# Lorentzian connections with parallel twistor-free torsion

Igor Ernst  and  Anton S. Galaev

###### Abstract.

We describe Lorentzian manifolds that admit metric connections with parallel
torsion having zero twistorial component and non-zero vectorial component. We
also describe Lorentzian manifolds admitting metric connections with closed
parallel skew-symmetric torsion.

Keywords: Lorentzian manifold; parallel torsion; twistor-free torsion; skew-
symmetric torsion; holonomy.

AMS Mathematics Subject Classification 2020: 53B30; 53C29.

1Department of Mathematics and Statistics, Masaryk University, Faculty of
Science, Kotlářská 2, 611 37 Brno, Czech Republic

2University of Hradec Králové, Faculty of Science, Rokitanského 62, 500 03
Hradec Králové, Czech Republic
E-mail: anton.galaev(at)uhk.cz

##  1\. Introduction

Metric connection with torsion appear both in the context of differential
geometry and mathematical physics. Under the action of the pseudo-orthogonal
group, the torsion of a metric connection on a pseudo-Riemannian manifold
decomposes into three components: vectorial, twistorial, and skew-symmetric.
Riemannian connections with skew-symmetric torsion have been studied in many
works, e.g., [1, 2, 9]. Connections with skew-symmetric torsion appear in
certain supergravity theories, see, e.g., [12, 13, 18, 21]. We studied
Lorentzian metric connections with parallel skew-symmetric torsion in [11].

Recently Moroianu and Pilca [20] classified complete simply connected
Riemannian manifolds that admit metric connections with parallel torsion
having zero twistorial component and non-zero vectorial component. It is shown
that each such manifold is a warped product of the line and a complete
Riemannian manifold admitting a parallel 3-form $\tau$ satisfying the
condition $\tau(X)\cdot\tau=0$ for all vector fields $X$, see Theorem 1 below.
Then complete Riemannian manifolds carrying such 3-forms were classified, see
Theorem 2 below.

_In the present paper we consider the corresponding problem in the Lorentzian
signature_. The first result (Theorem 3) states that if a Lorentzian manifold
$(M,g)$ admits a metric connection with parallel torsion having zero
twistorial component and non-zero vectorial component, then $(M,g)$ cannot be
geodesically complete, and even more it cannot satisfy no weaker condition of
geodesic completeness: space-like, time-like, or light-like geodesic
completeness. We distinguish two cases depending on the vector field $\xi$
defining the vectorial component of the torsion: $\xi$ is isotropic and $\xi$
is not isotropic. If $\xi$ is non-isotropic and complete, then Theorem 4
provides a result similar to Theorem 1. Next, Theorem 5 describes Lorentzian
manifolds admitting parallel 3-forms $\tau$ satisfying the above condition.
_In fact, Theorem 5 describes Lorentzian manifolds admitting metric
connections $\nabla$ with closed $\nabla$-parallel skew-symmetric torsion_.
Next, we assume that the vector field $\xi$ is isotropic. If the norm of the
skew-symmetric component $S$ of the torsion is zero, then the manifold under
consideration may be described as a manifold $(M_{0},g_{0})$ admitting a
parallel isotropic vector field and a parallel 2-form on the corresponding
screen bundle (Theorem 6). Suppose that the norm of $S$ is non-zero. If the
dimension of the manifold is 3, then we prove in Theorem 7 that the manifold
is a Kundt space of a special form. If the dimension is greater than 3, then
the manifold under consideration is a locally warped product of the manifold
$(M_{0},g_{0})$ and a Riemannian manifold again admitting a 3-form with the
above properties.

A large class of metric connections with torsion appear as homogeneous
structures. A homogeneous structure on a pseudo-Riemannian manifold is a
metric connection $\nabla$ with $\nabla$-parallel curvature and torsion. By
the Ambrose-Singer Theorem, a complete pseudo-Riemannian manifold is reductive
homogeneous if and only if it admits a homogeneous structure [8, 22]. Let us
mention several results about homogeneous structures related to the results of
the present paper. In [14] it is shown that Lorentzian space forms do not
admit homogeneous structures of vectorial type. Next, a homogeneous Lorentzian
space admitting a homogeneous structure of isotropic vectorial type is a
singular homogeneous plane wave [19, 5]. The paper [17] provides a
classification of Lorentzian homogeneous structures with the torsion having
zero twistorial component and non-zero vectorial component. These structures
are exhausted by locally symmetric spaces and singular homogeneous plane
waves. We thus generalize these results to general connections with parallel
twistor-free torsion having non-zero vectorial component.

##  2\. Preliminaries

Let $(M,g)$ be a pseudo-Riemannian manifold. Denote by $\nabla^{g}$ the Levi-
Civita connection on $(M,g)$. A connection $\nabla$ on $(M,g)$ is called
metric if $\nabla g=0$. Using the metric $g$ we identify the tangent and
cotangent bundles, and we use the obvious identifications for tensors. In
particular, we identify a field of bivectors $X\wedge Y$ with the field of
endomorphism

$(X\wedge Y)Z=g(X,Z)Y-g(Y,Z)X,$

and for a 3-tensor $B$ by abuse of notation we will write

$B(X,Y,Z)=g(B(X,Y),Z)=g(B(X)Y,Z).$

Let $A\in\Gamma(TM)\otimes(\wedge^{2}\Gamma(TM))$. Then the equality

$\nabla_{X}Y=\nabla^{g}_{X}Y+A(X,Y),\quad X,Y\in\Gamma(TM),$

defines the metric connection $\nabla$ with the torsion

$T(X,Y,Z)=A(X,Y,Z)-A(Y,X,Z),$

see, e.g, [2]. Let $\xi\in\Gamma(TM)$ and $S\in\Omega^{3}(M)$, then the tensor

$A(X)=X\wedge\xi+\frac{1}{2}S(X)$

defines the metric connection

(1) $\nabla_{X}=\nabla_{X}^{g}+X\wedge\xi+\frac{1}{2}S(X),$

with the torsion

$T(X,Y)=(X\wedge Y)\xi+S(X,Y).$

The torsion $T$ has vectorial component $\xi$, skew-symmetric component $S$
and zero twistorial component. In particular, a 3-form $\tau$ defines the
metric connection

(2) $\nabla_{X}=\nabla_{X}^{g}+\frac{1}{2}\tau(X)$

with skew-symmetric torsion $\tau$.

Let $\nabla$ be given by (1). Suppose that $\nabla T=0$. It is clear that this
is equivalent to the conditions $\nabla\xi=0$ and $\nabla S=0$. Consider the
first Bianchi identity

$\underset{XYZ}{\mathfrak{S}}R(X,Y)Z=\underset{XYZ}{\mathfrak{S}}\\{T(T(X,Y),Z)+(\nabla_{X}T)(Y,Z)\\},$

where $\underset{XYZ}{\mathfrak{S}}$ denotes the cyclic sum with respect to
$X,Y,Z$. It is easy to check that the identity may be rewritten in the form

(3)
$\underset{XYZ}{\mathfrak{S}}R(X,Y)Z=\underset{XYZ}{\mathfrak{S}}S(S(X,Y),Z)+\underset{XYZ}{\mathfrak{S}}g(X,\xi)S(Y,Z)+\underset{XYZ}{\mathfrak{S}}S(\xi,Y,Z)X.$

For a 3-form $\tau$ define the 4-form $\sigma_{\tau}$ by the equality

$\sigma_{\tau}(X,Y,Z)=\underset{XYZ}{\mathfrak{S}}\tau(\tau(X,Y),Z).$

It holds that

(4) $(\tau(X)\cdot\tau)(Y,Z,V)=-\sigma_{\tau}(X,Y,Z,V).$

Consider the connection (2). It holds

(5)
$d\tau(X,Y,Z,V)=\underset{XYZ}{\mathfrak{S}}\\{(\nabla_{X}\tau)(Y,Z,V)\\}-(\nabla_{V}\tau)(X,Y,Z)+2\sigma_{\tau}(X,Y,Z,V),$

see, e.g., [13]. From (2) it follows that

(6) $\nabla\tau=\nabla^{g}\tau-\frac{1}{2}\sigma_{\tau}.$

If $\nabla\tau=0$, then the first Bianchi identity for the connection $\nabla$
may be written in the form

(7) $\underset{XYZ}{\mathfrak{S}}R(X,Y)Z=\sigma_{\tau}(X,Y,Z).$

Note that by (4), the condition $\sigma_{\tau}=0$ is equivalent to the
condition $\tau(X)\cdot\tau=0$ for all $X\in\Gamma(TM)$. From (5) and (6)
easily follows

######  Lemma 1.

Let $(M,g)$ be a pseudo-Riemannian manifold and $\nabla$ a metric connection
with skew-symmetric torsion $\tau$ on it. Then the following conditions are
equivalent

  * 1.

$\nabla\tau=0$ and $d\tau=0$;

  * 2.

$\nabla\tau=0$ and $\sigma_{\tau}=0$;

  * 3.

$\nabla^{g}\tau=0$ and $\sigma_{\tau}=0$.

The following theorem from [20] describes complete simply connected Riemannian
manifolds with parallel twistor-free torsion.

######  Theorem 1.

A complete simply connected Riemannian manifold $(M,g_{M})$ carries a metric
connection with parallel twistor-free torsion if and only if $(M,g_{M})$ is
homothetic to a warped product $(\mathbb{R}\times N,(dt)^{2}+e^{2t}g_{N})$,
where $(N,g_{N})$ is a complete simply connected Riemannian manifold carrying
a parallel 3-form $\tau\in\Omega^{3}(N)$ such that $\sigma_{\tau}=0$.

In [3] it is shown that an irreducible complete simply connected Riemannian
manifold of dimension greater than $4$ with parallel skew torsion $\tau$
satisfying $\sigma_{\tau}=0$ is a simple compact Lie group with bi-invariant
metric or its dual noncompact symmetric space. The following is a more general
theorem from [20].

######  Theorem 2.

Let $(N,g)$ be a complete simply connected Riemannian manifold carrying a
metric connection with parallel skew-symmetric torsion $\tau$ which satisfies
$\sigma_{\tau}=0$. Then $(N,g)$ is one of the following:

  * 1.

$(N,g)$ is an oriented 3-dimensional Riemannian manifold and $\tau$ is a
constant multiple of its Riemannian volume form;

  * 2.

$(N,g)$ is a simple Lie algebra with an $\operatorname{ad}$-invariant metric
$g$ and $\tau$ is a constant multiple of its canonical 3-form;

  * 3.

$(N,g)$ is an irreducible symmetric space of type II or type IV and $\tau$ is
a constant multiple of its canonical 3-form;

  * 4.

$(N,g)$ is a Riemannian manifold and $\tau=0$;

  * 5.

$(N,g)$ is a product of some of the manifolds as in 1–4.

Now we recall the definition of Kundt and Walker spaces.

A Kundt space is a Lorentzian manifold $(M,g)$ admitting an isotropic vector
field $p$ satisfying the following conditions:

(8)
$\nabla^{g}_{p}p=0,\quad\operatorname{tr}\nabla^{g}p=0,\quad||(\nabla^{g}p)^{sym}||_{g}=0,\quad||(\nabla^{g}p)^{alt}||_{g}=0,$

i.e., the vector field $p$ is geodesic, expansion-free, shear-free and twist-
free, see, e.g., [10]. Locally there exist coordinates $v,x^{1},\dots x^{n},u$
such that $g$ takes the form

(9) $g=2dvdu+h+2Adu+H(du)^{2},$

where $h=\sum_{i,j=1}^{n}h_{ij}(x^{1},\dots,x^{n},u)dx^{i}dx^{j}$ is a
$u$-family of local Riemannian metrics,
$A=\sum_{i=1}^{n}A_{i}(v,x^{1},\dots,x^{n},u)dx^{i}$ is a 1-form, and
$H=H(v,x^{1},\dots,x^{n},u)$ is a local function. The vector field $p$ is
proportional to $\partial_{v}$.

A Walker manifold (see, e.g., [7]) is a Lorentzian manifold $(M,g)$ admitting
a $\nabla^{g}$-parallel distribution of isotropic lines $\ell$. Locally $g$ is
given by (9) with the functions $A_{i}$ independent of $v$. If moreover the
induced connection in $\ell$ is flat, i.e., $(M,g)$ admits (local)
$\nabla^{g}$-parallel isotropic vector field $p$, then in the function $H$ may
be chosen to be independent of $v$.

Let $\nabla$ be a metric connection on a Lorentzian manifold $(M,g)$ and
suppose that $p$ is a $\nabla$-parallel isotropic vector field on $M$. Since
the metric $g$ is $\nabla$-parallel, the distribution $p^{\bot}$ is
$\nabla$-parallel. The bundle $E=p^{\bot}/\left<p\right>$ is called _the
screen bundle_ , see [15, 11] for details. There is the obvious projection
$p^{\bot}\to E$. The connection $\nabla$ induces a connection on $E$ (which we
denote by the same symbol): if $X$ is a vector field on $M$ and $Y$ is a
section of $E$, then $\nabla_{X}Y$ is the projection to $E$ of the vector
field $\nabla_{X}\tilde{Y}$, where $\tilde{Y}$ is an arbitrary section of
$p^{\bot}$ such that its projection to $E$ is $Y$.

We will consider a Witt basis $p,e_{1},\dots,e_{n},q$ of the Minkowski space
$\mathbb{R}^{1,n+1}$. The non-zero values of the Minkowski metric with respect
to such basis are $g(p,q)=g(q,p)=g(e_{i},e_{i})=1$. We will denote by
$\mathbb{R}^{n}$ the Euclidean subspace of $\mathbb{R}^{1,n+1}$ spanned by the
vectors $e_{1},\dots,e_{n}$. With respect to the basis
$p,e_{1},\dots,e_{n},q$, the subalgebra of $\mathfrak{so}(1,n+1)$ preserving
the isotropic line $\mathbb{R}p$ has the following matrix form:

$\mathfrak{so}(1,n+1)_{\mathbb{R}p}=\left\\{\left.\begin{pmatrix}a&-X^{t}&0\\\
0&A&X\\\ 0&0&-a\end{pmatrix}\right|\begin{matrix}a\in\mathbb{R}\\\
A\in\mathfrak{so}(n)\\\ X\in\mathbb{R}^{n}\end{matrix}\right\\}.$

The above matrix is identified with the bivector

$-ap\wedge q+A+p\wedge X,$

and we get the decomposition

$\mathfrak{so}(1,n+1)_{\mathbb{R}p}=(\mathbb{R}p\wedge
q\oplus\mathfrak{so}(n))\ltimes p\wedge\mathbb{R}^{n}.$

For the subalgebra of $\mathfrak{so}(1,n+1)$ annihilating the isotropic vector
$p$ we have

$\mathfrak{so}(1,n+1)_{p}=\mathfrak{so}(n)\ltimes p\wedge\mathbb{R}^{n}.$

##  3\. Main Results

Let $(M,g)$ be a Lorentzian manifold admitting a metric connection with
parallel twistor-free torsion having non-vanishing vectorial component $\xi$.

