text:
\section{Introduction}
In this paper we will handle Nowicki's conjecture from its classical formulation to to some related problems. Our intent is producing a survey toward this theme and adding a new brick concerning Nowicki's conjecture in a Poisson algebra setting.

Our first aim is introducing the so-called Weitzenb\"ock derivations. We start considering a field $K$ of characteristic zero and a finite set of variables $X_m=\{x_1,\ldots,x_m\}$; we denote by $KX_m$ the vector space over $K$ with basis $X_m$ and we shall denote by $K[X_m]$ the (commutative) unitary polynomial algebra generated by $X_m$. Let $\delta$ be a non-zero nilpotent linear map of $KX_m$ that can be extended to a derivation of $K[X_m]$, then $\delta$ is called \textit{Weitzenb\"ock}. This name is due to the classical result of Weitzenb\"ock \cite{W}, dating back to 1932, which states that the algebra $K[X_m]^{\delta}=\text{ker}(\delta)$ of constants of the derivation $\delta$ in the algebra $K[X_m]$ is
finitely generated.
If $\delta$ is a Weitzenb\"ock derivation, then it is locally nilpotent and the linear map
$\text{exp}(\delta)$ acting on the vector space $KX_m$
is unipotent. Hence the algebra $K[X_m]^{\delta}$ of constants of $\delta$ is equal to the algebra of invariants
\[K[X_m]^{\text{exp}(\delta)}=\{x\in K[X_m]\mid \text{exp}(\delta)(x)=x\}.\]
The algebra of invariants of $\text{exp}(\delta)$ is equal to the algebra of invariants $K[X_m]^{G}$, where the group $G$ consists of all elements $\text{exp}(c\delta)$, $c\in K$.
As an abstract group, the group $G$ is isomorphic to the unitriangular group consisting of all $2\times 2$ upper triangular matrices with 1's on the diagonal.
The group $G$ can be embedded into the general linear group $GL_m(K)$ via the representation corresponding
to the Jordan normal form of $\delta$.
Thus we are allowed to study the algebra of constants $K[X_m]^{\delta}$ with the methods of the classical invariant theory.
The computational aspects of algebras of constants and invariant theory are depicted in part in the books by Nowicki \cite{N}, Derksen and Kemper \cite{DK},
and Sturmfels \cite{St}.

We can say more about Weitzenb\"ock derivations. For instance, take a Weitzenb\"ock derivation, then its Jordan normal form consists of Jordan cells with zero diagonals.
Certainly, Weitzenb\"ock derivations are in one-to-one correspondence with the partitions of $m$, up to a linear change of variables. This means there are essentially finite number of  Weitzenb\"ock derivations for a fixed dimension $m$.
In particular, let $m=2d$, $d\geq1$, and assume that the Jordan form of $\delta$ contains
the Jordan cells of size $2\times2$ only, i.e.,
\[
J(\delta)=\left(\begin{matrix}
0&1&\cdots&0&0\\
0&0&\cdots&0&0\\
\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&\cdots&0&1\\
0&0&\cdots&0&0
\end{matrix}\right).
\]
We assume that $\delta(x_{2j})=x_{2j-1}$, $\delta(x_{2j-1})=0$, $j=1,\ldots,d$.
Nowicki conjectured in his book \cite{N} (Section 6.9, page 76) that in this case the algebra $K[X_{2d}]^{\delta}$
is generated by $x_1,x_3.\ldots,x_{2d-1}$, and $x_{2i-1}x_{2j}-x_{2i}x_{2j-1}$, $1\leq i<j\leq d$. The last conjecture will be called throughout the text \textit{classical Nowicki's conjecture}.

Nowicki's conjecture was proved by several authors with different techniques.
The first published proof appeared in 2004 by Khoury \cite{K1} in his PhD thesis,
followed by his paper \cite{K2}, where he makes use of a computational approach involving Gr\"obner basis techniques. Derksen and Panyushev applied ideas of classical invariant
theory in order to prove the Nowicki conjecture but their proof remained unpublished. Later in 2009,
Drensky and Makar-Limanov \cite{DML} confirmed the conjecture
by an elementary proof from undergraduate algebra, without involving invariant theory. In this paper we shall highlight the key points of the proof given by Drensky and Makar-Limanov. Moreover, another simple short proof was given by Kuroda \cite{Kuroda} in 2009, too.
Bedratyuk \cite{Bed} proved the Nowicki conjecture by reducing it to a well known problem of classical invariant theory. We were able to find three more proofs of Nowicki's conjecture: two of them are cited in the paper by Kuroda \cite{Kuroda}, and the latest proof was given by Drensky in \cite{D2}.

In this paper we would like to attempt to give a complete account on the state of the art of Nowicki's conjecture presenting a sketch of the proof obtained by Drensky and Makar-Limanov. We will also treat a bit a generalization of Nowicki's conjecture (see Sections 3 and 4 in the text). We will also focus on recent developments of the following problem, strictly related to the classical Nowicki's conjecture, but with a scent of theory of algebras with polynomial identities (PI-algebras). Let us explain briefly our framework: let $\delta$ be a Weitzenb\"ock derivation of the free algebra $F_{m}(\mathfrak V)$ of rank $m$ in a given variety of algebras $\mathfrak V$ (or relatively free algebra of rank $m$ of $\mathfrak{V}$), then \[\textit{give an explicit set of generators of the algebra of constants $F_{m}^{\delta}(\mathfrak V)$.}\]
In \cite{DG} Drensky and Gupta studied Weitzenb\"ock derivations $\delta$ acting on $F_m(\mathfrak{V})$ proving that if the algebra $UT_2(K)$ of $2\times 2$ upper triangular matrices over a field $K$ of characteristic zero belongs to a variety $\mathfrak{V}$, then $F_m^\delta(\mathfrak{V})$ is not finitely generated whereas if $UT_2(K)$ does not belong to $\mathfrak{V}$, then $F_m^\delta(\mathfrak{V})$ is finitely generated by a result of Drensky in \cite{D1}.
Recently, Dangovski et al. \cite{DDF,DDF1} gave some new results in this direction. They showed that the algebra of constants $(F_d'(\mathfrak{V}))^\delta$ in the commutator ideal $F_d'(\mathfrak{V})$ of $F_d(\mathfrak{V})$
is a finitely generated module of $K[X_d]^{\delta}$ and $K[U_d,V_d]^{\delta}$, respectively.
Here $\delta$ acts on $U_d$ and $V_d$ in the same way as on $X_d$. In \cite{CF} the authors give an explicit set of generators of the algebra of constants of the variety generated by the infinite dimensional Grassmann algebra and of the free metabelian associative algebra. A good part of those result is presented explicitly in the text (Sections 5 and 6).

In the second part of the paper (Section 7) we will show a substantially new result. We consider the free metabelian Poisson algebra $P_{2n}$ of rank $2n$ over the field $K$ and we give out an explicit set of generators of its algebra of constants under the action of a Weitzenb\"ock derivation (see Theorems \ref{KB_3} and \ref{KB_4} in the text) as a vector space.

We would like to stress the fact that in the study of varieties of (not necessarily associative) algebras over a field of characteristic zero, the metabelian identities play a crucial role. It is well known the identities of $UT_2(K)$ follow from the metabelian identity of order 4, so every variety $\mathfrak{V}$ contains $UT_2(K)$ or satisfies the Engel identity $[x_2,x_1,\ldots,x_1]\equiv0$.

 \section{Preliminaries}
 Let $K$ be a field of characteristic 0.
We consider the commutative and unitary polynomial algebra $K[X_m]$ with generating set $X_m=\{x_1,\ldots,x_m\}$, $m\geq2$ over $K$. Of course, $K[X_m]$ is the free algebra of rank
$m$ in the variety of unitary commutative algebras over $K$. Now we consider the general linear group $GL_m(K)$ and $H\leq GL_m(K)$ a subgroup. We say a polynomial $p\in K[X_m]$ is $H$-\textit{invariant} if it is preserved under the action of each element of $H$.  The vector space $K[X_m]^H$ of all $H$-invariants turns out to be an algebra that is called the algebra of $H$-invariants.

The question whether the algebra $K[X_n]^H$ of invariants is finitely generated for every subgroup $H$ of $GL_m(K)$ is a special case of the \textbf{Hilbert's fourteenth problem} suggested by the German mathematician David Hilbert in 1900 at the International Congress of Mathematicians in Paris.

Although the answer to Hilbert's question turned out to be negative in general, as showed by Nagata in 1958, some remarkable affirmative cases have been handled as well.
Among them is the approach of Weitzenb\"ock, where he considered the \textbf{locally nilpotent} linear derivations $\delta$ of the algebra $K[X_m]$.
Then the kernel $\text{\rm ker}(\delta)=K[X_m]^{\delta}$ of the derivation $\delta$ is an algebra called the \textit{algebra of constants} of $K[X_m]$. He showed that the algebra $K[X_m]^{\delta}$ is finitely generated as an algebra and is equal to the algebra $K[X_m]^{UT_2(K)}$ of invariants of
the unitriangular group \[{UT_2(K)}=\{\text{exp}(c\delta)\mid c\in K\},\] as already said in the introduction.

The next question is \[\textit{What is the explicit form of the generators of $K[X_n]^\delta$?}\]
Nowicki conjectured in his book that the algebra $K[X_m,Y_m]^{\delta}$
is generated by \[\text{$x_1,\ldots,x_m$ and $u_{i,j}:=x_iy_j-x_jy_i$, $1\leq i<j\leq m$,}\]
assuming that $\delta(y_{i})=x_{i}$, $\delta(x_{i})=0$, $i=1,\ldots,m$, where $K[X_m,Y_m]$ denotes the commutative algebra generated by the set $X_m\cup Y_m$.

This is what we shall call \textit{Nowicki's conjecture} until the end of the paper.

Let us introduce now the noncommutative setting in which we shall generalize Nowicki's conjecture. Let $X=\{x_1,x_2,\ldots\}$ be an infinite set of variables and let $\mathcal{C}$ denote a special class of algebras (for instance, associative algebras, Lie algebras, commutative algebras, etc.). We can consider the free $\mathcal{C}$-algebra $\mathcal{C}\{X\}$ generated by $X$ over $K$ and we shall call its elements \textit{$\mathcal{C}$-polynomials} or simply \textit{polynomials}. A polynomial $f=f(x_1,\ldots,x_n)$ is said to be a \textit{polynomial identity} for the algebra $A\in\mathcal{C}$ if $f(a_1,\ldots,a_n)=0$ for any $a_i\in A$.

The set of all polynomial identities of a given algebra $A$ is denoted by $T_{\mathcal{C}}(A)$ (or $Id_{\mathcal{C}}(A)$) and is invariant under all $\mathcal{C}$-endomorphisms of $\mathcal{C}\{X\}$, i.e, it is a $T_{\mathcal{C}}$-ideal called the \textit{$T_{\mathcal{C}}$-ideal} of $A$. The algebra $\mathcal{C}\{X\}/Id_{\mathcal{C}}(A)$ is called \textit{relatively free $\mathcal{C}$-algebra} of $A$.

We can generalize Nowicki's conjecture to relatively free algebras in the following way:  let us consider the free algebra $\mathcal{C}\{X_m,Y_m\}$ endowed with the derivation $\delta$ such that $\delta(x_i)=0$ and $\delta(y_i)=x_i$ for any $i$.  Chosen $A\in\mathcal{C}$, $Id(A)$ is, of course, $\delta$-invariant, then $\delta$ induces a derivation $\delta$ on the relatively free algebra $\mathcal{C}_n(A):=\mathcal{C}\{X_m,Y_m\}/Id(A)$.
\[\textit{Is the algebra of constants of $\mathcal{C}_n(A)$ finitely generated?}\]\[\textit{ Are the generators uniformly looking?}\]

    Let $A\in\mathcal{C}$ and consider the class \[\mathfrak{V}(A)=\{B\in \mathcal{C}|Id(A)\subseteq Id(B)\}\] that is called \textit{variety of $\mathcal{C}$-algebras generated by $A$}.
    Here is the answer to the first question:
\begin{theorem}[Drensky \cite{D1}, Drensky and Gupta \cite{DG}]
Let $\mathfrak{V}:=\mathfrak{V}(A)$ be a variety of associative or Lie algebras, where $A$ is endowed with the derivation $\delta$. Then the algebra of constants of $A$ is not finitely generated if and only if $\mathfrak{V}$ contains $UT_2(K)$, the algebra of upper triangular $2\times 2$ matrices with entries from $K$ .
\end{theorem}

\section{Drensky and Makar-Limanov's approach}
Herein we shall depict sketches of the proof of Nowicki's conjecture as showed by Drensky and Makar-Limanov in \cite{DML}.

Let $X_m=\{x_1,\ldots,x_m\}$, $Y_m=\{y_1,\ldots,y_m\}$ be two disjoint sets of commutative variables.
    Let $X'=\{x_1,\ldots,x_{m-1}\}$, $Y'=\{y_1,\ldots,y_{m-1}\}$, $U'=\{u_{i,j}|1\leq i,j\leq m-1\}$   (where $u_{i,j}=x_iy_j-x_jy_i$), $X=X_m$, $Y=Y_m$ and consider a constant $f=f(X',Y',x_m,y_m)$. Of course $f$ can be taken to be homogeneous in $(X,Y)$ and we carry on by induction on $m$ and on the total degree in $x_m,y_m$.

    \begin{remark}
    For a monomial $v\in K[X,Y]$ of degree $(d_1,d_2)$ with respect to $(X,Y)$, $\delta(v)$ is of degree $(d_1+1,d_2-1)$ with respect to $(X,Y)$.
    \end{remark}

    If $\underline{n=1}$, then by the previous remark, we get $K[x_1,y_1]^\delta=K[x_1]$ and we are done because in this case $u_{1,1}=0$.

We consider \underline{$m>1$} and we may suppose $\deg_{x_m,y_m}f>0$. We write \[f=(a_py_m^p+a_{p-1}x_my_m^{p-1}+\cdots+a_1x_m^{p-1}y_m+a_0x_m^{p})x_m^q,\] where $a_j\in K[X',Y']$, $a_p\neq0$.

\begin{remark}
Since both $f$ and $x_m^q$ are constants, the same holds for $f/x_m^q$. Hence we may assume $q=0$.
\end{remark}

\begin{remark}
Because \[0=\delta(f)\]\[=\delta(a_p)y_m^p+(pa_p+\delta(a_{p-1}))x_my_m^{p-1}+\cdots+(a_1+\delta(a_0))x_m^p\] we have $a_p$ is a constant and, by induction, has the form \[a_p=\sum x_{s_1}\cdots x_{s_k}b_s(U').\]
\end{remark}

It is easy to note that $\deg_X(f)\geq p+\deg_{Y'}(a_p)$, then, by the last remark, we may write $f$ as \[f=\sum x_{s_1}\cdots x_{s_p}c_s(X',U')y_m^p+\sum_{i=0}^{p-1}a_i(X',Y')x_m^{p-i}y_m^i.\]

\begin{remark}
This is very important: \[x_{s_j}y_m=(x_{s_j}y_m-x_my_{s_j})+x_my_{s_j}=u_{s_j,m}+x_my_{s_j}.\]
\end{remark}

    Then we get \[f=\sum c_s(X',U')\prod_{j=1}^pu_{s_j,m}+x_mf_1(X',Y',x_m,y_m)\] and $\deg_{x_m,y_m}(f_1)<\deg_{x_m,y_m}(f)$ but $f$, $x_m$, $c_s(X',U')\prod_{j=1}^pu_{s_j,m}$ are all constants, then $f_1$ is a constant and we are done by induction.

