# Conformally Einstein Lorentzian Lie groups with Heisenberg symmetry

E. Calviño-Louzao, E. García-Río, I. Gutiérrez-Rodríguez,
R. Vázquez-Lorenzo  ECL: I.E.S Eduardo Pondal, 15706 Santiago de Compostela,
Spain<EMAIL_ADDRESS>EGR: Department of Mathematics, CITMAga,
University of Santiago de Compostela, 15782 Santiago de Compostela, Spain
<EMAIL_ADDRESS>IDGR: Facultad de Ciencias de la Educación y del
Deporte, University of Vigo, 36005 Pontevedra, Spain
<EMAIL_ADDRESS>RVL: I.E.S de Ribadeo Dionisio
Gamallo, 27700 Ribadeo, Spain<EMAIL_ADDRESS>
###### Abstract.

We describe all Lorentzian semi-direct extensions of the Heisenberg group
which are conformally Einstein. As a by side result, Bach-flat left-invariant
Lorentzian metrics on semi-direct extensions of the Heisenberg group are
classified, thus providing new background solutions in conformal gravity.

###### Key words and phrases:

Conformal Gravity, Bach tensor, conformally Einstein, Heisenberg group, semi-
direct extension

Supported by projects PID2019-105138GB-C21 and PID2021-127075NA-I00
(AEI/FEDER, Spain) and ED431C 2019/10, ED431F 2020/04 (Xunta de Galicia,
Spain).

##  1\. Introduction

Despite the great success of General Relativity, there are still some open
issues regarding big distance scales which suggest the necessity of
modifications in the theory of gravity itself. Conformal Geometry is one of
the different approaches to modify gravity, which belongs to the category of
higher-order derivatives in the action.

The equation of motion for the metric in General Relativity is derived by a
functional variation, $I=I_{EH}+I_{\wedge}+I_{m}$, of the Einstein-Hilbert
action:

$I:g\mapsto I(g)=-\frac{1}{16\pi G}\int d^{4}x\sqrt{-g}(\tau-2\wedge)+I_{m}$

where $\tau$ is the scalar curvature, $I_{m}$ is the matter part and
$I_{\wedge}$ denotes the action for the cosmological constant $\wedge$. The
corresponding equations of motion are given by $(\rho-\frac{1}{2}\tau
g)-\wedge g=-8\pi GT$, where $T$ is the energy-momentum tensor, and $\rho$ is
the Ricci tensor. It was originally expected that a modification of gravity
should reproduce these equations, but this is not necessarily the case.

Conformal gravity refers to gravity theories that are invariant under
conformal transformations in the pseudo-Riemannian geometrical sense. The
simplest theory in this category has the square of the norm of the Weyl tensor
as the Lagrangian [27, 28]

$I:g\mapsto I(g)=\int d^{4}x\sqrt{-g}\,\|W\|^{2}+I_{m},$

which has to be contrasted with the usual Einstein-Hilbert action, where the
Lagrangian is just the scalar curvature. The equation of motion upon varying
the metric is $\mathfrak{B}=0$, which is of fourth-order, where $\mathfrak{B}$
is the Bach tensor.

A four-dimensional pseudo-Riemannian manifold $(M^{n},g)$ is said to be _Bach-
flat_ if the Bach tensor

(1)
$\mathfrak{B}=\operatorname{div}_{2}\operatorname{div}_{4}W+\tfrac{1}{2}W[\rho]$

vanishes identically, where $W$ denotes the Weyl conformal curvature tensor
and $W[\rho]$ is the contraction of the Weyl tensor and the Ricci tensor
$W[\rho]_{ij}=W_{iajb}\rho^{ab}$. Locally conformally flat metrics have $W=0$
and thus they are trivially Bach-flat. Einstein metrics are also Bach-flat
since the Weyl tensor is traceless and $\operatorname{div}W=0$ in the Einstein
case. The Bach tensor is the gradient of the quadratic curvature functional
given by the $L^{2}$-norm of the Weyl tensor, which is preserved by conformal
transformations in dimension four [2]. Hence four-dimensional conformally
Einstein metrics are also Bach-flat.

A pseudo-Riemannian manifold $(M^{n},g)$ is _conformally Einstein_ if there is
an Einstein representative of the conformal class $[g]$. Equivalently, a
conformally related metric $\overline{g}=\varphi^{-2}g$ is Einstein if and
only if there exists a nowhere zero solution of the overdetermined PDE

(2)
$(n-2)\operatorname{Hes}_{\varphi}+\varphi\rho=\tfrac{1}{n}\\{(n-2)\Delta\varphi+\varphi\tau\\}g,$

where $\operatorname{Hes}_{\varphi}$ is the Hessian tensor of $\varphi$ and
$\Delta\varphi=\operatorname{tr}_{g}\operatorname{Hes}_{\varphi}$ is the
Laplacian of the (locally defined) conformal factor $\varphi$. Equation (2),
which is trivial in dimension two, was originally considered by Brinkmann [5].
Despite its apparent simplicity, the integration of (2) is surprisingly
difficult.

If a conformal metric $\overline{g}=e^{-2\sigma}g$ is Einstein, then
$\overline{\operatorname{div}}\,\overline{W}=0$, and thus
$\operatorname{div}_{4}W-W(\,\cdot\,,\,\cdot\,,\,\cdot\,,\nabla\sigma)=0$. A
pseudo-Riemannian manifold is _conformally Cotton-flat_ if there is a
conformal metric $\overline{g}=e^{-2\sigma}g$ which is Cotton-flat.
Equivalently, there is a function $\sigma$ so that

(3) $\operatorname{div}_{4}W-W(\,\cdot\,,\,\cdot\,,\,\cdot\,,\nabla\sigma)=0.$

More generally one says that $(M,g)$ is a _conformal $C$-space_ if there is a
(not necessarily gradient) vector field $X$ so that
$\operatorname{div}_{4}W-W(\,\cdot\,,\,\cdot\,,\,\cdot\,,X)=0$ (see [20]). The
special significance of the conformally Cotton-flat property was given in
[22], where it is shown that a weakly-generic Bach-flat manifold is
conformally Einstein if and only if it is conformally Cotton-flat, where being
weakly generic means that the Weyl tensor, viewed as a map
$W:TM\rightarrow\otimes^{3}TM$, is injective.

Spacetimes admitting a null parallel vector field $\ell$ have been widely
studied in General Relativity, where they are called _$pp$ -waves_ in the
transversally flat case, i.e., if the curvature endomorphism satisfies
$R(\ell^{\perp},\ell^{\perp})=0$ [18], where $R$ denotes the curvature tensor.
Furthermore, the spacetime is a _plane wave_ if, in addition, the curvature
tensor is transversally parallel (i.e., $\nabla_{\ell^{\perp}}R=0$). Four-
dimensional $pp$-waves were discovered in a mathematical context by Brinkmann
[6]. In Physics, plane waves and $pp$-waves appeared in General Relativity,
where they play an important role (see [4, 18]). It was shown in [8] that
four-dimensional $pp$-waves are conformally Einstein if and only if the Cotton
tensor vanishes, or equivalently, $\operatorname{div}W=0$. (See also [26] for
a description of Bach-flat $pp$-waves). Consequently plane waves are
conformally Einstein.

Connected and simply connected four-dimensional Lie groups are either products
$SU(2)\times\mathbb{R}$, $\widetilde{SL}(2,\mathbb{R})\times\mathbb{R}$, or
one of the solvable semi-direct extensions of three-dimensional unimodular Lie
groups $\widetilde{E}(2)\rtimes\mathbb{R}$, $E(1,1)\rtimes\mathbb{R}$,
$H_{3}\rtimes\mathbb{R}$ or $\mathbb{R}^{3}\rtimes\mathbb{R}$, where
$\widetilde{E}(2)$, $E(1,1)$, $H_{3}$ and $\mathbb{R}^{3}$ denote the simply
connected Euclidean, Poincaré, Heisenberg and Abelian three-dimensional Lie
groups, respectively. Bach-flat left-invariant Riemannian metrics on four-
dimensional Lie groups constitute a small class (see [1, 13]). The situation
is very different in the Lorentzian setting where there are plenty of Bach-
flat left-invariant metrics. A general classification of these metrics seems
quite an unfeasible task at this time. That is why in this paper we focus on a
special family of Lie groups, the semi-direct extensions
$H_{3}\rtimes\mathbb{R}$ of the Heisenberg group. Hence we restrict to four-
dimensional simply connected Lie groups which have the Heisenberg group as a
normal subgroup acting with cohomogeneity one. Left-invariant Einstein metrics
on $H_{3}\rtimes\mathbb{R}$ were described in [11], where it is shown that
they are Ricci-flat plane waves, or of constant non-positive sectional
curvature. We analyze all left-invariant Lorentzian metrics and obtain
classification results for the Bach-flat and conformally Einstein ones. Since
the Bach tensor and the conformally Einstein property are invariant by
homotheties, we work at the homothetical level to simplify the discussion. It
is important to emphasize that homotheties need not to be given by
isomorphisms of the Lie groups. Hence in order to preserve the Lie group
structure we work up to isomorphic homotheties.

###  1.1. Summary of results

Indecomposable Lorentzian symmetric spaces are irreducible (and hence of
constant sectional curvature) or locally isometric to a Cahen-Wallach
symmetric space [9]. Four-dimensional Cahen-Wallach symmetric spaces are
isometric to $\mathbb{R}^{4}$ with coordinates $(u,v,x^{1},x^{2})$ and metric
given by $g=dudv+H(v,x^{1},x^{2})dvdv+dx^{1}dx^{1}+dx^{2}dx^{2}$, with
$H(v,x^{1},x^{2})=\sum\lambda_{i}(x^{i})^{2}$. Being plane waves, Cahen-
Wallach symmetric spaces are conformally Einstein and thus Bach-flat.
Moreover, products of lower dimensional symmetric spaces are Bach-flat (indeed
conformally Einstein) unless they are products of surfaces $N^{2}(c_{1})\times
N^{2}(c_{2})$ with constant sectional curvature $c_{1}^{2}\neq c_{2}^{2}$.
Henceforth we consider Lorentzian Lie groups $H_{3}\rtimes\mathbb{R}$ which
are not locally symmetric.

As usual, the Lie algebra $\mathfrak{h}_{3}$ of the Heisenberg group $H_{3}$
is described by a basis $\\{v_{1},v_{2},v_{3}\\}$ with Lie bracket
$[v_{1},v_{2}]=v_{3}$. One-dimensional semi-direct extensions
$\mathfrak{g}=\mathfrak{h}_{3}\rtimes\mathfrak{r}$ are determined by
derivations of the Heisenberg algebra $\mathfrak{h}_{3}$, that becomes an
ideal in $\mathfrak{h}_{3}\rtimes\mathbb{R}$. Furthermore if
$\langle\cdot,\cdot\rangle$ is a Lorentzian inner product on $\mathfrak{g}$,
its restriction to $\mathfrak{h}_{3}$ may be degenerate, of Riemannian
signature, or of Lorentzian signature. This motivates a separate study of the
three distinct, although not necessarily disjoint, situations above.

######  Remark 1.1.

A special case of the analysis in this paper correspond to the _oscillator
algebra_ and the corresponding _oscillator group_ (see, for example, [30,
32]), and some particular metrics on semi-direct extensions
$H_{3}\rtimes\mathbb{R}$ (see [10]).

Another special situation is that of direct extensions $H_{3}\times\mathbb{R}$
so that the semi-direct extension reduces to the product Lie group. Rahmani
showed in [31] that there exist three non-homothetic classes of left-invariant
Lorentzian metrics in the Heisenberg group $H_{3}$. Kondo and Tamaru have
recently shown in [21] that there exist exactly six non-homothetic classes of
left-invariant Lorentzian metrics on $H_{3}\times\mathbb{R}$ up to
automorphisms, which are described by the Lie algebra structures

$[e_{1},e_{2}]=-(\alpha e_{1}-e_{4}),\quad[e_{2},e_{3}]=\beta(\alpha
e_{1}-e_{4}),\quad[e_{2},e_{4}]=\alpha(\alpha e_{1}-e_{4}),$

where $\\{e_{1},e_{2},e_{3},e_{4}\\}$ is an orthonormal basis of
$\mathfrak{h}_{3}\times\mathbb{R}$ with $e_{4}$ timelike, and the parameters
$(\alpha,\beta)\in\\{(0,0),(1,0),(1,1),(2,0),(2,\sqrt{3}),(2,2)\\}$.

A direct calculation shows that $(\alpha,\beta)=(1,0)$ determines a flat
metric. In the other cases the metric is Bach-flat if and only if
$(\alpha,\beta)=(1,1)$, in which case it is locally conformally flat and
locally symmetric, or $(\alpha,\beta)=(2,\sqrt{3})$, in which case the Ricci
operator and the Weyl curvature operator acting on the space of two-forms are
two-step nilpotent. Furthermore, in the latter case, the Ricci tensor is
parallel but the metric is not locally symmetric although the curvature tensor
is modeled on a symmetric space. This metric corresponds to a plane wave whose
derived algebra $[\mathfrak{g},\mathfrak{g}]$ is spacelike and the restriction
of the metric to the center $\mathfrak{z}(\mathfrak{g})$ is of signature
$(+,-,0)=(1,0,1)$. Bach-flat $pp$-wave left-invariant metrics on non-product
semi-direct extensions $H_{3}\rtimes\mathbb{R}$ are discussed in $\S$1.1.3.

####  1.1.1. Non-trivial conformally Einstein metrics

A conformally Einstein metric is said to be _non-trivial_ if it is neither
Einstein, nor locally conformally flat, nor a plane wave. The main result of
this paper describes all non-trivial conformally Einstein semi-direct
Lorentzian extensions of the Heisenberg group as follows.

######  Theorem 1.2.

Let $(H_{3}\rtimes\mathbb{R},\langle\cdot,\cdot\rangle)$ be a non-symmetric
semi-direct extension of the Heisenberg group equipped with a left-invariant
Lorentzian metric. Then, the metric is non-trivial conformally Einstein if and
only if it is isomorphically homothetic to one of the following:

  1. (D)

The restriction of the metric to $\mathfrak{h}_{3}$ is degenerate and the
metric is determined by

    1. (D.i)

$[u_{1},u_{4}]=u_{1}$, $[u_{2},u_{3}]=u_{1}$, $[u_{2},u_{4}]=u_{2}$.

    2. (D.ii)

$[u_{1},u_{3}]=u_{1}$, $[u_{1},u_{4}]=\varepsilon u_{1}$,
$[u_{2},u_{3}]=\alpha u_{1}$, $[u_{2},u_{4}]=\varepsilon u_{2}$, with
$\varepsilon^{2}=1$ and $\alpha\geq 0$.

Here $\\{u_{1},u_{2},u_{3},u_{4}\\}$ denotes a pseudo-orthonormal basis of
$\mathfrak{h}_{3}\rtimes\mathfrak{r}$ with $\langle u_{1},u_{1}\rangle=\langle
u_{2},u_{2}\rangle=\langle u_{3},u_{4}\rangle=1$.

  2. (R)

The restriction of the metric to $\mathfrak{h}_{3}$ is Riemannian and the
metric is determined by

$[e_{1},e_{2}]=e_{3},\quad[e_{1},e_{4}]=\alpha
e_{1}+e_{3},\quad[e_{3},e_{4}]=\alpha e_{3},\quad\text{with}\quad\alpha>0,$

where $\\{e_{1},e_{2},e_{3},e_{4}\\}$ denotes an orthonormal basis of
$\mathfrak{h}_{3}\rtimes\mathfrak{r}$ with $e_{4}$ timelike.

  3. (L)

The restriction of the metric to $\mathfrak{h}_{3}$ is Lorentzian and the
metric is determined by

    1. (L.i)

$[e_{1},e_{3}]=-\alpha e_{2}$, $[e_{1},e_{4}]=e_{1}+\alpha e_{2}$,
$[e_{2},e_{4}]=e_{2}$, where $\alpha>0$ and $\\{e_{1},e_{2},e_{3},e_{4}\\}$
denotes an orthonormal basis of $\mathfrak{h}_{3}\rtimes\mathfrak{r}$ with
$e_{3}$ timelike.

    2. (L.ii)

$[u_{1},u_{3}]=-u_{2}$, $[u_{1},u_{4}]=4u_{1}+\alpha u_{2}$,
$[u_{2},u_{4}]=4u_{2}$, where $\alpha\in\mathbb{R}$ and
$\\{u_{1},u_{2},u_{3},u_{4}\\}$ denotes a pseudo-orthonormal basis of
$\mathfrak{h}_{3}\rtimes\mathfrak{r}$ with $\langle u_{1},u_{2}\rangle=\langle
u_{3},u_{3}\rangle=\langle u_{4},u_{4}\rangle=1$.

    3. (L.iii)

$[u_{1},u_{3}]=-u_{2}$, $[u_{1},u_{4}]=-u_{1}$, $[u_{2},u_{4}]=3u_{2}$,
$[u_{3},u_{4}]=4u_{3},$ where $\\{u_{1},u_{2},u_{3},u_{4}\\}$ denotes a
pseudo-orthonormal basis of $\mathfrak{h}_{3}\rtimes\mathfrak{r}$, with
$\langle u_{1},u_{2}\rangle=\langle u_{3},u_{3}\rangle=\langle
u_{4},u_{4}\rangle=1$.

Moreover, in all cases but (L.iii) the left-invariant metrics are conformally
equivalent to a Ricci-flat $pp$-wave.

Considering the eigenvalue structure of the Ricci operator for the different
cases in Theorem 1.2 given in Remark 1.3, Remark 1.4, and Remark 1.5, one has
that all of them correspond to different homothetical classes except possibly
(D.ii) and (R) for $\varepsilon=-1$ (resp., (D.ii) and (L.i) for
$\varepsilon=1$). Moreover, a direct calculation of $W[\rho]$ and the
eigenvalues of the associated $(1,1)$-tensor field shows that the above
mentioned cases cannot be homothetic. Alternatively, the non-existence of
homotheties between the classes above also follows from the work of [15], just
considering the corresponding orthonormal bases of eigenvectors of the Ricci
operators.

######  Remark 1.3.

The Ricci operator of the left-invariant metric in Theorem 1.2-(D.i) is
diagonalizable with eigenvalues $\\{1$, $1$, $-1$, $-1\\}$, so that the scalar
curvature $\tau=0$. The Weyl curvature operator
$W:\Lambda^{2}\rightarrow\Lambda^{2}$ is two-step nilpotent and hence not
weakly generic.

The space of scalar quadratic curvature invariants of a pseudo-Riemannian
manifold is generated by $\\{\tau^{2},\|\rho\|^{2},\|R\|^{2},\Delta\tau\\}$.
Hence any quadratic curvature functional is a linear combination of the
corresponding $L^{2}$-norms: $g\mapsto\Phi_{abc}(g)=\int
dx^{4}\sqrt{-g}\,\,\\{a\|\rho\|^{2}+b\|R\|^{2}+c\tau^{2}\\}$. Due to the four-
dimensional Gauss-Bonnet Theorem, any quadratic curvature functional in
dimension four is equivalent to one of

$\mathcal{S}:g\mapsto\mathcal{S}(g)=\int
dx^{4}\sqrt{-g}\,\tau^{2}\quad\text{or}\quad\mathcal{F}_{t}:g\mapsto\mathcal{F}_{t}(g)=\int
dx^{4}\sqrt{-g}\,\left\\{\|\rho\|^{2}+t\tau^{2}\right\\}\,.$

In particular, one has that the functional given by the $L^{2}$-norm of the
Weyl conformal curvature tensor is equivalent to $\mathcal{F}_{-1/3}$.
Einstein metrics are critical for all quadratic curvature functionals in
dimension three and four (but not necessarily in higher dimensions). Moreover,
a four-dimensional metric is critical for all quadratic curvature functionals
if and only if it is critical for two-distinct quadratic curvature
functionals. In particular, any Bach-flat metric with vanishing scalar
curvature is critical for all quadratic curvature functionals. Since plane
waves are Bach-flat and have two-step nilpotent Ricci operator, they are
critical for all quadratic curvature functionals. The metric in Theorem
1.2-(D.i) has vanishing scalar curvature. Hence it is critical for all
quadratic curvature functionals although it is not a plane wave nor locally
symmetric (even not modeled on a symmetric space).

The Ricci operator of metrics in Theorem 1.2-(D.ii) has four different real
eigenvalues $\\{\varepsilon(-1\pm\sqrt{\alpha^{2}+2})$,
$\varepsilon(-2\pm\sqrt{\alpha^{2}+1})\\}$ so that the scalar curvature
$\tau=-6\varepsilon$. Moreover, fixing $\varepsilon$, $\alpha\geq 0$
determines the homothetic class, since the homothetic invariant
$\tau^{-2}\|\rho\|^{2}=\frac{1}{9}(\alpha^{2}+4)$. Metrics are not weakly
generic since the Weyl curvature operator acting on the space of two-forms is
two-step nilpotent.