Assuming that $(M,g)$ is simply connected and complete and that
$g(\xi,\xi)=\varepsilon=\pm 1$, one may apply all arguments from the proof of
Theorem 2 and conclude that $(M,g)$ must be isometric to the manifold given by
(10). But such Lorentzian manifold is not complete [4, 6] and we see that the
statement of Theorem 2 cannot be extended to complete Lorentzian manifolds
with non-isotropic $\xi$. Since assuming geodesic completeness is usually too
strong in Lorentzian geometry, one may consider weaker notions of geodesic
completeness, such as space-like geodesic completeness time-like geodesic
completeness, or light-like geodesic completeness (to be short, we use the
notion of $\eta$-completeness, where $\eta=1,-1,0$). It turned out that
assuming a weaker notion of geodesic completeness still does not allow to
prove an analogue of Theorem 2 for Lorentzian manifolds and finally we prove

######  Theorem 3.

Let $(M,g)$ be a Lorentzian manifold admitting a metric connection with
parallel twistor-free torsion having non-vanishing vectorial component $\xi$.
Then $(M,g)$ is not $\eta$-complete for $\eta=1,-1,0$.

Theorem 3 generalizes the following two results. Lorentzian space forms do not
admit homogeneous structures of vectorial type [14]. A homogeneous Lorentzian
space admitting a homogeneous structure of isotropic vectorial type is a
singular homogeneous plane wave [19, 5].

We prove an analogue of Theorem 1 for Lorentzian manifolds if we just require
the completeness of the vectorial component of the torsion in the case when
the vectorial component is non-isotropic.

######  Theorem 4.

Let $(M,g)$ be a simply connected Lorentzian manifold. Let $\varepsilon=\pm
1$. Then $(M,g)$ admits a metric connection with parallel twistor-free torsion
having the vectorial component given by a non-isotropic complete vector field
$\xi$ with $\varepsilon g(\xi,\xi)>0$ if and only if $(M,g)$ is homothetic to
the warped product

(10) $(\mathbb{R}\times N,\varepsilon(dt)^{2}+e^{2\varepsilon t}g_{N}),$

where $(N,g_{N})$ is a Riemannian (if $\varepsilon=-1$) or a Lorentzian (if
$\varepsilon=1$) manifold carrying a parallel 3-form $\tau\in\Omega^{3}(N)$
that satisfies $\sigma_{\tau}=0$.

######  Example 1.

The flat slicing coordinates on an open subset of the de Sitter space
$dS^{n+2}$ allow one to write the de Sitter metric in the form

$g=-(dt)^{2}+e^{-2t}\sum_{i=1}^{n+1}(dy^{i})^{2}.$

The vector field $\xi=\partial_{t}$ defines a metric connection $\nabla$ with
vectorial torsion. This connection is a homogeneous structure on an open
subset of $dS^{n+2}$ and, according to [14], $\nabla$ cannot be extended to
the entire $dS^{n+2}$.

Then we prove an analogue of Theorem 2 for the Lorentzian signature.

######  Theorem 5.

Let $(N,g)$ be a Lorentzian manifold carrying a parallel 3-form $\tau$ with
$\sigma_{\tau}=0$. Then $(N,g)$ is locally isometric to one of the following
manifolds:

  * 1.

$\dim N=3$, $(N,g)$ is an oriented 3-dimensional Lorentzian manifold and
$\tau$ is a constant multiple of its Lorentzian volume form;

  * 2.

$\dim N\geqslant 4$, $(N,g)$ is a Lorentzian manifold carrying a parallel
isotropic vector field $p$, and $\tau=p^{\flat}\wedge\omega$, where $\omega$
is a parallel 2-form on the screen bundle $p^{\bot}/\left<p\right>$.

  * 3.

$(N,g)$ is a Lorentzian manifold and $\tau=0$;

  * 4.

$(N,g)$ locally is a product of a Lorentzian manifold from one of the cases
1–3 with a Riemannian manifold from Theorem 2.

From Lemma 1 it follows that Theorem 5 describes Lorentzian manifolds
admitting metric connections $\nabla$ with closed $\nabla$-parallel skew-
symmetric torsion.

Then we consider the case when the vectorial part of the torsion is isotropic.
In that case the $g$-norm of the torsion coincides with the $g$-norm
$||S||_{g}$ of the skew-symmetric component $S$. We say that the component $S$
is degenerate if $||S||_{g}=0$, otherwise we say that $S$ is non-degenerate.
The first theorem deals with the case of degenerate skew-symmetric component
$S$.

######  Theorem 6.

Let $(M_{0},g_{0})$ be a Lorentzian manifold with a $\nabla^{g_{0}}$-parallel
isotropic vector field $p_{0}$ and a $\nabla^{g_{0}}$-parallel form
$\tau_{0}=p_{0}^{\flat}\wedge\omega$, where $\omega$ is a
$\nabla^{g_{0}}$-parallel 2-form on the screen bundle
$p^{\bot}/\left<p\right>$. Suppose that there exists a function $\varphi$ on
$M_{0}$ such that $p_{0}^{\flat}=de^{\varphi}$. Let $\xi=e^{-\varphi}p_{0}$
and $S=e^{-\varphi}\tau_{0}$. Then the connection $\nabla$ on $M_{0}$ given by
(1) has parallel twistor-free torsion with isotropic vectorial component $\xi$
and the skew-symmetric component $S$.

Conversely, each simply connected Lorentzian manifold admitting a metric
connection with parallel twistor-free torsion having isotropic vectorial
component and degenerate skew-symmetric component is globally equivalent to a
just constructed one.

######  Remark 1.

Let $(M_{0},g_{0})$ be a Lorentzian manifold with a $\nabla^{g_{0}}$-parallel
isotropic vector field $p_{0}$. Then $dp^{\flat}_{0}=0$, and there exists a
function $f$ such that $p^{\flat}_{0}=df$. Such a function is defined up to a
constant. The condition in the above construction is satisfied whenever each
such function is bounded from below. For example, let $(B,b)$ be a Riemannian
manifold. Consider the Walker manifold

$(M_{0}=\mathbb{R}\times B\times\mathbb{R}_{+},\quad
g_{0}=2dvdu+b+H(du)^{2}),$

where $v$ and $u$ are the coordinates on $\mathbb{R}$ and $\mathbb{R}_{+}$,
respectively, and $H$ is any function on $B\times\mathbb{R}_{+}$. Then
$p_{0}=\partial_{v}$ is a parallel isotropic vector, and
$p^{\flat}_{0}=du=de^{\varphi}$, $\varphi=\ln u$. The condition on
$p^{\flat}_{0}$ from the above construction is satisfied.

######  Example 2.

Consider the metric

$g_{0}=2dvdu+\sum_{i=1}^{n}(dx^{i})^{2}+(H+2v)(du)^{2},\quad\partial_{v}H=0.$

The vector field $p_{0}=-e^{-u}\partial_{v}$ is $\nabla^{g_{0}}$-parallel and
it satisfies $p_{0}^{\flat}=de^{-u}$. Let

$\xi=-\partial_{v},\quad S=du\wedge\sum_{i<j}F_{ij}dx^{i}\wedge dx^{j},$

where $F=(F_{ij})$ is a constant skew-symmetric matrix. Then the connection
$\nabla$ given by (1) is a metric connection with parallel twistor-free
torsion having isotropic vectorial component $\xi$ and skew-symmetric
component $S$. The connection $\nabla$ is a homogeneous structure if and only
if $\nabla R=0$. According to [5, 17, 19], this is the case if and only if

$H=A(e^{-uF}x,e^{-uF}x),$

i.e, $g$ is a singular homogeneous plane-wave metric. Here $A$ is a constant
symmetric bilinear form.

The following two theorems give a complete solution for non-degenerate $S$.

######  Theorem 7.

Let $(M,g)$ be a 3-dimensional Lorentzian manifold. Then $(M,g)$ admits a
metric connection with parallel twistor-free torsion having isotropic
vectorial component $p$ and non-degenerate skew-symmetric component $S$ if and
only if $M$ is oriented and admits an isotropic vector field $p$ such that

(11)
$\nabla^{g}_{X}p=-g(X,p)p-\frac{1}{2}S(X,p),\quad\forall\,X\in\Gamma(TM),$

where

$S=a\text{\rm Vol}_{g},\quad a\in\mathbb{R},\quad a\neq 0.$

In particular, such $(M,g)$ is a Kundt space, and locally $g$ may be written
as

$g=2dvdu+2avdxdu+(dx)^{2}+(-2ve^{-ax}+C(x,u))(du)^{2},$

where $C(x,u)$ is an arbitrary function and $p=e^{-ax}\partial_{v}$.

