---
abstract: 'We show that a Lorentzian homogeneous space admitting a homogeneous structure of type $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$ is either a (locally) symmetric space or a singular homogeneous plane wave.'
address: 'Physics Department, Theory, CERN, Geneva, Switzerland'
author:
- Patrick Meessen
title: 'Lorentzian homogeneous spaces admitting a homogeneous structure of type $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$'
---

[^1]

A theorem by Ambrose and Singer [@art:ambrose], generalized to arbitrary signature in [@art:oubinna], states that on a reductive homogeneous space, there exists a metric connection $\overline{\nabla}=\nabla -S$, with $\nabla$ the Levi-Cività connection, that parallelizes the Riemann tensor $R$, and the $(1,2)$-tensor $S$, [*i.e.*]{} $\overline{\nabla}g
=\overline{\nabla}R =\overline{\nabla}S =0$. Since a $(1,2)$-tensor in $D\geq 3$ decomposes into 3 irreps of $\mathfrak{so}(D)$, one can classify the reductive homogeneous spaces by the occurrence of one of these irreps in the tensor $S$ [@boek:tricerri; @art:gadea]. This leads to 8 different classes, which range from the maximal, denoted by $\mathcal{T}_{1}\oplus\mathcal{T}_{2}\oplus\mathcal{T}_{3}$, to the minimal $ \{ 0 \}$. Clearly, homogeneous spaces of type $ \{ 0 \}$ are just symmetric spaces. Moreover, also the homogeneous spaces admitting a homogeneous structure of type $\mathcal{T}_{i}$ ($i=1,2$ or $3$) have been characterized. For the case at hand it is worth knowing that the homogeneous spaces with a $\mathcal{T}_{3}$ structure, for which $S$ corresponds to a 3-form, are naturally reductive spaces [@boek:tricerri; @art:gadea] and that strictly Riemannian homogeneous $\mathcal{T}_{1}$ spaces are locally symmetric spaces [@boek:tricerri]. Since a homogeneous structure of type $\mathcal{T}_{1}$ is defined by an invariant vector field $\xi$, one must distinguish between two cases in the Lorentzian setting: the non-degenerate case, for which $\xi$ is a space- or time-like vector, and the degenerate case, when $\xi$ is a null vector. In the former case, Gadea and Oubiña [@art:gadea] proven that, analogously to the strictly Riemannian case, the space is locally symmetric. In the degenerate case, A. Montesinos Amilibia [@art:amilibia] showed that a homogeneous Lorentzian space admitting a degenerate $\mathcal{T}_{1}$ structure is a time-independent singular homogeneous plane wave [@Blau:2002js]. A small calculation shows that a generic, [*i.e.*]{} time-dependent, singular homogeneous plane wave admits a degenerate $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$ structure, see [*e.g.*]{} Appendix \[sec:SingHomWave\]. (By a (non-)degenerate $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$ structure, we mean that the vector field $\xi$ characterizing the $\mathcal{T}_{1}$ contribution has (non-)vanishing norm.) This then automatically leads to the question of whether the singular homogeneous plane waves exhaust the degenerate case in the $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$ class. As we will see, the answer is affirmative.

In the $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$ case the homogeneous structure is given by [@boek:tricerri] $$%  \label{eq:DefHomStrucT1T3}
  \overline{\nabla}_{X}Y \ -\ \nabla_{X}Y \ =\ - S_{X}Y \ =\ -T_{X}Y \
  -\ g (X,Y )\xi \ +\ \alpha ( Y ) X \; ,$$ where we have defined $\alpha  ( X ) = g ( \xi ,X )$, and $T_{X}Y (= -T_{Y}X)$ is the $\mathcal{T}_{3}$ contribution. Since the metric and $S$ are parallel under $\overline{\nabla}$, and $\xi$ is the contraction of $S$, it follows that $\overline{\nabla}\xi = 0$ or, written differently: $$%  \label{eq:DerXi}
  \nabla_{X}\xi \; =\; T_{X}\xi \ +\ \alpha ( X)\xi \ -\ \alpha (\xi ) X\; .$$ This equation, together with the fact that $T$ is a 3-form, implies that $\nabla_{\xi}\xi =0$, [*i.e.*]{} $\xi$ is a geodesic vector.

