# On generalized derivations of polynomial vector fields Lie algebras

Princy Randriambololondrantomalala  Sania Asif  Département de Mathématiques
et Informatique, Université d’Antananarivo, Madagascar<EMAIL_ADDRESS>School of Mathematics and Statistics, Nanjing University of information
science and technology, Nanjing, Jianngsu Province, PR China.
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

In this paper, we study the generalized derivation of a Lie subalgebra of the
Lie algebra of polynomial vector fields on $\mathbb{R}^{n}$, containing all
constant vector fields and the Euler vector field, under some conditions on
this Lie subalgebra.

###### keywords:

Lie Algebra, generalized derivation, quasiderivation, m-derivation,
quasicentroid, centroid, polynomial vector fields, Euler vector field.

††thanks: We would think Pr. Ivan Kaygorodov of Federal University of ABC, Sao
Paolo, Brazil, in his interest about this study. Many thanks to the referee
for the valuable suggestions in this paper.

17B40, 17B66, 17B70. 30 2 https://doi.org/10.46298/cm.10386

##  1 Introduction

In [sam], we have studied the derivations of a Lie subalgebra $\mathfrak{P}$
of the Lie algebra of vector fields $\chi\left(\mathbb{R}^{n}\right)$ on
$\mathbb{R}^{n}$ for $n\geq 1$. The subalgebra $\mathfrak{P}$ contains all
constant vector fields and the Euler vector field. In [pri], we explore that
each $m$-derivation $D$ of $\mathfrak{P}$ with $m\geq 2$ is an endomorphism of
$\mathfrak{P}$ such that $\forall X_{1},X_{2},\dots,X_{m}\in\mathfrak{P},$ we
have

$\displaystyle D$
$\displaystyle\left[X_{1},\left[X_{2},\dots,\left[X_{m-1},X_{m}\right]\dots\right]\right]$
(1)
$\displaystyle=\left[D\left(X_{1}\right),\left[X_{2},\dots\left[X_{m-1},X_{m}\right]\dots\right]\right]+\left[X_{1},\left[D\left(X_{2}\right),\dots,\left[X_{m-1},X_{m}\right]\dots\right]\right]$
(2)
$\displaystyle{\quad}+\dots+\left[X_{1},\left[X_{2},\dots,\left[D\left(X_{m-1}\right),X_{m}\right]\dots\right]\right]+\left[X_{1},\left[X_{2},\dots,\left[X_{m-1},D\left(X_{m}\right)\right]\dots\right]\right].$
(3)

It’s clear that every derivation (i.e 2-derivation) of $\mathfrak{P}$ is a
$m$-derivation where $m\geq 2$ but the converse is not true in general. We say
that $m$-derivations for $(m\geq 2)$ are a generalization of derivations. Now,
we study another generalization of derivations of $\mathfrak{P}$, that are
called as generalized derivations in [luk]. However, in [luk], only study for
the finite dimensional Lie algebra has made, but study to the generalized
derivations of finite and infinite dimensional Lie algebras can be made on Lie
algebra $\mathfrak{P}$. We prove this fact in the form of examples for finite
and infinite dimensional $\mathfrak{P}$, in the later sections. Let us recall
that a _monomial vector field_ is a vector field on $\mathbb{R}^{n}$ having
one and only one monomial component in one and only one element of the
canonical basis of $\chi\left(\mathbb{R}^{n}\right)$. A Lie algebra of
polynomial vector fields $\mathfrak{P}$ is called _separated_ if
$\mathfrak{P}$ is spanned by monomial vector fields. Moreover, a _diagonal
linear vector fields_ is an element of the Cartan subalgebra $\left\langle
x^{1}\frac{\partial}{\partial x^{1}},\dots,x^{n}\frac{\partial}{\partial
x^{n}}\right\rangle_{\mathbb{R}}$ where $\left(x^{i}\right)_{i=1,\dots,n}$ is
the usual coordinate system of $\mathbb{R}^{n}$. In this research we show that
if $\mathfrak{P}$ is separated and contains all diagonal linear vector fields
with some additional condition, then there are some examples of further Lie
algebras structures that admit generalized derivations that were not shown in
[luk] even for the finite dimensional Lie algebras. The results present in
this paper are explicit and advance in the sense that we completely describe
the centroid, the quasicentroid and the quasiderivations that arise in the
study of generalized derivations of Lie algebras. We adapt our results in
[pri] and [sam], to study generalized derivations of infinite dimensional Lie
algebras. We do not only present abstract theorems but we also give examples
to illustrate all theorems of this paper.This paper consists of four sections
after the present section for introduction. The second section is a general
result from [sam] on $\mathfrak{P}$ about its graded algebra structure and its
first space of cohomology of Chevalley-Eilenberg. The third section focuses on
the study of centroid and quasicentroid of $\mathfrak{P}$ and proves that
these two sets are equal. The next section describes the quasiderivation of
$\mathfrak{P}$ using the fact that this quasiderivation is a sum of
homogeneous degree $i\in\mathbb{Z}$ quasiderivations of $\mathfrak{P}$. The
final section uses the results from the previous sections and presents the
main theorem of this paper. According to which, generalized derivations of
$\mathfrak{P}$ are sum of derivations, quasicentroid elements, and two special
endomorphisms of $\mathfrak{P}$. As stated earlier, it would be of great
interest to explore generalized derivations of more general subalgebra
$\mathfrak{P}.$ Throughout this paper all Lie algebras are equipped as
algebras by the usual bracket notation $[,]$.

##  2 The Lie algebra $\mathfrak{P}$

Let us denote the Lie algebra of vector fields on $\mathbb{R}^{n}$ with the
coordinates system $\left(x^{i}\right)_{i}$, by
$\chi\left(\mathbb{R}^{n}\right)$. Throughout this section, we consider a Lie
algebra $\mathfrak{P}$ of polynomial vector fields on $\mathbb{R}^{n}.$ This
Lie algebra $\mathfrak{P}$ contains the _constant fields_
$\frac{\partial}{\partial x^{i}}$ for all $i$ and the _Euler vector field_
$E={\sum_{i}}x^{i}\frac{\partial}{\partial x^{i}}.$ Let
$\mathfrak{N}=\\{X\in\chi\left(\mathbb{R}^{n}\right)/\left[X,\mathfrak{P}\right]\subset\mathfrak{P}\\}$
denote the _normalizer_ of $\mathfrak{P}$ in $\chi\left(\mathbb{R}^{n}\right)$
and let $\mathfrak{H}_{i}$ be the vector space of _homogeneous polynomial
vector fields_ of degree $i\in\mathbb{N}\cup\\{-1\\}$. Thus
$X\in\mathfrak{H}_{i}$ if and only if

$\left[E,X\right]=iX.$ (4)

Denote $\mathfrak{H}^{d}_{0}$ the Lie algebra generated by all diagonal linear
fields. A _diagonal linear field_ is identified by the diagonal matrix with
coefficients in $\mathbb{R}$ of order $n$. We adopt the Einstein convention on
the summation index unless expressly mentioned. Here are two examples of
$\mathfrak{P}$ in $\chi\left(\mathbb{R}^{2}\right)$ with coordinate system
$(x,y)$:

  1. 1.

The Lie algebra spanned by $\frac{\partial}{\partial
x},\frac{\partial}{\partial y},x\frac{\partial}{\partial
x}+y\frac{\partial}{\partial y},x^{2}\frac{\partial}{\partial x}$ is of
_finite dimension_ and

  2. 2.

The Lie algebra spanned by $\frac{\partial}{\partial
x},\frac{\partial}{\partial y},x\frac{\partial}{\partial
x}+y\frac{\partial}{\partial y},x^{2}\frac{\partial}{\partial
x},x^{3}\frac{\partial}{\partial x}$ is of _infinite dimension._

The bracket of two vector fields $X=X^{i}\frac{\partial}{\partial x^{i}}$ and
$Y=Y^{j}\frac{\partial}{\partial x^{j}}$ of $\chi\left(\mathbb{R}^{n}\right)$
is the Lie algebra of vector fields on $\mathbb{R}^{n},$ given by:

$\left[X,Y\right]=\left(X^{i}\frac{\partial Y^{j}}{\partial
x^{i}}-Y^{i}\frac{\partial X^{j}}{\partial
x^{i}}\right)\frac{\partial}{\partial x^{j}}$ (5)

The following lemma is clear:

######  Lemma 2.1.

A Lie algebra of polynomial vector fields
$\mathfrak{P}=\bigoplus_{i\geq-1}\mathfrak{P}_{i}$ is graded with the
following
$\left[\mathfrak{P}_{i},\mathfrak{P}_{j}\right]\subset\mathfrak{P}_{i+j},$
where $\mathfrak{P}_{i}=\mathfrak{P}\cap\mathfrak{H}_{i}$.

We recall two theorems of [sam] on the centralizer and the cohomology space of
$\mathfrak{P}$:

######  Proposition 2.2.

Let $\mathfrak{C}\left(\mathfrak{P}\right)$ be the centralizer of
$\mathfrak{P}$,
$\mathfrak{C}\left(\mathfrak{P}\right)=\left\\{X\in\chi\left(\mathbb{R}^{n}\right)/\left[X,\mathfrak{P}\right]=\\{0\\}\right\\}$.
Here $\mathfrak{C}\left(\mathfrak{P}\right)$ is zero.

######  Definition 2.3.

We denote $Der\left(\mathfrak{P}\right)$ the Lie algebra of derivations of
$\mathfrak{P}$, $L_{X}=\left[X,.\right]$ is the Lie derivative with respect to
$X\in\mathfrak{P}$, it is called inner derivation. $ad_{\mathfrak{P}}$ is the
Lie algebra of inner derivations of $\mathfrak{P}$. The first Chevalley-
Eilenberg cohomology space is denoted
$\mathrm{H}^{1}\left(\mathfrak{P}\right)=Der\left(\mathfrak{P}\right)/ad_{\mathfrak{P}}$.

######  Theorem 2.4.

$\mathrm{H}^{1}\left(\mathfrak{P}\right)\cong\mathfrak{N}/\mathfrak{P}\cong\mathfrak{M}$
with $\mathfrak{M}$ a Lie subalgebra of $\mathfrak{H}_{0}$. If all linear
diagonal fields are in $\mathfrak{P}$ then
$\mathrm{H}^{1}\left(\mathfrak{P}\right)$ is zero .

##  3 Centroid and quasicentroid of $\mathfrak{P}$

######  Definition 3.1.

The centroid $C\left(L\right)$ of a Lie algebra $L$ is the Lie algebra of
endomorphisms $f$ in $L$ such that

$f\left[X,Y\right]=\left[fX,Y\right]=\left[X,fY\right],$ (6)

######  Definition 3.2.

The quasicentroid $QC\left(L\right)$ of a Lie algebra $L$ is the Lie algebra
of endomorphism $g$ in $L$ such that

$\left[g\left(X\right),Y\right]=\left[X,g\left(Y\right)\right],$ (7)

for all $X,Y\in L$.

Note that in general $C\left(L\right)\subset QC\left(L\right)$. In this
section, we prove that $C\left(L\right)=QC\left(L\right)$ for
$L=\mathfrak{P}$.

