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Let $W$ be the Weyl group of the root system of type $C_n$. Then $W$ can be identified with the group of signed permutations on $1, 2, ... , n$. Let $S = \{s_1, ... , s_n\}$, where $s_i$ swaps $i$ and $i+1$ for $1 \leq i \leq n -1$, and $s_n$ sends $n$ to $-n$.

Then $(W,S)$ is a Coxeter system. The long element $w_{\ell}$ of this system is the permutation which sends $i$ to $-i$ for all $i$. Is it possible to give a formula for a reduced decomposition of $w_{\ell}$ which works for all $n$?

D_S
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  • Related: https://mathoverflow.net/questions/54926/longest-element-of-weyl-groups , https://math.stackexchange.com/questions/59783/properties-of-the-longest-element-in-a-weyl-group and James Humphreys' notes http://people.math.umass.edu/~jeh/pub/longest.pdf – Torsten Schoeneberg Nov 09 '17 at 20:59

1 Answers1

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Yes. By realizing $C_n$ as the root system of $\textrm{Sp}_{2n}$, one can calculate the length of $w_0$ as the number of positive roots. This is

$$\sum\limits_{i=1}^{n-1} i + \sum\limits_{i=1}^n i = \frac{(n-1)n}{2} + \frac{n(n+1)}{2} = n^2$$

A decomposition for the Weyl group element which switches $i$ and $n$ is

$$(i \space n) = s_i \cdots s_{n-2}s_{n-1}s_{n-2} \cdots s_i$$

The permutation which sends $i$ to $-i$ then has a decomposition

$$(i \space n)s_n(i \space n) = (s_i \cdots s_{n-1} \cdots s_i)s_n(s_i \cdots s_{n-1} \cdots s_i) = s_i \cdots s_{n-1}s_ns_{n-1} \cdots s_i$$

where we have used the fact that $s_i$ commutes with $s_n$ for $i = 1, ... , n-2$. Then our long element is the product (in any order)

$$w_{\ell} = \prod\limits_{i=1}^n s_i \cdots s_{n-1}s_ns_{n-1} \cdots s_i$$

The length of this decomposition is

$$\sum\limits_{i=1}^n [2(n-1 - (i-1)) + 1] = n + 2 \sum\limits_{i=1}^n(n-i) = n + 2n^2 - 2\frac{n(n+1)}{2} = n^2$$

which is the correct length.

D_S
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