Given an isometry algebra $\mathfrak{g}$ ([*i.e.*]{} the Lie algebra of a Lie group acting transitively by isometries on a given homogeneous space), with a reductive split $\mathfrak{g}=\mathfrak{m}+\mathfrak{h}$, where $\mathfrak{h}\subseteq
\mathfrak{so}(1,n+1)$ is the isotropy subalgebra, it is possible, and usually done, to identify $\mathfrak{m}$ with $\mathbb{R}^{1,n+1}$; the action of $\mathfrak{h}$ on $\mathfrak{m}$ can then be given by the vector representation of $\mathfrak{so}(1,n+1)$ [@boek:kobayashi]. This identification enables one to express the algebra in terms of $S$ and the curvature $\overline{R}$ as, limiting ourselves to the $\mathfrak{m}\times\mathfrak{m}$ commutator, $$\label{eq:MMalg}
   [ X,Y ] \; =\; S_{X}Y \ -\ S_{Y}X \ +\ \overline{R} (
  X,Y ) \; ,$$ where $S$ and $\overline{R}$ are evaluated at some point $p$. In the above formula, $\overline{R}$ signals the presence of $\mathfrak{h}$ in $ [\mathfrak{m},\mathfrak{m} ]$. From now on, we only consider this Lie algebra and all the relevant tensor fields are evaluated at a specific point, even though this is not stated explicitly.

Up to this point not too much has been said about $\mathfrak{h}$, and in fact not too much can be said. It is known, however [@boek:kobayashi], that a tensor field parallelized by $\overline{\nabla}$, when evaluated at a point corresponds to an $\mathfrak{h}$-invariant tensor. Since in this article we take $\xi$ (an $\mathfrak{h}$-invariant vector field as $\overline{\nabla}\xi =0$) to be non-vanishing, this means that $\mathfrak{h}\subseteq\mathfrak{so}(n+1)$ when $\xi$ is light-like, $\mathfrak{h}\subseteq\mathfrak{so}(1,n)$ when $\xi$ is space-like, and $\mathfrak{h}\subseteq\mathfrak{iso}(n)$ when $\xi$ is null.

Let us briefly outline the manner in which we arrive at our results: given a reductive homogeneous space with reductive split $\mathfrak{g}=\mathfrak{m}+\mathfrak{h}$, the subalgebra $\mathfrak{g}^{\prime}=\mathfrak{m}+ [ \mathfrak{m},\mathfrak{m} ]=\mathfrak{m}+\mathfrak{h}^{\prime}$ is an ideal of $\mathfrak{g}$. It is this ideal, which is the Lie algebra of a Lie group still acting transitively, that we will consider; we will say that an element of $\mathfrak{h}$ appears in the algebra if it is an element of $\mathfrak{h}^{\prime}$. Given the homogeneous structure, we can then, following Eq. (\[eq:MMalg\]), write down the maximal form of the algebra compatible with the homogeneous structure. Since we are dealing with a Lie algebra, we can then use the Jacobi identities to constrain the structure constants; after a redefinition of some generators in $\mathfrak{m}$, corresponding to the choice of a different reductive split, this leads to a recognizable result. Since the non-degenerate case is far less involved than the degenerate case, and gives a better idea of the manipulations used, it will be discussed before the degenerate case.

