# On Homogeneous 3-dimensional Spacetimes:
Focus on plane waves

S. Allout  Faculty of mathematics, Ruhr-Universitaet Bochum, Germany
<EMAIL_ADDRESS>,  A. Belkacem  Department of mathematics, University of
Batna 2, Algeria<EMAIL_ADDRESS>and  A. Zeghib  UMPA,
CNRS, ENS de Lyon, France<EMAIL_ADDRESS>
###### Abstract.

We revisit the classification of Lorentz homogeneous spaces of dimension $3$,
fixing statements in the literature and relaxing usual completeness
assumptions. In particular, non-unimodular elliptic plane waves, and only
them, are neither locally symmetric nor locally isometric to a left invariant
Lorentz metric on a $3-$dimensional Lie group. We characterize homogeneous
plane waves in dimension $3$, and prove they are non-extendable, and
geodesically complete only if they are symmetric. Finally, only one non-flat
plane wave has a compact model.

###### Contents

  1. 1 Introduction.
  2. 2 What’s new in the non-Riemannian case.
  3. 3 The Heisenberg extensions.
  4. 4 Proof of Theorem 1.6 in the solvable case.
  5. 5 Proof of Theorem 1.6 in the non-solvable case.
  6. 6 Homogeneous locally symmetric spaces, proof of Theorem 1.7.
  7. 7 Proof of Theorem 1.1.
  8. 8 Plane waves, proof of Theorem 1.2.
  9. 9 Global coordinates, proof of Theorem 1.3.
  10. 10 Proof of the completeness statement in Theorem 1.5.
  11. 11 Proof of the non-extendibility statement in Theorem 1.5.
  12. 12 Compact models.
  13. 13 Appendix: Rosen and Brinkmann coordinates

##  1\. Introduction.

In dimension two, homogeneous pseudo-Riemannian spaces have constant curvature
and so they have the maximal possible symmetries. This is no longer true in
higher dimensions, some homogeneous spaces are more homogeneous than others!
This applies to any pseudo-Riemannian type, but let us focus on the Lorentzian
case in dimension three. We have two (non-disjoint) distinguished classes:

\- On one hand, the symmetric spaces. Among them, the most symmetric are those
of constant sectional curvature, they are three, up to scaling, and they have
isometry groups of maximal dimension equal to six. We, then have three
decomposable symmetric spaces, and two reducible non-decomposable (Cahen-
Wallach) spaces having solvable isometry groups of dimension four.

\- On the other hand, there is the wide world of Lorentz groups, i.e. Lie
groups of dimension three endowed with a left invariant Lorentz metric. They,
generically, have an isometry group equal to the supporting group itself, but
they can, sometimes, have extra isometries.

It was believed that any homogeneous Lorentz space of dimension $3$ fits into
one of these two classes, that is, any homogeneous Lorentz space of dimension
three is either (locally) symmetric or isometric to a left invariant metric on
a Lie group of dimension three. This was early proven in the Riemannian case
in [25] and stated in the Lorentzian case relatively recently, in particular
in [5].
The initial motivations of the present work were, on one hand, to provide a
simple proof of this fact in the Lorentzian case, say a Lie theoretical proof
instead of using tools from the homogeneous structures theory, which involves
complicated computations (see for instance [28, 7], and for curvature
homogeneity [25, 26], and also [26, 22], for comparison between homogeneity
and curvature homogeneity). On the other hand, we aimed to fix some
completeness as well as local vs global issues in the literature. Trying this,
we realized there is one forgotten (in [5]) family of examples of homogeneous
Lorentz spaces that are neither locally symmetric nor locally isometric to
Lorentz groups! In [11], too, the same family was ruled out incorrectly in the
proofs of the classification of homogeneous Lorentz geometries having compact
models, which we fix here without altering statements (see §1.2.2).

###  1.1. Results on plane waves

The forgotten examples turn out to be plane waves (of elliptic type). Our
first result concerns the global algebraic structure of these spaces. Next
results will describe their global geometry. Notice that most studies on the
subject, have a mathematical-physical origin and usually consider
infinitesimal and local aspects, e.g. Killing fields,… [1, 2, 21], instead of
global isometry groups as in our present work. Actually [1] contains
classification results on homogeneous plane waves of arbitrary dimension, but
from the point of view of theoretical physics.

####  1.1.1. Heisenberg extensions

Let ${\mathsf{Heis}}$ be the Heisenberg group of dimension three and consider
a semi-direct product $G_{\rho}=\mathbb{R}\ltimes{\mathsf{Heis}}$, where
$\mathbb{R}$ acts on ${\mathsf{Heis}}$ via a representation
$t\to\rho(t)\in\sf{Aut}({\mathsf{Heis}})$.

######  Theorem 1.1.

Let $I$ be any non-central one parameter subgroup of ${\mathsf{Heis}}$
generated by a vector $W\in{\mathfrak{heis}}$, and put
$\mathsf{P}_{\rho}=G_{\rho}/I$. Then we have

  1. (1)

The $G_{\rho}$-action on $\mathsf{P}_{\rho}$ preserves a Lorentz metric if and
only if the projection of $W$ in ${\mathfrak{heis}}/{\sf{Z}}$ is not an
eigenvector of the action of $\rho(t)$ on ${\mathfrak{heis}}/{\sf{Z}}$ where
${\sf{Z}}$ is the center of ${\mathfrak{heis}}$. In addition this Lorentz
metric is unique up to automorphisms, fixing $W$, and scaling.

  2. (2)

There is no $3-$dimensional Lie group acting on $\mathsf{P}_{\rho}$
isometrically with an open orbit if, and only if, the action of $\rho(t)$ on
${\mathfrak{heis}}/{\sf{Z}}$ is by similarities (with non-real eigenvalues).

  3. (3)

The space $\mathsf{P}_{\rho}$ is symmetric exactly for those representations
$t\to\rho(t)\in\sf{Aut}({\mathsf{Heis}})$ fixing the center ${\sf{Z}}$, or
equivalently, $G_{\rho}$ is unimodular. In this case:

     * •

$\mathsf{P}_{\rho}$ is the Minkowski space if $\rho$ is unipotent,

     * •

$\mathsf{P}_{\rho}$ is a hyperbolic or elliptic Cahen-Wallach according to
that $\rho$ is hyperbolic or elliptic.

And besides the (global) symmetric cases, $\mathsf{P}_{\rho}$ is locally
symmetric exactly for the non-unimodular case for which $G_{\rho}$ is
isomorphic to $\sf{Aff}\ltimes\mathbb{R}^{2}$, where $\sf{Aff}$ is a copy in
${\sf{O}}(1,2)$ of the affine group $\sf{Aff}^{+}(\mathbb{R})$, preserving a
(degenerate) plane $\mathbb{R}^{2}\subset\mathbb{R}^{3}$. In this case,
$\mathsf{P}_{\rho}$ is globally isometric to half Minkowski.

  4. (4)

$\mathsf{P}_{\rho}$ is isometric to the Minkowski space exactly when $\rho$ is
unipotent. The group $G_{\rho}$, in this case, is isomorphic to
$N\ltimes\mathbb{R}^{3}\subset{\sf{O}}(1,2)\ltimes\mathbb{R}^{3}$, where $N$
is a unipotent one-parameter subgroup of ${\sf{O}}(1,2)$.

  5. (5)

There is only one flat non-complete case, isometric to half Minkowski, for
which the group $G_{\rho}$ is isomorphic to
$\sf{Aff}\ltimes\mathbb{R}^{2}\subset{\sf{O}}(1,2)\ltimes\mathbb{R}^{3}$. This
corresponds to any non-unimodular representation
$t\to\rho(t)\in\sf{Aut}({\mathsf{Heis}})$ having a fixed vector.

  6. (6)

Besides the Minkowski case, ${\sf{Iso}}^{0}(\mathsf{P}_{\rho})=G_{\rho}$.

  7. (7)

Two spaces $\mathsf{P}_{\rho}$ and $\mathsf{P}_{\rho^{\prime}}$ are isometric
if and only if $G_{\rho}$ and $G_{\rho^{\prime}}$ are isomorphic.

  8. (8)

Besides the two flat cases, two spaces $\mathsf{P}_{\rho}$ and
$\mathsf{P}_{\rho^{\prime}}$ are locally isometric if and only if $G_{\rho}$
and $G_{\rho^{\prime}}$ are isomorphic.

Observe that in order for $\mathsf{P}_{\rho}$ to admit a $G_{\rho}-$invariant
Lorentz metric, it is necessary and sufficient that $\rho$ does not act by
homotheties on ${\mathfrak{heis}}/{\sf{Z}}$.

####  1.1.2. Plane waves

They are special pp-waves which in turn are special Brinkmann spacetimes. All
these classes play a central role in General Relativity both from the
mathematical and applied sides, especially since the discovery of
gravitational waves.

A plane wave spacetime is defined by having a null parallel vector field $V$
such that there are local Brinkmann coordinates in which the metric has the
form:

$2dvdu+\sum H_{ij}(u)x^{i}x^{j}du^{2}+\sum_{i=1}^{n-2}(dx^{i})^{2}$

where $V$ corresponds to $\frac{\partial}{\partial v}$. They can also be
defined by having Rosen coordinates where the metric has the form:

$2dvdu+g_{u},$

where $g_{u}$ is a flat metric depending on $u$ of the form:
$g_{u}=g_{ij}(u)dx^{i}dx^{j}$. In a more intrinsic way, a plane wave is
characterized by having a null parallel vector field $V$, such that its
orthogonal distribution $V^{\perp}$ has flat leaves, and in addition, the
space is “almost symmetric”: $\nabla_{W}R=0$, for any $W$ tangent to
$V^{\perp}$, where $R$ is the Riemannian tensor.

There are many works on homogeneous plane waves, actually in any dimension,
see for instance [1]. In [15], the authors aim to know when a homogeneous pp-
wave is in fact a plane wave. The existing classifications are however so far
from being limpid and explicit, especially regarding the global isometry
group. We yield here a self-contained proof of the following:

######  Theorem 1.2.

Let $(M,g)$ be a simply connected homogeneous Lorentzian three-dimensional
manifold. Then $(M,g)$ is a plane wave if, and only if, it is globally
isometric to some $\mathsf{P}_{\rho}$.
If $M$ is not simply connected (and homogeneous), then it is either flat, or
isometric to a cyclic central quotient of one of the Cahen-Wallach spaces.

In fact, it is the centre of the Heisenberg algebra which plays the role of
the null parallel vector field. Observe, however, that the Heisenberg center
is central in $G_{\rho}$ exactly if $\mathsf{P}_{\rho}$ is symmetric (7.1).

####  1.1.3. Some terminology

It appears worthwhile for us to give names to the plane waves classes in the
following way:

  * •

The flat complete case corresponding to $\rho$ unipotent.

  * •

The Cahen-Wallach, hyperbolic or elliptic, case corresponding to $\rho$
unimodular semi-simple on ${\mathfrak{heis}}/{\sf{Z}}$.

  * •

In the remaining non-unimodular cases, let us call $\mathsf{P}_{\rho}$ a
hyperbolic (resp. elliptic, resp. parabolic) non-unimodular plane wave,
according to the action of $G_{\rho}$ on ${\mathfrak{heis}}/{\sf{Z}}$ being
diagonalizable with real eigenvalues (resp. non-real eigenvalues, resp. real
non-diagonalizable).

####  1.1.4. Global coordinates system

It is known that a Cahen-Wallach symmetric space has a global chart where the
metric has the form $g=2dudv\pm x^{2}du^{2}+dx^{2}$ (the minus sign
corresponds to the elliptic case and the plus to the hyperbolic one, see for
instance [16]). The global coordinate system, in the next theorem, already
appears in [1]:

######  Theorem 1.3.

Assume $\rho$ is non-unimodular, then $\mathsf{P}_{\rho}$ has a global
coordinate system $(u,v,x)\in\mathbb{R}^{+}\times\mathbb{R}\times\mathbb{R}$,
where the metric has the form:

$g=2dudv+\frac{b(\rho)x^{2}}{u^{2}}du^{2}+dx^{2}$

The invariant $b(\rho)$ is defined by $\rho(t)=e^{tA}$ where
$A=\left(\begin{array}[]{ccc}1&0&0\\\ 0&0&1\\\ 0&b&1\end{array}\right)$ acting
on the Lie algebra ${\mathfrak{heis}}$ (endowed with the basis $\\{Z,X,Y\\}$,
with $[X,Y]=Z$). Moreover, $G_{\rho}$ and $G_{\rho^{{}^{\prime}}}$ are
isomorphic if, and only if, $b(\rho)=b(\rho^{{}^{\prime}})$.

Observe that one can modify any derivation $A$, of ${\mathfrak{heis}}$ with
$A(Z)\neq 0$, by $\alpha A+\mathrm{ad}_{u}$, $u\in{\mathfrak{heis}}$, and gets
a matrix of the form above (the two derivations generate isomorphic semi-
direct products).

  * •

The case $b(\rho)=0$ corresponds to $G_{\rho}=\sf{Aff}\ltimes\mathbb{R}^{2}$,
where $\sf{Aff}\subset{\sf{SL}}(2,\mathbb{R})$ is the affine group acting by
its usual representation in $\mathbb{R}^{2}$. The space $\mathsf{P}_{\rho}$ in
this case is half Minkowski space $\\{(u,v,x),u>0\\}$ endowed with the metric
$2dudv+dx^{2}$.

  * •

For $b<-1/4$, the space $\mathsf{P}_{\rho}$ is elliptic, that is, the
$A_{\rho}$-action on ${\mathfrak{heis}}/{\sf{Z}}$ has complex eigenvalues and
for $b>-1/4$, the space $\mathsf{P}_{\rho}$ is hyperbolic, that is, $A_{\rho}$
has two different real eigenvalues.

  * •

For $b=-1/4$, $\mathsf{P}_{\rho}$ is the parabolic non-unimodular plane wave,
that is, $A_{\rho}$ is conjugate to: $\left(\begin{array}[]{ccc}1&0&0\\\
0&1/2&1\\\ 0&0&1/2\end{array}\right)$.

####  1.1.5. Compact models of homogeneous plane waves

######  Theorem 1.4.

There is exactly one non-flat plane wave which admits a compact model. This is
given in Brinkmann coordinates defined on the half space $\\{u>0\\}$ as:

$g=2dudv+\frac{2x^{2}}{u^{2}}du^{2}+dx^{2}.$

In Rosen coordinates, the metric takes the form: $g=2dudv+u^{-2}dx^{2}$.

Its isometry group contains a copy of the group
${\mathsf{SOL}}={\sf{SO}}(1,1)\ltimes\mathbb{R}^{2}$ acting properly. Any
lattice $\Gamma\subset{\mathsf{SOL}}$ gives a compact quotient
$\Gamma\backslash\mathsf{P}$. All compact models arise in this way.

####  1.1.6. Completeness and extendibility

######  Theorem 1.5.

The non-unimodular homogeneous plane waves are not complete. More precisely:

  * a)

Any null geodesic, which is not an orbit of the parallel field
$\frac{\partial}{\partial v}$, is incomplete.

  * b)

Any timelike geodesic is incomplete.

In addition, a homogeneous plane wave $\mathsf{P}_{\rho}$ is non-extendable.

###  1.2. Classification of homogeneous Lorentz 3-spaces

Coming back to the issue in the classification of homogeneous Lorentz
3-manifolds, we have:

######  Theorem 1.6.

Any simply connected homogeneous Lorentz $3-$manifold is globally isometric
to:

  * •

Either, a $3-$Lorentz group, i.e. a $3-$dimensional Lie group endowed with a
left invariant Lorentz metric,

  * •

Or, a globally symmetric space,

  * •

Or, a non-unimodular elliptic plane wave, i.e. $\mathsf{P}_{\rho}$ with $\rho$
acting on ${\mathfrak{heis}}/{\sf{Z}}$ by similarities with non-real
eigenvalues. In this case, $\mathsf{P}_{\rho}$ is neither locally symmetric
nor locally isometric to a Lorentz group.

In other words, one has to add elliptic non-unimodular plane waves to existing
classifications in the literature (e.g. in [5]). More importantly, we drop
here the completeness assumption usually made to get such a classification and
we also put some order in local vs global hypotheses.

####  1.2.1. Homogeneous locally symmetric case

It is natural to ask what exactly happens in the locally symmetric case.

######  Theorem 1.7.

Let $X$ be a Lorentz symmetric space of dimension $3$. Homogeneous (simply
connected) spaces $M$ modelled on $X$ correspond to open orbits of connected
closed subgroups $L\subset{\sf{Iso}}(X)$. Such $L$ may have dimension $3,4,5$
or $6$.

  * •

Let $L$ be a subgroup of ${\sf{Iso}}(X)$ of dimension $4$. Then $L$ has at
least one open orbit. It can have a discrete quantity of $2-$dimensional
orbits, which are furthermore (complete) totally geodesic lightlike surfaces
in $X$.

  * •

If $\dim L\geq 5$, then it acts transitively and $X$ has constant sectional
curvature.

####  1.2.2. Locally homogeneous compact Lorentz manifolds

In the vein of Thurston’s eight Riemannian homogeneous geometries in dimension
$3$, the authors of [11] study Lorentz geometries in dimension $3$ possessing
compact forms, that is, pairs $(G,X)$ with $X$ a Lorentzian $3-$manifold upon
which $G$ acts isometrically transitively and we want that some compact
manifold can be modelled on this $(G,X)$-structure. If the isotropy of $G$
acting on $X$ is compact, then $G$ preserves a Riemannian metric and we find
ourselves with Thurston’s geometries. It is then natural to assume the Lorentz
geometry to be of non-Riemannian type, that is, it has non-compact isotropy.

######  Theorem 1.8.

[11] Let $(M,g)$ be a locally homogeneous compact Lorentz $3-$manifold of non-
Riemannian type. Then, $M$ admits a Lorentz metric, probably different from
$g$, of (non-positive) constant sectional curvature. More precisely, up to
finite cover, $(M,g)$ is isometric to:

  * •

An anti de Sitter manifold $\Gamma\backslash\mathsf{AdS_{3}}$, where
$\mathsf{AdS_{3}}$ is ${\sf{PSL}}(2,\mathbb{R})$ endowed with its Killing
form, and
$\Gamma\subset{\sf{PSL}}(2,\mathbb{R})\times{\sf{PSL}}(2,\mathbb{R})$ acts
properly co-compactly on it.

  * •

Or $M$ is a quotient $\Gamma\backslash L$, $L$ a 3-dimensional Lie group,
$\Gamma\subset L$ a co-compact lattice, and metrically $L$ is endowed with a
left invariant Lorentz metric, which fits into one of the following cases:

    1. (1)

$L=\mathbb{R}^{3}$ endowed with a flat metric.

