Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.)

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.