The non-degenerate case
=======================

Let $\mathfrak{m}$ be spanned by the generators $V$ and $Z_{i}$ ($i=1,\ldots ,n$), which in this case we can take to satisfy $$\begin{array}{lclclcl}
      \langle V,V\rangle & =& \aleph   &\hspace{.3cm},\hspace{.3cm}& \alpha ( V ) & =& \lambda\ =\ \aleph |\lambda | \; ,\\
      \langle Z_{i},Z_{j}\rangle & =& \eta_{ij} & ,& \alpha ( Z_{i} ) & =& 0 \; , \\
      \langle V, Z_{i}\rangle & = & 0 & ,& & & 
  \end{array}$$ where $\aleph =\pm 1$ distinguishes between the time-like (for $\aleph =-1$) and the space-like (for $\aleph =1$) cases and $\eta =\mathrm{diag}( -\aleph , 1,\ldots ,1)$. As is mentioned in the introduction, $\mathfrak{h}$ is contained in either $\mathfrak{so}(n+1)$ (for $\aleph =-1$) or $\mathfrak{so}(1,n)$ (for $\aleph =1$) and the relevant non-vanishing commutation relations are $$%  \label{eq:SOComRel}
  \begin{array}{lcl}
   [ M_{ij},M_{kl} ] & =& \eta_{jk}\ M_{il}
                                  -\eta_{ik}\ M_{jl}
                                  +\eta_{jl}\ M_{ki}
                                  -\eta_{il}\ M_{kj} \; ,\\ {}
   [ M_{ij},Z_{k} ] & =&  \eta_{jk}\ Z_{i}
                                  -\eta_{ik}\ Z_{j} \; .{}
  \end{array}$$ Once again, let us stress that not every $M$ needs appear, but the elements of $\mathfrak{h}^{\prime}$ can be written as combinations of the $M$’s, and their commutation relations are induced by the ones above.

With respect to the chosen basis we can decompose $2 T_{V}Z_{i}= F_{i}{}^{j}Z_{j}$ and $2 T_{Z_{i}}Z_{j} = \aleph F_{ij}V \ +\ C_{ij}{}^{k}Z_{k}$, which allows us to write $$\begin{aligned}
   [ V,Z_{i} ] & =& \lambda\ Z_{i} \ +\ F_{i}{}^{j}Z_{j} \ +\ \overline{R} ( V,Z_{i} ) 
  \; , \nonumber \\{}
  & & \nonumber \\{}
   [ Z_{i},Z_{j} ] & =& \aleph F_{ij}V \ +\ C_{ij}{}^{k}Z_{k} \ +\ \overline{R} ( Z_{i},Z_{j} ) 
  \; . \nonumber {}\end{aligned}$$

Let us then, following the strategy outlined above, check the Jacobi identities. The first one is the $(V,Z_{i},Z_{j})$ identity, which leads to $F=0$ and $$\begin{aligned}
  \label{eq:nondegenTZZidentity1}
  \textstyle{\lambda\over 2} C_{ijk} & =& R_{jik} \; -\; R_{ijk} \, ,\\
  \label{eq:nondegenTZZidentity2}
  2\lambda\ S_{ij}{}^{mn} & =& C_{ij}{}^{k}\ R_{k}{}^{mn} \; , \end{aligned}$$ where we expanded $\overline{R} ( V,Z_{i} ) = R_{i}{}^{mn}\ M_{mn}$ and $\overline{R} ( Z_{i},Z_{j} ) = S_{ij}{}^{mn}\ M_{mn}$. Since $F=0$ we can redefine $$%  \label{eq:nondegenZredef}
  Y_{i} \; =\; Z_{i} \; +\; \lambda^{-1}\ R_{i}{}^{mn}M_{mn} \; ,$$ from which we trivially find $$\label{eq:nondegTY}
   [ V, Y_{i} ] \; =\; \lambda\ Y_{i} \; ,$$ which at once implies that $C=0$, by Eq. (\[eq:nondegenTZZidentity1\]), and also that $S=0$ thanks to Eq. (\[eq:nondegenTZZidentity2\]). So the, quite remarkable, result is that a Lorentzian homogeneous space admitting a non-degenerate homogeneous structure of type $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$, also admits a non-degenerate $\mathcal{T}_{1}$ structure. Combining this with the results of Gadea and Oubiña [@art:gadea], we have proven the following result.

\[prop1\] A connected homogeneous Lorentzian space admitting a non-degenerate $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$ structure is a locally symmetric space.

