Particular reduced expression of the longest element of Weyl group

Question

Let $I$ be the Dynkin diagram vertex set and $K$ be a proper nonempty subset of it. Let $w_0^K$ be the longest word of the Dynkin subdiagram $K$, which might be a disjoint union of connected Dynkin diagrams.

I saw in the paper "Preprojective algebras and partial flag varieties" by Geiss, Leclerc and Schroer the following statement:

“We can write $w_0 = w^K_0 v_K$ with $\ell(w^K_0) + \ell(v_K) = ℓ(w_0).$ Therefore there exist reduced words $i$ for $w_0$ starting with a factor $(i_1,\dotsc,i_{r_K})$ which is a reduced word for $w^K_0$.”

I could run this through examples, like writing the longest element $w_0=s_3s_2s_3s_1s_2s_3s_1s_2s_1$ of type $B$ as $w_0=\boldsymbol{s_1s_2s_1} {s_3s_2s_1s_3s_2s_3}$, if $K=\{1،2\}$.

I was wondering if there is an algorithm showing how to do this in general, at least if the size of $K$ is known, like $|K|=|I|-1$.

Answer 1

Whenever $\ell(wv)=\ell(w)+\ell(v)$, you can construct a reduced word for $wv$ by producing one for $w$ and one for $v$ and then concatenating them. So, if you know an algorithm for producing reduced words, you can just do that.

On the other hand, there’s actually a particularly nice interpretation in this case. Reduced words for $w_0$ are in bijection with convex orders on roots (orders such that if $\alpha,\beta,\alpha+\beta$ are all roots, then $\alpha+\beta$ is always between $\alpha$ and $\beta$). The bijection is obtained by sending $(i_1,\dots, i_m)$ to the order $\alpha_{i_1} < s_{i_{1}}\alpha_{i_{2}} < s_{i_1}s_{i_{2}}\alpha_{i_{3}}<\cdots $; the lowest $k$ roots in this order are those sent to negative roots by $s_{i_k}\cdots s_{i_1}$. So, the reduced word you want comes from choosing this order so that the roots in the span of $K$ are below all other roots.

There are various ways of doing this. For example, if we choose vectors $\mathbf{x}=(x_i,y_i) \in \mathbb{R}^2$ for each simple root such that $y_i>0$, we can extend linearly to assign a vector to every root $\mathbf{x}_{\alpha}$. If these are generic, every root will have a different slope, and we can order roots by slope from lowest to highest. If we assume that $x_i\leq 0$ for all $i\in K$, and $x_i\gg 0$ for $i\notin K$, then this slope ordering will give us what we want once the positive $x_i$’s are big enough.

In your $B_3$ example, we would choose, say $(x_1,y_1)=(-1,1), (x_2,y_2)=(0,1), (x_3,y_2)=(2,1)$. We would then get the order on positive roots:

$$\alpha_1 < \alpha_1+ \alpha_2 < \alpha_2 < \alpha_1+ \alpha_2+\alpha_3 <\alpha_2+\alpha_3 <\alpha_1+ \alpha_2+2\alpha_3 < \alpha_2+2\alpha_3 <\alpha_3$$ This corresponds to the reduced word $s_1s_2s_1s_3s_2 s_3s_1s_2s_3$ (Assuming I did my computations right).

Answer 2

As I mentioned in a comment, this is a special case of finding a distinguished representative for a coset of a parabolic subgroup. Let's inductively define elements $w_n$, with the convenient starting point $w_n = w_0$ when $n = 0$. Having defined $w_n$ in general, one of two things can happen: either $w_n^{-1}K$ consists of positive roots, or there is some $\alpha \in K$ such that $w_n^{-1}\alpha$ is negative. In the former case, we stop, and put $v_K = w_n$. In the latter case, we put $w_{n + 1} = s_\alpha w_n$. It is part of the general theory of length in Coxeter groups that $\ell(w_{n + 1}) = \ell(w_n) - 1$, and hence that $\ell(w_n) = \ell(w_0) - n$ for all natural numbers $n$ such that $w_n$ is defined.

Suppose that $w^K \mathrel{:=} w_0 v_K^{-1}$ is not the long element $w_0^K$ of the parabolic subgroup of the Weyl group generated by $K$. Then $\ell(w_0^K)$ is strictly greater than $\ell(w^K)$; but then $\ell(w_0^K v_K)$, which equals $\ell(w_0^K) + \ell(v_K)$ by Proposition 1.10(c) of Humphreys - Reflection groups …, is strictly larger than $\ell(w^K) + \ell(v_K) = \ell(w^K v_K) = \ell(w_0)$. This is a contradiction.