    2. (2)

$L={\mathsf{Heis}}$ or $L={\mathsf{SOL}}$. In both cases, $L$ has (up to
isomorphism) two left invariant metrics, with non-compact isotropy, one of
which is flat.

    3. (3)

$L={\sf{PSL}}(2,\mathbb{R})$. It has, up to isomorphism, two left invariant
metrics of non-compact isotropy, in addition to the $\mathsf{AdS}$-one which
is given by its Killing form.

In proofs of [11], some cases of Heisenberg extensions, actually corresponding
to elliptic plane waves, were ruled out incorrectly, because they were treated
as algebraic groups (while they are not)! The present proof (Theorem 1.4) of
the non-existence of compact models of elliptic plane waves fills this gap.

Let us finally observe that the algebraically most difficult case is the anti
de Sitter one. It is indeed delicate to describe discrete subgroups
$\Gamma\subset{\sf{PSL}}(2,\mathbb{R})\times{\sf{PSL}}(2,\mathbb{R})$ acting
properly co-compactly on ${\sf{PSL}}(2,\mathbb{R})$ (see [9, 24] for recent
results).

####  1.2.3. Organization of the article

We start (§2) by classifying homogeneous $3-$dimensional Riemannian manifolds,
essentially for two purposes: to provide a simpler proof (compared to the ones
in the literature, for instance in [25]), and to compare with the Lorentzian
situation to see what is new.
In §3 we define our family of Lorentz homogeneous spaces $\mathsf{P}_{\rho}$,
as quotients of Heisenberg extensions. These spaces will play the major role
in the two main parts of this paper, on one hand, they are the spaces of
discussion in the part about plane waves and, on the other hand, a sub-family
of them will constitute the ”new” spaces that were forgotten in previous
classifications.
Sections 4 and 5 will be devoted to the proof of the classification theorem
1.6. For easiness, the proof is divided into two parts: §4 treats the case
where the isometry group is solvable and §5 treats the non-solvable case.
Proof of Theorem 1.7, about the classification of homogeneous locally
symmetric Lorentz $3-$manifolds, is presented in §6.
Theorem 1.1 concerning the algebraic structure of the spaces
$\mathsf{P}_{\rho}$ is proven in §7.
Proofs of statements about plane waves start, essentially, from §8. Section 8
is devoted to the proof of Theorem 1.2 that characterizes all homogeneous
$3-$dimensional plane waves. Theorem 1.3, about the global coordinates for
non-unimodular plane-waves, is proven in §9. Proof of Theorem 1.5 is divided
into two parts: the completeness statement is proven in §10, and the non-
extendibility statement is discusses in §11. Finally, proof of the statement
of the (non-)existence of compact models (Theorem 1.4) is detailed in §12.

### Acknowledgments

We would like to warmly thank the referee for the valuable remarks and
corrections that greatly improved the content of the paper. We also thank Ines
Kath for answering our question on which spaces $\mathsf{P}_{\rho}$ are Cahen-
Wallach, which was a crucial step in the development of our work. The first
author is supported by the DFG SFB/TRR 191 Symplectic Structures in Geometry,
Algebra and Dynamics.

##  2\. What’s new in the non-Riemannian case.

As we mentioned before, it is known that any homogeneous, simply connected,
Riemannian manifold of dimension three is either symmetric or isometric to a
left invariant metric on a Lie group of dimension three [25]. Comparing this
to the Lorentz situation led us to new examples. This section is devoted to
this comparison.

###  2.1. The Riemannian case

We give here a (sketchy) proof of the classification of three dimensional
homogeneous, simply connected, Riemannian manifolds. This will lead us, later,
to the Lorentzian case where a new family of examples will be constructed due
to the fact that new considerations (unlike the Riemannian situation) have to
be adapted.

######  Theorem 2.1.

Any simply connected homogeneous Riemannian 3-manifold $(M,g)$ is (globally)
symmetric, or isometric to a 3-Riemannian group (i.e a Lie group with a left
invariant Riemannian metric).

###### Proof.

(Sketch) denote by $G$ the identity component of the full isometry group of
$(M,g)$, let $m$ be a point of $M$, and $I$ its stabilizer (which is
connected). The isotropy representation of $I$ in ${\sf{SO}}(3)$ is faithful,
so $\dim(I)\leq 3$. Since ${\sf{SO}}(3)$ has no two dimensional subgroups, $I$
is either three dimensional, one dimensional, or trivial. If the dimension is
zero then we are done ($(M,g)$ is a Riemannian Lie group). Suppose that $I$ is
three dimensional, then the sectional curvature depends only on $m$, but $M$
is homogeneous, so $M$ is globally isometric to one of the Riemannian models
with constant sectional curvature.

Suppose now that $I$ is one dimensional, or, equivalently $G$ is four-
dimensional.

######  Fact 2.2.

the radical of ${\mathsf{Lie}}(G)$, denoted by $\sf{R}({\mathsf{Lie}}(G))$, is
non-trivial.

###### Proof.

If not, then $G$ is, by definition, a semi-simple Lie group of dimension $4$,
but in dimension four there are no semi-simple Lie groups. ∎

If $\dim(\sf{R}({\mathsf{Lie}}(G)))=1$, we have, by Levi’s decomposition, two
possibilities, either ${\mathsf{Lie}}(G)=\mathfrak{su}(2)\ltimes\mathbb{R}$ or
${\mathsf{Lie}}(G)=\mathfrak{sl}(2,\mathbb{R})\ltimes\mathbb{R}$. But, semi-
simplicity of $\mathfrak{sl}(2,\mathbb{R})$ and $\mathfrak{su}(2)$ implies
that the above semi-direct product is indeed a direct product since they don’t
admit non-trivial one-dimensional representations. In both cases if
${\mathfrak{i}}:={\mathsf{Lie}}(I)\subset\mathfrak{su}(2)$ or
${\mathsf{Lie}}(I)\subset\mathfrak{sl}(2,\mathbb{R})$, we have
$I={\sf{SO}}(2)$ up to conjugacy, and the tangent space of $G/I$ at $I$ is
$\mathbb{R}\oplus({\mathfrak{s}}/{\mathfrak{i}})$ with ${\mathfrak{s}}$ equals
$\mathfrak{sl}(2,\mathbb{R})$ or $\mathfrak{su}(2)$. Hence $\mathbb{R}$ is
exactly the factor on which $I$ acts trivially and
${\mathfrak{s}}/{\mathfrak{i}}$ is orthogonal to $\mathbb{R}$ since this is
the unique $I-$invariant subspace transverse to ${\mathfrak{i}}$. This implies
that the metric on $\mathbb{R}\oplus({\mathfrak{s}}/{\mathfrak{i}})$ is a
product metric. Therefore, up to scaling, $(M,g)$ is isometric to the
symmetric spaces:

$\mathbb{R}\times({\sf{SL}}(2,\mathbb{R})/{\sf{SO}}(2))=\mathbb{R}\times\mathbb{H}^{2}\
\ \text{or}\ \
\mathbb{R}\times({\sf{SU}}(2)/{\sf{SO}}(2))=\mathbb{R}\times\mathbb{S}^{2}.$

And, if ${\mathsf{Lie}}(I)$ is transverse to the semi-simple factor, then
$(M,g)$ is isometric to a left invariant metric on
$\widetilde{{\sf{SL}}}(2,\mathbb{R})$ or ${\sf{SU}}(2)$.

If $\dim(\sf{R}({\mathsf{Lie}}(G)))\geq 2$ then ${\mathsf{Lie}}(G)$ is
solvable and we have that ${\mathsf{Lie}}(I)$ cannot be contained in
$[{\mathsf{Lie}}(G),{\mathsf{Lie}}(G)]$ because, if not, we would have that
$\mathrm{ad}_{h}$ is nilpotent for every $h\in{\mathsf{Lie}}(I)$, so it cannot
be the infinitesimal isometry of a Riemannian inner product (unless
$\mathrm{ad}_{h}$ is identically zero, but this is not the case since the
derivative representation of $I$ is faithful, so $I$ cannot be central). Using
this along with the fact that
${\mathsf{Lie}}(G)/[{\mathsf{Lie}}(G),{\mathsf{Lie}}(G)]$ is abelian, one can
find an ideal $L$ supplementary to ${\mathsf{Lie}}(I)$ by pull-backing a
hyperplane in ${\mathsf{Lie}}(G)/[{\mathsf{Lie}}(G),{\mathsf{Lie}}(G)]$
transverse to the projection of ${\mathsf{Lie}}(I)$. Since the normal subgroup
corresponding to $L$ is transverse to $I$, then it acts transitively and
locally freely on $M=G/I$. But $M$ is simply connected, which implies that the
action is in fact free, and $(M,g)$ is globally isometric to a left invariant
metric on this Lie group.

∎

###  2.2. The Lorentzian case

One may observe that many ideas in the Riemannian case can be generalized
immediately to the case of homogeneous Lorentzian three-manifolds, except, a
priory, two of them:

  1. (1)

The one concerning the fact that $I$ cannot be two dimensional. Here one
observes that ${\sf{SO}}(1,2)$ contains two-dimensional subgroups (the affine
group of $\mathbb{R}$ which is unique, up to conjugacy, inside
${\sf{SO}}(1,2)$). But, still, one manages to show that it cannot represent
the full isotropy group of a point in a Lorentzian manifold.

  2. (2)

The fact that non-zero nilpotent endomorphisms cannot be the infinitesimal
isometry of a Riemannian inner product, which is possible in the Lorentzian
setting. This is the main difference which leads to new examples. We will
investigate all of this in what follows.

Let us first give some general considerations. Suppose that $(M,g)$ is a
locally homogeneous Lorentz three-manifold, meaning that any $p,q\in M$ have
neighbourhoods $U$ and $V$ such that there exists $f:U\to V$ isometric with
$f(p)=q$. In particular such $M$ is analytic and it is known that if it is
simply connected, then any locally defined Killing field extends (uniquely)
globally [20]. In this case, let ${\mathfrak{g}}$ be this Lie algebra of (all)
Killing fields and $G$ its associated simply connected group. Choose a base
point $p_{0}$ and let ${\mathfrak{i}}$ be its (full) isotropy subalgebra, i.e.
those Killing fields vanishing at $p_{0}$. We have the following:

######  Proposition 2.3.

The isotropy subalgebra ${\mathfrak{i}}$ cannot have dimension two.

###### Proof.

(Sketch) The idea relies on the fact that if we have a representation of
${\sf{SL}}(2,\mathbb{R})$ on some finite dimensional vector space $V$ such
that there is $v\in V$ fixed by a two-dimensional (connected) subgroup
$H\subset{\sf{SL}}(2,\mathbb{R})$ (which is unique up to conjugacy) then $v$
is fixed by ${\sf{SL}}(2,\mathbb{R})$ (this is due to the fact that
${\sf{SL}}(2,\mathbb{R})/H\sim\mathbb{S}^{1}$ and ${\sf{SL}}(2,\mathbb{R})$
cannot have non-trivial compact orbits when acting linearly on $V$). We have
in our situation that the faithful representation, of local isometries fixing
$p$, contains a $2-$dimensional subgroup inside ${\sf{SO}}_{0}(1,2)$ that
fixes the Ricci tensor $\sf{Ric}_{p}$, and since ${\sf{SO}}_{0}(1,2)$ is
isomorphic to ${\sf{SL}}(2,\mathbb{R})$ one deduces from the previous
discussion that $\sf{Ric}_{p}$ is fixed by all ${\sf{SO}}_{0}(1,2)$. But any
quadratic form invariant by ${\sf{SO}}_{0}(1,2)$ is proportional to the metric
itself. So our Lorentz manifold is Einstein and it is known that, in dimension
three, this is equivalent to having constant sectional curvature which implies
that ${\mathfrak{i}}$ is not the full isotropy subalgebra. ∎

So, if $(M,g)$ does not have constant sectional curvature then $\dim(I)$ can
be $0$ or $1$ and, hence $\dim(G)=3$ or $4$. It is known that in a simply
connected Lie group $G$ of dimension $\leq 5$, any connected Lie subgroup is
closed. Indeed, if $G$ is solvable then this is true in every dimension (see
[18] page 187, last paragraph). If $\dim(G)=4$ and $G$ is not solvable, then
its Lie algebra is either $\mathfrak{su}(2)\oplus\mathbb{R}$ or
$\mathfrak{sl}(2,\mathbb{R})\oplus\mathbb{R}$ and in both cases we know that
any connected Lie subgroup is closed. If $\dim(G)=5$ and $G$ is not solvable,
then its Lie algebra is either $\mathfrak{su}(2)\oplus\mathbb{R}^{2}$, or
$\mathfrak{sl}(2,\mathbb{R})\oplus\mathbb{R}^{2}$, or
$\mathfrak{sl}(2,\mathbb{R})\ltimes\mathbb{R}^{2}$ and, again, the same holds
in this case. But, observe that this is no longer true in higher dimensions as
can be shown in dimension $6$, with $I$ a dense subgroup in
${\sf{SO}}(2)\times{\sf{SO}}(2)\subset{\sf{SU}}(2)\times{\sf{SU}}(2)$, which
is simply connected. Therefore, in our situation, the quotient $G/I$ exists
and $M$ is locally isometric to it.

Observe that one can also find simply connected Lorentz homogeneous spaces
$M=G/I$ with $\dim(G)=5$ and $\dim(I)=2$. But in this case, the Lie algebra
${\mathfrak{g}}$ of $G$ is not the full Killing algebra of $M$. The full
Killing algebra ${\mathfrak{g}}^{\prime}$ has dimension $6$ and $M$ has
constant curvature, but its associated group $G^{\prime}$, a priori, acts only
locally on $M$. This is because $M$ could be incomplete. However, one can
prove that a constant curvature manifold admitting an isometric action of a
$5-$dimensional group is complete (1.7).

##  3\. The Heisenberg extensions.

######  Definition 3.1.

We say that a semi-direct product $G=\mathbb{R}\ltimes{\mathsf{Heis}}$ is non-
real if the $\mathbb{R}$-action on $\mathbb{R}^{2}={\mathsf{Heis}}/{\sf{Z}}$
(where ${\sf{Z}}$ denotes the center of ${\mathsf{Heis}}$) has complex non-
real spectrum. Equivalently, this action is conjugate to a non-trivial
similarity action, i.e. a one-parameter group
$R\subset\mathbb{R}^{*}\times{\sf{SO}}(2)\subset{\sf{GL}}(2,\mathbb{R})$ with
non-trivial projection on ${\sf{SO}}(2)$. For example, if $R$ is contained in
${\sf{SO}}(2)$, then $G$ is the oscillator group [27, 3]. In the opposite
case, we say that $G$ is real.

######  Proposition 3.2.

Let $I$ be any non-central one parameter group of ${\mathsf{Heis}}$ with
${\mathsf{Lie}}(I)={\mathfrak{i}}$. Let $G$ be any semi-direct product
$G=\mathbb{R}\ltimes{\mathsf{Heis}}$. Then:

  1. (1)

$G/I$ admits a $G$-invariant Lorentz metric if and only if the
$\mathbb{R}$-action does not preserve the abelian group of rank $2$ generated
by $I$ and ${\sf{Z}}$.

  2. (2)

Assume $G/I$ has a $G$-invariant Lorentz metric. Then there is a subalgebra
${\mathfrak{l}}$ of dimension $3$ transverse to ${\mathfrak{i}}$ if, and only
if, $G$ is real.

######  Lemma 3.3.

A nilpotent (non-zero) endomorphism $a$ of a $3-$dimensional linear space is
an infinitesimal isometry of some Lorentz scalar product, if, and only if its
nilpotency order equals $3$, that is, $a^{3}=0$ but $a^{2}\neq 0$.

###### Proof.

Indeed, for $u$ generic, the system $\\{a^{2}(u),a(u),u\\}$ is a basis in
which $a$ has a matrix $\left(\begin{array}[]{ccc}0&1&0\\\ 0&0&1\\\
0&0&0\end{array}\right)$. This is skew symmetric with respect to the Lorentz
form $-2dxdz+dy^{2}$. Suppose now that $a$ has nilpotency order two and $g$ is
an invariant Lorentz metric. Then $g(a(X),Y)+g(X,a(Y))=0$ for every $X$ and
$Y$. This show that $\ker(a)$ is orthogonal to $\sf{Im}(a)$ (which is, by
hypothesis, one-dimensional and inside $\ker(a)$). But, for arbitrary $X$ we
have

$g(a(X),X)+g(X,a(X))=2g(a(X),X)=0$

So every $X$, not in $\ker(a)$, is orthogonal to $\sf{Im}(a)$, which is
impossible since $g$ is non-degenerate. We conclude that $a$ must have order
three.

∎

Proof of (1). $G$ preserves a Lorentz metric on $G/I$ if and only if the
$\mathrm{Ad}(I)$ action on the quotient vector space
${\mathfrak{g}}/{\mathfrak{i}}$ (identified to the tangent space at the base
point $I$) preserves a Lorentz scalar product. Equivalently, the infinitesimal
action $\mathrm{ad}_{a}$ is skew symmetric with respect to a Lorentz scalar
product on ${\mathfrak{g}}/{\mathfrak{i}}$, where $a$ generates
${\mathfrak{i}}$. Consider a standard generating system $\\{Z,X,Y\\}$ of the
Heisenberg algebra ${\mathfrak{heis}}$, with a unique non-vanishing bracket
$Z=[X,Y]$. The Lie algebra ${\mathfrak{g}}$ is generated by adding a fourth
element $T$ (with brackets $[T,W]\in{\mathfrak{heis}}$, for any
$W\in{\mathfrak{heis}}$). Since all one-dimensional subspaces different from
$\mathbb{R}Z$ are equivalent by automorphisms (in ${\mathsf{Heis}}$), we can
assume ${\mathfrak{i}}=\mathbb{R}X$. Now, $\mathrm{ad}_{X}^{3}=0$. The action
of $\mathrm{ad}_{X}$ on ${\mathfrak{g}}/\mathbb{R}X$ has exactly degree $3$ if
and only if $\mathrm{ad}_{X}(T)\notin\mathbb{R}X\oplus\mathbb{R}Z$. This
proves the first part of the proposition.

