# Kundt Three Dimensional Left Invariant Spacetimes

Aissa Meliani  Mohamed Boucetta  Abdelghani Zeghib  Laboratory of Nonlinear
Partial Differential Equations and History of Mathematics, B.P. 92, 16006-ENS
Kouba, Algeria.
e-mail<EMAIL_ADDRESS>Université Cadi-Ayyad
Faculté des sciences et techniques
BP 549 Marrakesh Morocco.
e-mail<EMAIL_ADDRESS>UMPA, CNRS, ENS de Lyon, France.
e-mail<EMAIL_ADDRESS>
###### Abstract

Kundt spacetimes are of great importance to General Relativity. We show that a
Kundt spacetime is a Lorentz manifold with a non-singular isotropic geodesic
vector field having its orthogonal distribution integrable and determining a
totally geodesic foliation. We give the local structure of Kundt spacetimes
and some properties of left invariant Kundt structures on Lie groups. Finally,
we classify all left invariant Kundt structures on three dimensional simply
connected unimodular Lie groups.

###### keywords:

Kundt spaces , Lorentzian Lie groups

###### MSC:

53C50 ,

###### MSC:

53Z05 ,

###### MSC:

22E20

##  1 Introduction

Kundt spacetimes are of great importance to General Relativity, as well as
alternative gravity theories. To begin with, let us say that Kundt spacetimes
constitute a natural generalization of pp-wave spacetimes. Roughly speaking, a
Kundt spacetime is defined by the fact that it supports a vector field all of
its scalar invariants vanish, without being Killing (see for instance [19,
Chapter 6]). One of our motivations here is to provide a coordinate-free
treatment of Kundt spacetimes, which is hard to find in the General Relativity
literature. More precisely, let us define a Kundt spacetime as a Lorentz
manifold $(M,g)$ having the following property (see for instance [7]):

  1. $(\mathrm{K}1)$

There exists a non-singular vector field $V$ on $M$ such that

$g(V,V)=g(\nabla_{V}V,\nabla_{V}V)={\mathrm{tr}}(A^{V})=g(B^{V},B^{V})=g(\mathrm{C}^{V},\mathrm{C}^{V})=0,$
(1)

where $\nabla$ is the Levi-Civita connection, $A^{V}:TM\longrightarrow TM$
denotes the $(1,1)$ tensor field given by $A^{V}(X)=\nabla_{X}V$, $B$ its
symmetric part, $C$ its skew-symmetric part and if $F:TM\longrightarrow TM$ is
a $(1,1)$-tensor field then $g(F,F)={\mathrm{tr}}(F\circ F^{*})$ where $F^{*}$
is its adjoint with respect to $g$.

It turns out (see Proposition 2.1) that this property is equivalent to:

  1. $(\mathrm{K}2)$

There exists a non-singular vector field $V$ on $M$ and a differential 1-form
$\alpha$ such that

$g(V,V)=0,\nabla_{X}V=\alpha(X)V\quad\mbox{and}\quad\nabla_{V}V=0$ (2)

for any vector field $X$ orthogonal to $V$.

A fundamental observation for us was (see Proposition 2.1 and more details in
Sections 2, 3 and 4) that Kundt properties $(\mathrm{K}1)$ or $(\mathrm{K}2)$
imply the following property:

  1. $(\mathrm{LK})$

There exists on $M$ a codimension one totally geodesic foliation which is
degenerate with respect to $g$. More precisely, there exists a codimension one
foliation ${\cal F}$ such that each leaf $L$ of ${\cal F}$ is a lightlike
(totally) geodesic hypersurface (that is $T^{\perp}L\subset TL$ and any
geodesic $\gamma:[a,b]\to M$, $a<0<b$, somewhere tangent to $L$, is locally
contained in $L$: if $\gamma^{\prime}(0)\in T_{\gamma(0)}L$ then there exists
$\epsilon>0$ such that $\gamma([-\epsilon,+\epsilon])\subset L$).

Actually, up to assuming the direction field $T^{\perp}{\cal F}$ orientable
(which is always possible up to passing to a double covering), the property
$(\mathrm{LK})$ is equivalent to:

  1. $(\mathrm{LKbis})$

there exists a non-singular vector field $V$ on $M$ and a differential 1-form
$\alpha$ such that

$g(V,V)=0,\nabla_{X}V=\alpha(X)V$ (3)

for any vector field $X$ orthogonal to $V$.

We will refer to a Lorentz manifold satisfying $(\mathrm{LK})$ as a local
Kundt spacetime. Indeed, $(\mathrm{LK})$ implies $(\mathrm{K}1)$ locally near
any point in $M$. In other words, a spacetime admitting a codimension one
lightlike geodesic foliation is locally Kundt.

Another major motivation to study Kundt Lorentz manifolds lies in their
relation with CSI-spaces, those having all of their scalar curvature
invariants are constant. The simplest one is the scalar curvature
$\mathsf{Scal}_{g}$, but one can also consider eigenvalues of the Ricci
operator $\mathrm{Ric}_{g}$ or the $g$-norm of the Riemann tensor
$\mathsf{Rm}_{g}$. All those are scalar curvature invariants of order 1.
Higher order ones are obtained by considering covariant derivatives of
$\mathsf{Rm}_{g}$. So CSI means, in particular, that all these quantities are
constant functions on $M$.

Locally homogeneous spaces are CSI, and the existence of CSI spaces that are
not locally-homogeneous spaces is a non-Riemannian phenomena which makes a one
major difference between the positive and the non-definite cases in pseudo-
Riemannian structures. A typical example is given by the (conformally flat)
plane wave metric $g=dx^{2}+dy^{2}-2dvdu-2f(u)(x^{2}+y^{2})du^{2}$, which is
VSI (i.e. has vanishing scalar invariants) for any $f$, but locally
homogeneous for only few $f$’s. This example is Kundt, in fact
$V=\frac{\partial}{\partial v}$ is a parallel vector field.

It is believed that a CSI Lorentz space, if it is not locally homogeneous,
must be of Kundt type! This conjecture has been proved in some cases, e.g. in
the lower dimensions 3 and 4, see for instance [6, 8]. In another direction,
there is a notion of $\mathcal{I}$-degenerate metrics, meaning that they have
non-trivial (i.e. non-isometric) deformations keeping all the scalar curvature
invariant functions the same (non-depending of the deformation parameter, but
maybe depending on the point of $M$). Those are believed to be Kundt too.

Not all locally homogeneous spacetimes are Kundt, neither all Kundt spacetimes
are locally homogeneous, but it is worthwhile to consider locally homogeneous
Kundt spaces as a special class of both the homogeneous and the Kundt
categories! Our project is to study Kundt structures on three dimensional Lie
groups $G$ endowed with a left invariant Lorentzian metric $g$. It is natural,
in this special homogeneous framework, to introduce a stronger Kundt property
as follows. We call $(G,g)$ a Kundt Lie group (resp. locally Kundt Lie group)
if it admits a non-singular left invariant vector field $V$ satisfying
$(\mathrm{K}2)$ (resp. $(\mathrm{LKbis})$). In other words, we assume here
compatibility between the $(\mathrm{K}2)$ or the $(\mathrm{LKbis})$ property
and the algebraic structure of $G$.

###  1.1 Results

One of our principal results, Theorem 4.1, states, essentially, that a three
dimensional Lorentz group which is Kundt as a spacetime, is in fact a Kundt
group. We also classify, up to isometric isomorphism, all unimodular three
dimensional Kundt groups.

The paper is organized as follows. In Section 2, we provide a synthetic
(coordinate-free) account on Kundt spacetimes emphasising on their
relationship with lightlike geodesic foliations. We introduce Kundt groups and
general facts about them in Section 3. The proof of Theorem 4.1 as well as
further results are given in Section 4. Section 5 contains the classification
up to automorphism of Kundt Lorentz groups.

##  2 Kundt spacetimes and geodesic foliations

Recall from the introduction that a Kundt spacetime is a Lorentz manifold
satisfying the property $(\mathrm{K}1)$. Let us prove that this property is
equivalent to $(\mathrm{K2})$ and implies $(\mathrm{LK})$. Moreover,
$(\mathrm{LK})$ implies $(\mathrm{K1})$ locally near any point of the Lorentz
manifold.

######  Proposition 2.1.

Let $(M,g)$ be a Lorentz manifold. Consider the following assertions:

  1. $(i)$

$(M,g)$ is a Kundt spacetime.

  2. $(ii)$

There exists on $M$ an isotropic non-singular vector field $V$ and a
differential 1-form $\alpha$ such that, for any $X\in\Gamma(V^{\perp})$,

$\nabla_{X}V=\alpha(X)V\quad\mbox{and}\quad\nabla_{V}V=0.$

  3. $(iii)$

There exists on $M$ a totally geodesic codimension one foliation which is
degenerate with respect to $g$. This means that there exists a vector sub-
bundle $F\subset TM$ of rank $(\dim M)-1$, where the restriction of $g$ to $F$
is degenerate and, for any $X,Y\in\Gamma(F)$, $\nabla_{X}Y\in\Gamma(F)$ where
$\nabla$ is the Levi-Civita connection of $g$.

Then $(i)$ and $(ii)$ are equivalent and both imply $(iii)$. Moreover, $(iii)$
implies that $(ii)$ holds in a neighbourhood of any point in $M$.

