Remark 1

If we only assume that $v$ is diagonalizable with integer eigenvalues, $u$ is not necessarily diagonalizable (for example, take $M =\mathbf{Z}^2$ and $u(x, y) =$ $(y, x)$ for all $(x, y)\in M$).

Let $\mathfrak{g},\mathfrak{h},R,\mathscr{H},\mathscr{G}^{\alpha},\mathscr{G},\mathscr{U}$ be as in no. 7, and assume that $\mathscr{G}$ is a Chevalley order in $(\mathfrak{g},\mathfrak{h})$.