Proof of (2). Assume, $\mathrm{ad}_{T}$ acting on
${\mathfrak{heis}}/\mathbb{R}Z\cong\mathbb{R}X\oplus\mathbb{R}Y$ has a non-
vanishing eigenvector $u$. Then $u\notin\mathbb{R}X\oplus\mathbb{R}Z$, since
we are in situation of existence of an $\mathrm{ad}_{\mathfrak{i}}-$invariant
scalar product. Thus
${\mathfrak{l}}=\mathbb{R}T\oplus\mathbb{R}u\oplus\mathbb{R}Z$ is a Lie
algebra transverse to ${\mathfrak{i}}$. Conversely, if such a ${\mathfrak{l}}$
exists, consider its intersection
${\mathfrak{l}}^{\prime}={\mathfrak{l}}\cap{\mathfrak{heis}}$. It has
dimension $2$ and must contain $Z$, and hence another element
$u\in\mathbb{R}X\oplus\mathbb{R}Y$. It also contain and element
$T^{\prime}=T+v$, $v\in{\mathfrak{heis}}$. Now
$[T,u]+[v,u]=[T^{\prime},u]\in{\mathfrak{l}}^{{}^{\prime}}=\mathbb{R}Z\oplus\mathbb{R}u$.
Since $[v,u]\in\mathbb{R}Z$, the class of $u$ is an eigenvector of
$\mathrm{ad}_{T}$ acting on ${\mathfrak{heis}}/\mathbb{R}Z$. Therefore this
last endomorphism has a real spectrum. $\Box$

###  3.1. Description of the Heisenberg extensions

Let $\rho:\mathbb{R}\to\sf{Aut}({\mathfrak{heis}})$ be a representation into
the automorphism group of the Heisenberg algebra. This one-parameter group is
of the form $\exp(tA_{\rho})$ where $A_{\rho}$ is a derivation of the
Heisenberg algebra. Put $G_{\rho}={\mathsf{Heis}}\rtimes_{\rho}\mathbb{R}$ the
associated Lie group with Lie algebra, denoted by ${\mathsf{Lie}}(G_{\rho})$,
generated by $Z,X,Y$ and $T$ such that:

$[X,Y]=Z\ \ \text{and}\ \ [T,W]=A_{\rho}(W)\ \ \text{for every}\ \
W\in{\mathfrak{heis}}.$

In particular,
$[{\mathsf{Lie}}(G_{\rho}),{\mathsf{Lie}}(G_{\rho})]\subset{\mathfrak{heis}}$
(more precisely, the derived subalgebra is generated by the span of the image
of $A_{\rho}$ and $Z$ ). Let now $V\in{\mathfrak{heis}}$ be an element which
is not central. Hence
$\mathrm{ad}_{V}:{\mathsf{Lie}}(G_{\rho})\to{\mathsf{Lie}}(G_{\rho})$ is
nilpotent (${\mathfrak{heis}}$ is contained in the nil-radical and we have
equality if $A_{\rho}$ is not nilpotent). Let now $I\subset G_{\rho}$ be the
one-parameter subgroup (necessarily closed) tangent to $V$ and consider the
homogeneous space $\mathsf{P}_{\rho}=G_{\rho}/I$. In what follows we will be
interested in finding all possible situations where $\mathsf{P}_{\rho}$
possesses a $G_{\rho}-$invariant Lorentz metric and classify them.

Observe that we always have that $A_{\rho}(Z)=\alpha Z$ and $A_{\rho}$ induces
a linear transformation:

$\widetilde{A_{\rho}}:{\mathfrak{heis}}/Z=\mathbb{R}^{2}\to\mathbb{R}^{2}$

Where ${\sf{tr}}(\widetilde{A_{\rho}})=\alpha$ .

####  3.1.1. The real diagonalizable case

Suppose that $\widetilde{A_{\rho}}$ is diagonalizable with real eigenvalues.
Then, up to conjugacy by automorphisms, we have:

$A_{\rho}=\begin{pmatrix}a+b&0&0\\\ 0&a&0\\\ 0&0&b\end{pmatrix}$

with respect to the basis $\\{Z,X,Y\\}$ where either $a$ or $b$ is non-zero
(since we are in the situation of the existence of elements in
${\mathfrak{heis}}$ with nilpotency order $3$). We can assume $a\neq 0$ and we
have in addition that $a\neq b$ (observe that in this case $G_{\rho}$ is
unimodular if and only if $a=-b$). Furthermore, we can assume that $a=1$
because scaling $A_{\rho}$ doesn’t change the one-parameter group. Thus we
have a one-parameter family of Lie algebras, denoted ${\mathsf{Lie}}(G_{b})$
this time, given by the derivations:

$A_{b}=\begin{pmatrix}1+b&0&0\\\ 0&1&0\\\ 0&0&b\end{pmatrix}$

for $b$ different from $1$ (observe that $b$ and $1/b$ give isomorphic
algebras for $b\neq 0$). the brackets of this Lie algebra are given by:

$[T,Z]=(1+b)Z,\ \ [T,X]=X,\ \ [T,Y]=bY,\ \ \text{and}\ \ [X,Y]=Z$

Let now $V\in{\mathfrak{heis}}\subset{\mathsf{Lie}}(G_{b})$ be a non-central
element. We have seen that $\mathrm{ad}_{V}$ is nilpotent and preserves a
Lorentz metric on ${\mathsf{Lie}}(G_{b})/\mathbb{R}V$ if and only if it has
order three, so we always assume this. Put $V=\alpha X+\beta Y+\gamma Z$. We
can assume, up to inner conjugacy in ${\mathfrak{heis}}$, that $\gamma=0$ and
observe that both $\alpha$ and $\beta$ must be different from zero otherwise
$\mathrm{ad}_{V}$ has order two. In addition, all such $V$ are equivalent by
automorphisms of ${\mathsf{Lie}}(G_{b})$, it suffices to consider an
automorphism $\varphi:{\mathsf{Lie}}(G_{b})\to{\mathsf{Lie}}(G_{b})$ given by
$\varphi(X)=t_{1}X$, $\varphi(Y)=t_{2}Y$, $\varphi(Z)=t_{1}t_{2}Z$, and
$\varphi(T)=T$.

Suppose now that $V=X+Y$ and put $m={\sf{Span}}(T,Y^{\prime}=X+bY,Z)$. Then
$m$ is $\mathrm{ad}_{V}-$invariant and we have:

$\mathrm{ad}_{V}(T)=-Y^{\prime},\ \ \mathrm{ad}_{V}(Y^{\prime})=(b-1)Z,\ \
\text{and}\ \ \mathrm{ad}_{V}(Z)=0$

So giving a $G_{b}-$invariant Lorentzian metric on $P_{b}=G_{b}/I$, where
$I=\exp(tV)$, is equivalent to giving a Lorentzian $\mathrm{ad}_{V}-$invariant
inner product $g(.,.)$ on $m$. That is:

$g(\mathrm{ad}_{V}(W),U)+g(W,\mathrm{ad}_{V}(U))=0$

for every $W,U\in m$. By direct evaluations we find:

$g(T,T)=\beta,\ \ g(Y^{\prime},Y^{\prime})=\alpha,\ \
g(T,Z)=\frac{\alpha}{b-1}$

and:

$g(Z,Z)=g(Y^{\prime},Z)=g(T,Y^{\prime})=0$

where $\alpha$ is different from zero. But, up to automorphisms fixing $V$ and
preserving $m$, we can assume $\beta=0$. it suffices to consider the
automorphism $\varphi:{\mathsf{Lie}}(G_{b})\to{\mathsf{Lie}}(G_{b})$ defined
by $\varphi(X)=X$, $\varphi(Y)=Y$, $\varphi(T)=T+\delta Z$, and $\varphi(Z)=Z$
with a suitable choice of $\delta$. This shows that the Lorentzian homogeneous
space $P_{b}$ is unique up to isometry and scaling.

####  3.1.2. The real parabolic case

Suppose now that $\widetilde{A_{\rho}}$ has real non-zero eigenvalues but not
diagonalizable. we have in this case that, up to conjugacy by automorphisms:

$A_{\rho}=\begin{pmatrix}2t&0&0\\\ 0&t&s\\\ 0&0&t\end{pmatrix}$

with respect to the basis $\\{Z,X,Y\\}$ where $t$ and $s$ are non-zero. We can
assume that $t=s=1$ up to scaling and automorphisms of the Heisenberg algebra.
Thus we have a Lie algebra ${\mathsf{Lie}}(G_{\rho})$ given by the derivation:

$A_{\rho}=\begin{pmatrix}2&0&0\\\ 0&1&1\\\ 0&0&1\end{pmatrix}$

The brackets of this Lie algebra are given by:

$[T,Z]=2Z,\ \ [T,X]=X,\ \ [T,Y]=X+Y,\ \ \text{and}\ \ [X,Y]=Z$

Let $V=\alpha X+\beta Y+\gamma Z$ be a non-central element whose nilpotency
order is $3$. We can always assume, up to inner conjugacy in
${\mathfrak{heis}}$, that $\gamma=0$. Observe that $\beta\neq 0$ otherwise
$\mathrm{ad}_{V}$ has order two. In addition, all such $V$ are equivalent by
automorphisms of ${\mathsf{Lie}}(G_{\rho})$, it suffices to consider
$\varphi:{\mathsf{Lie}}(G_{\rho})\to{\mathsf{Lie}}(G_{\rho})$ given by
$\varphi(X)=X$, $\varphi(Y)=Y+tX$, $\varphi(Z)=Z$, and $\varphi(T)=T$ with a
suitable value of $t$.

Suppose now that $V=Y$ and put $m={\sf{Span}}(T,Y^{\prime}=X+Y,Z)$. Then $m$
is $\mathrm{ad}_{V}-$invariant and we have:

$\mathrm{ad}_{V}(T)=-Y^{\prime},\ \ \mathrm{ad}_{V}(Y^{\prime})=-Z,\ \
\text{and}\ \ \mathrm{ad}_{V}(Z)=0$

Suppose $g(.,.)$ is a Lorentzian $\mathrm{ad}_{V}-$invariant inner product on
$m$. That is:

$g(\mathrm{ad}_{V}(W),U)+g(W,\mathrm{ad}_{V}(U))=0$

for every $W,U\in m$. By evaluations we find:

$g(T,T)=\beta,\ \ g(Y^{\prime},Y^{\prime})=\alpha,\ \ g(T,Z)=-\alpha$

and:

$g(Z,Z)=g(Y^{\prime},Z)=g(T,Y^{\prime})=0$

where $\alpha$ is different from zero. We can also assume $\beta=0$ by
considering the automorphism
$\varphi:{\mathsf{Lie}}(G_{\rho})\to{\mathsf{Lie}}(G_{\rho})$ defined by
$\varphi(X)=X$, $\varphi(Y)=Y$, $\varphi(T)=T+\delta Z$, and $\varphi(Z)=Z$
with a suitable choice of $\delta$. This shows, similar to the previous cases,
that the Lorentzian homogeneous space $\mathsf{P}_{\rho}=G_{\rho}/\exp(tV)$ is
unique up to isometry and scaling.

####  3.1.3. The non-real case

Consider the case where $\widetilde{A_{\rho}}$ has complex eigenvalues. In
this case $A_{\rho}$ preserves a plane transverse to the central direction and
acting on it by a similarity. Since all transverse planes are equivalent by
Heisenberg inner automorphisms, we can assume that, up to scaling:

$A_{\rho}=\begin{pmatrix}2\cos(\theta)&0&0\\\ 0&\cos(\theta)&-\sin(\theta)\\\
0&\sin(\theta)&\cos(\theta)\end{pmatrix}$

with respect to the basis $\\{Z,X,Y\\}$ where $\theta\in\left(0,\pi\right)$.
Here, one thinks of $A_{\rho}$ acting on the plane ${\sf{Span}}(X,Y)$ as a
multiplication by a complex number in the upper half of the unit circle. But,
in order to simplify the brackets, we can scale and assume that $A_{\rho}$
acts on ${\sf{Span}}(X,Y)$ by multiplying by a complex number of the form
$i+c$ with $c\in\mathbb{R}$. Thus, we have a one-parameter family of Lie
algebras, which we denote by ${\mathsf{Lie}}(G_{c})$, given by the derivation
(observe also that one can take only $c\geq 0$):

$A_{c}=\begin{pmatrix}2c&0&0\\\ 0&c&-1\\\ 0&1&c\end{pmatrix}$

The brackets of this Lie algebra are given by:

$[T,Z]=2cZ,\ \ [T,X]=Y+cX,\ \ [T,Y]=-X+cY,\ \ \text{and}\ \ [X,Y]=Z$

As before if $V=\alpha X+\beta Y+\gamma Z$ is a non-central element we can
always assume that $\gamma=0$. All such $V$ are equivalent to $V=X$ by
automorphisms $\varphi:{\mathsf{Lie}}(G_{c})\to{\mathsf{Lie}}(G_{c})$ of the
form $\varphi(X)=\alpha X+\beta Y$, $\varphi(Y)=-\beta X+\alpha Y$,
$\varphi(Z)=(\alpha^{2}+\beta^{2})Z$, and $\varphi(T)=T$ which send $X$ to
$V$.

Suppose now that $V=X$ and $m={\sf{Span}}(T,Y^{\prime}=Y+cX,Z)$. Then $m$ is
$\mathrm{ad}_{V}-$invariant and we have:

$\mathrm{ad}_{V}(T)=-Y^{\prime},\ \ \mathrm{ad}_{V}(Y^{\prime})=Z,\ \
\text{and}\ \ \mathrm{ad}_{V}(Z)=0$

Suppose $g(.,.)$ is a Lorentzian $\mathrm{ad}_{V}-$invariant inner product on
$m$. By evaluations, as before, we find:

$g(T,T)=\beta,\ \ g(Y^{\prime},Y^{\prime})=\alpha,\ \ g(T,Z)=\alpha$

and:

$g(Z,Z)=g(Y^{\prime},Z)=g(T,Y^{\prime})=0$

where $\alpha$ is different from zero. By considering an automorphism
$\varphi:{\mathsf{Lie}}(G_{b})\to{\mathsf{Lie}}(G_{b})$ of the form
$\varphi(X)=X$, $\varphi(Y)=Y$, $\varphi(T)=T+\delta Z$, and $\varphi(Z)=Z$ we
can assume $\beta=0$ . This shows the uniqueness of the Lorentzian homogeneous
space $P_{c}=G_{c}/\exp(tV)$ up to isometry and scaling.

####  3.1.4. The nilpotent case

Lastly, suppose that $\widetilde{A_{\rho}}$ has only zero eigenvalues but not
diagonalizable. We have in this case that, up to scaling, conjugacy by
automorphisms, and adding a derivation of the form $\mathrm{ad}_{u}$ with
$u\in{\mathfrak{heis}}$:

$A_{\rho}=\begin{pmatrix}0&0&0\\\ 0&0&1\\\ 0&0&0\end{pmatrix}$

with respect to a basis $\\{Z,X,Y\\}$. The brackets of this Lie algebra are
given by:

$[T,Y]=X,\ \ \text{and}\ \ [X,Y]=Z$

Let $V=\alpha X+\beta Y+\gamma Z$ be a non-central element whose nilpotency
order is $3$. We can always assume, up to inner conjugacy in
${\mathfrak{heis}}$, that $\gamma=0$. Observe that $\beta\neq 0$ otherwise
$\mathrm{ad}_{V}$ has order two. In addition, all such $V$ are equivalent by
automorphisms of ${\mathsf{Lie}}(G_{\rho})$, it suffices to consider
$\varphi:{\mathsf{Lie}}(G_{\rho})\to{\mathsf{Lie}}(G_{\rho})$ given by
$\varphi(X)=\beta X$, $\varphi(Y)=\beta Y+\alpha X$, $\varphi(Z)=\beta^{2}Z$,
and $\varphi(T)=T$ which sends $Y$ to $V$.

Suppose now that $V=Y$ and put $m={\sf{Span}}(T,X,Z)$. Then $m$ is
$\mathrm{ad}_{V}-$invariant and we have:

$\mathrm{ad}_{V}(T)=-X,\ \ \mathrm{ad}_{V}(X)=-Z,\ \ \text{and}\ \
\mathrm{ad}_{V}(Z)=0$

Suppose $g(.,.)$ is a Lorentzian $\mathrm{ad}_{V}-$invariant inner product on
$m$. That is:

$g(\mathrm{ad}_{V}(W),U)+g(W,\mathrm{ad}_{V}(U))=0$

for every $W,U\in m$. By evaluations we find:

$g(T,T)=\beta,\ \ g(X,X)=\alpha,\ \ g(T,Z)=-\alpha$

and:

$g(Z,Z)=g(X,Z)=g(T,X)=0$

where $\alpha$ is different from zero. We can also assume $\beta=0$ by
considering the automorphism
$\varphi:{\mathsf{Lie}}(G_{\rho})\to{\mathsf{Lie}}(G_{\rho})$ defined by
$\varphi(X)=X$, $\varphi(Y)=Y$, $\varphi(T)=T+\delta Z$, and $\varphi(Z)=Z$
with a suitable choice of $\delta$. This shows that the Lorentzian homogeneous
space $\mathsf{P}_{\rho}=G_{\rho}/\exp(tV)$ is unique up to isometry and
scaling.

##  4\. Proof of Theorem 1.6 in the solvable case.

Recall that we have shown in 2.3 that if $(M,g)$ is a homogeneous simply
connected Lorentzian manifold with $\dim({\sf{Iso}}(M))\geq 5$, then $M$ has
constant sectional curvature. In fact, one shows, in this case, that $M$ is
complete (see 1.7). Now , since the case $\dim({\sf{Iso}}(M))=3$ corresponds
to the fact that $M$ is a left invariant metric on ${\sf{Iso}}^{0}(M)$, then
all what remains to consider is the case $\dim({\sf{Iso}}(M))=4$.

######  Proposition 4.1.

Let $G/I$ be a Lorentz space with $G$ solvable and has dimension $4$. Then
there exists a Lie subalgebra ${\mathfrak{l}}$ transverse to ${\mathfrak{i}}$,
if, and only if $G$ is not a non-real semi-direct product
$\mathbb{R}\ltimes{\mathsf{Heis}}$.

###### Proof.

Observe first that ${\mathfrak{i}}$ is not central since we always assume that
the action of $G$ on $G/I$ is faithful. Let $[{\mathfrak{g}},{\mathfrak{g}}]$
be the derived subalgebra. Since ${\mathfrak{g}}$ is solvable we have that
$[{\mathfrak{g}},{\mathfrak{g}}]$ is a proper subalgebra of ${\mathfrak{g}}$,
and we have the natural projection
$\pi:{\mathfrak{g}}\to{\mathfrak{g}}/[{\mathfrak{g}},{\mathfrak{g}}]=\mathbb{R}^{k}$
for $k\geq 1$. Suppose that ${\mathfrak{i}}$ is not contained in
$[{\mathfrak{g}},{\mathfrak{g}}]$, then ${\mathfrak{i}}$ projects injectively,
to a line $\pi({\mathfrak{i}})$, inside $\mathbb{R}^{k}$. Let
$H\subset\mathbb{R}^{k}$ be any hyperplane transverse to $\pi({\mathfrak{i}})$
(if $k=1$ then $H$ is reduced to $\\{0\\}$). Hence, $\pi^{-1}(H)$ is an ideal
transverse to ${\mathfrak{i}}$ in ${\mathfrak{g}}$ which proves the
proposition in this case.

