Proposition 1

The restriction of $B_M$ to $E_{s,s'}$ is positive, and it is non-degenerate if and only if $m$ is finite.

Let $z = x e_s + y e_{s'}$, with $x, y \in \mathbf{R}$, be an element of $E_{s,s'}$. We have

which shows that $B_M$ is positive on $E_{s,s'}$, and that it is non-degenerate there if and only if $\sin \frac{\pi}{m} \neq 0$. The proposition follows.

The reflections $\sigma_s$ and $\sigma_{s'}$ leave $E_{s,s'}$ stable. We are going to determine the order of the restriction of $\sigma_s \sigma_{s'}$ to $E_{s,s'}$. We distinguish two cases:

$a) \ m = +\infty$.

Let $u = e_s + e_{s'}$. We have $B_M(u, e_s) = B_M(u, e_{s'}) = 0$, so $u$ is invariant under $\sigma_s$ and $\sigma_{s'}$. Moreover,

hence It follows that the restriction of $\sigma_s \sigma_{s'}$ to $E_{s,s'}$ is of infinite order.

$b) \ m$ is finite.

The form $B_H$ provides $E_{s,s'}$ with the structure of a euclidean plane. Since the scalar product of $e_s$ and $e_{s'}$ is equal to $-\cos \frac{\pi}{m} = \cos (\pi - \frac{\pi}{m})$, we can orient $E_{s,s'}$ so that the angle between the half-lines $\mathbf{R}_+ e_s$ and $\mathbf{R}_+ e_{s'}$ is equal to $\pi - \frac{\pi}{m}$. If D and D' denote the lines orthogonal to $e_s$ and $e_{s'}$,

Now, the restrictions $\overline{\sigma}_s$ and $\overline{\sigma}_{s'}$ of $\sigma_s$ and $\sigma_{s'}$ to $E_{s,s'}$ are the orthogonal symmetries with respect to D and D'. By the Cor. to Prop. 6 of § 2, no. 5, it follows that $\overline{\sigma}_s \overline{\sigma}_{s'}$ is the rotation with angle $\frac{2\pi}{m}$. In particular, its order is $m$.

We now return to E: