# Enhancing GAN-Based Vocoders with Contrastive Learning Under Data-limited
Condition

###### Abstract

Vocoder models have recently achieved substantial progress in generating
authentic audio comparable to human quality while significantly reducing
memory requirement and inference time. However, these data-hungry generative
models require large-scale audio data for learning good representations. In
this paper, we apply contrastive learning methods in training the vocoder to
improve the perceptual quality of the vocoder without modifying its
architecture or adding more data. We design an auxiliary task with mel-
spectrogram contrastive learning to enhance the utterance-level quality of the
vocoder model under data-limited conditions. We also extend the task to
include waveforms to improve the multi-modality comprehension of the model and
address the discriminator overfitting problem. We optimize the additional task
simultaneously with GAN training objectives. Our result shows that the tasks
improve model performance substantially in data-limited settings. Our analysis
based on the result indicates that the proposed design successfully alleviates
discriminator overfitting and produces audio of higher fidelity.

Index Terms: GAN, self-supervised learning, vocoder

##  1 Introduction

Generative Adversarial Networks (GANs) [2] have been widely used in vocoders
and have achieved the state-of-the-art in the domain [3, 4, 5]. However,
training GAN vocoders still meets two challenges, data insufficiency and
discriminator overfitting.

In the realm of single-speaker speech synthesis, the limited size of available
datasets poses a significant challenge. To enhance the performance of vocoders
operating under such constraints, we propose the use of unsupervised learning
techniques to extract additional self-supervised signals for training. Self-
supervised learning (SSL) methods have demonstrated efficacy in a diverse
array of speech domains, including representation learning [6, 7, 8, 9, 10],
synthesis [11, 12, 13, 14], and multi-modality [15, 16]. Drawing on the
exceptional transfer learning capabilities of SSL, we seek to harness this
power in the realm of Vocoder modeling, focusing specifically on the
application of contrastive learning. Although contrastive learning has been
explored in the context of speech recognition [6], we are unaware of any
previous efforts to apply this approach to Vocoder modeling. In this work, our
aim is to leverage contrastive learning as an auxiliary task to enhance the
vocoding performance of GAN generators under data-limited conditions.

The second challenge, discriminator overfitting, is also shown to be crucial,
especially on small dataset [17, 18, 19], and the convergence of GAN also
critically depends on the quality of discriminators [20]. Contrastive learning
on the discriminator has been proved to alleviate this problem in image
generation [21], and the method, in general, is also shown to increase model’s
performance and robustness on vision and language tasks [22, 23, 24, 25].
However, in speech synthesis, a naive approach of mel-spectrogram contrastive
learning will only involve the generator, which encodes mel-spectrograms, but
not the discriminator, which encodes the waveform. Therefore, we propose to
extend the training to the discriminator by using a multi-modal contrastive
task between mel-spectrograms and waveforms.

Our contributions can be summarized as the following.

  1. 1.

We propose a contrastive learning task with masked mel-spectrograms to improve
the performance on limited data.

  2. 2.

We design a novel contrastive learning task of matching mel-spectrogram to
waveforms to regularize the discriminator and improve the perceptual quality
of the generator.

  3. 3.

We implement a framework for integrating contrastive learning into the GAN
training pipeline.

  4. 4.

We provide experimental results and in-depth analysis of the methods’
effectiveness compared to the baseline.

##  2 Methods

In this section, we first introduce the auxiliary contrastive task that we
have designed for the GAN vocoder model. Subsequently, we explicate the
details of how we modified the task to train both the generator and the
discriminator of the vocoder model. Finally, we illustrate our proposed
training framework, which synergizes the contrastive task with GAN objectives.
It is worth noting that we have utilized the same model architecture as HiFi-
GAN [4]. However, it is pertinent to mention that our method can be applied to
other GAN frameworks for vocoders as well.

###  2.1 Mel-spectrogram Contrastive Learning

In our GAN model, the generator takes a mel-spectrogram as input and outputs a
raw waveform through a stack of convolutional layers. We use a learnable feed-
forward layer to project the features of the convolutional layers onto a
latent space $R^{D}$ , where elements of similar semantics are close to each
other through contrastive learning. For each anchor in a batch of $N$ samples,
we apply masking on randomly selected intervals in time and frequency to
create a positive sample, while all other $(N-1)$ input samples and $(N-1)$
masked samples are used as negative samples. Together, the method results in
$1$ positive pair and $2(N-1)$ negative pairs in the batch. We then adapt the
InfoNCE loss [26] used in CLIP [27] for our loss function as follows:

$\begin{split}\mathcal{L}_{cl}&=-\frac{1}{N}\sum_{i=1}^{N}\left(\log\frac{\text{exp}(\tau{\textbf{v}_{i}}\cdot{\textbf{v}_{k})}}{\sum_{j=1;i\neq
j}^{2N}\text{exp}(\tau{\textbf{v}_{i}}\cdot{\textbf{v}_{j})})}\right)\end{split}$
(1)

where $\textbf{v}_{k}\in R^{D}$ is the masked sample from $\textbf{v}_{i}\in
R^{D}$ and $\tau$ is a temperature parameter. This method is shown in Fig. 1.

Fig. 1: Illustration of Mel-spectrogram Contrastive Learning. The Mel Encoder
is the backbone of the generator. This method only trains the generator in a
GAN framework.

###  2.2 Mel-spectrogram Waveform Contrastive Learning

In addition to training solely the generator, we propose a novel task that
involves contrastive spectrogram-waveform matching. This task serves to train
both the generator and the discriminators, promoting rich semantic
representation and preventing overfitting of the discriminators to the real or
fake classification. The method is illustrated in Fig. 2. For a batch of pairs
of mel-spectrograms and waveforms, we assign the labels of the true pairs to
be positive and those of the other pairs to be negative, resulting in $N$
positive pairs and $N(N-1)$ negative pairs in a batch of $N$ samples. We use
the backbone of the generator to encode the mel-spectrogram and the backbone
of the discriminator to encode the waveform. Similar to the method in section
2.1, we use two separate feed-forward layers to project each encoded feature
to the same latent dimension $R^{D}$. Then, we perform the modified loss
function

$\begin{split}\mathcal{L}_{cl}&=-\frac{1}{N}\sum_{i=1}^{N}\left(\log\frac{\text{exp}(\tau{\textbf{v}_{i}}\cdot{\textbf{w}_{i})}}{\sum_{j=1;i\neq
j}^{N}\text{exp}(\tau{\textbf{v}_{i}}\cdot{\textbf{w}_{j})})}\right)\end{split}$
(2)

where $\textbf{w}_{i}\in R^{D}$ is the latent embedding of the waveform
corresponding to the $i$th mel-spectrogram, $\textbf{v}_{i}\in R^{D}$ is the
latent embedding of the $i$th mel-spectrogram, and $\tau$ is a temperature
parameter. HiFi-GAN contains multiple discriminators, so we calculate a
contrastive loss between the mel-spectrogram embedding and each of the
waveform embeddings and sum them up. For simplicity, we refer them as one
discriminator in this paper unless otherwise mentioned.

Fig. 2: Illustration of Mel-spectrogram & Waveform Contrastive Learning. The
Mel Encoder is the backbone of the generator, and the Wave Encoder is the
backbone of the discriminator. Therefore, this method trains both the
generator and discriminator.

###  2.3 Multi-tasking Framework

To integrate contrastive learning with GAN tasks, we adopt a multi-tasking
framework that makes auxiliary tasks a joint optimization objective with
original learning goals [28]. As illustrated in Fig. 3, we create additional
heads for the training generator and discriminator with auxiliary tasks. The
total loss for training the vocoder model thus becomes:

$\begin{split}\mathcal{L}_{G}=\mathcal{L}_{adv}+\lambda_{fm}\mathcal{L}_{fm}+\lambda_{mel}\mathcal{L}_{mel}+\lambda_{cl}\mathcal{L}_{cl}\end{split}$
(3)
$\begin{split}\mathcal{L}_{D}=\mathcal{L}_{adv}+\mathcal{I}_{disc}\lambda_{cl}\mathcal{L}_{cl}\end{split}$
(4)

where $\mathcal{L}_{G}$ is the total loss for the generator and
$\mathcal{L}_{D}$ is the total loss for the discriminator. $\mathcal{L}_{adv}$
is the adversarial loss, $\mathcal{L}_{fm}$ is the feature matching loss, and
$\mathcal{L}_{mel}$ is the mel-spectrogram reconstruction loss in the original
HiFi-GAN training pipeline. $\mathcal{L}_{mel}$ can be either of the
contrastive loss described in section 2.1 or 2.2, and $\mathcal{I}_{disc}$ is
an indicator of whether the latter is used. Each loss is weighted with a
$\lambda$ coefficient which can be set as hyperparameters. We use a
$\lambda_{fm}$ of 2, $\lambda_{mel}$ of 45 from the HiFi-GAN setting [4] and a
$\lambda_{cl}$ of 1.

Fig. 3: Illustration of our multi-tasking frameworks. GAN-based Vocoder models
[3, 4] follow an adversarial network (top) consisting of a generator that
generates raw waveforms from mel-spectrograms and a discriminator that aims to
distinguish real from generated waveform samples. To incorporate the auxiliary
contrastive learning task, we propose a multi-tasking (bottom) framework,
which we set the contrastive task as additional learning objectives along with
the original GAN optimization objectives. This framework applies to both
contrastive learning methods described in section 2.1 and 2.2.

##  3 Experiments

###  3.1 Experimental Setting

In this section, we describe the details of our experimental settings
including the dataset, model choice, hyperparameters and evaluation metrics.

####  3.1.1 Dataset

In order to have a fair comparison with other vocoder models, we train the
model on the LJSpeech dataset [29] which is also used in other vocoder works
like HiFi-GAN [4]. LJSpeech is a public single-speaker dataset with 13100
short English audio clips whose durations span from 1 second to 10 seconds. We
use the default data split with 12950 training samples and 150 validation
samples. We use the same preprocessing configurations with HiFi-GAN, including
80 bands of mel-spectrograms as input and FFT size of 1024, window size of
1024, and hop size of 256 for conversion from waveform to mel-spectrograms.[4]

####  3.1.2 Implementation details

For experimental comparison on audio quality, we choose the most powerful
HiFi-GAN V1 and the most lightweight HiFi-GAN V3 as the baseline methods, and
we use the same model architecture as the backbone to apply the contrastive
tasks described in section 2.1 and 2.2. Under the multi-tasking framework, we
train HiFi-GAN along with the contrastive learning methods with a batch size
of 16, an AdamW optimizer, and a learning rate of 0.0002. For the following
experiments on the full dataset, all models are trained for 400k steps (about
96 hours) on one Nvidia TITAN RTX GPU. The experiments on 20% of the dataset
train for 300k steps (about 72 hours) on the same device, and those on 4% of
the dataset train for 200k steps. The model inference time on GPU is about
70ms for V1 models and 32ms for V3 models.

