Quantum factorials

Lemma 3.4 in this interesting paper by A. Kleshchev and D. Nash requires a combinatorial result that I hadn’t seen before. Here is a version of it, converted into a typical mathematical game (i.e. only entertaining for people who have nothing better to do).

We work in the ring \mathbf{Z}[q,q^{-1}] where q is an indeterminate. Imagine a row of n coins, numbered from 1 at the far left to n at the far right. Your task is to remove all n coins. As you do this, you must keep track of an element f(q) \in \mathbf{Z}[q,q^{-1}]. At the start of the game f(q) = 1. If, after taking coin i, there are r coins to its left, and s empty gaps, then the rules of the game require you to multiply f(q) by q^{r-s}.

For example, if n=5 and we take coins in the order 5, 1, 3, 4, 2, then at the end of the game f(q) = q^4 \times q^0 \times q^0 \times q^{-1} \times q^{-1} = q^2.

Now let F(q) be the sum of the final outcomes f(q) over all n! orders in which we might remove the coins. The remarkable result is F(q) is the quantum generalization of the factorial function, i.e.

F(q) = [n]_q! = [n]_q [n-1]_q \ldots [2]_q [1]_q

where the quantum number [r]_q \in \mathbf{Z}[q,q^{-1}] is defined by

[r]_q = \frac{q^r - q^{-1}}{q-q^{-1}}.

It is often useful to write [r]_q in the alternative form

[r]_q = q^{r-1} + q^{r-3} + \cdots + q^{-(r-3)} + q^{-(r-1)}.

In this form the important property that [r]_q \rightarrow r as q \rightarrow 1 is obvious. It follows that \lim_{q \rightarrow 1} F(q) = n!. Correspondingly, if we play the game with q=1, then the final outcome is always 1, so this limit recovers the obvious fact that there are n! different ways in which we may remove the coins.

The result that F(q) = [n]_q! is easily proved by induction on n. If there is just one coin then the final outcome is always q^0 = 1 = [1]_q. Now suppose the result holds for n-1 coins, and suppose that we begin an n-coin game by removing coin k. This gives us an immediate contribution of q^{k-1}. Imagine that we ignore the gap at coin k and continue playing the game as if we had started with n-1 coins. By induction, the sum of F(q) over all possible continuations is [n-1]_q!. This has to be scaled by q^{-(n-k)} to take into account the gap we ignored when removing coins k+1, \ldots, n. (The order in which we remove these coins is irrelevant.) Hence

F(q) = [n-1]_q! \sum_{k=1}^n q^{k-1} q^{-(n-k)}.

Since \sum_{k=1}^n q^{2k-n-1} = [n]_q, it follows that

F(q) = [n-1]_q! [n]_q = [n]!_q,

as required.

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