######  Remark 1.4.

The Ricci operator of left-invariant metrics in Theorem 1.2-(R) has four
different real eigenvalues $\\{\alpha(2\alpha\pm 1)$,
$\alpha(\alpha\pm\sqrt{\alpha^{2}+1})\\}$, so that the scalar curvature
$\tau=6\alpha^{2}$. Moreover, $\alpha>0$ determines different homothetic
classes since the homothetic invariant
$\tau^{-2}\|\rho\|^{2}=\frac{1}{9}(\frac{1}{\alpha^{2}}+3)$. Metrics are not
weakly generic since the Weyl curvature operator acting on the space of two-
forms is two-step nilpotent.

######  Remark 1.5.

The Ricci operator of left-invariant metrics in Theorem 1.2-(L.i) has four
different real eigenvalues $\\{-2\pm\alpha$, $-1\pm\sqrt{\alpha^{2}+1}\\}$, so
that the scalar curvature $\tau=-6$. The metric is not weakly generic since
the Weyl curvature operator is two-step nilpotent. Moreover, the parameter
$\alpha>0$ determines the homothetic classes since
$\tau^{-2}\|\rho\|^{2}=\frac{1}{9}(\alpha^{2}+3)$.

Metrics in Theorem 1.2-(L.ii) have Ricci operator with eigenvalues $\\{0$,
$-32$, $-32$, $-32\\}$, so that the scalar curvature $\tau=-96$. The Ricci
operator is diagonalizable if $\alpha=0$, having a double root of the minimal
polynomial otherwise. Moreover, the metrics are not weakly generic since the
Weyl curvature operator acting on the space of two-forms is two-step
nilpotent.

The Ricci operator of the left-invariant metric in Theorem 1.2-(L.iii) is
diagonalizable, $\operatorname{Ric}=\operatorname{diag}[-6$, $-6$, $-24$,
$-18]$ and therefore $\tau=-54$. The Weyl curvature operator has eigenvalues
$\\{-4,2,2,-4,2,2\\}$, where the eigenvalue $2$ is a double root of the
minimal polynomial. Hence, it is weakly generic.

######  Remark 1.6.

The two-loop renormalization group flow (RG2 flow for short) is a perturbation
of the Ricci flow $\partial_{t}g_{t}=-2\rho[g_{t}]$, which mathematically is
described by $\partial_{t}g_{t}=-2RG[g_{t}]$. The symmetric $(0,2)$-tensor
field $RG=\rho+\frac{\Upsilon}{4}\check{R}$, where $\Upsilon$ denotes a
positive coupling constant and $\check{R}$ is the symmetric $(0,2)$-tensor
field given by $\check{R}_{ij}=R_{iabc}R_{j}{}^{abc}$. We refer to [14, 19]
and references therein for more information on the RG2 flow. Genuine fixed
points of the flow are provided by those manifolds where the tensor field $RG$
vanishes, i.e., $\rho+\frac{\Upsilon}{4}\check{R}=0$. Given a one-parameter
family $\psi_{t}$ of diffeomorphisms of $M$ (with
$\psi_{0}=\operatorname{Id}$), a solution of the form
$g(t)=\sigma(t)\psi_{t}^{*}g$ (where $\sigma$ is a real-valued function with
$\sigma(0)=1$) is said to be a _self-similar solution_. A triple $(M,g,X)$,
where $X$ is a vector field on $M$, is called an RG2 _soliton_ if
$\mathcal{L}_{X}g+RG=\lambda g$ for some $\lambda\in\mathbb{R}$. Further the
soliton is said to be _expanding_ , _steady_ or _shrinking_ if $\lambda<0$,
$\lambda=0$, or $\lambda>0$, respectively. Any self-similar solution of the
RG2 flow is an RG2 soliton just considering the vector field $X$ generated by
the one-parameter group of diffeomorphisms $\psi_{t}$. Since the two terms
comprising $RG$ behave differently under homotheties ($\rho[\kappa g]=\rho[g]$
and $\check{R}[\kappa g]=\frac{1}{\kappa}\check{R}[g]$), one has that the
converse holds only for steady solitons, in which case $\psi_{t}$ is the one-
parameter group of diffeomorphisms associated to the vector field $X$
determined by the soliton equation $\mathcal{L}_{X}g+RG=0$ and
$g(t)=\psi_{t}^{*}g$ is a self-similar solution (see [33]).

Let $G$ be a Lie group with left-invariant metric $\langle\cdot,\cdot\rangle$
and let $(\mathfrak{g},\langle\cdot,\cdot\rangle)$ denote the corresponding
Lie algebra. An RG2 _algebraic soliton_ is a derivation of the Lie algebra
$\mathfrak{g}$ given by $\mathfrak{D}=\widehat{RG}-\lambda\operatorname{Id}$,
where $\widehat{RG}$ is the $(1,1)$-tensor field metrically equivalent to $RG$
and $\lambda\in\mathbb{R}$. RG2 algebraic solitons give rise to RG2 solitons
(where the vector field $X$ is associated to a one-parameter group of
automorphisms of $G$ determined by the derivation $\mathfrak{D}$) as in the
Ricci flow case (see [24, 33]).

Now, a straightforward calculation shows that left-invariant metrics in
Theorem 1.2-(L.ii) are steady RG2 algebraic solitons with
$RG2=\rho+\frac{1}{32}\check{R}$. Therefore they are steady RG2 solitons and
thus also self-similar solutions of the flow with two-step nilpotent tensor
field $\widehat{RG}$. Moreover, they are not algebraic Ricci solitons.

In contrast with the previous situation, the left-invariant metric in Theorem
1.2-(L.iii) is a shrinking RG2 algebraic soliton with
$RG2=\rho+\frac{1}{10}\check{R}$. Therefore they are shrinking RG2 solitons
with diagonal tensor field $\widehat{RG}$, but not algebraic Ricci solitons.

####  1.1.2. Strictly Bach-flat metrics

We say that a Bach-flat metric is _strict_ if it is not conformally Einstein
(and thus not a plane wave). In contrast with Theorem 1.2 and the case of
$pp$-waves in Theorem 1.9, strictly Bach-flat semi-direct extensions
$H_{3}\rtimes\mathbb{R}$ are quite rare and they are described in the
following

######  Theorem 1.7.

Let $(H_{3}\rtimes\mathbb{R},\langle\cdot,\cdot\rangle)$ be a non-symmetric
semi-direct extension of the Heisenberg group equipped with a left-invariant
Lorentzian metric. Then, the metric is strictly Bach-flat if and only if it is
isomorphically homothetic to one of the following Lie algebras:

  1. (i)

$[v_{1},v_{2}]=-3\sqrt{14}v_{2}+2\varepsilon\sqrt{11}v_{3}$,
$[v_{1},v_{3}]=\sqrt{14}v_{3}$, $[v_{1},v_{4}]=-2\sqrt{14}v_{4}$,
$[v_{2},v_{3}]=4v_{4}$, where $\varepsilon^{2}=1$ and
$\\{v_{1},v_{2},v_{3},v_{4}\\}$ denotes an orthonormal basis of
$\mathfrak{h}_{3}\rtimes\mathfrak{r}$ with $v_{1}$ and $v_{4}$ spacelike and
$\langle v_{2},v_{2}\rangle=-\langle v_{3},v_{3}\rangle=\bar{\varepsilon}=\pm
1$.

  2. (ii)

$[u_{1},u_{3}]=-\tfrac{1}{\sqrt{-(\alpha+\beta^{2})}}u_{2}$,
$[u_{1},u_{4}]=\tfrac{1}{\sqrt{-(\alpha+\beta^{2})}}(\beta u_{1}+\gamma
u_{2}+u_{3})$,

$[u_{3},u_{4}]=\tfrac{1}{\sqrt{-(\alpha+\beta^{2})}}\left(\alpha
u_{1}-\frac{\alpha+3\beta^{2}}{4\alpha}u_{2}-\beta u_{3}\right)$, where the
constants $\alpha$, $\beta$, $\gamma$ satisfy
$\gamma=\tfrac{1}{4\alpha^{2}}\left(\\!(3\alpha+\beta^{2})\beta\pm
2\sqrt{-4\alpha^{4}-2(\alpha+\beta^{2})^{3}}\right)$, with
$-\frac{1}{2}\\!\leq\\!\alpha\\!<\\!0$ and
$\beta^{2}\leq-\alpha\left(\sqrt[3]{2\alpha}+1\right)$. Moreover
$\\{u_{1},u_{2},u_{3},u_{4}\\}$ denotes a pseudo-orthonormal basis of
$\mathfrak{h}_{3}\rtimes\mathfrak{r}$ with $\langle u_{1},u_{2}\rangle=\langle
u_{3},u_{3}\rangle=\langle u_{4},u_{4}\rangle=1$.

######  Remark 1.8.

The Ricci operator of left-invariant metrics in Theorem 1.7-(i) is
diagonalizable, $\operatorname{Ric}=\operatorname{diag}[-174$, $-138$, $42$,
$-120]$, so $\tau=-390$. Moreover, $\bar{\varepsilon}$ determines the
spacelike or timelike character of the eigenvectors associated to the
eigenvalues $-138$ and $42$. Hence the parameter $\bar{\varepsilon}$
determines two non-homothetic classes. Furthermore the Weyl curvature operator
$W:\Lambda^{2}\rightarrow\Lambda^{2}$ has six-distinct non-zero complex
eigenvalues and thus it is weakly generic.

The characteristic polynomial of the Ricci operator corresponding to metrics
in Theorem 1.7-(ii) is given by

$\operatorname{det}(\operatorname{Ric}-\lambda\operatorname{Id})=\lambda^{4}-\tfrac{3}{4}\lambda^{3}+\tfrac{3}{16}\lambda^{2}-\tfrac{9}{64}\lambda+\tfrac{\alpha^{4}}{8(\alpha+\beta^{2})^{3}}.$

A straightforward calculation shows that the discriminant, given by

$\begin{array}[]{l}\Delta=\tfrac{1}{262144(\alpha+\beta^{2})^{9}}\left(131072\alpha^{12}-74160\alpha^{8}(\alpha+\beta^{2})^{3}\right.\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\Delta=\tfrac{1}{262144(\alpha+\beta^{2})^{9}}\left(\right.}\left.+12312\alpha^{4}(\alpha+\beta^{2})^{6}-2187(\alpha+\beta^{2})^{9}\right),\end{array}$

is strictly negative considering the restrictions on $\alpha,\beta$ in Theorem
1.7. As a consequence, the Ricci operator has two real roots (with opposite
sign) and two complex conjugate roots. Moreover, the scalar curvature is
strictly positive, $\tau=\frac{3}{4}$. The Weyl curvature operator acting on
the space of two-forms is three-step nilpotent.

####  1.1.3. Bach-flat $pp$-wave metrics

It was shown in Remark 1.1 that all Bach-flat left-invariant metrics on the
product $H_{3}\times\mathbb{R}$ are plane waves. Bach-flat left-invariant
metrics on non-product semi-direct extensions $H_{3}\rtimes\mathbb{R}$ which
are $pp$-waves are now given as follows. It turns out that they are plane
waves. We refer to [3] for a classification of homogeneous plane waves.

######  Theorem 1.9.

Let $(H_{3}\rtimes\mathbb{R},\langle\cdot,\cdot\rangle)$ be a non-product
semi-direct extension of the Heisenberg group equipped with a left-invariant
Lorentzian metric. If $(H_{3}\rtimes\mathbb{R},\langle\cdot,\cdot\rangle)$ is
a non-symmetric Bach-flat $pp$-wave which is not locally conformally flat,
then it is isomorphically homothetic to one of the following:

  1. (D.i)

The left-invariant metric determined by

$\begin{array}[]{ll}[u_{1},u_{2}]=u_{3},&[u_{1},u_{4}]=\kappa_{1}u_{1}+\kappa_{2}u_{2}+\kappa_{4}u_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=-\kappa_{2}u_{1}+\kappa_{3}u_{2}+\kappa_{5}u_{3},&[u_{3},u_{4}]=(\kappa_{1}+\kappa_{3})u_{3},\end{array}$

where $\kappa_{1}$, $\kappa_{2}$, $\kappa_{3}$, $\kappa_{4}$,
$\kappa_{5}\in\mathbb{R}$, with $(\kappa_{1}-\kappa_{3})(2\kappa_{2}+1)\neq
0$.

  2. (D.ii)

The left-invariant metric determined by

$[u_{1},u_{3}]=u_{1},\quad[u_{2},u_{3}]=\kappa\,u_{1},\quad\kappa>0,$

where $\\{u_{1},u_{2},u_{3},u_{4}\\}$ is a pseudo-orthonormal basis of
$\mathfrak{h}_{3}\rtimes\mathfrak{r}$, with $\langle
u_{1},u_{1}\rangle=\langle u_{2},u_{2}\rangle=\langle u_{3},u_{4}\rangle=1$.

  1. (L)

The left-invariant metric determined by

$[u_{1},u_{3}]=-\varepsilon
u_{2},\quad[u_{1},u_{4}]=\kappa_{1}u_{2}+\kappa_{2}u_{3},\quad[u_{3},u_{4}]=\kappa_{3}u_{2},$

where $\\{u_{1},u_{2},u_{3},u_{4}\\}$ is a pseudo-orthonormal basis of
$\mathfrak{h}_{3}\rtimes\mathfrak{r}$, with $\langle
u_{1},u_{2}\rangle=\langle u_{3},u_{3}\rangle=\langle u_{4},u_{4}\rangle=1$,
and $\kappa_{1},\kappa_{2},\kappa_{3}\in\mathbb{R}$ with
$\kappa_{2}(\kappa_{2}+\kappa_{3})\neq 0$, and $\varepsilon=\pm 1$.

Moreover, all the cases above are conformally Einstein plane waves.

######  Remark 1.10.

Recall from Remark 1.1 that all Bach-flat metrics on $H_{3}\times\mathbb{R}$
have parallel Ricci tensor. A straightforward calculation shows that the Ricci
tensor of metrics corresponding to (D.i) is parallel if and only if
$(\kappa_{1}+\kappa_{3})(4\kappa_{1}\kappa_{3}+1)=0$. Hence in the generic
situation these metrics are not homothetic to any left-invariant metric on the
product $H_{3}\times\mathbb{R}$.

The Lie group in (D.ii) is not isomorphic to the product
$H_{3}\times\mathbb{R}$, although the Ricci tensor is parallel. Indeed, the
underlying Lie group is solvable but not two-step nilpotent in contrast with
the product $H_{3}\times\mathbb{R}$.

The Lie group in (L) is three-step nilpotent. Hence left-invariant metrics in
(L) are not isomorphically homothetic to any left-invariant metric on the
product $H_{3}\times\mathbb{R}$, although the Ricci tensor is parallel.

A straightforward calculation shows that the sectional curvature of metrics
corresponding to (D.i) does not depend on the parameters $\kappa_{4}$ and
$\kappa_{5}$. Hence it follows from the work in [23] that a left-invariant
metric (D.i) is homothetic (although not necessarily isomorphically
homothetic) to a metric with $\kappa_{4}=\kappa_{5}=0$. Analogously, the
sectional curvature of metrics (L) does not depend both on $\kappa_{1}$ and
$\varepsilon$. Hence one has that a metrics in (L) is homothetic (but not
isomorphically homothetic) to a metric with $\varepsilon=-1$ and
$\kappa_{1}=0$.

###  1.2. Bach tensor of left-invariant metrics and Gröbner bases

Let $(G,\langle\cdot,\cdot\rangle)$ be a four-dimensional Lorentzian Lie
group. It is now immediate that the Bach-flatness condition $\mathfrak{B}=0$
equals to a system of polynomial equations on the structure constants (given
by the components $\mathfrak{B}_{ij}$ of the Bach tensor) which one has to
solve in order to obtain a complete classification. When the system under
consideration is simple, it is an elementary problem to find all common roots,
but if the number of equations, unknowns and their degrees increase, it may
become a quite unmanageable task.

Given a set $\mathcal{S}$ of polynomials
$\mathfrak{B}_{ij}\in\mathbb{R}[x_{1},\dots,x_{n}]$, an $n$-tuple of real
numbers $\vec{a}=(a_{1},\dots,a_{n})$ is a solution of $\mathcal{S}$ if and
only if $\mathfrak{B}_{ij}(\vec{a})=0$ for all $i$, $j$. It is a fundamental
observation to recognize that $\vec{a}$ is a solution of $\mathcal{S}$ if and
only if it is a solution of $\mathcal{I}=\langle\mathfrak{B}_{ij}\rangle$, the
ideal generated by the $\mathfrak{B}_{ij}$’s: if two sets of polynomials
generate the same ideal, the corresponding zero sets must be identical. The
theory of Gröbner bases provides a well-known strategy to solve rather large
polynomial systems obtaining “better” polynomials that belong to the ideal
generated by the initial polynomial system (see [16] for more information on
Gröbner bases).

Let $x^{\alpha}=x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}$ with
$\alpha\in\mathbb{Z}^{n}_{\geq 0}$ be a monomial in
$\mathbb{R}[x_{1},\dots,x_{n}]$. A monomial ordering is any relation on the
set of monomials $x^{\alpha}$ with $\alpha\in\mathbb{Z}^{n}_{\geq 0}$
satisfying

  1. (1)

It is a total ordering on $\mathbb{Z}^{n}_{\geq 0}$.

  2. (2)

If $\alpha>\beta$ and $\gamma\in\mathbb{Z}^{n}_{\geq 0}$, then
$\alpha+\gamma>\beta+\gamma$.

  3. (3)

$\mathbb{Z}^{n}_{\geq 0}$ is well-ordered, so that every non-empty subset of
$\mathbb{Z}^{n}_{\geq 0}$ has a smallest element with respect to the given
ordering.

Establishing an ordering on $\mathbb{Z}^{n}_{\geq 0}$ will induce an ordering
on the monomials. For our purposes we will use the _lexicographical order_ and
the _graded reverse lexicographical order_. We say that $\alpha>_{lex}\beta$
if in the vector $\alpha-\beta\in\mathbb{Z}^{n}$ the leftmost non-zero entry
is positive and we say that $\alpha>_{grevlex}\beta$ if $|\alpha|>|\beta|$ or
$|\alpha|=|\beta|$ and the rightmost non-zero entry of
$\alpha-\beta\in\mathbb{Z}^{n}$ is negative. We would like to emphasize that
the Gröbner basis construction is very sensitive to the ordering on the
variables and the monomials. For a certain ordering, a simple Gröbner basis
can be obtained with a reduced number of polynomials, while for other
orderings both the number of polynomials and their form can be completely
unmanageable. Lexicographical order for the monomials is the most appropriate
in most cases to get simple bases. However, it is not always possible to use
such ordering by computational reasons and other orderings must be taken into
consideration. We therefore emphasize in each case the ordering under
consideration for the monomials and the variables.

Finally, it is worth to note that Gröbner bases are not unique for a given
ordering on the variables and the monomials, since they may also depend on the
algorithm used in the calculations. However, they are suitable to check the
ideal membership problem and to decide whether a given polynomial is in the
ideal under consideration. All the calculations in this paper have been done
with Singular [17] and doubly checked with Mathematica. The corresponding
files are available from the Authors under request.

###  1.3. Schedule of the paper

We analyze the existence of Bach-flat metrics in semi-direct extensions
$H_{3}\rtimes\mathbb{R}$ in Sections $\S$2, $\S$3 and $\S$4, depending on
whether the induced metric in $\mathfrak{h}_{3}$ is degenerate, Riemannian or
Lorentzian, respectively. The proof of Theorem 1.9 follows directly from this
analysis and Remark 1.1. In all the cases we pay special attention to obtain
simpler descriptions of the corresponding Bach-flat metrics by using suitable
isomorphisms within the homothetic class. As a consequence the conformally
Einstein equation (2) coupled with the conformally Cotton-flat equation (3)
become tractable, and Theorem 1.2 and Theorem 1.7 are finally proven in
Section 5 and Section 6, respectively.

##  2\. Semi-direct extensions with degenerate normal subgroup $H_{3}$

In this section we analyze left-invariant Lorentzian metrics which are
extensions of the three-dimensional unimodular Lie group $H_{3}$ equipped with
a degenerate metric. Hence, let
$\mathfrak{g}=\mathfrak{h}_{3}\rtimes\mathfrak{r}$ be a four-dimensional Lie
algebra with a Lorentzian inner product $\langle\cdot,\cdot\rangle$ which
restricts to a degenerate inner product on the subalgebra $\mathfrak{h}_{3}$.
Let $\mathfrak{h}_{3}^{\prime}=\operatorname{span}\\{v\\}$ be the derived
subalgebra of $\mathfrak{h}_{3}$,
$\mathfrak{h}_{3}^{\prime}=[\mathfrak{h}_{3},\mathfrak{h}_{3}]$. Since the
restriction of the metric to $\mathfrak{h}_{3}$ has signature $(+,+,0)$ the
vector $v$ may be spacelike or null (see [12]). Next we analyze the vanishing
of the Bach tensor in those two cases by separate.