######  Proposition 3.3.

Recall that $E$ is the Euler vector field and let $Y\in\mathfrak{P}_{m}$ with
$m\geq-1$. Each element $f$ of the centroid $C\left(\mathfrak{P}\right)$ is
homogeneous of degree $0.$ If we denote
$f\left(E\right)=\sum_{i\geq-1}E^{\prime}_{i}$ and
$f\left(Y\right)=\sum_{i\geq-1}Y_{i}$, where the $E^{\prime}_{i},Y_{i}$ are in
$\mathfrak{P}_{i}$, then $Y_{m\neq
0}=\left[\frac{1}{m}E^{\prime}_{0},Y\right]$. Moreover $Y_{0}$ is completely
determined by
$\left[-E^{\prime}_{0},\left[C,Y\right]\right]=\left[C,Y_{0}\right]\text{for
all $C\in{\mathfrak{P}_{-1}}$}$ with $\left[E^{\prime}_{0},Y\right]=0$ for
$m=0$.

######  Proof 3.4.

If $f\in C\left(\mathfrak{P}\right)$, then we have (6) for $X=E$ and $Y$ an
homogeneous element in $\mathfrak{P}_{m}$ with $m\geq-1$:

$f\left[E,Y\right]=\left[fE,Y\right]=\left[E,fY\right].$ (8)

Thus

$mf\left(Y\right)=\sum_{i\geq-1}\left[E^{\prime}_{i},Y\right]=\sum_{i\geq-1}iY_{i}.$
(9)

Then, we have $mf\left(Y\right)=\sum_{i\geq-1}mY_{i}=\sum_{i\geq-1}iY_{i}$. In
term of homogeneous components $mY_{i}=iY_{i}$ and for $m\neq i$, it yields
$Y_{i}=0$. Hence, $f\left(Y\right)=Y_{m}$ and $f$ is homogeneous of degree 0.
So, we get $f\left(E\right)=E^{\prime}_{0}$ ($E$ of degree 0, then by $f$ is
homogeneous of degree 0, $f\left(E\right)=E^{\prime}_{0}$). Moreover, in term
of homogeneous components of (8), we have

$\text{for all $m\neq 0$},\ Y_{m}=\left[\frac{1}{m}E^{\prime}_{0},Y\right]$
(10)

In addition, we have for a constant field $C\in\mathfrak{P}_{-1}$

$f\left[C,Y\right]=\left[fC,Y\right]=\left[C,fY\right].$ (11)

If $m=0$, by the previous result: $Y$ homogeneous of degree 0, then
$f(Y)=Y_{0}$ and $\left[C,Y\right]$ of degree -1. By (10)
$f\left[C,Y\right]=\left[-E^{\prime}_{0},\left[C,Y\right]\right]$ and
$fC=\left[-E^{\prime}_{0},C\right]$. Thus, from (11), we obtain

$\left[-E^{\prime}_{0},\left[C,Y\right]\right]=\left[\left[-E^{\prime}_{0},C\right],Y\right]=\left[C,Y_{0}\right].$
(12)

So by the Jacobi identity and (12),we have

$\left[C,\left[-E^{\prime}_{0},Y\right]\right]=0,\quad\forall
C\in\mathfrak{P}_{-1}.$ (13)

It yields $\left[E^{\prime}_{0},Y\right]=0$ for $m=0$. Moreover,

$\left[-E^{\prime}_{0},\left[C,Y\right]\right]=\left[C,Y_{0}\right]\text{for
all $C\in\mathfrak{P}_{-1}$}$ (14)

determines $Y_{0}$, because $Y_{0}$ is of the form
$\beta^{i}_{j}x^{j}\frac{\partial}{\partial x^{i}}$ with
$\beta^{i}_{j}\in\mathbb{R}$ and $C=\frac{\partial}{\partial x^{k}}$ with
$k=1,\dots,n$, then
$\left[C,Y_{0}\right]=\beta^{i}_{k}\frac{\partial}{\partial
x^{i}}=\left[-E^{\prime}_{0},\left[C,Y\right]\right]$ determines
$\beta^{i}_{k}$ for all $k$ and $i$.

######  Corollary 3.5.

If $\mathfrak{H}_{0}\subset\mathfrak{P}_{0}$ then
$C\left(\mathfrak{P}\right)=\left\langle Id\right\rangle_{\mathbb{R}}$, where
$Id$ is the identity map. If $\mathfrak{H}^{d}_{0}=\mathfrak{P}_{0}$,
$C\left(\mathfrak{P}\right)=\left\langle
Id,Id_{1},\dots,Id_{n}\right\rangle_{\mathbb{R}}$ where $Id_{i}$ with $i$
fixed is such that

$Id_{i}\left(x^{i}\frac{\partial}{\partial
x^{i}}\right)=x^{i}\frac{\partial}{\partial
x^{i}},Id_{i}\left(\frac{\partial}{\partial
x^{i}}\right)=\frac{\partial}{\partial x^{i}}$

and null otherwise on $\mathfrak{P}_{0}$ where
$Id_{i}\left(Y\in\mathfrak{P}_{m\neq
0}\right)=\left[\frac{1}{m}x^{i}\frac{\partial}{\partial x^{i}},Y\right]$.

######  Proof 3.6.

We can refer on the previous Proposition with its notions saying that if $f\in
C\left(\mathfrak{P}\right)$, then $\left[E^{\prime}_{0},Y\right]=0$ for $m=0$.
If we run $Y$ on the set of linear diagonal vector fields (that is to say $Y$
takes all values of all elements in this set), we have $E^{\prime}_{0}=0$ or
$E^{\prime}_{0}=cE+c^{\prime i}x^{i}\frac{\partial}{\partial x^{i}}$ with
$c\neq 0$ and $c^{\prime i}\in\mathbb{R}$. By (14), $\left[C,Y_{0}\right]=0$
or $\left[C,Y_{0}-cY\right]=0$ for all $C\in\mathfrak{P}_{-1}$ of the form
$\frac{\partial}{\partial x^{k}}$ with $k=1,\dots,n$ except for $\mathfrak{P}$
where $\mathfrak{P}_{0}=\left\langle x^{1}\frac{\partial}{\partial
x^{1}},\dots,x^{n}\frac{\partial}{\partial x^{n}}\right\rangle_{\mathbb{R}}$.
Then it yields $Y_{0}=0$ or $Y_{0}=cY$ and (10) $Y_{m\neq 0}=0$ or $Y_{m}=cY$
for all $m\geq-1$ except for $\mathfrak{P}_{0}=\left\langle
x^{1}\frac{\partial}{\partial x^{1}},\dots,x^{n}\frac{\partial}{\partial
x^{n}}\right\rangle_{\mathbb{R}},$ otherwise $\left[E^{\prime}_{0},Y\right]=0$
for all $Y\in\mathfrak{P}_{0}$ is not satisfied. If we have this last
condition on $\mathfrak{P}_{0}$, we can conclude following the decomposition
of $E^{\prime}_{0}$.

######  Example 3.7.

In $\mathbb{R}^{2}$ with the usual coordinates system $(x,y)$,
$\mathfrak{P}=\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial
y},x\frac{\partial}{\partial x},y\frac{\partial}{\partial
y}\right\rangle_{\mathbb{R}}$, $f=Id+Id_{2}$ in $C\left(\mathfrak{P}\right)$
is such that

$f\left(\frac{\partial}{\partial x}\right)=\frac{\partial}{\partial x},\quad
f\left(\frac{\partial}{\partial y}\right)=2\frac{\partial}{\partial y},\quad
f\left(x\frac{\partial}{\partial x}\right)=x\frac{\partial}{\partial x},\quad
f\left(y\frac{\partial}{\partial y}\right)=2y\frac{\partial}{\partial y}.$

######  Corollary 3.8.

If $\mathfrak{P}_{0}=\left\langle E\right\rangle_{\mathbb{R}}$, then
$C\left(\mathfrak{P}\right)=\left\langle Id\right\rangle_{\mathbb{R}}.$

######  Proof 3.9.

If we denote $f\left(E\right)=\sum_{i\geq-1}E^{\prime}_{i}$ with $f\in
C\left(\mathfrak{P}\right)$, then $E^{\prime}_{0}=cE$ where $c\in\mathbb{R}$.
By equations from Proposition 3.3, $f\left(Y\right)=cY$ for all
$Y\in\mathfrak{P}_{m\geq-1}$.

######  Corollary 3.10.

If $E^{\prime}_{0}=cE$ for $c\in\mathbb{R}$, then $f\in
C\left(\mathfrak{P}\right)$ is always of the form $cId.$

######  Proof 3.11.

This follows directly from the previous corollary.

######  Remark 3.12.

In general, $C\left(\mathfrak{P}\right)$ is neither $\left\langle
Id\right\rangle_{\mathbb{R}}$ nor $\left\langle
Id,Id_{1},\dots,Id_{n}\right\rangle_{\mathbb{R}}$, for which we can take an
example of $\mathfrak{P}=\left\langle\frac{\partial}{\partial
x},\frac{\partial}{\partial y},E,x\frac{\partial}{\partial
y}\right\rangle_{\mathbb{R}}$ in the usual coordinates $(x,y)$ on
$\mathbb{R}^{2}$ and $f\in C\left(\mathfrak{P}\right)$ such that
$f\left(\frac{\partial}{\partial x}\right)=\frac{\partial}{\partial
x}+\frac{\partial}{\partial y},f\left(\frac{\partial}{\partial
y}\right)=\frac{\partial}{\partial
y},f\left(E\right)=E+x\frac{\partial}{\partial
y},f\left(x\frac{\partial}{\partial y}\right)=x\frac{\partial}{\partial y}$.

######  Proposition 3.13.

The element $f$ of the quasicentroid $QC\left(\mathfrak{P}\right)$ is such
that $f\left(E\right)=E^{\prime}_{0}$ and
$f\left(Y\right)=\left[\frac{1}{m}E^{\prime}_{0},Y\right]$, where the
$\left(E^{\prime}_{0},Y\right)$ is in
$\mathfrak{H}_{0}\times\mathfrak{H}_{m\geq-1,m\neq 0}$. Moreover,
$\left[E^{\prime}_{0},Y\right]=0$ for $m=0$, then $Y_{0}$ is completely
defined by
$\left[-E^{\prime}_{0},\left[C,Y\right]\right]=\left[C,Y_{0}\right]\text{for
all $C\in\mathfrak{P}_{-1}$}$.

######  Proof 3.14.

Following the above notations and using the second part of the Eq. (8), We
obtain

$\left[f\left(E\right),E\right]=\left[E,f\left(E\right)\right],\text{so}f\left(E\right)\text{is
homogeneous of degree 0}$

and

$\sum_{i\geq-1}\left[E^{\prime}_{i},Y\right]=\sum_{i\geq-1}iY_{i}.$

In term of homogeneous components: if $i\leq m-2$, $Y_{i}=0$ and for
$i\geq-1$,

$\left[E^{\prime}_{i},Y\right]=(m+i)Y_{m+i}.$ (15)

Because only $E^{\prime}_{0}$ is the only term of $f\left(E\right)$, so
$Y_{m}$ is the only term of $f\left(Y\right)$ and for $m\neq 0$,
$Y_{m}=\left[\frac{1}{m}E^{\prime}_{0},Y\right]$. For $m+i=0$ in (15) for
$m=0$, we have $\left[E^{\prime}_{0},Y\right]=0$. Using Jacobi identity and
the last equality, we get (12) for $f$ in the quasicentroid and it yields the
last assertion of our theorem.

Now, we can conclude for the centroid and the quasicentroid of $\mathfrak{P}$:

######  Corollary 3.15.

For all $\mathfrak{P}$,
$C\left(\mathfrak{P}\right)=QC\left(\mathfrak{P}\right)$.

##  4 Quasiderivation of $\mathfrak{P}$

######  Definition 4.1.

The set of quasiderivation $QDer\left(L\right)$ of a Lie algebra $L$ is the
Lie algebra of endomorphism $f$ in $L$ such that there exists another
endomorphism $g$ of $L$ such that

$\left[f\left(X\right),Y\right]+\left[X,f\left(Y\right)\right]=g\left[X,Y\right]\
\forall X,Y\in L.$ (16)

It is more convenient to denote $f\in QDer\left(L\right)$ by $(f,f,g)$ and
$f\in QC\left(L\right)$ by $(f,-f,0)$. A quasiderivation of $\mathfrak{P}$ is
a generalization of a derivation of $\mathfrak{P}$. We compute
$QDer\left(\mathfrak{P}\right)$, it is clear that:

$QDer\left(\mathfrak{P}\right)=QDer_{0}\left(\mathfrak{P}\right)+\sum_{i\in\mathbb{Z}^{\ast}}QDer_{i}\left(\mathfrak{P}\right)$

where $QDer_{i}\left(\mathfrak{P}\right)$ is the set of quasiderivation of
degree $i$. Lemma 4.1 of [luk] can be adapted to our $\mathfrak{P}$ taking
account that it doesn’t need to specify the dimension of $\mathfrak{P}$ and
$\mathfrak{P}$ contains a torus $\left\langle E\right\rangle_{\mathbb{R}}$. So

######  Proposition 4.2.

We have the following description of $QDer\left(\mathfrak{P}\right)$:

$QDer\left(\mathfrak{P}\right)=QDer_{0}\left(\mathfrak{P}\right)+\sum_{i\in\mathbb{Z}}QDer^{\prime}_{i}\left(\mathfrak{P}\right)+ad_{\mathfrak{P}}$

where

$QDer^{\prime}_{i}\left(\mathfrak{P}\right)=\left\\{f\in
QDer_{i}\left(\mathfrak{P}\right)/f(E)=0\right\\}.$

In the rest of this paper, we assume $n\geq 2$.

######  Lemma 4.3.

Each element $(f,f,g)\in QDer_{0}\left(\mathfrak{P}\right)$ such that
$f(E)=0$, $f_{|\mathfrak{P}_{-1}}\equiv 0$ and
$g_{|\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]}\equiv 0$, verifies
$f\equiv 0$ and
$g_{|\left(\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)}$
is arbitrary with $g_{|\mathfrak{P}_{i\neq
0}+\left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]}\equiv 0$.

######  Proof 4.4.

Let us reason by induction knowing that $f$ is homogeneous of degree 0. If
$X\in\mathfrak{P}_{-1}$ then $f(X)=0$ by hypothesis, by (16) where $Y=E$, we
have:

$\left[f\left(X\right),E\right]+\left[X,f\left(E\right)\right]=g\left[X,E\right].$

Then $-f\left(X\right)=-g\left(X\right)$, because $f(E)=0$ and
$f\left(X\right)$ of degree -1. It yields $g(X)=0$ from $f(X)=0$. For all
$X\in\mathfrak{P}_{0}$ by (16) where $Y\in\mathfrak{P}_{-1}$:

$\left[f\left(X\right),Y\right]+\left[X,f\left(Y\right)\right]=g\left[X,Y\right].$

From the previous results on $g$ and $f$, the second term of the equality is
null and then $\left[f\left(X\right),Y\right]=0$ for all
$Y\in\mathfrak{P}_{-1}$. Thus
$f(X)\in\mathfrak{P}_{-1}\cap\mathfrak{P}_{0}=\\{0\\}$ for all
$X\in\mathfrak{P}_{0}$. For
$X\in\mathfrak{P}_{0}\cap\left[\mathfrak{P},\mathfrak{P}\right]$, we have two
cases.

  1. 1.

The first is $X=\left[Y,Z\right]$ with
$\left(Y,Z\right)\in\mathfrak{P}_{0}\times\mathfrak{P}_{0}$ where we deduce by
definition 4.1 and the previous result that
$0=\left[f\left(Y\right),Z\right]+\left[Y,f\left(Z\right)\right]=g\left[Y,Z\right]=g(X)$.

  2. 2.

The second case is made by the hypothesis $g\left(X\right)=0$ where
$X\in\left[\mathfrak{P}_{-1},\mathfrak{P}_{1}\right]$.

It yields
$g\left(X\in\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)$
is arbitrary. For all $X\in\mathfrak{P}_{1}$ by (16) where $Y$ runs in
$\mathfrak{P}_{-1}$:

$\left[f\left(X\right),Y\right]+\left[X,f\left(Y\right)\right]=g\left[X,Y\right].$

From $g\left(\left[\mathfrak{P}_{-1},\mathfrak{P}_{1}\right]\right)=0$, the
second term of the above equality is null. Because of $f(Y)=0$, we get
$\left[f\left(X\right),Y\right]=0$ for all $Y\in\mathfrak{P}_{-1}$. Thus
$f(X)\in\mathfrak{P}_{-1}\cap\mathfrak{P}_{1}=\\{0\\}$ for all
$X\in\mathfrak{P}_{1}$. But (16) where $Y=E$ says $f(X)=g(X)$, then $g(X)=0$
also. We set $k\geq 2$, we suppose that $f_{|\mathfrak{P}_{t}}\equiv
g_{|\mathfrak{P}_{t}}\equiv 0$ for all $t\geq k-1$. For
$X\in\mathfrak{P}_{k\geq 2}$, we write (16) with $Y\in\mathfrak{P}_{-1}$. Then
we obtain $f(X)\in\mathfrak{P}_{-1}\cap\mathfrak{P}_{k\geq 2}=\\{0\\}$ for all
$X\in\mathfrak{P}_{k}$ knowing that $f_{|\mathfrak{P}_{-1}}\equiv
g_{|\mathfrak{P}_{k-1}}\equiv 0$. Again by (16) with $Y=E$, we have
$0=-kf(X)=-kg(X)=0$ which yields $g(X)=0$ by $f(X)=0$.

######  Proposition 4.5.

Suppose that all elements of $\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]$
are also elements of $\left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]$. The
$(f,f,g)\in QDer_{0}\left(\mathfrak{P}\right)$ vanishing on $E$ is of the
following form: $\left(L_{X},L_{X},L_{X}+k\right)$, where
$X\in\mathfrak{H}_{0}$ and $k\in End\left(\mathfrak{P},\mathfrak{P}\right)$
such that $k$ is arbitrary on
$\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]$ and null
elsewhere.

######  Proof 4.6.

Let $(f,f,g)\in QDer_{0}\left(\mathfrak{P}\right)$, such that $f(E)=0$. In the
system of coordinates on $\mathbb{R}^{n}$ $(x^{1},\dots,x^{n})$, we write
$f\left(X_{0}\right)=X_{1}\in\mathfrak{P}_{-1}$ for
$X_{0}\in\mathfrak{P}_{-1}$, with $f(E)=0$. Like in the proof of Proposition
2.8 of [sam], there exists an unique $X\in\mathfrak{H}_{0}$ such that
$f\underset{\left\langle
E\right\rangle_{\mathbb{R}}+\mathfrak{P}_{-1}}{=}L_{X}$. Then we write the
following $(f,f,g)-\left(L_{X},L_{X},L_{X}\right)\in
QDer_{0}\left(\mathfrak{P}\right)$ with $f(E)=L_{X}\left(E\right)=0$. If we
denote $f^{\prime}=f-L_{X}$ and $g^{\prime}=g-L_{X}$, we have
$f^{\prime}(E)=0$, $f^{\prime}\underset{\mathfrak{P}_{-1}}{=}0$ and
$g^{\prime}_{|\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]}\equiv 0$ by the
first hypothesis in the statement of the present proposition. That is to say,
if $X_{2}\in\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]$, then there are
$(Y,Z)\in\mathfrak{P}_{0}$ such that $X_{2}=\left[Y,Z\right]$. Then

$g^{\prime}\left(X_{2}\right)=\left[f^{\prime}\left(Y\right),Z\right]+\left[Y,f^{\prime}\left(Z\right)\right].$

By the proof of Lemma 4.3,
$f^{\prime}\left(Y\right)=f^{\prime}\left(Z\right)=0$, thus
$g^{\prime}\left(X_{2}\right)=0$. By the same lemma,
$(f^{\prime},f^{\prime},g^{\prime})\equiv 0$ except in
$\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]$ where
$g^{\prime}$ is arbitrary.

######  Proposition 4.7.

Let $(f,f,g)\in QDer^{\prime}_{i\geq 1}\left(\mathfrak{P}\right)$ and adopt
all hypothesis of Lemma 4.3. Then $f\equiv g_{|\mathfrak{P}_{j\neq
0}+\left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]}\equiv 0$ and
$g_{|\left(\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)}$
is arbitrary. If $(f,f,g)\in QDer^{\prime}_{-1}\left(\mathfrak{P}\right)$, we
suppose all hypothesis of Lemma 4.3 except for $f_{|\mathfrak{P}_{-1}}\equiv
0$, we have the same result as the previous one.

######  Proof 4.8.

We prove this Proposition by induction. Let $(f,f,g)\in QDer^{\prime}_{i\geq
1}\left(\mathfrak{P}\right)$. By (16) where $X=E$ and $Y$ runs in
$\mathfrak{P}_{-1}$, we can say that $g_{|\mathfrak{P}_{-1}}\equiv 0$ knowing
$f_{|\mathfrak{P}_{-1}}\equiv 0$ and $f(E)=0$. If $X\in\mathfrak{P}_{0}$, (16)
with $Y=E$ where $f(E)=0$ gives $\left[fX,E\right]=0$. It yields $f(X)=0$
because $f(X)\in\mathfrak{P}_{i\neq 0}$. Then, in the turn of $g$, if
$X\in\left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]$, then Eq. (16) yields
$g(X)=0$. In addition, we have
$g_{|\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]}\equiv 0$ by hypothesis.
Thus if $X\in\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]$, $g(X)=0$. Next,
$X\in\mathfrak{P}_{1}$, if $Y$ runs in $\mathfrak{P}_{-1}$, in (16) we obtain
$\left[f\left(X\right),\mathfrak{P}_{-1}\right]=\\{0\\}$. But the degree of
$f$ is not -2, then $f\left(X\right)=0$. It follows that the equation (16)
with $Y=E$ and this $X$ permit us to say that $g(X)=0$. Now, we suppose that
$f=g=0$ on $\mathfrak{P}_{t\leq l}$ with $l\geq 1$ and $t\geq 1$. Consider
$X\in\mathfrak{P}_{l+1}$, if we run $Y\in\mathfrak{P}_{-1}$ on (16), we have
$\left[f\left(X\right),\mathfrak{P}_{-1}\right]=\\{0\\}$. It yields $f(X)=0$
because $f(X)$ is of degree $i+l+1$ different to $-1$. If we write $Y=E$ with
this $X$ on (16), we have $-(l+1)g(X)=\left[f(X),E\right]=-(i+l+1)f(X)$ which
is equal to 0 by the above result on $f$. Then, we get $g(X)=0$ because
$l+1\neq 0$. For $(f,f,g)\in QDer^{\prime}_{-1}\left(\mathfrak{P}\right)$, it
is obvious that $f_{|\mathfrak{P}_{-1}}=g_{|\mathfrak{P}_{-1}}\equiv 0.$ The
rest of the proof is the same as for $(f,f,g)\in QDer^{\prime}_{i\geq
1}\left(\mathfrak{P}\right)$.

