathfrak{g}$ of $G$ has codimension $2$.

  1. (i)

If $J_{\pm}\mathfrak{h}=\mathfrak{h}$, then $(M,J_{\pm})$ admits a generalized
Kähler metric if and only if $(M,J_{+})$ admits a Kähler metric.

  2. (ii)

Assume that $\mathfrak{h}$ is abelian and the complex structures $J_{\pm}$
satisfy $J_{+}\mathfrak{h}\neq\mathfrak{h}$ and
$J_{-}\mathfrak{h}=\mathfrak{h}$. If $(M,J_{\pm})$ admits a generalized Kähler
metric, then $(M,J_{-})$ admits also a Kähler metric.

###### Proof.

To prove _(i)_ , we observe that the implication from right to left is
obvious. Let us prove the other implication. By the Symmetrization process,
$M=\Gamma\backslash G$ admits a generalized Kähler structure $(J_{\pm},g)$ if
and only if $\mathfrak{g}$ admits a generalized Kähler structure
$(J_{\pm},\langle\cdot,\cdot\rangle)$. Then the result follows by applying
Theorem 2.4.
To prove _(ii)_ , again by the Symmetrization process, there must exists a
left invariant generalized Kähler structure
$(\langle\cdot,\cdot\rangle,J_{\pm})$ on the Lie algebra $\mathfrak{g}$ with
$J_{\pm}$ satisfying $J_{+}\mathfrak{h}\neq\mathfrak{h}$ and
$J_{-}\mathfrak{h}=\mathfrak{h}$. Then the statement follows by applying
Proposition 3.2. ∎

######  Corollary 6.2.

Let $M=\Gamma\backslash G$ be a $2n$-dimensional solvmanifold and let
$I_{\pm}$ be left invariant complex structures on $M$. Assume that the
nilradical $\mathfrak{h}$ of the Lie algebra $\mathfrak{g}$ of $G$ is abelian
and of codimension $2$. If $(M,I_{\pm})$ admits a generalized Kähler metric,
then $I_{\pm}\mathfrak{h}\neq\mathfrak{h}$.

Now, we exhibit new examples of SKT and generalized Kähler compact
solvmanifolds.

######  Example 6.3.

Let $\mathfrak{g}^{b}_{8}$ be the $8$-dimensional solvable Lie algebra
constructed in Example 5.1. We prove that for $b=2\pi$, the simply conncted
Lie group with Lie algebra $\mathfrak{g}^{2\pi}_{8}$ admits a compact
quotient. Let $G^{2\pi}_{8}$ be the corresponding connected and simply
connected Lie Group with group operation $\ast$ given by

$\tiny\begin{pmatrix}a_{1}\\\ a_{2}\\\ a_{3}\\\ a_{4}\\\ a_{5}\\\ a_{6}\\\
a_{7}\\\ a_{8}\end{pmatrix}\ast\begin{pmatrix}x_{1}\\\ x_{2}\\\ x_{3}\\\
x_{4}\\\ x_{5}\\\ x_{6}\\\ x_{7}\\\ x_{8}\end{pmatrix}=\begin{pmatrix}a_{1}\\\
a_{2}\\\ a_{3}\\\ a_{4}\\\ a_{5}\\\ a_{6}\\\ a_{7}\\\
a_{8}\end{pmatrix}+\begin{pmatrix}\cos(2\pi a_{7})&-\sin(2\pi
a_{7})&0&0&0&0&0&0\\\ \sin(2\pi a_{7})&\cos(2\pi a_{7})&0&0&0&0&0&0\\\
0&0&\cos(2\pi a_{8})&-\sin(2\pi a_{8})&0&0&0&0\\\ 0&0&\sin(2\pi
a_{8})&\cos(2\pi a_{8})&0&0&0&0\\\ 0&0&0&0&1&0&0&-a_{7}\\\
0&0&0&0&0&1&0&-a_{7}\\\ 0&0&0&0&0&0&1&0\\\ 0&0&0&0&0&0&0&1\\\
\end{pmatrix}\cdot\begin{pmatrix}x_{1}\\\ x_{2}\\\ x_{3}\\\ x_{4}\\\ x_{5}\\\
x_{6}\\\ x_{7}\\\ x_{8}\end{pmatrix}.$