######  Theorem 8.

Let $(M_{0},g_{0},p_{0},\tau_{0},\varphi)$ be a Lorentzian manifold as in
Theorem 6. Let $(N,g_{N})$ be a Riemannian manifold with a non-zero
$\nabla^{g_{N}}$-parallel 3-form $\tau_{N}$ satisfying $\sigma_{\tau_{N}}=0$.
Consider the manifold

$M=M_{0}\times N$

with the Lorentzian metric

$g=g_{0}+e^{2\varphi}g_{N}.$

Let

$\xi=e^{-\varphi}p_{0},\quad S=e^{-\varphi}\tau_{0}+e^{3\varphi}\tau_{N}.$

Then the connection $\nabla$ on $M$ given by (1) has parallel twistor-free
torsion with isotropic vectorial component $\xi$ and non-degenerate skew-
symmetric component $S$.

Conversely, each Lorentzian manifold of dimension different from 3 admitting a
metric connection with parallel twistor-free torsion having isotropic
vectorial component and non-degenerate skew-symmetric component is locally
equivalent to a just constructed one.

##  4\. Proof of Theorem 4

Let $(M,g)$ be a simply connected Lorentzian manifold. Suppose that $(M,g)$
admits a metric connection $\nabla$ with parallel twistor-free torsion having
vectorial component given by a non-isotropic complete vector field $\xi$. Let
us assume that $g(\xi,\xi)=\varepsilon=\pm 1$. The connection $\nabla$ is
given by (1). For the proof we use the ideas from [20], and we apply [16,
Prop. 8] to obtain the global decomposition of $M$.

Consider the 1-form $\eta$ given by

$\eta(X)=\varepsilon g(\xi,X).$

It is clear that $\eta(\xi)=1$. As in [20] it can be shown that

$d\eta=S(\xi)=0.$

Proposition 8 from [16] implies that all leaves of the foliation on $M$
tangent to the distribution $\ker\eta$ are pairwise diffeomorphic, and the
manifold $M$ is diffeomorphic to the product

$M\cong\mathbb{R}\times N,$

where $N$ is a leaf of the foliation. Under this diffeomorphism, the vector
field $\xi$ corresponds to $\partial_{t}$. From the definition of $\eta$ it
follows that the vectors tangent to the leaves of the foliation are orthogonal
to $\partial_{t}$. Consequently, the metric $g$ may be written in the form

$g=\varepsilon(dt)^{2}+h,$

where $h$ is a $t$-family of metrics on $N$. Consider the metric

$\tilde{g}=e^{-2\varepsilon t}g.$

Applying the formula for the Levi-Civita connection under the conformal
change, (1), and the fact that $S(\xi)=0$, we obtain

$\nabla^{\tilde{g}}_{X}\xi=-\varepsilon(Xt)\xi.$

This shows that the distribution generated by $\xi$ and the distribution
$\ker\eta$ are $\nabla^{\tilde{g}}$-parallel. The local version of the Wu
Theorem implies that

$g_{N}=e^{-2\varepsilon t}h$

is a metric on $N$ independent of $t$. Thus,

$g=\varepsilon(dt)^{2}+e^{2\varepsilon t}g_{N}.$

The equality $S(\xi)=0$ shows that $S$ is a $t$-family of 3-forms on the
manifold $N$. As in [20] it can be shown that the condition $\nabla S=0$ is
equivalent to the conditions

$S=e^{3\varepsilon
t}\tau_{N},\quad\tau_{N}\in\Omega^{3}(N),\quad\nabla^{g}\tau_{N}=0,\quad\sigma_{\tau_{N}}=0.$

This completes the proof of the theorem. ∎

##  5\. Proof of Theorem 5

Let $(N,g)$ be a Lorentzian manifold carrying a non-zero parallel 3-form
$\tau$ such that $\sigma_{\tau}=0$. Consider the metric connection

$\nabla=\nabla^{g}+\frac{1}{2}\tau$

with skew-symmetric torsion $\tau$. From Lemma 1 it follows that $\tau$ is
$\nabla$-parallel. In [11] we described holonomy, curvature and torsion of
Lorentzian connections with parallel skew-symmetric torsion. We have now to
consider the additional condition $\sigma_{\tau}=0$. We assume that $\dim
N=n+2\geqslant 3$ and denote by $\mathfrak{g}\subset\mathfrak{so}(1,n+1)$ the
holonomy algebra of the connection $\nabla$. We fix a point $x\in N$. The
tangent space $T_{x}N$ may be identified with the Minkowski space
$\mathbb{R}^{1,n+1}$.

Recall that a subalgebra $\mathfrak{g}\subset\mathfrak{so}(1,n+1)$ is called
weakly irreducible if it does not preserve any proper non-degenerate subspace
of $\mathbb{R}^{1,n+1}$. The geometry $(N,g,\nabla)$ is called _reducible_ if
the holonomy algebra $\mathfrak{g}\subset\mathfrak{so}(1,n+1)$ of the
connection $\nabla$ is _not_ weakly irreducible, i.e., $\mathfrak{g}$
preserves a proper non-degenerate subspace of the tangent space. In this case
there exists a non-trivial $\mathfrak{g}$-invariant orthogonal decomposition
of the tangent space

(12) $T_{x}M=L\oplus E.$

The geometry $(N,g,\nabla)$ is called _decomposable_ if the holonomy algebra
$\mathfrak{g}\subset\mathfrak{so}(1,n+1)$ preserves an orthogonal
decomposition (12) such that it holds

$\tau_{x}\in\wedge^{3}L\oplus\wedge^{3}E.$

Otherwise we say that the geometry is _indecomposable_. If the geometry is
decomposable, then it is a product of two other geometries, i.e., $(N,g)$ is
as in the case 4 from the statement of the theorem. Thus we may assume that
the geometry is indecomposable.

In Section 3 from [11] we proved that if $\nabla$ is a metric connection on a
Lorentzian manifold $(N,g)$ with parallel skew-symmetric torsion $\tau$ and
weakly irreducible holonomy algebra, then $\tau$ automatically satisfies the
condition $\sigma_{\tau}=0$, moreover, $(N,g)$ is as in the case 1 or 2 from
the statement of the theorem.

Now we assume that the geometry $(N,g,\nabla)$ is reducible and
indecomposable. Then the holonomy algebra $\mathfrak{g}$ preserves a
decomposition (12). We may assume that the induced representation of
$\mathfrak{g}$ in $L$ is weakly irreducible. We consider several cases
depending on the dimension of $L$ and use the description of $\tau$ from [11].
We will denote by $\tau$ also the value of the field $\tau$ at the point $x$.
Since the field $\tau$ is $\nabla$-parallel, it is enough to check that its
value at the point $x$ satisfies the condition

(13) $\tau(X)\cdot\tau=0\quad\text{for all}\quad X\in T_{x}N.$

If $\dim L\geqslant 2$, we denote by $p,e_{1},\dots,e_{k},q$ a Witt basis in
$L$.