Model | MAE | MCD | MOS (CI)
---|---|---|---
Ground Truth | - | - | 4.32 ($\pm 0.05$)
HiFi-GAN V1 | 0.111 | 4.203 |  4.21 ($\pm 0.05$)
\+ Mel CL | 0.114 | 4.289 | 4.18 ($\pm 0.06$)
\+ Mel-Wave CL | 0.113 | 4.228 | 4.20 ($\pm 0.05$)
HiFi-GAN V3 | 0.203 | 7.786 | 4.10 ($\pm 0.05$)
\+ Mel CL | 0.204 | 7.766 | 4.13 ($\pm 0.07$)
\+ Mel-Wave CL | 0.203 | 7.723 | 4.09 ($\pm 0.06$)
Table 1: Objective and subjective evaluation results for models with mel-spectrogram contrastive loss (Mel CL) and mel-spectrogram contrastive loss (Mel-Wave CL). Models are trained on the full training set. CI is 95% confidence interval of the MOS score. Model | MAE $\downarrow$ | MCD $\downarrow$ | MOS $\uparrow$ (CI)
---|---|---|---
Ground Truth | - | - | 4.32 ($\pm 0.05$)
HiFi-GAN V1 (20% data) |  0.113 $(\uparrow 0.002)$ | 4.352 $(\uparrow 0.149)$ | 4.13 $(\downarrow 0.08)$ ($\pm 0.06$)
\+ Mel CL (20% data) | 0.116 $(\uparrow 0.002)$ | 4.430 $(\uparrow 0.139)$ | 4.11 $(\downarrow 0.07)$ ($\pm 0.07$)
\+ Mel-Wave CL (20% data) |  0.113 $(\uparrow 0.000)$ |  4.295 $(\uparrow 0.067)$ |  4.16 $(\downarrow 0.04)$ ($\pm 0.06$)
Hifi-GAN V3 (20% data) | 0.212 $(\uparrow 0.009)$ | 8.157 $(\uparrow 0.371)$ | 3.88 $(\downarrow 0.22)$ ($\pm 0.06$)
\+ Mel CL (20% data) |  0.207 $(\uparrow 0.003)$ |  7.960 $(\uparrow 0.206)$ | 3.95 $(\downarrow 0.18)$ ($\pm 0.06$)
\+ Mel-Wave CL (20% data) |  0.207 $(\uparrow 0.004)$ | 7.974 $(\uparrow 0.251)$ |  4.04 $(\downarrow 0.05)$ ($\pm 0.07$)
Table 2: Objective and subjective evaluation results for models trained with
20% of the training set. The number in parenthesis indicates the difference
from the results when trained on the full dataset.

####  3.1.3 Evaluation metrics

To objectively evaluate our models compared to the baseline, we measure the
mean average error (MAE) and mel-cepstral distortion (MCD) [30] on mel-
spectrograms. On both metrics, lower scores indicate closer alignment with the
ground truth. We also include a 5-scale mean opinion score (MOS) on audio
quality as a subjective evaluation performed on 50 samples excluded from the
training set.

###  3.2 Results

We present the results of models trained on full data with the multi-tasking
framework in Table 1. Below, we refer Mel CL as the mel-spectrogram
contrastive learning in section 2.1, and Mel-Wave CL as the mel-spectrogram
waveform contrastive learning in section 2.2. For V1 models, the baseline
performs slightly better than the proposed methods by margins of 0.02 on MAE,
0.025 on MCD, and 0.01 on MOS. For V3 models, on the objective tests, we
observe that the model trained with mel-spectrogram contrastive loss has
comparable performance with the baseline, while the one trained with mel-
spectrogram waveform contrastive loss achieves the highest scores on both
metrics. The results show that our proposed methods have at least comparable
performance to the baseline HiFi-GAN when training on the full dataset. On the
subjective tests, the V3 model with Mel CL achieves the highest MOS score,
0.03 above the V3 baseline. The model with Mel-Wave CL has a similar MOS score
with the baseline on the full dataset. Overall, when trained on the full
dataset, the proposed methods have limited gains on top of the baseline.

To investigate how each model performs under data limitation, we train the
three models on 20% of the dataset and evaluate them with the same validation
set. We present the results in Table 2. With less data, the baseline HiFi-GAN
V3 suffers a significant performance degradation across all metrics, including
0.371 on MCD and 0.22 on MOS. Meanwhile, the V3 model trained with Mel CL
experiences an increase of 0.194 on MCD and a drop of 0.18 on MOS. The V3
model trained with Mel-Wave CL has an increase of 0.251 on MCD and a drop of
only 0.05 on MOS. It suggests Mel-Wave CL is most resistant to data
insufficiency. The two proposed methods have comparable scores on the
objective evaluation, but the model with Mel-Wave CL obtains a significantly
higher score on the subjective test, 0.16 higher than the V3 baseline. The
findings align with our hypothesized alleviation of discriminator overfitting
by Mel-Wave CL, which is a more severe problem on the small training dataset.
Both of the proposed methods perform substantially better than the baseline by
0.07 and 0.16 respectively.

A similar trend exists in the HiFi-GAN V1 experiments, where Mel-Wave CL
achieves the best scores and the least performance drop on all metrics. One
slightly surprising finding is that the larger model V1 often experiences a
smaller performance drop compared to the smaller model V3 when trained on 20%
data. Typically, a larger model is expected to be more prone to overfitting
when trained on less data, which should lead to a larger performance drop. In
this specific case, however, HiFi-GAN V1 has a larger generator but the same
discriminator as HiFi-GAN V3 [4], which is our suspected reason for the
finding. Overall, the results show the benefits of additional supervision
signals from contrastive learning in data-limited situations and the superior
performance of Mel-Wave CL on a small dataset.

##  4 Conclusion

This paper describes our proposed contrastive learning framework to improve
GAN vocoders. Our results show the legacy of using contrastive learning as an
auxiliary task that facilitates vocoder training without adding more data or
modifying model architecture. We demonstrate that the proposed framework is
significant especially when training on limited data by extracting additional
supervision signals and reducing discriminator overfitting.

For future work, we plan to repeat the experiments on different model
architectures and datasets to test our method’s generalizability. In
particular, we want to test its extension to multi-speaker datasets, another
domain where data insufficiency is critical. We will also explore other
metrics to evaluate the discriminator overfitting problem more holistically.

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#  On the semi-symmetric Lorentzian spaces.

Abderrazzak Benroummane  Department of Computer Engineering and Mathematics
National School of Applied Sciences of Berrechid
Hassan 1st University, Morocco.
e-mail<EMAIL_ADDRESS>
###### Abstract

We give some properties of semisymmetric pseudo-Riemannian manifolds. These
are foliated manifolds and for the Lorentzian metric, Ricci operator has only
real eigenvalues.

###### keywords:

Lorentzean manifolds , Semi-symmetric pseudo-Riemannian manifolds , ,

##  1 Introduction

A pseudo-Riemannian manifold $(M,g)$ is said globally symmetric (respectively,
locally symmetric ) if any point $m\in M$ is a fixed point of a non-trivial
involutive isometric $\gamma_{m}$ (respectively, any point $m\in M$ admits a
neighborhood $N_{m}$ and the symmetric geodesic map $s_{m}:N_{m}\rightarrow
N_{m}$ is an isometry, where $s_{m}$ is defined by $x\mapsto
s_{m}(x)=\gamma_{x}(-1)$ such that $\gamma_{x}$ is the geodesic curve from
$m=\gamma_{x}(0)$ to $x=\gamma_{x}(1)$ for all $x\in N_{m}$) [See [5]].

It is necessary that a globally symmetrical manifold is locally symmetric but
is not sufficient unless the manifold is simply connected. So, that a manifold
$(M,g)$ is locally symmetric if and only if its Riemann curvature tensor $R$
is parallel, that is to say that $\nabla R=0$ where $\nabla$ is the Levi-
Civita connection associeted to metric $g$. Indeed; the necessary condition is
obvious from the fact that $(ds_{m})_{m}=-Id_{T_{m}M}$ and for the sufficient
condition it is a result of the following theorem:

######  Theorem 1.1.

[5] Let $(M,g)$ be a pseudo-Riemannian manifold. Let $\nabla$ and $R$ be the
Levi Civita connection and the Riemann curvature tensor associated to the
metric $g$ respectively.

Let $(x,y)\in M^{2}$ and $\tau:T_{x}M\rightarrow T_{y}M$ be an isomorphism.
Suppose that $\tau.R_{x}=R_{y}$ and $\nabla R=0$. Then, there exists a
neighborhood $V_{x}$ of $x$ and a diffeomorphism $\varphi:V_{x}\rightarrow
V_{y}$ with values in a neighborhood $V_{y}$ of $y$ such that
$(d\varphi)_{x}=\tau$ .

E. Cartan in ([8, 7]) showed that a locally symmetric Riemannian manifold
$(M,g)$ is locally homogeneous, more precisely, $M=G/H_{0}$ where $G$ is the
connected component of $id_{M}$ in $Iso(M)$ the Lie group of isometries of $M$
such that $G$ acts transitively on $M$. $H_{0}$ is the isotropic subgroup at
the point $m\in M$. This structure is independent of the chosen point $m$.

On the other hand, ${\mathfrak{g}}:={{\mathfrak{h}}}_{0}+T_{m}M$ is the Lie
algebra of $G$, where ${\mathfrak{h}}_{0}$ is the Lie algebra of $H_{0}$,
$T_{m}M$ is the tangent space of $M$ at the point $m$. The Lie bracket on
${\mathfrak{g}}$ is given by:

$\begin{array}[]{cccc}\left[X,Y\right]&:=&R_{m}(X,Y),&\;\;\forall X,\;Y\in
T_{m}M,\\\ \left[A,X\right]&:=&A(X),&\;\;\forall X\in T_{m}M,\;\forall
A\in{\mathfrak{h}}_{0},\\\ \left[A,B\right]&:=&AB-BA,&\;\;\forall
A,\;B\in{\mathfrak{h}}_{0}.\end{array}$ (1)

Moreover, $R$ checked

$\left[R_{m}(x,y),R_{m}(z,t)\right]=R_{m}(R_{m}(x,y)z,t)+R_{m}(z,R_{m}(x,y)t),\;\forall(x,y,z,t)\in(T_{m}M)^{4},$
(2)

######  Definition 1.1.

  1. a)

Let $(V,g)$ be a pseudo-Riemannian vector space, ${\mathfrak{h}}$ be a Lie
subalgebra of $so(V)$ and $R$ be a curvature tensor on $V$. If
$R(x,y)\in{\mathfrak{h}}$ for all $(x,y)\in V^{2}$, the triple
$\left[V,{\mathfrak{h}},R\right]$ is said holonomy sysmtem. If moreover $R$
satisfies the relationship

$(A.R)(x,y)=[A,R(x,y)]-R(A(x),y)-R(x,A(y)=0,\;\forall(x,y)\in V^{2},$ (3)

the holonomy sysmtem $\left[V,{\mathfrak{h}},R\right]$ is said to be semmetric
(see[13]).

  2. b)

A pseudo-Riemannian manifold $(M,g)$ is said to be semi-symmetric if, its
Riemann curvature tensor satisfies the equation (2).

Note that if, $(M,g)$ is a semi-symmetric pseudo-Riemannian manifold, then, at
any point $m\in M$, the triple
$\left[T_{m}M,{\mathfrak{h}}(R_{m}),R_{m}\right]$ is symmetric, where
${\mathfrak{h}}(R_{m})$ is the vector space spanned by all $R(x,y)$ for all
$x$ and $y$ in $T_{m}M$, moreover ${\mathfrak{h}}(R_{m})$ is a Lie algebra.
This result allows K. Nomizu in 1968 ([12]) to conjecture that any semi-
symmetric Riemannian manifold $(M,g)$ of dimension greater than or equal to
$3$ is locally symmetric (ie, $\nabla R=0$). But, in 1972, H. Takagi ([14])
gave a counter example of a Riemannian manifold of dimension $3$ satisfying
$R(X,Y).R=0$ and not $\nabla R=0$.

On the other hand, semi-symmetry is a generalization of two-symmetry
($\nabla^{2}R=0$) and of local symmetry. Several studies are done on semi-
symmetry. In the Riemannian case, Z. I Szabo gave the complete classification
of semi-symmetric Riemannian manifolds ([15, 16, 17]). For the other strictly
pseudo-Riemannian metrics, there is no general study, the few cases studied
are with additional conditions. We can mention the complete classification of
four dimensional semisymmetric homogeneous Lorentzian manifolds given by the
auther , M. boucetta and A. Ikemakhen (See [2]). In [3], the auther gave the
complete classification of four dimensional semisymmetric homogeneous neutral
manifolds. In [9], G. Calvaruso has studied the three-dimensional semi-
symmetric homogeneous Lorentzian manolds. G. Calvaruso and B. De Leo have
studied the semi-symmetric Lorentzian three-manifolds admitting a parallel
degenerate line field (See [10]). In [11], A. Haji-Badali and A. Zaeim gave a
complete classification of Semi-symmetric four dimensional neutral Lie groups.

