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Let $B_n$ denote the group of signed permutations on $n$ letters. Is there a good explanation or understandable way to see why $$ \sum_{w\in B_n}q^{\text{inv}(w)}=(2n)_q(2n-2)_q\cdots(2)_q? $$

I've been thinking about it on and off while reading through Taylor's Geometry of the Classical Groups, but don't understand why this identity holds. I appreciate any explanation. Thanks!

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You are using lots of terms/notations not everyone understands. Meet us halfway. – Gerry Myerson Feb 16 '12 at 10:50
@GerryMyerson Terribly sorry about that, I was unaware the notation is not standard. I will add those in soon. – Dani Hobbes Feb 16 '12 at 23:53

I can only offer a rough idea (and hope that I have the same definition of inv as you do). The proof in type A is on page 36 of the PDF version of EC 1 available on Stanley's website (here). Basically, any permutation can be encoded via its inversion table, a sequence $(a_1,\ldots,a_n)$, where $0\leq a_i\leq n-1$, and $\mathrm{inv}(w)=a_1+\cdots+a_n$, so the sum $\sum_{w \in S_n}q^{\mathrm{inv}(w)}$ can be converted to a sum over inversion tables.

One should be able to define the inversion table of a signed permutation similarly and push a similar proof through, but I can't get the right definition of inversion table. (The identity suggests one needs only $n$ entries in the table, which makes perfect sense, and that they can range between 0 and $2n-1$, which also makes sense, but I can't put the pieces together, nor find a reference.)

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There are multiple ways to generalize inv to $B_n$, and the literature is not always consistent. Here is one such generalization (denoted $\mathrm{inv_B}$ for clarity), and a corresponding proof. As in hoyland's answer, the approach is to show a bijection between permutations and inversion sequences, where the corresponding statistic is much simpler to count.

Let $\bar w$ be a signed permutation, and $w$ be the same permutation stripped of signs. Let $\sigma_i$ be 1 if the $i$th position in $w$ is signed, 0 if it is unsigned. Let $\mathrm{inv}(w)$ be the standard inversion statistic on unsigned permutations.

Define $$\mathrm{inv}_B(\bar w)=\mathrm{inv}(w)+\sum_{i=1}^n i\sigma_i$$

For $w\in S_n$, the inversion sequence of $w$ is defined as $a=(a_1,\cdots,a_n)$, where $a_i=|\{0\le j\le i-1\mid w_j>w_i\}|$. It is easy to see that $$|a|\equiv\sum_{i=1}^n a_i=\mathrm{inv}(w).$$ This produces a bijection between permutations and inversion sequences, although I will not prove that here. Now for $\bar w\in B_n$, let the inversion sequence of $\bar w$ be $(b_1,⋯,b_n)$, where $b_i=a_i+i\sigma_i$. Thus we have $0≤b_i≤2i−1$. By the way we've defined things it follows that $$|b|\equiv\sum_{i=1}^n b_i=\mathrm{inv_B}(w).$$ It's straightforward to see that the modified inversion sequences are also bijective assuming the original inversion sequences were. Denoting the set of inversion sequences of length $n$ as $I_n$, we have $$\sum_{w\in B_n}q^{\mathrm{inv_B}(w)}=\sum_{b\in I_n}q^{|b|}=\prod_{i=1}^n [2i]_q.$$ I hope this definition of inv is equal, or at least equivalent to the one you were using.

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