\section{Generalized Nowicki's conjecture}

    Let us take an integral domain $D$ over $K$, a set of variables $Y_m:=\{y_1,\ldots,y_m\}$ and the algebra $A:=D[Y_m]$. A derivation $\Delta$ on $A$ is said \textit{elementary} if $\Delta(D)=0$ and $\Delta(y_i)\in D$ for any $i$. \ We consider also the following objects  \[u_{i,j}:=\left|\begin{array}{cc}
       \Delta(y_i)  & y_j \\
       \Delta(y_j)  & y_i
    \end{array}\right|.\]
They are called \textit{determinants} and they belong to $D[y_m]^\Delta$.

\begin{question}[Generalized Nowicki's conjecture]
Is the $D$-algebra $D[Y_m]^\Delta$ generated by the determinants?
\end{question}

    Of course the previous question is precisely the Nowicki's conjecture when $\Delta$ is such that $\Delta(D)=0$ and $\Delta(y_i)=x_i$ for any $i$.
    \ In the first proof of Nowicki's conjecture by Khoury, the author considered an elementary derivation $\Delta$ sending $\Delta(y_i)=x_i^{m_i}$. Then the algebra of constants $K[X_m,Y_m]^\Delta$ is generated by $X_m$ and the $u_{i,j}$'s.
    The more general result in this direction is due to Kuroda:
    \begin{theorem}
    Let $\Delta$ be an elementary derivation of $D[Y_m]$ such that $\Delta(y_i)=f_i$, for any $i$. Suppose further the $f_i$'s are algebraically independent over $K$. Then $D[Y_m]^\Delta$ is generated by the determinants if $D$ is \textbf{flat} over $K[f_1,\ldots,f_m]$.
    \end{theorem}

  Notice that this result is deeply different from the original version of Nowicki's conjecture.\ \textit{Nowicki's conjecture can be restated in the language of \textbf{classical invariant theory}} but Kuroda's result in case the $f_i$'s are not linear!

  Anyway, using Kuroda's result, Drensky \cite{D2000} proved the next:

  \begin{theorem}
  Let $\Delta$ be an elementary derivation of $D[Y_m]$ such that $\Delta(y_i)=f_i=f_i(x_i)$, for any $i$, where the $f_i$'s are non-constant polynomials. Then $D[Y_m]^\Delta$ is generated by $X_m$ and the determinants.
  \end{theorem}

  Notice also that in all of the result above, the determinants generating the algebra of constants are not free generators. A prominent part of those works deals with finding a free set of generators.

\section{The free metabelian associative algebras}

We will focus on another generalization of Nowicki's conjecture. The environment now is that of relatively free algebras. The complete proofs of the results presented in the next two section can be found in the paper \cite{CF}.

We start off our investigation with the relatively free algebra of the associative metabelian algebra.
Let $K$ be a field of characteristic zero, $P_{2d}$ be the free unitary associative algebra of rank ${2d}$
over $K$, $P_{2d}'=P_{2d}[P_{2d},P_{2d}]P_{2d}$ be its commutator ideal generated by all elements of the form
\[
[x,y]=xy-yx, \quad x,y\in P_{2d}.
\]
Let us consider the quotient
algebra $F_{2d}=P_{2d}/(P_{2d}')^2$. The algebra $F_{2d}$ is the free algebra of rank $2d$ in the
variety of all associative algebras satisfying the polynomial identity $[x,y][z,t]\equiv0$.
Let the free associative metabelian algebra $F_{2d}$ be generated by $X_{2d}=\{x_1,\ldots,x_{2d}\}$.
We assume that all Lie commutators are left normed; i.e.,
\[
[x,y,z]=[[x,y],z], \quad x,y,z\in F_{2d}.
\]
The commutator ideal $F_{2d}'$ of $F_{2d}$
has the following basis (see e.g. \cite{B, D})
\[
x_1^{\xi_1}\cdots x_{2d}^{\xi_{2d}}[x_{i},x_{j},x_{j_1},\ldots,x_{j_m}],\quad \xi_i\geq 0, \quad i>j\leq j_1\leq\cdots\leq j_m\leq 2d.
\]
As a consequence of the metabelian identity in $F_{2d}'$ we have the following identity:
\[
x_{i_{\pi(1)}}\cdots x_{i_{\pi(m)}}[x_i,x_j,x_{j_{\sigma(1)}},\ldots,x_{j_{\sigma(n)}}]
\equiv x_{i_1}\cdots x_{i_m}[x_{i},x_{j},x_{j_1},\ldots,x_{j_n}],
\]
for any permutation $\pi\in S_m$ and $\sigma\in S_n$. Thus the commutator ideal
$F_{2d}'$ can be ``seen'' as a module of polynomial algebra from both sides via the associative (left side) and Lie (right side) multiplication.

We recall now some of the results and constructions given in \cite{DDF1}. Let
$U_{2d}=\{u_1,\ldots,u_{2d}\}$ and $V_{2d}=\{v_1,\ldots,v_{2d}\}$ be two sets of commuting variables
and let $K[U_{2d},V_{2d}]$ be the polynomial algebra acting on $F_{2d}'$ as follows.
If $f\in F_{2d}'$, then
\[
fu_i=x_if, \quad fv_i=[f,x_i], \quad i=1,\ldots,2d.
\]
This action defines a $K[U_d,V_d]$-module structure on the vector space $F_{2d}'$.

We are going to construct a wreath product which is the same as the one used in \cite{DDF1}. It is a particular case
of the construction of Lewin \cite{L} given in \cite{DG} and is similar to the construction of
Shmel'kin \cite{Sh} in the case of free metabelian Lie algebras as appeared in \cite{DDF}.

Let $Y_{2d} = \{y_1, \ldots, y_{2d}\}$ and $V_{2d}'=\{v_1',\ldots,v_{2d}'\}$ be sets of commuting variables and let
$A_{2d}=\{a_1,\ldots,a_{2d}\}$ in $2d$ variables.
Now let $M_{2d}$ be the free $K[U_{2d},V_{2d}']$-module generated by $A_{2d}$
equipped with with trivial multiplication
$M_{2d}\cdot M_{2d}=0$. We endow $M_{2d}$ with a structure of a free $K[Y_{2d}]$-bimodule
structure via the action
\[
y_ja_i = a_iu_j, \quad a_iy_j = a_iv_j', \quad i,j = 1, \ldots, 2d.
\]
The wreath product $W_{2d}=K[Y_{2d}]\rightthreetimes M_{2d}$
is an algebra satisfying the metabelian identity. As well as in \cite{L} $F_{2d}$ can be embedded into $W_{2d}$. In fact we have the following result.

\begin{proposition}\label{embedding}
The mapping $\varepsilon :x_j\to y_j+a_j$, $j=1,\ldots,2d$, extends to an embedding $\varepsilon$ of $F_{2d}$ into $W_{2d}$.
\end{proposition}

We identify $v_i=v_i'-u_i$, $i=1,\ldots,2d$, and get
\begin{equation}\label{commutator element}
\varepsilon(x_{i_1}\cdots x_{i_m}[x_{i},x_{j},x_{j_1},\ldots,x_{j_n}])
=(a_iv_j-a_jv_i)v_{j_1}\cdots v_{j_n}u_{i_1}\cdots u_{i_m}
\end{equation}
Thus we may assume that $M_{2d}$ is a free $K[U_{2d},V_{2d}]$-module. Clearly the commutator ideal
$F_{2d}'$ is embedded into $M_{2d}$, too.
An element $\sum a_if(U_{2d},V_{2d})\in M_{2d}$
is an image of some element from $F_{2d}'$ if and only if $\sum v_if(U_{2d},V_{2d})=0$,
as a consequence of (\ref{commutator element}).

Let $\delta$ be the Weitzenb\"ock derivation of $F_{2d}$ acting on the
variables $U_{2d},V_{2d}$, as well as explained in the Introduction on $X_{2d}$. By \cite{DDF1} we know that
the vector space $M_{2d}^{\delta}$ of the constants of $\delta$ in the $K[U_{2d},V_{2d}]$-module $M_{2d}$
is a $K[U_{2d},V_{2d}]^{\delta}$-module.
The following results are particular cases of \cite{D1} (Proposition 3) and \cite{DDF1}.

\begin{theorem}\label{theorem for finite generation}
The vector spaces $(F_{2d}')^{\delta}$ and $M_{2d}^{\delta}$ are finitely generated $K[U_{2d},V_{2d}]^{\delta}$-modules.
\end{theorem}

In the sequel we shall write down an explicit set of generators for the algebra of constants $K[U_{2d},V_{2d}]^{\delta}$ as a consequence of
the Nowicki conjecture \cite{N} (proved in \cite{K1, K2, DML, Bed, Kuroda}), and one of the results by Drensky and Makar-Limanov \cite{DML}.
Let the Weitzenb\"ock derivation $\delta$ act on $X_{2d},A_{2d},U_{2d},V_{2d}$
by the rule
\[
\delta(x_{2i})=x_{2i-1}, \delta(x_{2i-1})=0,\quad \delta(a_{2i})=a_{2i-1}, \delta(a_{2i-1})=0,
\]
\[
\delta(u_{2i})=u_{2i-1}, \delta(u_{2i-1})=0,\quad\delta(v_{2i})=v_{2i-1}, \delta(v_{2i-1})=0,
\]
for $i=1,\ldots,d$. Then the algebra of constants $K[U_{2d},V_{2d}]^{\delta}$ is generated by
\[
u_1,u_3,\ldots,u_{2d-1},v_1,v_3,\ldots,v_{2d-1}
\]
and the determinants
\[
\alpha_{pq}=u_{2p-1}u_{2q}-u_{2p}u_{2q-1}=\begin{vmatrix}
u_{2p-1} & u_{2p} \\
u_{2q-1} &u_{2q}
\end{vmatrix},\quad 1\leq p<q\leq d,
\]
\[
\beta_{pq}=v_{2p-1}v_{2q}-v_{2p}v_{2q-1}=\begin{vmatrix}
v_{2p-1} & v_{2p} \\
v_{2q-1} &v_{2q}
\end{vmatrix},\quad1\leq p<q\leq d,
\]
\[
\gamma_{pq}=u_{2p-1}v_{2q}-u_{2p}v_{2q-1}=\begin{vmatrix}
u_{2p-1} & u_{2p} \\
v_{2q-1} &v_{2q}
\end{vmatrix}, \quad p,q=1,\ldots,d,
\]
with the following defining relations
\begin{equation} \label{S1}
u_{2i-1}\alpha_{jk}-u_{2j-1}\alpha_{ik}+u_{2k-1}\alpha_{ij}=0,\quad 1\leq i<j<k\leq d,
\end{equation}
\begin{equation} \label{S2}
u_{2i-1}\gamma_{jk}-u_{2j-1}\gamma_{ik}+v_{2k-1}\alpha_{ij}=0,\, 1\leq i<j\leq d,\, 1\leq k\leq d,
\end{equation}
\begin{equation} \label{S3}
u_{2i-1}\beta_{jk}-v_{2j-1}\gamma_{ik}+v_{2k-1}\gamma_{ij}=0,\, 1\leq i\leq d,\, 1\leq j<k\leq d,
\end{equation}
\begin{equation} \label{S4}
v_{2i-1}\beta_{jk}-v_{2j-1}\beta_{ik}+v_{2k-1}\beta_{ij}=0,\quad 1\leq i<j<k\leq d,
\end{equation}
\begin{equation}  \label{R1}
\alpha_{ij}\alpha_{kl}-\alpha_{ik}\alpha_{jl}+\alpha_{il}\alpha_{jk}=0,\, 1\leq i<j<k<l\leq d,
\end{equation}
\begin{equation}  \label{R2}
\alpha_{ij}\gamma_{kl}-\alpha_{ik}\gamma_{jl}+\gamma_{il}\alpha_{jk}=0,\, 1\leq i<j<k\leq d,\, 1\leq l\leq d,
\end{equation}
\begin{equation}  \label{R3}
\alpha_{ij}\beta_{kl}-\gamma_{ik}\gamma_{jl}+\gamma_{il}\gamma_{jk}=0,\,1\leq i<j\leq d,\,1\leq k<l\leq d,
\end{equation}
\begin{equation}  \label{R4}
\gamma_{ij}\beta_{kl}-\gamma_{ik}\beta_{jl}+\gamma_{il}\beta_{jk}=0,\,1\leq i\leq d,\,1\leq j<k<l\leq d,
\end{equation}
\begin{equation}  \label{R5}
\beta_{ij}\beta_{kl}-\beta_{ik}\beta_{jl}+\beta_{il}\beta_{jk}=0,\quad 1\leq i<j<k<l\leq d.
\end{equation}
The vector space $K[U_{2d},V_{2d}]^{\delta}$
has a {\it canonical} linear basis consisting of the elements of the form
\begin{equation}  \label{B1}
v_{2i_1-1}\cdots v_{2i_m-1}\beta_{p_1q_1}\cdots \beta_{p_rq_r}
\gamma_{p'_1q'_1}\cdots \gamma_{p'_sq'_s}\alpha_{p''_1q''_1}\cdots \alpha_{p''_tq''_t}u_{2j_1-1}\cdots u_{2j_n-1}
\end{equation}
such that among the generators $\beta_{pq}$, $\gamma_{p'q'}$, and $\alpha_{p''q''}$ there is no \textit{intersection},
and no one \textit{covers} $v_{2i_k-1}$ or $u_{2j_l-1}$.

Note that each $\beta_{pq}$, $\gamma_{p'q'}$, and $\alpha_{p''q''}$ is identified with the open interval
$(p+d,q+d)$, $(p',q'+d)$, and $(p'',q'')$, respectively, on the real line. The generators \textit{intersect} each other if the corresponding
open intervals have a nonempty intersection and are not contained in each other. On the other hand, the generators
$v_{2i-1}$ and $u_{2j-1}$ are identified with the points $i+d$ and $j$, respectively.
We say also that a generator among $\beta_{pq}$, $\gamma_{p'q'}$, or $\alpha_{p''q''}$ \textit{covers}
$v_{2i-1}$ or $u_{2j-1}$ if the corresponding open interval covers the corresponding point.

The pairs of indices are ordered in the following way: $p_1 \leq \cdots \leq p_r$ and if $p_{\xi}=p_{{\xi}+1}$, then $q_{\xi}\leq q_{{\xi}+1}$;
$p'_1 \leq \cdots \leq p'_s$ and if $p'_{\mu}=p'_{{\mu}+1}$, then $q'_{\mu}\leq q'_{{\mu}+1}$;
$p''_1 \leq \cdots \leq p''_t$ and if $p''_{\sigma}=p''_{{\sigma}+1}$, then $q''_{\sigma}\leq q''_{{\sigma}+1}$;
and $i_1 \leq \cdots \leq i_m$, $j_1 \leq \cdots \leq j_n$.