###### Proof.

$(i)\Longrightarrow(ii).$ Assume that $(M,g)$ is a Kundt spacetime. This means
that there exists a non-singular vector field $V$ satisfying (1). Fix a point
$p\in M$ and denote by $A$ to the endomorphism given by $Au=\nabla_{u}V$ for
any $u\in T_{p}M$. Choose an isotropic vector $U\in T_{p}M$ such that
$g(U,V)=1$ and an orthonormal basis $(e_{1},\ldots,e_{n-2})$ of
$\left\\{U,V_{p}\right\\}^{\perp}$. Note that

$g(A(V),A(V))=g(V_{p},V_{p})=g(A(V),V)=0.$

Thus the vector subspace $\mathrm{span}\\{A(V),V\\}$ is totally isotropic,
hence its dimension equals 1 which means $A(V)=\alpha_{0}V$ for some
$\alpha_{0}\in\hbox{\bb R}$. On the other hand, since $g(V,V)=0$, for any
$u\in T_{p}M$, $g(A(u),V)=g(\nabla_{u}V,V)=0$ hence $A(T_{p}M)\subset
V^{\perp}$. This implies that for any $u\in T_{p}M$, $g(A(u),A(u))\geq 0$ and
$g(A(u),A(u))=0$ if and only if $A(u)=\alpha(u)V$. Now, from (1),

$0={\mathrm{tr}}(A^{*}A)=2g(A(V),AU)+\sum_{i=1}^{n-2}g(A(e_{i}),A(e_{i}))=\sum_{i=1}^{n-2}g(A(e_{i}),A(e_{i})).$

Moreover, we have

$0={\mathrm{tr}}(A)=g(AV,U)+g(AU,V)+\sum_{i=1}^{n-2}g(Ae_{i},e_{i})=\alpha_{0}$

hence $\nabla_{V}V=0$. This completes the proof of $(i)\Longrightarrow(ii)$.

Let us prove $(ii)\Longrightarrow(i)$. Fix a point $p$ and consider $A$ and
$(U,V,e_{2},\ldots,e_{n-2})$ as defined above. We have
$A(T_{p}M)\subset\hbox{\bb R}V$ hence $V^{\perp}\subset\ker A^{*}$. With this
fact in mind, we get

$\displaystyle{\mathrm{tr}}(A)$
$\displaystyle=g(AU,V)+g(AV,U)+\sum_{i=2}^{n}g(Ae_{i},e_{i})=0,$
$\displaystyle{\mathrm{tr}}(BB^{*})$
$\displaystyle=2g(BU,BV)+\sum_{i=2}^{n}g(Be_{i},Be_{i})=0,$
$\displaystyle{\mathrm{tr}}(CC^{*})$
$\displaystyle=2g(CU,CV)+\sum_{i=2}^{n}g(Ce_{i},Ce_{i})=0.$

This completes the proof of $(i)\Longrightarrow(ii)$.

Let us prove now that $(ii)\Longrightarrow(iii)$. Let $F=V^{\perp}$. We have,
for any $X,Y\in\Gamma(F)$,

$g(\nabla_{X}Y,V)=-g(Y,\nabla_{X}V)=-\alpha(X)g(Y,V)=0$

hence $\nabla_{X}Y\in\Gamma(F)$. This shows that $F$ is integrable and defines
a degenerate codimension one totally geodesic foliation.

Now, we prove that if $(iii)$ holds then, for any $p\in M$, there exists a
vector field $V$ near $p$ satisfying (1). So, suppose that there exists an
integrable degenerate codimension one sub-bundle $F\subset TM$ which defines a
totally geodesic foliation. Fix a point $p\in M$ and choose a non-singular
vector field $V\in\Gamma(F^{\perp})\subset\Gamma(F)$ near $p$. It is obvious
that $V$ is isotropic. Moreover, since $\nabla_{V}V\in\Gamma(F)$ and
$g(\nabla_{V}V,V)=0$ then $\nabla_{V}V=\alpha V$. As above, denote by $A$ the
endomorphism $A_{p}^{V}$ and choose a basis $(u,V_{p},e_{1},\ldots,e_{n-2})$.
We have, obviously, that $A(T_{p}M)\subset F_{p}$. Moreover, since $F$ is
totally geodesic, for any $X,Y\in\Gamma(F)$,

$0=g(\nabla_{X}Y,V)=-g(Y,\nabla_{X}V)$

which implies that $A(F_{p})\subset\hbox{\bb R}V$. So far, we have shown that
locally near $p$, for any $X\in\Gamma(F)$,

$AV=\alpha_{0}V,\,A(X)=\alpha(X)V.$

To finish the proof, we look for a vector field $V^{\prime}=e^{f}V$ where $f$
is a function such that $V^{\prime}$ satisfies $(ii)$. This is equivalent to
$V(f)=-\alpha_{0}$ and such a function exists locally. ∎

###  2.1 Kundt coordinates

Another way to compare the Kundt property with the existence of a codimension
one lightlike geodesic foliation is given by the following fact which asserts
the existence of adapted local coordinates associated to lightlike geodesic
foliations, where the metric has a special form. The same adapted coordinates
are known to characterize Kundt spacetimes.

######  Proposition 2.2.

Let $(M,g)$ be a Lorentz manifold satisfying $(iii)$ of Proposition 2.1. Then
near any point in $M$ there exists a local coordinates system
$(v,u,x=(x^{2},\ldots x^{n}))$ where the metric has the form:

$g=2dudv+H(v,u,x)du^{2}+\sum_{i=2}^{n}W_{i}(v,u,x)dudx^{i}+\sum_{i,j}h_{ij}(u,x)dx^{i}dx^{j}.$

######  Remarks 2.1.

\- Observe that the functions $h_{ij}$ do not depend on $v$.

\- The foliation in Proposition 2.1 corresponds to the (local) $u$-levels.

\- One can also show the converse, that a foliation admitting an adapted chart
where the metric has such a form is lightlike geodesic.

###### Proof.

Suppose that there exists a vector sub-bundle $F$ of $TM$ of rank $(\dim M)-1$
such that the restriction of the metric to $F$ is degenerate and the
$\Gamma(F)$ is stable by the Levi-Civita product.

Fix a point $p\in M$ and let $\Sigma$ be a local hypersurface containing $p$
and transversal to $F^{\perp}$. Then $F_{\Sigma}=F\cap T\Sigma$ determines a
foliation on $\Sigma$ hence there exists a coordinates system
$({{x^{2}}},\ldots,{x^{n}},{u})$ on $\Sigma$ such that the leaves of
$F_{\Sigma}$ are the ${u}$-levels. There is a section
$T:\Sigma\longrightarrow(F^{\perp})_{|\Sigma}$ such that
$g(T,\frac{\partial}{\partial{u}})=2$. Choose an injective immersion
$\phi:\hbox{\bb R}^{n-1}\longrightarrow M$ such that $\phi(\hbox{\bb
R}^{n-1})=\Sigma$. Then there exists $\epsilon>0$ such that the map
$\Phi:\hbox{\bb R}^{n-1}\times(-\epsilon,\epsilon)\longrightarrow M$ given by
$\Phi(t,s)=\exp_{\phi(t)}(sT)$ is a diffeomorphism into its image. Denote by
$V$ the image by $\Phi$ of the vector field $\frac{\partial}{\partial s}$.
Since $F^{\perp}$ is totally geodesic, then $V$ is tangent to $F^{\perp}$
hence satisfies $g(V,V)=0$. By construction, we have $\nabla_{V}V=0$ and
according to the proof of Proposition 2.1, for any $X\in\Gamma(F)$,
$\nabla_{X}V=\alpha(X)V$.

On the other hand, the vector fields
$\frac{\partial}{\partial{u}},\frac{\partial}{\partial{x^{2}}},\ldots,\frac{\partial}{\partial{x^{n}}}$
on $\Sigma$ define a family of vector fields on $\hbox{\bb R}^{n-1}$ hence
define a family of vector fields on $\hbox{\bb
R}^{n-1}\times(-\epsilon,\epsilon)$ which commute with
$\frac{\partial}{\partial s}$. Let $U,X_{2},\ldots,X_{n}$ be their images by
$\Phi$. We deduce that $V,U,X_{2},\ldots,X_{n}$ are commuting, and give rise
to a local coordinate system $(v,u,x^{2},\ldots,x^{n})$ on $M$ such that

$V=\frac{\partial}{\partial v},\;U=\frac{\partial}{\partial
u}\quad\mbox{and}\quad X_{i}=\frac{\partial}{\partial x_{i}},\quad
i=2,\ldots,n.$

Observe now that, for any vector field $Z$ commuting with $V$, the scalar
product $g(V,Z)$ is constant along the $V$-trajectories. Indeed, since
$g(V,V)=0$ and $[Z,V]=0$ we get

$V.g(V,Z)=g(\nabla_{V}V,Z)+g(V,\nabla_{V}Z)=g(V,\nabla_{Z}V)=0.$

We deduce that for any $i=2,\ldots,n$, $g(V,X_{i})$ and $g(U,V)$ are constant
along the trajectories of $V$ and since they are constant along $\Sigma$ we
get that $g(U,V)=2$ and $g(V,X_{i})=0$. Moreover, we have

$V.g(X_{i},X_{j})=g(\nabla_{V}X_{i},X_{j})+g(X_{i},\nabla_{V}X_{j})=g(\nabla_{X_{i}}V,X_{j})+g(X_{i},\nabla_{X_{j}}V)=\alpha(X_{i})g(V,X_{j})+\alpha(X_{j})g(X_{i},V)=0.$

This completes the proof. ∎

######  Remark 1.

For a general codimension one lightlike foliation, not necessarily geodesic,
we have similar adapted coordinates, but with the functions $h_{ij}$ depending
also on $v$.

