$\psi_j$ is similar to and simpler than that of the above cases when $N_0$ is *of type $I^e$* and so we skip it.\

2.  Assume that $M_0$ is *of type $I^e$* and $L_j$ is *of type $I^o$*. We write $M_0=N_0\oplus L_j$, where $N_0$ is unimodular with odd rank so that it is *of type $I^o$*. Then we can write $N_0=(\oplus_{\lambda'}H_{\lambda'})\oplus (a)$ and $L_j=(\oplus_{\lambda''}H_{\lambda''})\oplus (a')$ by Theorem \[210\], where $H_{\lambda'}=H(0)=H_{\lambda''}$ and $a, a'\in A$ such that $a, a' \equiv 1$ mod 2. Thus we write $M_0=(\oplus_{\lambda}H_{\lambda})\oplus (a)\oplus (a')$, where $H_{\lambda}=H(0)$. For this choice of a basis of $L^j=\bigoplus_{i \geq 0} M_i$, the image of a fixed element of $F_j$ in the special fiber of the smooth integral model associated to $L^j$ is $$\begin{pmatrix} id&0 &0\\ 0 &\begin{pmatrix} 1+ 2 z_j^{\ast}  \end{pmatrix}  &0 \\ 0& 0 &id \end{pmatrix}.$$ Here, $id$ in the $(1,1)$-block corresponds to the direct summand $(\oplus_{\lambda}H_{\lambda})\oplus (a)$ of $M_0$ and the diagonal block $\begin{pmatrix} 1+ 2 z_j^{\ast} \end{pmatrix} $ corresponds to the direct summand $(a')$ of $M_0$.

    Let $(e_1, e_2)$ be a basis for the direct summand $(a)\oplus (a')$ of $M_0$. Since this is *unimodular of type $I^e$*, we can choose another basis $(e_1, e_1+e_2)$ such that the associated Gram matrix is $A(a, a+a', a)$, where $a+a'\in (2)$, so that $$M_0=(\oplus_{\lambda}H_{\lambda})\oplus A(a, a+a', a).$$

    For this basis, the image of a fixed element of $F_j$ in the special fiber of the smooth integral model associated to $L^j$ is $$\label{e4.3}
    \begin{pmatrix} id&0 &0\\ 0 &\begin{pmatrix} 1 & -2 z_j^{\ast}\\ 0 & 1+2 z_j^{\ast} \end{pmatrix}  &0 \\ 0& 0 &id \end{pmatrix}.$$ Here, the diagonal block $\begin{pmatrix}  1 & -2 z_j^{\ast}\\ 0 & 1+2 z_j^{\ast} \end{pmatrix} $ corresponds to $A(a, a+a', a)$ with a basis $(e_1, e_1+e_2)$ and $id$ in the $(1,1)$-block corresponds to the direct summand $\oplus_{\lambda}H_{\lambda}$ of $M_0$.

    We now describe the image of a fixed eleme