 I*. If $L_i$ is *free of type I*, then $\pi^i f_{i,i}$ is of the form $$\xi^{(i-1)/2}\cdot \pi\begin{pmatrix} a_i&\pi b_i& e_i\\ -\sigma(\pi \cdot {}^tb_i) &\pi^3f_i&1+\pi d_i \\
    -\sigma({}^te_i) &-\sigma(1+\pi d_i) &\pi+\pi^3c_i \end{pmatrix}.$$ Here, the diagonal entries of $a_i$ are are divisible by $\pi^3$, where $a_i$ is an $(n_i-2) \times (n_i-2)$-matrix with entries in $B\otimes_AR$, etc. If $L_i$ is *bound of type I*, then each entry of $$\delta_{i-1}(0,\cdots, 0, 1)\cdot f_{i-1,i}+\delta_{i+1}(0,\cdots, 0, 1)\cdot f_{i+1,i}$$ lies in the ideal $(\pi)$.

7.  Since $\begin{pmatrix}\pi^{max\{i,j\}}f_{i,j}\end{pmatrix}$ is a hermitian matrix, its diagonal entries are fixed by the nontrivial Galois action over $E/F$ and hence belong to $R$.\

The functor $\underline{H}$ is represented by a flat $A$-scheme which is isomorphic to an affine space of dimension $n^2$. The proof of this is similar to that in Section 3C of [@C2] and so we skip it.

Now suppose that $R$ is any (not necessarily flat) $A$-algebra. We again use $\sigma$ to mean the automorphism of $B\otimes_AR$ given by $b\otimes r \mapsto \sigma(b)\otimes r$. Note that $\sigma(\pi)=-\pi$. By choosing a $B$-basis of $L$ as explained in Theorem \[210\] and Remark \[r211\].(a), we describe each element of $\underline{H}(R)$ formally as a matrix $\begin{pmatrix}\pi^{max\{i,j\}}f_{i,j}\end{pmatrix}$ with matrices $f_{i,i}^{\ast}$ satisfying the following:

1.  When $i\neq j$, $f_{i,j}$ is an $(n_i \times n_j)$-matrix with entries in $B\otimes_AR$ and $(-1)^{max\{i,j\}}\sigma({}^tf_{i,j})=f_{j,i}$.

2.  Assume that $i=j$ is even. Then $$\pi^if_{i,i}=\left\{
      \begin{array}{l l}
      \xi^{i/2}\begin{pmatrix} a_i&\pi b_i\\ \sigma(\pi\cdot {}^t b_i) &1+2\gamma_i +4c_i \end{pmatrix}    & \quad  \textit{if  $L_i$ is \textit{of type $I^o$}};\\
      \xi^{i/2}\begin{pmatrix} a_i&b_i&\pi e_i\\ \sigma({}^tb_i) &1+2f_i&1+\pi d_i \\ \sigma(\pi \cdot {}^te_i) &\sigma(1+\pi d_i) &2\gamma_i+4c_i \end{pmatrix}    & \quad  \textit{if  $L_i$ is \textit{of type $I^e