###  2.1. $\boldsymbol{\mathfrak{h}^{\prime}_{3}=\operatorname{\bf
span}\\{v\\}}$ is a null subspace

Setting $u_{3}=v$ we can take a pseudo-orthonormal basis
$\\{u_{1},u_{2},u_{3},u_{4}\\}$ of
$\mathfrak{g}=\mathfrak{h}_{3}\rtimes\mathfrak{r}$, with $\langle
u_{1},u_{1}\rangle=\langle u_{2},u_{2}\rangle=\langle u_{3},u_{4}\rangle=1$,
so that $\mathfrak{h}_{3}=\operatorname{span}\\{u_{1},u_{2},u_{3}\\}$ and
$\mathfrak{r}=\operatorname{span}\\{u_{4}\\}$. Since
$\operatorname{ad}(u_{4})$ acts on $\mathfrak{h}_{3}$ as a derivation, one has
the Lie brackets

$[u_{1},u_{2}]=\lambda_{1}u_{3},\quad[u_{1},u_{3}]=\lambda_{2}u_{3},\quad[u_{2},u_{3}]=\lambda_{3}u_{3},\quad\underset{\underset{(i=1,2,3)}{}}{[u_{i},u_{4}]}=\sum_{j=1}^{3}\alpha_{i}^{j}u_{j},$

for certain $\alpha_{i}^{j}\in\mathbb{R}$, where at least one of
$\lambda_{1}$, $\lambda_{2}$ and $\lambda_{3}$ is non-zero. Next, considering
the endomorphism determined by the matrix $(\alpha_{i}^{j})$, or equivalently
the Jacobi identity, one is led to the following different possibilities
depending on the $\lambda_{i}$’s.

We will show that Bach-flat left-invariant metrics in this case are locally
conformally flat or plane waves, thus not providing new examples of strictly
Bach-flat structures.

####  2.1.1. Case $\lambda_{2}=\lambda_{3}=0$

If $\lambda_{2}=\lambda_{3}=0$, then necessarily $\lambda_{1}\neq 0$ and

$\begin{array}[]{ll}[u_{1},u_{2}]=\lambda_{1}u_{3},&[u_{1},u_{4}]=\gamma_{1}u_{1}+\gamma_{2}u_{2}+\gamma_{3}u_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=\gamma_{4}u_{1}+\gamma_{5}u_{2}+\gamma_{6}u_{3},&[u_{3},u_{4}]=(\gamma_{1}+\gamma_{5})u_{3},\end{array}$

where $\gamma_{1}$, $\dots$, $\gamma_{6}\in\mathbb{R}$. In this case, without
further assumptions, a direct calculation shows that the Bach tensor vanishes.
Moreover, $u_{3}$ is a null recurrent vector field and the curvature tensor
satisfies $R(x,y)=0$ and $\nabla_{x}R=0$ for all $x,y\in
u_{3}^{\perp}=\operatorname{span}\\{u_{1},u_{2},u_{3}\\}$. Finally, the Ricci
tensor of the above metric is determined by
$\rho_{44}=\frac{1}{2}\\{\lambda_{1}^{2}+4\gamma_{1}\gamma_{5}-(\gamma_{2}+\gamma_{4})^{2}\\}$,
which implies that the Ricci operator is isotropic and therefore the
underlying structure is a plane wave (see [25]).

Rotating the spacelike vectors $\\{u_{1},u_{2}\\}$ one may assume that
$\gamma_{4}=-\gamma_{2}$. Furthermore, rescaling the vectors $u_{k}$ by
$\frac{1}{\lambda_{1}}u_{k}$ one may set $\lambda_{1}=1$ remaining in the same
homothety class. Finally a straightforward calculation shows that the metric
is locally conformally flat if and only if
$(\gamma_{1}-\gamma_{5})(2\gamma_{2}+1)=0$, and the metric is locally
symmetric if and only if $\gamma_{1}=\gamma_{5}=0$, or $2\gamma_{2}+1=0$ and
$(\gamma_{1}+\gamma_{5})(4\gamma_{1}\gamma_{5}+1)=0$. It corresponds to
Assertion (D.i) in Theorem 1.9.

####  2.1.2. Case $\lambda_{2}=0$, $\lambda_{3}\neq 0$

In this case, the Lie algebra structure is given by

$\begin{array}[]{lll}[u_{1},u_{2}]=\lambda_{1}u_{3},&[u_{1},u_{4}]=\gamma_{1}\lambda_{3}u_{1}+(\gamma_{1}-\gamma_{2})\lambda_{1}u_{3},&[u_{2},u_{3}]=\lambda_{3}u_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=\gamma_{3}u_{1}+\gamma_{4}u_{3},&[u_{3},u_{4}]=\gamma_{2}\lambda_{3}u_{3},\end{array}$

where $\gamma_{1}$, $\dots$, $\gamma_{4}\in\mathbb{R}$. A straightforward
calculation shows that the only non-zero component of the Bach tensor,
$\mathfrak{B}_{ij}=\mathfrak{B}(u_{i},u_{j})$, corresponds to
$\mathfrak{B}_{44}=\frac{1}{4}(\gamma_{1}^{2}\lambda_{3}^{2}+\gamma_{3}^{2})\lambda_{3}^{2}$.
Hence, since $\lambda_{3}\neq 0$, necessarily $\gamma_{1}=\gamma_{3}=0$.
Finally, one directly checks that the metric is locally conformally flat and
locally symmetric.

####  2.1.3. Case $\lambda_{2}\neq 0$

If $\lambda_{2}\neq 0$, then one has

$\begin{array}[]{l}[u_{1},u_{2}]=\lambda_{1}u_{3},\quad[u_{1},u_{3}]=\lambda_{2}u_{3},\quad[u_{2},u_{3}]=\lambda_{3}u_{3},\quad[u_{3},u_{4}]=\gamma_{4}\lambda_{2}u_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{1},u_{4}]=-\gamma_{1}\lambda_{2}\lambda_{3}u_{1}+\gamma_{1}\lambda_{2}^{2}u_{2}+\gamma_{2}\lambda_{2}u_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=-\gamma_{3}\lambda_{3}u_{1}+\gamma_{3}\lambda_{2}u_{2}+(\gamma_{1}\lambda_{1}\lambda_{3}-(\gamma_{3}-\gamma_{4})\lambda_{1}+\gamma_{2}\lambda_{3})u_{3},\end{array}$

where $\gamma_{1}$, $\dots$, $\gamma_{4}\in\mathbb{R}$. As in the previous
case, the Bach tensor is determined by just one component,
$\mathfrak{B}_{44}=\frac{1}{4}(\gamma_{1}^{2}\lambda_{2}^{2}+\gamma_{3}^{2})(\lambda_{2}^{2}+\lambda_{3}^{2})^{2}$,
which implies $\gamma_{1}=\gamma_{3}=0$. Again, a direct calculation shows
that the metric is locally conformally flat and locally symmetric.

###  2.2. $\boldsymbol{\mathfrak{h}^{\prime}_{3}=\operatorname{span}\\{v\\}}$
is a spacelike subspace

In this case, we set $u_{1}=\frac{v}{\|v\|}$ and consider a pseudo-orthonormal
basis $\\{u_{1},u_{2},u_{3},u_{4}\\}$ of
$\mathfrak{g}=\mathfrak{h}_{3}\rtimes\mathfrak{r}$, with $\langle
u_{1},u_{1}\rangle=\langle u_{2},u_{2}\rangle=\langle u_{3},u_{4}\rangle=1$,
where $\mathfrak{h}_{3}=\operatorname{span}\\{u_{1},u_{2},u_{3}\\}$ and
$\mathfrak{r}=\operatorname{span}\\{u_{4}\\}$, so that

$[u_{1},u_{2}]=\lambda_{1}u_{1},\quad[u_{1},u_{3}]=\lambda_{2}u_{1},\quad[u_{2},u_{3}]=\lambda_{3}u_{1},\quad\underset{\underset{(i=1,2,3)}{}}{[u_{i},u_{4}]}=\sum_{j=1}^{3}\alpha_{i}^{j}u_{j},$

for certain $\alpha_{i}^{j}\in\mathbb{R}$ and where at least one of
$\lambda_{1}$, $\lambda_{2}$ and $\lambda_{3}$ is non-zero. We proceed as in
Section 2.1, to have the following different possibilities depending on the
$\lambda_{i}$’s.

####  2.2.1. Case $\lambda_{1}=\lambda_{2}=0$

In this case, necessarily $\lambda_{3}\neq 0$ and the Lie algebra structure is
given by

$\begin{array}[]{l}[u_{1},u_{4}]=\gamma_{1}u_{1},\quad[u_{2},u_{3}]=\lambda_{3}u_{1},\quad[u_{2},u_{4}]=\gamma_{2}u_{1}+\gamma_{3}u_{2}+\gamma_{4}u_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{3},u_{4}]=\gamma_{5}u_{1}+\gamma_{6}u_{2}+(\gamma_{1}-\gamma_{3})u_{3},\end{array}$

where $\gamma_{1}$, $\dots$, $\gamma_{6}\in\mathbb{R}$. Since $\lambda_{3}\neq
0$, we consider the orthogonal basis $\hat{u}_{i}=\frac{1}{\lambda_{3}}u_{i}$
so that we can assume $\lambda_{3}=1$ working in the homothetic class of the
initial metric. A direct analysis of the components of the Bach tensor let us
to clear some of the structure constants and completely determine the Bach-
flat metrics. We proceed as follows.

First, we compute
$\mathfrak{B}_{33}=\frac{1}{12}(\gamma_{6}^{2}-8\gamma_{5}^{2}+16\gamma_{2})$,
which implies $\gamma_{2}=\frac{1}{16}(8\gamma_{5}^{2}-\gamma_{6}^{2})$, and
as a consequence we get

$\small\begin{array}[]{l}\phantom{..}-4\,\mathfrak{B}_{12}=\gamma_{6}^{2}(3\gamma_{5}\gamma_{6}+2\gamma_{1}-3\gamma_{3}),\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr-128\,\mathfrak{B}_{11}=21\gamma_{6}^{4}+144\gamma_{5}^{2}\gamma_{6}^{2}+\gamma_{5}\gamma_{6}(64\gamma_{1}-288\gamma_{3})-80\gamma_{1}^{2}+240\gamma_{3}^{2}-160\gamma_{1}\gamma_{3}+96\gamma_{4}\gamma_{6},\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{-}128\,\mathfrak{B}_{34}=63\gamma_{6}^{4}+144\gamma_{5}^{2}\gamma_{6}^{2}+\gamma_{5}\gamma_{6}(192\gamma_{1}-192\gamma_{3})+80\gamma_{1}^{2}+80\gamma_{3}^{2}-160\gamma_{1}\gamma_{3}+32\gamma_{4}\gamma_{6}.\par\end{array}$

Note that if $\gamma_{6}\neq 0$ then
$\gamma_{3}=\frac{1}{3}(3\gamma_{5}\gamma_{6}+2\gamma_{1})$, and a direct
calculation shows that
$\mathfrak{B}_{11}+3\,\mathfrak{B}_{34}=\frac{1}{48}(40\gamma_{1}^{2}+63\gamma_{6}^{4})\neq
0$. Hence, necessarily $\gamma_{6}=0$ and, moreover,
$\mathfrak{B}_{34}=\frac{5}{8}(\gamma_{1}-\gamma_{3})^{2}$, which implies
$\gamma_{3}=\gamma_{1}$.

At this point we have $\gamma_{2}=\frac{1}{2}\gamma_{5}^{2}$, $\gamma_{6}=0$
and $\gamma_{3}=\gamma_{1}$, and a final straightforward calculation shows
that the Bach tensor is determined by

$\mathfrak{B}_{24}=\tfrac{3}{4}\gamma_{1}(\gamma_{1}\gamma_{5}+\gamma_{4})\quad\text{and}\quad\mathfrak{B}_{44}=-\tfrac{3}{4}(\gamma_{1}\gamma_{5}+\gamma_{4})(3\gamma_{1}\gamma_{5}+\gamma_{4}).$

We conclude that $\gamma_{4}=-\gamma_{1}\gamma_{5}$ and thus the left-
invariant metric corresponds to

$[u_{1},u_{4}]=\gamma_{1}u_{1},\,\,\,[u_{2},u_{3}]=u_{1},\,\,\,[u_{2},u_{4}]=\tfrac{1}{2}\gamma_{5}^{2}u_{1}+\gamma_{1}u_{2}-\gamma_{1}\gamma_{5}u_{3},\,\,\,[u_{3},u_{4}]=\gamma_{5}u_{1}.$

Let $\zeta=-1$, $0$, $1$, depending on whether $\gamma_{1}$ is negative, zero,
or positive. Taking the basis

$\tilde{u}_{1}=u_{1},\quad\tilde{u}_{2}=u_{2}-\gamma_{5}u_{3},\quad\tilde{u}_{3}=u_{3},\quad\tilde{u}_{4}=\gamma_{5}u_{2}-\tfrac{1}{2}\gamma_{5}^{2}u_{3}+u_{4},$

if $\gamma_{1}=0$, and

$\tilde{u}_{1}=\tfrac{1}{|\gamma_{1}|^{\frac{1}{2}}}u_{1},\,\,\,\,\tilde{u}_{2}=\tfrac{\zeta}{|\gamma_{1}|^{\frac{1}{2}}}\left(u_{2}-\gamma_{5}u_{3}\right),\,\,\,\,\tilde{u}_{3}=\zeta
u_{3},\,\,\,\,\tilde{u}_{4}=\tfrac{1}{\gamma_{1}}\left(\gamma_{5}u_{2}-\tfrac{1}{2}\gamma_{5}^{2}u_{3}+u_{4}\right)$

if $\gamma_{1}\neq 0$, a direct calculations shows that the Lie bracket
transforms into

$[\tilde{u}_{1},\tilde{u}_{4}]=|\zeta|\tilde{u}_{1},\qquad[\tilde{u}_{2},\tilde{u}_{3}]=\tilde{u}_{1},\qquad[\tilde{u}_{2},\tilde{u}_{4}]=|\zeta|\tilde{u}_{2},$

while the inner product remains invariant if $\zeta=0$ and it is given by
$\frac{1}{|\gamma_{1}|}\langle\cdot,\cdot\rangle$ if $\zeta\neq 0$. Since we
are working at the homothetic level we can maintain the initial inner product
in any case, remaining in the same homothetic class. Now, a straightforward
calculation shows that the above metric is not locally symmetric, locally
conformally flat or Einstein. The Ricci operator of the above metric has
eigenvalues $\\{\zeta,\zeta,-\zeta,-\zeta\\}$ and the geometry of the
corresponding spacetime is completely different depending on $\zeta$ vanishes
or not. If $\zeta\in\\{-1,1\\}$, the Ricci operator is diagonalizable and the
metric corresponds to that in Theorem 1.2-(D.i). Otherwise, if $\zeta=0$, the
Ricci operator is $2$-step nilpotent and a straightforward calculation shows
that $\tilde{u}_{4}$ is a null parallel vector field and the curvature tensor
satisfies $R(x,y)=0$ and $\nabla_{x}R=0$ for all
$x,y\in\tilde{u}_{4}^{\perp}=\operatorname{span}\\{\tilde{u}_{1},\tilde{u}_{2},\tilde{u}_{4}\\}$.
Thus, the underlying structure is a left-invariant plane wave on the product
$H_{3}\times\mathbb{R}$, and thus isomorphically homothetic to the Bach-flat
metric discussed in Remark 1.1.

####  2.2.2. Case $\lambda_{1}=0$, $\lambda_{2}\neq 0$

The Lie algebra structure is then given by

$\begin{array}[]{lll}[u_{1},u_{3}]=\lambda_{2}u_{1},&[u_{1},u_{4}]=\gamma_{1}\lambda_{2}u_{1},&[u_{2},u_{3}]=\lambda_{3}u_{1},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=(\gamma_{1}-\gamma_{2})\lambda_{3}u_{1}+\gamma_{2}\lambda_{2}u_{2},&[u_{3},u_{4}]=\gamma_{3}u_{1}+\gamma_{4}u_{2},\end{array}$

where $\gamma_{1}$, $\dots$, $\gamma_{4}\in\mathbb{R}$. Since $\lambda_{2}\neq
0$, we consider the orthogonal basis $\hat{u}_{i}=\frac{1}{\lambda_{2}}u_{i}$
so that we can assume $\lambda_{2}=1$ working in the homothetic class of the
initial metric. In this case, we compute a Gröbner basis of the ideal
generated by the components of the Bach tensor,
$\langle\mathfrak{B}_{ij}\rangle\subset\mathbb{R}[\lambda_{3},\gamma_{1}$,
$\gamma_{2}$, $\gamma_{3}$, $\gamma_{4}]$, with respect to the lexicographical
order (which consists of $48$ polynomials) and get that the polynomials

$\mathbf{g}_{1}=\gamma_{4}^{5}(\gamma_{3}^{2}+\gamma_{4}^{2})\quad\text{and}\quad\mathbf{g}_{2}=(20\gamma_{3}^{2}-7\gamma_{4}^{2})(\gamma_{3}^{2}+\gamma_{4}^{2})^{2}$

belong to the ideal. Thus, necessarily $\gamma_{3}=\gamma_{4}=0$. Now, a
direct calculation shows that
$\mathfrak{B}_{33}=\frac{2}{3}(\gamma_{1}-\gamma_{2})(2\lambda_{3}^{4}+3\lambda_{3}^{2}+1)$,
which implies $\gamma_{2}=\gamma_{1}$. Under these assumptions, one easily
checks that the Bach tensor vanishes and the corresponding left-invariant
metric is given by

$[u_{1},u_{3}]=u_{1},\quad[u_{1},u_{4}]=\gamma_{1}u_{1},\quad[u_{2},u_{3}]=\lambda_{3}u_{1},\quad[u_{2},u_{4}]=\gamma_{1}u_{2}.$

Let $\\{\tilde{u}_{i}\\}$ be the basis given by the initial one
($\tilde{u}_{i}=u_{i}$) if $\gamma_{1}=0$, or

$\tilde{u}_{1}=\tfrac{1}{|\gamma_{1}|^{\frac{1}{2}}}u_{1},\quad\tilde{u}_{2}=\tfrac{1}{|\gamma_{1}|^{\frac{1}{2}}}u_{2},\quad\tilde{u}_{3}=u_{3},\quad\tilde{u}_{4}=\tfrac{1}{|\gamma_{1}|}u_{4}$

if $\gamma_{1}\neq 0$. Let $\zeta=-1$, $0$, $1$, depending on whether
$\gamma_{1}$ is negative, zero, or positive. One easily checks that the Lie
bracket transforms into

(4)
$[\tilde{u}_{1},\tilde{u}_{3}]=\tilde{u}_{1},\qquad[\tilde{u}_{1},\tilde{u}_{4}]=\zeta\tilde{u}_{1},\qquad[\tilde{u}_{2},\tilde{u}_{3}]=\lambda_{3}\tilde{u}_{1},\qquad[\tilde{u}_{2},\tilde{u}_{4}]=\zeta\tilde{u}_{2},$

while the inner product obviously remains invariant if $\gamma_{1}=0$ or,
otherwise, it is given by $\frac{1}{|\gamma_{1}|}\langle\cdot,\cdot\rangle$.
Since we are working at the homothetic level, we can maintain the initial
inner product $\langle\cdot,\cdot\rangle$ in any case, remaining in the same
homothetic class. Now, a straightforward calculation shows that the above
metric is not locally conformally flat nor Einstein, and it is locally
symmetric if and only if $\lambda_{3}=\zeta=0$. Moreover,
$(\tilde{u}_{1},\tilde{u}_{2},\tilde{u}_{3},\tilde{u}_{4})\mapsto(\tilde{u}_{1},-\tilde{u}_{2},\tilde{u}_{3},\tilde{u}_{4})$
defines an isometry interchanging $(\zeta,\lambda_{3})$ and
$(\zeta,-\lambda_{3})$ and hence we can restrict the parameter $\lambda_{3}$
to $\lambda_{3}\geq 0$. A direct calculation shows that the eigenvalues of the
Ricci operator of the above metric are given by
$\\{\zeta(-1\pm\sqrt{\lambda_{3}^{2}+2}),\zeta(-2\pm\sqrt{\lambda_{3}^{2}+1})\\}$.
If $\zeta\in\\{-1,1\\}$ then the four eigenvalues are different and the
metrics correspond to Theorem 1.2-(D.ii). If $\zeta=0$ then the metric is
never locally conformally flat nor locally symmetric. Moreover, the Ricci
operator is $2$-step nilpotent and $\tilde{u}_{4}$ is a null parallel vector
field so that $R(x,y)=0$ and $\nabla_{x}R=0$ for all
$x,y\in\tilde{u}_{4}^{\perp}=\operatorname{span}\\{\tilde{u}_{1},\tilde{u}_{2},\tilde{u}_{4}\\}$.
Thus, the underlying structure is a plane wave as in Theorem 1.9-(D.ii).