##  3 Kundt Groups

A Lorentz Lie group is a Lie group $G$ endowed with a left invariant
Lorentzian metric $g$. Denote by ${\mathfrak{g}}$ the Lie algebra of $G$ and
$\langle\;,\;\rangle=g(e)$. We call $({\mathfrak{g}},\langle\;,\;\rangle)$ a
Lorentz Lie algebra. The Levi-Civita product is the product $\bullet$ on
${\mathfrak{g}}$ given by

$2\langle u\bullet
v,w\rangle=\langle[u,v],w\rangle+\langle[w,u],v\rangle+\langle[w,v],u\rangle,\quad
u,v,w\in{\mathfrak{g}}.$ (4)

A Kundt Lie group (resp. locally Kundt Lie group) is a Lorentz Lie group
$(G,g)$ having an isotropic left invariant vector field satisfying (2) (resp
(3)).

######  Proposition 3.1.

Let $(G,g)$ be a connected Lorentz Lie group. Then the following are
equivalent:

  1. $(i)$

$(G,g)$ is a locally Kundt Lie group.

  2. $(ii)$

There exists a codimension one subalgebra ${\mathfrak{h}}$ of ${\mathfrak{g}}$
which is degenerate and stable by the Levi-Civita product.

Moreover, if $(G,g)$ is a locally Kundt Lie group then its is a Kundt Lie
group if and only if, for any generator $e$ of ${\mathfrak{h}}^{\perp}$,
$e\bullet e=0$.

###### Proof.

Let us prove that $(i)$ implies $(ii)$. $(G,g)$ is a locally Kundt Lie group
if and only if there exists a left invariant vector field $V$ satisfying (3).
The codimension one subspace ${\mathfrak{h}}=V(e)^{\perp}$ is degenerate and,
for any $u,v\in{\mathfrak{h}}$, denote by $u^{l}$ and $v^{l}$ the
corresponding left invariant vector fields. Then

$g(\nabla_{u^{l}}v^{l},V)=u^{l}.g(v^{l},V)-g(v^{l},\nabla_{u^{l}}V)=-\alpha(u^{l})g(v^{l},V)=0$

hence $\nabla_{u^{l}}v^{l}(e)=u\bullet v\in{\mathfrak{h}}$. This means that
${\mathfrak{h}}$ is stable by the Levi-Civita product, which completes the
proof of $(i)\Longrightarrow(ii)$.

Le us show now that $(ii)\Longrightarrow(i)$. Suppose there exists
${\mathfrak{h}}$ a codimension one degenerate subalgebra of ${\mathfrak{g}}$
which is stable by the Levi-Civita product and consider $v$ a generator of
${\mathfrak{h}}^{\perp}$. Denote by $V$ the left invariant vector field
associated to $v$. Then, according to the proof of Proposition 2.1, we have,
for any $x\in{\mathfrak{h}}$,

$\nabla_{x^{l}}V=\alpha(x^{l})V\quad\mbox{and}\quad\nabla_{V}V=\alpha_{0}V$

where $\alpha_{0}$ is a constant. The last assertion is obvious. ∎

######  Definition 3.1.

Let ${\mathfrak{g}}$ be a Lie algebra. A Kundt pair on ${\mathfrak{g}}$ is a
pair $(\langle\;,\;\rangle,{\mathfrak{h}})$ where $\langle\;,\;\rangle$ is a
Lorentzian product on ${\mathfrak{g}}$ and ${\mathfrak{h}}$ is a
$\langle\;,\;\rangle$-degenerate codimension one subalgebra stable by the
Levi-Civita product $\bullet$ given by (4) and for any
$e\in{\mathfrak{h}}^{\perp}$, $e\bullet e=0$.

There is a large class of Lie groups which cannot carry a locally Kundt group
structure.

######  Lemma 3.1.

Let ${\mathfrak{g}}$ be a semi-simple compact Lie algebra. Then
${\mathfrak{g}}$ cannot have a codimension one subalgebra.

###### Proof.

Suppose that ${\mathfrak{g}}$ has a codimension one Lie subalgebra
${\mathfrak{h}}$. Since ${\mathfrak{g}}$ is compact, it carries a bi-invariant
scalar product $\langle\;,\;\rangle$, i.e., ${\sf{ad}}_{x}$ is skew-symmetric
for any $x\in{\mathfrak{g}}$. For any $x\in{\mathfrak{h}}$, ${\sf{ad}}_{x}$
leaves ${\mathfrak{h}}$ invariant and since it is skew-symmetric it leaves
${\mathfrak{h}}^{\perp}=\hbox{\bb R}e$ invariant, so $[x,e]=0$ and $e$ is
central which contradicts the fact that ${\mathfrak{g}}$ is semi-simple. ∎

######  Corollary 3.1.

Let $G$ be a compact semi-simple Lie group. Then $G$ cannot carry any
structure of locally Kundt Lie group.

The oscillator group named so by Streater in [20], as a four-dimensional
connected, simply connected Lie group, whose Lie algebra (known as the
oscillator algebra) coincides with the one generated by the differential
operators, acting on functions of one variable, associated to the harmonic
oscillator problem. The oscillator group has been generalized to any even
dimension $2n\geq 4$, and several aspects of its geometry have been
intensively studied, both in differential geometry and in mathematical physics
(see [2, 4, 9, 10, 13, 15]). Oscillator Lie groups have a natural left
invariant Kundt structure.

######  Example 1.

For $n\in\hbox{\bb N}^{*}$ and
$\lambda=(\lambda_{1},\ldots,\lambda_{n})\in\hbox{\bb R}^{n}$ with
$0<\lambda_{1}\leq\cdots\leq\lambda_{n}$, the $\lambda$-oscillator group,
denoted by $G_{\lambda}$, is the Lie group with the underlying manifold
$\hbox{\bb R}^{2n+2}=\hbox{\bb R}\times\hbox{\bb R}\times\hbox{\bb C}^{n}$ and
product

$(t,s,z).(t^{\prime},s^{\prime},z^{\prime})=\Bigl{(}t+t^{\prime},s+s^{\prime}+\frac{1}{2}\sum_{j=1}^{n}\mbox{Im
}[\bar{z}_{j}\exp(it\lambda_{j})z_{j}^{\prime}],\ldots,z_{j}+\exp(it\lambda_{j})z^{\prime}_{j},\ldots\Bigr{)}.$

Its Lie algebra ${\mathfrak{g}}_{\lambda}$ is $\hbox{\bb R}\times\hbox{\bb
R}\times\hbox{\bb C}^{n}$ with its canonical basis $\hbox{\bb
B}=\left\\{e_{-1},e_{0},e_{j},\check{e}_{j}\right\\}_{j=1,\ldots,n}$ such that

$e_{-1}=(1,0,0),e_{0}=(0,1,0),\;e_{j}=(0,0,(0,\ldots,1,\ldots,0))\quad\mbox{and}\quad\check{e}_{j}=(0,0,(0,\ldots,\imath,\ldots,0)).$

and the Lie brackets are given by

$[e_{-1},e_{i}]=\lambda_{i}\check{e}_{i},\qquad[e_{-1},\check{e}_{i}]=-\lambda_{i}e_{i},\qquad[e_{i},\check{e}_{i}]=e_{0},$
(5)

for $i=1,\ldots,n$, the unspecified products are either given by antisymmetry
or zero. For $x\in{\mathfrak{g}}_{\lambda}$, let

$x=x_{-1}e_{-1}+x_{0}e_{0}+\sum_{i=1}^{n}(x_{i}e_{i}+\check{x}_{i}\check{e}_{i}).$

The non-degenerate symmetric bilinear form

$\textbf{k}_{\lambda}(x,x):=2x_{-1}x_{0}+\sum_{j=1}^{n}\frac{1}{\lambda_{j}}(x_{j}^{2}+\check{x}_{j}^{2})$
(6)

satisfies

$\textbf{k}_{\lambda}([x,y],z)+\textbf{k}_{\lambda}(y,[x,z])=0,\quad\mbox{for
any}\quad x,y,z\in{\mathfrak{g}}_{\lambda}$

and hence defines a Lorentzian bi-invariant metric $g$ on $G_{\lambda}$. The
Levi-Civita connection of $g$ is given by $\nabla_{X}Y=\frac{1}{2}[X,Y]$ for
any left invariant vector fields $X,Y$. Hence the left invariant vector field
associated to $e_{0}$ is parallel and defines a Kundt group structure on
$G_{\lambda}$.

##  4 Kundt Spacetimes vs Kundt Groups

In dimension 3, we have the following result.

######  Theorem 4.1.

Let $G$ be a Lie group of dimension 3 endowed with a left invariant Lorentz
metric $g$. Assume there exists a degenerate totally geodesic hypersurface
$\Sigma$ in $(G,g)$ (not necessarily complete).

  1. $(i)$

Then, either $(G,g)$ has a constant sectional curvature, or $(G,g)$ is a
locally Kundt Lie group.

  2. $(ii)$

Furthermore, if the isotropy group of 1 in the isometry group
$\mathsf{Iso}(G,g)$ is non-compact, then such a hypersurface $\Sigma$ exists.

To prove this theorem, we need the following lemma which also appears in [14,
18, 23].

######  Lemma 4.1.

Let $(G,g)$ be a Lorentz Lie group of dimension 3 . If there are three non-
tangent (i.e. having different tangent planes at 1) lightlike geodesic
hypersurfaces through 1, then $(G,g)$ has constant curvature.

###### Proof.

Assume there exist three different lightlike geodesic hypersurfaces
$\Sigma_{1}$, $\Sigma_{2}$ and $\Sigma_{3}$, through $1\in G$. Let
$T_{1},T_{2},T_{3}$ be their tangent spaces and $V_{1},V_{2},V_{3}$, non-
vanishing vectors in their orthogonals $T_{1}^{\perp},T_{2}^{\perp}$ and
$T_{3}^{\perp}$.

Since the $\Sigma_{i}$’s are geodesic, each $T_{i}$ is invariant under the
Riemann curvature: if $u,v,w\in T_{i}$, then $R(u,v)w\in T_{i}$.