######  Proposition 4.9.

Consider all hypothesis of Proposition 4.5. Every $(f,f,g)$ in the set
$QDer_{i\geq-1,\neq 0,1}\left(\mathfrak{P}\right)$ is of the form

$\left(L_{\frac{-1}{i}{f(E)}_{i}},L_{\frac{-1}{i}{f(E)}_{i}},L_{\frac{-1}{i}{f(E)}_{i}}+k\right),$

where ${f(E)}_{i}$ the homogeneous term $f(E)$ of degree $i$ and $k$ is an
endomorphism of $\mathfrak{P}$, arbitrary on
$\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]$ and null
elsewhere.

######  Proof 4.10.

Let $(f,f,g)\in QDer_{i\geq-1,\neq 0,1}\left(\mathfrak{P}\right)$ be such
quasiderivation, we have

$\left(f^{\prime}=f-L_{\frac{-1}{i}{f(E)}_{i}},f^{\prime}=f-L_{\frac{-1}{i}{f(E)}_{i}},g^{\prime}=g-L_{\frac{-1}{i}{f(E)}_{i}}\right)\in
QDer^{\prime}_{i\geq-1,\neq 0,1}\left(\mathfrak{P}\right).$

Moreover, by using $X=E$ and $Y\in\mathfrak{P}_{0}$ in Eq. (16), we have

$g^{\prime}\left[E,Y\right]=\left[E,f^{\prime}(Y)\right].$

This yields $0=\left[E,f^{\prime}(Y)\right]$ and $f^{\prime}(Y)=0$ because the
degree of $f^{\prime}(Y)$ is nonzero. Then
$g^{\prime}_{|\left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]}=\\{0\\}$ from
(16). Knowing that all elements of
$\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]$ are elements of
$\left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]$, we have
$g^{\prime}_{\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]}\equiv 0$. Next,
we will check whether $f^{\prime}_{\mathfrak{P}_{-1}}\equiv 0$. If $i=-1$, it
is obvious that
$f^{\prime}_{\mathfrak{P}_{-1}}=g^{\prime}_{\mathfrak{P}_{-1}}\equiv 0$. If
$i\geq 2$, we take $X=\frac{\partial}{\partial x^{t}}\in\mathfrak{P}_{-1}$
where $1\leq t\leq n$ is fixed. We write
$f^{\prime}\left(X\right)=\underset{k\neq t}{P^{k}}\frac{\partial}{\partial
x^{k}}+P^{t}\frac{\partial}{\partial x^{t}}$ where all $P^{s}$ is a polynomial
of degree $i$, in which all $P^{s}$ doesn’t depend on $x^{s}$ because of the
hypothesis that each linear diagonal field doesn’t belong to
$\left[\mathfrak{P},\mathfrak{P}\right]$. When
$Y=x^{t}\frac{\partial}{\partial x^{t}}$ in Eq. (16),
$0=g^{\prime}\left[\frac{\partial}{\partial
x^{t}},x^{l}\frac{\partial}{\partial
x^{l}}\right]=\left[f^{\prime}(X),x^{l}\frac{\partial}{\partial x^{l}}\right]$
with a fixed $l\neq t$. Then

$P^{l}\frac{\partial}{\partial x^{l}}+\beta^{k}\underset{k\neq
t,l}{P^{k}}\frac{\partial}{\partial
x^{k}}+\alpha^{t}P^{t}\frac{\partial}{\partial x^{t}}=0,$

where $\alpha^{t},\beta^{k}\in\mathbb{R}$. It yields $P^{l}=0$ for all $l\neq
t$. Thus,

$g^{\prime}\left(\frac{\partial}{\partial
x^{t}}\right)=g^{\prime}\left[\frac{\partial}{\partial
x^{t}},x^{t}\frac{\partial}{\partial
x^{t}}\right]=\left[f^{\prime}\left(\frac{\partial}{\partial
x^{t}}\right),x^{t}\frac{\partial}{\partial
x^{t}}\right]=f^{\prime}\left(\frac{\partial}{\partial x^{t}}\right).$

When in Eq.(16), $X\in\mathfrak{P}_{l\geq-1}$ and $Y=E$, we get

$\frac{-l}{i}g(X)=L_{\frac{-1}{i}{f(E)}_{i}}(X)-\frac{i+l}{i}f(X).$

Then
$\left(f-L_{\frac{-1}{i}{f(E)}_{i}}\right)(X)=\frac{f\left(X\right)-g\left(X\right)}{i}$
where $X\in\mathfrak{P}_{-1}$. If $i\geq 2$,
$f^{\prime}\left(X\right)-g^{\prime}\left(X\right)=0$ for
$X\in\mathfrak{P}_{-1}$ by the above result, then
$f\left(X\right)-g\left(X\right)=0$. Then by Proposition 4.7, we have
$f^{\prime}\equiv g^{\prime}_{|\mathfrak{P}_{j\neq
0}+\left(\mathfrak{P}_{0}\cap\left[\mathfrak{P},\mathfrak{P}\right]\right)}\equiv
0$ and
$g^{\prime}_{|\left(\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)}$
is arbitrary.

######  Proposition 4.11.

We suppose all hypothesis of Proposition 4.5. Every $(f,f,g)$ in the set
$QDer_{1}\left(\mathfrak{P}\right)$ is of the form

$\left(-L_{{f(E)}_{1}},-L_{{f(E)}_{1}},-L_{{f(E)}_{1}}+k\right)+\left(f^{\prime\prime},f^{\prime\prime},0\right)$

where ${f(E)}_{1}$ is the homogeneous term of $f(E)$ of degree $1$, $k$ is
arbitrary on $\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]$
and null elsewhere, $f^{\prime\prime}_{|\mathfrak{P}_{l\neq-1}}$ is null and
$f^{\prime\prime}\left(\mathfrak{P}_{-1}\right)$ is a subset of $\left\langle
x^{1}\frac{\partial}{\partial x^{1}},\dots,x^{n}\frac{\partial}{\partial
x^{n}}\right\rangle_{\mathbb{R}}$.

######  Proof 4.12.

Let $(f,f,g)\in QDer_{1}\left(\mathfrak{P}\right)$ be such quasiderivation, we
write for $i=1$:

$\left(f^{\prime}=f-L_{\frac{-1}{i}{f(E)}_{i}},f^{\prime}=f-L_{\frac{-1}{i}{f(E)}_{i}},g^{\prime}=g-L_{\frac{-1}{i}{f(E)}_{i}}\right)\in
QDer^{\prime}_{i\neq 0}\left(\mathfrak{P}\right).$

We can assume all results of the first part of the previous proof ending on
$g^{\prime}_{|\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]}\equiv 0$ with
$i=1$. In this result, we have $f^{\prime}_{|\mathfrak{P}_{0}}\equiv 0$. Let
us take $X\in\mathfrak{P}_{-1}$. When $Y=E$ in (16),
$g^{\prime}(X)=\left[f^{\prime}(X),E\right]=0$ because $f^{\prime}(X)$ of
degree 0 and $f^{\prime}(E)=0$. Let us precise $X=\frac{\partial}{\partial
x^{j}}$, $\left[\frac{\partial}{\partial x^{j}},x^{j}\frac{\partial}{\partial
x^{j}}\right]=X$ and by Eq. (16) with $Y=x^{j}\frac{\partial}{\partial
x^{j}}$,
$g^{\prime}\left(X\right)=\left[f^{\prime}(X),x^{j}\frac{\partial}{\partial
x^{j}}\right]=0$ knowing that $f^{\prime}(x^{j}\frac{\partial}{\partial
x^{j}})=0$. For all $k\neq j$ fixed, $\left[\frac{\partial}{\partial
x^{j}},x^{k}\frac{\partial}{\partial x^{k}}\right]=0$, then (16) where
$Y=x^{k}\frac{\partial}{\partial x^{k}}$ yields
$\left[f\left(\frac{\partial}{\partial
x^{j}}\right),x^{k}\frac{\partial}{\partial x^{k}}\right]=0$ for all $k\neq
j$. We can conclude that $f^{\prime}\left(\frac{\partial}{\partial
x^{j}}\right)=\alpha^{t}x^{t}\frac{\partial}{\partial x^{t}}$. If we do
$g^{\prime}\left[\frac{\partial}{\partial x^{j}},\frac{\partial}{\partial
x^{l}}\right]=\left[\alpha^{t}x^{t}\frac{\partial}{\partial
x^{t}},\frac{\partial}{\partial x^{l}}\right]+\left[\frac{\partial}{\partial
x^{j}},\beta^{s}x^{s}\frac{\partial}{\partial x^{t}}\right]$, we obtain
$\alpha^{l}=\beta^{j}=0$ for all $j\neq l$. Then
$f^{\prime}\left(\frac{\partial}{\partial
x^{j}}\right)=\alpha^{j}x^{j}\frac{\partial}{\partial x^{j}}$. Thus, let
$f^{\prime\prime}\in End\left(\mathfrak{P}\right)$ be such that
$f^{\prime\prime}$ is null except on $\mathfrak{P}_{-1}$ where the value is on
$\left\langle x^{1}\frac{\partial}{\partial
x^{1}},\dots,x^{n}\frac{\partial}{\partial x^{n}}\right\rangle_{\mathbb{R}}$
such that $f^{\prime}\left(\frac{\partial}{\partial
x^{j}}\right)=f^{\prime\prime}\left(\frac{\partial}{\partial x^{j}}\right)$.
It is easy to verify that $(f^{\prime\prime},f^{\prime\prime},0)$ is in
$QDer^{\prime}_{1}\left(\mathfrak{P}\right)$. We get

$\left(f^{\prime\prime\prime}=f^{\prime}-f^{\prime\prime},f^{\prime\prime\prime}=f^{\prime}-f^{\prime\prime},g^{\prime}\right)\in
QDer^{\prime}_{1}\left(\mathfrak{P}\right)$

verifying $f^{\prime\prime\prime}(E)=0$,
$f^{\prime\prime\prime}_{|\mathfrak{P}_{-1}}\equiv 0$ and
$g^{\prime}_{|\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]}\equiv 0$. Then
we can use Proposition 4.7 and have the following results:
$f^{\prime\prime\prime}=0$, $g^{\prime}_{|\mathfrak{P}_{t\neq
0}+\left(\mathfrak{P}_{0}\cap\left[\mathfrak{P},\mathfrak{P}\right]\right)}\equiv
0$ and
$g^{\prime}_{|\left(\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)}$
is arbitrary. It follows the final results.

######  Definition 4.13.

For a fixed $i_{0}$, $j_{1}<j_{2}\dotsb<j_{p}$ with $1\leq p\leq n$,
$\left(\alpha_{j_{1}},\dots,\alpha_{j_{p}}\right)\in\mathbb{N}^{p}$ and
$j_{l}\in\\{1,\dots,n\\}$, the vector field
${\left(x^{j_{1}}\right)}^{\alpha_{j_{1}}}\dots\left(x^{j_{p}}\right)^{\alpha_{j_{p}}}\frac{\partial}{\partial
x^{i_{0}}}$ is called monomial. We recall that $\mathfrak{P}$ is separated, if
all homogeneous elements of degree $k\geq-1$ are spanned by monomial vector
field in $\mathfrak{P}$.