####  2.2.3. Case $\lambda_{1}\neq 0$

In this case, the Lie algebra structure becomes

$\begin{array}[]{l}[u_{1},u_{2}]=\lambda_{1}u_{1},\quad[u_{1},u_{3}]=\lambda_{2}u_{1},\quad[u_{1},u_{4}]=\gamma_{1}\lambda_{1}u_{1},\quad[u_{2},u_{3}]=\lambda_{3}u_{1},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=\lambda_{1}\gamma_{2}u_{1}-\gamma_{3}\lambda_{1}\lambda_{2}u_{2}+\gamma_{3}\lambda_{1}^{2}u_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{3},u_{4}]=-(\gamma_{3}\lambda_{2}\lambda_{3}-\gamma_{2}\lambda_{2}+(\gamma_{1}-\gamma_{4})\lambda_{3})u_{1}-\gamma_{4}\lambda_{2}u_{2}+\gamma_{4}\lambda_{1}u_{3},\end{array}$

where $\gamma_{1}$, $\dots$, $\gamma_{4}\in\mathbb{R}$. As in the previous
case, we use Gröbner bases. First, we compute a Gröbner basis of the ideal
$\langle\bar{\mathfrak{B}}_{ij}\rangle\subset\mathbb{R}[\lambda_{1}$,
$\lambda_{2}$, $\lambda_{3}$, $\gamma_{1}$, $\gamma_{2}$, $\gamma_{3}$,
$\gamma_{4}]$ with respect to the graded reverse lexicographical order, where
$\bar{\mathfrak{B}}_{ij}$ are obtained from the components $\mathfrak{B}_{ij}$
after simplifying the parameter $\lambda_{1}\neq 0$ whenever possible. The
Gröbner basis, consisting of $2183$ polynomials, is obtained after a long
calculation (more than 12 hours with a last generation home computer) and
contains the polynomial

$\mathbf{g}=\gamma_{4}^{7}\lambda_{1}^{5}(\lambda_{2}^{2}+\lambda_{3}^{2}).$

Since $\lambda_{1}\neq 0$, it follows that either $\lambda_{2}=\lambda_{3}=0$
or $\gamma_{4}=0$. If $\lambda_{2}=\lambda_{3}=0$ a direct calculation shows
that $\mathfrak{B}_{11}=\frac{1}{6}\lambda_{1}^{4}$ and therefore the metric
cannot be Bach-flat. Thus, necessarily $\gamma_{4}=0$. Secondly, assuming this
last condition, we compute another Gröbner basis for
$\langle\bar{\mathfrak{B}}_{ij}\rangle\subset\mathbb{R}[\gamma_{1}$,
$\gamma_{2}$, $\gamma_{3}$, $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}]$ with
respect to the graded reverse lexicographical order and get the polynomials

$\begin{array}[]{l}\mathbf{g}^{\prime}_{1}=\lambda_{1}^{4}(\lambda_{1}+2\gamma_{1}\lambda_{2})(\lambda_{1}+2\lambda_{2}(\gamma_{3}\lambda_{2}+\gamma_{1})),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathbf{g}^{\prime}_{2}=\lambda_{1}^{2}(\lambda_{1}+2\gamma_{1}\lambda_{2})^{2}(2\gamma_{2}\lambda_{2}^{2}-\lambda_{1}\lambda_{3}),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathbf{g}^{\prime}_{3}=\gamma_{3}\lambda_{1}^{4}((\lambda_{1}+2\gamma_{1}\lambda_{2})(\lambda_{1}+2\gamma_{2}\lambda_{3})+2\gamma_{3}\lambda_{1}(\lambda_{2}^{2}+\lambda_{3}^{2})),\end{array}$

among the $216$ polynomials of the basis. The expressions of
$\mathbf{g}^{\prime}_{1}$ and $\mathbf{g}^{\prime}_{2}$ imply that either
$\lambda_{1}+2\gamma_{1}\lambda_{2}=0$ or, otherwise,
$\lambda_{1}+2\lambda_{2}(\gamma_{3}\lambda_{2}+\gamma_{1})=0$,
$2\gamma_{2}\lambda_{2}^{2}-\lambda_{1}\lambda_{3}=0$. Note that, in any case,
$\lambda_{1}\neq 0$ implies $\lambda_{2}\neq 0$. Next we show that the first
condition implies the second one in the Bach-flat setting. Indeed, if
$\lambda_{1}+2\gamma_{1}\lambda_{2}=0$, or equivalently
$\gamma_{1}=-\frac{\lambda_{1}}{2\lambda_{2}}$, then
$\mathbf{g}^{\prime}_{3}=2\gamma_{3}^{2}\lambda_{1}^{6}(\lambda_{2}^{2}+\lambda_{3}^{2})$.
Hence necessarily $\gamma_{3}=0$ and, in that case, a direct calculation shows
that
$\mathfrak{B}_{11}=-\frac{5(2\gamma_{2}\lambda_{2}^{2}-\lambda_{1}\lambda_{3})^{4}}{96\lambda_{2}^{4}}$.
Thus $2\gamma_{2}\lambda_{2}^{2}-\lambda_{1}\lambda_{3}=0$ and, since
$\gamma_{3}=0$, also
$\lambda_{1}+2\lambda_{2}(\gamma_{3}\lambda_{2}+\gamma_{1})=0$. As a
consequence, we conclude that if $\gamma_{4}=0$ and the metric is Bach-flat,
then $\lambda_{1}+2\lambda_{2}(\gamma_{3}\lambda_{2}+\gamma_{1})=0$ and
$2\gamma_{2}\lambda_{2}^{2}-\lambda_{1}\lambda_{3}=0$, or equivalently
$\gamma_{1}=-\frac{\lambda_{1}+2\gamma_{3}\lambda_{2}^{2}}{2\lambda_{2}}$ and
$\gamma_{2}=\frac{\lambda_{1}\lambda_{3}}{2\lambda_{2}^{2}}$. Finally, a
straightforward calculation shows that, under these assumptions, the Bach
tensor vanishes and the corresponding left-invariant metric is given by

$\begin{array}[]{lll}[u_{1},u_{2}]=\lambda_{1}u_{1},&[u_{1},u_{3}]=\lambda_{2}u_{1},&[u_{1},u_{4}]=-\frac{\lambda_{1}(\lambda_{1}+2\gamma_{3}\lambda_{2}^{2})}{2\lambda_{2}}u_{1},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{3}]=\lambda_{3}u_{1},&[u_{3},u_{4}]=\frac{\lambda_{1}\lambda_{3}}{\lambda_{2}}u_{1}&[u_{2},u_{4}]=\frac{\lambda_{1}^{2}\lambda_{3}}{2\lambda_{2}^{2}}u_{1}-\gamma_{3}\lambda_{1}\lambda_{2}u_{2}+\gamma_{3}\lambda_{1}^{2}u_{3}.\end{array}$

At this point, we introduce variables
$\lambda_{3}^{\prime}=\frac{\lambda_{3}}{\lambda_{2}}$ and
$\gamma_{3}^{\prime}=-\gamma_{3}\lambda_{1}\lambda_{2}^{2}$. Moreover, let
$\zeta=-1$, $0$, $1$, depending on whether $\gamma_{3}^{\prime}$ is negative,
zero, or positive. Considering the basis $\\{\tilde{u}_{i}\\}$ defined by

$\tilde{u}_{1}=u_{1},\quad\tilde{u}_{2}=u_{2}-\tfrac{\lambda_{1}}{\lambda_{2}}u_{3},\quad\tilde{u}_{3}=\tfrac{1}{\lambda_{2}}u_{3},\quad\tilde{u}_{4}=\lambda_{1}u_{2}-\tfrac{\lambda_{1}^{2}}{2\lambda_{2}}u_{3}+\lambda_{2}u_{4}$

if $\gamma_{3}^{\prime}=0$, and

$\begin{array}[]{ll}\tilde{u}_{1}=\tfrac{1}{|\gamma_{3}^{\prime}|^{\frac{1}{2}}}u_{1},&\quad\tilde{u}_{2}=\tfrac{1}{|\gamma_{3}^{\prime}|^{\frac{1}{2}}}\left(u_{2}-\tfrac{\lambda_{1}}{\lambda_{2}}u_{3}\right),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\tilde{u}_{3}=\tfrac{1}{\lambda_{2}}u_{3},&\quad\tilde{u}_{4}=\tfrac{1}{|\gamma_{3}^{\prime}|}\left(\lambda_{1}u_{2}-\tfrac{\lambda_{1}^{2}}{2\lambda_{2}}u_{3}+\lambda_{2}u_{4}\right)\end{array}$

if $\gamma_{3}^{\prime}\neq 0$, a direct calculation shows that the Lie
bracket transforms into

$[\tilde{u}_{1},\tilde{u}_{3}]=\tilde{u}_{1},\qquad[\tilde{u}_{1},\tilde{u}_{4}]=\zeta\tilde{u}_{1},\qquad[\tilde{u}_{2},\tilde{u}_{3}]=\lambda_{3}^{\prime}\tilde{u}_{1},\qquad[\tilde{u}_{2},\tilde{u}_{4}]=\zeta\tilde{u}_{2},$

while the inner product remains invariant if $\zeta=0$ or, otherwise, it is
given by $\frac{1}{|\gamma_{3}^{\prime}|}\langle\cdot,\cdot\rangle$. Since we
are working at the homothetic level, we can maintain the initial inner product
$\langle\cdot,\cdot\rangle$ in any case, remaining in the same homothetic
class. Hence, the setting is exactly the same as in $\S$2.2.2 (see Equation
(4)) and therefore the above metric is a plane wave if $\zeta=0$
(corresponding to Theorem 1.9-(D.ii)) while, otherwise, it corresponds to
Theorem 1.2–(D.ii).

##  3\. Semi-direct extensions with Riemannian normal subgroup $H_{3}$

In this section we consider left-invariant Lorentzian metrics which are
obtained as extensions of the three-dimensional unimodular Riemannian
Heisenberg Lie group $H_{3}$. Left-invariant Riemannian metrics on three-
dimensional unimodular Lie groups were described by Milnor [29] using the
self-dual structure tensor $L$ given by $L(X\times Y)=[X,Y]$, where “$\times$”
denotes the vector-cross product $\langle X\times
Y,Z\rangle=\operatorname{det}(X,Y,Z)$. Self-duality of $L$ ensures the
existence of an orthonormal basis of $\mathfrak{h}_{3}$ diagonalizing the
structure tensor in the positive definite case. Therefore any Riemannian left-
invariant metric on $H_{3}$ is homothetic to the one determined at the Lie
algebra level by an orthonormal basis
$\\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}\\}$ of $\mathfrak{h}_{3}$
such that (see [29])

$[\mathbf{v}_{3},\mathbf{v}_{2}]=0,\quad[\mathbf{v}_{3},\mathbf{v}_{1}]=0,\quad[\mathbf{v}_{1},\mathbf{v}_{2}]=\lambda\mathbf{v}_{3},\qquad\lambda\neq
0.$

The algebra of all derivations of $\mathfrak{h}_{3}$ is given, with respect to
the basis $\\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}\\}$, by

(5)
$\operatorname{der}(\mathfrak{h}_{3})=\left\\{\left(\begin{array}[]{ccc}\alpha_{11}&\alpha_{12}&0\\\
\alpha_{21}&\alpha_{22}&0\\\
\alpha_{31}&\alpha_{32}&\alpha_{11}+\alpha_{22}\end{array}\right);\,\alpha_{ij}\in\mathbb{R}\right\\}.$

For any semi-direct extension $\mathfrak{h}_{3}\rtimes\mathfrak{r}$, there is
a basis $\\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4}\\}$ so
that
$\mathfrak{g}=\mathfrak{h}_{3}\rtimes\mathfrak{r}=\operatorname{span}\\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}\\}\oplus\mathbb{R}\mathbf{v}_{4}$.
Since $\mathbb{R}\mathbf{v}_{4}$ is timelike but not necessarily orthogonal to
$\mathfrak{h}_{3}$, we set
$\bar{e}_{4}=\mathbf{v}_{4}-\sum_{i}\langle\mathbf{v}_{4},\mathbf{v}_{i}\rangle\mathbf{v}_{i}$
and normalize it to get an orthonormal basis $\\{e_{1},\dots,e_{4}\\}$ of
$\mathfrak{g}=\mathfrak{h}_{3}\oplus\mathbb{R}$ with $e_{4}$ timelike, where
$e_{i}=\mathbf{v}_{i}$ $(i=1,2,3)$, so that

$\begin{array}[]{ll}[e_{1},e_{2}]=\lambda
e_{3},&[e_{1},e_{4}]=\gamma_{1}e_{1}+\gamma_{2}e_{2}+\gamma_{3}e_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[e_{2},e_{4}]=\gamma_{4}e_{1}+\gamma_{5}e_{2}+\gamma_{6}e_{3},&[e_{3},e_{4}]=(\gamma_{1}+\gamma_{5})e_{3},\end{array}$

where $\lambda\neq 0$ and $\gamma_{1}$, $\dots$, $\gamma_{6}\in\mathbb{R}$.

In what follows we determine the Bach-flat metrics analyzing two cases by
separate, depending on whether $\gamma_{5}$ vanishes or not.

###  3.1. Case $\boldsymbol{\gamma_{5}=0}$

Since $\lambda\neq 0$, we can assume $\lambda=1$ working in the homothetic
class of the initial metric, just taking the orthogonal basis
$\hat{e}_{i}=\frac{1}{\lambda}e_{i}$. Considering the ideal
$\langle\mathfrak{B}_{ij}\rangle\subset\mathbb{R}[\gamma_{2}$, $\gamma_{3}$,
$\gamma_{4}$, $\gamma_{6}$, $\gamma_{1}]$, where
$\mathfrak{B}_{ij}=\mathfrak{B}(e_{i},e_{j})$ are the polynomials given by the
components of the Bach tensor in the orthonormal basis $\\{e_{i}\\}$, and
fixing the lexicographical order we make use of Gröbner bases. Thus we get 50
polynomials, being one of them the polynomial

$\mathbf{g}=\gamma_{1}^{4}\gamma_{6}^{2}(\gamma_{1}^{2}+1)(\gamma_{1}^{2}+2)(\gamma_{1}^{2}+4)(8\gamma_{1}^{2}+5)(16\gamma_{1}^{2}+25)(25\gamma_{1}^{2}+1).$

Hence, either $\gamma_{1}=0$ or $\gamma_{1}\neq 0$ and $\gamma_{6}=0$.
Moreover, if $\gamma_{1}=0$, we get that

$\mathbf{g}^{\prime}_{1}=\gamma_{4}^{4}\quad\text{and}\quad\mathbf{g}^{\prime}_{2}=\gamma_{2}^{2}-8\gamma_{3}^{2}+\gamma_{4}^{2}-8\gamma_{6}^{2}+8$

are polynomials in the ideal $\langle\mathfrak{B}_{ij}\rangle$, and the same
occur with the polynomials

$\mathbf{g}^{\prime\prime}_{1}=\gamma_{3}\gamma_{4}^{4}\quad\text{and}\quad\mathbf{g}^{\prime\prime}_{2}=\gamma_{1}\gamma_{2}\gamma_{3}(\gamma_{1}^{2}+1)$

if $\gamma_{6}=0$. Next we examine the different possibilities separately.

####  3.1.1. Case $\gamma_{1}=\gamma_{4}=0$,
$\gamma_{2}^{2}-8\gamma_{3}^{2}-8\gamma_{6}^{2}+8=0$

In this case, a direct calculation shows that
$\mathfrak{B}_{11}=\frac{9}{128}\gamma_{2}^{2}(7\gamma_{2}^{2}+16\gamma_{3}^{2})$,
which implies $\gamma_{2}=0$, and the Bach tensor vanishes under this last
assumption. Thus, the corresponding left-invariant metric is given by

$[e_{1},e_{2}]=e_{3},\qquad[e_{1},e_{4}]=\gamma_{3}e_{3},\qquad[e_{2},e_{4}]=\varepsilon\sqrt{1-\gamma_{3}^{2}}\,e_{3},$

where $\gamma_{3}\in[-1,1]$ and $\varepsilon^{2}=1$. The Ricci operator of the
above metric is $2$-step nilpotent and
$\xi=\varepsilon\sqrt{1-\gamma_{3}^{2}}\,e_{1}-\gamma_{3}e_{2}+e_{4}$ is a
null parallel vector field. Moreover, the curvature tensor satisfies
$R(x,y)=0$ and $\nabla_{x}R=0$ for all
$x,y\in\xi^{\perp}=\operatorname{span}\\{\xi,e_{3},\theta\\}$, where

$\theta=\left\\{\begin{array}[]{ll}e_{2}&\text{if $\gamma_{3}=0$},\\\ \vskip
6.0pt plus 2.0pt minus 2.0pt\cr
e_{1}+\varepsilon\sqrt{1-\gamma_{3}^{2}}\,e_{4}&\text{if $\gamma_{3}\neq
0$}.\end{array}\right.$

Thus, the underlying structure is a plane wave. Moreover, considering the
basis

$\tilde{e}_{1}=e_{1},\quad\tilde{e}_{2}=e_{2},\quad\tilde{e}_{3}=e_{3},\quad\tilde{e}_{4}=\varepsilon\sqrt{1-\gamma_{3}^{2}}e_{1}-\gamma_{3}e_{2}+e_{4},$

one has that the only non-zero bracket is
$[\tilde{e}_{1},\tilde{e}_{2}]=\tilde{e}_{3}$, which shows that the underlying
Lie group is the product $H_{3}\times\mathbb{R}$. It is therefore
isomorphically homothetic to the Bach-flat metric discussed in Remark 1.1.

####  3.1.2. Case $\gamma_{1}\neq 0$, $\gamma_{3}=\gamma_{6}=0$

By a straightforward calculation one gets the following components of the Bach
tensor

$\small\begin{array}[]{l}-24\,\mathfrak{B}_{33}=(\gamma_{2}+\gamma_{4})^{2}(4(\gamma_{2}+\gamma_{4})^{2}-12\gamma_{2}\gamma_{4}+3)-\gamma_{1}^{2}((\gamma_{2}+\gamma_{4})^{2}+12\gamma_{2}\gamma_{4}-28)+20,\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr-24\,\mathfrak{B}_{44}=(\gamma_{2}+\gamma_{4})^{2}(12(\gamma_{2}+\gamma_{4})^{2}-36\gamma_{2}\gamma_{4}+1)+\gamma_{1}^{2}(13(\gamma_{2}+\gamma_{4})^{2}-36\gamma_{2}\gamma_{4}-4)-4,\end{array}$

which implies

$\begin{array}[]{l}-12(\mathfrak{B}_{33}+3\,\mathfrak{B}_{44})=(\gamma_{2}+\gamma_{4})^{2}(20(\gamma_{2}+\gamma_{4})^{2}-60\gamma_{2}\gamma_{4}+3)\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{-12(\mathfrak{B}_{33}+3\,\mathfrak{B}_{44})=}+\gamma_{1}^{2}(19(\gamma_{2}+\gamma_{4})^{2}-60\gamma_{2}\gamma_{4}+8)+4.\end{array}$

Clearly this last expression is always strictly positive and therefore the
metric cannot be Bach-flat in this case.

####  3.1.3. Case $\gamma_{1}\gamma_{3}\neq 0$,
$\gamma_{2}=\gamma_{4}=\gamma_{6}=0$

A direct calculation shows that, in this case,
$\mathfrak{B}_{22}=-\frac{1}{6}(3\gamma_{1}^{2}+\gamma_{3}^{2}+3)(\gamma_{3}^{2}-1)$.
Hence $\gamma_{3}=\varepsilon$, with $\varepsilon^{2}=1$, and a
straightforward calculation shows that the Bach tensor vanishes. Thus, the
left-invariant metric corresponds to

(6) $[e_{1},e_{2}]=e_{3},\qquad[e_{1},e_{4}]=\gamma_{1}e_{1}+\varepsilon
e_{3},\qquad[e_{3},e_{4}]=\gamma_{1}e_{3},$

which is not locally symmetric, locally conformally flat nor Einstein.
Moreover, $(e_{1},e_{2},e_{3},e_{4})\mapsto(-e_{1},-e_{2},e_{3},-e_{4})$ and
$(e_{1},e_{2},e_{3},e_{4})\mapsto(e_{1},e_{2},e_{3},-e_{4})$ define isometries
interchanging $(\varepsilon,\gamma_{1})$ with $(\varepsilon,-\gamma_{1})$ and
$(\varepsilon,\gamma_{1})$ with $(-\varepsilon,-\gamma_{1})$, respectively,
and hence we can restrict the parameters to $\varepsilon=1$ and
$\gamma_{1}>0$. This corresponds to the metric in Theorem 1.2–(R).