Let $e_{3}$ be a unit (spacelike) vector generating $T_{1}\cap T_{2}$, and
consider $A_{e_{3}}:V\to R(e_{3},V)e_{3}$. Then $\langle
A_{e_{3}}(V_{1}),e_{3}\rangle=\langle R(e_{3},V_{1})e_{3},e_{3}\rangle=0$,
that is $A_{e_{3}}(V_{1})$ is orthogonal to $e_{3}$. But it is also orthogonal
to $V_{1}$, by curvature-invariance of $V_{1}^{\perp}$. Therefore,
$A_{e_{3}}(V_{1})$ is collinear to $V_{1}$, say
$A_{e_{3}}(V_{1})=\lambda_{12}V_{1}$. Similarly
$A_{e_{3}}(V_{2})=\lambda_{21}V_{2}$.

Consider $\langle R(e_{3},V_{1})e_{3},V_{2}\rangle$ , which equals $\langle
R(e_{3},V_{2})e_{3},V_{1}\rangle$. It also equals $\lambda_{12}\langle
V_{1},V_{2}\rangle=\lambda_{21}\langle V_{1},V_{2}\rangle$. Since $V_{1}$ and
$V_{2}$ are both null, $\langle V_{1},V_{2}\rangle\neq 0$ then
$\lambda_{12}=\lambda_{21}$. In conclusion $A_{e_{3}}$ is a homothety on
$(e_{3})^{\perp}$, of ratio, say $\lambda$ ($=\lambda_{12}=\lambda_{21}$).

Since $A_{e_{3}}(e_{3})=0$, we conclude that
$R(e_{3},W)e_{3}=\lambda\big{(}W\langle e_{3},e_{3}\rangle-\langle
W,e_{3}\rangle e_{3}\big{)}$, for any $W$.

In a similar way, one defines $e_{2}$ and $e_{1}$ unit vectors generating
$T_{1}\cap T_{3}$ and $T_{2}\cap T_{3}$, respectively, and deduce a similar
formula for $R(e_{i},W)e_{i}$, with the same $\lambda$.

In fact, up to orientation, the $e_{i}$ are uniquely defined. More precisely,
in a Lorentz linear 3-space $(E,\langle\;,\;\rangle)$, there is up to
isometry, a unique system of three null directions. Indeed, with the notation
above, we can choose generators $V_{1},V_{2}$ and $V_{3}$ of these directions
such that $\langle V_{i},V_{j}\rangle=1$, for any $i\neq j$, so the matrix
coefficients $\langle V_{i},V_{j}\rangle_{ij}$ are fully given. This implies
that any two such systems are related by an isometry.

Now, the system $\\{e_{1},e_{2},e_{3}\\}$ is given (up to orientation) by
$\\{V_{1},V_{2},V_{3}\\}$, so all scalar products $\langle e_{i},e_{j}\rangle$
are given, i.e. computable by means of $\langle\;,\;\rangle$.

It follows that $R(e_{i},e_{j})e_{i}$ are given for any $i,j$.

Consider now $X=R(e_{i},e_{j})e_{k}$, with $k\neq i,j$ and $i\neq j$. Then
$\langle X,e_{k}\rangle=0$, and $\langle X,e_{i}\rangle=\langle
R(e_{i},e_{j})e_{i},e_{k}\rangle$ and, hence, computable. Similarly for
$\langle X,e_{j}\rangle$, and therefore $X$ is computable.

From all this, it follows that all the curvaturse $R(e_{i},e_{j})e_{k}$ are
computable, exactly as in the case of a space of constant curvature $\lambda$.
∎

#### Proof of Theorem 4.1

###### Proof.

  1. $(i)$

It follows form the lemma that if $(G,g)$ does not have constant curvature,
then, through any point pass exactly one or two (germs of) lightlike geodesic
hypersurfaces.

For the sake of clarity, let us consider first the case where there exists
exactly one germ of such hypersurfaces. More precisely, there exists a
geodesic lightlike hypersurface $\Sigma$ containing 1. Uniqueness means that
for any $S$ a geodesic lightlike hypersurface, with $1\in S$, then $\Sigma\cap
S$ is a neighbourhood of 1 in both $\Sigma$ and $S$.

For any $x\in G$, the translated hypersurface $x\Sigma$ is the unique geodesic
lightlike hypersurface passing through $x$.

Let us see that the tangent space of $\Sigma$ is left invariant: if
$x\in\Sigma$, then $T_{x}\Sigma=xT_{1}\Sigma$ (the last notation means the
left translation by $x$ of $T_{1}\Sigma$). Indeed, both $\Sigma$ and $x\Sigma$
are geodesic lightlike hypersurfaces containing $x$ hence they coincide near
$x$, and thus have same tangent space:
$T_{x}\Sigma=T_{x}(x\Sigma)=xT_{1}\Sigma$.

This left invariance of $T\Sigma$, means that $\Sigma$ is a “local subgroup”.
Its “maximal extension” will be a subgroup which is a geodesic lightlike
hypersurface. To be more formal, one defines a plane field $E$ on $G$, with
$E(x)$ being the tangent space of the unique geodesic lightlike hypersurface
through $x$. So, $E(x)=T_{x}(x\Sigma)$ which implies $E$ is left invariant.
Uniqueness implies $E$ is integrable: $\Sigma$ is a local leaf of $E$ trough
$1$. The global leaf of $E$ is a subgroup.

Let us now consider the case where we have two geodesic lightlike
hypersurfaces $\Sigma^{1}$ and $\Sigma^{2}$ through 1, which do not coincide
near 1. this is equivalent to $E(1)=T_{1}\Sigma^{1}\neq T_{1}\Sigma^{2}=F(1)$.

Let $x\in\Sigma^{1}$. Then through $x$, we have three geodesic lightlike
hypersurfaces: $\Sigma^{1},x\Sigma^{1}$ and $x\Sigma^{2}$. Therefore, two
among them coincide locally. Let us assume $x$ is close to 1 and deduce that
$\Sigma^{1}$ and $x\Sigma^{2}$ can not coincide near $x$. For this, it is
enough to show they have different tangent spaces $A=T_{x}\Sigma^{1}\neq
T_{x}(x\Sigma^{2})=B$. If they were equal, they would have the same
translation to 1, $x^{-1}A=x^{-1}B$. On one hand,
$x^{-1}B=T_{x}\Sigma^{2}=F(1)$, and on the other hand,
$x^{-1}A=x^{-1}T_{x}\Sigma$ is close to $E(1)=T_{1}\Sigma^{1}$, by continuity
of the tangent space of $\Sigma^{1}$ and the fact that $x$ is close to 1.
Since $E(1)$ and $F(1)$ are transversal, the same is true for $x^{-1}A$ and
$x^{-1}B$, for $x$ sufficiently close to 1, and in particular they are not
equal, hence $A\neq B$. From all of this, we infer that $\Sigma^{1}$ and
$x\Sigma^{1}$ coincide near $x$.

As in the case of a unique geodesic lightlike hypersurface, one deduces that
$\Sigma^{1}$ is a “local group”. More precisely, one defines two left
invariant plane fields $E$ and $F$, extending $E(1)$ and $F(1)$ respectively.
Our previous argument implies that $\Sigma^{1}$ is a local leaf at 1 of $E$,
and so $E$ is integrable, and the same applies for $F$. The $E$ and $F$-leaves
of 1 are therefore two subgroups which are geodesic lightlike hypersurfaces.

  2. $(ii)$

Now, assume, $I$, the isotropy group of 1 in the isometry group
${\sf{Iso}}(G,g)$ is non-compact. Let $f_{n}\in I$ be a diverging sequence (it
has non-convergent sub-sequence) and consider their graphs
$F_{n}=Gr(f_{n})\subset G\times G$. Endow $G\times G$ with the metric
$g\oplus(-g)$, then the $F_{n}$’s are isotropic and totally geodesic. Then we
will consider a limit $L$ of a subsequence of the $F_{n}$. To give a formal
meaning of this, consider a small convex neighbourhood $C$ of $(1,1)$ in
$G\times G$. This means any two points of $C$ can be joined by a unique
geodesic segment contained in $C$. Consider $F_{n}\cap C$ and note $F_{n}^{0}$
the connected component of $(1,1)$ in $F_{n}\cap C$. Now, one can give sense
to convergence of $F_{n}$, exactly as in the situation of affine subspaces in
an affine flat space. More precisely, $F_{n}^{0}$ converge to $L$, if the
tangent spaces $T_{(1,1)}F^{0}_{n}$ converge to $T_{(1,1)}L$.

Such a limit $L$ is a geodesic isotropic submanifold in $G\times G$ of
dimension equal to $\dim G$, but it is no longer a graph of some map $f:G\to
G$, since otherwise, $f_{n}$ will converge to $f$. Thus $L$ intersects non-
trivially the vertical $\\{1\\}\times G$, and hence projects onto a degenerate
geodesic submanifold $\Sigma$ in $G\times\\{1\\}$. The intersection
$L\cap(\\{1\\}\times G)$ has dimension one, since it is isotropic, therefore
$\Sigma$ is a hypersurface.

∎

###  4.1 Comments on the constant curvature case

It is natural to ask what happens if $(G,g)$ has constant sectional curvature,
say $c$? Let us give here some examples and hints, details will appear
elsewhere. The proof of Theorem 4.1 does not apply since there are infinitely
many “germs” of lightlike geodesic hypersurfaces trough 1: any lightlike
hyperplane in $T_{1}G$ is tangent to a lightlike geodesic hypersurface exactly
as in the universal Lorentz space $\tilde{M}(c)$ of constant curvature $c$.

