\newpage

\begin{lem}
  Let \( \lambda \in \pdiw \).
  Let \( \alpha^\vee, \beta^\vee, \gamma^\vee, \delta_1^\vee, \delta_2^\vee \in \shape{\lambda} \) satisfy
  \begin{itemize}
    \item
    \( \alpha^\vee \horder{} \delta_i^\vee \horder{} \beta^\vee \,\,\,(i=1,2) \);
    \item
    \( \alpha^\vee + \beta^\vee = \delta_1^\vee + \delta_2^\vee \);
    \item
    \( \langle \gamma,\, \delta_1^\vee \rangle = 0 , \langle \gamma,\, \delta_1^\vee \rangle = 1 \);
  \end{itemize}
\end{lem}

  (Subcase (1B)) \quad
  Since \( \langle \delta_1,\, \delta_2^\vee \rangle = 0 \),
  we have \( \langle \delta_2,\, \beta^\vee \rangle = 1 \).

  Now, we note that \( 2\delta_2^\vee - \gamma^\vee = -s_{\delta_2}\gamma^\vee \in \rroot \).
  We consider the following two subsubcases:
  \begin{enumerate}
    \item[(1Ba)]
    \( 2\delta_2^\vee - \gamma^\vee > 0 \).
    \item[(1Bb)]
    \( 2\delta_2^\vee - \gamma^\vee < 0 \).
  \end{enumerate}

  (Subsubcase (1Ba)) \quad
  Put
  \begin{equation*}
    \begin{split}
    \eta_3^\vee &:= \delta_1^\vee - \alpha_i^\vee = s_i \delta_1^\vee \in \rroot_+^\vee , \\
    \eta_2^\vee &:= \alpha_i^\vee \in \rroot_+^\vee , \\
    \eta_1^\vee &:= \gamma^\vee - \delta_1^\vee - \alpha_i^\vee = s_i s_{2\delta_2^\vee-\gamma^\vee} \delta_2^\vee \in \rroot_+^\vee , \\
    \eta_0^\vee &:= 2\delta_2^\vee - \gamma^\vee \in \rroot_+^\vee .
    \end{split}
  \end{equation*}
  Then, it is straightforward to see
  \begin{equation*}
    \MATd{
    \langle \eta_3 ,\, \eta_3^\vee \rangle & \langle \eta_3 ,\, \eta_2^\vee \rangle & \langle \eta_3 ,\, \eta_1^\vee \rangle & \langle \eta_3 ,\, \eta_0^\vee \rangle
    \\
    \langle \eta_2 ,\, \eta_3^\vee \rangle & \langle \eta_2 ,\, \eta_2^\vee \rangle & \langle \eta_2 ,\, \eta_1^\vee \rangle & \langle \eta_2 ,\, \eta_0^\vee \rangle
    \\
    \langle \eta_1 ,\, \eta_3^\vee \rangle & \langle \eta_1 ,\, \eta_2^\vee \rangle & \langle \eta_1 ,\, \eta_1^\vee \rangle & \langle \eta_1 ,\, \eta_0^\vee \rangle
    \\
    \langle \eta_0 ,\, \eta_3^\vee \rangle & \langle \eta_0 ,\, \eta_2^\vee \rangle & \langle \eta_0 ,\, \eta_1^\vee \rangle & \langle \eta_0 ,\, \eta_0^\vee \rangle
  }
  =
  \MATd{
     2 & -1 &  0 &  0 \\
    -1 &  2 & -1 &  0 \\
     0 & -1 &  2 & -2 \\
     0 &  0 & -1 &  2
  } ,
  \end{equation*}
  which is a cartan matrix of type \( B_4 \).
  We have
  \( \gamma^\vee = 2 \eta_2^\vee + 2 \eta_1^\vee + \eta_0^\vee \) and
  \( \beta^\vee = \eta_3^\vee + \eta_2^\vee + \eta_1^\vee + \eta_0^\vee \).
  This contradicts \( \beta^\vee < \gamma^\vee \).

  (Subsubcase (1Bb)) \quad
  Put
  \begin{equation*}
    \begin{split}
    \eta_3^\vee &:= \delta_1^\vee - \alpha_i^\vee = s_i \delta_1^\vee \in \rroot_+^\vee , \\
    \eta_2^\vee &:= \alpha_i^\vee \in \rroot_+^\vee , \\
    \eta_1^\vee &:= \delta_2^\vee - \alpha_i^\vee = s_i \delta_2^\vee \in \rroot_+^\vee , \\
    \eta_0^\vee &:= \gamma^\vee - 2\delta_2^\vee \in \rroot_+^\vee .
    \end{split}
  \end{equation*}
  Then, it is straightforward to see
  \begin{equation*}
    \MATd{
    \langle \eta_3 ,\, \eta_3^\vee \rangle & \langle \eta_3 ,\, \eta_2^\vee \rangle & \langle \eta_3 ,\, \eta_1^\vee \rangle & \langle \eta_3 ,\, \eta_0^\vee \rangle
    \\
    \langle \eta_2 ,\, \eta_3^\vee \rangle & \langle \eta_2 ,\, \eta_2^\vee \rangle & \langle \eta_2 ,\, \eta_1^\vee \rangle & \langle \eta_2 ,\, \eta_0^\vee \rangle
    \\
    \langle \eta_1 ,\, \eta_3^\vee \rangle & \langle \eta_1 ,\, \eta_2^\vee \rangle & \langle \eta_1 ,\, \eta_1^\vee \rangle & \langle \eta_1 ,\, \eta_0^\vee \rangle
    \\
    \langle \eta_0 ,\, \eta_3^\vee \rangle & \langle \eta_0 ,\, \eta_2^\vee \rangle & \langle \eta_0 ,\, \eta_1^\vee \rangle & \langle \eta_0 ,\, \eta_0^\vee \rangle
  }
  =
  \MATd{
     2 & -1 &  0 &  0 \\
    -1 &  2 & -1 &  0 \\
     0 & -1 &  2 & -2 \\
     0 &  0 & -1 &  2
  } ,
  \end{equation*}
  which is a cartan matrix of type \( B_4 \).
  We have
  \( \gamma^\vee = 2 \eta_2^\vee + 2 \eta_1^\vee + \eta_0^\vee \) and
  \( \beta^\vee = \eta_3^\vee + \eta_2^\vee + \eta_1^\vee \).
  This contradicts \( \beta^\vee < \gamma^\vee \).