###  3.2. Case $\boldsymbol{\gamma_{5}\neq 0}$

Taking the orthogonal basis $\hat{e}_{i}=\frac{1}{\gamma_{5}}e_{i}$ we can
assume $\gamma_{5}=1$ continuing in the homothetic class of the initial
metric. We fix the polynomial ring $\mathbb{R}[\gamma_{2}$, $\gamma_{3}$,
$\gamma_{4}$, $\gamma_{6}$, $\gamma_{1}$, $\lambda]$. Considering the ideal
$\langle\overline{\mathfrak{B}}_{ij}\rangle$, where
$\overline{\mathfrak{B}}_{ij}$ are the components of the Bach tensor where we
have simplified the parameter $\lambda\neq 0$ whenever possible, and computing
a Gröbner basis with respect to the graded reverse lexicographical order
(consisting of 211 polynomials) we get that the polynomial

$\mathbf{g}_{1}=((12\gamma_{2}+5\gamma_{4})\gamma_{3}^{2}-(5\gamma_{2}+12\gamma_{4})\gamma_{6}^{2}-17(\gamma_{1}-1)\gamma_{3}\gamma_{6})(\lambda^{2}+(\gamma_{1}+1)^{2})\lambda^{10}$

belongs to the ideal $\langle\bar{\mathfrak{B}}_{ij}\rangle$. Hence, since
$\lambda\neq 0$, it follows that the polynomial

(7)
$\mathbf{\tilde{g}}_{1}=(12\gamma_{2}+5\gamma_{4})\gamma_{3}^{2}-(5\gamma_{2}+12\gamma_{4})\gamma_{6}^{2}-17(\gamma_{1}-1)\gamma_{3}\gamma_{6}$

must vanish. Now, for the ideal
$\langle\overline{\mathfrak{B}}_{ij}\cup\\{\mathbf{\tilde{g}}_{1}\\}\rangle$,
we consider the lexicographical order and obtain, after computing a Gröbner
basis, that

$\begin{array}[]{l}\mathbf{g}_{2}=-\gamma_{6}(\lambda^{2}-(\gamma_{1}+1)\gamma_{6}^{2})(\lambda^{2}+(\gamma_{1}+1)^{2})(\lambda^{2}+25(\gamma_{1}+1)^{2})(2\lambda^{2}+(\gamma_{1}+1)^{2})\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\mathbf{g_{2}}=}\times(4\lambda^{2}+(\gamma_{1}+1)^{2})(5\lambda^{2}+8(\gamma_{1}+1)^{2})(25\lambda^{2}+16(\gamma_{1}+1)^{2})\end{array}$

belongs to the ideal. Thus, either $\gamma_{6}=0$ or $\gamma_{6}\neq 0$ and
$\lambda^{2}-(\gamma_{1}+1)\gamma_{6}^{2}=0$. Next we examine these two cases
separately.

####  3.2.1. Case $\gamma_{6}=0$

In this case we show that there is no Bach-flat metrics. We make use of
Gröbner bases considering the lexicographical order in all the cases. First,
we consider the polynomial ring $\mathbb{R}[\gamma_{2}$, $\gamma_{3}$,
$\gamma_{4}$, $\gamma_{1}$, $\lambda]$. Taking the ideal
$\langle\overline{\mathfrak{B}}_{ij}\rangle$ we get, after computing a Gröbner
basis, that

$\mathbf{g}^{\prime}_{1}=\gamma_{3}\lambda^{6}(\lambda^{2}+4)(5\lambda^{2}+2)(16\lambda^{2}+7225)(32\lambda^{2}+9)(256\lambda^{2}+169)$

belongs to $\langle\bar{\mathfrak{B}}_{ij}\rangle$, which implies
$\gamma_{3}=0$. Adding this new condition, we repeat the process in the
polynomial ring $\mathbb{R}[\gamma_{2}$, $\gamma_{4}$, $\gamma_{1}$,
$\lambda]$ to obtain the polynomial

$\begin{array}[]{l}\mathbf{g}^{\prime}_{2}=-\lambda^{2}(7\lambda^{2}-2(\gamma_{1}+1)^{2})(\lambda^{2}+(\gamma_{1}+1)^{2})(\lambda^{2}+2(\gamma_{1}+1)^{2})\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\mathbf{g}^{\prime}_{2}=}\times(4\lambda^{2}+(\gamma_{1}+1)^{2})(5\lambda^{2}+8(\gamma_{1}+1)^{2}).\end{array}$

Thus, $\mathbf{\tilde{g}}^{\prime}_{2}=7\lambda^{2}-2(\gamma_{1}+1)^{2}$ must
vanish. Finally, we consider the polynomials in the ring
$\mathbb{R}[\gamma_{1}$, $\lambda$, $\gamma_{2}$, $\gamma_{4}]$ and compute a
last Gröbner basis for the ideal
$\langle\overline{\mathfrak{B}}_{ij}\cup\\{\mathbf{\tilde{g}}^{\prime}_{2}\\}\rangle$.
As a result, we get that

$\begin{array}[]{l}\mathbf{g}^{\prime}_{3}=(\gamma_{4}^{2}+1)^{4}(441\gamma_{2}^{4}+2\gamma_{2}^{2}(641\gamma_{4}^{2}+770)+(7\gamma_{4}^{2}+22)(63\gamma_{4}^{2}+22)\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\mathbf{g}^{\prime}_{3}=(\gamma_{4}^{2}+1)^{4}}-24\gamma_{2}\gamma_{4}(35\gamma_{2}^{2}+35\gamma_{4}^{2}+88))\end{array}$

belongs to the ideal. Since $\mathbf{g}^{\prime}_{3}$ does not vanish, we
conclude that there is no Bach-flat metrics in this case.

####  3.2.2. Case $\gamma_{6}\neq 0$,
$\lambda^{2}-(\gamma_{1}+1)\gamma_{6}^{2}=0$

To solve this case we work in the polynomial ring $\mathbb{R}[\gamma_{1}$,
$\gamma_{6}$, $\gamma_{4}$, $\gamma_{2}$, $\gamma_{3}$, $\lambda]$ and use the
lexicographical order for computing Gröbner bases. We start with the ideal
$\langle\overline{\mathfrak{B}}_{ij}\cup\\{\lambda^{2}-(\gamma_{1}+1)\gamma_{6}^{2}\\}\rangle$
and get that

$\begin{array}[]{l}\mathbf{g}^{\prime\prime}_{1}=-\lambda^{6}(\lambda^{2}+1)(4\lambda^{2}+9)(25\lambda^{2}+4)(45\lambda^{2}+98)(225\lambda^{2}+256)(256\lambda^{2}+1369)\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\mathbf{g}^{\prime\prime}_{1}=}\times(968\lambda^{2}+1521)(2304\lambda^{4}+46441425\lambda^{2}+1028805625)\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\mathbf{g}^{\prime\prime}_{1}=}\times(3211264\lambda^{4}+34731953\lambda^{2}+35796289)(\gamma_{2}^{2}\lambda^{2}-(\gamma_{2}^{2}+1)\gamma_{3}^{2})\end{array}$

belongs to the ideal. Since $\lambda\neq 0$, only the last factor,
$\mathbf{\tilde{g}}^{\prime\prime}_{1}=\gamma_{2}^{2}\lambda^{2}-(\gamma_{2}^{2}+1)\gamma_{3}^{2}$,
may vanish. Finally, we use again the polynomial $\mathbf{\tilde{g}}_{1}$
given by Equation (7) and considering the ideal
$\langle\overline{\mathfrak{B}}_{ij}\cup\\{\lambda^{2}-(\gamma_{1}+1)\gamma_{6}^{2}$,
$\mathbf{\tilde{g}}^{\prime\prime}_{1}$, $\mathbf{\tilde{g}}_{1}\\}\rangle$ we
obtain that the polynomials

$\mathbf{g}^{\prime\prime}_{2}=(\gamma_{4}-\gamma_{2})\lambda^{4},\quad\mathbf{g}^{\prime\prime}_{3}=(\gamma_{2}\gamma_{6}-\gamma_{3})\lambda^{4}\quad\text{and}\quad\mathbf{g}^{\prime\prime}_{4}=(79\gamma_{1}+11\gamma_{2}^{2}-90\gamma_{2}\gamma_{4})\lambda^{2}$

belong to the ideal. These expressions, together with
$\lambda^{2}-(\gamma_{1}+1)\gamma_{6}^{2}=0$, imply

$\gamma_{4}=\gamma_{2},\quad\gamma_{3}=\gamma_{2}\gamma_{6},\quad\gamma_{1}=\gamma_{2}^{2},\quad\lambda=\varepsilon\gamma_{6}\sqrt{\gamma_{2}^{2}+1},$

where $\varepsilon^{2}=1$, and a straightforward calculation shows that, under
these conditions, the Bach tensor vanishes. Hence, the left-invariant metric
is determined by

$\begin{array}[]{ll}[e_{1},e_{2}]=\varepsilon\gamma_{6}\sqrt{\gamma_{2}^{2}+1}e_{3},&[e_{1},e_{4}]=\gamma_{2}^{2}e_{1}+\gamma_{2}e_{2}+\gamma_{2}\gamma_{6}e_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[e_{2},e_{4}]=\gamma_{2}e_{1}+e_{2}+\gamma_{6}e_{3},&[e_{3},e_{4}]=(\gamma_{2}^{2}+1)e_{3}.\end{array}$

Let $\gamma_{6}^{\prime}=\frac{\sqrt{\gamma_{2}^{2}+1}}{\gamma_{6}}$ and
consider a new basis $\\{\tilde{e}_{i}\\}$ defined by

$\begin{array}[]{ll}\tilde{e}_{1}=\frac{\gamma_{6}^{\prime}}{(\gamma_{2}^{2}+1)^{\frac{3}{2}}}\left(\gamma_{2}e_{1}+e_{2}\right),&\quad\tilde{e}_{2}=-\frac{\gamma_{6}^{\prime}}{(\gamma_{2}^{2}+1)^{\frac{3}{2}}}\left(e_{1}-\gamma_{2}e_{2}\right),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\tilde{e}_{3}=\frac{\varepsilon\gamma_{6}^{\prime}}{\gamma_{2}^{2}+1}e_{3},&\quad\tilde{e}_{4}=\frac{\gamma_{6}^{\prime}}{\gamma_{2}^{2}+1}e_{4}.\end{array}$

Now, a direct calculation shows that the Lie bracket transforms into

$[\tilde{e}_{1},\tilde{e}_{2}]=\tilde{e}_{3},\qquad[\tilde{e}_{1},\tilde{e}_{4}]=\gamma_{6}^{\prime}\tilde{e}_{1}+\varepsilon\tilde{e}_{3},\qquad[\tilde{e}_{3},\tilde{e}_{4}]=\gamma_{6}^{\prime}\tilde{e}_{3},$

while the inner product becomes
$\left(\frac{\gamma_{6}^{\prime}}{\gamma_{2}^{2}+1}\right)^{2}\langle\cdot,\cdot\rangle$.
Since we are working at the homothetic level, we can maintain the initial
inner product remaining in the same homothetic class. Thus, we recover the
case in $\S$3.1.3 (see Equation (6)) and therefore the above metric
corresponds to the situation given by Theorem 1.2-(R).

##  4\. Semi-direct extensions with Lorentzian normal subgroup $H_{3}$

In this section we deal with left-invariant Lorentzian metrics whose
restriction to the three-dimensional unimodular Lie group $H_{3}$ is of
Lorentzian signature. Left-invariant Lorentzian metrics on the Heisenberg
group were described in [31] by using Milnor type frames. Let $L(X\times
Y)=[X,Y]$, where “$\times$” denotes the vector-cross product $\langle X\times
Y,Z\rangle=\operatorname{det}(X,Y,Z)$, be the structure tensor. Self-duality
of $L$ holds true in the unimodular case as well as in the Riemannian case,
but due to the Lorentzian signature, $L$ may have non-trivial Jordan normal
form. Since $L$ must have eigenvalues $\\{0,0,\lambda_{3}\\}$ one has that the
only possible Jordan normal forms are as follows:

  1. Ia.

$L$ is real diagonalizable. Hence there exists an orthonormal basis
$\left\\{e_{1},e_{2},e_{3}\right\\}$, where we assume $e_{3}$ to be timelike,
so that $L(e_{i})=\lambda_{i}e_{i}$ with $\lambda_{1}=\lambda_{2}=0$.

  2. II.

$L$ has a double root of its minimal polynomial. Then it is two-step nilpotent
and there exists a pseudo-orthonormal basis
$\left\\{u_{1},u_{2},u_{3}\right\\}$ so that

$L=\begin{pmatrix}0&0&0\\\ \pm 1&0&0\\\
0&0&0\end{pmatrix},\quad\text{where}\quad\langle u_{1},u_{2}\rangle=\langle
u_{3},u_{3}\rangle=1\,.$

Therefore inner products on $\mathfrak{h}_{3}\rtimes\mathfrak{r}$ are
described as in Section 3 by considering their restriction to the subalgebra
$\mathfrak{h}_{3}$, which is of type Ia (distinguishing the two possibilities
whether the kernel of the structure operator is positive definite or
Lorentzian) or two-step nilpotent as in II. Next we analyze the vanishing of
the Bach tensor in those three cases separately.

###  4.1. The structure operator is diagonalizable of rank one with positive
definite kernel

In this case, using (5) and proceeding as in Section 3, one gets that there
exists an orthonormal basis $\\{e_{1},e_{2},e_{3},e_{4}\\}$ of
$\mathfrak{g}=\mathfrak{h}_{3}\rtimes\mathfrak{r}$, with $e_{3}$ timelike,
where $\mathfrak{h}_{3}=\operatorname{span}\\{e_{1},e_{2},e_{3}\\}$ and
$\mathfrak{r}=\operatorname{span}\\{e_{4}\\}$, so that

$\begin{array}[]{ll}[e_{1},e_{2}]=-\lambda
e_{3},&[e_{1},e_{4}]=\gamma_{1}e_{1}+\gamma_{2}e_{2}+\gamma_{3}e_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[e_{2},e_{4}]=\gamma_{4}e_{1}+\gamma_{5}e_{2}+\gamma_{6}e_{3},&[e_{3},e_{4}]=(\gamma_{1}+\gamma_{5})e_{3},\end{array}$

where $\lambda\neq 0$ and $\gamma_{1}$, $\dots$, $\gamma_{6}\in\mathbb{R}$.
The analysis of this case is analogous to the one carried out in Section 3.
However, the different behavior of the restriction of the inner product to the
subalgebra $\mathfrak{h}_{3}$ (Lorentzian instead of Riemannian) is crucial
and leads to the non-existence of Bach-flat metrics. Next we schematize the
process distinguishing the cases $\gamma_{5}=0$ and $\gamma_{5}\neq 0$, but
omitting the details which coincide with the study developed in Section 3.

####  4.1.1. Case $\gamma_{5}=0$

Since $\lambda\neq 0$, we can assume $\lambda=1$ and work in the homothetic
class of the initial metric. Proceeding as in Section 3.1 we consider the
ideal generated by the polynomials
$\mathfrak{B}_{ij}\in\mathbb{R}[\gamma_{2},\gamma_{3},\gamma_{4},\gamma_{6},\gamma_{1}]$
and compute a Gröbner basis (consisting of 50 polynomials) with respect to the
lexicographical order to get that

$\mathbf{g}=\gamma_{1}^{4}\gamma_{6}^{2}(\gamma_{1}^{2}+1)(\gamma_{1}^{2}+2)(\gamma_{1}^{2}+4)(8\gamma_{1}^{2}+5)(16\gamma_{1}^{2}+25)(25\gamma_{1}^{2}+1)$

belongs to the ideal $\langle\mathfrak{B}_{ij}\rangle$. Hence, either
$\gamma_{1}=0$ or $\gamma_{1}\neq 0$, $\gamma_{6}=0$. If $\gamma_{1}=0$, then

$\mathbf{g}^{\prime}_{1}=\gamma_{2}^{2}+8\gamma_{3}^{2}+\gamma_{4}^{2}+8\gamma_{6}^{2}+8$

is a polynomial in $\langle\mathfrak{B}_{ij}\rangle$ and therefore there is no
Bach-flat metric in this case. Moreover, if $\gamma_{6}=0$, we obtain the
polynomials

$\mathbf{g}^{\prime\prime}_{1}=\gamma_{3}\gamma_{4}^{4}\quad\text{and}\quad\mathbf{g}^{\prime\prime}_{2}=\gamma_{1}\gamma_{2}\gamma_{3}(\gamma_{1}^{2}+1).$

Now, if $\gamma_{3}=0$, proceeding as in Section 3.1.2 we get

$\begin{array}[]{l}12(\mathfrak{B}_{33}+3\,\mathfrak{B}_{44})=(\gamma_{2}+\gamma_{4})^{2}(20(\gamma_{2}+\gamma_{4})^{2}-60\gamma_{2}\gamma_{4}+3)\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{-12(\mathfrak{B}_{33}+3\,\mathfrak{B}_{44})=}+\gamma_{1}^{2}(19(\gamma_{2}+\gamma_{4})^{2}-60\gamma_{2}\gamma_{4}+8)+4,\end{array}$

while if $\gamma_{3}\neq 0$ then necessarily $\gamma_{2}=\gamma_{4}=0$ and a
direct calculation shows that
$\mathfrak{B}_{11}=\frac{1}{6}(\gamma_{3}^{2}+1)(5\gamma_{1}^{2}+3\gamma_{3}^{2}+3)$.
Thus, in any case, the Bach tensor does not vanish.

####  4.1.2. Case $\gamma_{5}\neq 0$

Taking the orthogonal basis $\hat{e}_{i}=\frac{1}{\gamma_{5}}e_{i}$ we can
assume $\gamma_{5}=~{}1$. Consider the ideal generated by the polynomials
$\bar{\mathfrak{B}}_{ij}\in\mathbb{R}[\gamma_{2},\gamma_{3},\gamma_{4},\gamma_{6},\gamma_{1},\lambda]$,
where $\bar{\mathfrak{B}}_{ij}$ are obtained from the components
$\mathfrak{B}_{ij}$ after simplifying $\lambda\neq 0$ whenever possible. We
proceed as in Section 3.2 to construct a Gröbner basis (which consists of 211
polynomials) with respect to the graded reverse lexicographical order and get
that

$\mathbf{g}_{1}=((12\gamma_{2}+5\gamma_{4})\gamma_{3}^{2}-(5\gamma_{2}+12\gamma_{4})\gamma_{6}^{2}-17(\gamma_{1}-1)\gamma_{3}\gamma_{6})(\lambda^{2}+(\gamma_{1}+1)^{2})\lambda^{10}$

is a polynomial in the ideal $\langle\bar{\mathfrak{B}}_{ij}\rangle$.
Therefore the polynomial

(8)
$\mathbf{\tilde{g}}_{1}=(12\gamma_{2}+5\gamma_{4})\gamma_{3}^{2}-(5\gamma_{2}+12\gamma_{4})\gamma_{6}^{2}-17(\gamma_{1}-1)\gamma_{3}\gamma_{6}$

must vanish. Now, considering the ideal
$\langle\bar{\mathfrak{B}}_{ij}\cup\\{\mathbf{\tilde{g}}_{1}\\}\rangle$ and
the lexicographical order, we obtain that the polynomial

$\begin{array}[]{l}\mathbf{g}_{2}=\gamma_{6}(\lambda^{2}+(\gamma_{1}+1)\gamma_{6}^{2})(\lambda^{2}+(\gamma_{1}+1)^{2})(\lambda^{2}+25(\gamma_{1}+1)^{2})(2\lambda^{2}+(\gamma_{1}+1)^{2})\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\mathbf{g_{2}}=}\times(4\lambda^{2}+(\gamma_{1}+1)^{2})(5\lambda^{2}+8(\gamma_{1}+1)^{2})(25\lambda^{2}+16(\gamma_{1}+1)^{2})\end{array}$

belongs to the ideal. Thus, either $\gamma_{6}=0$ or $\gamma_{6}\neq 0$,
$\lambda^{2}+(\gamma_{1}+1)\gamma_{6}^{2}=0$. If $\gamma_{6}=0$, proceeding as
in Section 3.2.1 we get exactly the same polynomials
$\mathbf{g}^{\prime}_{1}$, $\mathbf{g}^{\prime}_{2}$ and
$\mathbf{g}^{\prime}_{3}$, so we conclude that there is no Bach-flat metric in
this case. Now, if $\gamma_{6}\neq 0$ and
$\lambda^{2}+(\gamma_{1}+1)\gamma_{6}^{2}=0$, we proceed as in Section 3.2.2
to get the polynomial

$\begin{array}[]{l}\mathbf{g}^{\prime\prime}_{1}=\lambda^{6}(\lambda^{2}+1)(4\lambda^{2}+9)(25\lambda^{2}+4)(45\lambda^{2}+98)(225\lambda^{2}+256)(256\lambda^{2}+1369)\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\mathbf{g}^{\prime\prime}_{1}=}\times(968\lambda^{2}+1521)(2304\lambda^{4}+46441425\lambda^{2}+1028805625)\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\mathbf{g}^{\prime\prime}_{1}=}\times(3211264\lambda^{4}+34731953\lambda^{2}+35796289)(\gamma_{2}^{2}\lambda^{2}+(\gamma_{2}^{2}+1)\gamma_{3}^{2})\end{array}$

in the ideal
$\langle\bar{\mathfrak{B}}_{ij}\cup\\{\lambda^{2}+(\gamma_{1}+1)\gamma_{6}^{2}\\}\rangle$.
Hence,
$\mathbf{\tilde{g}}^{\prime\prime}_{1}=\gamma_{2}^{2}\lambda^{2}+(\gamma_{2}^{2}+1)\gamma_{3}^{2}$
must vanish. Thus, $\gamma_{2}=\gamma_{3}=0$ and the polynomial
$\mathbf{\tilde{g}}_{1}$ given in Equation (8) reduces to
$\mathbf{\tilde{g}}_{1}=-12\gamma_{4}\gamma_{6}^{2}$, which implies
$\gamma_{4}=0$ since $\gamma_{6}\neq 0$. Finally, using
$\gamma_{2}=\gamma_{3}=\gamma_{4}=0$ and
$\lambda^{2}+(\gamma_{1}+1)\gamma_{6}^{2}=0$, a straightforward calculation
shows that

$\begin{array}[]{l}\frac{24\gamma_{6}^{6}}{\lambda^{2}+\gamma_{6}^{2}}\,\mathfrak{B}_{11}=4(5\gamma_{6}^{2}+4)\lambda^{4}+3(4\gamma_{6}^{4}-3\gamma_{6}^{2}+16)\gamma_{6}^{2}\lambda^{2}-(4\gamma_{6}^{4}-11\gamma_{6}^{2}+16)\gamma_{6}^{4},\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\frac{24\gamma_{6}^{6}}{\lambda^{2}+\gamma_{6}^{2}}\,\mathfrak{B}_{44}=-4(\gamma_{6}^{2}-4)\lambda^{4}-(4\gamma_{6}^{4}+5\gamma_{6}^{2}-48)\gamma_{6}^{2}\lambda^{2}+(12\gamma_{6}^{4}-33\gamma_{6}^{2}+48)\gamma_{6}^{4},\end{array}$

which implies

$\begin{array}[]{l}\frac{3\gamma_{6}^{6}}{\lambda^{2}(\lambda^{2}+\gamma_{6}^{2})}(3\,\mathfrak{B}_{11}+\mathfrak{B}_{44})=(7\gamma_{6}^{2}+8)\lambda^{2}+4(\gamma_{6}^{4}-\gamma_{6}^{2}+6)\gamma_{6}^{2}.\end{array}$

Since $\lambda\gamma_{6}\neq 0$, we conclude that the Bach tensor does not
vanish.

