Background
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When constructing the exterior algebra of a (finite-dimensional, complex) vector space $V$, there are two equivalent pictures. The first is the quotient picture. First you define the tensor algebra $T(V)$, define $\mathcal{J}$ to be the 2-sided ideal generated by elements of the form $x\otimes y + y \otimes x$, and then define the exterior algebra to be the quotient $\Lambda(V) = T(V)/\mathcal{J}$.


The other viewpoint is via embedding the exterior algebra in the tensor algebra. This is done as follows. There is an action $\rho$ of the symmetric group $S\_n$ on $V^{\otimes n}$ for each $n$, given by 
$$\rho\_\pi (v\_1 \otimes \dots \otimes v\_n) = v\_{\pi(1)} \otimes \dots \otimes v\_{\pi(n)}.$$
Then the map
$$A\_n = \frac{1}{n!} \sum\_{\pi \in S\_n}sgn(\pi) \rho\_\pi$$
is idempotent, i.e. satisfies $A\_n^2 = A\_n$. If we define 
$$ A = \bigoplus\_{n=0}^\infty A\_n$$
on $T(V)$, then it turns out that the kernel of $A$ is equal to $\mathcal{J}$, so that $\Lambda(V) \simeq \mathrm{im}(A)$.


Why I care
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I'm trying to understand the quantum analogue of this, where $V$ is the fundamental representation/vector representation of $U\_q(\mathfrak{sl}\_N)$. The problem is that there is no longer an action of the symmetric group on $V^{\otimes n}$; instead there is an action of the braid group. Both the quotient picture and the embedding picture have analogues in the quantum setting. In particular, denote by $\sigma$ the braiding on $V \otimes V$, and let $\sigma\_i$ be the automorphism of $V^{\otimes n}$ given by $\sigma$ acting in the $i$ and $i+1$ spots.


One constructs the $q$-antisymmetrizer as follows. Define $\tau\_i$ to be the adjacent transpositions $(i, i+1)$ in $S\_n$. For a permutation $\pi$, write
$\pi= \tau\_{i\_1} \dots \tau\_{i\_k},$
where $k$ is the minimal number of adjacent transpositions needed. Then define
$$\sigma\_\pi = \sigma\_{i\_1} \dots \sigma\_{i\_k}.$$
It is a theorem that this is well-defined, i.e. that any two minimal decompositions of $\pi$ as products of adjacent transpositions can be transformed into one another using only the braid relations. Anyway, once this is known, you can define the $q$-analogue of the antisymmetrizer map as
$$A\_n = \frac{q^{\binom{n}{2}}}{[n]!} \sum\_{\pi \in S\_n} sgn(\pi) q^{-\ell(\pi)} \sigma\_\pi,$$
where $[n]$ is the $q$-number and $\ell(\pi)$ is the length of a minimal decomposition of $\pi$. The significance of the $\binom{n}{2}$ is that it is the length of the longest word in $S\_n$. As far as I can tell, this was first defined by Jimbo in his 1986 paper "A $q$-analogue of $U(\mathfrak{gl}\_{N+1})$, Hecke Algebra, and the Yang-Baxter Equation". He states that $A\_n^2 = A\_n$, and refers to Gyoja's paper "A $q$-Analogue of Young Symmetrizer." Gyoja's paper certainly does the trick, but I find that something is lost in the abstraction.


I've verified that $A\_n^2 = A\_n$ for $n = 3$, which already involves 36 terms, and it wasn't that enlightening. The problem is that the braid generators $\sigma\_i$ aren't idempotent, they instead satisfy $\sigma\_i^2 = 1 + (q - q^{-1})\sigma\_i$.


The question
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Does anybody know a nice way of seeing that the $q$-antisymmetrizer is idempotent, or a nice presentation of it somewhere?



Idempotency of the q-antisymmetrizer
This element has many expressions. It is characterised up to a scalar multiple by the property that $\sigma\_iA\_n=-q^{-1}A\_n$ for $i=1,2,\ldots ,n-1$. It also satisfies
$A\_n\sigma\_i=-q^{-1}A\_n$ for $i=1,2,\ldots ,n-1$. In particular it is central.


Using this property you can calculate $A\_n^2$. Note that 
$\sigma\_\pi A\_n=(-q)^{-\ell(\pi)}= sgn(\pi)q^{-\ell(\pi)}$. So
$$A\_n.\sum\_\pi sgn(\pi)q^{-\ell(\pi)}\sigma\_\pi=A\_n.\sum\_\pi q^{-2\ell(\pi)}$$


You fix the scalar factor by the condition $A\_n^2=A\_n$. Alternatively in the one dimensional representation $\sigma\_i \mapsto -q^{-1}$ you require $A\_n \mapsto 1$. 


