ot h\cdot m=h$ with $i\neq j$ are trivial and the nondiagonal blocks of $\sigma({}^tm)\cdot h\cdot m=h$ are also trivial. The $(j, j)$-block of $\sigma({}^tm)\cdot h\cdot m$ is $$\left\{
\begin{array}{l l}
\pi^j\cdot\begin{pmatrix}a_j&0\\0&(1+\sigma(2 z_j^{\ast}))\cdot (1+2\bar{\gamma}_j)\cdot (1+2 z_j^{\ast}) \end{pmatrix}
& \quad \textit{$L_j$ : of type $I^o$ with j even};\\
\pi^j\cdot\begin{pmatrix}a_j&0&0\\0&(1+\sigma(\pi x_j))(1+\pi x_j)&(1+\sigma(\pi x_j))(1+2 z_j^{\ast})\\
0&(1+\sigma(2 z_j^{\ast}))(1+\pi x_j)&(1+2 z_j^{\ast})\sigma(2 z_j^{\ast})+2 z_j^{\ast}+2\bar{\gamma}_j \end{pmatrix} & \quad
\textit{$L_j$ : of type $I^e$ with j even};\\
\pi^j\cdot\begin{pmatrix}a_j&0&0\\0&\pi^3\bar{\gamma}_j+\pi^3\left((z_j)_1+(z_j)_1^2\right)&1+\sigma(\pi x_j)+\pi\cdot\sigma(\pi z_j)  \\
0&-1-\pi x_j+\pi^2 z_j&\pi \end{pmatrix} & \quad
\textit{$L_j$ : free of type $I$ with j odd}.
\end{array}\right.$$ We write $z_j^{\ast}=(z_j^{\ast})_1+\pi (z_j^{\ast})_2$, $x_j=(x_j)_1+\pi (x_j)_2$, and $z_j=(z_j)_1+\pi (z_j)_2$, where $(z_j^{\ast})_1, (z_j^{\ast})_2,$ $(x_j)_1,$ $(x_j)_2,$ $(z_j)_1,$ $(z_j)_2 \in R \subset R\otimes_AB$ and $\pi$ stands for $1\otimes \pi\in R\otimes_AB$. When $L_j$ is *of type $I^o$*, by considering the $(2, 2)$-block of the matrix above, we obtain the equation $$(z_j^{\ast})_1+(z_j^{\ast})_1^2=0.$$ Therefore, in this case, $F_j$ is isomorphic to $ \mathbb{A}^{1} \times \mathbb{Z}/2\mathbb{Z}$ as a $\kappa$-variety.

When $L_j$ is *of type $I^e$*, by considering the $(2, 2)$-block of the matrix above, we obtain the equation $$(x_j)_1^2=0.$$ We also consider the $(2, 3)$-block of the matrix above, and we obtain two equations $$(x_j)_1=0, ~~~ (x_j)_2+(z_j^{\ast})_1=0.$$ By considering the $(3, 3)$-block of the matrix above, we obtain the equation $$(z_j^{\ast})_1+(z_j^{\ast})_1^2=0.$$ By combining all these, we see that $F_j$ is isomorphic to $ \mathbb{A}^{1} \times \mathbb{Z}/2\mathbb{Z}$ as a $\kappa$-variety.

By using a similar argument used above, when $L_j$ is *free of type $I$* with $i$ odd, we