###  4.2. The structure operator is diagonalizable of rank one with Lorentzian
kernel

In this setting, it is possible to choose an orthonormal basis
$\\{e_{1},e_{2},e_{3},e_{4}\\}$ of
$\mathfrak{g}=\mathfrak{h}_{3}\rtimes\mathfrak{r}$, with $e_{3}$ timelike,
where $\mathfrak{h}_{3}=\operatorname{span}\\{e_{1},e_{2},e_{3}\\}$ and
$\mathfrak{r}=\operatorname{span}\\{e_{4}\\}$, so that the left-invariant
metrics are described by

$\begin{array}[]{ll}[e_{1},e_{3}]=-\lambda
e_{2},&[e_{1},e_{4}]=\gamma_{1}e_{1}+\gamma_{2}e_{2}+\gamma_{3}e_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[e_{2},e_{4}]=\gamma_{4}e_{2},&[e_{3},e_{4}]=\gamma_{5}e_{1}+\gamma_{6}e_{2}-(\gamma_{1}-\gamma_{4})e_{3},\end{array}$

where $\lambda\neq 0$ and $\gamma_{1}$, $\dots$, $\gamma_{6}\in\mathbb{R}$. In
this case we study the vanishing of the Bach tensor analyzing the cases
$\gamma_{4}=0$ and $\gamma_{4}\neq 0$ separately.

####  4.2.1. Case $\gamma_{4}=0$

Since $\lambda\neq 0$, we consider the orthogonal basis
$\hat{e}_{i}=\frac{1}{\lambda}e_{i}$ so that we can assume $\lambda=1$ working
in the homothetic class of the initial metric. In this case, we compute a
Gröbner basis of the ideal
$\langle\mathfrak{B}_{ij}\rangle\subset\mathbb{R}[\gamma_{2}$, $\gamma_{3}$,
$\gamma_{5}$, $\gamma_{6}$, $\gamma_{1}]$ with respect to the lexicographical
order (consisting of 25 polynomials) and get that the polynomials

$\mathbf{g}_{1}=\gamma_{1}^{4},\quad\mathbf{g}_{2}=\gamma_{3}^{4}\quad\text{and}\quad\mathbf{g}_{3}=\gamma_{5}^{4}$

belong to the ideal. Thus, necessarily $\gamma_{1}=\gamma_{3}=\gamma_{5}=0$
and a direct calculation shows
$\mathfrak{B}_{22}=-\frac{5}{6}(\gamma_{6}^{2}-\gamma_{2}^{2}+1)^{2}$, which
implies $\gamma_{2}=\varepsilon\sqrt{\gamma_{6}^{2}+1}$, where
$\varepsilon^{2}=1$. Under this assumption the Bach tensor vanishes and the
corresponding left-invariant metric is given by

$[e_{1},e_{3}]=-e_{2},\qquad[e_{1},e_{4}]=\varepsilon\sqrt{\gamma_{6}^{2}+1}\,e_{2},\qquad[e_{3},e_{4}]=\gamma_{6}e_{2}.$

Moreover, the Ricci operator is $2$-step nilpotent and
$\xi=-\gamma_{6}e_{1}+\varepsilon\sqrt{\gamma_{6}^{2}+1}\,e_{3}+e_{4}$ is a
null parallel vector field so that $R(x,y)=0$ and $\nabla_{x}R=0$ for all
$x,y\in\xi^{\perp}=\operatorname{span}\\{\xi$, $e_{2}$,
$e_{1}+\gamma_{6}e_{4}\\}$. Thus, the underlying structure is a plane wave.
Considering the basis

$\tilde{e}_{1}=e_{1},\quad\tilde{e}_{2}=e_{2},\quad\tilde{e}_{3}=-e_{3},\quad\tilde{e}_{4}=-\gamma_{6}e_{1}+\varepsilon\sqrt{\gamma_{6}^{2}+1}\,e_{3}+e_{4},$

one has that the only non-zero bracket is
$[\tilde{e}_{1},\tilde{e}_{3}]=\tilde{e}_{2}$. Hence the underlying group
structure is the product $H_{3}\times\mathbb{R}$ and the metric is
isomorphically homothetic to the Bach-flat metric discussed in Remark 1.1.

####  4.2.2. Case $\gamma_{4}\neq 0$

Taking the orthogonal basis $\hat{e}_{i}=\frac{1}{\gamma_{4}}e_{i}$ we can
assume $\gamma_{4}=1$ working in the homothetic class of the initial metric.
We consider the ideal $\langle\mathfrak{B}_{ij}\rangle$ in the polynomial ring
$\mathbb{R}[\gamma_{2}$, $\gamma_{3}$, $\gamma_{5}$, $\gamma_{6}$,
$\gamma_{1}$, $\lambda]$ and use the lexicographical order to compute a
Gröbner basis. As a consequence, we get 61 polynomials, being two of them

$\begin{array}[]{l}\mathbf{g}_{1}=(\gamma_{3}+\gamma_{5})\gamma_{6}(\lambda^{2}+1)(\lambda^{2}+25)(4\lambda^{2}+1)\quad\text{and}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathbf{g}_{2}=(\gamma_{3}+\gamma_{5})(7\lambda^{2}-2)(\lambda^{2}+1)(\lambda^{2}+2)(\lambda^{2}+25)(4\lambda^{2}+1).\end{array}$

Hence, either $\gamma_{3}+\gamma_{5}=0$ or $\gamma_{3}+\gamma_{5}\neq 0$,
$\gamma_{6}=0$, $7\lambda^{2}-2=0$. Next we analyze these two cases by
separate.

#####  4.2.2.1 _Case $\gamma_{3}+\gamma_{5}=0$_

Continuing in the polynomial ring $\mathbb{R}[\gamma_{2}$, $\gamma_{3}$,
$\gamma_{5}$, $\gamma_{6}$, $\gamma_{1}$, $\lambda]$ with the lexicographical
order, for the ideal
$\langle\mathfrak{B}_{ij}\cup\\{\gamma_{3}+\gamma_{5}\\}\rangle$ we obtain the
polynomials

$\begin{array}[]{l}\mathbf{g}^{\prime}_{1}=-((\gamma_{1}-1)\lambda^{2}-\gamma_{6}^{2})(\lambda^{2}+1)(2\lambda^{2}+1)(4\lambda^{2}+1)(5\lambda^{2}+8)(25\lambda^{2}+16)\quad\text{and}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathbf{g}^{\prime}_{2}=((\gamma_{1}-1)\gamma_{2}-\gamma_{5}\gamma_{6})(2\lambda^{2}+1)(5\lambda^{2}+8)(25\lambda^{2}+16).\end{array}$

Hence,
$\mathbf{\tilde{g}}^{\prime}_{1}=(\gamma_{1}-1)\lambda^{2}-\gamma_{6}^{2}$ and
$\mathbf{\tilde{g}}^{\prime}_{2}=(\gamma_{1}-1)\gamma_{2}-\gamma_{5}\gamma_{6}$
must vanish. Repeating the process for the ideal
$\langle\mathfrak{B}_{ij}\cup\\{\gamma_{3}+\gamma_{5}$,
$\mathbf{\tilde{g}}^{\prime}_{1}$, $\mathbf{\tilde{g}}^{\prime}_{2}\\}\rangle$
we get the polynomials

$\begin{array}[]{l}\mathbf{g}^{\prime}_{3}=(\gamma_{5}^{2}-\gamma_{1}(\gamma_{1}-1))(5\lambda^{2}+8),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathbf{g}^{\prime}_{4}=8(\gamma_{5}^{2}-\gamma_{1}(\gamma_{1}-1))+\gamma_{2}^{2}-\gamma_{1}\lambda^{2}\quad\text{and}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathbf{g}^{\prime}_{5}=8\gamma_{5}(\gamma_{5}^{2}-\gamma_{1}(\gamma_{1}-1))+\gamma_{2}\gamma_{6}-\gamma_{5}\lambda^{2}.\end{array}$

Thus, $\gamma_{2}^{2}-\gamma_{1}\lambda^{2}=0$ and
$\gamma_{2}\gamma_{6}-\gamma_{5}\lambda^{2}=0$, and these relations, together
with $\gamma_{3}+\gamma_{5}=0$ and
$(\gamma_{1}-1)\lambda^{2}-\gamma_{6}^{2}=0$, imply

$\gamma_{3}=-\gamma_{5},\quad\gamma_{1}=\tfrac{\lambda^{2}+\gamma_{6}^{2}}{\lambda^{2}},\quad\gamma_{2}=\varepsilon\sqrt{\lambda^{2}+\gamma_{6}^{2}},\quad\gamma_{5}=\tfrac{\gamma_{2}\gamma_{6}}{\lambda^{2}},$

where $\varepsilon^{2}=1$. A straightforward calculation shows that the Bach
tensor vanishes for the corresponding left-invariant metric, given by

$\begin{array}[]{ll}[e_{1},e_{3}]=-\lambda
e_{2},&[e_{1},e_{4}]=\frac{\lambda^{2}+\gamma_{6}^{2}}{\lambda^{2}}e_{1}+\varepsilon\sqrt{\lambda^{2}+\gamma_{6}^{2}}\,e_{2}-\frac{\varepsilon\gamma_{6}\sqrt{\lambda^{2}+\gamma_{6}^{2}}}{\lambda^{2}}e_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[e_{2},e_{4}]=e_{2},&[e_{3},e_{4}]=\frac{\varepsilon\gamma_{6}\sqrt{\lambda^{2}+\gamma_{6}^{2}}}{\lambda^{2}}e_{1}+\gamma_{6}e_{2}-\frac{\gamma_{6}^{2}}{\lambda^{2}}e_{3}.\end{array}$

Now, considering a new basis defined by

$\tilde{e}_{1}=\tfrac{\varepsilon\sqrt{\lambda^{2}+\gamma_{6}^{2}}}{\lambda}\,e_{1}-\tfrac{\gamma_{6}}{\lambda}e_{3},\quad\tilde{e}_{2}=e_{2},\quad\tilde{e}_{3}=-\tfrac{\gamma_{6}}{\lambda}e_{1}+\tfrac{\varepsilon\sqrt{\lambda^{2}+\gamma_{6}^{2}}}{\lambda}\,e_{3},\quad\tilde{e}_{4}=e_{4},$

the inner product remains invariant while the Lie bracket transforms into

$[\tilde{e}_{1},\tilde{e}_{3}]=-\lambda\tilde{e}_{2},\qquad[\tilde{e}_{1},\tilde{e}_{4}]=\tilde{e}_{1}+\lambda\tilde{e}_{2},\qquad[\tilde{e}_{2},\tilde{e}_{4}]=\tilde{e}_{2},$

and a straightforward calculation shows that this metric is not locally
symmetric, locally conformally flat or Einstein. Moreover,
$(\tilde{e}_{1},\tilde{e}_{2},\tilde{e}_{3},\tilde{e}_{4})\mapsto(\tilde{e}_{1},-\tilde{e}_{2},\tilde{e}_{3},\tilde{e}_{4})$
defines an isometry interchanging $\lambda$ and $-\lambda$, and therefore we
can restrict the parameter $\lambda$ to $\lambda>0$. This case corresponds to
the metrics in Theorem 1.2–(L.i).

#####  4.2.2.2 _Case $\gamma_{3}+\gamma_{5}\neq 0$, $\gamma_{6}=0$,
$7\lambda^{2}-2=0$_

For this setting we consider the ideal
$\langle\mathfrak{B}_{ij}\cup\\{7\lambda^{2}-2\\}\rangle$ in the polynomial
ring $\mathbb{R}[\gamma_{2}$, $\gamma_{3}$, $\gamma_{5}$, $\gamma_{1}$,
$\lambda]$. In this case, we use the graded reverse lexicographical order to
see that the polynomials

$\begin{array}[]{l}\mathbf{g}^{\prime\prime}_{1}=\gamma_{2}\gamma_{5},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathbf{g}^{\prime\prime}_{2}=\gamma_{2}\gamma_{3},\\\ \vskip 6.0pt
plus 2.0pt minus
2.0pt\cr\mathbf{g}^{\prime\prime}_{3}=8\gamma_{1}^{2}+21\gamma_{2}^{2}-8\gamma_{1}+8\gamma_{3}\gamma_{5}-6\quad\text{and}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathbf{g}^{\prime\prime}_{4}=-22\gamma_{1}^{2}+21\gamma_{3}^{2}+21\gamma_{5}^{2}+22\gamma_{1}+20\gamma_{3}\gamma_{5}\end{array}$

belong to the ideal. Since $\gamma_{3}+\gamma_{5}\neq 0$, clearly
$\gamma_{2}=0$. Now, $\mathbf{g}^{\prime\prime}_{3}=0$ and
$\mathbf{g}^{\prime\prime}_{4}=0$, together with $7\lambda^{2}-2=0$, imply

$\lambda=\varepsilon_{1}\sqrt{\tfrac{2}{7}},\quad\gamma_{5}=\varepsilon_{2}\sqrt{\tfrac{11}{14}}-\gamma_{3},\quad\gamma_{1}=\tfrac{1}{2}+\varepsilon_{3}\sqrt{\gamma_{3}^{2}+1-\varepsilon_{2}\sqrt{\tfrac{11}{14}}\,\gamma_{3}},$

where $\varepsilon_{1}^{2}=\varepsilon_{2}^{2}=\varepsilon_{3}^{2}=1$. A
straightforward calculation shows that the corresponding left-invariant
metric, given by

$\begin{array}[]{l}[e_{1},e_{3}]=-\varepsilon_{1}\sqrt{\frac{2}{7}}\,e_{2},\qquad[e_{1},e_{4}]=\left(\frac{1}{2}+\varepsilon_{3}\sqrt{\gamma_{3}^{2}+1-\varepsilon_{2}\sqrt{\frac{11}{14}}\,\gamma_{3}}\right)e_{1}+\gamma_{3}e_{3},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[e_{2},e_{4}]=e_{2},\quad\quad[e_{3},e_{4}]=\left(\varepsilon_{2}\sqrt{\frac{11}{14}}-\gamma_{3}\right)e_{1}+\left(\frac{1}{2}-\varepsilon_{3}\sqrt{\gamma_{3}^{2}+1-\varepsilon_{2}\sqrt{\frac{11}{14}}\,\gamma_{3}}\right)e_{3},\end{array}$

is Bach-flat.

Let $\gamma_{3}^{\prime}=\sqrt{\frac{14}{11}}\varepsilon_{2}\gamma_{3}$. A
crucial observation is that it is possible to reduce this case eliminating the
parameter $\gamma_{3}^{\prime}$. We proceed as follows. First, we consider a
new basis $\\{\tilde{e}_{i}\\}$ defined by

$\tilde{e}_{1}=2\sqrt{14}\,e_{1},\qquad\tilde{e}_{2}=2\sqrt{14}\,\varepsilon_{1}\varepsilon_{2}e_{2},\qquad\tilde{e}_{3}=2\sqrt{14}\,\varepsilon_{2}e_{3},\qquad\tilde{e}_{4}=2\sqrt{14}\,e_{4}.$

Then, the Lie bracket transforms into

$\begin{array}[]{l}[\tilde{e}_{1},\tilde{e}_{3}]=-4\tilde{e}_{2},\qquad[\tilde{e}_{2},\tilde{e}_{4}]=2\sqrt{14}\,\tilde{e}_{2},\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[\tilde{e}_{1},\tilde{e}_{4}]=\left(\sqrt{14}+2\varepsilon_{3}\sqrt{11\gamma_{3}^{\prime}(\gamma_{3}^{\prime}-1)+14}\right)\tilde{e}_{1}+2\sqrt{11}\gamma_{3}^{\prime}\tilde{e}_{3},\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[\tilde{e}_{3},\tilde{e}_{4}]=-2\sqrt{11}(\gamma_{3}^{\prime}-1)\tilde{e}_{1}+\left(\sqrt{14}-2\varepsilon_{3}\sqrt{11\gamma_{3}^{\prime}(\gamma_{3}^{\prime}-1)+14}\right)\tilde{e}_{3},\end{array}$

while the inner product is given by $56\langle\cdot,\cdot\rangle$. Since we
are working at the homothetic level, we can maintain the initial inner product
$\langle\cdot,\cdot\rangle$ remaining in the same homothetic class.