The degenerate case
===================

In the degenerate case we can choose the generators $U$, $V$ and $Z_{i}$ ($i=1,\ldots , n$) spanning $\mathfrak{m}$ such that $\alpha  ( U ) = \lambda\neq 0$, $\alpha ( V ) =\alpha ( Z_{i} ) =0$. The invariant norm is then $\langle U,V\rangle =1$ and $\langle Z_{i},Z_{j}\rangle = \delta_{ij}$ and we decompose the $\mathcal{T}_{3}$ contribution to $S$ as $$\label{eq:DegenerateSplit}
  \begin{array}{lclclcl}
      2T ( U,V,Z_{i} ) & =& W_{i} &\hspace{.2cm},\hspace{.2cm}&
      2T ( U,Z_{i},Z_{j} ) & =& F_{ij} \; , \\
      & & & & & & \\
      2T ( Z_{i},Z_{j},Z_{k} ) & =& C_{ijk} &\hspace{.2cm},\hspace{.2cm}&
      2T ( V,Z_{i},Z_{j} ) & =& \aleph_{ij} \; , 
  \end{array}$$ where $F$, $\aleph$ and $C$ are totally antisymmetric. Given these abbreviations we can write the most general $\mathfrak{m}\times\mathfrak{m}$ commutators as $$\begin{aligned}
  \label{eq:DegMostGen1}
   [ U,V ] & =& \lambda\ V \, +\, W^{i}\ Z_{i}\, +\, \overline{R} ( U,V ) \; , \nonumber \\
%
  & & \nonumber \\
%
  \label{eq:DegMostGen2}
   [ U,Z_{i} ] & =& \lambda\ Z_{i} \, +\, F_{i}{}^{j}\ Z_{j} \, -\, W_{i}\ U
                             \, +\, \overline{R} ( U,Z_{i} ) \; , \nonumber \\
%
  & & \nonumber \\
%
   \label{eq:DegMostGen3}
   [ V,Z_{i} ] & =& W^{i}\ V \, +\, \aleph_{i}{}^{j}\ Z_{j}\, +\, \overline{R} ( V,Z_{i} ) \; , \nonumber \\
%
  & & \nonumber \\
%
  \label{eq:DegMostGen4}
   [ Z_{i},Z_{j} ] & =& \aleph_{ij}\ U \, +\, F_{ij}\ V \, +\, C_{ijk}\ Z^{k}
                                 \, +\, \overline{R} ( Z_{i},Z_{j} ) \; ,\nonumber
%\end{aligned}$$ where the various $\overline{R}$ need to be expanded in terms of the generators of $\mathfrak{h}$. Since $\xi$ is null, we see that $\mathfrak{h}\subseteq\mathfrak{iso}(n)$, which we take to be spanned by $\overline{Z}_{i}$ and $M_{ij}$ with commutation relations $$\begin{aligned}
 \label{eq:IsoAction}
   [ M_{ij},M_{kl} ] & =& \delta_{jk}M_{il}
                                  -\delta_{ik}M_{jl}
                                  +\delta_{jl}M_{ki}
                                  -\delta_{il}M_{kj} \; , \nonumber \\ {}
   [ M_{ij},\overline{Z}_{k} ] & =&
                                  \delta_{jk}\overline{Z}_{i}
                                  -\delta_{ik}\overline{Z}_{j} \; ,\nonumber \\{}
   [ M_{ij},Z_{k} ] & =&
                                  \delta_{jk}Z_{i} -\delta_{ik}Z_{j}
                                  \; ,\nonumber \\ {}
   [ U,\overline{Z}_{i} ] & =& Z_{i} \; , \nonumber \\ {}
   [ Z_{i},\overline{Z}_{j} ] & =& -\delta_{ij}V \; , \nonumber {}\end{aligned}$$ where it should be kept in mind that not all elements of $\mathfrak{iso}(n)$ need appear.

We can then once again start to recover the information contained in the Jacobi identities: the $ ( U,V,Z )$ Jacobi identity reads $$\begin{aligned}
  \label{eq:PreUVZ}
  0 & =& -2\lambda W_{i}\ V \, -\,  \{
                                    \lambda\aleph_{ij} +F_{i}{}^{k}\aleph_{kj} + F_{j}{}^{k}\aleph_{ik} + W^{k}C_{kij}  
                                   \}\ Z^{k} \nonumber \\
%
 & & \nonumber \\
%
    & & - [ \overline{R} ( U,V ),Z_{i} ] \ -\  [ \overline{R} ( V,Z_{i} ),U ] \nonumber \\
%
 & & \nonumber \\
%    
 & & +\aleph_{i}{}^{j}\overline{R} ( U,Z_{i} ) -2\lambda\overline{R} ( V,Z_{i} )
        -F_{i}{}^{j}\overline{R} ( V,Z_{i} ) + W^{j}\overline{R} ( Z_{i},Z_{j} ) \; .\end{aligned}$$ Cancellation of the $V$ contribution then means that $\overline{R} ( U,V ) = -2\lambda W^{i}\overline{Z}_{i} + Y^{ij}M_{ij}$, which at once means that $W$ can only be non-zero for those directions for which a $\overline{Z}$ appears. Specifically, should none appear, then $W=0$. Let us then split the index $i$ into some indices $a$ and $I$, such that the $\overline{Z}_{a}$ do appear whereas the $\overline{Z}_{I}$ do not.