In order to detect the constants in the commutator ideal $F_{2d}'$ it is sufficient to work in
the $K[U_{2d},V_{2d}]^{\delta}$-submodule $C_{2d}^{\delta}$ of $M_{2d}^{\delta}$, which is generated by
\[
a_1,a_3,\ldots,a_{2d-1}
\]
and the determinants
\[
w_{pq}=a_{2p-1}v_{2q}-a_{2q}v_{2p-1}=\begin{vmatrix}
a_{2p-1} & a_{2q} \\
v_{2p-1} &v_{2q}
\end{vmatrix}, \quad p,q=1,\ldots,d.
\]
and spanned as a vector space on the elements of the form
\begin{equation}  \label{sp1}
a_{2i_0-1}v_{2i_1-1}\cdots v_{2i_m-1}\beta_{p_1q_1}\cdots \beta_{p_rq_r}
\gamma_{p'_1q'_1}\cdots \gamma_{p'_sq'_s}\alpha_{p''_1q''_1}\cdots \alpha_{p''_tq''_t}u_{2j_1-1}\cdots u_{2j_n-1}
\end{equation}
\begin{equation}  \label{sp2}
w_{p_0q_0}v_{2i_1-1}\cdots v_{2i_m-1}\beta_{p_1q_1}\cdots \beta_{p_rq_r}
\gamma_{p'_1q'_1}\cdots \gamma_{p'_sq'_s}\alpha_{p''_1q''_1}\cdots \alpha_{p''_tq''_t}u_{2j_1-1}\cdots u_{2j_n-1}
\end{equation}
for each $i_0,p_0,q_0=1\ldots,d$.

We also have the following relations in the algebra $C_{2d}^{\delta}$ as a consequence of \cite{DML}.
\begin{equation} \label{S}
a_{2i-1}\beta_{jk}-w_{ik}v_{2j-1}+w_{ij}v_{2j-1}=0,\quad 1\leq i\leq d, \,1\leq j<k\leq d,
\end{equation}
\begin{equation}  \label{R}
w_{ij}\beta_{kl}-w_{ik}\beta_{jl}+w_{il}\beta_{jk}=0,\quad 1\leq i\leq d, \,1\leq j<k<l\leq d.
\end{equation}

We denote by $L$ the $K[U_{2d},V_{2d}]^{\delta}$-submodule of $C_{2d}^{\delta}$
generated by the following elements
\begin{equation} \label{g1}
w_{ii},\quad \quad 1\leq i\leq d,
\end{equation}
\begin{equation} \label{g2}
w_{ij}+w_{ji},\quad 1\leq i<j\leq d,
\end{equation}
\begin{equation}\label{g3}
a_{2i-1}v_{2j-1}-a_{2j-1}v_{2i-1},\quad 1\leq i<j\leq d,
\end{equation}
\begin{equation}\label{g4}
a_{2i-1}\beta_{pq}-w_{pq}v_{2i-1},\, 1\leq i\leq d, \, 1\leq p<q\leq d,
\end{equation}
\begin{equation}\label{g5}
w_{ij}\beta_{pq}-w_{pq}\beta_{ij}, \, 1\leq i<j\leq d,  \, 1\leq p<q\leq d,
\end{equation}
\begin{equation} \label{g6}
a_{2i-1}\beta_{jk}-a_{2j-1}\beta_{ik}+a_{2k-1}\beta_{ij},\quad 1\leq i<j<k\leq d,
\end{equation}
\begin{equation}  \label{g7}
w_{kl}\gamma_{ij}-w_{jl}\gamma_{ik}+w_{jk}\gamma_{il},\,1\leq i\leq d,\,1\leq j<k<l\leq d.
\end{equation}
\begin{equation} \label{g8}
w_{jk}u_{2i-1}-a_{2j-1}\gamma_{ik}+a_{2k-1}\gamma_{ij},\, 1\leq i\leq d,\, 1\leq j<k\leq d.
\end{equation}
One can observe that the generating elements (\ref{g1})-(\ref{g8}) of $L$ are
the images of some elements in the commutator ideal $F_{2d}'$ of the free associative algebra $F_{2d}$.

\begin{theorem}
The $K[U_{2d},V_{2d}]^{\delta}$-submodule $L$ of $C_{2d}^{\delta}$ consists of all commutator elements in $C_{2d}^{\delta}$;
i.e., the images in $(F_{2d}')^{\delta}\subset F_{2d}'$.
\end{theorem}

As a consequence of the previous result it is possible to get the full list of generators of $(F'_{2d})^\delta$ as a $K[U_{2d},V_{2d}]$-module.

\begin{corollary}\label{generators in the ideal}
The $K[U_{2d},V_{2d}]^{\delta}$-module $(F_{2d}')^{\delta}$ is generated by the following polynomials:
\[
g_1(i)=[x_{2i-1},x_{2i}],\quad \quad 1\leq i\leq d,
\]
\[
g_2(i,j)=[x_{2i-1},x_{2j-1}],\quad 1\leq i<j\leq d,
\]
\[
g_3(i,j)=[x_{2i-1},x_{2j}]+[x_{2j-1},x_{2i}],\quad 1\leq i<j\leq d,
\]
\[
g_4(i,p,q)=[x_{2i-1},x_{2p-1},x_{2q}]-[x_{2i-1},x_{2p},x_{2q-1}],\, 1\leq i\leq d, \, 1\leq p<q\leq d,
\]
\[
g_5(i,j,k)=[x_{2i-1},x_{2j-1},x_{2k}]-[x_{2i-1},x_{2k-1},x_{2j}]+[x_{2j-1},x_{2k-1},x_{2i}],
\]
\[
1\leq i<j<k\leq d,
\]
\[
g_6(i,j,p,q)=[x_{2i-1},x_{2p-1},x_{2j},x_{2q}]+[x_{2i},x_{2p},x_{2j-1},x_{2q-1}]
\]
\[
-[x_{2i-1},x_{2p},x_{2j},x_{2q-1}]-[x_{2i},x_{2p-1},x_{2j-1},x_{2q}],
\]
\[
1\leq i<j\leq d,\,1\leq p<q\leq d,
\]
\[
g_7(i,j,k,l)=x_{2i}[x_{2j-1},x_{2k-1},x_{2l}]+x_{2i-1}[x_{2j},x_{2k},x_{2l-1}]
\]
\[
-x_{2i}[x_{2j-1},x_{2k},x_{2l-1}]-x_{2i-1}[x_{2j},x_{2k-1},x_{2l}]
\]
\[
1\leq i\leq d,\,1\leq j<k<l\leq d,
\]
\[
g_8(i,j,k)=x_{2i}[x_{2j-1},x_{2k-1}]-x_{2i-1}[x_{2j},x_{2k-1}],
\]
\[
1\leq i\leq d,\, 1\leq j<k\leq d.
\]
\end{corollary}

We have to add for the generating set of the whole algebra
$(F_{2d})^\delta$ the constants $x_{2i-1}$ and $x_{2i-1}x_{2j}-x_{2i}x_{2j-1}$, $1\leq i<j\leq d$,
which are needed for the generation
of the factor algebra of $(F_{2d})^\delta$ modulo the commutator ideal of $F_{2d}$.
These generators are the ones lifted from the algebra $K[X_{2d}]^{\delta}$ of constants
of the polynomial algebra to the algebra $(F_{2d})^\delta$  by the fact stated in Corollary 4.3 of the paper \cite{DG} by Drensky and Gupta.

The following result gives an infinite generating set of the subalgebra of constants $(F_{2d})^\delta$ of the free metabelian associative algebra $F_{2d}$ as an algebra.

\begin{corollary}
The algebra $(F_{2d})^\delta$ of the constants is generated by
\[
x_1,x_3,\ldots,x_{2d-1},
\]
\[
x_{2i-1}x_{2j}-x_{2i}x_{2j-1},
\]
\[
g_1f_1,\ldots,g_8f_8,
\]
where $1\leq i<j\leq d$, $f_1,\ldots,f_8\in K[U_{2d},V_{2d}]^{\delta}$, and $g_i$s are as in Corollary \ref{generators in the ideal}.
\end{corollary}

\section{The variety $\mathcal{G}$ generated by the Grassmann algebra}
In this section we shall show results concerning Nowicki's conjecture of the relatively free algebra of the infinite dimensional Grassmann algebra.

The variety $\mathcal{G}$ consists of all associative unitary algebras satisfying
the polynomial identity $[z_1,z_2,z_3]\equiv0$.
We shall set $F_{2l}:=F_{2l}(\mathcal{G})$ for $0\leq l\leq d$.
As noted in the Introduction, $\mathcal{G}$ is the variety generated by
the infinite dimensional Grassmann algebra $G$ and $F_{2d}$ coincides
with the $2d$-generated relatively free algebra in the variety $\mathcal{G}$.
The identities of $G$ and related topics have been studied by several authors.
See for example the paper \cite{KR} by Krakovski and Regev about the ideal of polynomial identities of $G$.

Let $X=\{x_1,x_2,\ldots,x_{d}\}$, $Y=\{y_1,y_2,\ldots,y_{d}\}$ be two disjoint sets of variables. Of course the relatively free algebra of $\mathfrak{G}$ of rank $2d$ in the variables from $U:=X\cup Y$ is isomorphic to $F_{2d}$ and we consider the following order inside $U$: \[x_1<y_1<x_2<\cdots<x_d<y_d.\]

Moreover we say a polynomial $f$  is \textit{homogeneous in the set of variables} $S=\{u_{1},\ldots,u_s\}$ if for each monomial $m$ appearing in $f$ we have $\sum_{i=1}^s\deg_{u_i}m$ is the same.

It is well known (see for example Theorem 5.1.2 of \cite{D}) $F_{2d}$ has a basis consisting of all
\[
x_1^{a_1}y_1^{b_1}\cdots x_{d}^{a_{2d}}y_{d}^{b_{2d}}[u_{i_1},u_{i_2}]\cdots[u_{i_{2c-1}},u_{i_{2c}}],\]\[a_i,b_j\geq0,\ ,u_{i_l}\in U,\ u_{i_1}<u_{i_2}<\cdots<u_{i_{2c}},\ c\geq0.\]

We recall the identity $[z_1,z_2,z_3]\equiv0$ implies the identity
\begin{equation}\label{proof Grass}
[z_1,z_2][z_3,z_4]\equiv-[z_1,z_3][z_2,z_4]
\end{equation}
in $F_{2d}$.

Consider the following Weitzenb\"ock derivation $\delta$ of $F_{2d}$ acting on
 $U$ such that
\[
\delta(y_{i})=x_{i}, \ \delta(x_i)=0,\quad 1\leq i\leq d.
\]

Let $\alpha=(\alpha_1,\cdots,\alpha_{d-1})\in K^{d-1}$ and consider the algebra endomomorphism $\phi_{\alpha}$ of $F_{2d}$ such that \[\phi_\alpha(x_i)=x_i,\ \ \phi_\alpha(y_i)=y_i,\ \ i=1,\ldots,d-1,\]
\[\phi_\alpha(x_d)=\sum_{i=1}^{d-1}\alpha_ix_i,\ \ \phi_\alpha(y_d)=\sum_{i=1}^{d-1}\alpha_iy_i.\] Notice that $\phi_\alpha$ commutes with $\delta$. Hence if $f\in F_{2d}^\delta$, then $\phi_\alpha(f)\in F_{2d-2}^\delta$ too. Moreover, if $f\in F_{2d}$ is homogeneous with respect to the set $\{x_d,y_d\}$ and $\phi_\alpha(f)=0$ for some non-zero $\alpha\in K^{d-1}$, then $f$ is in the left ideal generated by \[\omega_\alpha:=\left(\sum_{i=1}^{d-1}\alpha_ix_i\right)y_d-\left(\sum_{i=1}^{d-1}\alpha_iy_i\right)x_d,\ \ \ [\omega_\alpha,u],\ \ u\in U,\]\[\mu_\alpha:=\left[x_d,\sum_{i=1}^{d-1}\alpha_ix_i\right],\ \ \nu_\alpha:=\left[y_d,\sum_{i=1}^{d-1}\alpha_iy_i\right].\]

We define the following objects inside
$F_{2d}$:
\[
v_{ij}:=x_iy_j-y_ix_j, \quad 1\leq i,j\leq d,
\]
\[w_{ijk}:=y_i[x_j,x_k]-x_i[y_j,x_k], \quad 1\leq i,j,k\leq d,\]
\[z_{ijkl}:=y_i[x_j,v_{k,l}]-x_i[y_j,v_{k,l}], \quad 1\leq i,j,k,l\leq d,\]

Notice that the $v_{ij}$'s, the $w_{ijk}$'s and the $z_{ijkl}$'s are constants of $F_{2d}$ and starting from these objects we shall construct three subsets of elements of $F_{2d}$. We set \[V:=\{v_{ij}|1\leq i,j\leq d\},\]\[W_0:=\{w_{ijk}|1\leq i,j\leq k\leq d\}\] and \[Z_0:=\{z_{ijkl}|1\leq i\leq j\leq k\leq l\leq d\}.\] Suppose $s>0$ and we set \[W_s:=\{y[x,w]-x[y,w]|w\in W_{s-1},\ x\in X,\ y\in Y\},\] \[Z_s:=\{y[x,z]-x[y,z]|z\in Z_{s-1}\ x\in X,\ y\in Y\}.\]
As above, both the $W_s$'s and the $Z_s$'s are subsets of constants. Moreover it can be easily seen for $s\geq d$ both $V_s$ and $W_s$ are 0. We shall denote by $\mathcal{C}$ the algebra generated by the non-zero $X,V,W_l,Z_l$. Here we have the main result. We recall the proof of the following follows the main steps of the proof of the classical Nowicki's conjecture presented in Section 3. We also point out this proof cannot be adapted to algebras having a non $T$-prime ideal of their polynomial identities.

\begin{theorem}
The algebra of constants $F^\delta_{2d}$ is finitely generated as an algebra, and its generators are:
\begin{equation}\label{grass1} X;\end{equation}\begin{equation}\label{grass2} V,\end{equation}
\begin{equation}\label{grass3} W_{l},\ l\leq d-1; \end{equation}
\begin{equation}\label{grass4} Z_{l},\ l\leq d-1. \end{equation}
\end{theorem}

\begin{proof}
It is easy to see that the elements from (\ref{grass1})-(\ref{grass4}) are constants of ${F}_{2d}$.