Let $\dim L=1$. Then

$T_{x}M=\mathbb{R}e_{-}\oplus E,$ $\tau=e_{-}\wedge\theta+\omega_{E},$

where $g(e_{-},e_{-})=-1$, $\theta\in\wedge^{2}E$ and
$\omega_{E}\in\wedge^{3}E$ are annihilated by $\mathfrak{g}$ and it holds
$\theta\cdot\omega_{E}=0$. The last condition may be written in the form

$\underset{XYZ}{\mathfrak{S}}\omega_{E}(\theta(X),Y,Z)=0\quad\text{for
all}\quad X,Y,Z\in E.$

Let $X\in E$. Then

$\tau(X)=\theta(X)\wedge e_{-}+\omega_{E}(X),$
$\tau(X)\cdot\tau=\theta(X)\wedge\theta+\omega_{E}(X)\cdot\omega_{E}+e_{-}\wedge(\omega_{E}(X)\cdot\theta)=0.$

This is equivalent to

(14) $\displaystyle\theta(X)\wedge\theta+\omega_{E}(X)\cdot\omega_{E}$
$\displaystyle=0,$ (15) $\displaystyle\omega_{E}(X)\cdot\theta$
$\displaystyle=0.$

We have

$0=\omega_{E}(X)\cdot\theta(Y,Z)=-\theta(\omega_{E}(X,Y),Z)-\theta(Y,\omega_{E}(X,Z))=\\\
=\omega_{E}(X,Y,\theta(Z))-\omega_{E}(X,Z,\theta(Y))=\omega_{E}(\theta(Z),X,Y)+\omega_{E}(\theta(Y),Z,X)=\\\
=-\omega_{E}(\theta(X),Y,Z),$

which means that $\theta\in\wedge^{2}(\ker\omega_{E})$. Since the geometry is
indecomposable, this implies that $\omega_{E}=0$. Equation (14) now reads as

$\theta(X)\wedge\theta=0,$

for all $X\in E$. This is possible only if $\operatorname{rk}\theta\leqslant
2$. Thus the indecomposability implies that the dimension of $N$ is 3.

Let $\dim L=2$. In this case $\tau$ has the form

$\tau=p\wedge q\wedge v+\theta\wedge v+\omega_{E_{1}},$

where $v\in E$ is a non-zero vector, $E=\mathbb{R}v\oplus E_{1}$ is an
orthogonal decomposition, $\mathfrak{g}v=0$,
$\omega_{E_{1}}\in\wedge^{3}E_{1}$, $\theta\in\wedge^{2}E_{1}$,
$\mathfrak{g}\cdot\omega_{E_{1}}=0$, $\mathfrak{g}\cdot\theta=0$. Consider the
condition (13). It holds

$\tau(p)=-p\wedge v,$ $\tau(p)\cdot\tau=g(v,v)\theta\wedge v=0.$

Hence, $\theta=0$. The indecomposability again implies that the dimension of
$N$ is 3.

Assume that $\dim L=3$. Then, given an arbitrary Witt basis $p,e_{1},q$ of
$L$, the torsion $\tau$ has the form

$\tau=p\wedge(\alpha e_{1}\wedge q+e_{1}\wedge v+\lambda)+\theta\wedge
v+\omega_{E_{1}},$

where $v\in E$ is a vector, $E_{1}$ is the orthogonal complement of $v$ in
$E$, $\mathfrak{g}v=0$, $\omega_{E_{1}}\in\wedge^{3}E_{1}$,
$\theta\in\wedge^{2}E_{1}$ and $\lambda\in\wedge^{2}E$ are annihilated by
$\mathfrak{g}$. It holds that

$\tau(e_{1})=-\alpha p\wedge q-p\wedge v,$
$\tau(e_{1})\cdot\tau=p\wedge(\alpha\lambda+g(v,v)\theta)=0.$

Consequently,

$\alpha\lambda+g(v,v)\theta=0.$

First suppose that $\alpha\neq 0$. Let us set

$\tilde{v}=v-\frac{g(v,v)}{\alpha}p,\quad\tilde{q}=q+\frac{1}{\alpha}\left(v-\frac{1}{2\alpha}g(v,v)p\right),\quad\tilde{E}=\mathbb{R}\tilde{v}\oplus
E,\quad\tilde{L}=\left<p,e_{1},\tilde{q}\right>.$

Then $p,e_{1},\tilde{q}$ is a Witt basis in $\tilde{L}$ and
$T_{x}M=\tilde{L}\oplus\tilde{E}$ is a new $\mathfrak{g}$-invariant orthogonal
decomposition. The torsion $\tau$ now may be written as

$\tau=\alpha p\wedge
e_{1}\wedge\tilde{q}+\theta\wedge\tilde{v}+\omega_{E_{1}}$

and we see that the geometry is decomposable.

Suppose now that $\alpha=0$. The condition (13) may be rewritten in the
following way:

$\theta\wedge
v=0,\quad\lambda(v)=0,\quad\lambda\cdot\omega_{E_{1}}=0,\quad\omega_{E_{1}}(X)\cdot\omega_{E_{1}}=0,\quad\forall
X\in E_{1}.$

The expressions for the curvature tensor from [11, Theorem 5] imply that

$\mathfrak{g}=\mathbb{R}p\wedge e_{1}\oplus\mathfrak{b},$

where $\mathfrak{b}$ is the projection of $\mathfrak{g}$ onto
$\mathfrak{so}(E_{1})$. The above equalities show that $\omega_{E_{1}}$
defines a Lie algebra structure on $E_{1}$, and $\lambda$ is a derivation of
the Lie algebra $E_{1}$. Let $E_{0}=\ker\omega_{E_{1}}\subset E_{1}$ and let
$E_{1}^{\prime}$ be the orthogonal complement to $E_{0}$ in $E_{1}$. Then
$E_{0}$ is a commutative Lie algebra, and $E_{1}^{\prime}$ is either the
trivial or a semisimple Lie algebra. Therefore,

$\lambda=\omega_{E_{1}}(U_{0})+\lambda_{0},$

where $U_{0}\in E^{\prime}_{1}$ and $\lambda_{0}\in\wedge^{2}E_{0}$. Since
$\mathfrak{g}$ annihilates $\lambda$ and $\omega_{E_{1}}$, it annihilates the
vector $U_{0}$.

Let

$\tilde{E}=\\{\tilde{U}=U+g(U,U_{0})p\,|\,U\in
E\\},\quad\tilde{q}=q-U_{0}-\frac{1}{2}g(U_{0},U_{0})p,\quad\tilde{L}=\left<p,e_{1},\tilde{q}\right>.$

Then, $T_{x}M=\tilde{L}\oplus\tilde{E}$ is a $\mathfrak{g}$-invariant
orthogonal decomposition. Now,

$\tau=p\wedge(e_{1}\wedge\tilde{v}+\tilde{\lambda}_{0})+\omega_{\tilde{E}_{1}},$

where in the new notation $\tilde{E}=\mathbb{R}\tilde{v}+\tilde{E}_{1}$,
$\tilde{E}_{1}=\tilde{E}_{0}\oplus\tilde{E}^{\prime}_{1}$,
$\tilde{\lambda}_{0}\in\wedge^{2}\tilde{E}_{0}$,
$\omega_{\tilde{E}_{1}}\in\wedge^{3}\tilde{E}^{\prime}_{1}$. The
indecomposability implies that $\tilde{E}^{\prime}_{1}=0$. Thus, $(N,g)$ is as
in the case 1 or 2 from the statement of the theorem.

Finally let us assume that $\dim L\geqslant 4$. Then the torsion $\tau$ has
the form

(16)
$\tau=p\wedge(\omega+\sum_{i=1}^{k}e_{i}\wedge\mu_{i}+\lambda)+\omega_{E},$

where $\mu_{i}\in E$, $\omega_{E}\in\wedge^{3}E$,
$\omega\in\wedge^{2}\mathbb{R}^{k}$, $\lambda\in\wedge^{2}E$, and it holds
that $\lambda\cdot\omega_{E}=0$. Let us consider the condition (13):

$\tau(e_{i})=-p\wedge\mu_{i}-p\wedge\omega(e_{i}),$
$\tau(e_{i})\cdot\tau=-p\wedge\omega_{E}(\mu_{i})=0,$
$\tau(U)\cdot\tau=\omega_{E}(U)\cdot\omega_{E}=0,\quad U\in E.$

The condition $\tau(q)\cdot\tau=0$ is equivalent to
$\lambda\cdot\omega_{E}=0$. From these equalities we get
$\omega_{E}(\mu_{i})=0$. Again, $\omega_{E}$ defines a Lie algebra structure
on $E$, and as in the previous case, the indecomposability of the geometry
implies that $\omega_{E}=0$, and $(N,g)$ is as in the case 1 or 2 from the
statement of the theorem. ∎

##  6\. Proof of Theorem 7

Let $(M,g)$ be a 3-dimensional Lorentzian manifold. Suppose that $(M,g)$
admits a metric connection $\nabla$ with parallel twistor-free torsion having
isotropic vectorial component $\xi$ and non-zero skew-symmetric component $S$.