$\bullet$ It turns out that if the isotropy group $I$ is non-compact and has
dimension 1 or 2, then $(G,g)$ is a Kundt group. Indeed, the proof of Theorem
4.1 can be adapted well to this situation where $\dim I=1$ or $2$. Consider
for this the derivative action of $I$ on ${\mathfrak{g}}=T_{1}G$. It preserves
exactly one or two hyperplanes which are in fact lightlike. Indeed, let $L$ be
a closed connected non-compact subgroup of ${\sf O}(1,2)$. If $\dim L=1$, then
it is either a hyperbolic one parameter group, in this case it preserves
exactly two lightlike hyperplanes, or it is a unipotent one parameter group
and in this case it preserves exactly one lightlike hyperplane. In case $\dim
L=2$, it is conjugate to the triangular subgroup of ${\sf{SL}}(2,\hbox{\bb
R})$ and preserves exactly one lightlike hyperplane. All these claims can be
confirmed by a direct check-in. In summary, if $\dim I=1,2$, the argument in
the proof of Theorem 4.1 can be adapted and yields a left invariant plane
field $E$ and, thus, a lightlike geodesic subgroup $H$.

Let us give the following example with $c=0$. Consider on $\hbox{\bb R}^{3}$,
the Lorentz metric $dx^{2}+dydz$. The plane $E=\\{z=0\\}$ is lightlike. Its
linear stabilizer is a subgroup $S$ of dimension 2 in ${\sf{SO}}(1,2)$ and,
hence, its stabilizer in the full Poincaré group
${\sf{SO}}(1,2)\ltimes\hbox{\bb R}^{3}$ is $S\ltimes E$. It contains in
particular the 3-dimensional (non-unimodular) group $G$ of elements
$(t,a,b)\in\hbox{\bb R}\times E$ acting by $(x,y,z)\to(x+a,e^{t}y+b,e^{-t}z)$.
This action is free and transitive on the upper half space $\\{z>0\\}$ and,
hence, $G$ inherits a left invariant (non-complete) flat metric. It is Kundt,
since its (abelian) subgroup $E$ has lightlike geodesic orbits. Observe here
that the full isometry group of this left invariant metric is $S\ltimes E$. In
particular the isotropy group has dimension one.

$\bullet$ Assume now that the isotropy group has dimension $3$. Thus
$\dim{\sf{Iso}}(G,g)=6$, and $(G,g)$ is locally isometric to the universal
space $\tilde{M}(c)$ of constant curvature $c$. However, a subgroup of
dimension 3 in ${\sf O}(1,2)$ contains at least its identity component, and
also, a subgroup of dimension 6 in ${\sf{Iso}}(\tilde{M}(c))$ contains at
least its identity component ${\sf{Iso}}^{0}(\tilde{M}(c))$. Therefore, as a
homogeneous space $(G,g)$ is globally isometric to
$\tilde{M}(c)={\sf{Iso}}^{0}(\tilde{M}(c))/{\sf O}^{0}(1,2)$. In other words,
$G$ acts transitively and freely on $\tilde{M}(c)$ (or equivalently, $(G,g)$
has constant curvature $c$ and is complete).

Let us give the example of the Euclide group $\mathsf{Euc}_{2}$. Its universal
cover $\widetilde{\mathsf{Euc}_{2}}$ acts simply transitively isometrically on
$(\hbox{\bb R}^{3},dx^{2}+dy^{2}-dz^{2})$ by:
$((x,y),z)\to\big{(}(R_{t}(x,y)+(a,b)),z+t\big{)}$, where $R_{t}$ is the
rotation of angle $t$ and $(t,a,b)\in\widetilde{\mathsf{Euc}_{2}}$. Its unique
2-dimensional subgroup is $\hbox{\bb R}^{2}$, which acts by translations
$(x,y,z)\to(x+a,y+b,z)$. It has spacelike geodesic orbits. Therefore
$\widetilde{\mathsf{Euc}_{2}}$ is a flat complete Lorentz group that is not a
Kundt group.

We believe this is the unique complete Lorentz group of constant curvature
which is not a Kundt group?

$\bullet$ Finally, there are examples of flat groups $(G,g)$ with isotropy
group $I$ of dimension 0, which are not Kundt groups. To see an example,
consider as above the metric $dx^{2}+dydz$. Let $S\subset{\sf{SO}}(1,2)$ be
the stabilizer of the isotropic direction $\hbox{\bb
R}\frac{\partial}{\partial y}$ and $T$ the subgroup of translations in this
direction. Take $G=S\ltimes T$. It has an open orbit on which it acts freely
which allows one to endow it with a flat (non-complete) metric. One can show
it is not a Kundt group.

##  5 Classification of three dimensional unimodular simply-connected Kundt
Lie groups

In this section, we give a complete classification of Kundt Lie group
structures on three dimensional unimodular Lie groups. According to
Proposition 3.1, the classification of Kundt Lie group structures on a simply
connected Lie group $G$ is equivalent to the classification of Kundt pairs
$({\mathfrak{h}},\langle\;,\;\rangle)$ on its Lie algebra ${\mathfrak{g}}$.
Two Kundt pairs $({\mathfrak{h}}_{1},\langle\;,\;\rangle_{1})$ and
$({\mathfrak{h}}_{2},\langle\;,\;\rangle_{2})$ are called equivalent if there
exists an automorphism of Lie algebra
$\phi:{\mathfrak{g}}\longrightarrow{\mathfrak{g}}$ such that
$\phi({\mathfrak{h}}_{1})={\mathfrak{h}}_{2}$ and
$\phi^{*}(\langle\;,\;\rangle_{2})=\langle\;,\;\rangle_{1}$.

In dimension 3, we have the following useful characterization of Kundt pairs.

######  Proposition 5.1.

Let $({\mathfrak{g}},\langle\;,\;\rangle)$ be a Lorentzian Lie algebra and let
${\mathfrak{h}}$ be a codimension one subalgebra. Then:

  1. $(i)$

If ${\mathfrak{h}}$ is abelian then $({\mathfrak{h}},\langle\;,\;\rangle)$ is
a Kundt pair if and only if ${\mathfrak{h}}$ is degenerate and if $e$ is a
generator of ${\mathfrak{h}}^{\perp}$ then
${\sf{ad}}_{e}({\mathfrak{g}})\subset{\mathfrak{h}}$.

  2. $(ii)$

If ${\mathfrak{h}}$ is non-abelian then $({\mathfrak{h}},\langle\;,\;\rangle)$
is a Kundt pair if and only if
${\mathfrak{h}}^{\perp}=[{\mathfrak{h}},{\mathfrak{h}}]$ and if $e$ is a
generator of ${\mathfrak{h}}^{\perp}$ then
${\sf{ad}}_{e}({\mathfrak{g}})\subset{\mathfrak{h}}$.

###### Proof.

Let $e$ a generator of ${\mathfrak{h}}^{\perp}$. Then
$({\mathfrak{h}},\langle\;,\;\rangle)$ is a Kundt pair if and only if
$e\in{\mathfrak{h}}^{\perp}$, $\langle e,e\rangle=0$, $e\bullet e=0$ and for
any $u,v\in{\mathfrak{h}}$,

$0=2\langle u\bullet v,e\rangle=\langle[e,u],v\rangle+\langle[e,v],u\rangle.$
(7)

Note first that $e\bullet e=0$ and $\langle e,e\rangle=0$ if and only if, for
any $u\in{\mathfrak{g}}$,

$0=\langle e\bullet e,x\rangle=\langle[u,e],e\rangle$

which is equivalent to ${\sf{ad}}_{e}({\mathfrak{g}})\subset{\mathfrak{h}}$.

If ${\mathfrak{h}}$ is abelian then (7) holds trivially.

Suppose now that $\dim{\mathfrak{g}}=3$ and ${\mathfrak{h}}$ is not abelian.
Then there exists a basis $(u,v)$ of ${\mathfrak{h}}$ such $\langle
u,v\rangle=0$ and $[u,v]=u$. Put $e=au+bv$. Then from the relation above, we
get

$0=-b\langle u,v\rangle+a\langle u,u\rangle=a\langle
u,u\rangle\quad\mbox{and}\quad 0=\langle[e,u],u\rangle=b\langle u,u\rangle.$

This implies that $\langle u,u\rangle=0$ and hence
${\mathfrak{h}}^{\perp}=[{\mathfrak{h}},{\mathfrak{h}}]$. The converse is
obviously true.

∎

There are five simply connected three dimensional unimodular non abelian Lie
groups:

  1. 1.

The nilpotent Lie group $\mathrm{Nil}$ known as Heisenberg group whose Lie
algebra will be denoted by $\mathfrak{n}$. We have

$\mathrm{Nil}=\left\\{\left(\begin{matrix}1&x&z\\\ 0&1&y\\\
0&0&1\end{matrix}\right),x,y,z\in\hbox{\bb
R}\right\\}\quad\mbox{and}\quad\mathfrak{n}=\left\\{\left(\begin{matrix}0&x&z\\\
0&0&y\\\ 0&0&0\end{matrix}\right),x,y,z\in\hbox{\bb R}\right\\}.$

The Lie algebra ${\mathfrak{n}}$ has a basis $\hbox{\bb
B}_{0}=(X_{1},X_{2},X_{3})$ where

$X_{1}=\left(\begin{matrix}0&1&0\\\ 0&0&0\\\
0&0&0\end{matrix}\right),X_{2}=\left(\begin{matrix}0&0&0\\\ 0&0&1\\\
0&0&0\end{matrix}\right)\quad\mbox{and}\quad
X_{3}=\left(\begin{matrix}0&0&1\\\ 0&0&0\\\ 0&0&0\end{matrix}\right)$

where the non-vanishing Lie bracket is $[X_{1},X_{2}]=X_{3}$.

  2. 2.