Let $f=f(X',Y',x_d,y_d)\in F_{2d}^\delta$, then we may assume $f$ being homogeneous in $U$ and in $\{x_d,y_d\}$. We shall prove the theorem by induction on $d$ and on the total degree with respect to the set $\{x_d,y_d\}$. Suppose $d=1$, then \[f=\sum_{k=0}^n\alpha_k x^ky^{n-k}+\sum_{j=1}^{n-1}\beta_j x^{j-1}y^{n-j-1}[x,y].\] This means \[0=\delta(f)=\sum_{k=0}^{n-1}\alpha_k(n-k)x^{k+1}y^{n-k-1}\]\[+[x,y]\sum_{k=1}^{n-2}(n-k-1)\left(\alpha_k\frac{n-k}{2}x^{k-1}y^{n-k-1}+\beta_k x^ky^{n-k-2}\right).\] The previous relation gives us $\alpha_k=0$ for $k\leq n-1$, then $\beta_j=0$ for $1\leq j\leq n-2$. Thus the only possibility is $f=\alpha_nx^n+\beta_{n-1}x^{n-2}[x,y]$ and we are done.

Assume now $d>1$ and the result true for $F_{2d-2}$. We set $U'=X'\cup Y'$. If $v\in F_{2d-2}^\delta$ is homogeneous in $U$, then we shall study only the case $\deg_{x_d,y_d}(f)>0$.

Remark we can write $f$ in the following way:\[f=a_py_d^p+\sum_{u_0}a'_{p,u_0}[u_0,y_d]y_d^{p-1}+a_{p-1}x_dy_d^{p-1}+\sum_{u^1_1}a'_{p-1,u^1_1}[u^1_1,x_d]y_d^{p-1}\]\[+\sum_{u^2_1}a^{''}_{p-1,u^2_1}[u^2_1,y_d]x_dy_d^{p-2}+a^{'''}_{p-1}[x_d,y_d]y_d^{p-2}+\cdots+a_0x_d^p+\sum_{u_p}a'_{0}[u_p,x_d]x_d^{p-1},\] where the $u^i_j$'s are in $U$ whereas the $a_j$'s, the $a'_{r,u^i_j}$'s, the $a^{''}_{s,u^i_j}$ and the $a^{'''}_{t,u^i_j}$'s all belong to $F_{2d-2}$. Deriving $f$ we get $\delta(a_p)=0$, hence:

\begin{equation}\label{obs1}
a_p\in F_{2d}^\delta.
\end{equation}
We also have $\frac{p(p-1)}{2}a_p+\delta(a^{'''}_{p-1})=0$, then by (\ref{obs1}) we get $\delta(a^{'''}_{p-1})=0$. This means:
\begin{equation}\label{obs2}
a^{'''}_{p-1}\in F_{2d}^\delta.
\end{equation}
Suppose now $u_0=y_s$, $s\leq d-1$, then $a'_{p,u_0}=0$. So we can assume $u_0=x_s$, $s\leq d-1$. In this case we get $\delta(a'_{p,x_s})=0$, i.e.:
\begin{equation}\label{obs3}
a'_{p,x_s}\in F_{2d}^\delta.\end{equation}
By (\ref{obs1}), (\ref{obs2}) and (\ref{obs3}), using induction, we get \[a_p=\sum b_1(X',V',W',Z'),\]\[a^{'''}_{p-1}=\sum b_2(X',V',W',Z'),\]\[a'_{p,x_i}=\sum b_{3,i}(X',V',W',Z').\]

If $\phi_\alpha(f)=0$ for all $\alpha=(\alpha_1,\ldots,\alpha_{d-1})$, then we get $d=2$ and $f$ is in the left ideal generated by $v_{12}$ and $[v_{12},u]$, $u\in U$. Hence $f=gv_{12}+\sum_{u\in U}g_u[v_{12},u]$ and we apply inductive arguments to $g$ and $g_u$'s and we are done. Now we consider the case $\phi_\alpha(f)\neq0$ for some non-zero $\alpha\in K^{n-1}$. The next two remarks are crucial. First we have \[\deg_X(f)=\deg_{X'}(\phi_\alpha(f))\geq \deg_{Y'}(\phi_\alpha(f))=\deg_{Y}(f).\] Hence \[\deg_X(f)=\deg_{X'}(a_p)\geq\deg_Y(f)=p+\deg_{Y'}(a_p)\] so \[a_p=\sum b'_1(X',V',W',Z')x_{k_1}\cdots x_{k_p}.\] We can argue analogously and obtain \[a^{'''}_{p-1}=\sum b'_2(X',V',W',Z')x_{l_1}\cdots x_{l_p},\]\[a'_{p,x_i}=\sum b_{3,i}(X',V',W',Z')x_{m_1}\cdots x_{m_p}.\]

Now we rewrite $a_py_d^p$, $a^{'''}_{p-1}[x_d,y_d]y_d^{p-2}$ and $a'_{p,x_i}[x_i,y_d]y_d^{p-1}$ and we get \[f=c+gx_d+\sum_{i=1}^{d-1}m_{p,i}y_d[x_i,x_d],\] where $c\in\mathcal{C}$. We can rewrite the last summand as $g_1x_d+\sum_{i=1}^{d-1}h_i[x_i,x_d]$, where $\deg_{x_d}h_i=0$ and we get finally \[f=c+g'x_d+\sum_{i=1}^{d-1}h_i[x_i,x_d].\] Hence $g'$ and the $h_i$'s are constants too because $x_d$ and $[x_i,x_d]$ are. Now we are allowed to apply induction on $g'$ and the $h_i$'s because their degrees with respect to the set $\{x_d,y_d\}$ are strictly smaller than the one of $f$ and we conclude the proof.
\end{proof}

\section{The Nowicki conjecture for the free metabelian Poisson algebras}
This section presents a new result and is devoted to the study of Nowicki's conjecture in a nonassociative setting. In particular, we will handle the case of a free metabelian Poisson algebra and we shall give an explicit set of basis elements of the vector space of its constants.

Let $K$ be a field of characteristic zero. Let $(P,\cdot)$ and $(P,[\ ,\ ])$ be a commutative associative algebra and a Lie algebra structure on the $K$-vector space $P$, respectively, over $K$ such that the identity
\[
[a,b\cdot c]=[a,b]\cdot c+b\cdot [a,c]=[a,b]\cdot c+[a,c]\cdot b
\]
holds in $P$. Then $(P,\cdot,[\ ,\ ])$ is called a \textit{Poisson algebra}.

We will consider throughout the section only left normed commutators, i.e., $[a,b,c]:=[[a,b],c]$.
If $a\cdot b=[a,b]=0$ holds in $P$, then the Poisson algebra $P$ is called \textit{abelian} and any extension of $P$ by an abelian Poisson algebra is called \textit{metabelian} (see e.g. \cite{AM}).

Let us consider the free metabelian Poisson algebra $P_{2n}$ over $K$ of rank $2n$ generated by the set
\[
Z_{2n}=\{z_1,\ldots,z_{2n}\}=\{x_1,\ldots,x_n,y_1,\ldots,y_n\}=X_n\cup Y_n.
\]
It is well known (see e.g. \cite{Zh}) that the following identities hold in $P_{2n}$:
\begin{align}
0&=a\cdot b\cdot c\cdot d=[a\cdot b, c\cdot d]=[[a,b], c\cdot d]\nonumber\\
&=[a,b]\cdot c \cdot d=[a,b]\cdot[c,d]=[[a,b],[c,d]],\nonumber
\end{align}
which imply the following ones:
\[
[a,b,c]=[a,c,b]-[b,c,a]
\]
\[
[a,b,c]\cdot d=[d,b,c]\cdot a-[c,b,d]\cdot a
\]
\[
[a,b]+[b,a]=0 \ , \ [a,b,c,d]=[a,b,d,c]
\]
\[
[a,b,c]\cdot d=-[a,b,d]\cdot c \ , \ [a,b,c]\cdot c=0.
\]

We put an order "$<$" in the generating set so that
\[
z_1<\cdots<z_{2n}
\]
or
\[
x_1<\cdots<x_n<y_1<\cdots<y_n.
\]
It is well known (see \cite{Zh}) ${\mathcal B}_{1}\cup {\mathcal B}_{2}\cup {\mathcal B}_{3}\cup {\mathcal B}_{4}$
forms a $K$-basis for the free metabelian Poisson algebra $P_{2n}$,
where
\begin{align}
&{\mathcal B}_{1}=Z_{2n}\cup\{[z_{i_1},\dots,z_{i_k}]\mid i_1>i_2\leq i_3\leq\cdots\leq i_k, 2\leq k \},\nonumber\\
&{\mathcal B}_{2}=\{z_i\cdot z_j\ ,\ z_i\cdot z_j\cdot z_k\mid 1\leq i\leq j\leq k\leq 2n\},\nonumber\\
&{\mathcal B}_{3}=\{[z_i, z_j]\cdot z_k\mid 1\leq j\leq i\leq 2n \ , \ 1\leq k\leq 2n\},\nonumber\\
&{\mathcal B}_{4}=\{[z_i, z_j, z_k]\cdot  z_l \mid 1\leq j<i\leq l\leq 2n\ ,\ 1\leq j\leq k<l\leq 2n\}.\nonumber
\end{align}

Now consider the linear nilpotent derivation $\delta:y_i\to x_i\to0$, $i=1,\ldots,n$. Clearly the spaces $K{\mathcal B}_{i}$, $i=1,2,3,4$,
are $\delta$-invariant, i.e., $(K{\mathcal B}_{i})^{\delta}\subset K{\mathcal B}_{i}$, then
\[
P_{2n}^{\delta}=(K{\mathcal B}_{1})^{\delta}\oplus (K{\mathcal B}_{2})^{\delta}\oplus (K{\mathcal B}_{3})^{\delta}\oplus (K{\mathcal B}_{4})^{\delta}.
\]
Note that $K{\mathcal B}_{1}=L_{2n}$ is the free metabelian Lie algebra; moreover, by \cite{DF1}, we know the generators of the algebra $(L_{2n}')^{\delta}$
as a $K[X_n,Y_n]^{\delta}$-module:

\begin{theorem}\label{principal}
The $K[X_n,Y_n]^{\delta}$-module $(L_{2n}')^{\delta}$ is generated by the following elements
\[
[x_i,y_i],\quad \quad 1\leq i\leq n,
\]
\[
[x_i,x_j],\quad 1\leq i<j\leq n,
\]
\[
[x_i,y_j]+[x_j,y_i],\quad 1\leq i<j\leq n,
\]
\[
[x_i,x_p,y_q]-[x_i,y_p,x_q],\, 1\leq i\leq n, \, 1\leq p<q\leq n,
\]
\[
[x_i,x_j,y_k]-[x_i,x_k,y_j]+[x_j,x_k,y_i],\quad 1\leq i<j<k\leq n,
\]
and
\[
[x_i,x_p,y_j,y_q]+[y_i,y_p,x_j,x_q]-[x_i,y_p,y_j,x_q]-[y_i,x_p,x_j,y_q],
\]
where $1\leq i<j\leq n$, $1\leq p<q\leq n$.
\end{theorem}

By Nowicki's conjecture we get $(K{\mathcal B}_{2})^{\delta}$ is spanned on $X_n$ and the elements of the form
\[
u_{i,j}=x_i\cdot y_j-x_j\cdot y_i \ , \ 1\leq i<j<k\leq n
\]
and
\[
u_{i,j}\cdot x_k \ , \ 1\leq i<j<k\leq n \ , \ 1\leq k\leq n
\]
with the following relations:
\[
x_i\cdot u_{j,k}-x_j\cdot u_{i,k}+x_k\cdot u_{i,j}=0 \ , \ 1\leq i<j<k\leq n.
\]

We want to highlight the fact that once again some "determinants" (the $u_{i,j}$ above) appear as generators of the algebra of constants as in the classical Nowicki's conjecture and its generalization.

In the sequel we are going to show an explicit basis of $KB_3^\delta$ and $KB_4^\delta$ as vector spaces. Then the explicit basis of $P_{2n}^\delta$ will be an obvious corollary.

\begin{theorem}\label{KB_3}
The $K$-vector space $(K{\mathcal B}_3)^{\delta}$ is of a basis consisting of elements of the form
\[
[x_i, x_j]\cdot x_k \ , \  j<i,
\]
\[
[y_i, x_i]\cdot x_j \ , \  i,j,
\]
\[
[x_i, x_j]\cdot  y_k+[y_j, x_i]\cdot  x_k \ , \  j<i<k,
\]
\[
[y_i, x_j]\cdot  x_k+[y_j, x_i]\cdot  x_k \ , \  j<i<k.
\]
\end{theorem}

\begin{proof}
Let us set
\begin{align}
A_1=&\{[x_i, x_j]\cdot x_k\mid 1\leq j<i\leq n \ , \ 1\leq k\leq n \},\nonumber\\
A_2=&\{[y_i, y_j]\cdot y_k\mid 1\leq j<i\leq n \ , \ 1\leq k\leq n\},\nonumber\\
A_3=&\{[x_i, x_j]\cdot y_k\mid 1\leq j<i\leq n \ , \ 1\leq k\leq n\},\nonumber\\
A_4=&\{[y_i, y_j]\cdot x_k\mid 1\leq j<i\leq n \ , \ 1\leq k\leq n\},\nonumber\\
A_5=&\{[y_i, x_j]\cdot y_k\mid 1\leq i,j,k\leq n\},\nonumber\\
A_6=&\{[y_i, x_j]\cdot x_k\mid 1\leq i,j,k\leq n\}.\nonumber
\end{align}
Indeed $K{\mathcal B}_3=KA_1\oplus KA_2\oplus KA_3\oplus KA_4\oplus KA_5\oplus KA_6$,
while $(KA_1)^{\delta}=KA_1$,
\[
(KA_2)^{\delta}\subset KA_4\oplus KA_5 \ , \ (KA_3)^{\delta}\subset KA_1,
\]
\[
(KA_4)^{\delta}\subset KA_6 \ , \ (KA_5)^{\delta}\subset KA_3\oplus KA_6 \ , \ (KA_6)^{\delta}\subset KA_1.
\]
Thus
\[
(K{\mathcal B}_3)^{\delta}=KA_1\oplus(KA_2)^{\delta}\oplus(KA_3\oplus KA_6)^{\delta}\oplus(KA_4\oplus KA_5)^{\delta}.
\]
Now let
\[
p_2=\sum_{j<i}\alpha_{ijk}[y_i, y_j]\cdot y_k
\]
be an element in $(KA_2)^{\delta}$, then $\delta(p_2)=0$ and
\begin{align}
0=&\sum_{j<i}\alpha_{ijk}([x_i, y_j]\cdot y_k+[y_i,x_j]\cdot y_k+[y_i, y_j]\cdot x_k)\nonumber\\
=&\sum_{j<i}\alpha_{ijk}(-[y_j,x_i]\cdot y_k+[y_i,x_j]\cdot y_k)+\sum_{j<i}\alpha_{ijk}[y_i, y_j]\cdot x_k.\nonumber
\end{align}
Notice that $\sum_{j<i}\alpha_{ijk}[y_i, y_j]\cdot x_k$ belongs to $KA_4$, then
\[
0=\sum_{j<i}\alpha_{ijk}[y_i, y_j]\cdot x_k.
\]
We immediately get any $\alpha_{ijk}$ is 0 and $p_2$ turns out to be 0, too. This proves that there is no nonzero constant in $KA_2$.