######  Remark 4.14.

In the proof of the previous theorem, the following fact is true,
$x^{j}\frac{\partial}{\partial
x^{j}}\notin\left[\mathfrak{P},\mathfrak{P}\right]$ if we adopt the hypothesis
on $\mathfrak{P}$:
$\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]\subset\left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]$.
In the same proof, we take $X=\frac{\partial}{\partial x^{j}}$. It is supposed
that $\mathfrak{P}$ is separated. We can reason as the following, if there
exists $Y=x^{k}\frac{\partial}{\partial x^{j}}\in\mathfrak{P}_{0}$ or
$Y=x^{j}\frac{\partial}{\partial x^{k}}\in\mathfrak{P}_{0}$ for $j\neq k$,
$0=g^{\prime}\left[X,Y\right]=\left[f^{\prime}\left(X\right),Y\right]$ from
(16) knowing $f^{\prime}(Y)=0$ and $g^{\prime}_{|\mathfrak{P}_{-1}}\equiv 0$.
It yields $\alpha^{j}=0$.

######  Example 4.15.

We are in $\mathbb{R}^{2}$ with the habitual coordinates $(x,y)$,
$\mathfrak{P}=\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial
y},x\frac{\partial}{\partial x},y\frac{\partial}{\partial
y}\right\rangle_{\mathbb{R}}$. Define an endomorphism $f^{\prime\prime}$ on
$\mathfrak{P}$, such that $f^{\prime\prime}\left(\frac{\partial}{\partial
x}\right)=x\frac{\partial}{\partial x}$,
$f^{\prime\prime}\left(\frac{\partial}{\partial
y}\right)=y\frac{\partial}{\partial y}$ and null otherwise. Then
$(f^{\prime\prime},f^{\prime\prime},0)$ is an example of quasiderivation of
type Proposition 4.11, which is neither a derivation nor an element of
$QC\left(\mathfrak{P}\right)$.

######  Proposition 4.16.

Let $(f,f,g)\in QDer^{\prime}_{i\leq-2}\left(\mathfrak{P}\right)$, then
$f_{|\mathfrak{P}_{t\neq-(i+1)}}\equiv g_{|\mathfrak{P}_{t\neq-(i+1)}}\equiv
0$ with $f(X)=(i+1)g(X)$ for all $X\in\mathfrak{P}_{-(i+1)}$ and
$g_{|\left(\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)}$
is arbitrary.

######  Proof 4.17.

Let $(f,f,g)\in QDer^{\prime}_{i\leq-2}\left(\mathfrak{P}\right)$ a such
quasiderivation. By the fact that $\mathfrak{P}_{t\leq-2}=\\{0\\}$, $f=g\equiv
0$ on $\mathfrak{P}_{t\leq 0}$. We reason by induction, we suppose that
$f=g=0$ on $\mathfrak{P}_{t\leq 0}$. Consider $X\in\mathfrak{P}_{t+1}$ with
$t\geq 0$, if we write $Y=E$ on (16), we have

$-(t+1)g(X)=\left[f(X),E\right]=-(i+t+1)f(X).$

We take (16) and run $Y$ on $\mathfrak{P}_{-1}$, we obtain
$\left[f\left(X\right),Y\right]=\\{0\\}$. Hence, if $i+t\neq-2$, then
$f\left(X\right)=0$ and $g(X)=0$ by the previous equation. Otherwise,
$f(X)=-(t+1)g(X)$ for all $X\in\mathfrak{P}_{t+1}$, that is to say in the case
where $t+1=-1-i$.

######  Proposition 4.18.

If each element of $\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]$ is an
element of $\left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]$ and $\mathfrak{P}$
is separated with all linear diagonal fields in $\mathfrak{P}$, then
$(f,f,g)\in QDer^{\prime}_{i\leq-2}\left(\mathfrak{P}\right)$ is null except
$g_{|\left(\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)}$
which is arbitrary.

######  Proof 4.19.

To fix our idea, we cannot take $X=x^{1}P(x^{t\neq 1})\frac{\partial}{\partial
x^{1}}\in\mathfrak{P}_{l\geq 1}$ because in this case
$x^{1}\frac{\partial}{\partial
x^{1}}\in\mathfrak{P}_{0}\cap\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]\
\setminus\ \left[\mathfrak{P}_{0},\mathfrak{P}_{0}\right]$. So we are forced
to take $X=P(x^{t\neq 1})\frac{\partial}{\partial x^{1}}\in\mathfrak{P}_{l\geq
1}$. By using $f(X)=(i+1)g(X)$ for $X\in\mathfrak{P}_{l=-(i+1)}$ where
$-(i+1)\geq 1$ from Proposition 4.16 and when $Y=x^{1}\frac{\partial}{\partial
x^{1}}$ in the Eq. (16) with $g(X)=\alpha^{j}\frac{\partial}{\partial x^{j}}$
where $\alpha^{j}\in\mathbb{R}$: we obtain

$\alpha^{j}\frac{\partial}{\partial
x^{j}}=(i+1)\alpha^{1}\frac{\partial}{\partial x^{1}}.$

It yields $\alpha^{j}=0$ for all $j\neq 1$ and $\alpha^{1}=0$ because
$i+1\leq-1$. Thus $g(X)$ and $f(X)$ are zero.

Next, by using result on odd-derivations in [pri], we can prove that
$(f,f,g)\in QDer^{\prime}_{-2}\left(\mathfrak{P}\right)$ is an odd-derivations
of $\mathfrak{P}$ in the case $\mathfrak{P}_{0}=\left\langle
E\right\rangle_{\mathbb{R}}\subsetneq\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]$.

######  Proposition 4.20.

Let $(f,f,g)\in QDer^{\prime}_{-2}\left(\mathfrak{P}\right)$. Then
$g_{|\mathfrak{P}_{0}\cap\left[\mathfrak{P},\mathfrak{P}\right]}\equiv 0$ and
$g_{|\left(\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)}$
is arbitrary. In addition, $g_{|\mathfrak{P}_{i\neq 0,1}}\equiv 0$ with
$g\left(X\right)=-f\left(X\right)$ for $X\in\mathfrak{P}_{1}$,
$f\left[Y,X^{\prime}\right]=\left[fX^{\prime},Y\right]$ for all
$\left(X^{\prime},Y\right)\in\mathfrak{P}_{2}\times\mathfrak{P}_{-1}$ and
$f_{|\mathfrak{P}_{i\neq 1,2}}\equiv 0$.

######  Proof 4.21.

If $f$ have degree $-2$, then $f_{|\mathfrak{P}_{i\leq 0}}\equiv 0$. Thus,
applying (16) with $(X,Y)\in\mathfrak{P}_{1}\times\mathfrak{P}_{-1}$ and
$(X,Y)\in\mathfrak{P}_{0}\times\mathfrak{P}_{0}$ resp., knowing that $f$ is of
degree $-2$, we obtain that $g\left[X,Y\right]=0$ but
$g_{|\left(\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]\right)}$
is arbitrary. Now, let’s take $X\in\mathfrak{P}_{1}$ and $Y=E$ in (16), we
deduce that $g\left(X\right)=-f\left(X\right)$ for $f$ of degree $-2$. Next,
we consider $X\in\mathfrak{P}_{2}$ and $Y=E$ in Eq. (16), we find $g(X)=0$
with $f(X)$ is determined by the same relation, if $X\in\mathfrak{P}_{2}$ and
$Y\in\mathfrak{P}_{-1}$, we find that $f\left[Y,X\right]=\left[f(X),Y\right]$
taking account the value of $f$ and $g$ on $\mathfrak{P}_{1}$. If
$X\in\mathfrak{P}_{k\geq 3}$ and $Y$ runs in $\mathfrak{P}_{-1}$, by induction
we have $f\left(X\right)=0$ and $g\left(X\right)=0$ by Eq. (16) when $Y=E$.

Let us recall the Theorem 3.10 of [pri]:

######  Theorem 4.22.

We suppose that $m\geq 3$ is odd. A linear map $D$ of degree $-2$ on
$\mathfrak{P}$, null on $\mathfrak{P}\ominus\mathfrak{P}_{1}$ is an
$m$-derivation, if and only if:

$\left[D\left(\mathfrak{P}_{1}\right),\mathfrak{P}-\mathfrak{P}_{0}\right]=\\{0\\};$
(17) $D\left[X,Y\right]=\left[D\left(Y\right),X\right]\
\forall\left(X,Y\right)\in\mathfrak{P}_{0}\times\mathfrak{P}_{1};$ (18)
$\left[D\left(\mathfrak{P}_{1}\right),\left[\mathfrak{P},\mathfrak{P}\right]\cap\mathfrak{P}_{0}\right]=\\{0\\};$
(19)

if
$\left[X_{1},\left[X_{2},\dots,\left[X_{i},\dots,\left[X_{m-1},X_{m}\right]\dots\right]\right]\right]\in\mathfrak{P}_{1}$
where $i$ is the first index with $X_{i}\in\mathfrak{P}_{1}$ with the
existence of $1\leq j<i$ such that
$X_{j}\in\mathfrak{P}_{-1}\cup\mathfrak{P}_{t\geq 2}$, then

$D\left[X_{1},\left[X_{2},\dots,\left[X_{i},\dots,\left[X_{m-1},X_{m}\right]\dots\right]\right]\right]=0$
(20)

######  Proposition 4.23.

If $(f,f,g)\in QDer^{\prime}_{-2}\left(\mathfrak{P}\right)$ and if for all
$X\in\mathfrak{P}_{2}$ there exists $X^{\prime}\in\mathfrak{P}_{-1}$ such that
$\left[X,X^{\prime}\right]=0$. Moreover, suppose that $\mathfrak{P}$ is
separated and $\mathfrak{P}_{0}=\left\langle
E\right\rangle_{\mathbb{R}}\subsetneq\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]$.
Then the endomorphism $f$ is an odd-derivation of $\mathfrak{P}$, the
endomorphism $g=-f$ on $\mathfrak{P}_{1}$ and vanishes otherwise except on
$\left\langle E\right\rangle_{\mathbb{R}}$ where $g$ is arbitrary.

######  Proof 4.24.

The $f$ is an $\mathbb{R}$-linear map of degree $-2$ on $\mathfrak{P}$ with
$(f,f,g)\in QDer^{\prime}_{-2}\left(\mathfrak{P}\right)$. For all
$X^{\prime}\in\mathfrak{P}_{2}$, $\left[X^{\prime},E\right]=-2X^{\prime}$,
then
$\left[f\left(X^{\prime}\right),E\right]+\left[X^{\prime},f(E)\right]=g\left[X^{\prime},E\right]=-2g\left(X^{\prime}\right)=0$
by (16) and Proposition 4.20 saying $f(E)=0$ and $g(X^{\prime})=0$. It yields
that there exists $c\in\mathbb{R}$ such that $f\left(X^{\prime}\right)=cE$. In
addition, for all
$\left(X^{\prime},Y\right)\in\mathfrak{P}_{2}\times\mathfrak{P}_{-1}$, we have
$f\left[Y,X^{\prime}\right]=\left[fX^{\prime},Y\right]$ using Proposition
4.20. By hypothesis, there is a $Y\in\mathfrak{P}_{-1}$ such that
$\left[Y,X^{\prime}\right]=0$. We can choose $Y=\frac{\partial}{\partial
x^{j}}\in\mathfrak{P}_{-1}$ for a fixed $j$. It conducts to
$0=\left[fX^{\prime},Y\right]=-c\frac{\partial}{\partial x^{j}}$ and $c=0$. We
can conclude that $f_{|\mathfrak{P}_{2}}\equiv 0$. Now, if we look at Theorem
4.22, we must check the $4$ relations in that theorem for $f$. Let us take
$X\in\mathfrak{P}_{1},\ Y\in\mathfrak{P}_{i\neq 0},\ Z\in\mathfrak{P}_{0}$:

  1. 1.