The connection $\nabla$ is given by (1). Since $S$ is parallel, it is non-
vanishing, and, consequently, $M$ is orientable. Fix an orientation of $M$,
and let $\text{\rm Vol}_{g}$ be the volume form defined by $g$. The torsion is
$\nabla$-parallel if and only if $\nabla p=0$ and $\nabla S=0$.

It holds

$0=\nabla S=\nabla^{g}S,$

i.e., $S$ is proportional to the volume form,

$S=a\text{\rm Vol}_{g},\quad a\in\mathbb{R},\quad a\neq 0.$

Now, the torsion is parallel if and only if $\nabla p=0$. By (1), the last
condition is equivalent to (11). It is obvious that the vector field $p$
satisfies the conditions (8), i.e., $(M,g)$ is a Kundt space. Consequently,
the metric $g$ may be locally written in the form

$g=2dvdu+2A(v,x,u)dxdu+(dx)^{2}+H(v,x,u)(du)^{2}.$

The vector field $p$ satisfies $p=\varphi\partial_{v}$, for a function
$\varphi=\varphi(x,u)$. Consider the local field of Witt frames

$p=\varphi\partial_{v},\quad e=\partial_{x},\quad
q=\frac{1}{\varphi}\left(\partial_{u}-A\partial_{x}-\frac{1}{2}H\partial_{v}\right).$

The condition (11) is equivalent to

$\nabla^{g}_{p}p=0,\quad\nabla^{g}_{e}p=-\frac{1}{2}ap,\quad\nabla_{q}^{g}p=-p+\frac{1}{2}ae.$

It is easy to check that

$\nabla^{g}_{e}p=\left(\partial_{x}\varphi+\frac{1}{2}\varphi\partial_{v}A\right)\partial_{v}.$

This implies that

$\frac{\partial_{x}\varphi}{\varphi}+\frac{1}{2}\partial_{v}A=-\frac{1}{2}a.$

Next,

$\nabla^{g}_{q}p=\left(\frac{1}{\varphi}(\partial_{u}\varphi-A\partial_{x}\varphi)+\frac{1}{2}\partial_{v}H-A\partial_{v}A\right)\partial_{v}+\frac{1}{2}\partial_{v}A\partial_{x}.$

We conclude that $\partial_{v}A=a$, i.e., $A=av+B(x,u)$, and
$\partial_{x}\varphi=-\varphi,$ i.e., $\varphi=c(u)e^{-ax}$. A simple
coordinate transformation allows to assume that $B(x,u)=0$ and $c(u)=1$.
Finally we get that $H=-2e^{-ax}v+C(x,u)$. ∎

##  7\. Proof of Theorems 6 and 8

Let $(M,g)$ be a Lorentzian manifold admitting a metric connection $\nabla$
with parallel twistor-free torsion having isotropic vectorial component $p$
and skew-symmetric component $S$. If $\dim M=3$, then thanks to Theorem 7 we
will assume that $||S||_{g}=0$; since in dimension 3 each 3-form is
proportional to the (local) volume form, this implies that $S=0$. Denote by
$\mathfrak{g}\subset\mathfrak{so}(1,n+1)$ the holonomy algebra of the
connection $\nabla$ at a point $x\in M$. By abuse of notation, we denote by
$p$, $S$, $R$ the values of the corresponding tensor fields at the point $x$.
Then $\mathfrak{g}$ preserves the vector $p\in\mathbb{R}^{1,n+1}=T_{x}M$,
i.e., $\mathfrak{g}\subset\mathfrak{so}(1,n+1)_{p}$.

######  Lemma 2.

[11, Lemma 1] If $\mathfrak{g}\subset\mathfrak{so}(1,n+1)_{p}$ is a weakly
irreducible subalgebra, then each $S\in\wedge^{3}\mathbb{R}^{1,n+1}$
annihilated by $\mathfrak{g}$ is of the form $S=p\wedge\omega$ for a bivector
$\omega$ on $p^{\bot}/\left<p\right>$.

######  Lemma 3.

If the holonomy algebra $\mathfrak{g}\subset\mathfrak{so}(1,n+1)$ of the
connection $\nabla$ is not weakly irreducible, then $\mathfrak{g}$ preserves
an orthogonal decomposition

$\mathbb{R}^{1,n+1}=L\oplus E,$

where

$L=\mathbb{R}p\oplus\mathbb{R}^{k}\oplus\mathbb{R}q,\quad 0\leqslant
k\leqslant n,$

such that

$S=p\wedge\omega+\omega_{E},$

where $\omega\in\wedge^{2}\mathbb{R}^{k}$ and $\omega_{E}\in\wedge^{3}E$.
Moreover, $\sigma_{\omega_{E}}=0$ and $\ker\omega_{E}=0$.

Note that the statement of the lemma includes the case $E=0$.

Proof of Lemma 3. If $\mathfrak{g}\subset\mathfrak{so}(1,n+1)$ is not weakly
irreducible, then it is clear that $\mathfrak{g}$ preserves an orthogonal
decomposition

$\mathbb{R}^{1,n+1}=L\oplus E,\quad\dim L\geqslant 1$

such that the induced representation of $\mathfrak{g}$ in $L$ is weakly
irreducible.

First assume that $\dim L\geqslant 4$. By [11, Lemma 7],

$S=p\wedge\omega+\omega_{E},$

where

$\omega=\omega_{\mathbb{R}^{k}}+\mu+\lambda,$

here $\omega_{\mathbb{R}^{k}}\in\wedge^{2}\mathbb{R}^{k}$,
$\mu\in\mathbb{R}^{k}\wedge E$, $\lambda\in\wedge^{2}E$. The Bianchi identity
(3) written for the vectors $U,V\in E$ and $q$ takes the form

$R(U,V)q+R(V,q)U+R(q,U)V=-(\omega\cdot\omega_{E})(U,V)+\lambda(U,V)p+\omega_{E}(U,V).$

Since $E$ is holonomy-invariant, $R(V,q)U,R(q,U)V\in E$. Hence, multiplying
the above equality scalarly by $q$, we get

$\lambda=0.$

This implies that $\omega\cdot\omega_{E}=0$, and

$R(V,q)U+R(q,U)V=\omega_{E}(U,V).$

Since the first prolongation of $\mathfrak{so}(E)$ is trivial, the equation
$R(V,q)U+R(q,U)V=0$, for all $U,V\in E$, has only trivial solution. This shows
that

$R(q,U)V=\frac{1}{2}\omega_{E}(U,V),\quad\forall\,U,V\in E.$

Since $\mathfrak{g}$ annihilates $S$, and $R$ takes values in $\mathfrak{g}$,
we get

$(R(q,U)\cdot S)(U_{1},U_{2},U_{3})=0,\quad\forall\,U,U_{1},U_{2},U_{3}\in E.$

This implies that

$\omega_{E}(U)\cdot\omega_{E}=0,\quad\forall\,U\in E,$

i.e.,

$\sigma_{\omega_{E}}=0.$

We conclude that the equality

$g([U_{1},U_{2}],U_{3})=\omega_{E}(U_{1},U_{2},U_{3}),\quad\forall\,U_{1},U_{2},U_{3}\in
E$

defines a Lie bracket on the vector space $E$. Let $E_{0}=\ker\omega_{E}$, and
let $E^{\prime}$ be the orthogonal complement to $E_{0}$ in $E$. Then $E_{0}$
and $E^{\prime}$ are commutative and semisimple ideals in $E$, respectively.

