ined in Lemma \[l46\] and that the dimension of $\tilde{G}^1$ is given in Theorem \[ta4\].

-   $\beta$ is the number of integers $j$ such that $L_j$ is of type I and $L_{j+2}, L_{j+3}, L_{j+4}$ (resp. $L_{j-1},$ $L_{j+1},$ $L_{j+2}, L_{j+3}$) are of type II if $j$ is even (resp. odd).

Lemma \[la1\] and Theorem \[ta4\] imply that $\mathrm{Ker~}\varphi/\tilde{G}^1 $ represents the functor $R\mapsto \mathrm{Ker~}\varphi(R)/\tilde{G}^1(R)$. Recall that $\mathrm{Ker~}\varphi/\tilde{G}^1$ is a closed subgroup scheme of $\mathrm{Ker~}\tilde{\varphi}/\tilde{M}^1$ as explained at the paragraph just before Theorem \[ta4\]. Let $m= \begin{pmatrix} \pi^{max\{0,j-i\}}m_{i,j} \end{pmatrix} \mathrm{~with~}z_i^{\ast}, m_{i,i}^{\ast}, m_{i,i}^{\ast\ast}$ be an element of $(\mathrm{Ker~}\tilde{\varphi}/\tilde{M}^1)(R)$ such that $m$ belongs to $(\mathrm{Ker~}\varphi/\tilde{G}^1)(R)$. We want to find equations which $m$ satisfies. Note that the entries of $m$ involve $(B\otimes_AR)/(\pi\otimes 1)(B\otimes_AR)$ as explained in Remark \[ra5\].

Recall that $h$ is the fixed hermitian form and we consider it as an element in $\underline{H}(R)$ as explained in Remark \[r33\].(2). We write it as a formal matrix $h=\begin{pmatrix} \pi^{i}\cdot h_i\end{pmatrix}$ with $(\pi^{i}\cdot h_i)$ for the $(i,i)$-block and $0$ for the remaining blocks. We choose a representative $1+x\in \mathrm{Ker~}\varphi(R)$ of $m$ so that $h\circ (1+x)=h$. Any other representative of $m$ in $\mathrm{Ker~}\tilde{\varphi}(R)$ is of the form $(1+x)(1+\pi y)$ with $y \in \underline{M}'(R)$ and we have $h\circ (1+x)(1+\pi y)=h\circ (1+\pi y)$. Notice that $h\circ (1+\pi y)$ is an element of $\underline{H}(R)$ so we express it as $(f_{i,j}', a_i' \cdots f_i')$. We also let $h=(f_{i,j}, a_i \cdots f_i)$. Here, we follow notation from Section \[h\], the paragraph just before Remark \[r33\]. Recall that $h=(f_{i,j}, a_i \cdots f_i)$ is described explicitly in Remark \[r33\].(2). Now, $1+\pi y$ is an element of $\tilde{M}^1(R)$ and so we can use our result (Equations (\