The purpose of this work is to give a few characterizations of semi-symmetric
Lorentzian manifolds where we give the following results:

  1. 1.

Any semi-symmetric Lorentzian manifold, its Ricci operator has only real
eigenvalues.

  2. 2.

A semi-symmetric pseudo-Riemannian manifold $(M,g)$ is foliated manifold. In
the case where the metric is Lorentzian, the restriction of the Ricci operator
on each leaf has at most one non-zero real eigenvalue.

  3. 3.

If $(M,g)$ is a simple leaf semi-symmetric Lorentzian manifold, then the Ricci
opertor is diagonalizable or isotropic. Moreover, if $\lambda$ is a non-zero
eigenvalue of Ricci and $V_{1}:=\ker({\mathrm{Ric}}-\lambda id_{TM})$ is
Riemannian subspace of dimonsion $\geq 3$, then,
$1\leq\dim(\ker({\mathrm{Ric}}))\leq 2$.

This paper is organized in the following way. In the second paragraph, we give
generalities on curvature tensors on a pseudo-Riemannian vector space $V$.
Thus, characterizations of a semi-symmetric curvature tensor by decomposing
the space $V$ as a direct sum of the characteristic subspaces of the Ricci
operator. We also give another so-called primitive decomposition under the
action of primitive holonomy algebra. In third paragraph, we give some
proprties of _Ricci decomposition_ and the _primitive decomposition_ of
tangent bundle of the semi-symmetric pseudo-Riemannian manifold. The last
paragph is devoted to studying semi-symmetric Lorentzian manifolds which is a
simple leaf where Ricci is diagonalizabe or isotropic and if Ricci admits a
non-zero eigenvalue then it is diagonalizable. Moreover, if this eigenvalue
has a multiplicity hose greater than or equal to $3$ with Riemannian
eigenspace, we will have $1\leq\dim(\ker({\mathrm{Ric}}))\leq 2$.

At first, we give some definitions that we will use later

######  Definition 1.2.

Let $(M,g)$ be a locally connected pseudo-Riemannian manifold.

  1. 1.

$(M,g)$ is said to be irreducible if, at each point $m\in M$, the only
subspace of $T_{m}M$ which are invariant under the action of the holonomy
group $H_{m}(M)$ are $\\{0\\}$ and $T_{m}M$, and reducible otherwise.

  2. 2.

$(M,g)$ is said to be indecomposable if, at any point $m\in M$, the only
nondegenerate subspaces of $T_{m}M$ which are invariant under the action of
the holonomy Lie group $H_{m}(M)$ are $\\{0\\}$ and $T_{m}M$.

  3. 3.

$(M,g)$ is said to be weakly irreducible if it is reducible and
indecomposable.

##  2 Semisymmetrical curvature tensor on a pseudo-Riemannian vector space

###  2.1 Curvature tensor

Let $(V,\langle\;,\;\rangle)$ be a $n$-dimensional pseudo-Riemannian vector
space. We identify $V$ and its dual $V^{*}$ by the means of
$\langle\;,\;\rangle$. This implies that the Lie algebra $V\otimes V^{*}$ of
endomorphisms of $V$ is identified with $V\otimes V$, the Lie algebra
$\mathrm{so}(V,\langle\;,\;\rangle)$ of skew-symmetric endomorphisms is
identified with $V\wedge V$ and the space of symmetric endomorphisms is
identified with $V\vee V$ (the symbol $\wedge$ is the outer product and $\vee$
is the symmetric product). For any $u,v\in V$,

$(u\wedge v)w=\langle v,w\rangle u-\langle u,w\rangle
v\quad\mbox{and}\quad(u\vee v)w=\frac{1}{2}\left(\langle v,w\rangle u+\langle
u,w\rangle v\right).$

On the other hand, $V\wedge V$ carries also a nondegenerate symmetric product
also denoted by $\langle\;,\;\rangle$ and given by

$\langle u\wedge v,w\wedge t\rangle:=\langle u\wedge v(w),t\rangle=\langle
v,w\rangle\langle u,t\rangle-\langle u,w\rangle\langle v,t\rangle.$

We identify $V\wedge V$ with its dual by means of this metric.

We consider Bianchi’s linear mapping of the space $P=\vee^{2}(\wedge^{2}V)$
given by:

$B\bigg{(}(a\wedge b)\vee(c\wedge d)\bigg{)}=(a\wedge b)\vee(c\wedge
d)+(b\wedge c)\vee(a\wedge d)+(c\wedge a)\vee(b\wedge d).$ (4)

Let ${\mathfrak{g}}$ be a subalgebra of $so(V)$, we set:

$R({\mathfrak{g}}):=\ker(B)=\\{T\in{\mathfrak{g}}\vee{\mathfrak{g}}/\leavevmode\nobreak\
\leavevmode\nobreak\ B(T)=0\\}$

et

${\mathfrak{g}}_{sym}=\\{T\in R({\mathfrak{g}})/{\mathfrak{g}}.T=0\\}.$

The set $R({\mathfrak{g}})$ is called the space of all curvature tensors of
type ${\mathfrak{g}}$ and any element of $R(so(V))$ is called curvature tensor
on $V$. The set ${\mathfrak{g}}_{sym}$ is called the space of symmetric
curvature tensors of type ${\mathfrak{g}}$.

According identifications cited above, we obtained;

######  Lemma 2.1.

Let $(V,\langle\;,\;\rangle)$ be a pseudo-Riemannian vector space. Any
curvature tensor ${\mathrm{K}}\in R(so(V))$ can be identified with an element
of $\otimes^{4}V^{*}$ (i.e., a covariant 4-tensor on V) satisfying:

  1. i)

${\mathrm{K}}(a,b,u,v)=-{\mathrm{K}}(b,a,u,v)$,

  2. ii)

${\mathrm{K}}(a,b,u,v)=-{\mathrm{K}}(a,b,v,u)$,

  3. iii)

${\mathrm{K}}(a,b,u,v)+{\mathrm{K}}(b,u,a,v)+{\mathrm{K}}(u,a,b,v)=0,$

  4. iv)

${\mathrm{K}}(a,b,u,v)={\mathrm{K}}(u,v,a,b)$,

where $a,b,u,v\in V$. Note that $iv)$ is a result of $i)$, $ii)$ et $iii)$.

######  Lemma 2.2.

Let $(V,\langle\;,\;\rangle)$ be a pseudo-Riemannian vector space. Any
curvature tensor ${\mathrm{K}}\in R(so(V))$ can be identified with a symmetric
bilinear map also denoted by ${\mathrm{K}}:V\wedge V\longrightarrow V\wedge V$
satisfying:

  1. 1.

for all $u$, $v\in V$, ${\mathrm{K}}(u\wedge v)=-{\mathrm{K}}(v\wedge u)$,

  2. 2.

for all $u,v,w\in V$, ${\mathrm{K}}(u\wedge v)w+{\mathrm{K}}(v\wedge
w)u+{\mathrm{K}}(w\wedge u)v=0$.

These relationships lead to:

$\langle{\mathrm{K}}(a\wedge b)u,v\rangle=\langle{\mathrm{K}}(u\wedge
v)a,b\rangle,\quad a,b,u,v\in V.$ (5)

We often set:

${\mathrm{K}}(u,v):={\mathrm{K}}(u\wedge v)$

Let $(V,\langle\;,\;\rangle)$ be a pseudo-Riemannian vector space and let
${\mathrm{K}}\in R(so(V))$ be a curvature tensor $V$. We set:

${\mathfrak{h}}({\mathrm{K}}):={\mathrm{K}}(\wedge^{2}V)=\\{{\mathrm{K}}(u,v)/\;u,\;v\in
V\\},$

${\mathfrak{h}}({\mathrm{K}})$ is a vector subspace of $so(V)$. We set
$\bar{{\mathfrak{h}}}({\mathrm{K}})$ the smallest Lie subalgebra of $so(V)$
containing ${\mathfrak{h}}({\mathrm{K}})$ called primitive holonomy algebra of
${\mathrm{K}}$, its Lie groupe is noted $\bar{\cal H}({\mathrm{K}})$ and
called primitive holonomy group. The action of
$\bar{{\mathfrak{h}}}({\mathrm{K}})$ and $\bar{H}({\mathrm{K}})$) on the
curvature tensor ${\mathrm{K}}$ are given respectively by:

$(A.{\mathrm{K}})(a,b):=[A,{\mathrm{K}}(a,b)]-{\mathrm{K}}(A(a),b)-{\mathrm{K}}(a,A(b)),$
(6)

and

$(\sigma.{\mathrm{K}})(a,b):=\sigma\circ{\mathrm{K}}(\sigma^{-1}(a),\sigma^{-1}(b))\circ\sigma^{-1},$
(7)

where $A\in\bar{{\mathfrak{h}}}({\mathrm{K}})$,
$\sigma\in\bar{H}({\mathrm{K}})$ and $a$, $b\in V$.

The Ricci curvature associated to $\mathrm{K}$ is the symmetric bilinear form
on $V$ given by $\mathfrak{ric}_{{\mathrm{K}}}(u,v)=trace(\tau(u,v))$, where
$\tau(u,v):V\longrightarrow V$ is given by $\tau(u,v)(a)=\mathrm{K}(u,a)v$.
The Ricci operator is the symmetric endomorphism
$\mathrm{Ric}_{K}:V\longrightarrow V$ given by
$\langle\mathrm{Ric}_{{\mathrm{K}}}(u),v\rangle=\mathfrak{ric}_{{\mathrm{K}}}(u,v),\,\,\text{
for all }u,v\in V$. We call ${\mathrm{K}}$ Einstein (resp. Ricci isotropic) if
${\mathrm{Ric}}_{{\mathrm{K}}}=\lambda\mathrm{Id}_{V}$ (resp.
${\mathrm{Ric}}_{{\mathrm{K}}}\not=0$ and
${\mathrm{Ric}}_{{\mathrm{K}}}^{2}=0$).

######  Example 1.

if ${{\mathrm{K}}}=(u\wedge v)\vee(w\wedge t)$ then,

${\mathrm{ric}}_{K}=\langle u,w\rangle t\vee v+\langle v,t\rangle u\vee
w-\langle v,w\rangle t\vee u-\langle u,t\rangle v\vee w.$

Note that, if there is no ambiguity, we put
${\mathrm{Ric}}={\mathrm{Ric}}_{{\mathrm{K}}}$ and
$\mathfrak{ric}=\mathfrak{ric}_{{\mathrm{K}}}$.

###  2.2 Semi-symmetrical curvature tensors

####  2.2.1 Primitive decompotion

######  Definition 2.1.

Let $(V,\langle\;,\;\rangle)$ be a $n$-dimensional pseudo-Riemannian vector
space and let ${\mathrm{K}}$ a curvature tensor on $V$. The curvature tensor
${\mathrm{K}}$ is called semi-symmetric if ${\mathfrak{h}}({\mathrm{K}})$ is a
Lie algebra and ${\mathrm{K}}$ is a symmetric curvature tensor of type
${\mathfrak{h}}({\mathrm{K}})$, i.e.
${\mathrm{K}}\in{\mathfrak{h}}({\mathrm{K}})_{sym}.$

######  Remark 1.

In a pseudo-Riemannian space $(V,\langle\;,\;\rangle)$ of small dimension, to
give the semi-symmetric curvature tensors on $V$, it suffices to give the Lie
subalgebras ${\mathfrak{g}}$ of $so(V)$ satisfying,
${\mathfrak{g}}={\mathfrak{g}}_{sym}$ (For example, see [2, 3]).

It is obvious that if the curvature ${\mathrm{K}}$ is constant, i.e.
${\mathrm{K}}=\lambda Id_{V\wedge V}$, then it is semi-symmetric and if
$\lambda\neq 0$, we get $\mathfrak{h}({\mathrm{K}})=\mathrm{so}(V)$. In
particular, any curvature tensor on the space pseudo-riemannian of dimension
$2$ is semi-symmetric.