If we consider $\Gamma=\\{(a_{1},\dots,a_{8})\in G^{2\pi}_{8}\ |\
a_{i}\in\mathbb{Z}\\}\cong\mathbb{Z}^{8}$, then it is straightforward to note
that $\Gamma$ is a discrete subgroup of $G^{2\pi}_{8}$ of maximal rank. The
corresponding solvmanifold $(\Gamma\backslash G^{2\pi}_{8},J)$ admits a SKT
(non-flat) Chern Ricci flat metric. Note that $(\Gamma\backslash
G^{2\pi}_{8},J)$ does not admit any balanced metric. Indeed, if
$(\Gamma\backslash G^{2\pi}_{8},J)$ admits a balanced metric, then by the
symmetrization process, $(\mathfrak{g}^{2\pi}_{8},J)$ admits a balanced
metric, moreover since the Lie algebra has an abelian ideal of codimension
$2$, $\mathfrak{g}^{2\pi}_{8}$ would also admit a Kähler metric ([13]), giving
a contradiction.

######  Example 6.4.

Let $\mathfrak{s}$ be any of the Lie algebras listed in (16) (Example 5.3).
For $a=0,2\pi$ the corresponding connected and simply connected Lie groups
admit compact quotients.
For $a=0$ the statement is trivial. Indeed, in this case $\mathfrak{s}$ is
isomorphic to $\mathfrak{h}\oplus\mathbb{R}^{2}$. Moreover, since
$\mathfrak{h}$ is isomorphic to either
$\mathfrak{h}_{2},\mathfrak{h}_{4},\mathfrak{h}_{5},\mathfrak{h}_{8}$
depending on the values of $\rho,\gamma,\delta$, we always have that
$\mathfrak{h}$ admits a basis with rational structure constants.
For $a=2\pi$, since
$\mathfrak{s}\cong(\mathfrak{h}\rtimes_{A}\mathbb{R}\langle
X\rangle)\oplus\mathbb{R}$, we may restrict to prove that
$\mathfrak{g}=\mathfrak{h}\rtimes_{A}\mathbb{R}\langle X\rangle$ admits
compact quotients for $a=2\pi$. In particular, $\mathfrak{h}$ is the
nilradical of $\mathfrak{h}\rtimes_{A}\mathbb{R}\langle X\rangle$. Let
$G=H\rtimes_{\mu}\mathbb{R}$ be the corresponding connected and simply
connected almost nilpotent Lie group.
By [10], if there exists $0\neq t_{1}\in\mathbb{R}$ and a rational basis
$\\{Y_{1},\dots,Y_{6}\\}$ of $\mathfrak{h}$ such that the coordinate matrix of
$d_{e}\mu(t_{1})=\text{exp}(t_{1}A)$ in such a basis is integer, then $G$
admits a lattice $\Gamma$ of splitting type, i.e.,
$\Gamma=\Gamma_{H}\rtimes\Gamma_{\mathbb{R}}$. Let $t_{1}=1$, then
$d_{e}\mu(1)=\text{exp}(A)=Id_{6}.$ Hence, $d_{e}\mu(1)$ is the identity of
$\mathfrak{h}$ and its coordinate matrix is integer for any chosen basis of
$\mathfrak{h}$. In particular, this is true for any rational basis
$\\{Y_{1},\dots,Y_{6}\\}$ of $\mathfrak{h}$, which exists by previous
observations. Therefore, the connected and simply connected solvable Lie group
$G\times\mathbb{R}\cong H\rtimes\mathbb{C}$ corresponding to $\mathfrak{s}$
admits a lattice of splitting type
$\Gamma^{\prime}=(\Gamma_{H}\rtimes\Gamma_{\mathbb{R}})\times\mathbb{Z}=\Gamma_{H}\rtimes\Gamma_{\mathbb{C}}$,
where $\Gamma_{\mathbb{C}}=\Gamma_{\mathbb{R}}\times\mathbb{Z}$.
We also claim that the solvmanifold $\Gamma^{\prime}\backslash
H\rtimes\mathbb{C}$ does not admit any balanced metric. Indeed, one can easily
prove that the complex structure $J$ on $\Gamma^{\prime}\backslash
H\rtimes\mathbb{C}$ is of splitting type (see [3] for further details). The
only non trivial check is that the Dolbeault cohomology of
$(\Gamma_{H}\backslash H,J_{\mathfrak{h}})$ is the left invariant one.
However, this follows by [39, Corollary 3.10].
Now, assume by contradiction that $(\Gamma^{\prime}\backslash
H\rtimes\mathbb{C},J)$ admits a balanced metric. Then by [3, Proposition 2.1],
$(\mathfrak{h},J_{\mathfrak{h}})$ admits a balanced metric, and so
$(\mathfrak{h},J_{\mathfrak{h}})$ admits both a SKT and a balanced metric. By
[26], $\mathfrak{h}$ would be abelian, giving a contradiction.