$\left[f\left(X\right),Y\right]=g\left[X,Y\right]-\left[X,f(Y)\right]$ by
definition of quasiderivation. Since we have
$\left[X,Y\right]\notin\mathfrak{P}_{1}$, then
$\left[f\left(X\right),Y\right]=-\left[X,f(Y)\right]$ by Proposition 4.20
saying $g_{|\mathfrak{P}_{0}\cap\left[\mathfrak{P},\mathfrak{P}\right]}\equiv
0$ and $g_{|\mathfrak{P}_{i\neq 0,1}}\equiv 0$. Then if
$Y\notin\mathfrak{P}_{1}$,
$\left[f\left(X\right),Y\right]=-\left[X,f(Y)\right]=0$ by Proposition 4.20
and the above result on $f_{|\mathfrak{P}_{2}}\equiv 0$. But if
$Y\in\mathfrak{P}_{1}$, because of $\left[f\left(X\right),Y\right]$ and
$\left[X,f(Y)\right]$ are in $\mathfrak{P}_{0}=\left\langle
E\right\rangle_{\mathbb{R}}\subsetneq\left[\mathfrak{P}_{1},\mathfrak{P}_{-1}\right]$,
we have $\left[f\left(X\right),Y\right]=\left[X,f(Y)\right]=0.$ So we can
resume $\left[f\left(X\right),Y\right]=0$ for all $Y\in\mathfrak{P}_{i\neq
0}$.

  2. 2.

We have $g\left[Z,X\right]=\left[Z,f\left(X\right)\right]$, taking into
account that $f=-g$ on $\mathfrak{P}_{1}$, we obtain
$f\left[Z,X\right]=\left[f\left(X\right),Z\right]$.

  3. 3.

     * •

(i) First, consider $X^{\prime},X^{\prime\prime}\in\mathfrak{P}_{0}$,
$\left[f\left(X\right),\left[X^{\prime},X^{\prime\prime}\right]\right]=g\left[X,\left[X^{\prime},X^{\prime\prime}\right]\right].$
Then it is equal to
$g\left[\left[X,X^{\prime}\right],X^{\prime\prime}\right]+g\left[X^{\prime},\left[X,X^{\prime\prime}\right]\right]$
by Jacobi identity. The property of $f$ conducts this value to
$\left[f\left[X,X^{\prime}\right],X^{\prime\prime}\right]+\left[X^{\prime},f\left[X,X^{\prime\prime}\right]\right]$
and by $f=-g$ in $\mathfrak{P}_{1}$, it is of the following form

$\left[-g\left[X,X^{\prime}\right],X^{\prime\prime}\right]+\left[X^{\prime},-g\left[X,X^{\prime\prime}\right]\right].$

Then this expression becomes
$-\left[\left[fX,X^{\prime}\right],X^{\prime\prime}\right]-\left[X^{\prime},\left[fX,X^{\prime\prime}\right]\right]$.
By the identity of Jacobi, it is

$-\left[\left[fX,X^{\prime\prime}\right],X^{\prime}\right]-\left[\left[X^{\prime},fX\right],X^{\prime\prime}\right]-2\left[fX,\left[X^{\prime},X^{\prime\prime}\right]\right]$

and then

$-\left[g\left[X,X^{\prime\prime}\right],X^{\prime}\right]-\left[g\left[X^{\prime},X\right],X^{\prime\prime}\right]-2g\left[X,\left[X^{\prime},X^{\prime\prime}\right]\right].$

By $f=-g,$ we have

$g\left[\left[X,X^{\prime\prime}\right],X^{\prime}\right]+g\left[\left[X^{\prime},X\right],X^{\prime\prime}\right]-2g\left[X,\left[X^{\prime},X^{\prime\prime}\right]\right]$

and by the Jacobi identity, we have

$-g\left[X,\left[X^{\prime},X^{\prime\prime}\right]\right]-2g\left[X,\left[X^{\prime},X^{\prime\prime}\right]\right].$

By the first relation, it is equal to
$g\left[X,\left[X^{\prime},X^{\prime\prime}\right]\right]$, thus
$g\left[X,\left[X^{\prime},X^{\prime\prime}\right]\right]=0$ and
$\left[f\left(X\right),\left[X^{\prime},X^{\prime\prime}\right]\right]=0.$

     * •

(ii) Second, consider
$\left(X^{\prime},X^{\prime\prime}\right)\in\mathfrak{P}_{1}\times\mathfrak{P}_{-1}$,

$\left[f\left(X\right),\left[X^{\prime},X^{\prime\prime}\right]\right]=g\left[X,\left[X^{\prime},X^{\prime\prime}\right]\right].$

And it becomes

$\displaystyle
g\left[\left[X,X^{\prime}\right],X^{\prime\prime}\right]+g\left[X^{\prime},\left[X,X^{\prime\prime}\right]\right]$
$\displaystyle=\left[f\left[X,X^{\prime}\right],X^{\prime\prime}\right]+\left[fX^{\prime},\left[X,X^{\prime\prime}\right]\right]$
$\displaystyle=\left[\left[fX^{\prime},X\right],X^{\prime\prime}\right]+\left[X,\left[fX^{\prime},X^{\prime\prime}\right]\right].$

This right term of the above equality becomes
$\left[\left[fX^{\prime},X\right],X^{\prime\prime}\right]$ because the degree
of $f$ is $-2$. Thus
$\left[\left[fX^{\prime},X\right],X^{\prime\prime}\right]=0$ by (1). So we
conclude that

$\left[f\left(\mathfrak{P}_{1}\right),\left[\mathfrak{P},\mathfrak{P}\right]\cap\mathfrak{P}_{0}\right]=\\{0\\}.$

  4. 4.

Consider $2k+1$ elements $X_{1},\dots,X_{j},\dots,X_{i},\dots,X_{2k+1}$ on
$\mathfrak{P}$ as in Theorem 4.22, and knowing that $f=-g$ on
$\mathfrak{P}_{1}$, we have

$\displaystyle
f\left[X_{1},\left[X_{2},\dots,\left[X_{i},\dots,\left[X_{m-1},X_{m}\right]\dots\right]\right]\right]$
$\displaystyle=\left[X_{1},\left[X_{2},\dots,-f\left[X_{j},\dots,\left[X_{i},\dots,\left[X_{m-1},X_{m}\right]\dots\right]\right]\right]\right].$

With the property of $g$, it is

$\left[X_{1},\left[X_{2},\dots,\left[X_{j},f\left[X_{j+1},\dots,\left[X_{i},\dots,\left[X_{m-1},X_{m}\right]\dots\right]\right]\right]\right]\right].$

By the (1), this expression vanishes.

We conclude that $f$ is a $(2k+1)$-derivation on $\mathfrak{P}$ like in
Theorem 4.22. We have $f=-g$ except on $\left\langle
E\right\rangle_{\mathbb{R}}$ where $g$ is arbitrary.

######  Remark 4.25.

In general, the $(2k+1)$-derivation of the Proposition 4.23 is not null. We
can see it with one example from [pri]: In $\mathbb{R}^{3}$, the Lie algebra
$\mathfrak{P}$ spanned by $\frac{\partial}{\partial x}$,
$\frac{\partial}{\partial y}$, $\frac{\partial}{\partial z}$,$E$,
$x\frac{\partial}{\partial z}$ and $(x)^{2}\frac{\partial}{\partial z}$ admits
a $3$-derivation of $\mathfrak{P}$, $D_{0}$ defined by
$D_{0}\left((x)^{2}\frac{\partial}{\partial z}\right)=\frac{\partial}{\partial
z}$ and null otherwise. It is easy to check that $D_{0}$ is in
$QDer^{\prime}_{-2}\left(\mathfrak{P}\right)$. That is to say
$(D_{0},D_{0},-D_{0})\in QDer^{\prime}_{-2}\left(\mathfrak{P}\right)$.

######  Proposition 4.26.

If we have the hypothesis of Proposition 4.5 and if all linear diagonal fields
are in $\mathfrak{P}$, then every elements of $(f,f,g)\in
QDer_{0}\left(\mathfrak{P}\right)$ is a sum of an element of $\left\langle
Id,Id_{1},\dots,Id_{n}\right\rangle_{\mathbb{R}}$ and a quasiderivation of the
type Proposition 4.5.

######  Proof 4.27.

If we take such $f$, we can apply Eq.(16) with $X=E$ and $Y$ a linear diagonal
fields and find that $\left[f\left(E\right),Y\right]=0$ for all $Y$. It
conducts to the following facts, there exists $Y^{\prime}$ a non-null linear
diagonal field such that $f(E)=Y^{\prime}$. We take $h\in\left\langle
Id,Id_{1},\dots,Id_{n}\right\rangle_{\mathbb{R}}$ with $\left\langle
Id,Id_{1},\dots,Id_{n}\right\rangle_{\mathbb{R}}\supset
QC\left(\mathfrak{P}\right)$ by corollary 3.5 such that
$h\left(E\right)=Y^{\prime}$. This implies that $(f-h,f+h,g)\in
QDer^{\prime}_{0}\left(\mathfrak{P}\right)$ and by Proposition 4.5, we have
the result.

##  5 The main result on generalized Lie derivations of $\mathfrak{P}$ and
examples

######  Definition 5.1.

The set of all generalized derivations of a Lie algebra $L$ is denoted by
$GenDer\left(L\right)$. It is the Lie algebra of endomorphisms $f$ in $L$ such
that for every $f$ there exists another two endomorphisms $g,h$ in $L$ such
that

$\left[f\left(X\right),Y\right]+\left[X,h\left(Y\right)\right]=g\left[X,Y\right]\
$ (21)

for all $X,Y\in L.$

######  Definition 5.2.

The center of a Lie algebra $L$ is the set $Z(L)=\\{X\in
L\mid\left[X,L\right]=\\{0\\}\\}$.

By [luk], It is clear that $QDer\left(L\right)$ and $GenDer\left(L\right)$ are
two sets of Lie algebras and satisfy $QDer\left(L\right)\subset
GenDer\left(L\right)$. Moreover by assuming that a quasiderivation is a
generalization of derivation and a generalized derivation is a generalization
of quasiderivation, we have following result for a centerless Lie algebra $L$:

$ad_{L}\subseteq Der\left(L\right)\subseteq QDer\left(L\right)\subseteq
GenDer\left(L\right)\subseteq\mathfrak{gl}\left(L\right).$

As we know that the center of $\mathfrak{P}$ is null by Proposition 2.2, it is
interesting to find whether above inclusions verifies for $\mathfrak{P}$ or
not?. For our convenience, we denote $f\in GenDer\left(L\right)$ by $(f,h,g)$,
$f\in QDer\left(L\right)$ by $(f,f,g)$ and $f\in QC\left(L\right)$ by
$(f,-f,0)$. Study in [luk], holds only for finite dimensional Lie algebras.
But the following equation

$(f,h,g)=\left(\frac{f+h}{2},\frac{f+h}{2},g\right)+\left(\frac{f-h}{2},\frac{h-f}{2},0\right)$

is always holds for our $L=\mathfrak{P}.$ It further yields the following
results.