Having made the split, we can investigate the implication of having the null-boosts in the algebra. Let us start by looking at the $ ( U, Z_{i},\overline{Z}_{a} )$ Jacobi: a small calculation then shows that this implies $$\begin{aligned}
  \label{eq:PreUZbarZ}
  0 & =& -\aleph_{ia}\ U \ -\delta_{ia}W^{i}\ Z_{i} \ +\ W_{i}Z_{a} \ +\ C_{aik}\ Z^{k} \nonumber \\
%
 & & \nonumber \\
%
 & & - [ \overline{R} ( U,Z_{i} ), \overline{Z}_{a} ]
     \ -\ \delta_{ia}\overline{R} ( U,V )
     \ -\ \overline{R} ( Z_{i},Z_{a} ) \; . \nonumber\end{aligned}$$ In order for the above to be true we must have that $\aleph_{ai}=C_{aij}=0$ and that $W$ can be non-zero only if no or only one $\overline{Z}$ appears in $\mathfrak{h}$. As was said above, the no-case already implies that $W=0$, so we had better have a look at the case of one appearing null boost. For this we are helped by the $\mathfrak{h}$-part of the above equation. Clearly in the case when we are dealing with only one $\overline{Z}$, this amounts to the statement that $ [ \overline{R} ( U,Z_{a} ),\overline{Z}_{a} ] = -\overline{R} ( U,V )$, which, since there is no rotation in $\mathfrak{so}(n)$ that can take $Z_{a}$ to $Z_{a}$, means that $\overline{R} ( U,V )=0$, and hence that $W_{a}=0$. This then means that in all cases we have $W=0$.

Continuing with the analysis, one can see that the $ ( Z_{i},Z_{j},\overline{Z}_{a} )$ Jacobi leads to $$\begin{aligned}
  \aleph_{ij}\ Z_{a} & =& \delta_{ja}\aleph_{i}{}^{k}Z_{k} \ -\ \delta_{ia}\aleph_{j}{}^{k}Z_{k}\; , \nonumber \\ {}
%
 & & \nonumber \\ {}
%
   [\overline{R} ( Z_{i},Z_{j} ),\overline{Z}_{a} ] & =& 
      \delta_{ja}\overline{R} ( U,Z_{i} )
     \ -\ \delta_{ia}\overline{R} ( U,Z_{j} ) \; . \nonumber {}\end{aligned}$$ Then, using the fact that $\aleph_{ia}=0$, one then sees that $\aleph_{IJ}=0$ and that hence $\aleph_{ij}=0$ when $\mathfrak{h}$ includes some null boost. In the case when there is no $\overline{Z}$, the relevant information can be obtained by picking out the $V$ component in the $ ( V,Z_{i},Z_{j} )$ Jacobi: this implies that $\lambda \aleph_{ij} = F_{i}{}^{k}\aleph_{kj} + F_{j}{}^{k}\aleph_{ik}$, which after contraction leads to $\lambda \aleph_{ij}\aleph^{ij}=0$ and thus implies that $\aleph =0$.

The $\mathfrak{h}$-part of Eq. (\[eq:PreUVZ\]) then implies that $2\lambda \overline{R} ( V,Z_{i} ) = - F_{i}{}^{j}\overline{R} ( V,Z_{j} )$, so that $\overline{R} ( V,Z_{i} ) =0$. In order to then identically satisfy Eq. (\[eq:PreUVZ\]) we must have $ [ \overline{R} ( U,V ),Z_{i} ] =0$, so that $\overline{R} ( U,V ) =0$.