######  Proposition 2.1.

Let $(V,\langle\;,\;\rangle)$ be a pseudo-Riemannian vector space of dimension
$n$ and let ${\mathrm{K}}$ be a curvature tensor on $V$ treated as an
symmetrical endomorphism, ${\mathrm{K}}\;:\>a\wedge
b\in\wedge^{2}V\mapsto{\mathrm{K}}(a\wedge b)\in\wedge^{2}V$. Then,

${\mathrm{K}}$ is semi-symmetric iff ${\mathrm{K}}.{\mathrm{K}}=0$, i.e.
${\mathrm{K}}$ checks;

$[{\mathrm{K}}(u,v),{\mathrm{K}}(a,b)]={\mathrm{K}}({\mathrm{K}}(u,v)a,b)+{\mathrm{K}}(a,{\mathrm{K}}(u,v)b),\quad
u,v,a,b\in V.$ (8)

Let ${\mathrm{K}}$ be a semisymmetric curvature tensor on $n-$dimensional
pseudo-Riemannian vector space $(V,\langle\;,\;\rangle)$. Then
${\mathfrak{h}}({\mathrm{K}})=\bar{{\mathfrak{h}}}({\mathrm{K}})$ is the
primitive holonomy Lie algebra of ${\mathrm{K}}$ and we set ${\cal
H}({\mathrm{K}}):=\bar{\cal H}({\mathrm{K}})$ its the primitive holonomy Lie
group. Moreover, ${\mathrm{K}}$ is invariant by ${\cal H}({\mathrm{K}})$ and
it is parallel under the action of ${\mathfrak{h}}({\mathrm{K}})$, i.e;

$\sigma^{*}{\mathrm{K}}={\mathrm{K}}\Longleftrightarrow(\sigma^{*}{\mathrm{K}})(a,b)u=\sigma\big{(}{\mathrm{K}}(\sigma^{-1}(a),\sigma^{-1}(b))(\sigma^{-1}(u))\big{)}={\mathrm{K}}(a,b)(u)$

and

${\mathrm{K}}(a,b).{\mathrm{K}}=0\Longleftrightarrow[{\mathrm{K}}(a,b),{\mathrm{K}}(u,v)]={\mathrm{K}}({\mathrm{K}}(a,b)(u),v)+{\mathrm{K}}(u,{\mathrm{K}}(a,b)(v)),$

for all $\sigma\in{\cal H}({\mathrm{K}})$ and for all $(a,b,u,v)\in V^{4}$.

So the action of the primitive holonomy Lie algebra on the space $V$
introduces a weakly decomposition of $V$;

$V=V_{0}+V_{1}+\ldots+V_{s}+V_{0}^{\prime}$ (9)

where $V_{0}=\Big{\\{}x\in V\;/\;\hbox{\bb K}(u,v)x=0,\forall(u,v)\in
V^{2}\Big{\\}}$, $V_{0}^{\prime}$ is the dual of the subspace
$V_{0}\cap(V_{1}+\ldots+V_{s})$ and each subspace $V_{i}+(V_{i}\cap
V_{i}^{\bot})^{\prime}$ is indecomposable subspace under the action of
${\mathfrak{h}}({\mathrm{K}})$ for all $i\geq 1$.

######  Definition 2.2.

The decomposition (9) is called primitive decompostion of $V$.

Moreover, the primitive decomposition satisfies the following properties;

  1. 1.

for any $i=0,\ldots,s$, $V_{i}$ is ${\mathfrak{h}}({\mathrm{K}})$-invariant,

  2. 2.

for any $i,j=0,\ldots,s$ with $i\not=j$, ${\mathrm{K}}(V_{i},V_{j})=0$,

####  2.2.2 Ricci decomposition

######  Lemma 2.3.

Let ${\mathrm{K}}$ be a semi-symmetrical curvature tensor on the pseudo-
Riemannian space $(V,\langle\;,\;\rangle)$. Then, its Ricci operator commutes
with all the endomorphisms ${\mathrm{K}}(u,v)$, that is:

${\mathrm{K}}(u,v)\circ\mathrm{Ric}=\mathrm{Ric}\circ{\mathrm{K}}(u,v),\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \forall\leavevmode\nobreak\ u,v\in
V.$ (10)

###### Proof.

Soit $u$, $v$, $w$, $z\in V$.

$\displaystyle\langle{\mathrm{Ric}}({\mathrm{K}}(u,v)w,z\rangle$
$\displaystyle=$ $\displaystyle\mathfrak{trace}(\tau(z,{\mathrm{K}}(u,v)w))$
$\displaystyle=$
$\displaystyle\mathfrak{trace}(t\mapsto{\mathrm{K}}(z,t)({\mathrm{K}}(u,v)w))$
$\displaystyle=$
$\displaystyle\mathfrak{trace}(t\mapsto{\mathrm{K}}(u,v)({\mathrm{K}}(z,t)w))-\mathfrak{trace}(t\mapsto{\mathrm{K}}(z,{\mathrm{K}}(u,v)t)w)$
$\displaystyle-\mathfrak{trace}(t\mapsto{\mathrm{K}}({\mathrm{K}}(u,v)z,t)w)$
$\displaystyle=$
$\displaystyle\mathfrak{trace}([{\mathrm{K}}(u,v),\tau(z,w)])-\mathfrak{trace}(\tau({\mathrm{K}}(u,v)z,w))$
$\displaystyle=$
$\displaystyle\langle{\mathrm{K}}(u,v)\circ{\mathrm{Ric}}(w),z\rangle.$

∎

A Ricci operator ${\mathrm{Ric}}$ satisfying the equation (10) is said semi-
symmetric. In particular, for a pseudo-Riemannian manifold whose Ricci
operator is parallel ($\nabla{\mathrm{Ric}}=0$), ${\mathrm{Ric}}$ is semi-
symmetric. This type of manifolds was studied by C. Boubel (See [6]) by giving
the following classification theorem:

######  Theorem 2.1.

([6]) Let $(M,g)$ be a pseudo-Riemannian manifold with a parallel Ricci
curvature ${\mathrm{Ric}}$(i.e. $\nabla.{\mathrm{Ric}}=0$) and let $\chi$ be
the minimal polynomial of ${\mathrm{Ric}}$. Then:

  1. 1.

$\chi=\Pi_{i}P_{i}$ where:

    1. $\bullet$

$\forall i\neq j$, $P_{i}\wedge P_{j}=1$ (i.e, $P_{i}$ et $P_{j}$ are mutually
prime),

    2. $\bullet$

$\forall i$, $P_{i}$ is irreducible or $P_{i}=X^{2}$.

  2. 2.

There is a canonical family $(M_{i})_{i}$ of pseudo-Riemannian manifolds such
that the minimal polynomial of ${\mathrm{Ric}}_{i}$ on each $M_{i}$ is
$P_{i}$, and a local isometry $f$ mapping the Riemarmian product $\Pi M_{i}$
onto $M$. $f$ is unique up to composition with a product of isometries of each
factor $M_{i}$. If $M$ is complete and simply connected, $f$ is an isometry.

For the proof of the first result of this theorem above, C. Boubel only used
the following hypothesis: On the tangent space $T_{x}M$ at each point $x\in
M$, the Ricci operator ${\mathrm{Ric}}_{x}$ commutes with the endomorphisms
$R_{x}(u,v)$ for all $u$, $v\in T_{x}M$. A result which remains valid for
spaces provided with semi-symmetrical curvatures and more particularly, for
spaces with semi-symmetrical Ricci. Thus, the following result:

######  Theorem 2.2.

Let ${\mathrm{K}}$ be a semisymmetric curvature tensor on a pseudo-Riemannian
space $(V,\langle\;,\;\rangle)$ and let $\chi$ be the minimal polynomial of
its Ricci operator ${\mathrm{Ric}}$. Then the following properties are
checked:

  1. 1.

$\chi=\Pi_{i}P_{i}$ where:

    1. $\bullet$

$\forall i\neq j$, $P_{i}\wedge P_{j}=1$,(i.e, $P_{i}$ et $P_{j}$ are mutually
prime),

    2. $\bullet$

$\forall i$, $P_{i}$ is irreducible or $P_{i}=X^{2}$.

  2. 2.

$V$ splits orthogonally as

$V=E_{0}\oplus E_{1}\oplus\ldots\oplus E_{r},$ (11)

where $E_{0}=\ker((\mathrm{Ric}^{2}))$ and $E_{i}=\ker(P_{i}(\mathrm{Ric}))$,

  3. 3.

for any $u,v\in V$ and $i=0,\ldots,r$, $E_{i}$ is
${\mathfrak{h}}({\mathrm{K}})$-invariant,

  4. 4.

for any $i,j=0,\ldots,r$ with $i\not=j$, ${\mathrm{K}}_{|E_{i}\wedge
E_{j}}=0$,

  5. 5.

for any $i=1,\ldots,r$, $\dim E_{i}\geq 2$.

  6. 6.

The primitive holonomy algebra ${\mathfrak{h}}({\mathrm{K}})$ is satisfied;

${\mathfrak{h}}({\mathrm{K}})={\mathfrak{h}}_{0}({\mathrm{K}})+{\mathfrak{h}}_{1}({\mathrm{K}})+...+{\mathfrak{h}}_{r}({\mathrm{K}}),$

where ${\mathfrak{h}}_{i}(R):=\\{{\mathrm{K}}(X,Y)\;/\;X,Y\in E_{i}\\}$ is a
Lie subalgebra for all $0\leq i\leq r$.

######  Definition 2.3.

The decomposition (11) is called Ricci decompostion of $V$.

######  Remark 2.

If $P_{0}=X$, we get that $\ker(\mathrm{Ric})=\ker(\mathrm{Ric}^{2})$. That’s
why we always set $E_{0}=\ker((\mathrm{Ric}^{2}))$.

###### Proof.

  1. 1.

In the proof of the theorem 2.1, C. Boubel only used the commutativity of the
Ricci operator with the each endomorphisms $R(u,v)$ which remains valid for
the first result of our theorem.

  2. 2.

Let $\chi=\Pi_{i}P_{i}$ be the irreducible decomposition of $\chi$.

Then, $V$ splits orthogonally as:

$V=\bigoplus_{i}E_{i},\leavevmode\nobreak\ \leavevmode\nobreak\ \text{ where
}\leavevmode\nobreak\ \leavevmode\nobreak\ E_{i}=\ker(P_{i}({\mathrm{Ric}})).$

  3. 3.

Let $(u,v)\in V^{2}$. So ${\mathrm{K}}(u,v)$ commutes with ${\mathrm{Ric}}$,
then each $E_{i}$ is invariant by ${\mathrm{K}}(u,v)$.

  4. 4.

Let $u\in E_{i}$, $v\in E_{j}$ and $a,b\in V$ such that $i\neq j$. Since
${\mathrm{K}}(a,b)(E_{i})\subset E_{i}$ and $\langle E_{i},E_{j}\rangle=0$, we
obtain

$0=\langle{\mathrm{K}}(a,b)u,v\rangle\stackrel{{\scriptstyle\eqref{eq1-5}}}{{=}}\langle{\mathrm{K}}(u,v)a,b\rangle$

and then, ${\mathrm{K}}(u,v)=0$.

  5. 5.

