Computing the longest element of the Weyl group

Question

I want to compute the longest element $w_0$ of the Weyl group $W$ for $A_2$, $B_2$ and $G_2$. I saw this has already been asked before here for the case of $G_2$, but the answers are still not very clear to me. Indeed, I would like to prove this without using any notion about Coxeter groups and only using the properties of the longest element $w_0$: for instance, I know that $w_0$ is the unique element in $W$ such that $w_0$ sends all positive roots into negative ones and that $w_0^2 = id$. Finally, I know that every reduced expression of $w_0$ must contain all simple reflections (possibly appearing more than once). Any hints on how to proceed?

Answer 1

How are you describing the elements of these Weyl groups? Take $B_2$. I've drawn the standard picture below. I've picked a couple of simple reflections, $s_\alpha$ and $s_\beta$. They bound the fundamental Weyl chamber, which contains the point $v$. (Ok, I've added a square since I think of this group as symmetries of a square, but you can delete the square if you want.) I've applied the simple reflections $s_\alpha, s_\beta$ to $v$ in all possible ways, recording the shortest expression(s) for each point in the orbit. We can just read off that there is a unique longest expression, $s_\alpha s_\beta s_\alpha s_\beta(v) = s_\beta s_\alpha s_\beta s_\alpha(v)$. This is hence the longest element.