Suppose now $p_3+p_6\in (KA_3\oplus KA_6)^{\delta}$ for some $p_3\in KA_3$ and $p_6\in KA_6$.
First observe that elements of the form $[y_i, x_i]\cdot x_k$ are constants in $KA_6$. Then we may assume
 \[
p_3=\sum_{j<i}\alpha_{ijk}[x_i, x_j]\cdot y_k \ \in KA_3 \ \ \ \text{\rm and} \ \ \ p_6=\sum_{i\neq j}\beta_{ijk}[y_i, x_j]\cdot x_k \ \in KA_6.
\]
Hence
\begin{align}
\delta(p_3+p_6)=0=&\sum_{j<i}\alpha_{ijk}[x_i, x_j]\cdot x_k+\sum_{j<i}\beta_{ijk}[x_i, x_j]\cdot x_k+\sum_{i<j}\beta_{ijk}[x_i, x_j]\cdot x_k\nonumber\\
=&\sum_{j<i}(\alpha_{ijk}+\beta_{ijk}-\beta_{jik})[x_i, x_j]\cdot x_k\nonumber
\end{align}
and $\beta_{jik}=\alpha_{ijk}+\beta_{ijk}$ for $j<i$. This gives us $(KA_3\oplus KA_6)^{\delta}$
is spanned on the elements of the form
\[
[x_i, x_j]\cdot y_k+[y_j, x_i]\cdot x_k \ \ \ \text{\rm and} \ \ \ [y_i, x_j]\cdot x_k+[y_j, x_i]\cdot x_k
\]
for $j<i$.

We simply need to show $(KA_4\oplus KA_5)^{\delta}=\{0\}$. Let $p_4+p_5\in (KA_4\oplus KA_5)^{\delta}$ such that
 \[
p_4=\sum_{j<i}\alpha_{ijk}[y_i, y_j]\cdot x_k
\]
and
\begin{align}
p_5=&\sum_{}\beta_{ijk}[y_i, x_j]\cdot y_k=\sum_{j<i}(\beta_{ijk}[y_i, x_j]\cdot y_k+\beta_{jik}[y_j, x_i]\cdot y_k)+\sum_{}\beta_{iik}[y_i, x_i]\cdot y_k.\nonumber
\end{align}
We have
\[
\delta(p_4)=\sum_{j<i}\alpha_{ijk}([x_i, y_j]\cdot x_k+[y_i,x_j]\cdot x_k),
\]
\begin{align}
\delta(p_5)=&\sum_{j<i}\left(\beta_{ijk}([x_i, x_j]\cdot y_k+[y_i, x_j]\cdot x_k)-\beta_{jik}([x_j, x_i]\cdot y_k+[y_j, x_i]\cdot x_k)\right)\nonumber\\
&+\sum_{}\beta_{iik}[y_i, x_i]\cdot x_k.\nonumber
\end{align}
Then $\delta(p_4+p_5)=0$ gives that
\[
0=\sum_{j<i}(\alpha_{ijk}+\beta_{ijk})[y_i,x_j]\cdot x_k+(-\alpha_{ijk}+\beta_{jik})[y_j,x_i]\cdot x_k \ \in KA_6
\]
\[
0=\sum_{}\beta_{iik}[y_i, x_i]\cdot x_k \ \in KA_6 \ , \  0=\sum_{j<i}(\beta_{ijk}-\beta_{jik})[x_i, x_j]\cdot y_k \ \in KA_3
\]
that is
\[
\alpha_{ijk}+\beta_{ijk}=0 \ , \ -\alpha_{ijk}+\beta_{jik}=0 \ , \ \beta_{iik}=0 \ , \ \beta_{ijk}-\beta_{jik}=0;
\]
hence
\[
\alpha_{ijk}=\beta_{ijk}=\beta_{jik}=\beta_{iik}=0
\]
for $j<i$. Therefore, $p_4=p_5=0$. Now the proof follows by direct computations showing that the constants found are linearly independent.
\end{proof}

\begin{theorem}\label{KB_4}
The $K$-vector space $(K{\mathcal B}_4)^{\delta}$ is of a basis consisting of elements of the form
\[
[x_i, x_j, x_k]\cdot  x_l \ , \  j<i\leq l\ ,\ j\leq k<l,
\]
\[
[x_i, x_j, x_j]\cdot  y_j \ , \ [x_i, x_j, x_i]\cdot  y_i \ , \ [x_i, x_j, x_i]\cdot  y_j+[x_i, x_j, x_j]\cdot  y_i \ , \ j<i,
\]
\[
[x_i, x_j, x_k]\cdot  y_k \ , \ [x_k, x_j, x_i]\cdot  y_i \ , \ [x_i, x_j, x_i]\cdot  y_k+[x_i, x_j, x_k]\cdot  y_i\ , \ j<i<k,
\]
\[
[x_i, x_j, x_i]\cdot  y_k+[x_k, x_i, x_i]\cdot  y_j\ , \ [x_i, x_j, x_k]\cdot  y_j-[x_k, x_j, x_i]\cdot  y_j\ , \  j<i<k,
\]
\[
[x_i, x_j, x_k]\cdot  y_j+[x_i, x_j, x_j]\cdot  y_k\ , \  [x_i, x_j, x_k]\cdot  y_j+[x_k, x_j, x_j]\cdot  y_i\ , \ j<i<k,
\]
\[
[x_k, x_j, x_k]\cdot  y_i+[x_k, x_j, x_i]\cdot  y_k \ , \ [x_k, x_j, x_k]\cdot  y_i-[x_k, x_i, x_k]\cdot  y_j\ , \  j<i<k,
\]
\[
[x_k, x_j,x_l]\cdot y_i+[x_i, x_j,x_k]\cdot y_l-[x_k, x_i,x_l]\cdot y_j  \ , \  j<i<k<l,
\]
\[
[x_k, x_j,x_i]\cdot y_l-[x_i, x_j,x_k]\cdot y_l+[x_k, x_i,x_l]\cdot y_j  \ , \  j<i<k<l,
\]
\[
[x_l, x_i,x_k]\cdot y_j+[x_i, x_j,x_k]\cdot y_l-[x_k, x_i,x_l]\cdot y_j  \ , \  j<i<k<l,
\]
\[
[x_i, x_j,x_l]\cdot y_k+[x_i, x_j,x_k]\cdot y_l \ , \ [x_l, x_j,x_i]\cdot y_k+[x_k, x_i,x_l]\cdot y_j \ , \ j<i<k<l,
\]
\[
[x_l, x_j,x_k]\cdot y_i-[x_k, x_i,x_l]\cdot y_j \ , \ j<i<k<l,
\]
\[
[y_i, x_i, x_i]\cdot  y_i \ , \ [y_j, x_j, x_i]\cdot  y_i \ , \ [y_j, x_i, x_i]\cdot  y_i+[y_i, x_j, x_i]\cdot  y_i \ , \ j<i,
\]
\[
[y_j, x_j, x_i]\cdot  y_j+[y_j, x_j, x_j]\cdot  y_i \ , \ [y_j, x_i, x_i]\cdot  y_j-2[x_i, x_j, y_j]\cdot  y_i+[y_i, x_j, x_j]\cdot  y_i \ , \ j<i,
\]
\[
[y_i, x_j, x_k]\cdot  y_i+[x_i, x_j, y_i]\cdot  y_k+[x_k, x_i, y_j]\cdot  y_i +[y_j, x_i, x_i]\cdot  y_k \ , \ j<i<k,
\]
\[
[y_j, x_i, x_k]\cdot  y_i-[x_i, x_j, y_i]\cdot  y_k-[x_k, x_i, y_j]\cdot  y_i +[y_i, x_j, x_i]\cdot  y_k \ , \ j<i<k,
\]
\[
[x_k, x_j, y_i]\cdot  y_k+[x_k, x_i, y_j]\cdot  y_k -[y_j, x_k, x_k]\cdot  y_i -[y_k, x_j, x_i]\cdot  y_k \ , \ j<i<k,
\]
\[
[y_i, x_j, x_k]\cdot  y_k+[y_j, x_i, x_k]\cdot  y_k \ , \ [y_j, x_j, x_k]\cdot  y_i+[y_j, x_j, x_i]\cdot  y_k \ , \ j<i<k,
\]
\[
[x_i, x_j, y_j]\cdot  y_k+[x_k, x_j, y_j]\cdot  y_i -[y_i, x_j, x_j]\cdot  y_k-[y_j, x_i, x_k]\cdot  y_j \ , \ j<i<k,
\]
\[
[x_l, x_k, y_j]\cdot  y_i +[x_i, x_j, y_k]\cdot  y_l+[y_j, x_i, x_k]\cdot  y_l+[y_i, x_j, x_l]\cdot  y_k \ , \ j<i<k<l,
\]
\[
[y_j, x_k, x_l]\cdot  y_i+[y_k, x_j, x_i]\cdot  y_l-[x_l, x_i, y_j]\cdot  y_k-[x_k, x_j, y_i]\cdot  y_l  \ , \ j<i<k<l,
\]
\[
[y_j, x_i, x_l]\cdot  y_k+[y_i, x_j, x_k]\cdot  y_l +[y_j, x_i, x_k]\cdot  y_l+[y_i, x_j, x_l]\cdot  y_k \ , \ j<i<k<l,
\]
\[
[x_l, x_j, y_i]\cdot  y_k+[x_k, x_i, y_j]\cdot  y_l -[y_j, x_i, x_k]\cdot  y_l-[y_i, x_j, x_l]\cdot  y_k-[x_l, x_i, y_j]\cdot  y_k-[x_k, x_j, y_i]\cdot  y_l.
\]
\end{theorem}

\begin{proof}
It is easy to see
\[
K{\mathcal B}_4=KB_1\oplus KB_2\oplus KB_3\oplus KB_4\oplus KB_5\oplus KB_6,
\]
where
\begin{align}
B_1=&\{[x_i, x_j,x_k]\cdot x_l\mid 1\leq j<i\leq l\leq n \ , \ 1\leq j\leq k<l \leq n \},\nonumber\\
B_2=&\{[y_i, y_j,y_k]\cdot y_l\mid 1\leq j<i\leq l\leq n \ , \ 1\leq j\leq k<l \leq n \},\nonumber\\
B_3=&\{[x_i, x_j,x_k]\cdot y_l\mid 1\leq j<i \leq n \ , \ 1\leq j\leq k \leq n\},\nonumber\\
B_4=&\{[x_i, x_j,y_k]\cdot y_l\mid 1\leq j<i\leq n \ , \ 1\leq k<l \leq n\},\nonumber\\
B_5=&\{[y_i, x_j,x_k]\cdot y_l\mid 1\leq i\leq l\leq n \ , \ 1\leq j\leq k\leq n\},\nonumber\\
B_6=&\{[y_i, x_j,y_k]\cdot y_l\mid 1\leq i\leq l\leq n \ , \ 1\leq k<l \leq n\}.\nonumber
\end{align}
We have $(KB_1)^{\delta}=KB_1$,
\[
(KB_2)^{\delta}\subset KB_6 \ , \ (KB_3)^{\delta}\subset KB_1,
\]
\[
(KB_4)^{\delta}\subset KB_3 \ , \ (KB_5)^{\delta}\subset KB_3 \ , \ (KB_6)^{\delta}\subset KB_4\oplus KB_5
\]
and we get
\[
(K{\mathcal B}_4)^{\delta}=KB_1\oplus(KB_2)^{\delta}\oplus(KB_3)^{\delta}\oplus(KB_4\oplus KB_5)^{\delta}\oplus(KB_6)^{\delta}.
\]
We divide the proof in four cases: we study separately $(KB_2)^{\delta}$, $(KB_3)^{\delta}$, $(KB_4\oplus KB_5)^{\delta}$, and finally $(KB_6)^{\delta}$.

\

\noindent\underline{Case 1} Let $p_2\in (KB_2)^{\delta}$ be of the form
\begin{align}
p_2=&\sum_{j<i\leq l, j\leq k<l}\alpha_{ijkl}[y_i, y_j,y_k]\cdot y_l.\nonumber
\end{align}
Note that the condition $j<i\leq l \ , \ j\leq k<l$ can be divided into six parts:
\begin{align}
&j=k<i=l \ , \ j<k<i<l \ , \ j<i<k<l,\nonumber\\
&j<i=k<l \ , \ j<k<i=l \ , \ j=k<i<l.\nonumber
\end{align}
Hence
\begin{align}
p_2=&\sum_{j<i}\alpha_{ijji}[y_i, y_j,y_j]\cdot y_i+\sum_{j<k<i<l}\alpha_{ijkl}[y_i, y_j,y_k]\cdot y_l+\sum_{j<i<k<l}\alpha_{ijkl}[y_i, y_j,y_k]\cdot y_l\nonumber\\
&+\sum_{j<i<l}\alpha_{ijil}[y_i, y_j,y_i]\cdot y_l+\sum_{ j<k<i}\alpha_{ijki}[y_i, y_j,y_k]\cdot y_i+\sum_{j<i<l}\alpha_{ijjl}[y_i, y_j,y_j]\cdot y_l\nonumber\\
=&\sum_{j<i}\alpha_{ijji}[y_i, y_j,y_j]\cdot y_i+\sum_{j<i<k<l}(\alpha_{kjil}[y_k, y_j,y_i]\cdot y_l+\alpha_{ijkl}[y_i, y_j,y_k]\cdot y_l)\nonumber\\
&+\sum_{j<i<k}(\alpha_{ijik}[y_i, y_j,y_i]\cdot y_k+\alpha_{kjik}[y_k, y_j,y_i]\cdot y_k+\alpha_{ijjk}[y_i, y_j,y_j]\cdot y_k).\nonumber
\end{align}
Let us rewrite $p_2$ as
\[
p_2=p_{22}+p_{23}+p_{24},
\]
where
\[
p_{22}=\sum_{j<i}\alpha_{ijji}[y_i, y_j,y_j]\cdot y_i \ , \ p_{24}=\sum_{j<i<k<l}(\alpha_{kjil}[y_k, y_j,y_i]\cdot y_l+\alpha_{ijkl}[y_i, y_j,y_k]\cdot y_l),
\]
\[
p_{23}=\sum_{j<i<k}(\alpha_{ijik}[y_i, y_j,y_i]\cdot y_k+\alpha_{kjik}[y_k, y_j,y_i]\cdot y_k+\alpha_{ijjk}[y_i, y_j,y_j]\cdot y_k).
\]
Let $V_{2m}$, $m=2,3,4$, be the subspace of $KB_6$ spanned on monomials $[y_i, x_j,y_k]\cdot y_l$ with $|\{i,j,k,l\}|=m$.
Then $V_{22}\cap V_{23}=\{0\}$, $V_{22}\cap V_{24}=\{0\}$, $V_{23}\cap V_{24}=\{0\}$,
and $\delta(p_{22})\in V_{22}$, $\delta(p_{23})\in V_{23}$, $\delta(p_{24})\in V_{24}$.
Therefore $\delta(p_2)=0$ implies that $\delta(p_{22})=\delta(p_{23})=\delta(p_{24})=0$.

