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](http://aif.cedram.org/item?id=AIF_2008__58_3_825_0)" 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 $|K|=|I|-1$.



Particular reduced expression of the longest element of Weyl group
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).


