Longest element of Weyl groups. - MathOverflow

Longest element of Weyl groups.

2011-02-09T20:53:45Z

What is a reduced expression of the longest element of each type of Weyl group. For type $A_n$ it is just $s_n(s_ns_{n-1})...(s_n...s_1)$. I know for type $B_n,C_n,E_7,E_8$,$G_2$ and $D_n$ (n even) it is just $-id$, although I don't have an explicit reduced expression for them. For type $D_n$ (n odd) and type $E_6$ I don't know what are the longest elements. Any reference where it is written explicitely ?

Answer by Petra Schwer for Longest element of Weyl groups.

2011-02-09T21:57:27Z

A good reference to try is Bourbaki "Lie groups and Lie Algebras, Chapters 4-6" Look at the plates at the end of the book, which contain all kinds of useful information about each one of the types.

Answer by Steven Sam for Longest element of Weyl groups.

2011-02-09T22:47:03Z

You can get some reduced expressions using LiE: http://www-math.univ-poitiers.fr/~maavl/LiE/

via the command long_word(Xn) where Xn is the Dynkin diagram.

Answer by Jim Humphreys for Longest element of Weyl groups.

2011-02-09T22:54:53Z

EDIT: This is a belated attempt (motivated by a question from Yongjun Xu) to answer the question more precisely than I did at first, with more emphasis on careful choice of a Coxeter element when the Coxeter number $h$ is even (as happens for all irreducible Weyl groups except those of type $A_\ell$ with $\ell$ even). As Allen points out, for even type $A$ the algorithm needs to be modified somewhat.

I think the best conceptual approach (leaving even type $A$ aside) is based on the Coxeter element, treated in detail in Bourbaki's Chapter V and later in Chapter 3 of my 1990 book Reflection Groups and Coxeter Groups. My exercise 2 at the end of 3.19 deals with the longest element explicitly, but requires a special choice of the Coxeter element $w$ as in the treatment of the Coxeter plane in 3.17: here you start by dividing the simple reflections into two sets, each consisting of mutually orthogonal reflections. For instance, with the usual numbering in type $A_3$ you can take $w = s_1 s_3 s_2$ but not $w =s_1 s_2 s_3$ (whose square is non-reduced). Then you automatically get a reduced expression $w_\circ = w^{h/2}$ for the longest element.

Bourbaki discusses all of this in section 6.2: see Proposition 2, where it is understood that $W$ is irreducible and the Coxeter element (denoted $c$ ) has the special format fixed earlier in the section. (The wording of my exercise was corrected a long time ago in the list of revisions which I keep on my homepage, but my original answer here overlooked that correction. After the 1992 reprinting of my book, publishers began using print-on-demand technology involving photocopies, so no further corrections can be made.)

As in Bourbaki's Chapter V, the treatment actually involves arbitrary irreducible finite reflection groups, not just Weyl groups. Besides even rank $A_\ell$ , the odd rank dihedral groups have to be omitted because their Coxeter numbers are odd.

Answer by Allen Knutson for Longest element of Weyl groups.

2011-02-10T02:21:32Z

2-color your Dynkin diagram, black and white. Let $w$ be the product of the white simple reflections, $b$ the product of the black. Note that $w$ and $b$ are well-defined, as the reflections you're multiplying to make them, commute. You'll have to pick the order if you want an actual word, in what follows.

If $G$ is not $A_{even}$: the affinization of the diagram is also 2-colorable, so you can choose the affine vertex to be white. Let $\chi = w b$, a Coxeter element. The Coxeter number $h$ is even, and $\chi^{h/2} = w_0$. So you get a reduced word $wbwbwb\ldots wb$, where the total number of letters is $h$ (and each letter is a product of commuting reflections).

If $G$ is, unfortunately, $A_{even}$: you have to pick $w$ vs. $b$, and the diagram automorphism shows that the choice is unavoidable. The Coxeter number is odd. But you still get a reduced word, $wbwb\ldots bw$, again with $h$ letters.