We assume that it exists $i\in\\{1,\ldots,r\\}$ such that $\dim E_{i}=1$. We
choose an element $e$ in $E_{i}$ checking $\langle e,e\rangle=\epsilon$ with
$\epsilon^{2}=1$ and we complete it to an orthonormal basis
$(e,e_{1},\ldots,e_{n-1})$ with $\langle e_{i},e_{i}\rangle=\epsilon_{i}$ and
$\epsilon_{i}^{2}=1$. For all $a,b\in V$, we get that ${\mathrm{K}}(a,b)$ is
skewsymmetric endomorphism which leaves $E_{i}$ invariant. Then
${\mathrm{K}}(a,b)e=0$, while

$\epsilon\alpha_{i}=\langle\mathrm{Ric}(e),e\rangle=\epsilon\langle{\mathrm{K}}(e,e)e,e\rangle+\sum_{i=1}^{n-1}\epsilon_{i}\langle{\mathrm{K}}(e,e_{i})e,e_{i}\rangle=0,$

which is absurd and this completes the proof of the theorem. ∎

This proposition reduces the determination of semi-symmetric curvature tensors
on pseudo-riemannian vector spaces to the determination of two classes of
semi-symmetric curvature tensors:

  1. 1.

Einstein semi-symmetric curvature tensors ( with complexification in the case
of non-real complex eigenvalues)

  2. 2.

Semi-symmetric curvature tensors with Ricci operator is satisfied
${\mathrm{Ric}}^{2}=0$.

###  2.3  Semi-symmetrical Lorentzian space

######  Proposition 2.2.

([2]) Let ${\mathrm{K}}$ be a semi-symmetric curvature tensor on a Lorentzian
vector space $(V,\langle\;,\;\rangle)$. Then all eigenvalues of
$\mathrm{Ric}_{K}$ are real. Denoted by $\alpha_{1},\ldots,\alpha_{r}$ the non
null eigenvalues and $E_{1},\ldots,E_{r}$ the corresponding eigenspaces. Then:

  1. 1.

$V$ splits orthogonally as $V=E_{0}\oplus E_{1}\oplus\ldots\oplus E_{r}$,
where $E_{0}=\ker(\mathrm{Ric}^{2})$,

  2. 2.

for any $i=0,\ldots,r$, $E_{i}$ is ${\mathfrak{h}}({\mathrm{K}})$-invariant,

  3. 3.

for any $i,j=0,\ldots,r$ with $i\not=j$, ${\mathrm{K}}_{|E_{i}\wedge
E_{j}}=0$,

  4. 4.

for any $i=1,\ldots,r$, $\dim E_{i}\geq 2$.

Moreover, the primitive holonomy algebra ${\mathfrak{h}}({\mathrm{K}})$ splits
orthogonally:

${\mathfrak{h}}({\mathrm{K}})={\mathfrak{h}}_{0}({\mathrm{K}})+{\mathfrak{h}}_{1}({\mathrm{K}})+...+{\mathfrak{h}}_{r}({\mathrm{K}}),$

where ${\mathfrak{h}}_{i}({\mathrm{K}}):=\\{{\mathrm{K}}(u,v)\;/\;u,v\in
E_{i}\\}$ is a Lie subalgebra for all $0\leq i\leq r$.
More precisely, one of the following two situations occurs:

  1. a)

Ricci is diagonlizable; $E_{0}=\ker({\mathrm{Ric}})$.

  2. b)

Ricci is of isotropic type: $\ker({\mathrm{Ric}})\subsetneq
E_{0}=\ker({\mathrm{Ric}}^{2})$.

###### Proof.

This is a special case of theorem.2.2 and it suffices to show that Ricci has
only real eigenvalues, this is equivalent to showing that Ricci’s minimal
polynomial has no non-real roots.

Suppose that ${\mathrm{Ric}}$ admits a non-real eigenvalue $z=a+ib$ where
$b\neq 0$. Then, $z$ and $\bar{z}$ are roots of an irreducible factor of
degree $2$ of the minimal Ricci polynomial.
Let $e$ and $\overline{e}$ in $V$ such that

$\langle
e,e\rangle=-\langle\overline{e},\overline{e}\rangle=1,\;\langle{e},\overline{e}\rangle=0,\;\mathrm{Ric}(e)=ae-b\overline{e}\quad\mbox{and}\quad\mathrm{Ric}(\overline{e})=be+a\overline{e}.$

Then, the subspace $E=\mathrm{span}\\{e,\overline{e}\\}$ and its orthogonal
subspace $E^{\perp}$ are stable by ${\mathrm{Ric}}$ and by endomorphisms
${\mathrm{K}}(u,v)$ for all $u$ and $v$ in $V$. Let $\hbox{\bb
B}=(e_{1},\ldots,e_{n-2})$ be a orthonormally basis of $E^{\perp}$.

Then, we have:

$\displaystyle b$ $\displaystyle=$
$\displaystyle\langle\mathrm{Ric}(\overline{e}),e\rangle$ $\displaystyle=$
$\displaystyle\langle{\mathrm{K}}(\overline{e},e)e,e\rangle-\langle{\mathrm{K}}(\overline{e},\overline{e})e,\overline{e}\rangle+\sum_{i=1}^{n-2}\langle{\mathrm{K}}(\overline{e},e_{i})e,e_{i}\rangle$
$\displaystyle=$ $\displaystyle 0,$

which contradicts the fact that $b\not=0$. ∎

######  Remark 3.

In [2], the author had demonstrated this proposition with different
techniques.

######  Lemma 2.4.

Let ${\mathrm{K}}$ be a semi-symmetric curvature tensor on the Lorentzian
vector space $(V,\langle\;,\;\rangle)$. Then the prmitive decomposition of $V$
is given as;

$V=\Big{(}V_{0}+V_{1}+V_{0}^{\prime}\Big{)}\oplus V_{2}\oplus\ldots\oplus
V_{s},$

where $V_{0}^{\prime}$ is the dual subspace of $V_{0}\cap V_{1}$.

Therefore, $V_{i}$ is a Riemannian subspace of dimension greater than or equal
to $2$ for any $i\geq 2$.
And so,

  1. 1.

If $E_{0}$ is a riemannian space, then $V_{0}=E_{0}$ is also riemannian space
and for all $i\geq 1$, the $V_{i}$ is Einstein space.

  2. 2.

If ${\mathrm{K}}$ has a tensor Ricci of isotropic type, i;e,
$E_{0}=\ker({\mathrm{Ric}}^{2})$, then $E_{0}=V_{0}+V_{1}+V_{0}^{\prime}$ such
that $V_{0}^{\prime}$ is of dimension $1$ and for all $i\geq 2$, the $V_{i}$
is Einstein Riemannain space.

##  3 Semi-symmetric pseudo-Riemannian differential manifolds

Let $(M,g)$ be a pseudo-Riemannian differential manifold of dimension $n$. Let
$\nabla$, $R$, $\mathfrak{ric}$ and ${\mathrm{Ric}}$ are respectively, the
Levi-Civita connection, the Riemann curvature, the tensor and the Ricci
operator associated to the metric $g$. We set ${\cal X}(M)$ the set of vector
fields on $M$.

So $(M,g)$ is semisymmetric manifold. Then, at each point $m\in M$, the
restriction $R_{m}$ of $R$ on $T_{m}M$ is a semi-symmetric curvature tensor
and the minimal polynomial of ${\mathrm{Ric}}_{m}$ is of the form ${\cal
X}=\prod_{i=0}^{r}P_{i}$ where the polynomials $(P_{i})_{i}$ are mutually
prime and for any $i$, $P_{i}$ is irreducible or $P_{i}=X^{2}$. We can assume
that $P_{i}$ is irreducible for all $i\geq 1$. We define the distributions:

$E_{0}(m):=\ker({\mathrm{Ric}}_{m}^{2})\leavevmode\nobreak\
\leavevmode\nobreak\ \text{ et}\leavevmode\nobreak\ \leavevmode\nobreak\
E_{i}(m):=\ker(P_{i}({\mathrm{Ric}}_{m}))\leavevmode\nobreak\
\leavevmode\nobreak\ \text{ for }\leavevmode\nobreak\ \leavevmode\nobreak\
i\geq 1,$

and we have the following proposition:

######  Proposition 3.1.

The distributions $(E_{i})_{i}$ checked the following properties:

For all, $i$, $j\geq 1$ whit $i\neq j$, we have:

$\nabla_{E_{j}}E_{i}\subset E_{i},\;\nabla_{E_{i}}E_{i}\subset
E_{0}+E_{i},\;\nabla_{E_{0}}E_{i}\subset E_{i},\;\nabla_{E_{0}}E_{0}\subset
E_{0},\;\nabla_{E_{i}}E_{0}\subset E_{0}+E_{i}.$ (12)

###### Proof.

Let $m\in M$. The theorem.2.2 induces that:

$T_{m}M=E_{0}(m)\oplus E_{1}(m)\oplus...\oplus E_{r}(m).$ (13)

In the first step, we will show that for $i\geq 1$ and $X\in E_{i}^{\perp}$,
$\nabla_{X}E_{i}\subset E_{i}$;
Let $i\geq 1$ and $X\in E_{i}^{\perp}$. First, we show
$\nabla_{X}(R(E_{i},E_{i})E_{i})\subset E_{i}$:

Let us take $Y,\;Z,\;T\in E_{i}$, by the second Bianchi identity, we get

$\displaystyle\nabla_{X}R(Y,Z,T)$ $\displaystyle:=$
$\displaystyle(\nabla_{X}R)(Y,Z)T$ $\displaystyle=$
$\displaystyle-\nabla_{Y}R(Z,X,T)-\nabla_{Z}R(X,Y,T)$ $\displaystyle=$
$\displaystyle-\nabla_{Y}(R(Z,X)T)+R(\nabla_{Y}Z,X)T+R(Z,\nabla_{Y}X)T+R(Z,X)\nabla_{Y}T$
$\displaystyle-\nabla_{Z}(R(Y,X)T)+R(\nabla_{Z}Y,X)T+R(Y,\nabla_{Z}X)T+R(Y,X)\nabla_{Z}T$
$\displaystyle=$ $\displaystyle
R(\nabla_{Y}Z,X)T+R(Z,\nabla_{Y}X)T+R(\nabla_{Z}Y,X)T+R(Y,\nabla_{Z}X)T.$

By theorem.2.2, we get; $R(V,V)(E_{i})\subset E_{i}$ and
$\nabla_{X}R(Y,Z,T)\in E_{i}$.

On the other hand,

$\displaystyle\nabla_{X}R(Y,Z,T)$ $\displaystyle=$
$\displaystyle\nabla_{X}(R(Y,Z)T)-R(\nabla_{X}Y,Z)T-R(Y,\nabla_{X}Z)T-R(Y,Z)\nabla_{X}T$
$\displaystyle=$
$\displaystyle\nabla_{X}(R(Y,Z)T)-R(\nabla_{X}Y,Z)T-R(Y,\nabla_{X}Z)T$
$\displaystyle+R(Z,\nabla_{X}T)Y+R(\nabla_{X}T,Y)Z.$

which proves that, $\nabla_{X}(R(Y,Z)T)\in E_{i}$.

Now we will show that $\nabla_{X}{\mathrm{Ric}}(Y)\in E_{i}$.
We choose an orthogonal basis $(e_{1},...,e_{n})$ associeted to the
decomposition(13) where $\epsilon_{k}=\langle e_{k},e_{k}\rangle$ such that
$\epsilon_{k}^{2}=1$.
Let $U\in E_{i}^{\perp}$. If $e_{k}\in E_{i}$, we have already seen that
$\nabla_{X}(K(Y,e_{k})e_{k})\in E_{i}$ and if $e_{k}\in E_{i}^{\perp}$, we get
$R(Y,e_{k})=0$. Consequently, we have

$\displaystyle\langle\nabla_{X}({\mathrm{Ric}}(Y)),U\rangle$ $\displaystyle=$
$\displaystyle-\langle{\mathrm{Ric}}(Y),\nabla_{X}U\rangle$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\epsilon_{k}\langle
R(Y,e_{k})e_{k},\nabla_{X}U\rangle$ $\displaystyle=$
$\displaystyle-\sum_{k=1}^{n}\epsilon_{k}\langle\nabla_{X}(R(Y,e_{k})e_{k}),U\rangle$
$\displaystyle=$ $\displaystyle 0.$

then $\nabla_{X}\big{(}{\mathrm{Ric}}(Y)\big{)}\in E_{i}$.