$\mathrm{SU}(2)=\left\\{\left(\begin{matrix}a+bi&-c+di\\\
c+di&a-bi\end{matrix}\right),a^{2}+b^{2}+c^{2}+d^{2}=1\right\\}\quad\mbox{and}\quad\mathfrak{su}(2)=\left\\{\left(\begin{matrix}iz&y+ix\\\
-y+xi&-zi\end{matrix}\right),x,y,z\in\hbox{\bb R}\right\\}.$ The Lie algebra
$\mathfrak{su}(2)$ has a basis $\hbox{\bb B}_{0}=(X_{1},X_{2},X_{3})$

$X_{1}=\frac{1}{2}\left(\begin{matrix}0&i\\\
i&0\end{matrix}\right),\;X_{2}=\frac{1}{2}\left(\begin{matrix}0&1\\\
-1&0\end{matrix}\right)\quad\mbox{and}\quad
X_{3}=\frac{1}{2}\left(\begin{matrix}-i&0\\\ 0&i\end{matrix}\right)$

where the non-vanishing Lie brackets are

$[X_{1},X_{2}]=X_{3},\;[X_{2},X_{3}]=X_{1}\quad\mbox{and}\quad[X_{3},X_{1}]=X_{2}.$

  3. 3.

The universal covering group $\widetilde{\mathrm{PSL}}(2,\hbox{\bb R})$ of
$\mathrm{SL}(2,\hbox{\bb R})$ whose Lie algebra is $\mathrm{sl}(2,\hbox{\bb
R})$. The Lie algebra $\mathrm{sl}(2,\hbox{\bb R})$ has a basis $\hbox{\bb
B}_{0}=(e,f,h)$ where

$e=\left(\begin{matrix}0&1\\\ 0&0\end{matrix}\right),\
f=\left(\begin{matrix}0&0\\\ 1&0\end{matrix}\right),\
h=\left(\begin{matrix}1&0\\\ 0&-1\end{matrix}\right).$

where the non-vanishing Lie brackets are

$[e,f]=h,\;[h,e]=2e\quad\mbox{and}\quad[h,f]=-2f.$

  4. 4.

The solvable Lie group $\mathrm{Sol}=\left\\{\left(\begin{matrix}e^{x}&0&y\\\
0&e^{-x}&z\\\ 0&0&1\end{matrix}\right),x,y,z\in\hbox{\bb R}\right\\}$ whose
Lie algebra is $\mathfrak{sol}=\left\\{\left(\begin{matrix}x&0&y\\\ 0&-x&z\\\
0&0&0\end{matrix}\right),x,y,z\in\hbox{\bb R}\right\\}$. The Lie algebra
$\mathfrak{sol}$ has a basis $\hbox{\bb B}_{0}=(X_{1},X_{2},X_{3})$ where

$X_{1}=\left(\begin{matrix}1&0&0\\\ 0&-1&0\\\
0&0&0\end{matrix}\right),\;X_{2}=\left(\begin{matrix}0&0&1\\\ 0&0&0\\\
0&0&0\end{matrix}\right),\;X_{3}=\left(\begin{matrix}0&0&0\\\ 0&0&1\\\
0&0&0\end{matrix}\right)\quad\mbox{and}\quad$

where the non-vanishing Lie brackets are

$[X_{1},X_{2}]=X_{2}\quad\mbox{and}\quad[X_{1},X_{3}]=-X_{3}.$

  5. 5.

The universal covering group $\widetilde{\mathrm{E}_{0}}(2)$ of the Lie group

$\mathrm{E}_{0}(2)=\left\\{\left(\begin{matrix}\cos(\theta)&\sin(\theta)&x\\\
-\sin(\theta)&\cos(\theta)&y\\\
0&0&1\end{matrix}\right),\theta,x,y\in\hbox{\bb R}\right\\}.$

Its Lie algebra is

$\mathrm{e}_{0}(2)=\left\\{\left(\begin{matrix}0&\theta&x\\\ -\theta&0&y\\\
0&0&0\end{matrix}\right),\theta,y,z\in\hbox{\bb R}\right\\}.$

The Lie algebra $\mathrm{e}_{0}(2)$ has a basis $\hbox{\bb
B}_{0}=(X_{1},X_{2},X_{3})$ where

$X_{1}=\left(\begin{matrix}0&-1&0\\\ 1&0&0\\\
0&0&0\end{matrix}\right),\;X_{2}=\left(\begin{matrix}0&0&1\\\ 0&0&0\\\
0&0&0\end{matrix}\right),\;\quad\mbox{and}\quad
X_{3}=\left(\begin{matrix}0&0&0\\\ 0&0&1\\\ 0&0&0\end{matrix}\right)$

where the non-vanishing Lie brackets are

$[X_{1},X_{2}]=X_{3}\quad\mbox{and}\quad[X_{1},X_{3}]=-X_{2}.$

Let us find the 2-dimensional subalgebras of the 3-dimensional unimodular Lie
algebras.

######  Proposition 5.2.

  1. 1.

Let ${\mathfrak{h}}$ be a 2-dimensional subalgebra of $\mathfrak{n}$. Then
${\mathfrak{h}}=\mathrm{span}\\{X_{3},aX_{1}+bX_{2}\\}$, $(a,b)\not=(0,0)$.

  2. 2.

$\mathfrak{su}(2)$ has no subalgebra of dimension 2.

  3. 3.

Let ${\mathfrak{h}}$ be a 2-dimensional subalgebra of $\mathfrak{sol}$ then
either ${\mathfrak{h}}=\mathrm{span}\\{X_{2},X_{3}\\}$,
${\mathfrak{h}}=\mathrm{span}\\{X_{2},X_{1}+aX_{3}\\}$ or
${\mathfrak{h}}=\mathrm{span}\\{X_{3},X_{1}+aX_{2}\\}$ ($a\in\hbox{\bb R})$

  4. 4.

Let ${\mathfrak{h}}$ be a 2-dimensional subalgebra of $e_{0}(2)$ then
${\mathfrak{h}}=\mathrm{span}\\{X_{2},X_{3}\\}$.

  5. 5.

Let ${\mathfrak{h}}$ be 2-dimensional subalgebra of $\mathrm{sl}(2,\hbox{\bb
R})$. Then there exists an automorphism of $\mathrm{sl}(2,\hbox{\bb R})$ which
sends ${\mathfrak{h}}$ to $\mathrm{span}\\{h,e\\}$.

###### Proof.

  1. 1.

A 2-dimensional subalgebra ${\mathfrak{h}}$ of $\mathfrak{n}$ must be abelian
and contains the center. So
${\mathfrak{h}}=\mathrm{span}\\{X_{3},aX_{1}+bX_{2}\\}$ and $(a,b)\not=(0,0)$.

  2. 2.

It is a consequence of Lemma 3.1.

  3. 3.

Denote by ${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{3}\\}$. If
${\mathfrak{h}}$ is abelian and ${\mathfrak{h}}\not={\mathfrak{h}}_{0}$ then
${\mathfrak{h}}=\mathrm{span}\\{X_{1}+U,V\\}$ where $U,V\in{\mathfrak{h}}_{0}$
hence $[X_{1},V]=0$ which is impossible. So if ${\mathfrak{h}}$ is abelian
then ${\mathfrak{h}}={\mathfrak{h}}_{0}$.

If ${\mathfrak{h}}$ is not abelian then
${\mathfrak{h}}\not={\mathfrak{h}}_{0}$. Thus
${\mathfrak{h}}=\mathrm{span}\\{X_{1}+U,V\\}$ where
$U,V=aX_{2}+bX_{3}\in{\mathfrak{h}}_{0}$ and
$[{\mathfrak{h}},{\mathfrak{h}}]=\hbox{\bb R}V$. Now

$[X_{1}+U,V]=aX_{2}-bX_{3}.$

So the vectors $aX_{2}+bX_{3},aX_{2}-bX_{3}$ must be linearly dependent, hence
$ab=0$. Which complete the proof.

  4. 4.

We can use the same argument as above and get $a^{2}+b^{2}=0$.

  5. 5.

Note first that $[e,f]=h,\ [h,e]=2e,\ [h,f]=-2f$. Let ${\mathfrak{h}}$ be a
2-dimensional subalgebra of $\mathrm{sl}(2,\hbox{\bb R})$. Then there exists a
basis $(u,v)$ of ${\mathfrak{h}}$ such that $[u,v]=2v$. The endomorphism
${\sf{ad}}_{u}$ is skew-symmetric with respect to the Killing form hence
${\mathrm{tr}}({\sf{ad}}_{u})=0$. It has $2$ and $0$ as eigenvalues so the
third eigenvalue is $-2$. So there exists $w\in\mathrm{sl}(2,\hbox{\bb R})$
such that $[u,w]=-2w$. Now

$[u,[v,w]]=2[v,w]-2[v,w]=0$

hence $[v,w]=\alpha u$. By replacing $w$ by $\frac{1}{\alpha}w$ we get that
the automorphism $\phi$ which sends $(u,v,w)$ to $(h,e,f)$ sends
${\mathfrak{h}}$ to $\mathrm{span}\\{h,e\\}$ which completes the proof.

∎

######  Theorem 5.1.

Let $({\mathfrak{h}},\langle\;,\;\rangle)$ be a Kundt pair of $\mathfrak{n}$.
Then $({\mathfrak{h}},\langle,\rangle)$ is equivalent to
$({\mathfrak{h}}_{0},\langle,\rangle_{0})$ where either:

  1. 1.

$\langle,\rangle_{0}=\left[\begin{array}[]{ccc}1&0&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr 0&-1&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&\mu\end{array}\right],\mu>0$ and
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{1}\pm X_{2},X_{3}\\}$.

  2. 2.

$\langle,\rangle_{0}=\left[\begin{array}[]{ccc}1&0&0\\\ 0&0&1\\\
0&1&0\end{array}\right]$ and
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{1},X_{3}\\}$.

