u_i&1+\pi w_i \end{pmatrix}\in \mathrm{GL}_{n_i}(B\otimes_AR),$$ where $s_i$ is an $(n_i-2) \times (n_i-2)-$matrix, etc.\

3.  Let $i=2m$ be even. Then $g$ stabilizes $Z_i$ and induces the identity on $W_i/(X_i\cap Z_i)$. For the proof, it is easy to show that $g$ stabilizes $Z_i$. To prove the latter, we choose an element $w$ in $W_i$. Since $gw-w\in X_i$ by the above step (2), it suffices to show that $gw-w\in Z_i$. Recall that $Z_i$ is the sublattice of $A_i$ such that $Z_i/\pi B_i$ is the radical of the quadratic form $\frac{1}{2^{m+1}}q$ mod $2$ on $B_i/\pi B_i$, where $\frac{1}{2^{m+1}}q(a)=\frac{1}{2^{m+1}}h(a,a)$ for $a\in B_i$. The lattice $Z_i$ can also be described as the sublattice of $B_i$ such that $Z_i/\pi B_i$ is the kernel of the additive polynomial $\frac{1}{2^{m+1}}q$ mod $2$ on $Y_i/\pi B_i$. Now it is also easy to show that $gw-w\in Y_i$. Thus our claim that $gw-w\in Z_i$ follows from the following computation:

    $$\frac{1}{2}\cdot \frac{1}{2^m} q(gw-w)=\frac{1}{2}(2\cdot \frac{1}{2^m}q(w)- \frac{1}{2^m}(h(gw,w)+h(w,gw)))=$$ $$\frac{1}{2^m}(q(w)-\frac{1}{2}(h(w+x,w)+h(w,w+x))) =-\frac{1}{2^m}\cdot\frac{1}{2}(h(x,w)+h(w,x))=0 \textit{ mod } 2,$$ where $gw=w+x$ for some $x\in X_i$, since $\frac{1}{2^m}h(x,w)\in \pi B$ and $\pi + \sigma(\pi) \in 4A$. Recall that $\frac{1}{2^m}q(w)=\frac{1}{2^m}h(w,w)$.

    We interpret this in terms of matrices.

    -   If $L_i$ is *free of type II*, then $W_i=X_i=Z_i$ and so there is no contribution.

    -   If $L_i$ is *bound of type II*, then $W_i=X_i$ and $Z_i$ is as explained in Remark 2.11 of [@C2]. Matrix interpretation in this case is covered by matrix interpretation of Steps (2) and (4) below. Thus there is no new contribution.

    -   If $L_i$ is *of type I*, then $$z_i+\delta_{i-2}k_{i-2, i}+\delta_{i+2}k_{i+2, i} \in (\pi).$$ Here,

        -   $z_i$ is an entry of $g_{i,i}$ as described in *Step (2)*.

        -   $k_{i-2, i}$ (resp. $k_{i+2, i}$) is the $(n_{i-2}, n_i)^{th}$-entry (resp. $(n_{i+2}, n_i)^{th}$-entry) of the matrix $g_{i-2, i}