Secondly, we make another change of basis taking $\\{f_{i}\\}$ given by

$f_{1}=\tilde{e}_{4},\qquad
f_{2}=\tfrac{\tilde{f}_{2}}{\|\tilde{f}_{2}\|},\qquad
f_{3}=\tfrac{\tilde{f}_{3}}{\|\tilde{f}_{3}\|},\qquad f_{4}=\tilde{e}_{2},$

where

$\tilde{f}_{2}=\left\\{\begin{array}[]{ll}-2\sqrt{\frac{14}{11}}\,\tilde{e}_{1}+\tilde{e}_{3}&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(1,0)$},\par\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr\tilde{e}_{1}&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(1,1)$},\par\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr\tilde{e}_{3}&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(-1,0)$},\par\\\ \vskip 6.0pt plus
2.0pt minus
2.0pt\cr-\frac{1}{2}\sqrt{\frac{11}{14}}\,\tilde{e}_{1}+\tilde{e}_{3}&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(-1,1)$},\par\\\ \vskip 6.0pt plus
2.0pt minus
2.0pt\cr\tfrac{\sqrt{14}+\varepsilon_{3}\sqrt{11\gamma_{3}^{\prime}(\gamma_{3}^{\prime}-1)+14}}{\sqrt{11}(\gamma_{3}^{\prime}-1)}\,\tilde{e}_{1}+\tilde{e}_{3}&\text{if
$\gamma_{3}^{\prime}\notin\\{0,1\\}$,}\end{array}\right.$

and

$\tilde{f}_{3}=\left\\{\begin{array}[]{ll}-\frac{1}{2}\sqrt{\frac{11}{14}}\,\tilde{e}_{1}+\tilde{e}_{3}&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(1,0)$},\par\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr\tilde{e}_{3}&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(1,1)$},\par\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr\tilde{e}_{1}&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(-1,0)$},\par\\\ \vskip 6.0pt plus
2.0pt minus
2.0pt\cr-2\sqrt{\frac{14}{11}}\,\tilde{e}_{1}+\tilde{e}_{3}&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(-1,1)$},\par\\\ \vskip 6.0pt plus
2.0pt minus
2.0pt\cr-\tfrac{\sqrt{14}-\varepsilon_{3}\sqrt{11\gamma_{3}^{\prime}(\gamma_{3}^{\prime}-1)+14}}{\sqrt{11}\,\gamma_{3}^{\prime}}\,\tilde{e}_{1}+\tilde{e}_{3}&\text{if
$\gamma_{3}^{\prime}\notin\\{0,1\\}$.}\end{array}\right.$

A long but straightforward calculation shows that the inner product transforms
into $\operatorname{diag}[1,\varepsilon_{3},-\varepsilon_{3},1]$, while the
Lie bracket is given by

$\begin{array}[]{ll}[f_{1},f_{2}]=-3\sqrt{14}f_{2}+2\delta_{1}\sqrt{11}f_{3},&\qquad[f_{1},f_{3}]=\sqrt{14}f_{3},\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[f_{1},f_{4}]=-2\sqrt{14}f_{4},&\qquad[f_{2},f_{3}]=4\delta_{2}f_{4},\end{array}$

where

$(\delta_{1},\delta_{2})=\left\\{\begin{array}[]{ll}(1,1)&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(1,0)$},\par\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr(-1,-1)&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(1,1)$},\par\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr(-1,1)&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(-1,0)$},\par\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr(1,-1)&\text{if
$(\varepsilon_{3},\gamma_{3}^{\prime})=(-1,1)$},\par\\\ \vskip 6.0pt plus
2.0pt minus
2.0pt\cr(\operatorname{sign}(1-\gamma_{3}^{\prime}),\operatorname{sign}(1-\gamma_{3}^{\prime}))&\text{if
$\varepsilon_{3}=1$ and $\gamma_{3}^{\prime}\notin\\{0,1\\}$},\par\par\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr(\operatorname{sign}(\gamma_{3}^{\prime}),-\operatorname{sign}(\gamma_{3}^{\prime}))&\text{if
$\varepsilon_{3}=-1$ and
$\gamma_{3}^{\prime}\notin\\{0,1\\}$},\end{array}\right.$

so we have $\delta_{1}^{2}=\delta_{2}^{2}=1$.

Now, a final calculation shows that the metric above is not locally symmetric,
locally conformally flat or Einstein. One may further assume
$\delta_{2}=~{}1$, just considering the isometry given by
$(f_{1},f_{2},f_{3},f_{4})\mapsto(f_{1},f_{2},\delta_{2}f_{3},f_{4})$. This
corresponds to the metric Theorem 1.7–(i).

###  4.3. The structure operator is $2$-step nilpotent

In this last case, there exists a pseudo-orthonormal basis
$\\{u_{1},u_{2},u_{3},u_{4}\\}$ of
$\mathfrak{g}=\mathfrak{h}_{3}\rtimes\mathfrak{r}$, with $\langle
u_{1},u_{2}\rangle=\langle u_{3},u_{3}\rangle=\langle u_{4},u_{4}\rangle=1$,
where $\mathfrak{h}_{3}=\operatorname{span}\\{u_{1},u_{2},u_{3}\\}$ and
$\mathfrak{r}=\operatorname{span}\\{u_{4}\\}$, so that

$\begin{array}[]{ll}[u_{1},u_{3}]=-\varepsilon
u_{2},&[u_{1},u_{4}]=\gamma_{1}u_{1}+\gamma_{2}u_{2}+\gamma_{3}u_{3},\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=\gamma_{4}u_{2},&[u_{3},u_{4}]=\gamma_{5}u_{1}+\gamma_{6}u_{2}-(\gamma_{1}-\gamma_{4})u_{3},\end{array}$

with $\varepsilon^{2}=1$ and $\gamma_{1}$, $\dots$, $\gamma_{6}\in\mathbb{R}$.
A direct calculation shows that
$\mathfrak{B}_{34}=-\frac{3}{4}\varepsilon\gamma_{4}\gamma_{5}^{2}$. We
analyze by separate the cases $\gamma_{4}=0$ and $\gamma_{4}\neq 0$,
$\gamma_{5}=0$.

####  4.3.1. Case $\gamma_{4}=0$

In this case, we have

$\mathfrak{B}_{44}=\tfrac{1}{8}(3\gamma_{1}^{2}+(\gamma_{3}+4\gamma_{6})\gamma_{5})^{2}.$

If $\gamma_{5}=0$ then necessarily $\gamma_{1}=0$ and a straightforward
calculation shows that the Bach-flatness condition is satisfied. Hence, the
corresponding left-invariant metric is given by

$[u_{1},u_{3}]=-\varepsilon
u_{2},\qquad[u_{1},u_{4}]=\gamma_{2}u_{2}+\gamma_{3}u_{3},\qquad[u_{3},u_{4}]=\gamma_{6}u_{2},$

and $u_{2}$ is a null parallel vector field so that the curvature tensor
satisfies $R(x,y)=0$ and $\nabla_{x}R=0$ for all $x,y\in u_{2}^{\perp}=\langle
u_{2}$, $u_{3}$, $u_{4}\rangle$. Moreover, the only non-zero component of the
Ricci tensor is $\rho_{11}=-\frac{1}{2}(\gamma_{3}^{2}-\gamma_{6}^{2})$ which
implies that the Ricci operator is isotropic and therefore the underlying
structure is a plane wave. Furthermore the metric is locally conformally flat
or locally symmetric if and only if $\gamma_{3}(\gamma_{3}+\gamma_{6})=0$.
This corresponds to Theorem 1.9-(L).

If $\gamma_{5}\neq 0$, then
$\gamma_{6}=-\frac{3\gamma_{1}^{2}+\gamma_{3}\gamma_{5}}{4\gamma_{5}}$ and the
only non-zero component of the Bach tensor is determined by

$\begin{array}[]{l}\frac{64\gamma_{5}^{2}}{3}\mathfrak{B}_{11}=16\gamma_{5}^{4}\gamma_{2}^{2}-8\gamma_{1}(3(\gamma_{1}^{2}+\gamma_{3}\gamma_{5})-2\gamma_{1}^{2})\gamma_{5}^{2}\gamma_{2}\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\frac{64\gamma_{5}^{2}}{3}\mathfrak{B}_{11}=}\par+(\gamma_{1}^{2}+\gamma_{3}\gamma_{5})(8(\gamma_{1}^{2}+\gamma_{3}\gamma_{5})^{2}+9(\gamma_{1}^{2}+\gamma_{3}\gamma_{5})\gamma_{1}^{2}-12\gamma_{1}^{4})+4(\gamma_{1}^{6}+4\gamma_{5}^{4}),\end{array}$

which is a polynomial of degree two in the variable $\gamma_{2}$ whose
discriminant reduces to
$\Delta=-512\gamma_{5}^{4}(2\gamma_{5}^{4}+(\gamma_{1}^{2}+\gamma_{3}\gamma_{5})^{3})$.
Hence, if $\gamma_{1}^{2}+\gamma_{3}\gamma_{5}\geq 0$ we have $\Delta<0$ and
the metric cannot be Bach-flat. Thus, necessarily
$\gamma_{1}^{2}+\gamma_{3}\gamma_{5}<0$ and, in particular,
$\gamma_{3}\gamma_{5}<0$.

Let $\gamma_{1}^{\prime}$ and $\gamma_{5}^{\prime}$ be new variables defined
by $\gamma_{1}^{\prime}=\frac{\gamma_{1}}{\gamma_{3}^{2}}$ and
$\gamma_{5}^{\prime}=\frac{\gamma_{5}}{\gamma_{3}^{3}}$ so that the condition
$\gamma_{1}^{2}+\gamma_{3}\gamma_{5}<0$ reads
$\gamma_{3}^{4}((\gamma_{1}^{\prime})^{2}+\gamma_{5}^{\prime})<0$. In order to
simplify the writing, let
$\kappa=-((\gamma_{1}^{\prime})^{2}+\gamma_{5}^{\prime})$, $\kappa>0$. We make
a change of basis and from now on we consider a new basis
$\\{\tilde{u}_{i}\\}$ defined by

$\tilde{u}_{1}=\tfrac{\varepsilon}{\gamma_{3}\sqrt{\kappa}}u_{1},\qquad\tilde{u}_{2}=\tfrac{\varepsilon}{\gamma_{3}^{3}\sqrt{\kappa}}u_{2},\qquad\tilde{u}_{3}=\tfrac{\varepsilon}{\gamma_{3}^{2}\sqrt{\kappa}}u_{3},\quad\tilde{u}_{4}=\tfrac{1}{\gamma_{3}^{2}\sqrt{\kappa}}u_{4}.$

Hence, the Lie bracket transforms into

(9)
$\begin{array}[]{l}[\tilde{u}_{1},\tilde{u}_{3}]=-\tfrac{1}{\sqrt{\kappa}}\tilde{u}_{2},\qquad[\tilde{u}_{1},\tilde{u}_{4}]=\tfrac{1}{\sqrt{\kappa}}(\gamma_{1}^{\prime}\tilde{u}_{1}+\gamma_{2}\tilde{u}_{2}+\tilde{u}_{3}),\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[\tilde{u}_{3},\tilde{u}_{4}]=\tfrac{1}{\sqrt{\kappa}}\left(\gamma_{5}^{\prime}\tilde{u}_{1}-\frac{3(\gamma_{1}^{\prime})^{2}+\gamma_{5}^{\prime}}{4\gamma_{5}^{\prime}}\tilde{u}_{2}-\gamma_{1}^{\prime}\tilde{u}_{3}\right),\par\end{array}$

while the inner product is given by
$\frac{1}{\gamma_{3}^{4}\kappa}\langle\cdot,\cdot\rangle$. Since we are
working at the homothetic level we can maintain the initial inner product
remaining in the same homothetic class. With respect to the new basis the Bach
tensor is determined by

$\begin{array}[]{l}\frac{64(\gamma_{5}^{\prime})^{2}\kappa^{2}}{3}\mathfrak{B}_{11}=16(\gamma_{5}^{\prime})^{4}\gamma_{2}^{2}+8\gamma_{1}^{\prime}(\gamma_{5}^{\prime})^{2}\left(2(\gamma_{1}^{\prime})^{2}+3\kappa\right)\gamma_{2}\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\phantom{\frac{64(\gamma_{5}^{\prime})^{2}\kappa^{2}}{3}\mathfrak{B}_{11}=}\par+(12(\gamma_{1}^{\prime})^{4}+9\kappa(\gamma_{1}^{\prime})^{2}-8\kappa^{2})\kappa+4((\gamma_{1}^{\prime})^{6}+4(\gamma_{5}^{\prime})^{4})\end{array}$

and analyzing the Ricci operator of the metric, given by

$\operatorname{Ric}=\left(\begin{array}[]{cccc}\frac{2\gamma_{5}^{\prime}+3\kappa}{8\kappa}&\frac{(\gamma_{5}^{\prime})^{2}}{2\kappa}&-\frac{\gamma_{1}^{\prime}\gamma_{5}^{\prime}}{2\kappa}&0\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr-\frac{4(8\gamma_{1}^{\prime}\gamma_{2}+3)(\gamma_{5}^{\prime})^{2}-12\gamma_{5}^{\prime}\kappa-9\kappa^{2}}{32(\gamma_{5}^{\prime})^{2}\kappa}&\frac{2\gamma_{5}^{\prime}+3\kappa}{8\kappa}&-\frac{2\gamma_{2}(\gamma_{5}^{\prime})^{2}+3\gamma_{1}^{\prime}\kappa}{4\gamma_{5}^{\prime}\kappa}&-\frac{\gamma_{5}^{\prime}}{2\kappa}\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr-\frac{2\gamma_{2}(\gamma_{5}^{\prime})^{2}+3\gamma_{1}^{\prime}\kappa}{4\gamma_{5}^{\prime}\kappa}&-\frac{\gamma_{1}^{\prime}\gamma_{5}^{\prime}}{2\kappa}&-\frac{2\gamma_{5}^{\prime}+3\kappa}{4\kappa}&0\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr-\frac{\gamma_{5}^{\prime}}{2\kappa}&0&0&\frac{3}{4}\par\end{array}\right),$

it follows that
$\mathfrak{B}_{11}=\frac{8\kappa}{(\gamma_{5}^{\prime})^{2}}\det(\operatorname{Ric})+\frac{(\gamma_{5}^{\prime})^{2}}{\kappa^{2}}$.
Hence, if $\gamma_{5}^{\prime}\neq 0$,
$\kappa=-((\gamma_{1}^{\prime})^{2}+\gamma_{5}^{\prime})>0$ and
$\det(\operatorname{Ric})=-\frac{(\gamma_{5}^{\prime})^{4}}{8\kappa^{3}}$,
which is equivalent to

$\gamma_{2}=\tfrac{1}{4(\gamma_{5}^{\prime})^{2}}\left(\gamma_{1}^{\prime}(2\gamma_{5}^{\prime}-\kappa)+2\varepsilon^{\prime}\sqrt{2\left(\kappa^{3}-2(\gamma_{5}^{\prime})^{4}\right)}\right)\,\,\text{with}\,\,\kappa^{3}-2(\gamma_{5}^{\prime})^{4}\geq
0,\,(\varepsilon^{\prime})^{2}=1,$

then the left-invariant metric given by Equation (9) is Bach-flat and a
straightforward calculation shows that it has positive scalar curvature
$\tau=\frac{3}{4}$ and it is not locally symmetric, locally conformally flat
or Einstein. Note that the conditions $\gamma_{5}^{\prime}\neq 0$,
$\kappa=-((\gamma_{1}^{\prime})^{2}+\gamma_{5}^{\prime})>0$ and
$\kappa^{3}-2(\gamma_{5}^{\prime})^{4}\geq 0$ are equivalent to
$-\frac{1}{2}\leq\gamma_{5}^{\prime}<0$,
$(\gamma_{1}^{\prime})^{2}\leq-\gamma_{5}^{\prime}\left(\sqrt[3]{2\gamma_{5}^{\prime}}+1\right)$.
These metrics correspond to those in Theorem 1.7–(ii).

####  4.3.2. Case $\gamma_{4}\neq 0$, $\gamma_{5}=0$

In this case,
$\mathfrak{B}_{44}=\frac{1}{24}(3\gamma_{1}-\gamma_{4})^{2}(\gamma_{1}-\gamma_{4})(3\gamma_{1}+\gamma_{4})$.
Hence, $\gamma_{4}=3\gamma_{1}$, $\gamma_{4}=\gamma_{1}$ or
$\gamma_{4}=-3\gamma_{1}$. If $\gamma_{4}=3\gamma_{1}$ then a direct
calculation shows that the Bach-flatness condition is equivalent to
$\gamma_{6}=-\gamma_{3}$ and the corresponding metric is locally conformally
flat. Moreover, the metric is both locally symmetric and Einstein if and only
if $\gamma_{2}=0$. Next we analyze by separate the other two cases,
$\gamma_{4}=\gamma_{1}$ and $\gamma_{4}=-3\gamma_{1}$.

#####  4.3.2.1 _Case $\gamma_{4}=\gamma_{1}$_

A direct calculation shows that the non-zero components of the Bach tensor are
determined by

$\mathfrak{B}_{11}=\tfrac{1}{4}\gamma_{1}^{2}(\gamma_{3}-7\gamma_{6})(\gamma_{3}-\gamma_{6})\quad\text{and}\quad\mathfrak{B}_{13}=\tfrac{3}{4}\gamma_{1}^{3}(\gamma_{3}-\gamma_{6}).$

Since $\gamma_{1}\neq 0$, it follows that $\gamma_{6}=\gamma_{3}$ and the
left-invariant Bach-flat metric is given by

$\begin{array}[]{ll}[u_{1},u_{3}]=-\varepsilon
u_{2},&[u_{1},u_{4}]=\gamma_{1}u_{1}+\gamma_{2}u_{2}+\gamma_{3}u_{3},\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=\gamma_{1}u_{2},&[u_{3},u_{4}]=\gamma_{3}u_{2}.\end{array}$

Let
$\gamma_{2}^{\prime}=\frac{\gamma_{1}\gamma_{2}+\gamma_{3}^{2}}{|\gamma_{1}|}\in\mathbb{R}$
and let $\nu=-1,1,$ depending on whether $\gamma_{1}$ is negative or positive.
Define a new basis $\\{\tilde{u}_{i}\\}$ by

$\begin{array}[]{ll}\tilde{u}_{1}=-\tfrac{2}{|\gamma_{1}|^{\frac{1}{2}}}u_{1}+\tfrac{\gamma_{3}^{2}}{|\gamma_{1}|^{\frac{5}{2}}}u_{2}-\tfrac{2\nu\gamma_{3}}{|\gamma_{1}|^{\frac{3}{2}}}u_{3},&\quad\tilde{u}_{2}=-\tfrac{8}{|\gamma_{1}|^{\frac{3}{2}}}u_{2},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\tilde{u}_{3}=-\tfrac{4\nu\varepsilon\gamma_{3}}{\gamma_{1}^{2}}u_{2}+\tfrac{4\varepsilon}{|\gamma_{1}|}u_{3},&\quad\tilde{u}_{4}=\tfrac{4}{\gamma_{1}}u_{4}.\end{array}$

Now, a direct calculation shows that the Lie bracket transforms into

$[\tilde{u}_{1},\tilde{u}_{3}]=-\tilde{u}_{2},\qquad[\tilde{u}_{1},\tilde{u}_{4}]=4\tilde{u}_{1}+\gamma_{2}^{\prime}\tilde{u}_{2},\qquad[\tilde{u}_{2},\tilde{u}_{4}]=4\tilde{u}_{2},$

while the inner product is given by
$\frac{16}{\gamma_{1}^{2}}\langle\cdot,\cdot\rangle$. Since we are working at
the homothetic level we can maintain the initial inner product remaining in
the same homothetic class. A straightforward calculation shows that the above
metrics are not locally symmetric, locally conformally flat or Einstein. They
correspond to the metrics in Theorem 1.2–(L.ii).

#####  4.3.2.2 _Case $\gamma_{4}=-3\gamma_{1}$_

The Bach tensor is determined by

$\mathfrak{B}_{11}=-\tfrac{1}{4}\gamma_{1}^{2}(7\gamma_{3}^{2}+41\gamma_{6}^{2}+32\gamma_{1}\gamma_{2}-24\gamma_{3}\gamma_{6})\,,\quad\quad\mathfrak{B}_{13}=\tfrac{7}{4}\gamma_{1}^{3}(\gamma_{3}-5\gamma_{6}),$

and since $\gamma_{1}\neq 0$ we get $\gamma_{3}=5\gamma_{6}$ and
$\gamma_{2}=-\frac{3\gamma_{6}^{2}}{\gamma_{1}}$. Thus, the Bach-flat left-
invariant metric corresponds to

$\begin{array}[]{ll}[u_{1},u_{3}]=-\varepsilon
u_{2},&[u_{1},u_{4}]=\gamma_{1}u_{1}-\frac{3\gamma_{6}^{2}}{\gamma_{1}}u_{2}+5\gamma_{6}u_{3},\par\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr[u_{2},u_{4}]=-3\gamma_{1}u_{2},&[u_{3},u_{4}]=\gamma_{6}u_{2}-4\gamma_{1}u_{3}.\end{array}$

Let $\nu=-1$, $1$, depending on whether $\gamma_{1}$ is negative or positive.
Considering the basis $\\{\tilde{u}_{i}\\}$ defined by

$\begin{array}[]{ll}\tilde{u}_{1}=-\tfrac{1}{|\gamma_{1}|^{\frac{3}{2}}}\left(\gamma_{1}u_{1}-\tfrac{\gamma_{6}^{2}}{2\gamma_{1}}u_{2}+\gamma_{6}u_{3}\right),&\quad\tilde{u}_{2}=-\tfrac{1}{\gamma_{1}|\gamma_{1}|^{\frac{1}{2}}}u_{2},\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\tilde{u}_{3}=-\tfrac{\nu\varepsilon\gamma_{6}}{\gamma_{1}^{2}}u_{2}+\tfrac{\varepsilon}{|\gamma_{1}|}u_{3},&\quad\tilde{u}_{4}=-\tfrac{\nu}{|\gamma_{1}|}u_{4},\end{array}$

a direct calculation shows that the Lie bracket transforms into

$[\tilde{u}_{1},\tilde{u}_{3}]=-\tilde{u}_{2},\qquad[\tilde{u}_{1},\tilde{u}_{4}]=-\tilde{u}_{1},\qquad[\tilde{u}_{2},\tilde{u}_{4}]=3\tilde{u}_{2},\qquad[\tilde{u}_{3},\tilde{u}_{4}]=4\tilde{u}_{3},$

while the inner product is given by
$\frac{1}{\gamma_{1}^{2}}\langle\cdot,\cdot\rangle$. Since we are working at
the homothetic level we can maintain the initial inner product remaining in
the same homothetic class. Now, a straightforward calculation shows that the
above metric is not locally symmetric, locally conformally flat or Einstein.
It corresponds to the metric in Theorem 1.2–(L.iii).