So, it remains to consider the case where
${\mathfrak{i}}\subset[{\mathfrak{g}},{\mathfrak{g}}]$. Let
$a\in{\mathfrak{i}}$ be a generator of ${\mathfrak{i}}$, we have that
$\mathrm{ad}_{a}$ is nilpotent (recall that $\mathrm{ad}_{a}\neq 0$ since
${\mathfrak{i}}$ is not central). If $[{\mathfrak{g}},{\mathfrak{g}}]$ is
abelian then $\mathrm{ad}_{a}^{2}=0$, hence $\mathrm{ad}_{a}$ can not
infinitesimally preserves a Lorentz product (by Lemma 3.3), so
$[{\mathfrak{g}},{\mathfrak{g}}]$ is non-abelian. Being nilpotent and non-
abelian, $[{\mathfrak{g}},{\mathfrak{g}}]$ is therefore the Heisenberg algebra
(since $\dim([{\mathfrak{g}},{\mathfrak{g}}])\leq 3$) and this shows that the
group $G$ is, thus, a semi-direct product $\mathbb{R}\ltimes\mathsf{Heis}$
with $I\subset G$ is a non-central one-parameter group of ${\mathsf{Heis}}$,
which is already treated in Proposition 3.2. This completes the proof of
Proposition 4.1. ∎

######  Remark 4.2.

Observe from the proof that in all cases where $G$ is not a semi-direct
product of ${\mathsf{Heis}}$, we found ${\mathfrak{l}}$ transverse to
${\mathfrak{i}}$ which is an ideal.

###  4.1. Global identification

So far, we proved the existence of a sub-algebra ${\mathfrak{l}}$ transverse
to ${\mathfrak{i}}$ if $G$ is not a non-real semi-direct product of
${\mathsf{Heis}}$, but we want to prove that globally, $G/I$ is identified to
$L$, where $L$ is the group determined by ${\mathfrak{l}}$. This essentially
means that $L$ acts transitively on $G/I$, whereas the transversality means
that the $L$-orbit of the point $I$ in $G/I$ is open. It is likely, in our
setting here (e.g. $G$ solvable and $G/I$ of Lorentz type), that any open
orbit necessarily equals the full space. We will prove this for the $L$ we
found in the proofs of Propositions 4.1 and 3.2.

First, let us notice that for simply connected groups of dimension $4$, any
connected subgroup is closed (2.2). Let $L$ be the subgroup determined by the
subalgebra ${\mathfrak{l}}$. If $L$ is normal, then $G/L$ is isomorphic to
$\mathbb{R}$ as a group, and $I$ projects as a subgroup of it, and hence
isomorphically. That is $G/L=I$, or equivalently $G=LI$, and $I\cap
L=\\{1\\}$. In other words $L$ acts freely transitively on $G/I$.

From Remark 4.2, we can find $L$ normal unless $G$ is a semi-direct product
$\mathbb{R}\ltimes{\mathsf{Heis}}$. In this case let $I$ be the one parameter
group $\\{\exp(tX),t\in\mathbb{R}\\}\subset{\mathsf{Heis}}$ and the group $L$
transverse to $I$ as constructed in the proof of Propositions 3.2. The orbit
of a point $gI\in G/I$ under the action of $L$ is open if, and only if, $L$ is
transverse to $\text{Stab}(gI)=gIg^{-1}$ where $\text{Stab}(gI)$ is the
stabilizer of $gI$. We have that, as in the proof of 3.2, that
${\mathfrak{l}}=\mathbb{R}T\oplus\mathbb{R}u\oplus\mathbb{R}Z$ and
$\mathbb{R}u\oplus\mathbb{R}Z$ is an ideal in ${\mathfrak{g}}$. So the
conjugates of $X$ are never inside ${\mathfrak{l}}$ since they are never
inside the ideal
$\mathbb{R}u\oplus\mathbb{R}Z={\mathfrak{l}}\cap{\mathfrak{heis}}$. This shows
that $L$ is always transverse to the stabilizer of any point in $G/I$ which
implies that the $L-$orbit of every point $gI\in G/I$ is open. Hence $L$ acts
transitively locally freely on $G/I$ because $G/I$ is connected and the orbits
are open, and, by simple-connectedness, we have that the $L-$action is in fact
free. $\Box$

######  Theorem 4.3.

Let $G$ be a simply connected solvable group of dimension $4$ and $G/I$ a
Lorentz $3-$dimensional $G$ homogeneous space. Then there exists a simply
connected $3-$dimensional group $L$ acting isometrically, freely and
transitively on $G/I$ if, and only if, $G$ is not a non-real semi-direct
product $\mathbb{R}\ltimes{\mathsf{Heis}}$.

##  5\. Proof of Theorem 1.6 in the non-solvable case.

Let $G/I$ be a $3-$dimensional homogeneous Lorentz space with $G$ of dimension
$4$, non-solvable and simply connected. Necessarily, for the same reason that
we mentioned in the Riemannian case, $G$ is isomorphic to $\mathbb{R}\times
S$, with $S=\widetilde{{\sf{SL}}}(2,\mathbb{R})$ or ${\sf{SU}}(2)$. The
isotropy group $I$ cannot be the factor $\mathbb{R}$ since it is central (and
the $G$-action would be non-faithful). If $I$ is not contained in $S$, then
$S$ acts freely transitively and isometrically on $G/I$, since $S$ is normal
and transverse to $I$. In other words, the Lorentz space $G/I$ is identified
with a left invariant metric on $S$.

It remains to consider the case $I\subset S$, so $G/I=\mathbb{R}\times(S/I)$.
In this case,
${\mathfrak{g}}/{\mathfrak{i}}=\mathbb{R}\oplus({\mathfrak{s}}/{\mathfrak{i}})$.
Consider the case $S=\widetilde{{\sf{SL}}}(2,\mathbb{R})$. If
${\mathfrak{i}}\subset\mathfrak{sl}(2,\mathbb{R})$ is nilpotent, then it has
nilpotency of order $2$ when acting on ${\mathfrak{s}}/{\mathfrak{i}}$ (since
the dimension of this space is 2) and, hence, has the same order when acting
on $\mathbb{R}\times({\mathfrak{s}}/{\mathfrak{i}})$. Therefore, it cannot
infinitesimally preserve a Lorentz product. It follows that $I$ must be
hyperbolic or elliptic. In both cases, the factor $\mathbb{R}$ is exactly the
trivial factor of the $I$-action. Thus, it is orthogonal to
${\mathfrak{s}}/{\mathfrak{i}}$ (observe that $\mathbb{R}$ is not null).

One concludes from this that $G/I$ is a direct product
$\mathbb{R}\times(S/I)$. The metric on $\mathbb{R}$ can be positive or
negative. As for $S/I$, it can be: the sphere $\mathbb{S}^{2}$ if
$S={\sf{SU}}(2)$, the hyperbolic plane $\mathbb{H}^{2}$ or the de Sitter plane
$\mathbb{D}S^{2}=\widetilde{{\sf{SO}}}_{0}(1,2)/{\sf{SO}}_{0}(1,1)$ if
$S=\widetilde{{\sf{SL}}}(2,\mathbb{R})$. In particular, the space $G/I$ is
symmetric.

######  Remark 5.1.

In the case of $G=\mathbb{R}\times{\sf{SU}}(2)$, the unique $3-$dimensional
subgroup of $G$ is ${\sf{SU}}(2)$. Hence, if the isotropy group $I$ is
included in ${\sf{SU}}(2)$, then there is no Lie subgroup of $G$ transverse to
it, that is, there is no $3-$dimensional Lie subgroup of $G$ having an open
orbit. Whereas in the case of
$G=\mathbb{R}\times\widetilde{{\sf{SL}}}(2,\mathbb{R})$ and
$I\subset\widetilde{{\sf{SL}}}(2,\mathbb{R})$ hyperbolic, the subgroup
$L=\mathbb{R}\times\sf{Aff}$, where $\sf{Aff}$ is the group of upper
triangular matrices, has an open orbit in $G/I$ but it does not act
transitively.

##  6\. Homogeneous locally symmetric spaces, proof of Theorem 1.7.

### 6.1.

Let $M$ be modelled on $X$ and homogeneous. Let us begin with assuming that
$M$ is simply connected. Then it has a developing map $d:M\to X$, equivariant
with respect to a local isomorphism $\psi:G\to L\subset{\sf{Iso}}(X)$ (that
is, $d\circ g=\psi(g)\circ d$ for every $g\in G$). The $d$-image is an open
$L$-orbit. Conversely, any open orbit of a subgroup of ${\sf{Iso}}(X)$ is a
homogeneous space modelled on $X$.

Therefore, the proof of Theorem 1.7 reduces to the study of $L$-orbits, for
$L$ a subgroup of dimension $\geq 4$, essentially $\dim L=4$, since the
$5-$dimensional case follows. Observe first that in the case where $X$ is
symmetric, but not of constant curvature, $\dim({\sf{Iso}}(X))=4$ and there is
nothing to prove, and hence, we will assume that $X$ has constant curvature.

### 6.2.

The stabilizer of a line $\mathbb{R}v\in T_{x}X$ in
${\sf{SO}}(T_{x}X,g_{x})={\sf{SO}}(1,2)$ has dimension $2$ if $v$ is
isotropic, and dimension $1$ otherwise. It follows that a curve can be
preserved by a subgroup $\subset{\sf{Iso}}(X)$ of dimension at most $3$.
Hence, if $\dim(L)=4$, then any $L$-orbit is either a surface or is open.
Similarly, the stabilizer of a plane $F\subset T_{x}X$ has dimension $1$
unless $F$ is lightlike. It follows that if a connected surface $\Sigma$ is
preserved by a subgroup $L\subset{\sf{Iso}}(X)$ of dimension $4$, then it is
lightlike and homogeneous.

###  6.3. Action of $\sf{Aff}\subset{\sf{SO}}(1,2)$ near $0$ in
$\mathbb{R}^{1,2}$

Write the metric of $\mathbb{R}^{1,2}$ as $2dudv+dx^{2}$. The subgroup
$H\subset{\sf{SO}}(1,2)$ preserving the direction $\frac{\partial}{\partial
u}=(1,0,0)$ is isomorphic to the affine group $\sf{Aff}$. In fact any subgroup
in ${\sf{SO}}(1,2)$ of dimension $2$ is conjugate to $H$ and equals the
stabilizer of an isotropic direction.

Let $\Sigma_{0}$ be the lightlike affine $2-$plane tangent to
$\frac{\partial}{\partial u}^{\perp}=\mathbb{R}\frac{\partial}{\partial
u}\oplus\mathbb{R}\frac{\partial}{\partial x}$ at $0$. So,
$\Sigma_{0}=\\{v=0\\}$. The $H$-orbits of the $\Sigma_{0}-$points are
$\\{0\\}$, $\mathbb{R}(\frac{\partial}{\partial u})\setminus\\{0\\}$, and
(null) affine lines parallel to $\frac{\partial}{\partial u}$ defined
explicitly by $\\{v,=0,x=a\\}$ (for $a\in\mathbb{R}$). A level $E^{-}_{r}$
defined by

$2uv+x^{2}=-r^{2},\ \ r\neq 0$

is homothetic to a hyperbolic plane, and $H$ acts on it transitively.
Similarly, a level $E^{+}_{r}$ given by

$2uv+x^{2}=+r^{2},\ \ r\neq 0$

corresponds to a de Sitter plane. It intersects $\Sigma_{0}$ on the two lines
$\\{v=0,x=\pm r\\}$. Then $E^{+}_{r}$ minus these two lines is an $H$-orbit.
Finally $E_{0}-\Sigma_{0}$ is a $H$-orbit. This is the light cone with the
affine line tangent to $\frac{\partial}{\partial u}$ removed.

From all this, we infer that if a lightlike surface is $H$-invariant, then it
is either $E_{0}-\Sigma_{0}$ or an open subset of $\Sigma_{0}$.

###  6.4. $2-$dimensional $L$-orbits are lightlike and geodesic

Let $\Sigma$ be a $L$-orbit of dimension $2$. It is, thus, lightlike and has
isotropy of dimension $2$. In the Minkowski case, we can assume $0\in\Sigma$
and its isotropy is $H$. From our previous analysis, since $\Sigma$ is
lightlike and $H$-invariant, $\Sigma$ is $E_{0}-\Sigma_{0}$ or contained in
$\Sigma_{0}$.

Now, the subgroup of isometries preserving the cone $E_{0}$ is exactly
${\sf{SO}}(1,2)$, and in particular $E_{0}-\Sigma_{0}$ can not be preserved by
a $4-$dimensional group. It follows that $\Sigma$ is open in $\Sigma_{0}$ and
hence geodesic. In fact, the subgroup of elements preserving $\Sigma_{0}$ is
$H\ltimes T$, where $T\cong\mathbb{R}^{2}$ acts by translation on
$\Sigma_{0}$. It has dimension $4$ and, thus, coincides with $L$. We then see
that $L$ has $\Sigma_{0}$ as a unique $2-$dimensional orbit. It has two half-
Minkowski spaces as open orbits.

### 6.5.

In the non-flat case, one uses the exponential map at a point $p$ to get a
similar picture as that in Minkowski. One also uses Gauss lemma to see that
the metric levels at $p$ are orthogonal to the geodesic rays through $p$ and
hence are not lightlike, unless in the light cone. Similarly, one concludes
that $L$ is the stabilizer of a lightlike geodesic surface $\Sigma$ in $X$.

The de Sitter space $\mathsf{dS}_{3}$ has a model $2uv+x^{2}+y^{2}=1$ in
$\mathbb{R}^{4}$. A lightlike geodesic surface is given by:
$\\{(u,0,x,y)/x^{2}+y^{2}=+1\\}$. Its stabilizer $L$ is the stabilizer of the
direction $\mathbb{R}.(1,0,0,0)$ in ${\sf{SO}}(1,3)$. Thus, the exterior of
$\Sigma$ consists of two open orbits.

Consider now the anti de Sitter space $\mathsf{AdS}_{3}$. It is represented as
the level $2uv+x^{2}-y^{2}=-1$. Its intersection with a lightlike hyperplane
in $\mathbb{R}^{2,2}$ is up to isometry $\\{(u,0,x,y)/x^{2}-y^{2}=-1\\}$. It
has two connected components, and $\Sigma$ is one of them. In other words, the
stabilizer of $\Sigma$ has exactly another $2-$dimensional orbit, and has $2$
open orbits.

###  6.6. Case where $M$ is not simply connected

In the flat case, the group $L$ preserving a lightlike geodesic surface is, up
to conjugacy, $L=\sf{Aff}\ltimes\mathbb{R}^{2}$. If $M$, a flat homogeneous
manifold with $\dim{\sf{Iso}}(M)=4$, is not simply connected, then, its
isometry group is a non-trivial central quotient of $L$. But $L$ has trivial
center, and thus $M$ has to be simply connected.

In the case of $\mathsf{dS}_{3}$, the corresponding $L$ is a maximal parabolic
subgroup of ${\sf{SO}}(1,3)$, which is isomorphic to
$(\mathbb{R}\times{\sf{SO}}(2))\ltimes\mathbb{R}^{2}$. Here, too, $L$ has a
trivial center. The same conclusion applies to $\mathsf{AdS}_{3}$ as $L$ in
this case is isomorphic to $\sf{Aff}\times\sf{Aff}$. $\Box$

##  7\. Proof of Theorem 1.1.

######  Lemma 7.1.

Suppose that the representation $\rho:\mathbb{R}\to\sf{Aut}({\mathsf{Heis}})$
is unimodular. Then $\mathsf{P}_{\rho}$ is globally symmetric. Furthermore, if
$\rho$ is unipotent then $\mathsf{P}_{\rho}$ is isometric to Minkowski and, if
$\rho$ is hyperbolic (resp. elliptic) then $\mathsf{P}_{\rho}$ is a Cahen-
Wallach space.

###### Proof.

First, suppose that $\rho$ is unipotent, that is (3.1.4) $\rho$ is given by
the derivation:

$\mathrm{ad}_{T}=A_{\rho}=\begin{pmatrix}0&0&0\\\ 0&0&1\\\ 0&0&0\end{pmatrix}$

in the basis $\\{Z,X,Y\\}$. The isotropy in this case can be chosen to
correspond to the subalgebra ${\mathfrak{i}}=\mathbb{R}Y$. One sees that
${\sf{Span}}(T,X,Z)$ is an abelian subalgebra transverse to all conjugates of
${\mathfrak{i}}$. This shows (4.1) that $\mathsf{P}_{\rho}$ is isometric to a
translation invariant metric on $\mathbb{R}^{3}$ $i.e$ to Minkowski.

Suppose now that $\rho$ is hyperbolic, given by:

$A_{b}=\begin{pmatrix}0&0&0\\\ 0&1&0\\\ 0&0&-1\end{pmatrix}$

The brackets of ${\mathsf{Lie}}(G_{\rho})$ are given by:

$[T,X]=X,\ \ [T,Y]=-Y,\ \ \text{and}\ \ [X,Y]=Z$

And an inner Lorenz product (3.1.1) on $m={\sf{Span}}(T,Y^{\prime}=X-Y,Z)$
given by

$g(Y^{\prime},Y^{\prime})=1,\ \ g(T,Z)=-\frac{1}{2}$

and:

$g(T,T)=g(Z,Z)=g(Y^{\prime},Z)=g(T,Y^{\prime})=0$

Consider the automorphism of ${\mathsf{Lie}}(G_{\rho})$ given by:

$\varphi(T)=-T,\ \ \varphi(X)=Y,\ \ \varphi(Y)=X,\ \ \text{and}\ \
\varphi(Z)=-Z.$

So, $\varphi$ fixes ${\mathfrak{i}}=\mathbb{R}(X+Y)$ corresponding tho the
isotropy and acts on ${\mathsf{Lie}}(G_{\rho}/{\mathfrak{i}})$ by $-Id$.
Furthermore, $\varphi$ preserves $g$ (on $m$) and acts isometrically on
$\mathsf{P}_{\rho}=G_{\rho}/I$ by fixing the point $I$ with derivative $-Id$.
This shows that $\mathsf{P}_{\rho}$ is symmetric.

