of the $2$-signature of $M$, we define $\tilde f_2 M$ as follows: $$\label{lem2}
          \tilde f_2 M=
          \begin{cases}
          M X_{\bar3}(-1)^{(0,-1)}X_{\bar2}(-1)^{(0,1)}=M A_2(0)^{-1}
                \ \ \ \textup{if $X={X_{\bar3}(-1)}^{(0,b_{\bar3}^{-1})}$},\\
          MX_2(-1)^{(0,-1)}X_{3}(-1)^{(0,1)}=MA_2(-1)^{-1}
                 \ \textup{if $X={X_2(-1)}^{(0,b_2^{-1})}$}.
          \end{cases}$$ We define $$\label{lem4}
          \tilde f_2 M=M{X_2(-2)}^{(0,-1)}{X_{3}(-2)}^{(0,1)}
          =MA_2(-2)^{-1}$$ if no $0$ remains.

-   Depending on the component $X$ corresponding to the rightmost $1$ of the $2$-signature of $M$, we define $\tilde e_2 M$ as follows: $$\tilde e_2 M=
          \begin{cases}
          M X_{\bar3}(-1)^{(0,1)}X_{\bar2}(-1)^{(0,-1)}=MA_2(0)
                \quad\ \ \textup{if $X={X_{\bar2}(-1)}^{(0,b_{\bar2}^{-1})}$},\\
          M X_2(-1)^{(0,1)}X_{3}(-1)^{(0,-1)}=MA_2(-1)
                \quad \textup{if $X={X_{3}(-1)}^{(0,b_{3}^{-1})}$},\\
          M {X_2(-2)}^{(0,1)}{X_{3}(-2)}^{(0,-1)}=MA_2(-2)
                \quad \textup{if $X={X_3(-2)}^{(0,b_3^{-2})}$}.
          \end{cases}$$ We define $\tilde e_2 M=0$ if no $1$ remains.

We first show that the action of these operators is closed on $\M(\infty)$.

For $M\in\M(\infty)$, if the $i$-signature of $M$ contains at least one $0$, then the exponent of component ${X_j(-1)}^{(0,b_j^{-1})}$ corresponding to the left-most $0$ shows the property $(0,b_j^{-1})\ge
(0,1)$. In particular, in dealing with the action of $\tilde f_1$, if the left-most $0$ corresponds to ${X_3(-1)}^{(0,b_3^{-1})}$, it means that the exponent $b_0^{-1}$ of ${X_0(-1)}^{(0,b_0^{-1})}$, a component of $M$, is $0$, due to the $(0,1)$-pair cancellation rule. Thus the monomial $\fit M$ defined in  and is contained in $\M(\infty)$. In the case where $i$-signature of $M$ contains no $0$, the exponents of the components $X_i(-i)$ 