Remark that $\delta([y_i, y_j,y_j]\cdot y_i)$ can be written as a linear combination of basis elements as follows:
\begin{align}
\delta([y_i, y_j,y_j]\cdot y_i)=&[x_i, y_j,y_j]\cdot y_i+[y_i, x_j,y_j]\cdot y_i+[y_i, y_j,x_j]\cdot y_i+[y_i, y_j,y_j]\cdot x_i\nonumber\\
=&-[y_j,x_i, y_j]\cdot y_i+[y_i, x_j,y_j]\cdot y_i-[y_j,x_j,y_i]\cdot y_i+[y_i,x_j, y_j]\cdot y_i\nonumber\\
&+[x_i,y_j,y_j]\cdot y_i-[y_j, y_j,x_i]\cdot y_i\nonumber\\
=&-2[y_j,x_i, y_j]\cdot y_i+2[y_i, x_j,y_j]\cdot y_i.\nonumber
\end{align}
Thus
\[
\delta(p_{22})=0=2\sum_{j<i}\alpha_{ijji}(-[y_j,x_i, y_j]\cdot y_i+[y_i, x_j,y_j]\cdot y_i)
\]
and consequently
\[
\sum_{j<i}\alpha_{ijji}[y_j,x_i, y_j]\cdot y_i=0=\sum_{j<i}\alpha_{ijji}[y_i, x_j,y_j]\cdot y_i
\]
since the basis elements belong to different multilinear components for $j<i$.
We get $\alpha_{ijji}=0$ if $j<i$ that means $p_{22}=0$.

Similarly, because $\delta(p_{23})=0$ we have the sum
\[
\sum_{j<i<k}\left(\begin{array}{ll}\alpha_{ijik}(-2[y_j, x_i,y_i]\cdot y_k+[y_i, x_j,y_i]\cdot y_k+[y_i, x_i,y_j]\cdot y_k+[y_i, x_k,y_j]\cdot y_i)\\
+\alpha_{kjik}([y_k, x_j,y_i]\cdot y_k+[y_k, x_i,y_j]\cdot y_k-[y_j, x_k,y_i]\cdot y_k-[y_i, x_l,y_j]\cdot y_l)\\
+\alpha_{ijjk}(2[y_i, x_j,y_j]\cdot y_k-[y_j, x_i,y_j]\cdot y_k-[y_j, x_j,y_i]\cdot y_k-[y_j, x_k,y_j]\cdot y_i)\end{array}\right)\nonumber
\]
is zero, from which we deduce that each summand is zero; moreover,
\[
0=\sum_{j<i<k}\alpha_{ijik}[y_j, x_i,y_i]\cdot y_k=\sum_{j<i<k}\alpha_{kjik}[y_k, x_j,y_i]\cdot y_k=\sum_{j<i<k}\alpha_{ijjk}[y_i, x_j,y_j]\cdot y_k
\]
because the basis elements belong to different multilinear components. As above, we get $0=\alpha_{ijik}=\alpha_{kjik}=\alpha_{ijjk}$
for $j<i<k$ implying that $p_{23}=0$.

Again, writing $\delta(p_{24})$ as a linear combination of basis elements, we get
\begin{align}
0=&\sum_{j<i<k<l}\alpha_{kjil}\left(\begin{array}{ll}-[y_j, x_k,y_i]\cdot y_l+[y_k, x_j,y_i]\cdot y_l
-[y_j, x_i,y_k]\cdot y_l\\+[y_k, x_i,y_j]\cdot y_l-[y_i, x_l,y_j]\cdot y_k \end{array}\right)\nonumber\\
&+\sum_{j<i<k<l}\alpha_{ijkl}\left(\begin{array}{ll}-[y_j, x_i,y_k]\cdot y_l+[y_i, x_j,y_k]\cdot y_l-[y_j, x_k,y_i]\cdot y_l\\
+[y_i, x_k,y_j]\cdot y_l-[y_i, x_l,y_j]\cdot y_k+[y_j, x_l,y_i]\cdot y_k\end{array}\right)\nonumber
\end{align}
and we get $\alpha_{kjil}=\alpha_{ijkl}=0$ for $j<i<k<l$. Thus $p_{24}=0$, from which we get $p_2=0$.

\

\noindent\underline{Case 2} We study $(KB_3)^{\delta}$. Let
\[
p_3=\sum_{ j<i, \ j\leq k}\alpha_{ijkl}[x_i, x_j,x_k]\cdot y_l=p_{32}+p_{33}+p_{34}
\]
be a constant, where
\[
p_{32}=\sum_{j<i}\alpha_{ijii}[x_i, x_j,x_i]\cdot y_i+\alpha_{ijij}[x_i, x_j,x_i]\cdot y_j+\alpha_{ijji}[x_i, x_j,x_j]\cdot y_i+\alpha_{ijjj}[x_i, x_j,x_j]\cdot y_j,
\]
\[
p_{33}=\sum_{j<i<k}\left(\begin{array}{ll}\alpha_{ijik}[x_i, x_j,x_i]\cdot y_k+\alpha_{ijik}[x_i, x_j,x_i]\cdot y_k+\alpha_{ijik}[x_i, x_j,x_i]\cdot y_k\\
+\alpha_{ijkk}[x_i, x_j,x_k]\cdot y_k+\alpha_{ijki}[x_i, x_j,x_k]\cdot y_i+\alpha_{ijkj}[x_i, x_j,x_k]\cdot y_j\\
+\alpha_{kjik}[x_k, x_j,x_i]\cdot y_k+\alpha_{kjii}[x_k, x_j,x_i]\cdot y_i+\alpha_{kjij}[x_k, x_j,x_i]\cdot y_j\\
+\alpha_{ijjk}[x_i, x_j,x_j]\cdot y_k+\alpha_{kjji}[x_k, x_j,x_k]\cdot y_i+\alpha_{kiij}[x_k, x_i,x_i]\cdot y_j\end{array}\right),
\]
\[
p_{34}=\sum_{j<i<k<l}\left(\begin{array}{ll}\alpha_{ijkl}[x_i, x_j,x_k]\cdot y_l+\alpha_{ijlk}[x_i, x_j,x_l]\cdot y_k+\alpha_{kjli}[x_k, x_j,x_l]\cdot y_i\\
+\alpha_{kilj}[x_k, x_i,x_l]\cdot y_j+\alpha_{kjil}[x_k, x_j,x_i]\cdot y_l+\alpha_{ljik}[x_l, x_j,x_i]\cdot y_k\\
+\alpha_{ljki}[x_l, x_j,x_k]\cdot y_i+\alpha_{likj}[x_l, x_i,x_k]\cdot y_j\end{array}\right).
\]
As well as in the previous case, $\delta(p_{32})=\delta(p_{33})=\delta(p_{34})=0$. Let us study the case $\delta(p_{32})=0$. An easy computation shows the elements
\[
[x_i, x_j,x_i]\cdot y_i \ , \ [x_i, x_j,x_j]\cdot y_j \ , \ j<i
\]
are constants, then we get
\[
0=\sum_{j<i}(-\alpha_{ijij}+\alpha_{ijji})[x_i, x_j,x_j]\cdot x_i
\]
 and consequently $\alpha_{ijij}=\alpha_{ijji}$. Moreover, we get the elements of the form
\[
[x_i, x_j,x_i]\cdot y_j+[x_i, x_j,x_j]\cdot y_i \ , \ j<i
\]
are constants, too.

Notice also
\[
[x_i, x_j,x_k]\cdot y_k \ , \ [x_k, x_j,x_i]\cdot y_i \ , \ j<i<k.
\]
are constants, then $\delta(p_{33})=0$ may be rewritten as
\[
0=\sum_{j<i<k}\left(\begin{array}{ll}(\alpha_{ijik}-\alpha_{ijki}-\alpha_{kiij})[x_i, x_j,x_i]\cdot x_k\\
(-\alpha_{kjki}-\alpha_{kikj}+\alpha_{kjik})[x_k, x_j,x_i]\cdot x_k\\
(-\alpha_{ijkj}-\alpha_{kjij}+\alpha_{ijjk}+\alpha_{kjji})[x_i, x_j,x_j]\cdot x_k\end{array}\right).
\]
Hence we get the following system of equations for $j<i<k$:
\[\left\{
\begin{array}{ll}\alpha_{ijik}=\alpha_{ijki}+\alpha_{kiij}\\
\alpha_{kjki}=\alpha_{kjik}-\alpha_{kikj}\\
\alpha_{ijkj}=-\alpha_{kjij}+\alpha_{ijjk}+\alpha_{kjji}\end{array}\right..
\]
By the previous system, we get the next polynomials are constants, too.
\[
[x_i, x_j, x_i]\cdot  y_k+[x_i, x_j, x_k]\cdot  y_i\ , \ [x_i, x_j, x_i]\cdot  y_k+[x_k, x_i, x_i]\cdot  y_j
\]
\[
[x_k, x_j, x_k]\cdot  y_i+[x_k, x_j, x_i]\cdot  y_k \ , \ [x_k, x_j, x_k]\cdot  y_i-[x_k, x_i, x_k]\cdot  y_j
\]
\[
[x_i, x_j, x_k]\cdot  y_j-[x_k, x_j, x_i]\cdot  y_j\ , \ [x_i, x_j, x_k]\cdot  y_j+[x_i, x_j, x_j]\cdot  y_k
\]
\[
[x_i, x_j, x_k]\cdot  y_j+[x_k, x_j, x_j]\cdot  y_i \ , \ j<i<k.
\]
Let us rewrite now the relation $0=\delta(p_{34})$ as a linear combination of the basis elements.
\begin{align}
0=&\sum_{j<i<k<l}\left(\begin{array}{ll}\alpha_{ijkl}[x_i, x_j,x_k]\cdot x_l+\alpha_{ijlk}[x_i, x_j,x_l]\cdot x_k+\alpha_{kjli}[x_k, x_j,x_l]\cdot x_i\\
+\alpha_{kilj}[x_k, x_i,x_l]\cdot x_j+\alpha_{kjil}[x_k, x_j,x_i]\cdot x_l+\alpha_{ljik}[x_l, x_j,x_i]\cdot x_k\\
+\alpha_{ljki}[x_l, x_j,x_k]\cdot x_i+\alpha_{likj}[x_l, x_i,x_k]\cdot x_j\end{array}\right)\nonumber\\
=&\sum_{j<i<k<l}\left(\begin{array}{ll}(\alpha_{ijkl}-\alpha_{ijlk}+\alpha_{kilj}-\alpha_{ljik}+\alpha_{ljki})[x_i, x_j, x_k]\cdot  x_l\\
+(-\alpha_{kjli}-\alpha_{kilj}+\alpha_{kjil}+\alpha_{ljik}-\alpha_{ljki}-\alpha_{likj})[x_k, x_j, x_i]\cdot  x_l\end{array}\right).\nonumber
\end{align}
Hence we get
\[
\alpha_{ijkl}-\alpha_{ijlk}+\alpha_{kilj}-\alpha_{ljik}+\alpha_{ljki}=0
\]
\[
-\alpha_{kjli}-\alpha_{kilj}+\alpha_{kjil}+\alpha_{ljik}-\alpha_{ljki}-\alpha_{likj}=0
\]
for $j<i<k<l$, from which we have the following relations:
\[
\alpha_{ijkl}=\alpha_{ijlk}+\alpha_{kjli}-\alpha_{kjil}+\alpha_{likj}
\]
\[
\alpha_{kilj}=\alpha_{ljik}-\alpha_{ljki}-\alpha_{kjli}+\alpha_{kjil}-\alpha_{likj}.
\]
Substituting the relations above in the expression of $p_{34}$,
we get the following constants:
\begin{align}
&[x_i, x_j,x_l]\cdot y_k+[x_i, x_j,x_k]\cdot y_l,\nonumber\\
&[x_l, x_j,x_i]\cdot y_k+[x_k, x_i,x_l]\cdot y_j,\nonumber\\
&[x_l, x_j,x_k]\cdot y_i-[x_k, x_i,x_l]\cdot y_j,\nonumber\\
&[x_k, x_j,x_l]\cdot y_i+[x_i, x_j,x_k]\cdot y_l-[x_k, x_i,x_l]\cdot y_j,\nonumber\\
&[x_k, x_j,x_i]\cdot y_l-[x_i, x_j,x_k]\cdot y_l+[x_k, x_i,x_l]\cdot y_j,\nonumber\\
&[x_l, x_i,x_k]\cdot y_j+[x_i, x_j,x_k]\cdot y_l-[x_k, x_i,x_l]\cdot y_j.\nonumber
\end{align}