Summarizing the results obtained thus far, we find that the non-trivial $\mathfrak{m}\times\mathfrak{m}$-commutators, scaling $U$ in such a way that $\lambda =1$ and decomposing the various $\overline{R}$’s, are $$\begin{aligned}
  \label{eq:StartingPoint}
   [ U,V ] & =& V \; , \nonumber \\ {}
   [ U, Z_{i} ] & =&  ( F+\delta )_{ij}Z_{j} +
                              h_{ij}\overline{Z}_{j} +
                              \textstyle{1\over 2}R_{ijk}M_{jk} \; ,\nonumber \\ {}
   [ Z_{i},Z_{j} ] & =& F_{ij}V +
                              C_{ijk}Z_{k} + S_{ijk}\overline{Z}_{k} +
                              N_{ijkl}M_{kl} \; .\nonumber {}\end{aligned}$$ Let us then continue our analysis of the Jacobi identities: the $ ( U,Z_{i},Z_{j} )$ Jacobi implies $$\begin{aligned}
  h_{ij} & =& A_{(ij)} \, -\, \textstyle{1\over 2}F_{ij}  \; , \nonumber \\ %\label{eq:UZZ-1}\; , \\ 
  C_{ijk}h_{kl} & =&  ( F+\delta )_{ik} S_{kjl} 
                \ +\  ( F+\delta )_{jk} S_{ikl} \; , \nonumber \\ %\label{eq:UZZ-2}\; ,\\ 
  \textstyle{1\over 2}C_{ijk}R_{kmn} & =&  ( F+\delta )_{ik} N_{kjmn} 
                                     \ +\  ( F+\delta )_{jk} N_{ikmn} \label{eq:UZZ-3}\; ,\\ 
  S_{ijk} +  R_{ijk} -R_{jik} & =& \delta_{F}C_{ijk} \ +\ C_{ijk}  \label{eq:UZZ-4}\; ,\end{aligned}$$ where we defined $$\delta_{F}C_{ijk} \; =\; F_{il}C_{ljk} \ +\ F_{jl}C_{ilk} \ +\
  F_{kl}C_{ijl} \; .$$ From Eq. (\[eq:UZZ-4\]) one sees that $S$ must be totally antisymmetric. Denoting by $\mathfrak{S}_{(ijk)}$ the sum over the permutations $(ijk)$, $(jki)$ and $(kij)$, the $ ( Z_{i},Z_{j},Z_{k} )$ Jacobi results in $$\begin{aligned}
  0 & =& \mathfrak{S}_{(ijk)}\ C_{jkl}S_{ilm} \label{eq:ZZZ-1}\; , \nonumber \\
  0 & =& \mathfrak{S}_{(ijk)}\ C_{jkl}N_{ilmn} \label{eq:ZZZ-2}\; , \nonumber \\
  0 & =& \mathfrak{S}_{(ijk)} [ C_{jkl}C_{ilm} \ +\ 2N_{jkim}
   ] \label{eq:ZZZ-3}\; , \nonumber\end{aligned}$$ and also, since $S$ is totally antisymmetric, $$\label{eq:ZZZ-4}
  3 S \; =\; \delta_{F}C \; .$$ Of course, if a $\overline{Z}_{a}$ occurs in $ [
\mathfrak{m},\mathfrak{m} ]$, then the $ (
U,Z_{i},\overline{Z}_{a} )$ Jacobi implies that $$\begin{aligned}
  \label{eq:UZbarZ-1}
  C_{iaj} & =& 0 \; , \nonumber \\ 
  S_{iaj} & =& R_{iaj} \label{eq:UZbarZ-2}\; ,\\
  N_{iakl} & =& 0 \label{eq:UZbarZ-3}\; .\end{aligned}$$

Let us then, as before, split the indices $i$ into $(a,I)$, where the $\overline{Z}_{a}$’s occur but the $\overline{Z}_{I}$’s do not. This means by assumption that $h_{iI}=0$, which implies $2A_{aI}=F_{aI}$, $A_{IJ}=0=F_{IJ}$ and $S_{ijI}=0$, which implies that only $S_{abc}$ is non-zero. Furthermore, we then see that only $C_{IJK}$ is non-vanishing. Together with Eq. (\[eq:ZZZ-4\]), this then implies that $S=0$, and we get the extra constraint $$\label{eq:Restr-1}
    F_{aI}C_{IJK} \; =\; 0 \; .$$ This last constraint also follows from the $ (
Z_{i},Z_{j},\overline{Z}_{a} )$ Jacobi, which also tells us that $N_{ijal}=0$.