If $P_{i}(t)=t^{2}+at+b$ whit $b\neq 0$, then for all $Y\in E_{i}$, we have
$Y=-\frac{1}{b}({\mathrm{Ric}}^{2}(Y)+a{\mathrm{Ric}}(Y))$.
Consequently, $\nabla_{X}Y\in E_{i}.$

If $P_{i}(t)=t-\lambda_{i}$ whit $\lambda_{i}\neq 0$, then for all $Y\in
E_{i}$, we have $Y=\frac{1}{\lambda_{i}}{\mathrm{Ric}}(Y)\leavevmode\nobreak\
\leavevmode\nobreak\ .$
Consequently $\nabla_{X}Y\in E_{i}.$

So, $\nabla_{X}E_{i}\subset E_{i}$, which shows that
$\nabla_{E_{j}}E_{i}\subset E_{i}$ and $\nabla_{E_{0}}E_{i}\subset E_{i}$, for
all $i,\;j\geq 1$ whit $i\neq j$.

The other results are obtained immediately because the metric $g$ is
parallel(i,e. $\nabla g=0$).

∎

######  Corollary 3.1.

Let $(M,g)$ be a connected semi-symmetric pseudo-Riemannian manifold. Let
${\cal X}=\prod_{i}P_{i}$ be the minimal polynomial of ${\mathrm{Ric}}$. If we
set $E_{0}:=\ker({\mathrm{Ric}}^{2})$ and for all $i\geq 1$,
$E_{i}:=\ker(P_{i}({\mathrm{Ric}}))$. Then, for all $i\geq 1$, the
distributions $E_{0}$ et $E_{0}+E_{i}$ are involutive.
We set ${\cal N}_{0}$ the integral submanifolds of $E_{0}$.

######  Remark 4.

The distributions $E_{0}$ and $E_{0}+E_{i}$ are involutive but not necessarly
parallele.

On the other way, for a semisymmetrical pseudo-Riemannian manifolds $(M,g)$,
the action of the primitive holonomy Lie algebra ${\mathfrak{h}}(R_{m})$ gives
the primitive decomposition of tangent space $T_{m}M$ on point $m\in M$ as

$T_{m}M=V_{0}(m)+V_{1}(m)+\ldots+V_{s}(m)+V_{0}^{\prime}(m)$ (14)

where $V_{0}(m)=\Big{\\{}x\in T_{m}M\;/\;R(u,v)x=0,\;\text{ for all}\;(u,v)\in
T_{m}^{2}M\Big{\\}}$ and the space $V_{0}^{\prime}(m)$ is the dual subspace of
$V_{0}(m)\cap\\{V_{1}(m)+\ldots+V_{s}(m)\\}$.

With a similar proof of the proposition3.1, we check that the distributions
$V_{i}$ have the following properties;

######  Proposition 3.2.

For all, $i$, $j\geq 1$ whit $i\neq j$;

$\nabla_{V_{j}}V_{i}\subset V_{i},\;\nabla_{V_{i}}V_{i}\subset
V_{0}+V_{i},\;\nabla_{V_{0}}V_{i}\subset V_{i},\;\nabla_{V_{0}}V_{0}\subset
V_{0},\;\nabla_{V_{i}}V_{0}\subset V_{0}+V_{i}.$ (15)

Now we go back to the Ricci decomposition and for all $1\leq i\leq r$, we
consider the distributions $F_{i}$ spanned by the vector fields of the forms

$X_{1},\nabla_{X_{1}}X_{2},\nabla_{X_{1}}\nabla_{X_{2}}X_{3},...,\nabla_{X_{1}}...\nabla_{X_{l}}X_{l+1},...\text{ect}.,$
(16)

where the vector fields $(X_{i})_{i\geq 1}$ are belong to $E_{i}$.

In the same way, let us the subspaces; $F$, $\tilde{F}_{i}$, and ${F}_{0}$
checking:

$\displaystyle F$ $\displaystyle=$ $\displaystyle F_{1}+F_{2}+...+F_{r},$
$\displaystyle TM$ $\displaystyle=$ $\displaystyle F\oplus(F\cap
F^{\bot})^{\prime}\oplus{F}_{0},$ $\displaystyle\tilde{F}_{i}$
$\displaystyle=$ $\displaystyle=F_{i}\oplus\left(F_{i}\cap
F_{i}^{\bot}\right)^{\prime},$

where $\left(F\cap F^{\bot}\right)^{\prime}$ and $\left(F_{i}\cap
F_{i}^{\bot}\right)^{\prime}$ are respectively, the dual subspaces of $F\cap
F^{\bot}$ and $F_{i}\cap F_{i}^{\bot}$.

######  Remark 5.

it’s obvious that:

$\forall i\geq 1,\;\>F_{0}\cup\left(F\cap F^{\bot}\right)\cup\left(F\cap
F^{\bot}\right)^{\prime}\subseteq E_{0}\;\quad\mbox{and}\quad\;E_{i}\subseteq
F_{i}\subseteq E_{0}+E_{i}.$

For $i$, $j\geq 1$, we put
$X_{1},...,X_{k},(\text{resp.}Y_{1},...,Y_{l}...,\text{ect},)$ the vector
fields belong to $E_{i}$ (resp. $E_{j}$).
In the first step, we show the following lemma:

######  Lemma 3.1.

For all $\>0\leq i\neq j\neq 0$, the vector fields of the forms

$\nabla_{X_{1}}\nabla_{Y_{1}}\nabla_{Y_{2}}...\nabla_{Y_{l}}Y_{l+1}$

are tangent to $F_{j}$, i.e, we have

$\nabla_{E_{i}}F_{j}\subseteq F_{j}.$

As a consequence, for all $j\geq 1$, the distribution $F_{j}$ is parallel and
the integral submanifolds are totally geodesic.

###### Proof.

We can prove this lemma by induction:
Let $0\leq i\neq j\geq 1$. So $R(X_{1},Y_{1})Y_{2}=0$, then

$\nabla_{X_{1}}\nabla_{Y_{1}}{Y_{2}}=\nabla_{Y_{1}}\nabla_{X_{1}}{Y_{2}}+\nabla_{[X_{1},Y_{1}]}{Y_{2}}=\nabla_{Y_{1}}{Y_{2}}^{*}+\nabla_{Y_{1}^{*}}{Y_{2}}-\nabla_{X_{1}^{*}}{Y_{2}},$

where the vector fields ${Y_{2}}^{*}:=\nabla_{X_{1}}{Y_{2}}$ and
${Y_{1}}^{*}:=\nabla_{X_{1}}{Y_{1}}$ are tangent to $E_{j}$ and the vector
fields ${X_{1}}^{*}:=\nabla_{Y_{1}}{X_{1}}$ is tangent to $E_{i}$ for $i\geq
1$ and for $i=0$, we get ${X_{1}}^{*}\in E_{0}+E_{j}$ and
$\nabla_{X_{1}^{*}}{Y_{2}}$ belong to $F_{j}$. This assumes that the three
terms above are tangent to $F_{j}$. So $\nabla_{X_{1}}\nabla_{Y_{1}}{Y_{2}}\in
F_{j}$ is checked.
Now, in the genral case:

So $R(X_{1},Y_{1})=0$, we obtain;

$\displaystyle\nabla_{X_{1}}\nabla_{Y_{1}}...\nabla_{Y_{k}}Y_{k+1}$
$\displaystyle=$
$\displaystyle\nabla_{Y_{1}}\nabla_{X_{1}}\nabla_{Y_{2}}....\nabla_{Y_{k}}Y_{k+1}+\nabla_{[X_{1},Y_{1}]}\nabla_{Y_{2}}...\nabla_{Y_{k}}Y_{k+1}$
$\displaystyle=$
$\displaystyle\nabla_{Y_{1}}\nabla_{X_{1}}\nabla_{Y_{2}}....\nabla_{Y_{k}}Y_{k+1}+\nabla_{{Y_{1}}^{*}}\nabla_{Y_{2}}...\nabla_{Y_{k}}Y_{k+1}-\nabla_{{X_{1}}^{*}}\nabla_{Y_{2}}...\nabla_{Y_{k}}Y_{k+1}.$

As the vector field ${Y_{1}}^{*}:=\nabla_{X_{1}}{Y_{1}}$ is tangent to
$E_{j}$, and the vector field ${X_{1}}^{*}:=\nabla_{Y_{1}}{X_{1}}$ is tangent
to $E_{i}$, by the induction hypothesis, we get that
$\nabla_{X_{1}}\nabla_{Y_{1}}...\nabla_{Y_{k}}Y_{k+1}$ is tangent to $F_{j}$.
∎

######  Lemma 3.2.

The subspaces $F_{1}$, $F_{2}$,…and $F_{r}$ are pairwise orthogonal.

###### Proof.

By induction, we show that the vector field
$\nabla_{Y_{1}}\nabla_{Y_{2}}...\nabla_{Y_{k}}Y_{k+1},\;\;k\geq 0$ is
orthogonal to $E_{i}$.
For $k=0$ and $k=1$, it’s obvious.

Now, suppos that result is true for any vector fields of the form
$\nabla_{Y_{2}}...\nabla_{Y_{k}}Y_{k+1}$. Let $X$ be a vector fields in
$E_{i}$. Then, the vector fields $\nabla_{Y_{1}}{X}$ is also tangent to
$E_{i}$. By induction, we get

$g(X,\nabla_{Y_{1}}\nabla_{Y_{2}}...\nabla_{Y_{k}}Y_{k+1})=-g(\nabla_{Y_{1}}{X},\nabla_{Y_{2}}...\nabla_{Y_{k}}Y_{k+1})=0.$

This proves the statement.

Using induction again, we can prove that the vector fields on the form
$\nabla_{X_{1}}\nabla_{X_{2}}...\nabla_{X_{k}}X_{k+1}$ are orthogonal to those
which are of the form $\nabla_{Y_{1}}\nabla_{Y_{2}}...\nabla_{Y_{l}}Y_{l+1}$.

For the case $k=0$, the proof is given above. Now, for the vectors fields on
the form $\nabla_{X_{1}}\nabla_{X_{2}}...\nabla_{X_{k}}X_{k+1}$ are orthogonal
to the vector fields on the form
$\nabla_{Y_{1}}\nabla_{Y_{2}}...\nabla_{Y_{l}}Y_{l+1}$, so the induction
hypothesis and lemma(3.2), we get

$\displaystyle
g(\nabla_{X_{1}}\nabla_{X_{2}}...\nabla_{X_{k}}X_{k+1},\nabla_{Y_{1}}\nabla_{Y_{2}}...\nabla_{Y_{l}}Y_{l+1})$
$\displaystyle=$
$\displaystyle-g(\nabla_{X_{2}}...\nabla_{X_{k}}X_{k+1},\nabla_{X_{1}}\nabla_{Y_{1}}\nabla_{Y_{2}}...\nabla_{Y_{l}}Y_{l+1})$
$\displaystyle=$ $\displaystyle 0,$

whil the vector fields on the form
$\nabla_{X_{1}}\nabla_{Y_{1}}\nabla_{Y_{2}}...\nabla_{Y_{l}}Y_{l+1}$ are
tangent to $F_{j}$. This gives completely the proof of lemma. ∎

######  Corollary 3.2.

$(F\cap F^{\bot})^{\prime}\oplus{F}_{0}$ and
$\tilde{F}_{j}=F_{j}\oplus\left(F_{j}^{\bot}\cap F_{j}\right)^{\prime}$ are
involutive distributions, for all $j\geq 1$,

###### Proof.

We have already seen the relations $\nabla_{E_{i}}{F_{j}}\subseteq F_{j}$ only
for the cases $i\geq 0$ and $j\geq 1$. It remains to show the relationships;
$\nabla_{E_{j}}{F_{0}}\subseteq F_{0}$ for $j>0$. The first is obvious, since

$g(\nabla_{E_{i}}{F_{0}},F_{k})=-g(F_{0},\nabla_{E_{j}}F_{k})=-g(F_{0},F_{k})=0,\text{for
all}\;k>0.$

Finally, for the formula $\nabla_{E_{0}}{F_{0}}$, we get

$g(\nabla_{E_{0}}{F_{0}},F_{i})=-g(F_{0},\nabla_{E_{0}}F_{i})=g(F_{0},F_{i})=0.$

∎

######  Remark 6.

The distribution $F_{j}\oplus\left(F_{j}\cap(F_{j})^{\bot}\right)^{\prime}$ is
involutive non degenerated.