\begin{lem}[—\‘z]\label{lem:value3}
  Let \( \lambda \in \pdiw \).
  Let
  \( \gamma^\vee, \delta^\vee \in \shape{\lambda} \) satisfy \( \langle \delta ,\, \gamma^\vee \rangle \geq 3 \).
  Then we have
  \begin{enumerate}
    \item
    \( \delta^\vee \leq \gamma^\vee \).
    \item
    \( \langle \gamma ,\, \delta^\vee \rangle = 1 \).
    \item
    \( s_\delta \gamma^\vee > 0 \), i.e. \( \langle \delta ,\, \gamma^\vee \rangle \delta^\vee \leq \gamma^\vee \).
  \end{enumerate}
\end{lem}
\begin{proof}
  (1)\quad
  This follows from Lemma \ref{lem:value} and our assumption.

  (2)\quad
  By part (1) and Lemma \ref{lem:value}, we have \( \langle \gamma ,\, \delta^\vee \rangle = 1,2 \).
  If \( \langle \gamma ,\, \delta^\vee \rangle = 2 \),
   then, by Lemma \ref{lem:value}, we get \( \langle \delta ,\, \gamma^\vee \rangle = 1,2 \).
  This contradicts our assumption.
  Hence, we get \( \langle \gamma ,\, \delta^\vee \rangle = 1 \).

  (3)\quad
%  Suppose \( \langle \delta,\, \gamma^\vee \rangle \geq 4 \).
%
  If \( s_\delta (\gamma^\vee) < 0 \), then we have
  \begin{equation*}
    \begin{split}
    -1
    &\leq
    \langle \lambda ,\, -s_\delta (\gamma^\vee) \rangle
    =
    \langle \lambda ,\, - \gamma^\vee + \langle \delta ,\, \gamma^\vee \rangle \delta^\vee \rangle
    \\&=
    - \langle \lambda ,\,\gamma^\vee \rangle
    + \langle \delta ,\, \gamma^\vee \rangle \langle \lambda ,\, \delta^\vee \rangle
    =
    1 - \langle \delta ,\, \gamma^\vee \rangle .
    \end{split}
  \end{equation*}
  This contradicts \( \langle \delta ,\, \gamma^\vee \rangle \geq 3 \).
  Hence, we have
  \( \gamma^\vee -\langle \delta ,\, \gamma^\vee \rangle \delta^\vee = s_\delta (\gamma^\vee) > 0 \).
  
\end{proof}


\begin{lem}[—\‘z]
  \( \gamma^\vee, \delta^\vee \in \Phi_+ \) with
  \( p:= \langle \delta ,\, \gamma^\vee \rangle >0 \)
  and \( q:= \langle \gamma ,\, delta^\vee \rangle >0 \).
  Suppose \( pq \geq 4 \).
  Then we have
  \begin{enumerate}
    \item
    \( \delta^\vee \leq \gamma^\vee \).
    \item
    \( \langle \gamma ,\, \delta^\vee \rangle = 1 \).
    \item
    \( \langle \delta ,\, \gamma^\vee \rangle = 3 \).
  \end{enumerate}
\end{lem}
  \begin{equation*}
    \begin{array}{ll}
	\gamma^\vee					& \delta^\vee				\\
	\hline
	s_\delta(\gamma^\vee)				& s_\gamma(\delta^\vee)			\\
	= \gamma^\vee - p \delta^\vee			& = \delta^\vee - q \gamma^\vee		\\
	\hline
	s_\gamma s_\delta(\gamma^\vee)			& s_\delta s_\gamma(\delta^\vee)	\\
	= (pq-1)\gamma^\vee - p \delta^\vee		& = (pq-1)\delta^\vee - q \gamma^\vee	\\
	\hline
	s_\delta s_\gamma s_\delta(\gamma^\vee)		& s_\gamma s_\delta s_\gamma(\delta^\vee)	\\
	= (pq-1)\gamma^\vee - (pq-2)p \delta^\vee	& = (pq-1)\delta^\vee - (pq-2)q \gamma^\vee	\\
	\hline
	s_\gamma s_\delta s_\gamma s_\delta(\gamma^\vee)	& s_\delta s_\gamma s_\delta s_\gamma(\delta^\vee)	\\
	= ((pq)^2-3pq+1)\gamma^\vee - (pq-2)p \delta^\vee	& = ((pq)^2-3pq+1)\delta^\vee - (pq-2)q \gamma^\vee	\\
	\hline
	s_\delta s_\gamma s_\delta s_\gamma s_\delta(\gamma^\vee)	& s_\gamma s_\delta s_\gamma s_\delta s_\gamma(\delta^\vee)	\\
	= ((pq)^2-3pq+1)\gamma^\vee - ((pq)^2-4pq+3)p \delta^\vee	& = ((pq)^2-3pq+1)\delta^\vee - ((pq)^2-4pq+3)q \gamma^\vee	\\
    \end{array}
  \end{equation*}