\

\noindent\underline{Case 3} We study $(KB_4\oplus KB_5)^{\delta}$.
Let
\[
p_4=\sum_{j<i , \ k<l}\alpha_{ljki}[x_i, x_j,y_k]\cdot y_l \ \in KB_4 \ , \ p_5=\sum_{i\leq l , \ j\leq k}\beta_{ljki}[y_i, x_j,x_k]\cdot y_l\ \in KB_5
\]
such that $\delta(p_4+p_5)=0$.
We point out $[y_i, x_i,x_i]\cdot y_i \ , [y_j, x_j,x_i]\cdot y_i \in(KB_4\oplus KB_5)^{\delta}$, $1\leq j<i\leq n$, since
\begin{align}
&\delta([y_i, x_i,x_i]\cdot y_i)=[x_i, x_i,x_i]\cdot y_i+[y_i, x_i,x_i]\cdot x_i=0,\nonumber\\
&\delta([y_j, x_j,x_i]\cdot y_i)=[x_j, x_j,x_i]\cdot y_i+[y_j, x_j,x_i]\cdot x_i=0.\nonumber
\end{align}
Thus we may assume $p_4+p_5=p_{452}+p_{453}+p_{454}$,
then
\[
\delta(p_4+p_5)=\delta(p_{452})+\delta(p_{453})+\delta(p_{454})=0,
\]
where
\[
p_{452}=\sum_{j<i}\left(\begin{array}{ll}\alpha_{ijji}[x_i, x_j,y_j]\cdot y_i+\beta_{ijii}[y_i, x_j,x_i]\cdot y_i+\beta_{ijji}[y_i, x_j,x_j]\cdot y_i\\
+\beta_{jjji}[y_j, x_j,x_j]\cdot y_i+\beta_{jjij}[y_j, x_j,x_i]\cdot y_j+\beta_{jiii}[y_j, x_i,x_i]\cdot y_i\\
+\beta_{jiij}[y_j, x_i,x_i]\cdot y_j\end{array}\right),
\]
\[
p_{453}=\sum_{j<i<k}\left(\begin{array}{ll}\alpha_{ijik}[x_i, x_j,y_i]\cdot y_k+\alpha_{kiji}[x_k, x_i,y_j]\cdot y_i+\alpha_{ijjk}[x_i, x_j,y_j]\cdot y_k\\
+\alpha_{kjji}[x_k, x_j,y_j]\cdot y_i+\alpha_{kijk}[x_k, x_i,y_j]\cdot y_k+\alpha_{kjik}[x_k, x_j,y_i]\cdot y_k\\
+\beta_{ijik}[y_i, x_j,x_i]\cdot y_k+\beta_{ijki}[y_i, x_j,x_k]\cdot y_i+\beta_{jiki}[y_j, x_i,x_k]\cdot y_i\\
+\beta_{jiik}[y_j, x_i,x_i]\cdot y_k+\beta_{ijjk}[y_i, x_j,x_j]\cdot y_k+\beta_{jjik}[y_j, x_j,x_i]\cdot y_k\\
+\beta_{jjki}[y_j, x_j,x_k]\cdot y_i+\beta_{jikj}[y_j, x_i,x_k]\cdot y_j+\beta_{ijkk}[y_i, x_j,x_k]\cdot y_k\\
+\beta_{kjik}[y_k, x_j,x_i]\cdot y_k+\beta_{jikk}[y_j, x_i,x_k]\cdot y_k+\beta_{jkki}[y_j, x_k,x_k]\cdot y_i\end{array}\right),
\]
\[
p_{454}=\sum_{j<i<k<l}\left(\begin{array}{ll}\alpha_{ijkl}[x_i, x_j,y_k]\cdot y_l+\alpha_{kjil}[x_k, x_j,y_i]\cdot y_l+\alpha_{ljik}[x_l, x_j,y_i]\cdot y_k\\
+\alpha_{kijl}[x_k, x_i,y_j]\cdot y_l+\alpha_{lijk}[x_l, x_i,y_j]\cdot y_k+\alpha_{lkji}[x_l, x_k,y_j]\cdot y_i\\
+\beta_{ijkl}[y_i, x_j,x_k]\cdot y_l+\beta_{kjil}[y_k, x_j,x_i]\cdot y_l+\beta_{ijlk}[y_i, x_j,x_l]\cdot y_k\\
+\beta_{jikl}[y_j, x_i,x_k]\cdot y_l+\beta_{jilk}[y_j, x_i,x_l]\cdot y_k+\beta_{jkli}[y_j, x_k,x_l]\cdot y_i\\
\end{array}\right).
\]
Rewriting $\delta(p_{452})\in KB_3$ as a linear combination of basis elements, we have
\[
0=\sum_{j<i}\left(\begin{array}{ll}(\alpha_{ijji}+2\beta_{ijji})[x_i, x_j,x_j]\cdot y_i+(-\alpha_{ijji}-2\beta_{jiij})[x_i, x_j,x_i]\cdot y_j\\
(\beta_{ijii}-\beta_{jiii})[x_i, x_j,x_i]\cdot y_i+(\beta_{jjji}-\beta_{jjij})[x_i, x_j,x_j]\cdot y_j\end{array}\right),
\]
then $\alpha_{ijji}=-2\beta_{jiij}$, $\beta_{ijji}=\beta_{jiij}$, $\beta_{ijii}=\beta_{jiii}$, $\beta_{jjji}=\beta_{jjij}$.
Thus, for $j<i$, the next are constants:
\begin{align}
&[y_j, x_i,x_i]\cdot y_j-2[x_i, x_j,y_j]\cdot y_i+[y_i, x_j,x_j]\cdot y_i,\nonumber\\
&[y_j, x_i,x_i]\cdot y_i+[y_i, x_j,x_i]\cdot y_i \ , \ [y_j, x_j,x_i]\cdot y_j+[y_j, x_j,x_j]\cdot y_i.\nonumber
\end{align}
A careful check of the coefficients of $[x_i, x_j,x_i]\cdot y_k$, $[x_i, x_j,x_k]\cdot y_i$, $[x_k, x_j,x_i]\cdot y_i$, $[x_k, x_i,x_i]\cdot y_j$, $[x_k, x_j,x_i]\cdot y_k$,
$[x_k, x_j,x_k]\cdot y_i$, $[x_i, x_j,x_k]\cdot y_k$, $[x_k, x_i,x_k]\cdot y_j$, $[x_i, x_j,x_j]\cdot y_k$, $[x_i, x_j,x_k]\cdot y_j$, $[x_k, x_j,x_j]\cdot y_i$,
$[x_k, x_j,x_i]\cdot y_j$, respectively, in the expression of $\delta(p_{453})=0$, gives us the following homogeneous system of equation:
\[
\begin{array}{ll}
0=\alpha_{ijik}+\beta_{ijik}-\beta_{jiik} & 0=-\alpha_{kijk}+\beta_{ijkk}-\beta_{jikk}-\beta_{kjik}\\
0=-\alpha_{ijik}-\alpha_{kiji}-\beta_{ijik}+2\beta_{ijki}-\beta_{jiki} & 0=-\alpha_{kijk}-\beta_{jkki}\\
0=\alpha_{kiji}+\beta_{ijik}-\beta_{ijki} & 0=\alpha_{ijjk}+\beta_{ijjk}\\
0=-\alpha_{kiji}-\beta_{jiki}+\beta_{jiik}& 0=-\alpha_{ijjk}-\beta_{jjik}+\beta_{jjki}-\beta_{jikj}\nonumber\\
0=\alpha_{kjik}+\alpha_{kijk}+2\beta_{kjik} & 0=\alpha_{kjji}+\beta_{ijjk}\nonumber\\
0=-\alpha_{kjik}-\beta_{jkki} & 0=-\alpha_{kjji}+\beta_{jjik}-\beta_{jjki}-\beta_{jikj}\end{array}
\]
We have the next relations:
\[
\alpha_{ijik}=\alpha_{kiji}=\beta_{ijki}-\beta_{jiki} \ , \ \beta_{ijik}=\beta_{jiki} \ , \ \beta_{jiik}=\beta_{ijki}
\]
\[
\alpha_{kjik}=\alpha_{kijk}=-\beta_{jkki}=-\beta_{kjik} \ , \ \beta_{ijkk}=\beta_{jikk}
\]
\[
\alpha_{ijjk}=\alpha_{kjji}=-\beta_{ijjk}=-\beta_{jikj} \ , \ \beta_{jjki}=\beta_{jjik}.
\]
Now substituting $\beta_{ijki}$, $\beta_{jiki}$, $\alpha_{kjik}$, $\beta_{ijkk}$, $\beta_{jjki}$, $\alpha_{ijjk}$, respectively,
in the expression of $p_{453}$, we obtain the following constants for $j<i<k$:
\begin{align}
&[y_i, x_j, x_k]\cdot  y_i+[x_i, x_j, y_i]\cdot  y_k+[x_k, x_i, y_j]\cdot  y_i +[y_j, x_i, x_i]\cdot  y_k\nonumber\\
&[y_j, x_i, x_k]\cdot  y_i-[x_i, x_j, y_i]\cdot  y_k-[x_k, x_i, y_j]\cdot  y_i +[y_i, x_j, x_i]\cdot  y_k\nonumber\\
&[x_k, x_j, y_i]\cdot  y_k+[x_k, x_i, y_j]\cdot  y_k -[y_j, x_k, x_k]\cdot  y_i -[y_k, x_j, x_i]\cdot  y_k\nonumber\\
&[y_i, x_j, x_k]\cdot  y_k+[y_j, x_i, x_k]\cdot  y_k \ , \ [y_j, x_j, x_k]\cdot  y_i+[y_j, x_j, x_i]\cdot  y_k \nonumber\\
&[x_i, x_j, y_j]\cdot  y_k+[x_k, x_j, y_j]\cdot  y_i -[y_i, x_j, x_j]\cdot  y_k-[y_j, x_i, x_k]\cdot  y_j\nonumber
\end{align}
Similar arguments for $\delta(p_{454})=0$ give out the following relations:
\[
\alpha_{ijkl}=\alpha_{lkji} \ , \ \alpha_{kijl}=\alpha_{ljik} \ , \ \beta_{ijkl}=\beta_{jilk} \ , \ \beta_{kjil}=\beta_{jkli}
\]
\[
\beta_{jikl}=\beta_{jilk}+\alpha_{lkji}-\alpha_{ljik}=\beta_{ijlk} \ , \ \alpha_{lijk}=-\beta_{jkli}-\alpha_{ljik}=\alpha_{kjil}
\]
and substituting $\alpha_{lkji}$, $\beta_{jkli}$, $\beta_{jilk}$, $\alpha_{ljik}$, in the expression of $p_{454}$,
\[
[x_l, x_k, y_j]\cdot  y_i +[x_i, x_j, y_k]\cdot  y_l+[y_j, x_i, x_k]\cdot  y_l+[y_i, x_j, x_l]\cdot  y_k
\]
\[
[y_j, x_k, x_l]\cdot  y_i+[y_k, x_j, x_i]\cdot  y_l-[x_l, x_i, y_j]\cdot  y_k-[x_k, x_j, y_i]\cdot  y_l
\]
\[
[y_j, x_i, x_l]\cdot  y_k+[y_i, x_j, x_k]\cdot  y_l +[y_j, x_i, x_k]\cdot  y_l+[y_i, x_j, x_l]\cdot  y_k
\]
\[
[x_l, x_j, y_i]\cdot  y_k+[x_k, x_i, y_j]\cdot  y_l -[y_j, x_i, x_k]\cdot  y_l-[y_i, x_j, x_l]\cdot  y_k-[x_l, x_i, y_j]\cdot  y_k-[x_k, x_j, y_i]\cdot  y_l
\]
are constants, where $j<i<k<l$.

\

\noindent\underline{Case 4} Finally we have to show $(KB_6)^{\delta}=\{0\}$.
Let
\[
p_6=\sum_{i\leq l , \ k<l}\alpha_{ijkl}[y_i, x_j,y_k]\cdot y_l
\]
be a constant of $KB_6$.
We have $p_6=p_{62}+p_{63}+p_{64}$,
where
\[
p_{62}=\sum_{j<i}\alpha_{iiji}[y_i, x_i,y_j]\cdot y_i+\alpha_{ijji}[y_i, x_j,y_j]\cdot y_i+\alpha_{jiji}[y_j, x_i,y_j]\cdot y_i+\alpha_{jjji}[y_j, x_j,y_j]\cdot y_i,
\]
\[
p_{63}=\sum_{j<i<k}\left(\begin{array}{ll}
\alpha_{ikji}[y_i, x_k,y_j]\cdot y_i+\alpha_{kijk}[y_k, x_i,y_j]\cdot y_k+\alpha_{kjik}[y_k, x_j,y_i]\cdot y_k\\
+\alpha_{ikjk}[y_i, x_k,y_j]\cdot y_k+\alpha_{iijk}[y_i, x_i,y_j]\cdot y_k+\alpha_{ijjk}[y_i, x_j,y_j]\cdot y_k\\
+\alpha_{jkji}[y_j, x_k,y_j]\cdot y_i+\alpha_{jijk}[y_j, x_i,y_j]\cdot y_k+\alpha_{ijik}[y_i, x_j,y_i]\cdot y_k\\
+\alpha_{jkik}[y_j, x_k,y_i]\cdot y_k+\alpha_{jiik}[y_j, x_i,y_i]\cdot y_k+\alpha_{jjik}[y_j, x_j,y_i]\cdot y_k \end{array}\right),
\]
\[
p_{64}=\sum_{j<i<k<l}\left(\begin{array}{ll}
\alpha_{iljk}[y_i, x_l,y_j]\cdot y_k+\alpha_{ikjl}[y_i, x_k,y_j]\cdot y_l+\alpha_{kijl}[y_k, x_i,y_j]\cdot y_l\\
+\alpha_{kjil}[y_k, x_j,y_i]\cdot y_l+\alpha_{jlik}[y_j, x_l,y_i]\cdot y_k+\alpha_{jkil}[y_j, x_k,y_i]\cdot y_l\\
+\alpha_{jikl}[y_j, x_i,y_k]\cdot y_l+\alpha_{ijkl}[y_i, x_j,y_k]\cdot y_l\end{array}\right).
\]
Arguing as in the previous cases, we get all coefficients are 0 and we are done with the spanning elements. The final step proving that these spanning elements, all written as linear combinations of basis elements of the free metabelian Poisson algebra, is straightforward.
\end{proof}

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According to new research from the Centers for Disease Control and Prevention (CDC), hospitalizations from three common and preventable waterborne diseases – Legionnaire's disease, cryptosporidiosis and giardiasis – could cost the U.S an estimated $539 million dollars each year. These figures represent only a fraction of all waterborne disease costs. They underscore the large and hidden burden of waterborne disease including direct federal and state expenditures associated with contaminated drinking water and recreational water. Modest investment in waterborne disease prevention programs in the U.S. could lead to significant healthcare cost savings.
According to the study, "These cost data highlight that water-related diseases pose not only a physical burden to the thousands of people sickened by them each year, but also a substantial burden in health care costs, including direct government payments through Medicare and Medicaid," says Michael Beach of the Centers for Disease Control and Prevention.
Using data from a large insurance claims database between 2004 and 2007, Beach and his team estimated the total range of hospitalization costs for giardiasis to be $16-63 million; cryptosporidiosis to be $37-145 million; and Legionnaires' disease to be $101-321 million.
There are several modest changes that can be made to prevent these diseases and reduce health care costs associated with treating waterborne diseases. Examples of these changes include public education campaigns, appropriate maintenance and improvement of building water systems and regular inspection of pools and other recreational water facilities. This research illustrates the importance of properly maintaining water storage and distribution equipment and adjusting swimming pool water chemistry and addressing indoor cooling water systems.
Investment in wastewater and drinking water upgrades will be an important piece of ensuring safe water in the future. Monitoring for pathogens and indicators of safe water are necessary. Most U.S. water systems employ chlorine-based disinfectants to achieve residual protection throughout the distribution system mandated by the Safe Drinking Water Act, and new sensors and real-time assessment will make it possible to monitor and maintain water quality.
For more information about the public health benefits of chlorinated water, please visit our website.


text:
Q: Black body radiation I have a few questions related to the emission of electromagnetic radiation by black bodies.

Consider the following image:

On the above image I have drawn the rays of light that are emmited by black bodies assuming that they have only 8 points of emmision (They are marked with red dots).
Which of these two images shows the real situation? The single point emits the radiation in all directions or only in one direction?

If my intuition is not wrong and the image on the right is correct consider the next image where I assume that there is only one point of emmission on the object:

*

*Is the intensity of radiation the same in every direction?

*Does the radiation in each direction have the same intensity-wavelength distribution (presented in the image below)?

A: This is a difficult concept to talk about without using a few proper definitions.  Unfortunately these radiometric definitions all sound very similar and have similar meanings but important differences.
Intensity is the rate of energy transfer per unit area.  Intensity does not have direction and thus we cannot even talk about radiation having 'different' intensities in different directions by definition.
The radiant intensity, however, is the amount of energy transfer per unit solid angle.
The radiance is the amount of energy transfer per unit area per unit solid angle.
Lambert's cosine law states that the radiant intensity of an ideally diffuse emitter (e.g. a perfect black body) is proportional to $\cos^2\theta$, where $\theta$ is the angle from surface normal.
However, the apparent surface area of a flat object is also reduced by a factor of $\cos^2\theta$ when viewed from an angle $\theta$ from normal.  Thus the radiance does not vary with angle.
When you use the term intensity in your question, you probably mean radiance, which is the most natural term to talk about.  It is the same as talking about how bright an object appears.  To quote wikipedia:

Radiance is useful because it indicates how much of the power emitted by an emitting or reflecting surface will be received by an optical system looking at the surface from some angle of view.