For the Lie algebras constructed in Example 5.5, we have the following

######  Theorem 6.5.

Let $G^{2n}_{b,c,c^{\prime}}$ be the connected and simply connected Lie group
corresponding to the Lie algebra $\mathfrak{g}^{2n}_{b,c,c^{\prime}}$
constructed in Example 5.5. $G^{2n}_{b,c,c^{\prime}}$ admits a lattice
$\Gamma$ for some values of $b,c,c^{\prime}\in\mathbb{R}\setminus\\{0\\}$ and,
denoted by $M$ the corresponding compact solvmanifold, $M$ admits generalized
Kähler structure of split type for $n\geq 3$ and of non-split type for $n\geq
5$. Furthermore, we have that $b_{1}(M)=3$. In particular, $M$ does not admit
any Kähler metric.

###### Proof.

The Lie group $G^{2n}_{b,c,c^{\prime}}$ underlying the Lie algebra
$\mathfrak{g}^{2n}_{b,c,c^{\prime}}$ is the semi-direct product
$\mathbb{R}^{2n-2}\rtimes_{\phi}\mathbb{R}^{2}$. In the following we will
denote by $\alpha=\,^{t}(\alpha_{1},\dots,a_{2n-2})$ and $(a_{2n-1},a_{2n})$
the coordinates on $\mathbb{R}^{2n-2}$ and $\mathbb{R}^{2}$, respectively. The
multiplication $\ast$ is defined as

$\tiny\begin{pmatrix}\alpha\\\ a_{2n-1}\\\ a_{2n}\\\
\end{pmatrix}\ast\begin{pmatrix}\underline{x}\\\ x_{2n-1}\\\ x_{2n}\\\
\end{pmatrix}=\begin{pmatrix}\alpha+\phi(a_{2n-1},a_{2n})\cdot\underline{x}\\\
a_{2n-1}+x_{2n-1}\\\ a_{2n}+x_{2n}\\\ \end{pmatrix}$

where $\phi$ is the diagonal block matrix

$\tiny\phi(a_{2n-1},a_{2n})\cdot\underline{x}=\begin{pmatrix}e^{a_{2n-1}}\cdot
x_{1}\\\
e^{-{\frac{a_{2n-1}}{2}}}R(ca_{2n}-ba_{2n-1})\cdot\begin{pmatrix}x_{2}\\\
x_{3}\\\ \end{pmatrix}\\\ \vdots\\\
R(c^{\prime}a_{2n})\cdot\begin{pmatrix}x_{2l}\\\ x_{2l+1}\\\ \end{pmatrix}\\\
\vdots\\\ x_{2n-2}\end{pmatrix}$

with

$\tiny R(\theta)=\begin{pmatrix}\cos(\theta)&\sin(\theta)\\\
-\sin(\theta)&\cos(\theta)\\\ \end{pmatrix}.$