Similarly, if $\rho$ is elliptic (3.1.3) given by:

$A_{0}=\begin{pmatrix}0&0&0\\\ 0&0&-1\\\ 0&1&0\end{pmatrix}$

The brackets given by:

$[T,X]=Y,\ \ [T,Y]=-X,\ \ \text{and}\ \ [X,Y]=Z$

As before if $m={\sf{Span}}(T,Y,Z)$ (3.1.3). Then $m$ is
$\mathrm{ad}_{X}-$invariant and we have:

$\mathrm{ad}_{V}(T)=-Y,\ \ \mathrm{ad}_{V}(Y)=Z,\ \ \text{and}\ \
\mathrm{ad}_{V}(Z)=0$

with Lorentzian inner product

$g(Y,Y)=1,\ \ g(T,Z)=1$

and:

$g(T,T)=g(Z,Z)=g(Y,Z)=g(T,Y)=0$

And if we consider the automorphism:

$\varphi(T)=-T,\ \ \varphi(X)=X,\ \ \varphi(Y)=-Y,\ \ \text{and}\ \
\varphi(Z)=-Z.$

we obtain, the same as before, that $\mathsf{P}_{\rho}$ is symmetric. The last
two cases are known to be the only Cahen-Wallach spaces in dimension three
(see for instance [4, 16]). ∎

######  Lemma 7.2.

$\mathsf{P}_{\rho}$ has constant curvature if and only if it is flat. In
addition, if $\mathsf{P}_{\rho}$ is flat then it is either isometric to the
Minkowski space with $G_{\rho}$ isomorphic to
$\mathbb{R}\ltimes\mathbb{R}^{3}$ where $\mathbb{R}$ acts via a unipotent one-
parameter subgroup of ${\sf{SO}}(1,2)$, or $\mathsf{P}_{\rho}$ is isometric to
half Minkowski with $G_{\rho}$ non-unimodular isomorphic to
$\sf{Aff}\ltimes\mathbb{R}^{2}$.

###### Proof.

The first statement follows from the following fact: the (local) holonomy of a
non-flat three dimensional Lorentz manifold with constant curvature is
maximal. In particular, there cannot be a parallel null vector field, so if
$\mathsf{P}_{\rho}$ has constant curvature then it must be flat.
Suppose now that $\mathsf{P}_{\rho}$ is flat, then ${\mathsf{Lie}}(G_{\rho})$
has a faithful representation in ${\mathfrak{so}}(1,2)\ltimes\mathbb{R}^{3}$.
Consider the projection homomorphism:

$\pi:{\mathfrak{so}}(1,2)\ltimes\mathbb{R}^{3}\longrightarrow{\mathfrak{so}}(1,2)$

We know that ${\mathsf{Lie}}(G_{\rho})\cap\mathbb{R}^{3}$ is non-trivial and
equals the kernel of $\pi$ restricted to ${\mathsf{Lie}}(G_{\rho})$ and, also,
${\mathsf{Lie}}(G_{\rho})\cap{\mathfrak{so}}(1,2)$ has dimension at least $1$.
The case $\dim(\pi({\mathsf{Lie}}(G_{\rho})))=3$ is not possible since
$G_{\rho}$ is solvable. Suppose that $\dim(\pi({\mathsf{Lie}}(G_{\rho})))=1$
then necessarily
$\pi({\mathsf{Lie}}(G_{\rho}))={\mathsf{Lie}}(G_{\rho})\cap{\mathfrak{so}}(1,2)$.
In this case, $\mathbb{R}^{3}\subset{\mathsf{Lie}}(G_{\rho})$ and $G_{\rho}$
is isomorphic to $\mathbb{R}\ltimes\mathbb{R}^{3}$ where $\mathbb{R}$ must act
via a unipotent one parameter group (otherwise $G_{\rho}$ doesn’t contain a
copy of Heisenberg which is not the case). In this case, $\mathsf{P}_{\rho}$
is isometric to Minkowski (see 7.1).
Suppose now that $\dim(\pi({\mathsf{Lie}}(G_{\rho})))=2$, then
$\dim({\mathsf{Lie}}(G_{\rho})\cap\mathbb{R}^{3})=2$. So
${\mathsf{Lie}}(G_{\rho})$ is contained in the stabilizer of a (degenerate)
plane $\mathbb{R}^{2}$ inside ${\mathfrak{so}}(1,2)\ltimes\mathbb{R}^{3}$
(acting by conjugacy), which is isomorphic to $\sf{aff}\ltimes\mathbb{R}^{3}$.
Denote the basis of the Lie algebra $\sf{aff}\ltimes\mathbb{R}^{3}$ by
$\\{h,f,e_{1},e_{2},e_{3}\\}$ such that the stabilized plane is
${\sf{Span}}(e_{1},e_{2})$ and the brackets are given by:

$[h,f]=f\ \ \text{and}\ \ [e_{i},e_{j}]=0$

and the matrices of $\mathrm{ad}_{h},\mathrm{ad}_{f}$ acting on
$\mathbb{R}^{3}$ with respect to the basis $\\{e_{1},e_{2},e_{3}\\}$ are given
by:

$\mathrm{ad}_{f}=\begin{pmatrix}0&-1&0\\\ 0&0&1\\\ 0&0&0\end{pmatrix}\ \
\text{and}\ \ \mathrm{ad}_{h}=\begin{pmatrix}1&0&0\\\ 0&0&0\\\
0&0&-1\end{pmatrix}$

So we can suppose that ${\mathsf{Lie}}(G_{\rho})={\sf{Span}}(h+\beta
e_{3},f+\alpha e_{3},e_{1},e_{2})$. We have that $[h+\beta e_{3},f+\alpha
e_{3}]=f-\alpha e_{3}-\beta e_{2}$. Since the derived subalgebra must be
contained in ${\sf{Span}}(f+\alpha e_{3},e_{1},e_{2})$, we deduce that
$\alpha=0$. So:

${\mathsf{Lie}}(G_{\rho})={\sf{Span}}(h+\beta e_{3},f,e_{1},e_{2})$

It is easy to see that ${\sf{Span}}(h+\beta e_{3},f-\beta e_{2})$ is a
subalgebra transverse to the subalgebra ${\sf{Span}}(e_{1},e_{2})$, so
$G_{\rho}$ is isomorphic to $\sf{Aff}\ltimes\mathbb{R}^{2}$ and it preserves
half Minkowski (given by the stabilized plane for $\beta=0$) which must be
isometric to $\mathsf{P}_{\rho}$ since this latter is unique (3.1.1). ∎

This proves (4) and (5) in Theorem 1.1. $\Box$

######  Lemma 7.3.

Except the two flat cases (7.2), the following statements are equivalent:

  * 1)

Up to scaling, $\mathsf{P}_{\rho}$ is locally isometric to
$\mathsf{P}_{\rho^{{}^{\prime}}}$.

  * 2)

$G_{\rho}$ is isomorphic $G_{\rho^{{}^{\prime}}}$.

  * 3)

Up to scaling, $\mathsf{P}_{\rho}$ is globally isometric to
$\mathsf{P}_{\rho^{{}^{\prime}}}$.

###### Proof.

Let $G_{\rho}$ and $G_{\rho^{{}^{\prime}}}$ be the associated groups. We know
that their derived subgroups are exactly ${\mathsf{Heis}}$ (the only cases
where this is not true are the flat ones). Suppose that $\mathsf{P}_{\rho}$ is
locally isometric to $\mathsf{P}_{\rho^{{}^{\prime}}}$, then their Killing
algebras
$\sf{Kil}(\mathsf{P}_{\rho}),\sf{Kil}(\mathsf{P}_{\rho^{{}^{\prime}}})$ are
isomorphic. Taking into account that
${\mathsf{Lie}}(G_{\rho})\subset\sf{Kil}(\mathsf{P}_{\rho})$ and
${\mathsf{Lie}}(G_{\rho^{{}^{\prime}}})\subset\sf{Kil}(\mathsf{P}_{\rho^{{}^{\prime}}})$
and that the dimension of the Killing algebras is equal to $4$ (since we
excluded the flat cases 7.2), we deduce that
${\mathsf{Lie}}(G_{\rho})=\sf{Kil}(\mathsf{P}_{\rho})=\sf{Kil}(\mathsf{P}_{\rho^{{}^{\prime}}})={\mathsf{Lie}}(G_{\rho^{{}^{\prime}}})$
which means that $G_{\rho}$ and $G_{\rho^{\prime}}$ are isomorphic. Suppose
now that $G_{\rho}$ and $G_{\rho^{\prime}}$ are isomorphic and put
$\mathsf{P}_{\rho}=G_{\rho}/I$,
$\mathsf{P}_{\rho^{{}^{\prime}}}=G_{\rho^{{}^{\prime}}}/I^{{}^{\prime}}$. Let
$\phi:G_{\rho}\longrightarrow G_{\rho^{{}^{\prime}}}$ be a Lie group
isomorphism. Since $\phi$ sends ${\mathsf{Heis}}$ to ${\mathsf{Heis}}$ (the
derived subgroups), we have $\phi(I)\subset{\mathsf{Heis}}$ and
$G_{\rho^{{}^{\prime}}}/\phi(I)$ admits a $G_{\rho^{{}^{\prime}}}-$invariant
Lorentz metric isometric to $\mathsf{P}_{\rho}$. But, the uniqueness of the
Lorentz metric on $\mathsf{P}_{\rho^{{}^{\prime}}}$ (up to scaling and
automorphisms) shows that $G_{\rho^{{}^{\prime}}}/\phi(I)$, up to scaling, is
isometric to $\mathsf{P}_{\rho^{{}^{\prime}}}$. Hence $\mathsf{P}_{\rho}$ and
$\mathsf{P}_{\rho^{{}^{\prime}}}$ are isometric up to scaling. ∎

Hence, 7.3 together with 7.2 imply (6), (7), and (8) in Theorem 1.1. $\Box$

Proof of part (3) in Theorem 1.1. Suppose that $\mathsf{P}_{\rho}$ is locally
symmetric, then $\mathsf{P}_{\rho}$ is locally isomorphic to a (simply
connected) symmetric space $X$. Hence
$\sf{Kil}(\mathsf{P}_{\rho})=\sf{Kil}(X)$, that is, $\mathsf{P}_{\rho}$ and
$X$ have isomorphic Killing algebras. Assume that $\mathsf{P}_{\rho}$ is not
flat, then (7.2) we have
$\sf{Kil}(\mathsf{P}_{\rho})={\mathsf{Lie}}(G_{\rho})=\sf{Kil}(X)$. Let
${\mathfrak{i}}_{x}\subset\sf{Kil}(X)$ be the (infinitesimal) isotropy
subalgebra of a point $x\in X$. Then we have, since $X$ is symmetric,
${\mathsf{Lie}}(\sf{Hol}_{x})\subset{\mathfrak{i}}_{x}$ where $\sf{Hol}_{x}$
denotes the holonomy group at $x$. But, since $X$ is not flat, one has
${\mathsf{Lie}}(\sf{Hol}_{x})={\mathfrak{i}}_{x}$. This shows that the
$\sf{Hol}_{x}-$action on $T_{x}X$ is indecomposable because the
$\mathrm{ad}({\mathfrak{i}}_{x})-$action is indecomposable since it
corresponds to the $\mathrm{ad}-$action of the isotropy
${\mathfrak{i}}\subset{\mathsf{Lie}}(G_{\rho})$ which is nilpotent with
nilpotency order equals $3$ (3.3). Therefore, $X$ is indecomposable.
Furthermore, $X$ admits, locally, a parallel null vector field (coming from
$\mathsf{P}_{\rho}$) which extends, uniquely, to a global null parallel vector
field. Thus, $X$ is in fact a Cahen-Wallach space. But in dimension three, we
have only two such spaces corresponding to the elliptic and hyperbolic plane
waves with unimodular isometry groups $G_{\rho}$ [16]. So, from 7.3,
$\mathsf{P}_{\rho}$ is a symmetric Cahen-Wallach space with $G_{\rho}$
unimodular elliptic or hyperbolic. Finally, all this shows that if
$\mathsf{P}_{\rho}$ is locally symmetric not isometric to a Cahen-Wallach,
then it is necessarily flat, and in this case (7.2) we know that we have only
two possibilities corresponding to $G_{\rho}$ unimodular nilpotent which gives
the Minkowski space and $G_{\rho}$ non-unimodular isomorphic to
$\sf{Aff}\ltimes\mathbb{R}^{2}$ which gives half Minkowski. $\Box$

Lastly, let us mention that parts (1) and (2) in Theorem 1.1 are proven in
3.2. $\Box$

##  8\. Plane waves, proof of Theorem 1.2.

###  8.1. $\mathsf{P}_{\rho}$ is a plane wave.

Remember the definition of the Lorentz metric on $\mathsf{P}_{\rho}$ in
Proposition 3.2. From its true definition, one sees that the Killing field $Z$
is null, and that its orthogonal distribution is generated by the Heisenberg
algebra, i.e. the three vector fields $Z,X,Y$.

Let ${\mathfrak{l}}=\mathbb{R}Y+\mathbb{R}Z$. This is an abelian
$2-$dimensional Lie sub-algebra of ${\mathfrak{g}}$. In a neighbourhood of the
basis point $1I$, $Z$ and $Y$ are linearly independent (recall that $I$ is the
one parameter group generated by $X$). It follows in particular that
${\mathfrak{l}}$ generates the orthogonal distribution $Z^{\perp}$ (locally).

It is proven in [15] (Theorem 3) that if a Lorentz $3-$manifold admits a
degenerate distribution generated by two commuting Killing fields, then
Lorentz metric is a plane wave. This applies to our situation and yields that
$\mathsf{P}_{\rho}$ is a plane wave in a neighbourhood of the basis point
$1I$. This obviously implies that $\mathsf{P}_{\rho}$ is everywhere a plane
wave, by homogeneity. For easiness, let us give a proof of the Brinkmann
property, that is, $Z$ is a parallel field on $\mathsf{P}_{\rho}$.

######  Lemma 8.1.

Let $(M,g)$ be a Lorentzian manifold and let $A,B,$ and $C$ be three Killing
vector fields on $M$. Suppose that $A$ is null, $g(A,B)=0$, and:

$[B,A]=0,\ \ [C,A]=\alpha A,\ \ \text{and}\ \ [C,B]=\beta A+\gamma B$

Then we have $\nabla_{W}A=0$ for every $W=A,B,$ or $C$.

###### Proof.

The Koszul formula for three Killing vector fields $X,Y,Z$ is the following:

$2g(\nabla_{X}Y,Z)=g([X,Y],Z)+g([Y,Z],X)-g([Z,X],Y)$

By evaluations we find:

$g(\nabla_{A}A,A)=g(\nabla_{A}A,B)=g(\nabla_{B}A,A)=g(\nabla_{C}A,A)=g(\nabla_{B}A,B)=0$

We have also:

$2g(\nabla_{B}A,C)=g(-\beta A-\gamma B,A)-g(\alpha A,B)=0$
$2g(\nabla_{C}A,B)=g(\alpha A,B)+g(\beta A+\gamma B,A)=0$

This proves that $\nabla_{W}A=0$ for every $W=A,B,$ or $C$. In particular we
have:

$\nabla_{(fA+gB+hC)}A=0\ \ \text{for every}\ \ f,g,h\in C^{\infty}(M).$

∎

In our case, we apply the lemma for $A=Z,B=Y$ and $C=T$, and get that $Z$ is
parallel. $\Box$

The flatness of the $Z^{\perp}$-leaves follows straightforwardly from the fact
that $Z$ and $Y$ commute. The proof of the last property of plane waves, that
is $\nabla_{U}R=0$, for any $U\in Z^{\perp}$ is detailed in [15]. $\Box$

###  8.2. A homogeneous $3-$plane wave is isometric to some
$\mathsf{P}_{\rho}$

We are now going to prove that a simply connected homogeneous $3-$plane wave
$M$ is isometric to some $\mathsf{P}_{\rho}$. The Heisenberg group
${\mathsf{Heis}}$ preserves the flat metric $g_{0}=2dudv+dx^{2}$. It acts by
translations on $x$ and $v$, and by a linear one parameter group of unipotent
transformations:

$(v,x,u)\to(v+tx-\frac{t^{2}}{2}u,x-tu,u)$

Consider now more generally a metric of the form:
$g_{\delta}=2dudv+\delta(u)dx^{2}$ where $\delta$ is a function depending on
$u$. It is obviously invariant by translation on $x$ and $v$. It is also
invariant under a modification of the previous unipotent one parameter group
as follows:

$(v,x,u)\to(v+tx-\frac{t^{2}}{2}F_{\delta}(u),x-tF_{\delta}(u),u)$

where $F_{\delta}(u)$ is an anti-derivative of $\frac{1}{\delta(u)}$. The
obtained $3-$dimensional isometry group of $g_{\delta}$, is in fact isomorphic
to ${\mathsf{Heis}}$ (and acts trivially on $u$).

Remember (1.1.2) that a plane wave can be defined by having a local Rosen
coordinates where the metric has the form $g_{\delta}=2dudv+\delta(u)dx^{2}$.
It follows that it admits an isometric infinitesimal action of the Heisenberg
algebra. This action preserves individually the leaves of the orthogonal
distribution $\frac{\partial}{\partial v}^{\perp}$.
In fact, any Killing field $U$ tangent to the $\frac{\partial}{\partial
v}^{\perp}$-distribution belongs to ${\mathfrak{heis}}$. Indeed, let $F_{0}$
be a leaf of $\frac{\partial}{\partial v}^{\perp}$, then it is (locally)
affinely isomorphic to the flat $\mathbb{R}^{2}$ endowed with a degenerate
Riemannian metric $dy^{2}$. Transformations preserving such a structure have
the form $(z,y)\to(\lambda z+ty+a,y+b)$. If $\lambda\neq 1$, this can not be
the restriction of an isometry preserving individually the leaves of
$\frac{\partial}{\partial v}^{\perp}$, since such a transformation has a
transverse linear distortion $\lambda^{-1}$. But $\lambda=1$ exactly means
that the transformation belongs to ${\mathsf{Heis}}$. Therefore, the Killing
field $U$ coincides with a Killing field $W\in{\mathfrak{heis}}$ on $F_{0}$.
The Killing field $U-W$ vanishes on the hypersurface $F_{0}$ and hence equals
$0$ (at a singularity, a non-zero Killing form has a derivative, element of
${\mathfrak{so}}(1,2)$ and one easily sees that its $0-$set has a most
dimension $1$). From all this, one infers that ${\mathfrak{heis}}$ is an ideal
in the Killing algebra of the plane wave.
It follows that if the plane wave is locally homogeneous, its Killing algebra
contains at least one Killing field $T$ allowing one to pass from a
$\frac{\partial}{\partial v}^{\perp}$-leaf to another. Thus
${\mathfrak{g}}=\mathbb{R}T\oplus{\mathfrak{heis}}$ is an extension of the
Heisenberg algebra. The associated simply connected Lie group has the form
$G_{\rho}$, for some $\rho$, and furthermore the homogeneous plane wave $M$ is
locally isometric to $\mathsf{P}_{\rho}$.
We will assume $\mathsf{P}_{\rho}$ is not flat, since this particular case is
treated in 1.7. So we have a developing map $d:M\to\mathsf{P}_{\rho}$. The
developing map $d$ is equivariant with respect to a homomorphism $h:G\to
G_{\rho}$ where $G$ is the Lie group acting (transitively) on $M$. Since $M$
and $\mathsf{P}_{\rho}$ have same Killing algebra, and $G_{\rho}$ is simply
connected, it follows that $h$ is an isomorphism. By equivariance, $d$ is a
diffeomorphism. In other words, $M$ is identified with $\mathsf{P}_{\rho}$.

