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?
Computing the longest element of the Weyl group
abstract-algebradynkin-diagramsrepresentation-theoryroot-systemsweyl-group
Related Solutions
While Travis' answer gives a nice hands-on calculation, I like to point out two answers to related questions which put things in perspective:
Anton Geraschenko's answer here states, among other things, that the longest element in most simple types (actually, all except $A_{n \ge 2}, D_{2n+1}$ and $E_6$) is just $-id$. So this is also the case here, $w_0$ must be multiplication with $-1$ (which, since we are in two dimensions, is the same as rotating by $\pi$).
Allen Knutson's answer here on MathOverflow gives a nice general method to express $w_0$ as product of simple reflections. In the case of type $G_2$, we can choose $w=w_\alpha, b=w_\beta$, and the Coxeter number for type $G_2$ is $h=6$, so the general formula there gives $w_0 =(w_\alpha w_\beta)^{h/2} = (w_\alpha w_\beta)^{3}$. Switching the roles of $w$ and $b$, which is allowed, gives alternatively $w_0=(w_\beta w_\alpha)^{3}$, as in the other answer.
This proof works for any finite Coxeter group $W$ generated by the set $S$ of simple reflections (whence $W$ has the unique longest element $w_0$). We need the following ingredients for the proof:
- knowing that $l(ww')\leq l(w)+l(w')$ for all $w,w'\in W$;
- knowing that $l(w^{-1})=l(w)$ for all $w\in W$;
- knowing that $w_0^2=1_W$; and
- knowing that, if $w\in W$ is such that $l(w)<l(w_0)$, then there exists $s\in S$ for which $l(w)<l(sw)$.
Let $w\in W$ be arbitrary. Note that $w_0=w^{-1}(ww_0)$. Consequently, $$l(w_0)\leq l(w^{-1})+l(ww_0)\,.$$ Because $l(w^{-1})=l(w)$, we obtain $$l(ww_0)\geq l(w_0)-l(w^{-1})=l(w_0)-l(w)\,.$$
Now, we shall prove $l(ww_0)\leq l(w_0)-l(w)$ by induction on $k:=l(w_0)-l(w)$. If $k=0$, then $w=w_0$ and so $ww_0=w_0^2=1_W$. That is, $$l(ww_0)=l(1_W)=0=l(w_0)-l(w_0)=l(w_0)-l(w)=k\,.$$ Suppose now that $w\neq w_0$ and $k=l(w_0)-l(w)>0$. There exists $s\in S$ such that $l(w)<l(sw)$. Define $w':=sw$. Hence, $l(w_0)-l(w')<l(w_0)-l(w)=k$. Therefore, from $ww_0=s^{-1}(sww_0)=s(w'w_0)$ (recalling that $s^2=1_W$), we get $$l(ww_0)\leq l(w'w_0)+l(s)=l(w'w_0)+1.$$ By induction hypothesis, $l(w'w_0)\leq l(w_0)-l(w')$. Note that $l(w')=l(w)+1$. Therefore, $$l(ww_0)\leq \big(l(w_0)-l(w')\big)+1=l(w_0)-\big(l(w')-1\big)=l(w_0)-l(w)=k\,.$$
From the results above, we conclude that $l(ww_0)=l(w_0)-l(w)$. On the other hand, $$l(w_0w)=l\big((w_0w)^{-1}\big)=l(w^{-1}w_0^{-1})\,.$$ Because $w_0^{-1}=w_0$, we get $$l(w_0w)=l(w^{-1}w_0)=l(w_0)-l(w^{-1})=l(w_0)-l(w)=l(ww_0)\,.$$
Now, let $W:=\mathfrak{S}_n$, which is generated by $$S:=\big\{(1\;\;2),(2\;\;3),\ldots,({n-1}\;\;n)\big\}\,.$$ We want to show that $w_0$ is the order-reversing permutation $w_0'$ given by $$(1,2,3,\ldots,n-2,n-1,n)\mapsto (n,n-1,n-2,\ldots,3,2,1)\,.$$
For each $w\in W$, let $L(w)$ be the number of pairs $(i,j)$ such that $i,j\in\{1,2,\ldots,n\}$, $i<j$, and $w(i)>w(j)$. Then, for $s\in S$, $$L(sw)=L(w)\pm 1\,.$$ If we pick $s:=(i\;\;i+1)$ where $i\in\{1,2,\ldots,n-1\}$ satisfies $w(i)>w(i+1)$, then $L(sw)=L(w)-1$. From this observation, any reduced expression for $w$ must have exactly $L(w)$ simple reflections. Therefore, $l(w)=L(w)$ for every $w\in W$.
This argument also proves that $\displaystyle l(w)=L(w)\leq \binom{n}{2}$ for all $w\in W$ (because there are precisely $\displaystyle \binom{n}{2}$ pairs $(i,j)$ such that $i,j\in\{1,2,\ldots,n\}$ and $i<j$, so at most $\displaystyle \binom{n}{2}$ such pairs also satisfy $w(i)>w(j)$). Particularly, because $L(w_0')=\displaystyle \binom{n}{2}$, this means $w_0=w_0'$ and $l(w_0)=\displaystyle \binom{n}{2}$.
Best Answer
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.