The Bianchi identity written for the vectors $X\in\mathbb{R}^{k}$, $U\in E$
and $q$ easily implies that $g(R(q,U)X,Y)=0$ for all $Y\in\mathbb{R}^{k}$.
Hence, $R(q,U)X\in\mathbb{R}p$. Let us write the bivector $\mu$ in the form
$\mu=\sum_{i=1}^{k}e_{i}\wedge\mu_{i}$, where $\mu_{i}\in E$. Now,
$\mathfrak{g}$ annihilates the tensor $p\wedge\mu$, hence,

$0=R(q,U)\cdot(p\wedge\mu)=\sum_{i=1}^{k}p\wedge e_{i}\wedge
R(q,U)\cdot\mu=\frac{1}{2}\sum_{i=1}^{k}p\wedge e_{i}\wedge[U,\mu_{i}].$

This implies that $[U,\mu_{i}]=0$ for all $U\in E$. Consequently, $\mu_{i}\in
E_{0}$. Denoting now $L\oplus E_{0}$ by $L$ and $E^{\prime}$ by $E$, we see
that $S$ is just as in the statement of the lemma.

Assume now again that the representation of $\mathfrak{g}$ in $L$ is weakly
irreducible.

Suppose that $\dim L=3$. According to [11], $S$ has the form

$S=p\wedge\omega+\omega_{E}+\alpha p\wedge e_{1}\wedge q,$

where $\alpha\in\mathbb{R}$ and all other elements are as above. Using the
Bianchi identity for the vectors $U\in E$, $e_{1}$ and $q$, it is easy to
shown that $\alpha=0$, and further considerations are just as in the previous
case.

Since $\mathfrak{g}$ annihilates the isotropic vector $p$, it is not possible
that the representation of $\mathfrak{g}$ in $L$ is weakly irreducible and
$\dim L=2$. Finally, if $\dim L=1$, then since $\mathfrak{g}$ annihilates the
isotropic vector $p$, it annihilates a two-dimensional Lorentzian subspace of
$\mathbb{R}^{1,n+1}$. Again, it is not hard to show that the statement of the
lemma holds true. The lemma is proved. ∎

######  Corollary 1.

Under the current assumptions it holds $S(p)=0$ unless $\dim M=3$ and $(M,g)$
is as in Theorem 7.

Let us suppose that the manifold $M$ is simply connected. Consider the dual
1-form $\eta$ to the vector field $p$,

$\eta(X)=g(p,X),\quad\forall\,X\in\Gamma(TM).$

It holds

$d\eta(X,Y)=Xg(p,Y)-Yg(p,X)-g(p,[X,Y])\\\
=g(\nabla^{g}_{X}p,Y)+g(p,\nabla^{g}_{X}Y)-g(\nabla^{g}_{Y}p,X)-g(p,\nabla^{g}_{Y}X)-g(p,\nabla^{g}_{X}Y-\nabla^{g}_{Y}X)\\\
=g(\nabla^{g}_{X}p,Y)-g(\nabla^{g}_{Y}p,X)=g\left(-(X\wedge
p)p-\frac{1}{2}S(X,p),Y\right)-g\left(-(Y\wedge
p)p-\frac{1}{2}S(Y,p),X\right)=0.$

Then there exists a function $\varphi$ such that $\eta=d\varphi$.

Proof of Theorem 6. Suppose now that $S$ is degenerate. From the above lemmata
it follows that $S=p\wedge\omega$.

From (1) it follows that

$\nabla^{g}_{X}p=-g(X,p)p=-\eta(X)p.$

Consequently the vector field

$p_{0}=e^{\varphi}p$

is $\nabla^{g}$-parallel. Its dual 1-form $p_{0}^{\flat}$ is closed, and there
exists a function $u$ on $M$ such that $du=p_{0}^{\flat}$. We conclude that

$d\varphi=e^{-\varphi}du,$

i.e.,

$de^{\varphi}=du.$

Since the both functions $\varphi$ and $u$ are defined up to a constant, we
may assume that

$u=e^{\varphi},$

i.e., the function $u$ is positive. By (1), $\nabla^{g}\tau_{0}=0$, where
$\tau_{0}=e^{\varphi}S$. ∎

Proof of Theorem 8. Suppose that the manifold $M$ is simply connected. Suppose
that $||S||_{g}\neq 0$ and $\dim M\neq 3$. Let $L$ and $E$ be as in Lemma 3.
Since the subspaces $L$ and $E$ of $T_{x}M$ are $\mathfrak{g}$-invariant, they
define $\nabla$-parallel distributions $\mathcal{L}$ and $\mathcal{E}$ on $M$.
Moreover,

$S=S_{1}+S_{2},\quad S_{1}\in\wedge^{3}\Gamma(\mathcal{L}),\quad
S_{2}\in\wedge^{3}\Gamma(\mathcal{E}),\quad\nabla S_{1}=\nabla
S_{2}=0,\quad\sigma_{S_{1}}=\sigma_{S_{2}}=0.$

Consider the new metric

$h=e^{-2\varphi}g,$

where the function $\varphi$ is as above.

By the standard formula, for all $X,Y\in\Gamma(TM)$ it holds

$\nabla^{h}_{X}Y=\nabla^{g}_{X}Y-g(p,X)Y-g(p,Y)X+g(X,Y)p.$

Combining this equality with (1), we get

(17) $\nabla^{h}_{X}Y=\nabla_{X}Y-g(p,X)Y-\frac{1}{2}S(X,Y),$

or just

(18) $\nabla^{h}_{X}=\nabla_{X}-g(p,X)\mathop{\text{missing}}{\rm
id}\nolimits-\frac{1}{2}S(X).$

This implies that the distributions $\mathcal{L}$ and $\mathcal{E}$ are
$\nabla^{h}$-parallel. By the Wu Theorem, $(M,h)$ is locally a product of a
Lorentzian manifold $(M_{0},h_{0})$ and a Riemannian manifold $(N,g_{N})$,
where $M_{0}$ and $N$ are integral submanifolds of the distributions
$\mathcal{L}$ and $\mathcal{E}$, respectively. In particular, locally it holds

$h=h_{0}+g_{N}.$

The vector field $p$ is tangent to the distribution $\mathcal{L}$, i.e., $p$
is a family of vector fields on the manifold $M_{0}$ depending on the local
coordinates on the manifold $N$. If $U\in\Gamma(TN)$, then
$\nabla^{h}_{U}p=0$. Thus, $p$ is a vector field on $M_{0}$. If
$X\in\Gamma(TM_{0})$, then

$\nabla^{h_{0}}_{X}p=\nabla^{h}_{X}p=-g(X,p)p=-\eta(X)p=-d\varphi(X)p.$

This shows that $\varphi$ is a function on the manifold $M_{0}$. Consequently,

$g_{0}=e^{2\varphi}h_{0}$

is a Lorentzian metric on $M_{0}$ and it coincides with $g$ restricted to
$M_{0}$. We get that

$g=g_{0}+e^{2\varphi}g_{N}.$

If $U\in\Gamma(TN)$, then by (18),

$\nabla^{h}_{U}S_{1}=\nabla_{U}S_{1}=0,$

i.e., $S_{1}$ is a 3-form on $M_{0}$. Next, if $X\in\Gamma(TM_{0})$, then, by
(1),

$\nabla_{X}^{g_{0}}S_{1}=\nabla_{X}^{g}S_{1}=\nabla_{X}S_{1}=0.$

We conclude that the data

$(M_{0},g_{0},p_{0}=e^{\varphi}p,\tau_{0}=e^{\varphi}S_{1},\varphi)$

is as in Theorem 6.