In summary:
Is the radiance the same in every direction? Yes.
Does the radiance in each direction have the same spectral distribution?  Yes.

A: In the classical case a single point emits in all directions.
Obviously it only emits a single photon at a time and that photon only goes in one direction, but averaged over time a blackbody emits photons in all directions with equal probability

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← Jersey Boys closing January 15th!
And this year, there are TWO songs on it from Jersey Boys! One of them I wrote with Barry Anderson on tour last year. We're psyched it made the cut and hope you enjoy!
This entry was posted in Acting, Music, Press, Uncategorized. Bookmark the permalink.


Explore delightful fashion temptation when you carry the fabulous Bottega Veneta™ Intrecciato Flap Coin Purse Wallet.
Made of lambskin leather in the iconic intrecciato design throughout.
Custom leather lined interior features front zipper pocket, three slide compartments, and eight credit card slots for secure, easy storage.
We can't currently find any similar items for the Intrecciato Flap Coin Purse Wallet by Bottega Veneta. Sorry!


Arkoun means thank you in Khmer. It is also an Australian social enterprise that Georgie Demir started in 2017 at the age of thirteen. Profits from the sale of her soaps will help fund the building of wash houses in Cambodian schools.
In 2017 Georgie and her mum traveled to Cambodia and volunteered to build a school wash house to help solve this health problem. A wash house provides 400 children with clean water to brush their teeth and wash their hands.
Georgie saw the good it did and continues to do so she set up my own social enterprise in 2017 at the age of thirteen. Profits from the sale of her handmade soaps will help build future wash houses and save more lives.


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Intervals of clouds and sunshine. High 101F. Winds WNW at 10 to 15 mph..
A few clouds from time to time. A stray shower or thunderstorm is possible. Low 79F. Winds S at 5 to 10 mph.
Ex-UA track and field assistant coach Craig Carter was convicted of two counts of aggravated assault in March.
Arizona Daily Star/
athlete's civil action to stretch into 2020
State-funded legal fees for convicted ex-UA track coach surpass $2.3 million
By Caitlin Schmidt Arizona Daily Star
Caitlin Schmidt
The state of Arizona has spent $2.3 million to defend a former UA assistant track and field coach who admitted to assaulting a student-athlete — and the case is far from over.
Craig Carter was convicted by a Pima County jury last March of two counts of aggravated assault in connection with a May 2015 incident in his office at McKale Center. He admitted to campus police that he choked former UA thrower Baillie Gibson and threatened to cut her face with a box cutter. The two had been involved in a sexual relationship which Gibson says was not consensual.
Carter is currently serving five years in prison.
In November 2015, Gibson filed a lawsuit against Carter and the UA, saying that the school had knowledge of the illicit relationship nearly two years before the attack but failed to intervene. Because Carter was a state employee, his defense in the civil suit is being paid for by the Arizona Department of Administration's Risk Management Division.
The Star asked for an interview with ADOA officials to discuss whether there are any policies in place to limit spending, or if risk management has the authority to limit billing or provide input regarding legal tactics being used in the case.
"The state provides a defense for state employees when acting within their official job capacity, however, we aren't able to discuss ongoing litigation," Megan Rose, ADOA's assistant director of communications, said in an email response.
State Sen. David Bradley, whose district includes parts of Tucson, told the Star that he didn't think the $2.3 million is a good use of taxpayer funds.
"We have calls into ADOA to find the governing rule or statute that's in play here," Bradley told the Star on Thursday.
All indications are that the trial — and the expenses — are going to continue at least into next year. Pima County Superior Court Judge Charles Harrington has yet to set a date for a trial-setting conference. Last month, he set a Dec. 12 deadline to file motions in the case.
Seventeen people have already been deposed in connection with Gibson's case against Carter and the UA. At least 30 depositions requested by the UA's lawyers have yet to be completed; those witnesses are located across the country, according to Pima County Superior Court records. Attorneys have not yet interviewed any expert witnesses, who can be costly.
In the meantime, the bills continue to mount. As of Dec. 31, lawyers from the Tucson firm Munger, Chadwick and Denker had billed $1.44 million to defend Carter in the suit, according to ADOA billing totals obtained by the Star through a public-records request. Munger, Chadwick and Denker have billed between $2,175 to $76,572 per month, records show. John Munger, one of the firm's partners, served on the Arizona Board of Regents from 1992 to 2000.
Those aren't the only attorneys involved in defending the case. In October 2017, the UA opted to hire private lawyers instead of allowing the Arizona Attorney General's Office to continue to defend the school in the case at a reduced rate. Those attorneys, from the Tucson firm Rusing, Lopez and Lizardi, billed $813,949 through Nov. 12 for its defense of the UA, according to ADOA data. The last three months of billings have accounted for nearly half the total amount, with $342,678 of it coming between September and November.
Former UA athletic director Greg Byrne was initially named as a defendant in the suit and hired his own lawyers. The state paid attorneys from Thorpe Shwer $87,063 for work performed on Byrne's behalf between August 2017 and November 2018, ADOA data shows. Byrne was dropped from the civil lawsuit in December 2017.
The Star reached out to eight members of the Arizona Board of Regents last month to ask if they had the ability to intervene or provide input on the course of the case. None of them responded directly. Sarah Harper, ABOR's vice president of communications, then issued a statement on the regents' behalf. It read: "The State of Arizona's Department of Administration determines whether individual university employees are entitled to state coverage of legal costs and fees in civil actions. The terms and conditions around those decisions, including continuation of the coverage, are ultimately within the sole control and authority of the ADOA, not the Arizona Board of Regents.
"The Arizona Board of Regents takes all litigation that is brought against it and the universities for which it is responsible very seriously and regularly assesses the legal claims that are made, with the assistance and advice of legal counsel," Harper said. "In this, as in all other matters, the Board takes actions and makes decisions that it believes to be in the best interests of the citizens of the State of Arizona, and the universities and students whom it serves."
Harper said the regents couldn't comment further, citing pending litigation.
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Watch: Arizona Wildcats DB Demetrius Flannigan-Fowles reacts to Pro Day performance
Contact reporter Caitlin Schmidt at cschmidt@tucson.com or 573-4191. Twitter: @caitlincschmidt
UA creates new Department of Title IX, Equity and Inclusion as part of latest overhaul
Retired South Tucson Judge Ronald Wilson will run the Department of Title IX, Equity and Inclusion.
Police reports detail complaints of violence against women by UA football players
New Title IX director wants UA to be 'national model for how you can do it right'
UA's self-reported statistics paint incomplete, complicated picture of sexual misconduct
Misconduct claims led to probes of 16 University of Arizona athletes, staffers since 2012
Policy and staff changes follow review of UA's efforts against sex-abuse, violence
Turmoil in UA athletic department prompts review of policies involving sexual violence, discrimination
Pima College's JJ Nakai continues to climb Aztecs' scoring charts
Starting anew: UA football players among many taking advantage of NCAA transfer portal
Tucson's Tumamoc Hill to reopen Wednesday
More sex abuse, domestic violence cases in UA athletics than first claimed
Three in UA football program were disciplined for involvement in alleged sexual assault
The future of AZ criminal justice reform is unclear as several bills fail to advance in House
UA consortium to research and address gender-based violence
UA grad student creates path to healing for sexual assault survivors
University of Arizona reaches settlement with former athlete in years-long lawsuit
State paid $999,000 to settle lawsuit by former University of Arizona track athlete
I'm a watchdog reporter covering local government, the University of Arizona and sports investigations.
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A Tree Grows in Palestine
Rabbledocs
rabble docs A tree grows in Palestine final.mp3
Today's rabble doc is about a tree. A 5,000-year-old tree in the Israeli occupied West Bank. But what you'll hear isn't just about a tree. It's about a town, a community and the Israeli occupation of the West Bank and features first hand account of lives under siege.
Tourists come to Al-Walaja from around the world, or from nearby Bethlehem, to enjoy the lovely surrounding landscape. A huge, gnarly olive tree, reputedly over 5,000 years old, is a big draw. For political tourists, Israel's imposing and highly illegal security barrier, soon to enclose little Al-Walaja in a sort of a cage, is a must-see.
David toured Al-Walaja in the company of a congenial guy named Baha Hilo in June 2017. The history Baha shared is the story of the Palestinian people in microcosm.
David Kattenburg is one of this country's most prolific independent audio documentary producers. His online news magazine, The Green Planet Monitor, covers a wide range of social justice issues from around the world. Of all the stories Dave reports on, Palestine is one of his main areas of focus. He tries to do stories from there every year, most recently just a couple of months ago.
"A Tree Grows in Palestine" is from his most recent trip in June 2017.
David Kattenburg is the publisher of the online newsmagazine The Green Planet Monitor. You can read the print version of "A Tree Grows in Palestine" here.
Image: David Kattenburg: Heading Down to see a big old tree. Used with permission.
Like this podcast? rabble is reader/listener supported journalism.
israeli occupation


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Nakamba tipped to remain at Villa until season end
Soccer24 on 5 Dec, 2020
Ashley Preece of Birmingham Mail newspaper is backing Marvelous Nakamba to stay at Aston Villa until the end of the season.
The Zimbabwean midfielder has been linked with an exit in January after struggling to get game time. He has only played a total of thirteen minutes in the league so far.
Reports in Turkey suggest Galatasaray have reignited their interest in Nakamba and are considering a loan move.
But Preece believes Villa will not accept the offer as the player remains an integral back-up in the midfield.
He said: "Marvelous Nakamba is another player who's rarely featured having tasted just 13 minutes of top-flight action.
"Villa aren't, though, blessed with too many options in central midfield with the Zimbabwean set to stick around until the summer at the earliest."
Nakamba joined Villa from Club Brugge in 2019 and was a regular in his debut season at Villa Park. He made 34 appearances in all competitions but has since fallen out of favour.
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Marvelous Nakamba dropped for Watford clash
Nakamba turned down Warriors captaincy
Nakamba's squad number at Aston Villa revealed
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Movie Review: "Like a Boss" (2020)
Movie: "Like a Boss"
Director: Miguel Arteta
Mia and Mel are two lifelong best friends who founded a makeup company together. Unfortunately, they are over $400K in debt. To save their business, they take a deal from a successful cosmetics company run by Claire Luna, who does her best to destroy their friendship so she can steal their ideas and take over their company.
In "Like a Boss," Mia (Tiffany Haddish) and Mel (Rose Byrne), two business partners/best friends, hope to do business with a large cosmetics company run by Claire Luna (Salma Hayek). (Image Source)
"Like a Boss" (2020) is supposed to be a movie about female empowerment that was made for women that also stars women and has a story written by veteran comedy writer Danielle Sanchez-Witzel. Who better to write the screenplay than Adam Cole-Kelly and Sam Pitman, two men with almost zero experience? And who better to direct than Miguel Arteta, who has never made anything that has risen above "fine"? The story revolves around two lifelong best friends named Mel (Rose Byrne) and Mia (Tiffany Haddish), who are not only roommates but business partners and founders of a beauty company called, what else, Mel and Mia's. Mia is the front-end personality, makeup artist, and inventor of many of the company's most unique, colorful products. Mel handles most of the back-end financials and the business side of things. It turns out, they are now over $400,000 in debt. When cosmetic giant Claire Luna (Salma Hayek) offers to pay off all their debt for a 49% share of the company, Mel is quick to jump at the offer. Mia is reluctant and doesn't trust Claire, but she does trust Mel, so she agrees, and they take her proposal to save Mel and Mia's. There's just one catch. If either Mia or Mel quit the business, Claire Luna's ownership stake jumps to 51%, giving her the controlling share in their brand. Of course, Claire makes it her mission to drive a wedge between the two besties for her own self-interest.
After being fired from Mel (Rose Byrne) and Mia's (Tiffany Haddish) cosmetic company, Barrett (Billy Porter) stages a dramatic exit in the middle of a restaurant in a movie still for "Like a Boss" (2020). (Image Source)
We are big fans of both Tiffany Haddish and Rose Byrne. Both women have delivered some fantastic comedic performances in the past. Even though both of them have an occasional "miss" on their track record, we still had high hopes for "Like a Boss" (2020) given their overall talent. Those hopes were quickly dashed as the narrative of the film unfolded over time.
Byrne and Haddish don't give bad performances in this, though both of their characters' personalities can be occasionally grating. Mel is always way too eager to please, especially when Claire Luna enters the picture. Also, for someone who deals with the books of their business, she doesn't seem to understand what a business deal with Claire actually entails money wise. At one point, Mel doesn't know Claire will get $0.49 out of every dollar of profit they earn. It's business 101, and if she doesn't know that, it may have something to do with why they are hemorrhaging money. Mia is skeptical of every situation, even if she doesn't need to be. She's always at an 11, which we've come to expect from Haddish, but talking about "blue waffles" only goes so far when it doesn't really fit in the narrative. Also, old meme. The comedy in "Like a Boss" (2020) works best when it showcases Byrne and Haddish shooting the shit together since they play well off of one another and legitimately feel like long-time friends, but they aren't given a solid script to work with, and it puts a damper on the entire project. The supporting cast is hit and miss, too. The biggest laughs almost always come from Billy Porter's character Barrett, though his best scene is spoiled in the trailer. This just reaffirms what we already thought: Billy Porter needs to be in everything! Despite their best efforts, the rest of the cast and the characters they embody leave a lot to be desired. Even though we usually love everything she does, Salma Hayek's vicious beauty mogul antagonist didn't work for us, mostly because her performance felt very forced. Hayek failed to bring any laughs our way, which is a huge bummer. The satirical misogynist business rivals played by Ryan Hansen and Jimmy O. Yang also fell flat. What wound up being a stupid, basic as hell, irritating inclusion could have been a smart commentary on men in makeup in better scriptorial hands. Jennifer Coolidge has been doing the same routine for decades, and it's now tired in 2020. The way the story unfolds is too contrived and too formulaic to be engaging, and the humor just doesn't work most of the time. We see every "big moment" coming from ten miles away, and we could have overlooked this fact if we were too busy laughing to notice, but we weren't, at least not often enough to justify such poor execution.
Salma Hayek stars as Claire Luna, a makeup mogul with sinister intentions, in the comedy film "Like a Boss" (2020). (Image Source)
"Like a Boss" (2020) is basically what would happen if three men wrote and directed a movie about how they think women would act and talk and joke if they are boss business owners. Oh, wait...it is! There are a few fleeting moments that garnered big laughs from us, but, unfortunately, those instances never came together to make a cohesive, satisfying experience. All in all, this is a generic "female empowerment" movie we wish had worked a lot better than it did.
My Rating: 4/10
BigJ's Rating: 4/10
RT Rating: 20%
Do we recommend this movie: No.
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Posted by Lolo at Wednesday, January 22, 2020
Labels: 2020, 2020's, comedy, movie review


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Сан-Жорді-Дасбальш — село в Іспанії, у складі автономної спільноти Каталонія, у провінції Жирона. Муніципалітет займає площу 11,63 км2, а населення в 2014 році становило 708 осіб.

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