delta_{i+1}v_{i+1}\cdot m_{i+1, i}=\pi m_{i,i}^{\ast}$$ such that $m_{i,i}^{\ast} \in M_{1\times n_i}(B\otimes_AR)$. This equation is considered in $B\otimes_AR$ and $\pi$ stands for $\pi\otimes 1\in B\otimes_AR$. Here,

    -   $v_{i-1}=(0,\cdots, 0, 1)$ (resp. $v_{i-1}=(0,\cdots, 0, 1, 0)$) of size $1\times n_{i-1}$ if $L_{i-1}$ is *of type* $\textit{I}^o$ (resp. *of type* $\textit{I}^e$).

    -   $v_{i+1}=(0,\cdots, 0, 1)$ (resp. $v_{i+1}=(0,\cdots, 0, 1, 0)$) of size $1\times n_{i+1}$ if $L_{i+1}$ is *of type* $\textit{I}^o$ (resp. *of type* $\textit{I}^e$).

-   Assume that $i$ is odd and that $L_i$ is *bound of type I*. Then $$\delta_{i-1}v_{i-1}\cdot {}^tm_{i, i-1}+\delta_{i+1}v_{i+1}\cdot {}^tm_{i, i+1} = \pi m_{i,i}^{\ast\ast}$$ such that $ m_{i,i}^{\ast\ast} \in M_{1\times n_i}(B\otimes_AR)$. This equation is considered in $B\otimes_AR$ and $\pi$ stands for $\pi\otimes 1\in B\otimes_AR$. Here, $v_{i-1}$ (resp. $v_{i+1}$)$=(0,\cdots, 0, 1)$ of size $1\times n_{i-1}$ (resp. $1\times n_{i+1}$).\

Let $$\tilde{M_i}= \mathrm{GL}_{B/\pi B}(B_i/Y_i) \textit{ for all $i$}.$$

Let $s_i=m_{i,i}$ if $L_i$ is *of type II* or if $L_i$ is *bound of type I* with $i$ odd in the above description of an element of $\tilde{M}(R)$. Then $s_i$ mod $\pi\otimes 1$ is an element of $\tilde{M}_i(R)$. Therefore, we have a surjective morphism of algebraic groups $$r : \tilde{M} \longrightarrow \prod\tilde{M}_i, ~~~~~~~ m \mapsto \prod \left(\textit{$s_i$ mod $\pi\otimes 1$}\right)$$ defined over $\kappa$. We now have the following easy lemma:

\[la2\] The kernel of $r$ is the unipotent radical $\tilde{M}^+$ of $\tilde{M}$, and $\prod\tilde{M}_i$ is the maximal reductive quotient of $\tilde{M}$.\

Since $\prod\tilde{M}_i$ is a reductive group, we only have to show that the kernel of $r$ is a connected smooth unipotent group. By the description of the morphism $r$ in terms of matrices explained above, the kernel of $r$ is isomorphic to an affine space as an algebraic variety over $\kappa$. Therefore, it is a connected smooth unipo