######  Proposition 3.3.

Let $(M,g)$ be a semi-symmetric, locally connected pseudo-Riemannian manifold.
Then ,$M$ is a foliated manifold and the minimal polynomial of the restriction
of the Ricci operator on each leaf has one of the following forms $X$,
$X^{2}$, $P$, $XP$ or $X^{2}P$ where $P$ is an irreducible polynomial and it
is prime with the polynomial $X$.

In the case where the tangent bundle of $M$ admits only one involutive
subbundle nondegenerate, $M$ is called a simple leaf.

######  Lemma 3.3.

Let $(M,g)$ be a simple leaf semisymmetric, locally connected, pseudo-
Riemannian manifold. Then its a tangent bundle has one of the following forms:

  1. 1.

$TM={F_{1}}\oplus\left(F_{1}\cap\left(F_{1}\right)^{\bot}\right)^{\prime}=E_{0}\overset{\bot}{\oplus}E_{1}$.

  2. 2.

$TM=E_{0}$.

##  4 The simple leaf semi-symmetrical Lorentzian manifold

In this section we will give some properties of semi-symmetric Lorentzian
spaces

###  4.1 Ricci decomposition

In the Lorentzian case, the comparison of primitive and Ricci decompositions
produces the following Proposition;

######  Proposition 4.1.

Let $(M,g)$ be a simple leaf semisymmetric, locally connected, Lorentzian
manifolds. Then, we get one of the following situations:

  1. 1.

$E_{0}=V_{0}$, i.e, ${\cal N}_{0}$ is a plat submanifolds.

  2. 2.

The tangent bundle over $M$ is on the forme $E_{0}=V_{0}+V_{1}+(V_{0}\cap
V_{1})^{\prime},$ where $(V_{0}\cap V_{1})^{\prime}$ is the dual subspace of
$(V_{0}\cap V_{1})$. In this situation, we get one of the two following cases:

    1. (a)

$E_{0}=V_{0}\oplus V_{1}$,

    2. (b)

There exists an istropic vector fields $p$ with its dual vector fields $q$
such that $E_{0}=V_{0}+V_{1}+span\\{q\\}$. In this case, the tensor curvature
holds

$R^{2}=0.$

###### Proof.

The primitive and Ricci decompositions imply that

$V_{0}=E_{0}\;\text{ or}\;V_{0}\varsubsetneq E_{0}.$

If $V_{0}\varsubsetneq E_{0}$, this means that the curvature tensor is non-
zero on ${\cal N}_{0}$ and the action of the primitive holonomy algebra
${\mathfrak{h}}_{0}(R)$ on $E_{0}$ induces a non-trivial decomposition:

$E_{0}=V_{0}+V_{1}+(V_{0}\cap V_{1})^{\prime},,$

where $V_{0}:\\{x\in E_{0}/\;\forall
h\in{\mathfrak{h}}_{0}(R),\;h(x)=0\\},V_{1}:={\mathfrak{h}}_{0}(R)(E_{0})\neq\\{0\\}$
and $(V_{0}\cap V_{1})^{\prime}$ is the dual subspace of $V_{0}\cap V_{1}$.

By proposition3.2, we show that the distribution $V_{0}+V_{1}$ is involutive
and since $M$ is a simple leaf, we get that $TM=E_{0}$.

So, if $V_{0}\cap V_{1}\neq\\{0\\}$ is a non-trivial subspace, then it is
generated by an istropic parallel vector fields $p$ and according to the
classification of weakly irreducible holonomy algebras of Lorentzian manifolds
given by L. B. Bergery and A. Ikemakhen in [4], for any point $m\in M$,
$\;{\mathfrak{h}}_{0}(R_{m})$ is a Lie subalgebra of type $2$ or $4$.
Consequently, each element ${\mathrm{K}}\in{\mathfrak{h}}_{0}(R_{m})$ is
written in the form

${\mathrm{K}}={\mathrm{K}}_{0}-p_{m}\wedge X,$

where $\;X\in{\widehat{E}}_{0}(m)\cong{E_{0}(m)}_{/\\{p,q\\}}$ and
${\mathrm{K}}_{0}\in\mathfrak{so}({\widehat{E}_{0}(m)})\cong\wedge^{2}\hbox{\bb
R}^{r}$, such that $r=\dim(E_{0})-2$ and $q$ is an isotropic vector fields
which $g(p,q)=1$, .

Then the restriction of the curvature on the space $E_{0}(m)$ checks

$\forall(X,Y)\in E_{0}^{2}(m),\;\exists\;E\in{\widehat{E}}_{0}(m)\;\text{ such
that }\;R_{m}(X,Y)=\widehat{R}(\widehat{X},\widehat{Y})-p_{m}\wedge E,$

where $\widehat{X}$ and $\widehat{Y}$ are the projection respectively of the
$X$ and $Y$ on the subspace ${\widehat{E}}_{0}(m)$ and $\widehat{R}$ is the
restrection of $R_{m}$ on ${\widehat{E}}_{0}(m)$ and so $\widehat{R}$ is a
semisymmetric curvature tensor on the irreducible Riemannian space
${\widehat{E}}_{0}(m)$ where ${\mathrm{Ric}}_{m}=0$. Then

$\widehat{R}=0\quad\mbox{and}\quad R_{m}(X,Y)=-p_{m}\wedge E.$

thus

$R^{2}=0.$

∎

######  Proposition 4.2.

Let $(M,g)$ be a connected simple leaf semi-symmetric Lorentzian manifold.
Then the Ricci operator admits at most an one non-zero real eigenvalue. If
$\lambda$ is such an eigenvalue, then the tangent space of $M$ at any point
$m\in M$ will be of the form:

$T_{m}M=\ker\bigg{(}{\mathrm{Ric}}_{m}-\lambda(m)Id_{T_{m}M}\bigg{)}\oplus
V_{0}(m),$ (17)

Moreover, if $\dim(E_{1})\geq 3$, then the function $m\in M\mapsto\lambda(m)$
is of class $C^{\infty}$ and it depends only on ${\cal N}_{0}$ the flat
integral submanifolds of $V_{0}=E_{0}$.

###### Proof.

Let $(M,g)$ be a connected, semi-symmetric, simple leaf Lorentzian manifold.
Let $\lambda$ be a non-zero eigenvalue of ${\mathrm{Ric}}$.

The equation17 is a resultat of Lemma3.2 and Proposition4.1.

Now, we consider the codifferential $\delta$ on $M$ given by:

$\delta(\alpha)(Y_{1},...,Y_{r})=-\Sigma_{i}(\nabla_{X_{i}}\alpha)(X_{i},Y_{1},...,Y_{r}),$

where $\alpha$ is a $(r+1)-$differential form, $(X_{1},...,X_{n})$ is an
orthonormal fram on $M$ and $(Y_{1},...,Y_{r})$ a family of $r$ vector fields
(See [5], page 34).

The Ricci tensor verifies:

$\delta(\mathfrak{ric})=-\frac{1}{2}d(\mathfrak{s}),$ (18)

where $\mathfrak{s}$ is the scalar curvature, $d$ is the exterior differential
over $M$ (See proposition 1.94 [5] page 43).

Let $(X_{1},...,X_{n})$ be a orthonormal fram of $TM$ whit the $X_{i}$ are
tangent to $E_{0}$ for $i\leq\nu=\dim(E_{0})$ and for $i>\nu$, the $X_{i}$ are
tangent to $E_{1}$ .

Let us $j>\nu$. we get:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{1}{2}d(\mathfrak{s})(X_{j})+\delta(\mathfrak{ric})(X_{j})$
$\displaystyle=$
$\displaystyle\frac{1}{2}X_{j}(\mathfrak{s})-\Sigma_{i}(\nabla_{X_{i}}\mathfrak{ric})(X_{i},X_{j})$
$\displaystyle=$
$\displaystyle\frac{1}{2}X_{j}((n-\nu)\lambda)-\Sigma_{i}{X_{i}}.(\mathfrak{ric}(X_{i},X_{j}))-\mathfrak{ric}(\nabla_{X_{i}}X_{i},X_{j})-\mathfrak{ric}(X_{i},\nabla_{X_{i}}X_{j}),$
$\displaystyle=$
$\displaystyle\frac{n-\nu}{2}X_{j}(\lambda)-\Sigma_{i}({X_{i}}.\lambda)g(X_{i},X_{j})+\lambda{X_{i}}.g(X_{i},X_{j})-\mathfrak{ric}(\nabla_{X_{i}}X_{i},X_{j})-\mathfrak{ric}(X_{i},\nabla_{X_{i}}X_{j}),$
$\displaystyle=$ $\displaystyle(\frac{n-\nu}{2}-1)X_{j}(\lambda).$

So $n-\nu\geq 3$ implies that $X_{j}(\lambda)=0$ and therefore, $\lambda$ only
depends on the integral submanifolds of $E_{0}$.

∎

###  4.2 Basic formulas

In this subsection, we consider $(M,g)$ a simple leaf semi-symmetric
Lorentzian manifold of dimension $n$ such that the Ricci admits a non-zero
eigenvalue $\lambda$ of multiplicity $n-r$ such that $E_{0}$ is a Lorantzian
subspace.

So at any point $m\in M$, the tangent space splits as:

$T_{m}M=E_{0}(m)\oplus E_{1}(m),$

where

$E_{0}(m)=\ker({\mathrm{Ric}}_{m})\leavevmode\nobreak\ \text{ and
}\leavevmode\nobreak\
E_{1}(m)=\ker({\mathrm{Ric}}_{m}-\lambda(m)id_{T_{m}M}).$

and ${\cal N}_{0}$ the integral submanifold of $E_{0}$ is flat.

The real number $r=\dim(E_{0}(m))$ called the nullity index of the curvature
at the point $m$. The multiplicity of eingenvalue $\lambda$ also called the
co-nullity index.

So ${\cal N}_{0}$ the integral submanifold of $E_{0}$ is flat, then we can
shoose a fram $(e_{1},...,e_{r})$ of $E_{0}$ such that, for all $i\geq 2$,
$\varepsilon_{i}=g(e_{i},e_{i})=-g(e_{1},e_{1})=1,\quad\mbox{and}\quad
g(e_{i},e_{j})=0$, for all $\leavevmode\nobreak\ \leavevmode\nobreak\ 1\leq
i\neq j\leq r.$

Let us $(u_{1},...,u_{r})$ a local coordinate system associated to
$(e_{1},...,e_{r})$, i.e. $e_{i}=\frac{\partial}{\partial u_{i}}$ for all
$1\leq i\leq r$.

On the other hand, $R(E_{0},E_{0})E_{1}=0$, then we can also choose
$(X_{1},...,X_{n-r})$ an orthonormal fram of $E_{1}$ such that, for all $1\leq
j\leq n-r$, $X_{j}$ is parralel on ${\cal N}_{0}$, i.e.
$\nabla_{e_{i}}X_{j}=0$.

For all vector fields $e_{i}$ and for all vector fields $X$, we set

$\nabla_{X}e_{i}=A_{i}(X)+\sum_{j=1}^{r}B_{i}^{j}(X)e_{j},$ (19)

where $A_{i}(X)$ is the orthogonal projection of $\nabla_{X}e_{i}$ on $E_{1}$.

$A_{i}$ is $(1,1)-$tensor on $M$ which is zero on $E_{0}$ and $B_{i}^{j}$ are
covariant tensors on $M$ wich they have the value zero on $E_{0}$.