Consider the $3\crossproduct 3$ square matrix

$\tiny\begin{split}A(a_{2n-1},b)&=\begin{pmatrix}e^{a_{2n-1}}&0\\\
0&e^{-{\frac{a_{2n-1}}{2}}}\cdot
R(-ba_{2n-1})\end{pmatrix}=\exp\left(a_{2n-1}\cdot\begin{pmatrix}1&0&0\\\
0&-\frac{1}{2}&-b\\\ 0&b&-\frac{1}{2}\\\ \end{pmatrix}\right).\end{split}$

By [2] there exist $t_{0},\overline{b}\in\mathbb{R}\setminus\\{0\\}$ such that
$A:=A(t_{0},\overline{b})$ is similar to an integer matrix $\Lambda\in
SL(3,\mathbb{Z})$ via an invertible matrix $P=(p_{ij})$, i.e., there exists an
invertible matrix $P$ such that $AP=P\Lambda$. It is then easy to observe that
for each $n\in\mathbb{Z}$, $A^{n}=P\Lambda^{n}P^{-1}$, where by a
straightforward computation $A^{n}=A(t_{0}n,\overline{b}).$
If we set $b=\overline{b}$, $c=\frac{2\pi}{t_{0}}$ and
$c^{\prime}=\frac{\pi}{2t_{0}}$, we claim that the discrete set
$\Gamma:=P\mathbb{Z}^{3}\times(\mathbb{Z}^{2})^{n-3}\times\mathbb{Z}\times
t_{0}\mathbb{Z}^{2}$ is a lattice of
$G^{2n}_{\overline{b},\frac{2\pi}{t_{0}},\frac{\pi}{2t_{0}}}$. We first
observe that for such values of $b,c,c^{\prime}$

$\tiny\phi(t_{0}n,t_{0}m)\cdot\underline{x}=\begin{pmatrix}A^{n}\cdot\begin{pmatrix}x_{1}\\\
x_{2}\\\ x_{3}\\\ \end{pmatrix}\\\ \vdots\\\ R\left(\frac{\pi}{2}\cdot
m\right)\cdot\begin{pmatrix}x_{2l}\\\ x_{2l+1}\\\ \end{pmatrix}\\\ \vdots\\\
x_{2n-2}\end{pmatrix}$

Then, fixed $\,{}^{t}(Pz,w_{1},\dots,w_{n-3},q,t_{0}n,t_{0}m)$ and
$\,{}^{t}(Pz^{\prime},w^{\prime}_{1},\dots,w^{\prime}_{n-3},q^{\prime},t_{0}n^{\prime},t_{0}m^{\prime})\in\Gamma$

$\tiny\begin{pmatrix}Pz\\\ w_{1}\\\ \vdots\\\ w_{n-3}\\\ q\\\ t_{0}n\\\
t_{0}m\end{pmatrix}\ast\begin{pmatrix}Pz^{\prime}\\\ w^{\prime}_{1}\\\
\vdots\\\ w^{\prime}_{n-3}\\\ q\\\ t_{0}n^{\prime}\\\
t_{0}m^{\prime}\end{pmatrix}^{-1}=\begin{pmatrix}P(z-\Lambda^{(n-n^{\prime})}\cdot
z^{\prime})\\\ w_{1}-R(\frac{\pi}{2}\cdot(m-m^{\prime}))w_{1}^{\prime}\\\
\vdots\\\ w_{n-3}-R(\frac{\pi}{2}\cdot(m-m^{\prime}))w_{n-3}^{\prime}\\\
q-q^{\prime}\\\ t_{0}(n-n^{\prime})\\\ t_{0}(m-m^{\prime})\\\ \end{pmatrix}$

which is in $\Gamma$ as $\Lambda^{r}$ and $R\left(\frac{\pi}{2}\cdot r\right)$
are integer matrix for each $r\in\mathbb{Z}$.
The action induced by $\Gamma$ on
$G^{2n}_{\overline{b},\frac{2\pi}{t_{0}},\frac{\pi}{2t_{0}}}$ is generated by