For $U\in\Gamma(TN)$ it holds

$\nabla^{g_{N}}_{U}S_{2}=\nabla^{h}_{U}S_{2}=\nabla_{U}S_{2}=0.$

If $X\in\Gamma(TM_{0})$, then

$\nabla^{h}_{X}S_{2}=-g(p,X)\mathop{\text{missing}}{\rm id}\nolimits\cdot
S_{2}=3g(p,X)S_{2}=3d\varphi(X)S_{2}.$

Consequently,

$\nabla^{h}_{X}\tau_{N}=0,$

where $\tau_{N}=e^{-3\varphi}S_{2}$. Thus,

$S=e^{-\varphi}\tau_{0}+e^{3\varphi}\tau_{N},\quad\nabla^{g_{N}}\tau_{N}=0,\quad\sigma_{\tau_{N}}=0.$

This proves the theorem. ∎

##  8\. Proof of Theorem 3

Let $(M,g)$ be a Lorentzian manifold that admits a metric connection $\nabla$
with parallel twistor-free torsion having non-zero vectorial component $\xi$.
Passing to the universal covering, we may assume that the manifold is simply
connected.

First suppose that $g(\xi,\xi)=0$. Then $(M,g)$ is given by one of the
Theorems 6, 7, 8.

Let $(M,g)$ be as in Theorem 7. Let $\gamma(t)$ be a geodesic such that

(19) $g(\dot{\gamma}(0),p)=a\neq 0\text{ and
}g(\dot{\gamma}(0),\dot{\gamma}(0))=\eta.$

It is clear that $\eta$ may be chosen to take any of the values $1,-1,0$.
Consider the function $\alpha(t)=g(\dot{\gamma}(t),p)$ defined along the
geodesic $\gamma(t)$. It holds

$\dot{\alpha}(t)=\frac{d}{dt}\alpha(t)=\dot{\gamma}(t)\alpha(t)=\dot{\gamma}(t)g(\dot{\gamma}(u),p)=g(\nabla^{g}_{\dot{\gamma}(t)}\dot{\gamma}(t),p)+g(\dot{\gamma}(t),\nabla^{g}_{\dot{\gamma}(t)}p)\\\
=g\left(\dot{\gamma}(t),-g(\dot{\gamma}(t),p)p-\frac{1}{2}S(\dot{\gamma}(t),p)\right)=-g(\dot{\gamma}(t),p)^{2}-\frac{1}{2}S(\dot{\gamma}(t),p,\dot{\gamma}(t))=-\alpha^{2}(t).$

Thus,

$\dot{\alpha}(t)+\alpha^{2}(t)=0.$

This shows that

$\alpha(t)=\frac{1}{t+c},\quad c\in\mathbb{R}$

is not defined for all values $t\in\mathbb{R}$, i.e., geodesic $\gamma$ is not
complete.

Let now $(M,g)$ be as in Theorem 6 or 8, i.e., $(M,g)$ is as in Section 7. We
use the notation of this section. If $\dim M=3$, then we may assume that
$S=0$. Let $\gamma(t)$ be a geodesic satisfying (19) with $\eta$ having one of
the values $1,-1,0$. Since $p_{0}$ is $\nabla^{g}$-parallel, for each
$t\in\mathbb{R}$ it holds
$g_{\gamma(t)}(\dot{\gamma}(t),(p_{0})_{\gamma(t)})=a$. Let
$\alpha(t)=\varphi(\gamma(t))$ be a function along $\gamma$, where $\varphi$
is as in Section 7. It holds

$\frac{d}{dt}\alpha(t)=\dot{\gamma}(t)\varphi=d\varphi(\dot{\gamma}(t))=g(\dot{\gamma}(t),p)=e^{-\alpha(t)}g(\dot{\gamma}(t),p_{0})=ae^{-\alpha(t)}.$

We conclude that

$e^{\alpha(t)}=at+c,\quad c\in\mathbb{R},$

and wee see that $\alpha(t)$ cannot be defined for all $t\in\mathbb{R}$, i.e.,
$\gamma(t)$ is not complete.

Next we suppose that $g(\xi,\xi)\neq 0$. We may assume that
$g(\xi,\xi)=\varepsilon\pm 1$. As in the proof of Theorem (4), it can be shown
that $S(\xi)=0$. Let as above, $\gamma(t)$ be a geodesic satisfying

$g(\dot{\gamma}(0),\xi)\neq 0\text{ and
}g(\dot{\gamma}(0),\dot{\gamma}(0))=\eta$

with $\eta$ having one of the values $1,-1,0$. As above, it is easy to show
that the function $\alpha(t)=g(\dot{\gamma}(t),\xi)$ defined along $\gamma(t)$
satisfies the equation

$\dot{\alpha}(t)+\alpha^{2}(t)=\eta\varepsilon.$

As we have seen, if $\eta=0$, then the geodesic $\gamma(t)$ is not complete.
If $\eta\varepsilon=-1$, then $\alpha(t)$ satisfies

$\arctan(\alpha(t))=-t+c.$

Since the function $\arctan$ is bounded, $\gamma(t)$ is not complete.

Suppose that $\eta\varepsilon=1$, i.e., $\eta=\varepsilon$. As in the proof of
Theorem 4, it can be shown that $\nabla^{g}_{\xi}\xi=0$, i.e., $\xi$ is a
geodesic vector field. We conclude that $\xi$ is a complete vector field. This
allows us to apply Theorem 4 and to conclude that $(M,g)$ is isomorphic to the
warped product

$(\mathbb{R}\times N,\varepsilon(ds)^{2}+e^{2s}g_{N}),$

where $(N,g_{N})$ is a Riemannian (if $\varepsilon=-1$) or a Lorentzian (if
$\varepsilon=1$) manifold. A curve $\gamma(t)$ in $M$ is defined by a function
$s(t)$ and a curve $\gamma_{1}(t)$ in $N$. According to [4, 6], if $\gamma(t)$
is a geodesic, then the function $s(t)$ satisfies the equation

$\ddot{s}=\varepsilon e^{2s}g_{N}(\dot{\gamma}_{1},\dot{\gamma}_{1}).$

Next,

$g(\dot{\gamma},\dot{\gamma})=\varepsilon\dot{s}^{2}+e^{2s}g_{N}(\dot{\gamma}_{1},\dot{\gamma}_{1}).$

This implies

$\ddot{s}+\dot{s}^{2}=\varepsilon g(\dot{\gamma},\dot{\gamma}).$

Let $a>0$. Consider a geodesic $\gamma(t)$ such that
$g(\dot{\gamma}(0),\dot{\gamma}(0))=\varepsilon a^{2}$ and $\dot{s}(0)>a$. We
obtain the equation

$\ddot{s}+\dot{s}^{2}=a^{2}.$

If $\dot{s}(t_{0})=a$ for some $t_{0}\in\mathbb{R}$, then $s(t)=at+c$ is the
unique solution satisfying the conditions $s(t_{0})=at_{0}+c$,
$\dot{s}(t_{0})=a$. Then $\dot{s}(0)=a$, which contradicts the assumption
$\dot{s}(0)>a$. We conclude that $\dot{s}(t)>a$ for all $t$. Then

$\frac{\ddot{s}}{\dot{s}^{2}-a^{2}}=-1,$

and

$\ln\left|\frac{\dot{s}+a}{\dot{s}-a}\right|=2a(t-c),\quad c\in\mathbb{R}.$

Since $\dot{s}(t)>a$ for all $t$, the function $\dot{s}(t)$ is not defined for
$t=c$, and the geodesic $\gamma(t)$ is not complete. ∎

Theorems 3 and 4 imply

######  Corollary 2.

The Lorentzian warped product (10) is not $\eta$-complete for $\eta=1,-1,0$.

Acknowledgements. The authors are thankful to Andrei Moroianu, Thomas Leistner
and Eivind Schneider for useful discussions. The authors are grateful to the
Reviewer for the important comments. I.E. was supported by grant
MUNI/A/1099/2022 of Masaryk University. A.G. acknowledges institutional
support of University of Hradec Králové.

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