######  Proposition 5.3.

$GenDer\left(\mathfrak{P}\right)=QDer\left(\mathfrak{P}\right)+QC\left(\mathfrak{P}\right)$.

######  Theorem 5.4.

If $\mathfrak{P}$ verifies all hypotheses in the Proposition 4.18, then all
generalized derivations of $\mathfrak{P}$ are of the form

$h+\left(L_{X+Y},L_{X+Y},L_{X+Y}+k\right)+\left(f^{\prime\prime},f^{\prime\prime},0\right)$

with $h\in QC\left(\mathfrak{P}\right)\subset\left\langle
Id,Id_{1},\dots,Id_{n}\right\rangle_{\mathbb{R}}$ if

$\left\langle x^{1}\frac{\partial}{\partial
x^{1}},\dots,x^{n}\frac{\partial}{\partial
x^{n}}\right\rangle_{\mathbb{R}}\subset\mathfrak{P}_{0}$

and $h\in\left\langle Id\right\rangle_{\mathbb{R}}$ if

$\mathfrak{H}_{0}=\mathfrak{P}_{0}.$

Where $X,Y\in\mathfrak{P}$, $k\in End\left(\mathfrak{P}\right)$ is null except
on $\mathfrak{P}_{0}\ominus\left[\mathfrak{P},\mathfrak{P}\right]$ where $k$
is arbitrary, $f^{\prime\prime}$ is a homogeneous map of degree $1$, given in
Proposition 4.11. $f^{\prime\prime}$ is null if $\mathfrak{P}_{0}$ contains a
non diagonal linear vector field.

######  Proof 5.5.

By corollary 3.5, Propositions 5.3, 4.2, 4.5, 4.9, 4.11, 4.18 and 4.26, we
have the final result taking into account that $X\in\mathfrak{H}_{0}$ must be
in the normalizer $\mathfrak{N}$ of $\mathfrak{P}$, because $L_{X}$ is stable
in $\mathfrak{P}$. By Theorem 2.4, $\mathfrak{N}=\mathfrak{P}$ because all
linear diagonal vector fields are in $\mathfrak{P}$ which is also separated.

######  Example 5.6.

If we take the example of Example 4.15, we have

$GenDer\left(\mathfrak{P}\right)=\left\langle
Id,Id_{1},Id_{2}\right\rangle_{\mathbb{R}}+ad_{\mathfrak{P}}+G+K.$

Where $G$ is the space of $f^{\prime\prime}$ in Proposition 4.11 and $K$ is
the space of quasiderivations of the form $(0,0,g)$, where $g$ is arbitrary
and $g$ vanishes except on $\left\langle x\frac{\partial}{\partial
x},y\frac{\partial}{\partial y}\right\rangle_{\mathbb{R}}$.

######  Example 5.7.

In $\mathbb{R}^{2}$, we consider the habitual coordinate system $(x,y)$. The
Lie algebra $\mathfrak{P}$ is the infinite dimensional vector space

$\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial
y},x\frac{\partial}{\partial x},y\frac{\partial}{\partial
y},y\frac{\partial}{\partial x},y^{2}\frac{\partial}{\partial
x},y^{3}\frac{\partial}{\partial x},\dots\right\rangle_{\mathbb{R}},$

then

$GenDer\left(\mathfrak{P}\right)=\left\langle
Id\right\rangle_{\mathbb{R}}+ad_{\mathfrak{P}}+K.$

Where $K$ is same as in the previous example.

## References

  * [1] Paperpri Randriambololondrantomalala P. Les $m$-dérivations des algébres de Lie de champs de vecteurs polynomiaux Afrika Matematika 27 2016 5-6 1065-1077
  * [2] Papersam Ravelonirina H.S.G., Randriambololondrantomalala P. and Anona M.F. Sur les algébres de Lie des champs de vecteurs polynomiaux African Diaspora Journal of Mathematics 10 2010 2 87-95
  * [3] Paperluk Leger G. F., Luks E.M. Generalized derivations of Lie algebras Journal of Algebra 228 2000 165-203
  * [4]

January 04, 2022June 14, 2022Friedrich Wagemann


# Evaluation of Xilinx Deep Learning Processing Unit under Neutron Irradiation

Dimitris Agiakatsikas, Nikos Foutris, Aitzan Sari, Vasileios Vlagkoulis,
Ioanna Souvatzoglou, Mihalis Psarakis, Mikel Luján, Maria Kastriotou and Carlo
Cazzaniga Experiments at the ISIS Neutron and Muon Source were supported by a
beamtime allocation RB2000230 from the Science and Technology Facilities
Council. This work has been partially supported by the University of Piraeus
Research Center and the EU Horizon 2020 EuroEXA 754337 grant.Dimitris
Agiakatsikas (e-mail: agiakatsikas@gmail.com), Aitzan Sari, Vasileios
Vlagkoulis, Ioanna Souvatzoglou, and Mihalis Psarakis (e-mail:
<EMAIL_ADDRESS>are with the Dept. of Informatics, University of Piraeus,
Greece.Nikos Foutris and Mikel Luján are with the Dept. of Computer Science,
The University of Manchester, UK.Maria Kastriotou and Carlo Cazzaniga are with
the ISIS Facility, STFC, Rutherford Appleton Laboratory, Didcot OX110 QX,
UK.©2021 IEEE. Personal use of this material is permitted. Permission from
IEEE must be obtained for all other uses, in any current or future media,
including reprinting/republishing this material for advertising or promotional
purposes, creating new collective works, for resale or redistribution to
servers or lists, or reuse of any copyrighted component of this work in other
works. This paper has been accepted by the 2021 European Conference on
Radiation and Its Effects on Components and Systems (RADECS)

###### Abstract

This paper studies the dependability of the Xilinx Deep-Learning Processing
Unit (DPU) under neutron irradiation. It analyses the impact of Single Event
Effects (SEEs) on the accuracy of the DPU running the resnet50 model on a
Xilinx Ultrascale+ MPSoC.

###### Index Terms:

DPU, Data-center, Radiation-test, Neutrons, Reliability, DNN, AI

##  I Introduction

Field Programmable Gate Arrays (FPGAs) have evolved from glue-logic devices to
sophisticated heterogeneous computing platforms that pave the way to the new
Artificial Intelligence (AI) wave.

As a result, datacenter market leaders increasingly integrate FPGA System-on-
Chip (SoC) in their infrastructure to target complex AI applications.
Companies like Microsoft, Amazon Web Services, and Baidu scale up AI and high-
performance applications on hundreds of thousands of Intel and Xilinx FPGA
devices [1].

A popular application of AI is Convolutional Neural Networks (CNNs). A growing
body of research shows that deploying optimised CNN models on FPGAs achieves a
higher performance per watt than CPU and GPU solutions [2]. However, modern
FPGAs are vulnerable to Single Event Upsets (SEUs) due to their reliance on
SRAM memory to store their configuration and application data. SEUs in SRAM
FPGA-SoCs are not destructive but cause various failure modes, such as Silent
Data Corruption (SDC), application crashes and kernel panics when an OS is
used.

Previous works have explored the Architectural Vulnerability Factor (AVF) of
custom FPGA CNN designs with fault-injection campaigns and irradiation
experiments [3, 4]. These works have targeted relative simple CNN case studies
for edge computing that neither require an Operating System (OS) nor have
complex CNN topologies. Simple case studies serve well when focusing on
analysing the reliability tradeoffs of various CNN configurations, e.g., the
reliability of a CNN under different quantisation and model-compression
schemes. Exploring the reliability of large-scale datacenter CNN applications,
however, requires a different testing paradigm. The case studies should test
additional aspects of a system, such as the operating system’s stability,
while SEUs are occurring in resources of the CPU, e.g., L1 and L2 caches, on
the on-chip Ethernet controller and in custom logic implemented with
Programmable Logic (PL).

To this extent, this work analyses the dependability of the whole
computational stuck of a commercial CNN inference solution, namely the Xilinx
Vitis AI Deep Learning Processing Unit (DPU). In more detail, we implemented
the DPU on a Zynq Ultrascale+ XCZU9EG MPSoC, which executed image
classification with the resnet50 CNN model. By performing Neutron Irradiation
experiments, we observed that the FPGA-SoC OS did not crash due to errors in
L1 and L2 caches of the CPU, while the DPU application had a very low AVF.

##  II Background

###  II-A Vitis AI and the Deep Learning Processing Unit (DPU)

Xilinx has introduced a rich ecosystem of tools and Intellectual Property (IP)
cores to ease the development of AI applications. In more detail, Xilinx
provides the Vitis AI development environment that encompasses 1) AI
frameworks (e.g., Tensorflow), 2) pre-optimised AI models, 3) quantization and
model compression tools, and 4) the DPU, with all necessary Linux drivers to
seamlessly deploy a CNN application on Xilinx devices.

The DPU is a convolution neural network accelerator IP offered by Xilinx for
Zynq-7000 SoC and Zynq Ultrascale+ MPSoC devices. The DPU is implemented with
PL and is tightly interconnected via the AXI bus to the SoC processing system
(PS), as shown in Figure 1. The DPU executes special instructions that are
generated by the Vitis AI compiler. A typical Vitis AI development flow
involves 1) the optimisation and compilation of a CNN model to DPU
instructions and 2) the compilation of software running on the Application
Processing Unit (APU), i.e., CPU.

The APU pre- and post-processes DNN data, controls the DPU, and orchestrates
the movement of instructions and data between the DPU, the CPU, and the off-
chip DDR memory.

The DPU consists of an instruction scheduler and up to three on-chip BRAM
buffers and computing engines. The instruction scheduler fetches and decodes
DPU instructions from off-chip memory and controls the on-chip memories and
computing engines. The DPU is available in eight configurations, i.e., B512,
B800, B1024, B1152, B1600, B2304, B3136, and B4096. Each configuration
utilises a different number of computing engines and on-chip memories in order
to target different sized devices and support various DPU functionalities,
e.g., ReLU, RELU6, or Leaky ReLU.

Figure 1: Deep-learing acceleration with the Xilinx Deep Processing Unit (DPU)
on Zynq©-7000 SoC and Zynq© Ultrascale+TM MPSoC devices.

##  III Experimental Setup

###  III-A ChipIr Neutron Beam

ChipIr is an ISIS neutron and muon facility instrument at the Rutherford
Appleton Laboratory (UK), designed to deliver an atmospheric-like fast neutron
spectrum to test radiation effects on electronic components and devices [5],
[6]. The ISIS accelerator provides a proton beam of 800 MeV, 40 $\mu$A, 10 Hz,
impinging on the tungsten target of its Target Station 2, where ChipIr is
located. The spallation neutrons produced illuminate a secondary scatterer
which optimises the atmospheric-like spectrum arriving at ChipIr, with an
acceleration factor of up to 109 for ground-level applications. With a
frequency of 10 Hz, the beam pulses consist of two 70 ns wide bunches
separated by 360 ns. The beam fluence at the position of the Device Under Test
(DUT) was continuously monitored by a silicon diode, while the beam flux of
neutrons above 10 MeV during the experimental campaign was 5.6 x 106
neutrons/cm2/s. The beam size was set through the two sets of the ChipIr jaws
to 7cm x 7cm.

Figure 2: Neutron beam experiment at the ChipIr facility of RAL, UK.