###### Proof.

Let $({\mathfrak{h}},\langle\;,\;\rangle)$ be a Kundt pair on $\mathfrak{n}$.
According to [1, Theorem 3.1], there exists an automorphism $\phi$ of
$\mathfrak{n}$ such that the matrix of $(\phi^{-1})^{*}(\langle\;,\;\rangle)$
in the basis $(X_{1},X_{2},X_{3})$ has one of the following forms:

${\mathfrak{n}}_{1}=\left[\begin{array}[]{ccc}1&0&0\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr 0&-1&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&\mu\end{array}\right],\;{\mathfrak{n}}_{2}=\left[\begin{array}[]{ccc}1&0&0\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&1&0\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&0&-\mu\end{array}\right]\ \text{or}\
{\mathfrak{n}}_{3}=\left[\begin{array}[]{ccc}1&0&0\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr 0&0&1\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&1&0\end{array}\right],\quad\mu>0.$

By virtue of Proposition 5.2,
$\phi({\mathfrak{h}})={\mathfrak{h}}_{0}=\mathrm{span}\\{X_{3},aX_{1}+bX_{2}\\}$
with $(a,b)\not=(0,0)$. According to Proposition 5.1,
$(\phi({\mathfrak{h}}),(\phi^{-1})^{*}(\langle\;,\;\rangle))$ is a Kundt pair
if and only if ${\mathfrak{h}}_{0}$ is degenerate and
${\sf{ad}}_{e}({\mathfrak{n}})\subset{\mathfrak{h}}_{0}$ where $e$ is a
generator of ${\mathfrak{h}}_{0}^{\perp}$. Since
$[{\mathfrak{n}},{\mathfrak{n}}]\subset{\mathfrak{h}}_{0}$ then the last
condition holds.

Now, ${\mathfrak{h}}_{0}$ cannot be ${\mathfrak{n}}_{2}$-degenerate and it is
${\mathfrak{n}}_{3}$-degenerate if and only if $b=0$. Finally,
${\mathfrak{h}}_{0}$ is ${\mathfrak{n}}_{1}$-degenerate if and only if
$a^{2}-b^{2}=0$ which completes the proof. ∎

######  Theorem 5.2.

Let $({\mathfrak{h}},\langle\;,\;\rangle)$ be a Kundt pair of
$\mathfrak{sol}$. Then $({\mathfrak{h}},\langle,\rangle)$ is equivalent to
$({\mathfrak{h}}_{0},\langle\;,\;\rangle_{0})$ where either:

  1. 1.

$\langle\;,\;\rangle_{0}=\left(\begin{matrix}\lambda&0&0\\\ 0&0&-1\\\
0&-1&0&\end{matrix}\right),\lambda>0$ and
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{1}\\}$ or
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{3},X_{1}\\}$,

  2. 2.

$\langle\;,\;\rangle_{0}=\left(\begin{matrix}\lambda^{2}&0&0\\\ 0&\lambda&1\\\
0&1&0\end{matrix}\right)$ and
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{3},X_{1}\\}$,

  3. 3.

$\langle\;,\;\rangle_{0}=\left(\begin{matrix}0&0&-\frac{2}{b}\\\ 0&1&1\\\
-\frac{2}{b}&1&1\end{matrix}\right),\quad b>0$ and
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{3}\\}$.

  4. 4.

$\langle,\rangle_{0}=\left[\begin{array}[]{ccc}0&0&1\\\ 0&1&0\\\
1&0&0\end{array}\right]$ and
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{3}\\}$.

###### Proof.

Let $({\mathfrak{h}},\langle\;,\;\rangle)$ be a Kundt pair on
$\mathfrak{sol}$. Then according to [1, Theorem 3.4], there exists an
automorphism $\phi$ of $\mathfrak{sol}$ such that the matrix of
$(\phi^{-1})^{*}(\langle\;,\;\rangle)$ in the basis $(X_{1},X_{2},X_{3})$ has
one of the following forms:

$\begin{cases}\mathrm{sol}_{1}=\left(\begin{matrix}\frac{4}{u^{2}-v^{2}}&0&0\\\
0&1&\frac{u}{v}\\\
0&\frac{u}{v}&1\end{matrix}\right),v>0,u<v,\;\mathrm{sol}_{2}=\left(\begin{matrix}\frac{4}{v^{2}-u^{2}}&0&0\\\
0&\frac{u}{v}&-1\\\
0&-1&\frac{u}{v}&\end{matrix}\right),v>0,u<v,\;\mathrm{sol}_{3}=\left(\begin{matrix}\frac{1}{u+v}&0&0\\\
0&-\frac{v}{u}&1\\\ 0&1&1\end{matrix}\right),\;u>0,v>0,\\\
\mathrm{sol}_{4}=\left(\begin{matrix}\frac{1}{u}&0&0\\\ 0&-1&0\\\
0&0&1\end{matrix}\right),\;u>0,\;\par\mathrm{sol}_{5}=\left(\begin{matrix}0&0&-\frac{2}{b}\\\
0&1&1\\\ -\frac{2}{b}&1&1\end{matrix}\right),\quad
b>0,\;\par\mathrm{sol}_{6}=\left(\begin{matrix}\lambda^{2}&0&0\\\
0&\lambda&1\\\
0&1&0\end{matrix}\right),\quad\lambda\not=0,\;\mathrm{sol}_{7}=\left(\begin{matrix}0&0&1\\\
0&1&0\\\ 1&0&0\end{matrix}\right).\end{cases}$

By virtue of Proposition 5.2, $\phi({\mathfrak{h}})={\mathfrak{h}}_{0}$ where
either ${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{3}\\}$,
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{1}+aX_{3}\\}$ or
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{3},X_{1}+aX_{2}\\}$ ($a\in\hbox{\bb
R})$.

If ${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{3}\\}$ then it is abelian and
it is $\mathrm{sol}_{5}$-degenerate and $\mathrm{sol}_{7}$-degenerate. For
$\mathrm{sol}_{5}$, ${\mathfrak{h}}_{0}^{\perp}=\hbox{\bb R}(X_{2}-X_{3})$ and
we have ${\sf{ad}}_{X_{2}-X_{3}}(\mathfrak{sol})\subset{\mathfrak{h}}_{0}$. We
have the same situation for $\mathrm{sol}_{7}$. Thus
$({\mathfrak{h}}_{0},\mathrm{sol}_{5})$ and
$({\mathfrak{h}}_{0},\mathrm{sol}_{7})$ are Kundt pairs.

If ${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{1}+aX_{3}\\}$ then
$[{\mathfrak{h}}_{0},{\mathfrak{h}}_{0}]=\hbox{\bb R}X_{2}$. We have
obviously, ${\sf{ad}}_{X_{2}}(\mathfrak{sol})\subset{\mathfrak{h}}_{0}$ and,
according to Proposition 5.1, $({\mathfrak{h}}_{0},\langle\;,\;\rangle)$ is a
Kundt pair if and only if $\langle X_{2},X_{2}\rangle=\langle
X_{2},X_{1}+aX_{3}\rangle=0$. This is possible if and only if
$\langle\;,\;\rangle=\mathrm{sol}_{2}$ with $u=0$ and $a=0$.

If ${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{3},X_{1}+aX_{2}\\}$ then
$[{\mathfrak{h}}_{0},{\mathfrak{h}}_{0}]=\hbox{\bb R}X_{3}$. We have
obviously, ${\sf{ad}}_{X_{3}}(\mathfrak{sol})\subset{\mathfrak{h}}_{0}$ and,
according to Proposition 5.1, $({\mathfrak{h}}_{0},\langle\;,\;\rangle)$ is a
Kundt pair if and only if $\langle X_{3},X_{3}\rangle=\langle
X_{3},X_{1}+aX_{2}\rangle=0$. This is possible if and only if
$\langle\;,\;\rangle=\mathrm{sol}_{2}$, $u=0$ and $a=0$ or
$\langle\;,\;\rangle=\mathrm{sol}_{6}$ and $a=0$. ∎

######  Theorem 5.3.

Let $({\mathfrak{h}},\langle\;,\;\rangle)$ be a Kundt structure on $e_{0}(2)$.
Then $({\mathfrak{h}},\langle\;,\;\rangle)$ is equivalent to
$({\mathfrak{h}}_{0},\langle\;,\;\rangle_{0})$ where
$\langle\;,\;\rangle_{0}=\left[\begin{array}[]{ccc}0&1&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr 1&0&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&\mu\end{array}\right],\mu>0$ and
${\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{3}\\}$.