$\begin{split}&\gamma_{1}(x_{1},x_{2},\dots,x_{2n-2},x_{2n-1},x_{2n})=(x_{1}+p_{11},x_{2}+p_{21},x_{3}+p_{31},x_{4},\dots,x_{2n-1},x_{2n}),\\\
&\gamma_{2}(x_{1},x_{2},\dots,x_{2n-2},x_{2n-1},x_{2n})=(x_{1}+p_{12},x_{2}+p_{22},x_{3}+p_{32},x_{4},\dots,x_{2n-1},x_{2n}),\\\
&\gamma_{3}(x_{1},x_{2},\dots,x_{2n-2},x_{2n-1},x_{2n})=(x_{1}+p_{13},x_{2}+p_{23},x_{3}+p_{33},x_{4},\dots,x_{2n-1},x_{2n}),\\\
&\gamma_{l}(x_{1},x_{2},\dots,x_{2n-2},x_{2n-1},x_{2n})=(x_{1},x_{2},x_{3},\dots,x_{l}+1,\dots,x_{2n-1},x_{2n}),\
\text{$l=4,\dots,2n-2$}\\\
&\gamma_{2n-1}(x_{1},x_{2},\dots,x_{2n-2},x_{2n-1},x_{2n})=(A\cdot(x_{1},x_{2},x_{3}),x_{4},\dots,x_{2n-2},x_{2n-1}+t_{0},x_{2n})\\\
&\gamma_{2n}(x_{1},x_{2},\dots,x_{2n-2},x_{2n-1},x_{2n})=(x_{1},x_{2},x_{3},\dots,\underbrace{x_{2j+1},-x_{2j}}_{\begin{subarray}{l}\text{entries\
$2j$ \ and \ $2j+1$}\\\ \text{for each
$j=2,\dots,n-2$}\end{subarray}},\dots,x_{2n-2},x_{2n-1},x_{2n}+t_{0}).\end{split}$

It is immediate to see that the action is free and properly discontinuous,
which implies that $\Gamma$ is a lattice of
$G^{2n}_{\overline{b},\frac{2\pi}{t_{0}},\frac{\pi}{2t_{0}}}$. We compute now
the commutators $[\gamma_{r},\gamma_{s}]$ of $\Gamma$, for each pair
$r,s\in\\{1,\dots,2n\\}$. The only non-trivial commutators are

$\begin{split}&[\gamma_{2n-1},\gamma_{i}]=\gamma_{1}^{\Lambda_{1i}}\cdot\gamma_{2}^{\Lambda_{2i}}\cdot\gamma_{3}^{\Lambda_{3i}}\cdot\gamma_{i}^{-1},\
i=1,\dots,3,\quad[\gamma_{2n},\gamma_{2l}]=\gamma_{2l}^{-1}\cdot\gamma_{2l+1}^{-1},\
l=2,\dots,n-2,\\\
&[\gamma_{2n},\gamma_{2l+1}]=\gamma_{2l}\cdot\gamma_{2l+1}^{-1},\
l=2,\dots,n-2,\\\ \end{split}$

where $(\Lambda_{ij})$ is the integer matrix described previously. Then, since
$\Gamma$ is $2$-step solvable, it follows that $[\Gamma,\Gamma]$ is a torsion-
free abelian subgroup of $\Gamma$ of rank $2n-3$. Therefore, by

$\rank(\Gamma)=\rank(\Gamma/[\Gamma,\Gamma])+\rank([\Gamma,\Gamma]),$

$\Gamma/[\Gamma,\Gamma]=\mathbb{Z}^{3}$ and, accordingly, $b_{1}(M)=3$. ∎

######  Remark 6.1.

Observe that since the group $\Gamma\backslash
G^{2n}_{\overline{b},\frac{2\pi}{t_{0}},\frac{\pi}{2t_{0}}}$ is not completely
solvable, we cannot apply Hattori’s Theorem [34] to compute the De Rham
cohomology of $M$.

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