###  III-B Design Under Test (DUT): DPU-B4096

The Vivado DPU targeted reference design (TRD) [7] provided by Vitis AI v1.3.1
was implemented with Vivado 2020.2 for a ZCU102 development board. The ZCU102
board features the Zynq UltraScale+ XCZU9EG MPSoC. The B4096 configuration of
the DPU was synthesised with default settings, i.e., with RAM_USAGE_LOW,
CHANNEL_AUGMENTATION_ENABLE, DWCV_ENABLE, POOL_AVG_ENABLE,
RELU_LEAKYRELU_RELU6, Softmax.

The design was implemented with Vivado’s Performance_ExplorePostRoutePhysOpt*
run strategy because Vivado’s default run strategy resulted in time
violations. Table I shows the resource utilisation and operating frequency of
the DPU TRD. Due to the high resource utilisation of the TRD, Vivado reported
a relatively high percentage (41.45%) of essential bits – 59281993 out of the
143015456 total configuration bits were essential bits. Please recall that
essentials bits are configuration bits that, when corrupted have the potential
to cause functional errors.

TABLE I: Resource utilisation and operating frequency of the DPU targeted reference design Resource | Utilisation | Available | Utilisation % |  Frequency
---|---|---|---|---
LUT | 108,208 | 274,080 | 39.48 | 325 MHz
LUTRAM | 11,960 | 144,000 | 8.31 | 325 MHz
FF | 203,901 | 548,160 | 37.20 | 325 MHz
BRAM | 522 | 912 | 57.24 | 325 MHz
DSP | 1,395 | 2,520 | 55.36 | 650 MHz
IO | 7 | 328 | 2.13 | 325 MHz
BUFG | 6 | 404 | 1.49 | 325 MHz
MMCM | 1 | 4 | 25.00 | 325 MHz
PLL | 1 | 8 | 12.50 | 325 MHz
APU | 1 | 1 | 100.00 | 1200 MHz
DDR ctrl. | 1 | 1 | 100.00 | 533 MHz

Two important notes can be made for Table I.

1st note: All resources in the DPU operate at 325 MHz except for the DSPs,
which run at 2 x 325 MHz = 650 MHz. This is because the DPU design applies a
double data rate technique on DSP resources. Since DSPs are able to operate in
a much higher frequency than other PL resources, one can perform _N_ times
more computation by running the DSPs with _N_ times the frequency of the
surrounding logic while multiplexing and de-multiplexing their input and
output data, respectively.

2nd note: The design utilises 319, 55, 405, 4 and 1 LUT, LUTRAM, Flip-Flops
(FF), BRAM and DSP more resources, respectively, than the baseline DPU-TRD
design. This is because we included the Xilinx Soft Error Mitigation (SEM)
controller in the design to perform fault injection and validate our
experimental setup prior to the radiation tests. The clock of the SEM
controller was gated off during beamtime to replicate a simple _out-of-the-
box_ implementation scenario.

We used Petalinux 2020.2 to generate a Linux OS image for the ZCU102 by using
the default Board Support Package (BSP) provided by the DPU-TRD, except 1) the
nfs_utils package which was additionally added in RootFS to mount a Network
File Sharing (NFS) folder on Linux, and 2) the u-boot bootloader that mounted
an external SD EXT4 file system instead of INITRD RAM disk.

The CNN application that run on the DPU was the resnet50.xmodel, which was
also provided by the Vitis AI DPU-TRD. This resnet50 model is neither
compressed nor quantised but serves nice as a baseline application for
comparison with more optimised models that we aim to implement and test in
future work.

###  III-C Test procedure

Figure 3 shows the test setup of the radiation experiment. A laptop in
ChipIr’s control room orchestrated the test procedure of the DUT. The ZCU102
development board (hosting the DUT), an Ethernet-controlled Power Supply Unit
(PSU), and a USB device that remotely reset the ZCU102 (i.e., by electrically
shorting the SRTS_B and POR_B buttons of the board) was located in the beam
room.

The test of the DUT took place as follows. 1) An Experiment Control Software
(ECS) running on the laptop remotely resets the DUT and waits for the DUT to
boot, 2) the DUT Linux OS restarts, and 3) after a successful kernel boot, an
/etc/init.d/startup.sh script executes the following sub-tasks: 4a) the DUT
connects on an NFS folder located on the laptop, 4b) the DUT writes a sync.log
file in the shared NFS folder to notify the ECS of a successful boot, 4c) an
initial resnet50 classification takes place to warm-up the CPU caches, 4d) the
sync.log is updated to notify ECS that it is ready to start image
classifications, 4e) the /etc/init.d/startup.sh enters an infinite loop where
it continuously runs DPU classifications and stores the results in the NFS
folder to be checked by the ECS. The result checking (i.e. by the ECS) of each
classification iteration is synchronised with the DUT via a mutex stored in
the shared sync.log file. The ECS remotely resets the DUT when it detects a
boot timeout, a Critical Error (see Sec. IV) or a result timeout. It is worth
noting that for each classification iteration, the DUT saves the
classification result and the Linux dmesg.log for post-analysis.

Figure 3: Test setup at the ChipIr facility of RAL, UK.

##  IV Experimental Results

In this section, we discuss the impact of neutron radiation effects on the
reliability of the DPU accelerator. Given that the DPU comprises of a
heterogeneous architecture including the ARM SoC and the FPGA fabric, we first
present the cross-sections of the memories of the PS part (i.e. CPU caches)
and then discuss how the SEUs in the PL configuration memory affect the
behaviour of the system. Please note that the neutron radiation experiments
took place at ChipIr on May 2021.

Table II presents the cross-sections of the 32KB Level-1 Data (L1-D) Cache,
the 32KB Level-1 Instruction (L1-I) Cache, and the translation lookaside
buffer (TLB) – a two-level TLB with 512 entries that handles all translation
table operations of the CPU. Table III presents the cross-sections of the 1MB
Level-2 (L2) Cache and the Snoop Control Unit (SCU). The SCU has duplicate
copies of the L1 data-cache tags. It connects the APU cores with the device’s
accelerator coherency port (ACP) to enable hardware accelerators issue
coherent accesses to the L1 memory space. The upsets in the data and tag
arrays in both the L1 and L2 caches have been separately identified. The
cross-sections of the tag arrays have been calculated based on the tag sizes
of the caches, e.g., a 16-bit tag in the 16-way set associative 1MB L2 cache.
The cross sections have been calculated for a total fluence of 5.5x1010
neutrons/cm2 on a more than 3-hour radiation experiment. The results show that
the cross-sections of the tag arrays are slightly lower than those of the data
arrays. Our cross-section calculations for all caches (i.e., L1 and L2) are
very close to those reported in [4].

Figure 4 presents the proportion of detected upsets during cache accesses per
CPU core. As shown in the figure, the upsets in the L1 caches are balanced
between the four cores, while in the L2 cache, more upsets were observed in
Core 3. In future work, we aim to save the utilisation of all MPSoC cores and
OS running processes to better understand the unbalanced distribution of
detected upsets per core in L2.

All MPSoC caches are protected against SEUs with Error Correction Code (ECC)
mechanisms, e.g., L1-D and L2 caches incorporate ECC with Single Error
Correction Double Error Detection (SECDED) capability, while the L1-I cache
has parity that supports only SED. All these SED and SECDEC mechanisms in L1
and L2 caches mitigated soft errors in the CPU of the FPGA-SoC, therefore
resulting in a stable OS execution, with no application crashes or kernel
panics during the radiation tests. This was achieved either by having the ECC
scheme correct the error or by flushing and reloading the cache during
exceptions of the Linux EDAC driver (/sys/devices/system/edac/mc).

TABLE II: L1 Cache Cross Section | Upsets | Cross Section | Conf. Level 95%
---|---|---|---
|  | (cm2/bit) | Lower | Upper
L1-D Data | 32 | 2.20E-15 | 1.50E-15 | 3.11E-15
L1-D Tag | 3 | 3.47E-16 | 7.16E-17 | 1.02E-15
L1-D Total | 35 | 1.51E-15 | 1.05E-15 | 2.10E-15
L1-I Data | 25 | 1.72E-15 | 1.11E-15 | 2.54E-15
L1-I Tag | 4 | 4.89E-16 | 1.33E-16 | 1.25E-15
L1-I Total | 29 | 1.28E-15 | 8.54E-16 | 1.83E-15
L1 TLB | 9 | 9.90E-15 | 4.53E-15 | 1.88E-14
TABLE III: L2 Cache Cross Section | Upsets | Cross Section | Conf. Level 95%
---|---|---|---
|  | (cm2/bit) | Lower | Upper
L2 Data | 293 | 6.29E-16 | 5.59E-16 | 7.06E-16
L2 Tag | 20 | 8.59E-17 | 5.25E-17 | 1.33E-16
L2 Total | 313 | 4.48E-16 | 4.00E-16 | 5.01E-16
Snoop CU | 4 | 4.63E-16 | 1.26E-16 | 1.19E-15
TABLE IV: DPU Cross Section | Classification | Cross | Conf. Level
---|---|---|---
| runs | Section | 95%
| # | % | (cm2) | Lower | Upper
Correct runs | 2964 | 50.27% | - | - | -
Critical (C) | 46 | 0.78% | 8.29E-10 | 6.07E-10 | 1.11E-09
Tolerable (T) | 2886 | 48.95% | 5.20E-08 | 5.01E-08 | 5.39E-08
C+T errors | 2932 | 49.73% | 5.28E-08 | 5.09E-08 | 5.48E-08

Figure 4: Cache upsets per CPU Core.

Next, we discuss the impact of radiation effects on the DPU classification
accuracy. During the 3-hour experiment, the DPU performed 5896 classification
runs. Only a very small portion of runs (0.78%) resulted in misclassification.
Notice that the errors of the resnet50 classification results are categorized
similarly to [3] as a) Critical Errors, which lead to misclassification and b)
Tolerable Errors, where the errors observed in a result do not affect the
classification decision. Table IV presents the number of Critical Errors and
Tolerable Errors and their cross-sections. Based on the SEU vulnerability of
the configuration memory and the BRAM reported in [4] and the programmable
resources of the DPU (i.e. essential configuration bits and BRAMs used), we
calculate the rate of upsets affecting the DPU execution. These are 0.14 and
0.55 upsets per classification run in the configuration memory and BRAM
contents, respectively. This means that each classification run (which lasts
1.7 seconds), experienced, on average, 0.69 upsets. Notice that since
scrubbing was not supported in the experiment, the upsets in the configuration
memory were accumulated until the next reset cycle of the system. Moreover, we
estimated that more than one upsets were accumulated during the 38 seconds
boot and warm-up period of each reset cycle. Thus, for all classification
runs, the DPU circuit encountered more than one upsets in the PL memories. As
a worst-case analysis, we estimate that the AVF of the DPU accelerator is less
than 0.78%, assuming any single upset in the DPU leads to a critical error.

##  V Conclusion

The neutron radiation experiment demonstrated that the ECC and the
interleaving schemes integrated into the UltraScale+ MPSoC caches considerably
protect the software stack (OS, pre and post-processing of DNN data, data
movement) of the Xilinx Vitis AI DPU from radiation-induced SEUs. It was shown
that the most vulnerable part of the DPU is the logic implemented in the FPGA
PL. Due to the large amount of utilised PL resources by the DPU, we reasoned
that it is likely that the CNN application will be highly vulnerable to SEUs.
However, due to the inherent error resiliency of the neural network, only a
small portion of SEUs lead to classification errors, resulting in a
significantly small AVF. In future work, we aim to perform fault injection
experiments in the DPU to obtain a better understanding of its failure
mechanisms and propose efficient SEE mitigation approaches to further reduce
its AVF.

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