For the symmetric group $S_n$, an inversion of a permutation $\pi∈S_n$ is a pair $1\leq i<j\leq n$ such that $\pi(i)>\pi(j)$. It is known that the length $\ell(\pi)$ of a permutation (i.e. the least number of simple transpositions needed to express $\pi$) coincide with its number of inversions. This is useful to compute $$\sum_{\pi\in S_n}q^{\ell(\pi)} = (n)_q!$$ What happens if I replace $S_n$ by an arbitrary finite Coxeter group? Is there still a notion for an inversion? What is the above sum in this case?
Edit Okay, to be able say that I have my homework done, here some thoughts on the problem: A general definition of the inversion number for Weyl groups seems to be the following: The inversion number of a word $w$ is its number of positive roots that are mapped to negative roots. This still equals its length $\ell(w)$.
Example: $B_n$ For this group this seems to be easy (since I can draw pictures in my head for what it does…): It is simply the group $S_n$ with one additional generator that flips let's say the first sign in a sequence.
Apart from permuting elements, we may flip zero to $n$ signs, where we have $n$ possibilities for the first sign, $n-1$ for the second etc. So for $B_n$, the required polynomial should be
$$\sum_{w\in B_n}q^{\ell(w)} = (1+nq + (n-1)q^2 + \cdots + q^n)(n)_q!$$
which coincides with counting the elements explicitly (at least, for $B_2$)
- Is there a nicer way to express this number?
- Is there a general systematic, e.g. how to obtain this from the presentation?
- Is there a listing for other Coxeter groups?