# The longest word in Weyl group and positive roots.

How to write down a reduced decomposition of the longest word in a Weyl group? For example, how to write down a reduced decomposition of the longest word in type B3 Weyl group? For a decomposition of the longest word, how can we write down an ordering of positive roots? I am asking these questions because I am trying to understand Section 2.4 of http://arxiv.org/abs/math/0202148. Thank you very much.

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To address the ordering question: One common way of ordering the positive roots (independent of a reduced decomposition) is to scale the roots so that the sum of their coefficients is 1, and then use the lexicographic order. Given a reduced decomposition $s_1 \cdots s_{\ell}$, a natural ordering of the roots associated to that particular decomposition $\alpha_1 < s_1(\alpha_2) < s_1s_2(\alpha_3) < \cdots < s_1\cdots s_{\ell - 1}(\alpha_{\ell})$, where $\alpha_i$ is the simple root associated to the $i$-th generator of the word. –  Hugh Denoncourt Aug 10 '12 at 19:25
@user9791: I added my current best guess to the meaning of order. Can you comment? –  Jyrki Lahtonen Aug 10 '12 at 19:37
The ordering described by @HughDenoncourt is related to the one I had in mind (but not quite the same). May be I should scratch that part, and let him describe that in an answer? –  Jyrki Lahtonen Aug 10 '12 at 19:39
@JyrkiLahtonen: There are variations - my construction is recovered from your construction applied to $w^{-1} = s_{\ell} \cdots s_1$. (I succumbed to the order given in Bjorner and Brenti, though they order reflections instead of roots. I prefer your order because of the nice interpretation regarding when a positive root is sent negative.) I don't think I can say it better than you just did, so I favor leaving your answer in! –  Hugh Denoncourt Aug 10 '12 at 20:22

Here's an algorithm for finding such a reduced word decomposition.

1. Start with $w=e$ (=the empty word) and $\lambda=-\rho=-\sum_i\lambda_i$, where the $\lambda_i$ are the fundamental dominant weights determined by $\langle \lambda_i,\check{\alpha_j}\rangle=\delta_{ij}$. Here $\alpha_1,\ldots,\alpha_n$ are simple positive roots.
2. If there exists an index $i$ such that $\langle\lambda,\alpha_i\rangle <0$, let $i_0$ be one of those (for example the smallest index, so you don't need to check too many). If no such $i$ exists, then exit with output $w$.
3. Replace $w$ with $s_{i_0}w$ and $\lambda$ with $s_{i_0}(\lambda)$. Go back to step 2.

Here initially $\lambda$ is in the negative of the dominant Weyl chamber. Each iteration brings it one simple reflection closer to the dominant chamber, and in the end it will reach it. The number of simple reflections will be equal to the number of positive roots, so you can use that as a stopping condition as well.

Let's do an example run with $B_3$. Here $\alpha_3$ is the short simple root.

$$\begin{array}{c|l} \lambda& w\\ \hline (-1,-1,-1) &e\\ (1,-2,-1) &s_1\\ (-1,2,-5) &s_2s_1\\ (1,1,-5) &s_1s_2s_1\\ (1,-4,5) &s_3s_1s_2s_1\\ (-3,4,-3) &s_2s_3s_1s_2s_1\\ (3,1,-3) &s_1s_2s_3s_1s_2s_1\\ (3,-2,3) &s_3s_1s_2s_3s_1s_2s_1\\ (1,2,-1) &s_2s_3s_1s_2s_3s_1s_2s_1\\ (1,1,1) &s_3s_2s_3s_1s_2s_3s_1s_2s_1 \end{array}$$

We knew in advance that the root system of type $B_3$ has nine positive roots, so predictably we get a weight in the fundamental Weyl chamber after nine simple reflections. This presentation as a product of simple reflections is not unique (we could have used another simple reflection at multiple points). Undoubtedly you knew about that already.

I am not positive about the meaning of ordering of positive roots. I suspect (an educated guess based on the context) that it is the order in which the partial words building up the longest element map the positive roots to negative ones. Let $w_i$ be the element of the Weyl group that is the product of $i$ first simple reflections in our decomposition of the longest element. To such an element of the Weyl group we can associate the set $\Phi_i$ of those positive roots $\beta$ such that $w_i(\beta)$ is a negative root. It is known that $|\Phi_i|=i$. It is also known that if $w_{i+1}=s_{i+1}w_i$, then $\Phi_i\subset \Phi_{i+1}$ and that the positive root in $\Phi_{i+1}$ "missing" from $\Phi_i$ is then $\beta_{i+1}:=w_i^{-1}(\alpha_{i+1})$. Thus we get a vaguely natural ordering related to the decomposition: $\beta_1<\beta_2<\beta_3<\cdots$. With 2-dimensional root systems you then get all the positive roots "in a rotational order" (which is why I am guessing that this is what it means). For example in type $A_2$ you get first one of the simple roots, then the highest root (=the one between the simple roots), and the other simple root is the last to join in.

I am not too confident about this, because it is also possible to use partial words clipped from the other end of the decomposition, and build a slightly different interpretation but analogous ordering. If the paper you cited has worked out examples, then it should be possible to reverse engineer this. I don't have the time to delve deeper into this today. Does this fit at all?

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