at least one $0$ and $X$ is the component corresponding to the left-most $0$ in the $2$-signature of $M$, then we can obtain $$m_f=\textup{min}\{j\in\Z\ \vert\
    \textup{max}\{ \sum_{k\le j} y_2(k)\}\}=
\begin{cases}
0 &\text{for $X={X_{\bar 3}(-1)}^{(0,b_{\bar 3}^{-1})}$},\\
-1 &\text{for $X={X_2(-1)}^{(0,b_2^{-1})}$},
\end{cases}$$ where $y_2(k)$ is the exponent of $Y_2(k)$ appearing in $M$ given by expression .

If the $i$-signature of $M$ contains no $0$, $$m_f=\textup{min}\{j\in\Z\ \vert\
     \textup{max}\{ \sum_{k\le j} y_i(k)\}\}=-i,$$ where $y_i(k)$ is the exponent of $Y_i(k)$ appearing in $M$ given by expression .

In all cases, we can confirm that $m_f=m$, where $m$ is given through equations  to  stating $\fit M = MA_i(m)^{-1}$. Proof for the statements concerning $\eit$ may be done in a similar manner.

In the above lemma, we showed that the action of Kashiwara operators  and  on $\Me$ satisfy the following properties: $$\fit \M(\infty) \subset \M(\infty), \qquad \eit \M(\infty) \subset
\M(\infty) \cup \{0\} \qquad \text{for all} \ \ i\in I.$$ Thus we obtain the following result.

\[zaq\] The set $\M(\infty)$ forms a $\uq(G_2)$-subcrystal of $\Me$.

Figures \[fig:3\] illustrates the top part of crystal $\M(\infty)$ for finite type $G_2$. It was obtained by applying the Kashiwara actions introduced in Lemma \[anaysiskashi2\] on the new expression for elements of $\M(\infty)$. Readers 