$X_{\bar j}(m)^{(u,v)}$.

\[propnk\] Consider elements of $\Me$ having the form $$~\label{equ:jin18}
\begin{aligned}
      M=& X_1(-1)^{(2,-b^{-1}_2-b^{-1}_3-b^{-1}_0-
                      b^{-1}_{\bar3}-b^{-1}_{\bar2}-b^{-1}_{\bar1})}
          X_2(-1)^{(0,b_2^{-1})}
          X_3(-1)^{(0,b_3^{-1})}\\
        & \cdot
          X_0(-1)^{(0,b_0^{-1})}
          X_{\bar3}(-1)^{(0,b_{\bar3}^{-1})}
          X_{\bar2}(-1)^{(0,b_{\bar2}^{-1})}
          X_{\bar1}(-1)^{(0,b_{\bar1}^{-1})}\\
        & \cdot
          X_2(-2)^{(1,-b_3^{-2})}
          X_3(-2)^{(0,b_3^{-2})}
\end{aligned}$$ where $b_i^j\ge 0$ for all $i,j$ and $b_0^{-1}\le 1$. Each element of $\M(\infty)$ may be written uniquely in this form. Conversely, any element of this form is an element of $\M(\infty)$.

Given any monomial $$\begin{aligned}
      M=&{Y_1(-1)}^{(1,a_1^{-1})}{Y_1(0)}^{(0,a_1^0)}
        {Y_1(1)}^{(0,a_1^1)}{Y_1(2)}^{(0,a_1^2)}\\
        &\cdot {Y_2(-2)}^{(1,a_2^{-2})}{Y_2(-1)}^{(0,a_2^{-1})}
        {Y_2(0)}^{(0,a_2^0)}{Y_2(1)}^{(0,a_2^1)}\in\M(\infty),
\end{aligned}$$ through simple computation, we can obtain the expression $$~\label{equ:jin19}
\begin{aligned}
M=& X_1(-1)^{(2,a_1^{-1}+3a_2^{-2})}
    X_2(-1)^{(0,a_2^{-1}-a_2^{-2})}\\
  & \cdot X_3(-1)^{(0,t_3^{-1})}
    X_0(-1)^{(0,t_0^{-1})}
    X_{\bar 3}(-1)^{(0,t_{\bar3}^{-1})}\\
  & \cdot X_{\bar2}(-1)^{(0,-a_2^1)}
    X_{\bar1}(-1)^{(0,-a_1^2)}X_2(-2)^{(1,a_2^{-2})}
    X_3(-2)^{(0,-a_2^{-2})},
\end{aligned}$$ where either

-   $t_0^{-1}=0$,  $2t_3^{-1}=a_1^0+a_2^{-1}-a_2^{-2}$,  and $2t^{-1}_{\bar 3}=-a_1^1-a_2^1$,

or

-   $t^{-1}_0=1$,  $2t^{-1}_3=a_1^0+a_2^{-1}-a_2^{-2}-1$,  and $2t^{-1}_{\bar 3}=-a_1^1-a_2^1-1$.

Since $M\in\M(\infty)$, from the conditions given in , we obtain the form given in . Specifically, the element $M_{\infty}={Y_1(-1)}^{(1,0)}{Y_2(-2)}^{(1,0)}$ corresponds to ${X_1(-1)}^{(2,0)}{X_2(-2)}^{(1,0)}$.

Conversely, given any monomial of the form , we have $$\label{equ:jin7}
\begin{aligned}
M=&{Y_1(-1)}^{(1,-b^{-1}_2-b^{-1}_3-b^{-1}_0-b^{-1}_{\bar3}
                 -b^{-1}_{\bar2}-b^{-1}_{\bar1}+3b^{-2}_3)}\\
  &\cdot {Y_1(0)}^{(0,b^{-1}_0-b^{-1}_2+2b^{-1}_3)}
   {Y_1(1)}^{(0,-b^{-1}_0-2b^{-1}_{\bar 3}+b^{-1}_{\bar2})}
   {Y_1(2)}^{(0,-b^{-1}_{\bar1})}\\
  &\cdot {Y_2(-2)}^{(1,-b^{-2}_3)}
         {Y_2(-1)}^{(0,b^{-1}_2-b^{-2}_3)}
         {Y_2(0)}^{(0,b^{-1}_{\bar3}-b^{-1}_3)}
         {Y_2(1)}^{(0,-b^{-1}_{\bar2})}.
\end{aligned}$$ It is now straightforward to check that $M\in\M(\infty)$. We have thus shown that $\M(\infty)$ consists of elements of the form .

The uniqueness part may be proved through simple computation.

There are other ways to write each element of $\M(\infty)$ as products of the 