Eq. (\[eq:UZbarZ-2\]) then implies that only $R_{IJK}$ and $R_{aJK}$ are non-vanishing, and from Eq. (\[eq:UZbarZ-3\]) we find that only $N_{IJmn}$ can be non-zero. We can calculate $R_{aJK}$ from Eq. (\[eq:UZZ-4\]), which then gives $R_{aIJ} = F_{aK}C_{KIJ} = 0$ because of Eq. (\[eq:Restr-1\]). The same equation then states $R_{IJK}-R_{JIK}=C_{IJK}$, which by means of Eq. (\[eq:UZZ-3\]) then also implies that only the $N_{IJKL}$ can be non-vanishing.

Let us define the generator $$\label{eq:ZIRedef}
  Y_{I} \; =\; Z_{I} \ -\ F_{Ia}\overline{Z}_{a} \; ,$$ from which we can then derive that the algebra takes on the form $$\begin{aligned}
  \label{eq:NewAlgebra}
   [ U, Z_{a} ] & =&  ( F+\delta )_{ab}Z_{b} \ +\
                             ( A_{ab} -\textstyle{1\over
                            2}F_{ab} ) \overline{Z}_{b} \; ,\nonumber \\ {}
   [ Z_{a},Z_{b} ] & =& F_{ab}V\; , \nonumber  \\ {}
   [ U, Y_{I} ] & =& Y_{I} \ +\ \textstyle{1\over 2}R_{IJK}M_{JK} \; , \nonumber \\ {}
   [ Y_{I},Y_{J} ] & =& C_{IJK}Y_{K} \ +\ N_{IJKL}M_{KL} \; ,\nonumber {}\end{aligned}$$ so that the $a$- and the $I$-sectors decouple from each other.

Restricting ourselves to the $I$-sector and further defining $$\label{eq:YIRedef}
  W_{I} \; =\; Y_{I} \ +\ \textstyle{1\over 2}R_{IJK}M_{JK} \; ,$$ we immediately find $ [ U,W_{I} ] \ =\ W_{I}$; calculating the remaining commutator, we find $$[ W_{I},W_{J} ] \ =\  ( C_{IJK} - R_{IJK} +
         R_{JIK} ) Y_{K} \ + \ldots\ ,$$ where the $\ldots$ stands for terms in $M_{JK}$. Using now Eq. (\[eq:UZZ-4\]), we see that this redefinition trivializes $C$, and by way of Eq. (\[eq:UZZ-3\]), also $N$.

At this point, the only difference between the algebra we deduced and the generic singular homogeneous plane wave algebra in Eq. (\[eq:sing\_hom\_waves\_algebra\]) are the null boosts in the $I$-sector, that is a generator one would call $\overline{W}_{I}$. It is, however, always possible to extend our algebra to an algebra that does contain them; in fact this follows immediately from the consistency of the singular homogeneous plane wave algebra. Putting everything together, one sees that we obtain the isometry algebra of a generic singular homogeneous plane wave in Eq. (\[eq:sing\_hom\_waves\_algebra\]) by, basically, choosing a different reductive split of the same algebra. Thus we have proven the next theorem.

The underlying geometry of a connected homogeneous Lorentzian space that admits a degenerate $\mathcal{T}_{1}\oplus\mathcal{T}_{3}$ structure is that of a singular homogeneous plane wave.

[**Note added**]{}: The author recently became aware of [@art:pastore], where Proposition \[prop1\] is proven for the Riemannian case. The reasonings leading to Proposition \[prop1\], however, only depend on the non-degeneracy of the $\mathcal{T}_{1}$ contribution and not on the signature of the space. This means that Proposition \[prop1\] also holds in the pseudo-Riemannian case.

Acknowledgments {#acknowledgments .unnumbered}
===============

The author would like to thank J. Figueroa-O’Farrill, R. Hernández and S. Philip for very useful discussions, and S. Vascotto for improving the readability of the text.

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[^1]: CERN-PH-TH/2005-006
