Orientation in reflection groups

Question

Let R be a (crystallographic) root system on an Euclidian space $(E,\langle-,-\rangle)$ and $$W:=gen\{\sigma_r \mid r\in R\}=gen\{\sigma_r \mid r\in R^+\}$$ its associated reflection group. If $r_1$, $r_2$ are different elements of $R^+$, I know that $g=\sigma_{r_1}\sigma_{r_2}$ acts as a rotation on the plane generated by $r_1$ and $r_2$ by an angle of $2\pi/ord(g)$ (where $ord(g)$ is the order of $g$) and as the identity on the orthogonal complement. Considering the case $ord(g)\not=2$ (or in general if it is possible), I would like to know if under the assumption that there exist $r_3$, $r_4$ $\in$ $R^+$ such that $\sigma_{r_3}\sigma_{r_4}=g=\sigma_{r_1}\sigma_{r_2}$ the plane generated by $r_3$ and $r_4$ has the same orientation as the one generated by $r_1$ and $r_2$.

Answer 1

This question doesn't really have to do with root systems, it's really just about rotations and reflections. The composite $\sigma_{r_1} \sigma_{r_2}$ restricted to the plane $\langle r_1, r_2 \rangle$ is a rotation of angle twice the angle between $r_1$ and $r_2$ measured in the sense going from $r_2$ to $r_1$. Notice that as long as the angle isn't $2 \pi$, the rotation determines the plane in which it occurs. And if the angle is neither $2 \pi$ nor $\pi$, the sense of rotation is also determined.

So if $\sigma_{r_1} \sigma_{r_2} = \sigma_{r_3} \sigma_{r_4} = g$ and $g$ is neither the identity nor of order 2, then $\{r_1, r_2\}$ and $\{r_3, r_4\}$ span the same plane and determine the same orientation of it.