###  8.3. The non-simply connected case

If $M$ is homogeneous non-simply connected, then the previous discussion shows
that $\widetilde{M}=\mathsf{P}_{\rho}$ for some $\rho$, and we have, in this
case, $M=\Gamma\backslash\mathsf{P}_{\rho}$ where $\Gamma\subset G_{\rho}$ is
a normal subgroup (hence central). But the center of $G_{\rho}$ is trivial
unless $\rho$ is unimodular where the center is exactly the direction $Z$.
And, $\rho$ unimodular corresponds to either the Minkowski case, or to the two
Cahen-Wallach cases 7.1.

##  9\. Global coordinates, proof of Theorem 1.3.

Consider a plane wave metric given in Brinkmann as:

$g_{b}=2dudv+\frac{bx^{2}}{u^{2}}du^{2}+dx^{2}$

on $\\{(u,v,x),u>0\\}$. Let us say that most computations are actually done in
[1], but only at a Killing fields level. The metric $g_{b}$ obviously admits a
one parameter group of isometries (boosts)
$\phi:(u,v,x)\to(e^{t}u,e^{-t}v,x)$. These isometries act transitively on the
set of $u$-levels.
Now, as any plane wave, the Heisenberg algebra acts (locally) by preserving
$u$-levels and transitively on each of them. It follows that $g_{b}$ is
locally homogeneous and, if it is non-flat, its Killing algebra equals some
extension ${\mathfrak{g}}_{\rho}$ of the Heisenberg algebra (this will be
detailed).

####  9.0.1. Global action of $G_{\rho}$

We are now going to prove that the ${\mathfrak{g}}_{\rho}$ infinitesimal
action integrates to a global $G_{\rho}-$action. For this, we have to consider
the ${\mathfrak{heis}}$-action in Brinkmann coordinates. But understanding
such an action involves analyzing the transform Brinkmann to Rosen, which
sounds highly “transcendental”!

######  Fact 9.1.

Let a metric $g$ in Brinkmann form $g=2dudv+H(u)x^{2}du^{2}+dx^{2}$, defined
on $(v,x),\in\mathbb{R}^{2}$, and $u\in J$, $J$ an interval. Then
${\mathsf{Heis}}$ acts (globally, not merely locally) isometrically by
preserving the $u$-levels.

###### Proof.

Equivalence of Brinkmann and Rosen coordinates is done in Appendix 13 (see [2]
for more details).

Then, we describe the ${\mathsf{Heis}}$-action in the Rosen form
$g_{\delta}=2dudv+\delta(u)dx^{2}$ as given in 8.2. Let us focus on the
$\mathbb{R}^{2}$ ($\subset{\mathsf{Heis}}$)-action given by
$(v,x,u)\to(v+a,x+b,u)$ which is isometric and generates the horizontal
foliation $\mathcal{F}$, whose leaves are the $u$-levels.

All these leaves are thus $\mathbb{R}^{2}$-affine homogeneous spaces. In
particular, the geodesics of $g_{\delta}$ tangent to $\mathcal{F}$ do not
depend of $\delta$, that is, they are usual straight lines as in the case of
the Minkowski space corresponding to $g_{1}=2dudv+dx^{2}$.

The Killing field $W$ corresponding to the translation action on the
$x$-coordinate, $(v,x,u)\to(v,x+b,u)$ is thus parallel on each
$\mathcal{F}$-leaf. This Killing field is however not parallel for a general
$g_{\delta}$ (on the whole space). Let us say, there exits a map $u\to W_{u}$
(depending on $\delta$), such that, on each $\mathcal{F}_{u}$ , $W$ is
parallel to $W_{u}$ $i.e$ directed by $W_{u}$.

As for the Brinkmann form $g=2dudv+H(u)x^{2}du^{2}+dx^{2}$, the same Killing
field $W$ is parallel on each $u$-level. In order to prove that $W$ has a
(complete) global flow, it is enough to show that the $u$-levels are complete.
It turns out that geodesics of these levels are still straight lines in the
(Brinkmann) $(v,x)$-coordinates (see [8] where geodesic equations are
explained and shown to have affine solutions for horizontal initial
conditions) and since $(v,x)\in\mathbb{R}^{2}$, they are complete.

As for the action
$(v,x,u)\to(v+tx-\frac{t^{2}}{2}F_{\delta}(u),x-tF_{\delta}(u),u)$ of the
other one parameter subgroup of ${\mathsf{Heis}}$ in Rosen coordinates, it
does not have geodesic orbits, but it is affine on the $u$-levels endowed with
the Rosen $(v,x)$-coordinates. This is still affine in the Brinkmann
$(v,x)$-coordinates since the geodesics are straight lines. By completeness of
these $u$-levels, the flow acts globally. ∎

####  9.0.2. Identification of $\rho$ as a function of $b$

Let us now show that any $\mathsf{P}_{\rho}$ (for $\rho$ non-unimodular) is
isometric to a metric $g_{b}$, for some $b=b(\rho)$. For this, it is enough to
check that the Killing algebra of $g_{b(\rho)}$ equals
${\mathfrak{g}}_{\rho}$. Now, the computation of the Killing algebra of
$g_{b}$ is done in [1] (Formula 3.33). It is stated there that the Heisenberg
algebra ${\mathfrak{heis}}={\sf{Span}}(Z,X,Y)$, is extended by an element $T$,
such that $[T,X]=bY,$ $[T,Y]=X+Y$ and $[T,Z]=Z$, in other words
$\mathrm{ad}_{T}$ equals $A_{b}=\left(\begin{array}[]{ccc}1&0&0\\\ 0&0&1\\\
0&b&1\end{array}\right)$ in the basis $\\{Z,X,Y\\}$. The associated semi-
direct product is thus defined by $\rho(t)=e^{tA}$. Let now $A$ to be a
derivation of ${\mathfrak{heis}}$, with $A(Z)\neq 0$, that is $A(Z)=aZ$,
$a\neq 0$. Up to scaling (of $A$), we can assume $a=1$, and up to adding an
interior derivation $\mathrm{ad}_{u}$, $u\in{\mathfrak{heis}}$, we can assume
that $A$ preserves a supplementary of $Z$, say the plane
$\mathbb{R}X\oplus\mathbb{R}Y$. And up to conjugacy by automorphisms, $A$ has
the form $A_{b}$. This completes the proof of Theorem 1.3 ∎

####  9.0.3. Hyperbolic case

We also have an explicit global Rosen formula for hyperbolic plane waves
1.1.3:

$g_{\alpha}=2dudv+u^{2\alpha}dx^{2}$

defined on $(v,x)\in\mathbb{R}^{2},$ $u>0$ (considered, in particular, in
[21]). So ${\mathsf{Heis}}$ acts isometrically affinely in these coordinates.
Observe that the linear one parameter group $(v,x,u)\to(e^{t}v,e^{\alpha
t}x,e^{-t}u)$, also preserves $g_{\alpha}$. We know that ${\mathsf{Heis}}$ is
normal in ${\sf{Iso}}(g_{\alpha})$, and thus an extension $G_{\rho}$, acts
isometrically in an affine manner. So $G_{\rho}$ is in fact a subgroup of
$\sf{Aff}(\mathbb{R}^{3})$. We can show that the representation associated to
$\alpha$ is $\rho(t)=e^{tA}$, where $A$ is diagonal with entries
$(1,1-\alpha,\alpha)$. We deduce that any hyperbolic non-unimodular plane wave
$\mathsf{P}_{\rho}$ has a global Rosen model $g_{\alpha}$.

####  9.0.4. Parabolic (non-unimodular) case

$g_{b}=2dudv+\frac{x^{2}}{4u^{2}}du^{2}+dx^{2}$

#### Proof that $G_{\rho}$ is isomorphic to $G_{\rho^{{}^{\prime}}}$ if and
only if $b(\rho)=b(\rho^{{}^{\prime}})$.

In what follows, we denote $\mathfrak{g}_{\rho}$ by $\mathfrak{g}$. Since the
center $\mathfrak{z}$ of $\mathfrak{g}^{\prime}=[\mathfrak{g},\mathfrak{g}]$
is equal to $[\mathfrak{g}^{\prime},\mathfrak{g}^{\prime}]$, and
$[\mathfrak{g},\mathfrak{z}]=\mathfrak{z}$, the adjoint representation
$\mathrm{ad}$ of $\mathfrak{g}$ induces an action of
$\mathfrak{g}/\mathfrak{g}^{\prime}$ on $\mathfrak{z}$, and since this action
is not trivial ($\mathfrak{z}$ is not central), there is a unique
$[T]\in\mathfrak{g}/\mathfrak{g}^{\prime}$ that acts as the identity.
Moreover, $\mathrm{ad}$ induces an action $\overline{\mathrm{ad}}$ of
$\mathfrak{g}/\mathfrak{g}^{\prime}$ on $\mathfrak{g}^{\prime}/\mathfrak{z}$
(this is well-defined, thanks to the two properties mentioned at the beginning
of the proof), and ${\sf{det}}(\overline{\mathrm{ad}}_{[T]})=-b$, this shows
that $b$ is determined by the algebra structure of $\mathfrak{g}$. $\Box$

##  10\. Proof of the completeness statement in Theorem 1.5.

Fix a metric $g=2dudv+\frac{bx^{2}}{u^{2}}du^{2}+dx^{2}$, $b\neq 0$ and let
$\mathsf{P}_{\rho}$ the so defined plane wave and $G_{\rho}$ its isometry
group (or sometimes simply $\mathsf{P}$ and $G$).

###  10.1. Space of null geodesics

Let $\mathcal{G}^{0}$ be the space of all inextendible geometric (i.e. non-
parametric) isotropic geodesics of $\mathsf{P}_{\rho}$. Let $\mathcal{H}^{0}$
be the space of horizontal ones, i.e. those for which $u$ is constant and are
in fact orbits of the parallel field $\frac{\partial}{\partial v}$. The set of
vertical i.e. non-horizontal ones is denoted $\mathcal{V}^{0}$, so that
$\mathcal{G}^{0}=\mathcal{V}^{0}\cup\mathcal{H}^{0}$.

####  10.1.1. Example of an incomplete (vertical) null geodesic

Observe that $(v,x,u)\to(v,-x,u)$ is isometric and hence $\\{x=0\\}$ is
geodesic. The metric on the latter plane is $dudv$. Therefore, $t\to(0,0,t)$
is a null geodesic. Its maximal existence domain is $]0,+\infty[$. Let us
denote this geodesic $\gamma_{0}$.

####  10.1.2. Vertical null geodesics are non-complete

Indeed, for such a geodesic $\gamma(t)=(v(t),x(t),u(t))$, it is known that, up
to a scaling and a shift of time, one can choose $u(t)=t$, that is the
coordinate $t$ plays the role of (an affine) time. The (second order) equation
on $x(t)$ is linear, and $v(t)$ is an integral of it. More precisely, as
detailed in [8], $x(t)$ satisfies, up to a constant, an equation
$\ddot{x}(t)=\frac{x}{t^{2}}$ (after identifying $u$ with a multiple $ct$).
Solutions are, thus, defined on $]0,+\infty[$. Observe in fact, that this
applies to all non-horizontal geodesics, null or not, in particular to all
timelike geodesics.

####  10.1.3. Action of ${\sf{Iso}}(\mathsf{P}_{\rho})$ on $\mathcal{G}^{0}$

######  Proposition 10.1.

The action of $G_{\rho}$ on $\mathcal{G}^{0}$ has exactly two orbits
$\mathcal{V}^{0}$ and $\mathcal{H}^{0}$. In particular, any non-horizontal
null geodesic is congruent, mod $G_{\rho}$, to the vertical geodesic
$\gamma_{0}$. Also, any such geodesic is the orbit of a one parameter group of
$G_{\rho}$.

###### Proof.

Let $T^{0}$ be the space of null tangent directions of $\mathsf{P}_{\rho}$.
This is circle bundle over
$\\{(v,x,u)\in\mathbb{R}^{2}\times\mathbb{R}^{+}\\}$. The set $H^{0}$ of
horizontal null directions gives a section, and so is connected. The set
$V^{0}$ of non horizontal directions forms a bundle with fiber
$\mathbb{S}^{1}-1\sf{pt}\cong\mathbb{R}$, and thus is connected, too. Spaces
of geodesics $\mathcal{H}^{0}$ and $\mathcal{V}^{0}$ are just quotients of
$H^{0}$ and $V^{0}$ by the geodesic flow, and hence are connected. The
isotropy in $G_{\rho}$ is unipotent (of dimension $1$) fixing only horizontal
directions. It follows that $G_{\rho}$ acts freely on $V^{0}$ and has
$1-$dimensional isotropy on $H^{0}$. Since $\dim V^{0}=4$, and $\dim
G_{\rho}=4$, connectedness of $G_{\rho}$ implies that it acts transitively
(and freely) on $V^{0}$. In particular $\mathcal{V}^{0}$ is a homogeneous
space $G_{\rho}/L$, where $L$ is a one parameter group. From this follows that
any vertical null geodesic in an isometric image of $\gamma_{0}$, and that any
such geodesic is preserved by a one parameter group of isometries acting
freely (and hence transitively since the dimension is one) on it. As for
$\mathcal{H}^{0}$, it is a $G_{\rho}$ homogeneous surface homeomorphic to
$\mathbb{R}\times\mathbb{R}^{+}$. ∎

##  11\. Proof of the non-extendibility statement in Theorem 1.5.

In this section, $\rho$ is fixed once for all, and hence $\mathsf{P}_{\rho}$
will be simply denoted $\mathsf{P}$. An non-trivial extension of
$(\mathsf{P},g_{b})$ consists of a Lorentz manifold $(N,g)$ together with an
embedding $\phi:\mathsf{P}\to N$, such that $\phi^{*}g=g_{b}$, but $\phi$ is
not onto, so $\mathsf{P}^{\prime}=\phi(\mathsf{P})$ is an open proper subset
of $N$. We say that $\mathsf{P}$ is inextendible if such $\phi$ does not
exist, say alternatively if any isometric embedding is bijective. It is
important to note that this depends on the regularity of the metric $g$ on
$N$.

####  11.0.1. Analytic case

We can assume $N$ is simply connected, and then use that any locally defined
Killing field extends everywhere. The Lie algebra ${\mathfrak{g}}_{\rho}$ then
acts on $N$. The leaves can not be all of dimension $3$, since this means they
are all open, but then by connectedness $\mathsf{P}^{\prime}=N$. One can
exclude orbits of dimension one (6.2), and conclude that some orbit in the
closure of $\mathsf{P}^{\prime}$ has dimension $2$. The same argument applies
to the ${\mathfrak{heis}}$-action and yields that all its orbits have
dimension 2. Therefore, this 2-dimensional ${\mathfrak{g}}_{\rho}$ orbit
coincides with a ${\mathfrak{heis}}$-orbit. Now, these ${\mathfrak{heis}}$\-
leaves are affine $\mathbb{R}^{2}$ endowed with a degenerate form $dy^{2}$.
The automorphism group of such a structure is $\sf{Aff}\ltimes\mathbb{R}^{2}$,
where $\mathbb{R}^{2}$ acts by translation and $\sf{Aff}$, is the stabilizer
in ${\sf{GL}}(2,\mathbb{R})$ of the parallel direction
$\frac{\partial}{\partial v}$, identified to $(1,0)\in\mathbb{R}^{2}$ (more
explicitly $\sf{Aff}$ consists of matrices $\left(\begin{array}[]{cc}a&b\\\
0&1\par\end{array}\right)$, preserving the direction of $(1,0)$ and acting
isometrically transversely). We then deduce that the Lie algebra
${\mathfrak{g}}_{\rho}$ embeds in the Lie algebra of
$\sf{Aff}\ltimes\mathbb{R}^{2}$, but the letter corresponds to the flat metric
$g_{0}$. Let us mention that the non-extendibility in the analytic case can
also be derived from the main result of [10]. It says that if the Killing
algebra of an analytic Lorentz metric on a connected $3-$manifold $N$ has an
open orbit, then this manifold is locally homogeneous, i.e. its (full) Killing
algebra has one orbit. In our case, if $\mathsf{P}^{\prime}\neq N$, then the
Lie algebra of $N$ has dimension $\geq 5$, which implies (7.2) that $N$ has
constant curvature, hence flat.

####  11.0.2. Smooth complete case

It is shown in [14] that there exists a geodesic $\gamma(t)$ in $\mathsf{P}$,
$t\in]-\infty,1[$ and a parallel $2-$plane field $Q(t)$ along $\gamma$ such
that the sectional curvature of $Q(t)$ tends to $\infty$ when $t\to 1$. This
implies that $\mathsf{P}$ can not be embedded in a geodesically complete
space. Here the metric $g$ (on $N$) is assumed to be $C^{2}$ (and complete).

####  11.0.3. A synthetic proof

Let us give a brief outline of a proof that works in the $C^{2}$ case, but one
may expect to adapt it even to the $C^{1}$ (perhaps $C^{0}$) case (we hope to
give more details elsewhere). The lightlike geodesic foliation $\mathcal{F}$
tangent to $\frac{\partial}{\partial v}^{\perp}$ is defined on
$\mathsf{P}^{\prime}$. It has complete leaves. Let us now recall a Lipschitz
regularity of such foliations. For this, it might be better to give an example
brought out in particular from [29, 30]. Let $B^{3}$ be the unit open
Euclidean ball, and consider $\mathcal{L}_{s}$ a family of disjoint affine
$2-$dimensional subsets, which are relatively complete in $B^{3}$, to mean
that $\mathcal{L}_{s}$ equals the intersection with $B^{3}$ of a complete
affine 2-plane of $\mathbb{R}^{3}$ (so in $B^{3}$, $\mathcal{L}_{s}$ is as
complete as possible). Let $\Sigma=\cup_{s}\mathcal{L}_{s}$, be the support of
our disjoint family. The Lipschitz regularity fact is that the family is
(locally) Lipschitz and in particular extends to a similar disjoint geodesic
family on the closure $\overline{\Sigma}$. It turns out this generalizes to
disjoint families of geodesic hypersurfaces in general pseudo-Riemannian
manifolds (in fact manifolds with a connection), see [29, 30]. In our
situation, the foliation $\mathcal{F}$ of $\mathsf{P}^{\prime}$ extends to a
foliation on $\overline{\mathsf{P}^{\prime}}$. If $\tau$ is a transverse curve
to $\mathcal{F}$, then $\mathsf{P}\cap\tau$ is an open set of $\tau$, and
because $\dim\tau=1$, we infer that $\overline{\mathsf{P}^{\prime}}$ is a
regular open domain in $N$, i.e. a submanifold of codimension $0$ with regular
(in fact geodesic) boundary.
Let $F_{0}$ be a boundary leaf of $\mathcal{F}$. By continuity, as above
(§11.0.1), this is an affine $\mathbb{R}^{2}$ endowed with a degenerate
Riemannian metric. In order to get a contradiction as above, we need to prove
that $G_{\rho}$ acts on $F_{0}$ (preserving its structures).
For this goal, let $p$ a point of $\mathsf{P}^{\prime}$ close enough to
$F_{0}$, and $\phi\in G_{\rho}$ a small isometry of $\mathsf{P}^{\prime}$, say
$\phi$ is in a small neighbourhood of 1 in $G_{\rho}$ and denote $q=\phi(p)$.
Consider a cone of directions in $T_{p}\mathsf{P}^{\prime}$ whose geodesics
all cut $F_{0}$ transversely. These geodesics cut $F_{0}$ and also nearby
leaves $F$ of $\mathcal{F}$ in an open subset (of these leaves). For $\gamma$
such a geodesic, $F$ a leaf of $\mathcal{F}$, close to $F_{0}$, $\gamma\cap F$
is one point, and $\phi(\gamma\cap F)=\phi(\gamma)\cap\phi(F)$. This formula
allows one to define an extension of $\phi$ on some open subset of $F_{0}$ by
$\phi(\gamma\cap F_{0})=\phi(\gamma)\cap F_{0}$. Observe here that $\phi$
preserves $F_{0}$.
Now, to check that everything is coherent, that is, these extensions are well
defined and they give rise to a group, or at least an algebra action, observe
that since we have a true action on $\mathsf{P}^{\prime}$, the construction,
(in $\mathsf{P}^{\prime}$) does not depend on the choice of the point $p$.
Therefore, the extensions to the leaf $F_{0}$ coincide since they are limits
of actions on interior leaves in $\mathsf{P}^{\prime}$.
From all this, we infer that $G_{\rho}$ acts on $F_{0}$, which implies that it
embeds in the automorphism group of its structure,
$\sf{Aff}\ltimes\mathbb{R}^{2}$, as discussed in 11.0.1, which is impossible.

####  11.0.4. Conformal extensions

Observe that all plane waves (in dimension 3) are conformally flat. Indeed, in
Rosen coordinates $g=2dudv+\delta(u)dx^{2}=\delta(u)(2dwdv+dx^{2})$, where
$dw=\frac{du}{\delta(u)}$. Therefore, their conformal developing in the
Einstein universe $\sf{Ein}_{1,2}$ is a non-trivial conformal extension.

##  12\. Compact models.

Let us first make the following observation: suppose that $\mathsf{P}_{\rho}$
is not flat, then we have that ${\sf{Iso}}^{0}(\mathsf{P}_{\rho})=G_{\rho}$
(1.1). Let ${\sf{Iso}}(\mathsf{P}_{\rho})$ denote the full isometry group and
$I_{x}\subset{\sf{Iso}}(\mathsf{P}_{\rho})$ the full isotropy of a point
$x\in\mathsf{P}_{\rho}$ with identity component $I^{0}_{x}=I_{x}\cap G_{\rho}$
consisting of a unipotent one-parameter group of ${\sf{SO}}(1,2)$ when acting
on $T_{x}\mathsf{P}_{\rho}$. Furthermore, $I_{x}/I^{0}_{x}$ is finite since
the image of $I_{x}$ inside ${\sf{O}}(1,2)$ (by the faithful derivative
representation) is an algebraic group. This shows that
${\sf{Iso}}(\mathsf{P}_{\rho})/G_{\rho}$ is finite, which implies that if we
have a compact $({\sf{Iso}}(\mathsf{P}_{\rho}),\mathsf{P}_{\rho})-$manifold,
then it has a finite cover which is a $(G_{\rho},\mathsf{P}_{\rho})-$manifold.
So we restrict our attention only to compact
$(G_{\rho},\mathsf{P}_{\rho})-$manifolds.

######  Proposition 12.1.

Let $M$ be a closed $(G_{\rho},\mathsf{P}_{\rho})-$manifold. Then $M$ is
complete (as a $(G_{\rho},\mathsf{P}_{\rho})-$structure), that is, the
developing map $D:\widetilde{M}\to\mathsf{P}_{\rho}$ is a diffeomorphism.

###### Proof.

Let $Z\in{\mathsf{Lie}}(G_{\rho})$ be the vector that generates the central
direction in ${\mathfrak{heis}}$ and $V$ a generator of ${\mathsf{Lie}}(I)$
where $\mathsf{P}_{\rho}=G_{\rho}/I$. We have that $Z$ is
$\mathrm{ad}_{V}-$invariant $i.e$ $\mathrm{ad}_{V}(Z)=0$. This implies that
the vector field $L_{*}(Z)$ on $G_{\rho}$ obtained by left translating $Z$ is
$I-$right invariant. So $L_{*}(Z)$ projects to a $G_{\rho}-$invariant vector
field on $\mathsf{P}_{\rho}$ which we denote by $\mathcal{Z}$. Thus,
$\mathcal{Z}$ can be pulled back to a well defined vector field
$\mathcal{Z}_{M}$ on $M$. Let $\pi^{*}(\mathcal{Z}_{M})$ be the pull back of
$\mathcal{Z}_{M}$ on $\widetilde{M}$ where $\pi:\widetilde{M}\to M$ is the
covering projection. The developing map $D$ sends $\pi^{*}(\mathcal{Z}_{M})$
to $\mathcal{Z}$, so it sends a $\pi^{*}(\mathcal{Z}_{M})-$trajectory to a
$\mathcal{Z}-$trajectory where $D$, restricted to each trajectory, is a local
diffeomorphism into the corresponding $\mathcal{Z}-$trajectory. Since the
trajectories are one-dimensional, $D$ must be injective on each
$\pi^{*}(\mathcal{Z}_{M})-$trajectory. But, since $\pi^{*}(\mathcal{Z}_{M})$
is complete (because $\mathcal{Z}_{M}$ is complete) then $D$ sends a
$\pi^{*}(\mathcal{Z}_{M})-$trajectory bijectively onto a
$\mathcal{Z}-$trajectory.

Consider now the action of the Heisenberg group ${\mathsf{Heis}}\subset
G_{\rho}$ on $\mathsf{P}_{\rho}$. Since the stabilizer of every point
$x\in\mathsf{P}_{\rho}$ must be contained in ${\mathsf{Heis}}$ (because
${\mathsf{Heis}}$ is normal), then the orbits of this action define a
$G_{\rho}-$invariant foliation by (totally geodesics degenerate) surfaces of
$\mathsf{P}_{\rho}$. Thus, this foliation, denoted by $\mathcal{F}$, can be
pulled back to a well defined foliation $\mathcal{F}_{M}$ on $M$. Let
$\widetilde{\mathcal{F}}_{M}$ be the lift of the foliation $\mathcal{F}_{M}$
to the universal cover $\widetilde{M}$. The vector field
$\pi^{*}(\mathcal{Z}_{M})$ is everywhere tangent to
$\widetilde{\mathcal{F}}_{M}$, so each leaf $\widetilde{\mathcal{F}}_{M}(p)$
is foliated by the $\pi^{*}(\mathcal{Z}_{M})-$trajectories and the developing
$D$ map sends each leaf of $\widetilde{\mathcal{F}}_{M}$ equivariantly to the
$\mathcal{F}-$leaves. Let now
$Y\in{\mathfrak{heis}}\subset{\mathsf{Lie}}(G_{\rho})$ transverse to
$\mathbb{R}Z\oplus\mathbb{R}V$ inside ${\mathfrak{heis}}$, and $V$, as
indicated above, a generator of ${\mathsf{Lie}}(I)$. We have that
$\mathrm{ad}_{V}(Y)$ is proportional to $Z$. This implies that the flow of the
left invariant vector field $L_{*}(Y)$ on $G_{\rho}$ preserves the surfaces
$gK$ where $K$ is the subgroup of $G_{\rho}$ associated to the abelian
subalgebra ${\sf{Span}}(V,Z)$. Hence, since $G_{\rho}/K$ is nothing but the
space of all $\mathcal{Z}-$leaves on $\mathsf{P}_{\rho}$, we have that $Y$
generates a $G_{\rho}-$equivariant flow $\phi^{t}$ on the space of
$\mathcal{Z}-$leaves and preserving each Heisenberg leaf. This flow, thus, can
be pulled back to a complete flow (an $\mathbb{R}-$action), $\phi_{M}^{t}$, on
the space of the $\mathcal{Z}_{M}-$trajectories (completeness is due to the
compactness of $M$ 12.2). Moreover, $\phi_{M}^{t}$ can be lifted to a complete
flow, $\widetilde{\phi}_{M}^{t}$, on the space of all
$\pi^{*}(\mathcal{Z}_{M})-$trajectories on $\widetilde{M}$. Now, since $D$
sends a leaf $\widetilde{\mathcal{F}}_{M}(p)$ to the leaf $\mathcal{F}(D(p))$
by sending a $\pi^{*}(\mathcal{Z}_{M})-$trajectory bijectively to a
$\mathcal{Z}-$trajectory, we have then, for similar reasons, that $D$ sends a
$\widetilde{\phi}_{M}^{t}-$orbit to a $\phi^{t}-$orbit injectively (because
the flow $\phi^{t}$ is injective since the space of $\mathcal{Z}-$trajectories
inside each Heisenberg leaf is identified with $\mathbb{R}$ and $\phi_{t}$
acts on it properly freely) and surjectively (because
$\widetilde{\phi}_{M}^{t}$ is complete). So, $D$ sends each
$\widetilde{\mathcal{F}}_{M}(p)$ bijectively onto $\mathcal{F}(D(p))$.

Finally, let $T\in{\mathsf{Lie}}(G_{\rho})$ not inside ${\mathfrak{heis}}$. We
have that $\mathrm{ad}_{V}(T)\in{\mathfrak{heis}}$. this implies, with the
same ideas, that the flow of the left invariant vector field $L_{*}(T)$
preserves the cosets $gL$ where $L={\mathsf{Heis}}$. Thus, this flow is well
defined on $G_{\rho}/L$ which is the space of all $\mathcal{F}-$leaves. By
similar arguments, this defines a flow on the space of all
$\widetilde{\mathcal{F}}_{M}-$leaves on $\widetilde{M}$ which corresponds
bijectively to the flow on the space of the $\mathcal{F}-$leaves on
$\mathsf{P}_{\rho}$. This proves that $D$ is bijective, hence, a
diffeomorphism. ∎

######  Remark 12.2.

Observe that the space of $\mathcal{Z}_{M}-$trajectories, introduced in the
proof, is pathological in general, yet the $\mathbb{R}-$action,
$\phi_{M}^{t}$, is well defined on it and it can also have a wild behaviour
(like a transverse $\mathbb{R}-$action, on the leaves of a dense linear
foliation on the $2-$torus, has a dense stabilizer). But none of this is
important in our situation, in fact, one can even construct a vector field on
$M$, tangent to the Heisenberg leaves and transverse to the
$\mathcal{Z}_{M}-$orbits, and whose flow sends orbit to orbit realizing our
flow $\phi_{M}^{t}$ on the space of $\mathcal{Z}_{M}-$orbits.

######  Corollary 12.3.

If $M$ is a closed $(G_{\rho},\mathsf{P}_{\rho})-$manifold. Then
$M=\Gamma\backslash\mathsf{P}_{\rho}$ where $\Gamma\subset G_{\rho}$ is a
discrete subgroup acting freely and properly discontinuously on
$\mathsf{P}_{\rho}$

###  12.1. (Non-)existence of compact models:

We have seen so far that if $\mathsf{P}_{\rho}$ admits compact models, that
is, closed manifolds locally isometric to $\mathsf{P}_{\rho}$, then these
compact models are, up to finite covers, of the form
$\Gamma\backslash\mathsf{P}_{\rho}$, in the sense that they are quotients of
$\mathsf{P}_{\rho}$ by a discrete subgroup acting freely and properly 12. It
turns out that all non-flat $\mathsf{P}_{\rho}$, except one, do not support
compact models, even in the case where $\mathsf{P}_{\rho}$ is a Cahen-Wallach
space.

######  Theorem 12.4.

All Lorentz spaces $\mathsf{P}_{\rho}$, except two, do not support compact
quotients ($i.e$ of the form $\Gamma\backslash\mathsf{P}_{\rho}$). The two
exceptional cases correspond to left invariant metrics on $\sf{SOL}$ and
${\mathsf{Heis}}$, where the latter is globally isometric to the Minkowski
space.

###### Proof.

We will prove that there is no discrete subgroup $\Gamma$ of $G_{\rho}$ acting
freely and cocompactly on $\mathsf{P}_{\rho}$, except for two cases. this is
done by considering all the possible representations
$\rho:\mathbb{R}\to\sf{Aut}({\mathsf{Heis}})$ (up to equivalence) then giving
separate proofs of the (non-)existence in each case. The basis for
${\mathsf{Lie}}(G_{\rho})$ will be $\\{Z,X,Y,T\\}$ as usual where $Z,X,$ and
$Y$ generate ${\mathfrak{heis}}$ and $A_{\rho}=\mathrm{ad}_{T}$ acting on
${\mathfrak{heis}}$.

$\bullet$ Let $\mathbb{R}$ act on ${\mathsf{Heis}}$ via the representation
$\rho:\mathbb{R}\to\sf{Aut}({\mathsf{Heis}})$ generated by the derivation
(3.1.1):

$A_{\rho}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&0\end{pmatrix}$

and let $G_{\rho}=\mathbb{R}\ltimes_{\rho}{\mathsf{Heis}}$. We have seen
(3.1.1) that $G_{\rho}/I$ where ${\mathsf{Lie}}(I)={\mathfrak{i}}$ is not
included in ${\sf{Span}}(Z,X)$ or in ${\sf{Span}}(Z,Y)$, is a Lorentz space.
Let $\Gamma\subset G_{\rho}$ be a discrete subgroup acting freely and co-
compactly on $\mathsf{P}_{\rho}=G_{\rho}/I$. We have that the connected Lie
subgroup $[G_{\rho},G_{\rho}]$ is generated by ${\sf{Span}}(Z,X)$. Indeed
$[G_{\rho},G_{\rho}]$ is generated by the image of $A_{\rho}$ and $Z$. Hence
$[\Gamma,\Gamma]$ lies inside the abelian subgroup $H$ given by $Z$ and $X$.
Clearly $\Gamma$ is not contained in ${\mathsf{Heis}}$ since
${\mathsf{Heis}}\backslash\mathsf{P}_{\rho}\sim\mathbb{R}$ is not compact. So
there is $\gamma\in\Gamma$ of the form $\gamma=h\exp(tT)$ where $t\neq 0$ and
$h\in{\mathsf{Heis}}$. The adjoint action of $\gamma$ on $G_{\rho}$ (the
action by conjugacy) preserves $H$ and acts on it via the automorphism
$\mathrm{Ad}(\gamma)=\mathrm{Ad}(h)\circ\exp(tA_{\rho})$. Since
$\exp(tA_{\rho})$ restricted to $H$ is of the form $\begin{pmatrix}e^{t}&0\\\
0&e^{t}\\\ \end{pmatrix}$, then the action of $\mathrm{Ad}(\gamma)$ on $H$ is
a contraction (up to forward or backward iterations). But
$\mathrm{Ad}(\gamma)$ preserves $\Gamma\cap H$, so we necessarily have
$\Gamma\cap H=\\{1\\}$ since $\mathrm{Ad}(\gamma)$ is a contraction on $H$ and
$\Gamma\cap H$ is discrete. We conclude that $[\Gamma,\Gamma]=\\{1\\}$ which
means that $\Gamma$ is abelian. Consider now the quotient $\pi:G_{\rho}\to
G_{\rho}/Z$ where $Z$ refers to the center of ${\mathsf{Heis}}$. We have that
$G_{\rho}/Z=\mathbb{R}\ltimes\mathbb{R}^{2}$ where $\mathbb{R}$ acts on
$\mathbb{R}^{2}$ via the representation $t\to A_{t}=\begin{pmatrix}e^{t}&0\\\
0&1\\\ \end{pmatrix}$ with $\mathbb{R}^{2}$ endowed the basis
$\\{\widetilde{X},\widetilde{Y}\\}$ obtained by projecting $X$ and $Y$. The
restriction of $\pi$ to $\Gamma$ is injective since $\Gamma\cap H=\\{1\\}$.
Thus $\pi(\Gamma)$ is an abelian subgroup of
$\mathbb{R}\ltimes\mathbb{R}^{2}$. Furthermore, $\pi(\Gamma)$ is not contained
in the translation factor, $\mathbb{R}^{2}$, of
$\mathbb{R}\ltimes\mathbb{R}^{2}$ because
$\pi^{-1}(\mathbb{R}^{2})={\mathsf{Heis}}$ and $\Gamma$ is not contained in
${\mathsf{Heis}}$. So, there is an element $\alpha=A_{t}+v$ in $\pi(\Gamma)$
with $t\neq 0$. Suppose that $\alpha$ has a fixed point when acting on
$\mathbb{R}^{2}$, then, up to conjugacy, we can assume that
$A_{t}\in\pi(\Gamma)$. Since the line ${\sf{Span}}(\widetilde{Y})$ is exactly
the fixed subspace of $A_{t}$ and $\pi(\Gamma)$ is abelian, then all elements
of $\pi(\Gamma)$ preserve ${\sf{Span}}(\widetilde{Y})$ and, hence, they are
all of the form $A_{t}+v$ with $v\in{\sf{Span}}(\widetilde{Y})$. That is
$\pi(\Gamma)\subset\mathbb{R}\oplus{\sf{Span}}(\widetilde{Y})$ which implies
that $\Gamma\subset\pi^{-1}(\mathbb{R}\oplus{\sf{Span}}(\widetilde{Y}))=L$
where $L$ is the Lie subgroup generated by $Z,Y$, and $T$. But, we have seen
that (4.1) $\mathsf{P}_{\rho}$ as a homogeneous Lorentz space, is identified
with a left invariant metric on $L$. This shows that
$\Gamma\backslash\mathsf{P}_{\rho}=\Gamma\backslash L$ and $\Gamma\subset L$
is a co-compact lattice which is impossible since $L$ is not unimodular. It
remains now to treat the case where all elements of $\pi(\Gamma)$ are without
fixed points when acting on $\mathbb{R}^{2}$. Suppose that we are in this
situation and consider the quotient
$p:\mathbb{R}\ltimes\mathbb{R}^{2}\to(\mathbb{R}\ltimes\mathbb{R}^{2})/{\sf{Span}}(\widetilde{Y})$
(observe that ${\sf{Span}}(\widetilde{Y})$ is central in
$\mathbb{R}\ltimes\mathbb{R}^{2}$). Then:

$(\mathbb{R}\ltimes\mathbb{R}^{2})/{\sf{Span}}(\widetilde{Y})=\mathbb{R}\ltimes\mathbb{R}=\sf{Aff}$

and the projection $p(\pi(\Gamma))$ is an abelian subgroup which is not
contained in the normal translation factor $\mathbb{R}$ (because
$\pi^{-1}(p^{-1}(\mathbb{R}))={\mathsf{Heis}}$). Since every element of
$\sf{Aff}$, which is not in the normal factor, has exactly one fixed point
when acting on $\mathbb{R}$. We can assume then, up to conjugacy, that all
elements of $p(\pi(\Gamma))$ fix $0$. This implies that
$p(\pi(\Gamma))\subset\mathbb{R}$ (where $\mathbb{R}$ here refers to the
”linear” factor). This gives that, as before, that $\Gamma\subset L$ (where
$L$ is as above) and, for the same reason, this is impossible. Observe, in
this case, that $\mathsf{P}_{\rho}$ supports compact models ($i.e$ compact
manifolds locally isometric to $\mathsf{P}_{\rho}$) since it is flat (7.2),
but the previous discussion shows that $\mathsf{P}_{\rho}$ does not support
$(G_{\rho},\mathsf{P}_{\rho})-$compact models.