##  5\. Proof of Theorem 1.2

We consider the Bach-flat metrics obtained in Sections $\S$2, $\S$3 and $\S$4,
except those which are $pp$-waves already covered by Remark 1.1 and Theorem
1.9. Next we show that the metric is conformally Einstein in all cases of
Theorem 1.2 by analyzing each situation separately.

Recall that a metric is conformally Einstein if and only if there exists a
(locally defined) nowhere zero function $\varphi$ so that
$\overline{g}=\varphi^{-2}g$ satisfies Equation (2). Moreover, setting
$\varphi=e^{\sigma}$ one has that the gradient of the function $\sigma$
satisfies the conformal Cotton-flat equation (3) since
$\overline{g}=e^{-2\sigma}g$ is Cotton-flat. Set $\xi=\nabla\sigma$. Now a
straightforward calculation shows that $\nabla\varphi=\varphi\xi$ and

$\operatorname{Hes}_{\varphi}(X,Y)=\varphi\\{\langle X,\xi\rangle\langle
Y,\xi\rangle+\langle\nabla_{X}\xi,Y\rangle\\}.$

In order to analyze Equation (2) we consider the symmetric $(0,2)$-tensor
field

$\begin{array}[]{rcl}\mathfrak{CE}(X,Y)&=&2\operatorname{Hes}_{\varphi}(X,Y)+\varphi\rho(X,Y)-\frac{1}{4}\\{2\Delta\varphi+\varphi\tau\\}\langle
X,Y\rangle\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr&=&2\varphi\\{\langle
X,\xi\rangle\langle
Y,\xi\rangle+\langle\nabla_{X}\xi,Y\rangle\\}+\varphi\rho(X,Y)-\frac{1}{4}\\{2\Delta\varphi+\varphi\tau\\}\langle
X,Y\rangle\end{array}$

and evaluate it on the left-invariant vector fields obtained from the
corresponding basis of each Lie algebra in Theorem 1.2.

#### Proof of Theorem 1.2-(D.i)

Let $\\{U_{1},U_{2},U_{3},U_{4}\\}$ be the pseudo-orthonormal left-invariant
global frame obtained from the pseudo-orthonormal basis of the Lie algebras in
Theorem 1.2-(D).

A straightforward calculation shows that gradient vector fields solving the
conformally Cotton-flat equation (3) in case (D.i) are given by
$\xi=U_{3}+\lambda U_{4}$ for some smooth function satisfying
$d\lambda(U_{1})=d\lambda(U_{2})=d\lambda(U_{4})=0$. A direct calculation now
shows that, when evaluating on the basis $\\{U_{k}\\}$, one has

$\operatorname{Hes}_{\varphi}=\varphi\left(\begin{array}[]{cccc}\lambda&\frac{1}{2}&0&0\\\
\frac{1}{2}&\lambda&0&0\\\ 0&0&d\lambda(U_{3})+\lambda^{2}&\lambda\\\
0&0&\lambda&1\end{array}\right)\quad\text{and}\quad\rho=-\left(\begin{array}[]{cccc}0&1&0&0\\\
1&0&0&0\\\ 0&0&\frac{1}{2}&0\\\ 0&0&0&2\end{array}\right),$

from where it follows that $\Delta\varphi=4\lambda\varphi$ and the scalar
curvature vanishes. Hence the only non-zero component of the tensor field
$\mathfrak{CE}$ is given by

$\mathfrak{CE}(U_{3},U_{3})=\tfrac{1}{2}\varphi(4d\lambda(U_{3})+4\lambda^{2}-1).$

This shows that the conformal metric determined by the gradient vector field
$\xi=U_{3}+\lambda U_{4}$, given by a function $\lambda$ solving

$d\lambda(U_{1})=d\lambda(U_{2})=d\lambda(U_{4})=0\quad\text{and}\quad
4d\lambda(U_{3})+4\lambda^{2}=1,$

is Einstein.

#### Proof of Theorem 1.2-(D.ii)

Gradient vector fields solving the conformally Cotton-flat equation (3) in
case (D.ii) are given by $\xi=\varepsilon U_{3}+\lambda U_{4}$ for some smooth
function satisfying $d\lambda(U_{1})=d\lambda(U_{2})=d\lambda(U_{4})=0$. A
straightforward calculation now shows that

$\operatorname{Hes}_{\varphi}\\!=\\!\varphi\\!\left(\\!\\!\begin{array}[]{cccc}\varepsilon(\lambda+1)&\frac{1}{2}\varepsilon\alpha&0&0\\\
\frac{1}{2}\varepsilon\alpha&\varepsilon\lambda&0&0\\\
0&0&d\lambda(U_{3})+\lambda^{2}&\varepsilon\lambda\\\
0&0&\varepsilon\lambda&1\end{array}\\!\\!\right),\,\,\rho\\!=\\!-\\!\left(\\!\\!\begin{array}[]{cccc}3\varepsilon&\varepsilon\alpha&0&0\\\
\varepsilon\alpha&\varepsilon&0&0\\\
0&0&\frac{1}{2}\alpha^{2}+1&\varepsilon\\\
0&0&\varepsilon&2\end{array}\\!\\!\right),$

from where it follows that $\Delta\varphi=\varepsilon(4\lambda+1)\varphi$ and
the scalar curvature $\tau=-6\varepsilon$. Now, a straightforward calculation
shows that the only non-zero component of $\mathfrak{CE}$ is given by

$\mathfrak{CE}(U_{3},U_{3})=\varphi\left(2d\lambda(U_{3})+2\lambda^{2}-\tfrac{1}{2}\alpha^{2}-1\right).$

This shows that the conformal metric determined by the gradient vector field
$\xi=\varepsilon U_{3}+\lambda U_{4}$, given by a function $\lambda$ solving

$d\lambda(U_{1})=d\lambda(U_{2})=d\lambda(U_{4})=0\quad\text{and}\quad
2d\lambda(U_{3})+2\lambda^{2}-\tfrac{1}{2}\alpha^{2}=1,$

is Einstein.

#### Proof of Theorem 1.2-(R)

Let $\\{E_{i}\\}$ be the global orthonormal frame on $H_{3}\rtimes\mathbb{R}$
obtained by left-translating the orthonormal basis $\\{e_{i}\\}$ of the Lie
algebra. A straightforward calculation shows that the metric is conformal
Cotton-flat. Indeed, the vector field $\xi=-(\lambda+\alpha)E_{2}+\lambda
E_{4}$ solves the equation
$\operatorname{div}_{4}W-W(\,\cdot\,,\,\cdot\,,\,\cdot\,,\xi)=0$, and it is a
gradient if the smooth function $\lambda$ satisfies
$d\lambda(E_{1})=d\lambda(E_{3})=0$ and $d\lambda(E_{2})-d\lambda(E_{4})=0$.
The Hessian and the Ricci tensors on the global frame $\\{E_{i}\\}$ are
expressed as

$\begin{array}[]{rcl}\operatorname{Hes}_{\varphi}&=&\varphi\left(\begin{array}[]{cccc}\lambda\alpha&0&-\frac{1}{2}\alpha&0\\\
0&(\lambda+\alpha)^{2}-d\lambda(E_{4})&0&\lambda(\lambda+\alpha)-d\lambda(E_{4})\\\
-\frac{1}{2}\alpha&0&\alpha\lambda&0\\\
0&\lambda(\lambda+\alpha)-d\lambda(E_{4})&0&\lambda^{2}-d\lambda(E_{4})\end{array}\right)\quad\text{and}\\\
\vskip 12.0pt plus 4.0pt minus
4.0pt\cr\rho&=&\phantom{\varphi}\left(\begin{array}[]{cccc}2\alpha^{2}&0&\alpha&0\\\
0&-\frac{1}{2}&0&-\frac{1}{2}\\\ \alpha&0&2\alpha^{2}&0\\\
0&-\frac{1}{2}&0&-2\alpha^{2}-\frac{1}{2}\end{array}\right),\end{array}$

from where it follows that $\Delta\varphi=\alpha(4\lambda+\alpha)\varphi$ and
the scalar curvature $\tau=6\alpha^{2}$.

Hence the non-zero components of $\mathfrak{CE}$ are determined by

$\mathfrak{CE}(E_{2},E_{2})=\mathfrak{CE}(E_{4},E_{4})=\mathfrak{CE}(E_{2},E_{4})=-\tfrac{1}{2}\varphi\left(4d\lambda(E_{4})-4\lambda(\lambda+\alpha)+1\right),$

which shows that the vector field $\xi=-(\lambda+\alpha)E_{2}+\lambda E_{4}$,
given by the equations

$d\lambda(E_{1})=d\lambda(E_{3})=0,\quad
d\lambda(E_{2})-d\lambda(E_{4})=0\quad\text{and}\quad
d\lambda(E_{4})-\lambda(\lambda+\alpha)+\tfrac{1}{4}=0,$

is a gradient and the conformal metric induced by the potential function is
Einstein.

#### Proof of Theorem 1.2-(L.i)

Let $\\{E_{i}\\}$ be the global orthonormal frame on $H_{3}\rtimes\mathbb{R}$
obtained by left-translating the orthonormal basis $\\{e_{i}\\}$ of the Lie
algebra. The metric is conformally Cotton-flat and, moreover, gradient vector
fields satisfying
$\operatorname{div}_{4}W-W(\,\cdot\,,\,\cdot\,,\,\cdot\,,\xi)=0$ are given by
$\xi=\lambda E_{3}+(\lambda+1)E_{4}$, for some smooth function $\lambda$ on
$H_{3}\rtimes\mathbb{R}$ such that $d\lambda(E_{1})=d\lambda(E_{2})=0$ and
$d\lambda(E_{3})+d\lambda(E_{4})=0$. A straightforward calculation shows that,
on the orthonormal basis $\\{E_{i}\\}$, one has

$\begin{array}[]{rcl}\operatorname{Hes}_{\varphi}&=&\varphi\left(\begin{array}[]{cccc}\lambda+1&\frac{1}{2}\alpha&0&0\\\\[2.168pt]
\frac{1}{2}\alpha&\lambda+1&0&0\\\\[2.168pt]
0&0&d\lambda(E_{4})+\lambda^{2}&-d\lambda(E_{4})-\lambda(\lambda+1)\\\\[2.168pt]
0&0&-d\lambda(E_{4})-\lambda(\lambda+1)&d\lambda(E_{4})+(\lambda+1)^{2}\end{array}\right)\quad\text{and}\\\
\vskip 12.0pt plus 4.0pt minus
4.0pt\cr\rho&=&-\left(\begin{array}[]{cccc}2&\alpha&0&0\\\\[2.168pt]
\alpha&2&0&0\\\\[2.168pt]
0&0&\frac{1}{2}\alpha^{2}&-\frac{1}{2}\alpha^{2}\\\\[2.168pt]
0&0&-\frac{1}{2}\alpha^{2}&\frac{1}{2}\alpha^{2}+2\end{array}\right),\end{array}$

from where it follows that $\Delta\varphi=(4\lambda+3)\varphi$ and the scalar
curvature $\tau=-6$. Hence the only non-zero components of the tensor field
$\mathfrak{CE}$ are determined by

$\mathfrak{CE}(E_{3},E_{3})=\mathfrak{CE}(E_{4},E_{4})=-\mathfrak{CE}(E_{3},E_{4})=\tfrac{1}{2}\varphi\left(4d\lambda(E_{4})+4\lambda(\lambda+1)-\alpha^{2}\right).$

Thus we conclude that the conformal metric determined by the gradient vector
field $\xi=\lambda E_{3}+(\lambda+1)E_{4}$, given by a solution of the
equations

$d\lambda(E_{1})=d\lambda(E_{2})=0,\quad
d\lambda(E_{3})+d\lambda(E_{4})=0\quad\text{and}\quad
d\lambda(E_{4})+\lambda(\lambda+1)-\tfrac{1}{4}\alpha^{2}=0,$

is Einstein.

#### Proof of Theorem 1.2-(L.ii)

Let $\\{U_{i}\\}$ be the pseudo-orthonormal global frame obtained by left-
translating the vectors $\\{u_{i}\\}$ on the Lie algebra. The metrics in
Theorem 1.2-(L.ii) are conformally Cotton-flat and, moreover, gradient vector
fields satisfying
$\operatorname{div}_{4}W-W(\,\cdot\,,\,\cdot\,,\,\cdot\,,\xi)=0$ are given by
$\xi=\lambda U_{2}+4U_{4}$, for some smooth function $\lambda$ on
$H_{3}\rtimes\mathbb{R}$ satisfying $d\lambda(U_{2})=d\lambda(U_{3})=0$ and
$d\lambda(U_{4})=-4\lambda$. The Hessian and the Ricci tensors on the basis
$\\{U_{i}\\}$ are given by

$\operatorname{Hes}_{\varphi}=\varphi\left(\\!\\!\begin{array}[]{cccc}d\lambda(U_{1})+\lambda^{2}+4\alpha&16&0&0\\\
16&0&0&0\\\ 0&0&0&0\\\
0&0&0&16\end{array}\\!\\!\right)\quad\text{and}\quad\rho=-\left(\\!\\!\begin{array}[]{cccc}8\alpha&32&0&0\\\
32&0&0&0\\\ 0&0&0&0\\\ 0&0&0&32\end{array}\\!\\!\right),$

from where one has that $\Delta\varphi=48\varphi$ and the scalar curvature
$\tau=-96$. Hence the only non-zero component of $\mathfrak{CE}$ is given by
$\mathfrak{CE}(U_{1},U_{1})=2\varphi(d\lambda(U_{1})+\lambda^{2})$. Therefore
the conformal metric determined by the gradient vector field $\xi=\lambda
U_{2}+4U_{4}$, given by a solution of the equations

$d\lambda(U_{2})=d\lambda(U_{3})=0,\quad
d\lambda(U_{4})=-4\lambda\quad\text{and}\quad d\lambda(U_{1})=-\lambda^{2},$

is Einstein.

#### Proof of Theorem 1.2-(L.iii)

The Ricci operator of the left-invariant metric in Theorem 1.2-(L.iii) is
diagonalizable, $\operatorname{Ric}=\operatorname{diag}[-6$, $-6$, $-24$,
$-18]$, and therefore $\tau=-54$. The Weyl curvature operator acting on the
space of two-forms has eigenvalues $\\{-4,2,2,-4,2,2\\}$, where the eigenvalue
$2$ is a double root of the minimal polynomial, and hence the metric is weakly
generic. The metric is also conformally Cotton-flat with left-invariant
gradient vector field $\xi=3U_{4}$ satisfying
$\operatorname{div}_{4}W-W(\,\cdot\,,\,\cdot\,,\,\cdot\,,\xi)=0$. Hence it is
conformally Einstein by using the results in [22].

Alternatively one may proceed as in the previous cases and consider all
gradient solutions of (3) given by $\xi=\lambda U_{4}$ for any smooth function
satisfying $d\lambda(U_{1})=d\lambda(U_{2})=d\lambda(U_{3})=0$. Hence one has

$\operatorname{Hes}_{\varphi}=\varphi\left(\begin{array}[]{cccc}0&\lambda&0&0\\\
\lambda&0&0&0\\\ 0&0&4\lambda&0\\\
0&0&0&d\lambda(U_{4})+\lambda^{2}\end{array}\right)\quad\text{and}\quad\rho=-\left(\begin{array}[]{cccc}0&6&0&0\\\
6&0&0&0\\\ 0&0&24&0\\\ 0&0&0&18\end{array}\right),$

so that
$\Delta\varphi=\left(d\lambda(U_{4})+\lambda(\lambda+6)\right)\varphi$. Hence
the only non-zero terms in $\mathfrak{CE}$ are determined by

$\begin{array}[]{l}\mathfrak{CE}(U_{1},U_{2})=-\frac{1}{2}\varphi\left(d\lambda(U_{4})+(\lambda+5)(\lambda-3)\right),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathfrak{CE}(U_{3},U_{3})=-\frac{1}{2}\varphi\left(d\lambda(U_{4})+(\lambda-3)(\lambda-7)\right),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\mathfrak{CE}(U_{4},U_{4})=\phantom{-}\frac{3}{2}\varphi\left(d\lambda(U_{4})+(\lambda+1)(\lambda-3)\right),\end{array}$

from where it follows that the function $\lambda$ is constant $\lambda=3$.

######  Remark 5.1.

A straightforward calculation shows that all the conformally Einstein metrics
in Theorem 1.2 are indeed conformally Ricci-flat. While the conformally
Einstein metric in Theorem 1.2-(L.iii) is unique (up to scaling), the other
cases admit infinitely many conformally Einstein metrics, from where it
follows that they are conformally equivalent to a Ricci-flat $pp$-wave (see
[6]).

##  6\. Proof of Theorem 1.7

Considering the left-invariant metric in Theorem 1.7-(i), a straightforward
calculation shows that the Weyl curvature operator
$W:\Lambda^{2}\rightarrow\Lambda^{2}$ has six-distinct non-zero complex
eigenvalues and thus it is weakly generic. Next, we show that the metric does
not satisfies the conformal $C$-space condition, i.e., there does not exist a
(not necessarily gradient) vector field $X$ so that
$\operatorname{div}_{4}W-W(\,\cdot\,,\,\cdot\,,\,\cdot\,,X)=0$. Hence the
metric is strictly Bach-flat.

Let $\\{V_{i}\\}$ be the orthonormal frame on $H_{3}\rtimes\mathbb{R}$
obtained by left-translating the vectors $\\{v_{i}\\}$ at the Lie algebra and
set $X=\sum X^{j}V_{j}$. Now a straightforward calculation shows that

$\begin{array}[]{l}\operatorname{div}_{4}W(V_{4},V_{1},V_{1})-W(V_{4},V_{1},V_{1},X)=26X^{4},\\\
\operatorname{div}_{4}W(V_{4},V_{1},V_{2})-W(V_{4},V_{1},V_{2},X)=4\sqrt{14}X^{3},\\\
\operatorname{div}_{4}W(V_{4},V_{1},V_{3})-W(V_{4},V_{1},V_{3},X)=-4\sqrt{14}X^{2},\\\
\operatorname{div}_{4}W(V_{4},V_{1},V_{4})-W(V_{4},V_{1},V_{4},X)=54\sqrt{14}-26X^{1},\\\
\operatorname{div}_{4}W(V_{3},V_{2},V_{4})-W(V_{3},V_{2},V_{4},X)=144-4\sqrt{14}X^{1},\end{array}$

from where one has that the above equations have no solution, and thus
$H_{3}\rtimes\mathbb{R}$ is not a conformal $C$-space.

We proceed in an analogous way with metrics in Theorem 1.7-(ii). The Weyl
curvature operator acting on the space of two-forms is three-step nilpotent
and it is not a conformal $C$-space, which shows that these metrics are also
strictly Bach-flat. We omit the details that are completely analogous to the
previous case.

##  7\. Conclusions

Bach-flatness is a very restrictive condition for left-invariant Riemannian
metrics on four-dimensional Lie groups [1, 13]. The Lorentzian situation is
more subtle due to the fact that the restriction of the metric to the three-
dimensional normal subgroup $G$ may be a positive definite, Lorentzian or
degenerate metric.

We classify all left-invariant Bach-flat Lorentzian metrics on semi-direct
extensions $H_{3}\rtimes\mathbb{R}$. As a consequence it is shown that the
class of conformally Einstein metrics which are not locally conformally flat
reduces to plane waves and six generically non-homothetic classes (see Theorem
1.2). On the opposite, the class of strictly Bach-flat metrics, i.e., those
which are not conformally Einstein, reduces to two non-homothetic families
(see Theorem 1.7).

A special situation occurs when the semi-direct extension is a product. Bach-
flat Lorentzian metrics on the product Lie group $H_{3}\times\mathbb{R}$ are
locally conformally flat or a plane wave with parallel Ricci tensor (see
Remark 1.1).

Among the Bach-flat metrics there is a single one which is critical for all
quadratic curvature functionals without being neither symmetric nor a plane
wave (see Theorem 1.2–(D.i)). This is in sharp contrast with the Riemannian
situation [7]. Moreover, the conformally Einstein metric in Theorem
1.2–(L.iii) is a shrinking algebraic soliton for the RG2 flow, while the
family of metrics in Theorem 1.2–(L.ii) provides Bach-flat steady solitons for
the RG2 flow, which are therefore self-similar solutions of the flow.

Finally, note that all conformally Einstein semi-direct extensions
$H_{3}\rtimes\mathbb{R}$ are conformal to a $pp$-wave except metrics
corresponding to Theorem 1.2–(L.iii) whose Weyl curvature operator acting on
the space of two-forms is not nilpotent.

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