###### Proof.

Let $({\mathfrak{h}},\langle\;,\;\rangle)$ be a Kundt pair on $e_{0}(2)$. Then
according to [1, Theorem 3.5], there exists an automorphism $\phi$ of
$e_{0}(2)$ such that the matrix of $(\phi^{-1})^{*}(\langle\;,\;\rangle)$ in
the basis $(X_{1},X_{2},X_{3})$ has one of the following forms:

$\langle\;,\;\rangle_{1}=\left(\begin{matrix}0&1&0\\\ 1&u&0\\\
0&0&v\end{matrix}\right),\;u>0,v>0,\;\langle\;,\;\rangle_{2}=\left(\begin{matrix}0&1&0\\\
1&u&0\\\
0&0&v\end{matrix}\right),\;u>0,v>0,\;\langle\;,\;\rangle_{3}=\left(\begin{matrix}0&1&0\\\
1&0&0\\\ 0&0&u\end{matrix}\right),\quad u>0.$

By virtue of Proposition 5.2,
$\phi({\mathfrak{h}})={\mathfrak{h}}_{0}=\mathrm{span}\\{X_{2},X_{3}\\}$.
According to Proposition 5.1,
$(\phi({\mathfrak{h}}),(\phi^{-1})^{*}(\langle\;,\;\rangle))$ is a Kundt pair
if and only if ${\mathfrak{h}}_{0}$ is degenerate and
${\sf{ad}}_{e}({\mathfrak{n}})\subset{\mathfrak{h}}_{0}$ where $e$ is a
generator of ${\mathfrak{h}}_{0}^{\perp}$. Now, ${\mathfrak{h}}_{0}$ cannot be
$\langle\;,\;\rangle_{1}$-degenerate neither
$\langle\;,\;\rangle_{2}$-degenerate. Finally, ${\mathfrak{h}}_{0}$ is
$\langle\;,\;\rangle_{3}$-degenerate, ${\mathfrak{h}}_{0}^{\perp}=\hbox{\bb
R}X_{2}$ and ${\sf{ad}}_{X_{2}}(e_{0}(2))\subset{\mathfrak{h}}_{0}$. ∎

######  Theorem 5.4.

Let $({\mathfrak{h}},\langle\;,\;\rangle)$ be a Kundt pair on
$\mathrm{sl}(2,\hbox{\bb R})$. Then there exists an automorphism $\phi$ of
$\mathrm{sl}(2,\hbox{\bb R})$ such that
$\phi({\mathfrak{h}})=\mathrm{span}\\{e,h\\}$ and the matrix of
$\phi^{*}(\langle\;,\;\rangle)$ in the basis $(e,f,h)$ has one of the
following forms:

$\left[\begin{array}[]{ccc}0&4\,\alpha&0\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 4\,\alpha&0&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&8\,\beta\end{array}\right],\;\left[\begin{array}[]{ccc}0&4\,\alpha&0\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr 4\,\alpha&1&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr
0&0&8\,\beta\end{array}\right]\quad\mbox{or}\quad\left[\begin{array}[]{ccc}0&4\,\alpha&0\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr 4\,\alpha&0&2\,\sqrt{2}\\\ \vskip 6.0pt
plus 2.0pt minus 2.0pt\cr 0&2\,\sqrt{2}&8\,\alpha\end{array}\right],\
\beta>0,\ \alpha\in\mathbb{R}^{*}$

###### Proof.

Let $({\mathfrak{h}},\langle\;,\;\rangle)$ be a Kundt pair on
$\mathrm{sl}(2,\hbox{\bb R})$. According to Proposition 5.2, we can suppose
that ${\mathfrak{h}}=\mathrm{span}\\{e,h\\}$. A direct computation using the
software Maple shows that the automorphisms of $\mathrm{sl}(2,\hbox{\bb R})$
leaving ${\mathfrak{h}}$ invariant are of the form

$T=\left[\begin{array}[]{ccc}a&-ab^{2}&-2\,ab\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&a^{-1}&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&b&1\end{array}\right],\;a,b\in\hbox{\bb R}.$

Denote by $B$ the Killing form of $\mathrm{sl}(2,\hbox{\bb R})$. Its matrix in
the basis $(e,f,h)$ is given by

$M=\left[\begin{array}[]{ccc}0&4&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
4&0&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&0&8\end{array}\right].$

We consider the isomorphism $A$, symmetric with respect to $B$, and given by
$B(Au,v)=\langle u,v\rangle$ for any $u,v\in\mathrm{sl}(2,\hbox{\bb R})$.
According to Proposition 5.1, the pair $({\mathfrak{h}},\langle\;,\;\rangle)$
is Kundt if and only if $\langle e,e\rangle=\langle e,h\rangle=0$ and
${\sf{ad}}_{e}(\mathrm{sl}(2,\hbox{\bb R}))\subset{\mathfrak{h}}$. This is
equivalent to $Ae$ is a generator of the orthogonal of ${\mathfrak{h}}$ with
respect to $B$ which is equivalent to the existence of $\alpha\not=0$ such
that $Ae=\alpha e$. The normal form of isomorphisms which are symmetric with
respect to a Lorentzian scalar product are known. We give here the normal form
of those having an isotropic eigenvector. According to [16], there exits a
basis $\hbox{\bb B}=(f_{1},f_{2},f_{3})$ of $\mathrm{sl}(2,\hbox{\bb R})$ such
that:

  1. $(i)$

$Mat(A,\hbox{\bb B})=\left[\begin{array}[]{ccc}\beta&0&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr 0&\alpha&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&\alpha\end{array}\right]$ and $Mat(B,\hbox{\bb
B})=\left[\begin{array}[]{ccc}1&0&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&1\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&1&0\end{array}\right]$,

  2. $(ii)$

$Mat(A,\hbox{\bb B})=\left[\begin{array}[]{ccc}\beta&0&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr 0&\alpha&1\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&\alpha\end{array}\right]$ and $Mat(B,\hbox{\bb
B})=\left[\begin{array}[]{ccc}1&0&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&1\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 0&1&0\end{array}\right],$

  3. $(iii)$

$Mat(A,\hbox{\bb B})=\left[\begin{array}[]{ccc}\alpha&1&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr 0&\alpha&1\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&\alpha\end{array}\right]$ and $Mat(B,\hbox{\bb
B})=\left[\begin{array}[]{ccc}0&0&1\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&1&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr 1&0&0\end{array}\right]$.

Let $P$ be the passage matrix from $\hbox{\bb B}_{0}=(e,f,h)$ to 𝔹.

$\bullet$ The case $(i)$. The relation $A(e)=\alpha e$ implies that we can
choose $f_{2}=e$ and the relation $P^{t}MP=Mat(B,\hbox{\bb B})$ gives that
$P=\left[\begin{array}[]{ccc}-\sqrt{2}&1&-1\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&0&1/4\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
1/4\,\sqrt{2}&0&1/2\end{array}\right]$. The matrix of $\langle\;,\;\rangle$ in
the basis $B_{0}$ is given by $Mat(A,\hbox{\bb B}_{0})^{t}M$ and
$Mat(A,\hbox{\bb B}^{0})=PMat(A,\hbox{\bb B})P^{-1}$. So we get

$Mat(\langle\;,\;\rangle,\hbox{\bb
B}_{0})=\left[\begin{array}[]{ccc}0&4\,\alpha&0\\\ \vskip 6.0pt plus 2.0pt
minus 2.0pt\cr 4\,\alpha&32\,\beta-32\,\alpha&-16\,\beta+16\,\alpha\\\ \vskip
6.0pt plus 2.0pt minus 2.0pt\cr
0&-16\,\beta+16\,\alpha&8\,\beta\end{array}\right].$

Now the automorphism

$T_{1}=\left[\begin{array}[]{ccc}1&-4&-4\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&1&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&2&1\end{array}\right]$

satisfies

$T_{1}^{t}Mat(\langle\;,\;\rangle,\hbox{\bb
B}_{0})T_{1}=\left[\begin{array}[]{ccc}0&4\,\alpha&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr 4\,\alpha&0&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&8\,\beta\end{array}\right].$

$\bullet$ The case $(ii)$. We have

$P=\left[\begin{array}[]{ccc}-\sqrt{2}&1&-1\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&0&1/4\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
1/4\,\sqrt{2}&0&1/2\end{array}\right]\quad\mbox{and}\quad
T_{2}=\left[\begin{array}[]{ccc}4&-1&-4\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&1/4&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&1/2&1\end{array}\right]$ $T_{2}^{t}Mat(\langle\;,\;\rangle,\hbox{\bb
B}_{0})T_{2}=\left[\begin{array}[]{ccc}0&4\,\alpha&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr 4\,\alpha&1&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&0&8\,\beta\end{array}\right].$

$\bullet$ The case $(iii)$. We have

$P=\left[\begin{array}[]{ccc}1&-\sqrt{2}&-1\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&0&1/4\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&1/4\,\sqrt{2}&1/2\end{array}\right]\quad\mbox{and}\quad
T_{3}=\left[\begin{array}[]{ccc}4&-1&-4\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&1/4&0\\\ \vskip 6.0pt plus 2.0pt minus 2.0pt\cr
0&1/2&1\end{array}\right]$ $T_{3}^{t}Mat(\langle\;,\;\rangle,\hbox{\bb
B}_{0})T_{3}=\left[\begin{array}[]{ccc}0&4\,\alpha&0\\\ \vskip 6.0pt plus
2.0pt minus 2.0pt\cr 4\,\alpha&0&2\,\sqrt{2}\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr 0&2\,\sqrt{2}&8\,\alpha\end{array}\right].$

∎

###  5.1 Kundt vs Locally Kundt Lie Groups

In fact, it turns out from the previous proofs, we have shown that any
3-dimensional unimodular locally Kundt Lie group is in fact a Kundt Lie group.
This result is not true in general as the following example shows.

######  Example 2.

Consider $\hbox{\bb R}^{4}$ endowed with the Lie algebra structure where the
only non vanishing Lie bracket is given by $[e_{1},e_{2}]=e_{2}$ and the
Lorentzian scalar product given by

$\langle\;,\;\rangle=\left(\begin{array}[]{cccc}0&1&0&0\\\ 1&0&0&0\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right).$

The Lie subalgebra ${\mathfrak{h}}=\mathrm{span}\\{e_{1},e_{3},e_{4}\\}$ is
abelian and satisfies ${\mathfrak{h}}^{\perp}=\hbox{\bb R}e_{1}$ and hence,
according to Proposition 3.1, defines a local Kundt Lie group structure on the
corresponding simply connected Lie group. However,
${\sf{ad}}_{e_{1}}(\hbox{\bb R}^{4})\nsubset{\mathfrak{h}}$ and hence,
according to Proposition 5.1 this structure is not global.

We think however, it is worthwhile to investigate the natural question: is a
locally Kundt Lie group, without being a Kundt group, still a (globally) Kundt
spacetime?

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