\left(P^{s,\,t}_{s+t+m-k-l,\,n}-P^{s+t-m+k-n,\,l}_{s,\,t} \right)=
\left\{
\begin{aligned}
=0,&\quad\text{as}\quad m=k,\,\,n=l=t\\
\neq 0,&\quad\text{otherwise};\\
\end{aligned}
\right.
\label{13-7.8d}\\
\left[\widehat{P}^{k,\,[l]}_{m,\,[n]},\,P^{s,\,t}_{s,\,t} \right]=
\left(P^{s,\,t}_{m,\,s+t+n-k-l}-P^{k,\,s+t-m+l-n}_{s,\,t} \right)=
\left\{
\begin{aligned}
=0,&\quad\text{as}\quad n=l,\,\,m=k=s\\
\neq 0,&\quad\text{otherwise};\\
\end{aligned}
\right.
\label{13-7.8e}\\
\left[P^{k,\,l}_{m,\,n},\,P^{s,\,t}_{s,\,t} \right]=
\left\{
\begin{aligned}
P^{s,\,t}_{m,\,n},&\quad\text{as}\quad (k,l)=(s,t)\neq(m,n)\\
-P^{k,\,l}_{s,\,t},&\quad\text{as}\quad (m,n)=(s,t)\neq(k,l)\\
 0,&\quad\text{otherwise};\\
\end{aligned}
\right.
\qquad\qquad\qquad\label{13-7.8f}\end{aligned}$$

We remind that all generators are linearly independent. Then, from it follows that all coefficients at nonzero commutators must be equal to zero. This allows us to simplify the as follows $$\begin{gathered}
\label{13-7.9}
   \qquad\qquad z=d_0I+d_1\mathbb{N}_1+d_2\mathbb{N}_2+
    \sum_{k\geq 0}
    d^{(L)}_{k,\,t;\,k,\,t}\widehat{P}^{[k],\,t}_{[k],\,t}+
    \sum_{l\geq 0}
    d^{(R)}_{s,\,l;\,s,\,l}\widehat{P}^{s,\,[l]}_{s,\,[l]}
    +\\
    \sum_{{\begin{matrix}(k,l)\neq(s,t)\neq(m,n)\\(k,l)\neq(m,n)\end{matrix}}}
   \hspace{-1cm} d^{(D)}_{k,\,l;\,m,\,n}P^{k,\,l}_{m,\,n}+
    \sum_{k,\,l\geq 0}d^{(D)}_{k,\,l;\,k,\,l}P^{k,\,l}_{k,\,l}.\qquad\qquad\qquad\qquad\end{gathered}$$ We can put $(s,\,t)=(k,\,l)$ or $(s,\,t)=(m,\,n)$, because $s,\,t$ are arbitrary integer numbers. Then we can prove that all coefficients $d^{(D)}_{k,\,l;\,m,\,n}$ in the penultimate sum in the right-hand side of equals to zero. Thus become simpler and takes the form $$\label{13-7.10}
z=d_0I+d_1\mathbb{N}_1+d_2\mathbb{N}_2
+\sum_{k\geq 0}d^{(L)}_{k,\,t;\,k,\,t}\widehat{P}^{[k],\,t}_{[k],\,t}
+\sum_{l\geq 0}d^{(R)}_{s,\,l;\,s,\,l}\widehat{P}^{s,\,[l]}_{s,\,[l]}
+\sum_{k,\,l\geq 0}d^{(D)}_{k,\,l;\,k,\,l}P^{k,\,l}_{k,\,l}.$$ Let X be one of ladder operators, for example $\asp$. Then, from , and $$[\mathbb{N}_1,\asp]=-\asp\neq 0,\quad    [\mathbb{N}_2,\asp]=\asp\neq 0,$$ we conclude that all coefficients in the right-hand side of the equality , excluding $d_0$, are equal to zero. So, we have $z=d_0 I$ that is, $$\mathfrak{Z}(\mathfrak{A})=d_0 I,$$ which means that the center $\mathfrak{Z}(\mathfrak{A})$ of the algebra $\mathfrak{A}$ is one-dimensional.

 Maximal Abelian subalgebra of the algebra $\AG$
------------------------------------------------

We consider the commutative 