$\bullet$ Let $\mathbb{R}$ act on ${\mathsf{Heis}}$ via the representation
$\rho:\mathbb{R}\to\sf{Aut}({\mathsf{Heis}})$ generated by the derivation
(3.1.1):

$A_{\rho}=\begin{pmatrix}1+b&0&0\\\ 0&1&0\\\ 0&0&b\end{pmatrix}$

where we suppose $b>0$. In this case, $\mathbb{R}$ acts on ${\mathsf{Heis}}$
via:

$t\mapsto\begin{pmatrix}e^{t(1+b)}&0&0\\\ 0&e^{t}&0\\\
0&0&e^{tb}\end{pmatrix}$

in the usual basis $\\{Z,X,Y\\}$. So the $\mathbb{R}$-action on
${\mathsf{Heis}}$ is a contraction (for $t\to-\infty$), which implies that
$\Gamma\cap{\mathsf{Heis}}=\\{1\\}$ since $\Gamma\cap{\mathsf{Heis}}$ is
preserved by the conjugacy action (necessarily contracting up to forward or
backward iteration) of an element of $\Gamma$ which is not in
${\mathsf{Heis}}$. But $\Gamma\cap{\mathsf{Heis}}$ is discrete, which implies
$\Gamma\cap{\mathsf{Heis}}=\\{1\\}$. In particular $\Gamma$ is abelian.
Consider now the quotient $\pi:G_{\rho}\to G_{\rho}/Z$. Remark that
$G_{\rho}/Z=\mathbb{R}\ltimes\mathbb{R}^{2}\subset\sf{Aff}(\mathbb{R}^{2})$
where $\mathbb{R}$ acts on $\mathbb{R}^{2}$ via the representation $t\to
A_{t}=\begin{pmatrix}e^{t}&0\\\ 0&e^{bt}\\\ \end{pmatrix}$. We know that,
except translations, every element of $\mathbb{R}\ltimes\mathbb{R}^{2}$ has
exactly one fixed point. Moreover, we have that
$\pi(\Gamma)\cap\mathbb{R}^{2}=\\{1\\}$, because
$\pi^{-1}(\mathbb{R}^{2})={\mathsf{Heis}}$ and
$\Gamma\cap{\mathsf{Heis}}=\\{1\\}$. So (up to conjugacy), all elements of
$\pi(\Gamma)$ fix $0\in\mathbb{R}^{2}$, i.e $\pi(\Gamma)\subset\mathbb{R}$,
where $\mathbb{R}$ denotes the linear factor, which implies that
$\Gamma\subset\pi^{-1}(\mathbb{R})=H$ with $H$ generated by
${\sf{Span}}(Z,T)$. But $H\backslash\mathsf{P}_{\rho}\cong\mathbb{R}$ where
$H$ acts freely and properly on $\mathsf{P}_{\rho}$ since it doesn’t intersect
any conjugate of $I$. This can be seen easily, since $I\subset{\mathsf{Heis}}$
and ${\mathsf{Heis}}$ is a normal subgroup, moreover $H\cap{\mathsf{Heis}}=Z$.
We conclude that $\mathsf{P}_{\rho}$ doesn’t admit any compact quotient.

$\bullet$ Let $\mathbb{R}$ act on ${\mathsf{Heis}}$ via the representation
$\rho:\mathbb{R}\to\sf{Aut}({\mathsf{Heis}})$ generated by the derivation
(3.1.3):

$A_{\rho}=\begin{pmatrix}2c&0&0\\\ 0&c&-1\\\ 0&1&c\end{pmatrix}$

with $c\neq 0$, where $\mathbb{R}$ acts on $Z$ by $t\mapsto e^{2tc}$ and acts
by similarities on $\mathbb{R}X\oplus\mathbb{R}Y$ with non-trivial homothety
factor. A similar reasoning to the preceding case (using the same notations)
shows that $\Gamma$ is abelian and that, up to conjugacy, $\Gamma\subset H$
with $H$ given by ${\sf{Span}}(Z,T)$ which is impossible. Hence there can’t be
any compact quotient of $\mathsf{P}_{\rho}$.

$\bullet$ Let $\mathbb{R}$ act on ${\mathsf{Heis}}$ via the representation
$\rho:\mathbb{R}\to\sf{Aut}({\mathsf{Heis}})$ generated by the derivation
(3.1.1):

$A_{\rho}^{b}=\begin{pmatrix}1+b&0&0\\\ 0&1&0\\\ 0&0&b\end{pmatrix}$

with $b<0$. We can suppose that $b\in[-1,0[$, because $A_{\rho}^{b}$ is
similar to $A_{\rho}^{\frac{1}{b}}$. Observe that:

$G_{\rho}\ \ \text{is unimodular if, and only if}\ \ b=-1$

We know that (3.1.1) in order for $\mathsf{P}_{\rho}=G_{\rho}/I$ to carry a
Lorentz metric, it is necessary and sufficient that $I$ avoids
${\sf{Span}}(X,Z)\cup{\sf{Span}}(Y,Z)$. Let us assume, up to conjugacy, that
$I$ lives inside ${\sf{Span}}(X,Y)$ and generated by $X+Y$. The orbits of
$\mathbb{R}$ action on ${\sf{Span}}(X,Y)$ are hyperbolas, so the orbit of $I$
under this action covers two opposite quarters among the four quarters of
${\sf{Span}}(X,Y)\setminus(\mathbb{R}X\cup\mathbb{R}Y)$. This shows that all
the conjugates of $I$ cover two opposite quarters of the complimentary of
${\sf{Span}}(X,Z)\cup{\sf{Span}}(Y,Z)$ inside ${\mathsf{Heis}}$. Now in order
for $\Gamma$ to act freely, it must not intersect any conjugate of $I$, So
$\Gamma\cap{\mathsf{Heis}}$ must be contained in the other two quarters (with
their boundaries included). Consider the quotient $\pi:G_{\rho}\to
G_{\rho}/Z=\mathbb{R}\ltimes\mathbb{R}^{2}$, so
$\pi(\Gamma)\cap\mathbb{R}^{2}$ must be inside some line
$l\subset\mathbb{R}^{2}$ because, otherwise, the intersection contains a
lattice and this would contradict that it is inside two quarters. Since
$\pi(\Gamma)$ is not contained in $\mathbb{R}^{2}$, we can pick an element of
$\alpha\in\pi(\Gamma)\setminus\mathbb{R}^{2}$, of the form
$\alpha=\begin{pmatrix}e^{t}&0\\\ 0&e^{bt}\\\ \end{pmatrix}+v$. So the induced
action of $\alpha$ on $\mathbb{R}^{2}$ (by conjugacy) is equivalent to the
action of $\begin{pmatrix}e^{t}&0\\\ 0&e^{bt}\\\ \end{pmatrix}$. Hence either
$\pi(\Gamma)\cap\mathbb{R}^{2}\subset\mathbb{R}X$ or
$\pi(\Gamma)\cap\mathbb{R}^{2}\subset\mathbb{R}Y$. Suppose that
$\pi(\Gamma)\cap\mathbb{R}^{2}\subset\mathbb{R}X$ (the case where
$\pi(\Gamma)\cap\mathbb{R}^{2}\subset\mathbb{R}Y$ can be treated similarly)
and consider the projection:

$P:\mathbb{R}\ltimes\mathbb{R}^{2}\mapsto\mathbb{R}\ltimes\mathbb{R}^{2}/\mathbb{R}X\cong\sf{Aff}$

Since $P(\pi(\Gamma))$ is abelian, and it is not contained in the normal
factor, then, up to conjugacy, it is contained in the linear factor
$\mathbb{R}\subset\sf{Aff}$. This means that up to conjugacy we have
$\Gamma\subset L_{X}$ is a cocompact lattice with $L_{X}$ generated by
${\sf{Span}}(Z,X,T)$, but this is impossible since $L_{X}$ is not unimodular.
Thus it remains (similarly) that $\Gamma\subset L_{Y}$ with $L_{Y}$ generated
by ${\sf{Span}}(Z,Y,T)$. For $b=-1$, we have $G_{\rho}$ is unimodular but
$L_{Y}$ is not. In fact, $L_{Y}$ is unimodular, if and only if $b=-1/2$
(remember $b\in[-1,0[$), and in this case, $L_{Y}=\mathsf{SOL}$ which admits
cocompact lattices.

$\bullet$ Suppose now that (3.1.2) $A_{\rho}=\begin{pmatrix}2&0&0\\\ 0&1&1\\\
0&0&1\end{pmatrix}$. The $\mathbb{R}$-action on ${\mathsf{Heis}}$ is given by:

$t\mapsto\begin{pmatrix}e^{2t}&0&0\\\ 0&e^{t}&f(t)\\\ 0&0&e^{t}\end{pmatrix}$

Consider the projection $\pi:G_{\rho}\to
G_{\rho}/Z\cong\mathbb{R}\ltimes\mathbb{R}^{2}\subset\sf{Aff}(\mathbb{R}^{2})$.
For similar reasons as before we have that $\Gamma\cap{\mathsf{Heis}}=\\{1\\}$
since the $\mathbb{R}-$action is expanding on ${\mathsf{Heis}}$. So
$\pi(\Gamma)$ is abelian, since $\pi(\Gamma)$ cannot be contained in
$\mathbb{R}^{2}$. We conclude that $\pi(\Gamma)$ is (up to conjugacy)
contained in the linear factor, which is impossible since in this case
$\Gamma$ is inside $H$ given by ${\sf{Span}}(Z,T)$.

$\bullet$ Suppose now (3.1.3) $A_{\rho}=\begin{pmatrix}0&0&0\\\ 0&0&-1\\\
0&1&0\end{pmatrix}$. The $\mathbb{R}$-action fixes $Z$ and acts by rotations
on ${\sf{Span}}(X,Y)$. We have $I\subset{\mathsf{Heis}}\setminus Z$ and by
hypothesis, $\Gamma$ acts freely. So $\Gamma\cap{\mathsf{Heis}}\subset Z$
because the orbit of $I$ under the conjugacy action covers
${\mathsf{Heis}}\setminus Z$. Consider Now the projection $\pi:G_{\rho}\to
G_{\rho}/Z\cong\widetilde{\sf{Euc}}(\mathbb{R}^{2})=\mathbb{R}\ltimes\mathbb{R}^{2}$.
Then $\pi(\Gamma)\subset\widetilde{\sf{Euc}}(\mathbb{R}^{2})$ is abelian
(because $[\Gamma,\Gamma]\subset Z$) and it is not contained in the normal
factor $\mathbb{R}^{2}$. Since $\Gamma\cap{\mathsf{Heis}}\subset Z$, we deduce
that $\pi(\Gamma)\cap\mathbb{R}^{2}={0}$. Now, we know that, except the center
(which is isomorphic to $\mathbb{Z}$), every element of
$\widetilde{\sf{Euc}}(\mathbb{R}^{2})$ either has exactly one fixed point or
has no fixed point if it belongs $\mathbb{R}^{2}$ (the normal) which is not
our case since $\pi(\Gamma)\cap\mathbb{R}^{2}=\\{0\\}$. Suppose that there is
$\alpha\in\pi(\Gamma)$ with exactly one fixed point, then, up to conjugacy,
one can suppose that all elements fix the origin, that is,
$\pi(\Gamma)\subset\sf{Stab}(0)=\mathbb{R}$. So $\Gamma\subset H$ given by
${\sf{Span}}(Z,T)$ which is clearly impossible. Now if all elements of
$\pi(\Gamma)$ act trivially, then $\pi(\Gamma)$ is contained in the center of
$\widetilde{\sf{Euc}}(\mathbb{R}^{2})$, this means that $\Gamma$ is contained
in the subgroup generated by $Z$ and a discrete subgroup of $\mathbb{R}$
generated by $T$, which is impossible.

$\bullet$ Lastly, let $\mathbb{R}$ act on ${\mathsf{Heis}}$ via the
representation $\rho:\mathbb{R}\to\sf{Aut}({\mathsf{Heis}})$ generated by the
derivation (3.1.4):

$A_{\rho}=\begin{pmatrix}0&1&0\\\ 0&0&1\\\ 0&0&0\end{pmatrix}$

In this case, the $\mathbb{R}$-action on ${\mathsf{Heis}}$ is unipotent. Since
all vectors in the complement of
$\sf{Im}(A_{\rho})\oplus\mathbb{R}Z=\mathbb{R}X\oplus\mathbb{R}Z$ are
equivalent, we can fix $I$ to be the one parameter subgroup generated by $Y$,
in this case, the Lie subalgebra, isomorphic to ${\mathfrak{heis}}$, generated
by $Z,X,T$ is transverse to all the conjugates of $\mathbb{R}Y$. So
$\mathsf{P}_{\rho}$ is isometric to a left invariant metric on
${\mathsf{Heis}}$ which is globally isometric to the Minkowski space (7.1) and
admits compact quotients. ∎

######  Remark 12.5.

Non-existence of compact models in the unimodular case (that is on spaces that
are locally isometric to Cahen-Wallach spaces) follows from [17] Corollary 2,
and the non-existence of compact quotients of $3-$dimensional Cahen-Wallach
spaces follows from [16].

##  13\. Appendix: Rosen and Brinkmann coordinates

Consider a metric $g$ written in Rosen coordinates $(v,x,u),\ u>0$ as
$g=2dudv+u^{2a}dx^{2}$. Consider the following coordinate change:

$v=\overline{v}+\frac{a}{2}\overline{u}^{-1}\overline{x}^{2},\
x=\overline{u}^{-a}\overline{x},\ u=\overline{u}$

We have $\frac{\partial}{\partial\overline{x}}=(au^{a-1}x,u^{-a},0)$,
$\frac{\partial}{\partial\overline{u}}=(-\frac{a}{2}u^{2a-2}x^{2},-au^{-1}x,1)$,
and $\frac{\partial}{\partial\overline{v}}=(1,0,0)$. One verifies that

$g(\frac{\partial}{\partial\overline{x}},\frac{\partial}{\partial\overline{x}})=1,\
\
g(\frac{\partial}{\partial\overline{u}},\frac{\partial}{\partial\overline{u}})=(a^{2}-a)\overline{u}^{-2}\overline{x}^{2},\
\
g(\frac{\partial}{\partial\overline{v}},\frac{\partial}{\partial\overline{v}})=0$

And,

$g(\frac{\partial}{\partial\overline{x}},\frac{\partial}{\partial\overline{u}})=0,\
\
g(\frac{\partial}{\partial\overline{u}},\frac{\partial}{\partial\overline{v}})=1,\
\
g(\frac{\partial}{\partial\overline{x}},\frac{\partial}{\partial\overline{v}})=0$

Therefore, we obtain that the metric $g$ in the coordinates
$(\overline{v},\overline{x},\overline{u})$ has the form

$g=2d\overline{u}d\overline{v}+(a^{2}-a)\overline{u}^{-2}\overline{x}^{2}d\overline{u}^{2}+d\overline{x}^{2}$

This correspondence Rosen-Brinkmann covers cases of metrics of the form
$g=2d{u}d{v}+b{u}^{-2}{x}^{2}d{u}^{2}+d{x}^{2}$, with $b\geq-\frac{1}{4}$.

In general, for a metric $g=2dudv+\delta(u)dx^{2}$, one uses a coordinate
change of the form $v=\overline{v}+c(u)\overline{x}^{2},\
x=\delta(u)^{-1/2}\overline{x},\ u=\overline{u}$., where $c$ is a function to
be determined.

In order to get a Brinkmann form for the metric, we must have:
$g(\frac{\partial}{\partial\overline{x}},\frac{\partial}{\partial\overline{u}})=0$.
This is equivalent to $4c(u)+\delta(u)^{1/2}a(u)=0$, where $a(u)$ is the
derivative of $\delta(u)^{-1/2}$. One then verifies that indeed, the new form
of the metric is of Brinkmann type:
$g=2d\overline{u}d\overline{v}+d(\overline{u})\overline{x}^{2}d\overline{u}^{2}+d\overline{x}^{2}$,
where $d(u)=2c^{\prime}(u)+\delta(u)a(u)^{2}$.

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