Moreover, we get

$B_{i}^{j}(X)=\varepsilon_{j}g(\nabla_{X}e_{i},e_{j})=-\varepsilon_{j}g(e_{i},\nabla_{X}e_{j})=-\varepsilon_{j}\varepsilon_{i}B^{i}_{j}(X).$
(20)

######  Definition 4.1.

The field tensors $A_{i}$ and $B_{i}^{j}$ are called the second fundamental
forms corresponding to the system $\\{e_{1},...,e_{r}\\}$.

We define the $(0,2)$-tensor $B^{i}$ by:

$\begin{array}[]{ccccc}B^{i}(X,Y)&:=&-g(A_{i}(X),Y),&&\\\
B^{i}(X,e_{j})&:=&B^{i}(e_{j},X)&=&B^{i}(e_{i},e_{l})=0,\end{array}$ (21)

where $X$ and $Y$ are vectors fields tangent to $E_{1}$.

######  Lemma 4.1.

The second fundamental forms $A_{i}$ and $B_{j}^{i}$ and the curvature $R$
satisfy the following properties:

$2trace(A_{i})=-\frac{(n-r)}{\lambda}\frac{\partial\lambda}{\partial
u_{i}},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ 1\leq i\leq r,$ (22) $\nabla_{e_{i}}A_{j}(X)=-A_{j}\circ
A_{i}(X),$ (23) $(\nabla_{e_{i}}B_{j}^{k})(X)=-B_{j}^{k}(A_{i}(X)),$ (24)

$(\nabla_{e_{i}}R)(X,Y)=R(Y,A_{i}(X))+R(A_{i}(Y),X),$ (25)

$R(X,Y)A_{i}(Z)+R(Y,Z)A_{i}(X)+R(Z,X)A_{i}(Y)=0,$ (26)

for $X$, $Y$, $Z\in E_{1}$.

###### Proof.

The formula (22) is result of the formula(18). Indeed;

$\delta(\mathfrak{ric})(e_{i})=-\frac{1}{2}d(\mathfrak{s})(e_{i})$

Let us $X$, $Y$, $Z\in E_{1}$.
Formulas (23) and (24) follow from the equation $R(e_{i},X)e_{j}=0$.

The formulas (25) and (26) follow frome the second Bianchi identity, indeed:

$\displaystyle(\nabla_{e_{i}}R)(X,Y)$ $\displaystyle=$
$\displaystyle(\nabla_{X}R)(e_{i},Y)+(\nabla_{Y}R)(X,e_{i})$ $\displaystyle=$
$\displaystyle[\nabla_{X},R(e_{i},Y)]-R(\nabla_{X}e_{i},Y)-R(e_{i},\nabla_{X}Y)$
$\displaystyle+[\nabla_{Y},R(X,e_{i})]-R(\nabla_{Y}X,e_{i})-R(X,\nabla_{Y}e_{i})$
$\displaystyle=$ $\displaystyle-R(\nabla_{X}e_{i},Y)-R(X,\nabla_{Y}e_{i})$
$\displaystyle=$ $\displaystyle R(Y,A_{i}(X))+R(A_{i}(Y)+X)$

So $R(X,Y)(e_{i})=R(Z,X)(e_{i})=R(Y,Z)(e_{i})=0$,

we get

$\displaystyle 0$ $\displaystyle=$
$\displaystyle(\nabla_{X}R)(Y,Z)(e_{i})+(\nabla_{Y}R)(Z,X)(e_{i})+(\nabla_{Z}R)(X,Y)(e_{i})$
$\displaystyle=$
$\displaystyle-R(Y,Z)(\nabla_{X}e_{i})-R(Z,X)(\nabla_{Y}e_{i})-R(X,Y)(\nabla_{Z}e_{i})$

∎

######  Corollary 4.1.

$(\nabla_{e_{i}}R)$ is a curvature tensor on $M$ and
$\nabla_{e_{i}}R(X,Y)\in{\mathfrak{h}}_{1}(R)$ for all $X$, $Y\in E_{1}$ .

######  Proposition 4.3.

There exists a function $\mu_{i}$ of class $C^{\infty}$ on $M$ such that

$\nabla_{e_{i}}R(X,Y)=-2\mu_{i}R(X,Y),$ (27)

where $X$, $Y\in E_{1}$.

###### Proof.

Let $m$ be a point in $M$. So the primitive holonomy group ${\cal
H}_{1}(R_{m})$ acts irreducibly on $E_{1}(m)$. Therefore,
$[E_{1}(m),R_{m},{\cal H}_{1}(R_{m})]$ and
$[E_{1}(m),(\nabla_{e_{i}}R)_{m},{\cal H}_{1}(R_{m})]$ are two Riemannian
irreducible symmetric holonomy systems. According to the corollary of
Theorem.6 in ([13]), there exists a real $\mu_{i}(m)$ satisfying

$(\nabla_{e_{i}}R)_{m}=-2\mu_{i}(m)R_{m}.$

So the tensors $\nabla_{e_{i}}R$ and $R$ are of class $C^{\infty}$, then the
function $\mu_{i}$ is of class $C^{\infty}$. ∎

######  Lemma 4.2.

Any second fundamental form $A_{i}$ on $E_{1}$ has one of the following forms
$A_{i}=\mu_{i}I$ or else $n\in 2\hbox{\bb N}$ and
$A_{i}=\mu_{i}I+\lambda_{i}J$ with $J^{2}=-I$, where $\mu_{i}$ and
$\lambda_{i}$ are functions of class $C^{\infty}$. The skewsymmetric
endomorphism $J$ is a uniquely determined and is independent of $i$ and the
choice of the system $(e_{1},...,e_{r})$ and commutes with each elements of
${\cal H}_{1}(R_{m})$ at any point $m\in M$.

###### Proof.

The proof is similar as that of Lemma 4.4 in ([15]) considering the Lie
algebra ${\cal H}_{1}(R_{m})$ and the Riemannian irreducible symmetric
holonomy system $S_{m}:=[E_{1},R_{m},{\cal H}_{1}(R_{m})]$. ∎

From the formula22, we get

$\mu_{i}=-\frac{1}{2\lambda}\frac{\partial\lambda}{\partial
u_{i}},\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ 1\leq i\leq r.$ (28)

So, if $A_{i}=\mu_{i}I+\lambda_{i}J$, since $\nabla_{e_{i}}A_{i}=-A_{i}^{2}$,
we get

$\nabla_{e_{i}}J=0.$ (29) $\nabla_{e_{i}}\mu_{i}=\lambda_{i}^{2}-\mu_{i}^{2}.$
(30) $-2\lambda_{i}\mu_{i}=\frac{\partial\lambda_{i}}{\partial u_{i}}.$ (31)

Then

$\lambda\frac{\partial\lambda_{i}}{\partial
u_{i}}=\lambda_{i}\frac{\partial\lambda}{\partial u_{i}}$ (32)

######  Lemma 4.3.

If the co-nullity index satisfies $n-r\geq 3$, we get that

$r\in\\{1,2\\}.$

###### Proof.

By lemma3.3, we get

$TM={F_{1}}\oplus\left(F_{1}\cap\left(F_{1}\right)^{\bot}\right)^{\prime}=E_{0}\oplus
E_{1},$

and

$F_{1}=span\\{pr_{0}(\nabla_{X}Y)/\leavevmode\nobreak\ X,\leavevmode\nobreak\
Y\in E_{1}\\}$

where $pr_{0}\leavevmode\nobreak\ :TM\rightarrow{{E}}_{0}$ is the orthogonal
projection in ${{E}}_{0}$. We will show that

$r=\dim(span\\{pr_{0}(\nabla_{X}Y)/\leavevmode\nobreak\ X,\leavevmode\nobreak\
Y\in E_{1}\\}).$

Necessarily, we have $r\geq 1$. Let $(e_{1},...,e_{r})$ be an orthonormal fram
of ${{E}}_{0}$.

First, we show $\dim(span\\{pr_{0}(\nabla_{X}X)/\leavevmode\nobreak\ X\in
E_{1}\\})=1$;
Let us $X$, $Y\in E_{1}$. If $X$ and $Y$ are orthogonal unit field. Let
$A_{i}=\mu_{i}I+\lambda_{i}J$ be a second fundamental form, where $J$ is skew-
symmetric satisfying $J^{2}=-I$. Then

$g(\nabla_{X}X,e_{i})=-\mu_{i}g(X,X)=-\mu_{i}g(Y,Y)=g(\nabla_{Y}Y,e_{i}).$

consequently, $pr_{0}(\nabla_{X}X)=pr_{0}(\nabla_{Y}Y)$.

If $X$ and $Y$ are not orthogonal. Since $\dim(E_{1})\geq 3$, we can choose
$Z$ orthogonal to both $X$ and $Y$.

Then, $pr_{0}(\nabla_{X}X)=pr_{0}(\nabla_{Z}Z)=pr_{0}(\nabla_{Y}Y)$, thus

$\dim(span\\{pr_{0}(\nabla_{X}X)/\leavevmode\nobreak\ X\in E_{1}\\})=1.$ (33)

Moreover, there is no isotropic vector field in $F_{1}$ unless
$\dim(E_{0})=2$.

If any second fundamental forms are of the form $A_{i}=\mu_{i}I$, we get that

$\dim(span\\{pr_{0}(\nabla_{X}Y)/\leavevmode\nobreak\ X,\leavevmode\nobreak\
Y\in E_{1}\\})=1.$

Indeed, for all orthogonal fields $X$ and $Y$ in $E_{1}$, we have

$g(\nabla_{X}Y,e_{i})=-\mu_{i}g(X,Y)=g(\nabla_{Y}X,e_{i})=0.$

Hence $span\\{pr_{0}(\nabla_{X}Y)/\leavevmode\nobreak\ X,\leavevmode\nobreak\
Y\in E_{1}\\}=span\\{pr_{0}(\nabla_{X}X)/\leavevmode\nobreak\ X\in E_{1}\\}.$

If there is a second fundamental form $A_{i}=\mu_{i}I+\lambda_{i}J$ where
$\lambda_{i}\neq 0$ and $J$ is skewsymmetric endomorphism checking $J^{2}=-I$.
Then $\dim(E_{1})=2l$ and we can choose
$\\{X_{1},Y_{1}=J(X_{1}),...,X_{l},Y_{l}=J(X_{l})\\}$ an orthonormal fram of
$E_{1}$.

Let $A_{k}=\mu_{k}I+\lambda_{k}J$ be the second fundamental form associted to
the vector fields $e_{k}$. For all $i\neq j$, we get

$\begin{array}[]{ccccc}g(\nabla_{X_{i}}Y_{i},e_{k})&=g(\nabla_{X_{j}}Y_{j},e_{k})&=&-\lambda_{k}&\\\
g(\nabla_{X_{i}}Y_{j},e_{k})&=g(\nabla_{Y_{j}}X_{i},e_{k})&=&g(\nabla_{X_{j}}X_{i},e_{k})&=g(\nabla_{Y_{j}}Y_{i},e_{k})=0.\end{array}$
(34)

Thus

$\dim\bigg{(}span\bigg{\\{}pr_{0}(\nabla_{X}Y)/\leavevmode\nobreak\ X,Y\in
E_{1},\;g(X,Y)=0\bigg{\\}}\bigg{)}=1$

and $F_{1}=span\\{pr_{0}(\nabla_{X}Y)/\leavevmode\nobreak\ X,Y\in
E_{1},\;g(X,Y)=0\\}+span\\{pr_{0}(\nabla_{X}X)/\leavevmode\nobreak\ X\in
E_{1}\\}.$

Consequently, we get

$r=\dim\bigg{(}span\bigg{\\{}pr_{0}(\nabla_{X}Y)/\leavevmode\nobreak\
X,\leavevmode\nobreak\ Y\in E_{1}\bigg{\\}}\bigg{)}\leq 2.$

∎

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