Idempotency of the q-antisymmetrizer - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:50:54Z http://mathoverflow.net/feeds/question/33724 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33724/idempotency-of-the-q-antisymmetrizer Idempotency of the q-antisymmetrizer MTS 2010-07-28T22:26:44Z 2010-07-29T08:48:51Z <h2>Background</h2> <p>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 $\Lambda(V) = T(V)/\mathcal{J}$.</p> <p>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 $\Lambda(V) \simeq \mathrm{im}(A)$.</p> <h2>Why I care</h2> <p>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.</p> <p>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.</p> <p>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 $\sigma_i^2 = 1 + (q - q^{-1})\sigma_i$.</p> <h2>The question</h2> <p>Does anybody know a nice way of seeing that the $q$-antisymmetrizer is idempotent, or a nice presentation of it somewhere?</p> http://mathoverflow.net/questions/33724/idempotency-of-the-q-antisymmetrizer/33734#33734 Answer by Bruce Westbury for Idempotency of the q-antisymmetrizer Bruce Westbury 2010-07-28T23:59:54Z 2010-07-29T08:48:51Z <p>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.</p> <p>Using this property you can calculate $A_n^2$. Note that $\sigma_\pi A_n=(-q)^{-\ell(\pi)}= sgn(\pi)q^{-\ell(\pi)}$. So $$A_n.\sum_\pi sgn(\pi)q^{-\ell(\pi)}\sigma_\pi=A_n.\sum_\pi q^{-2\ell(\pi)}$$</p> <p>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 $A_n \mapsto 1$. </p> http://mathoverflow.net/questions/33724/idempotency-of-the-q-antisymmetrizer/33739#33739 Answer by Sammy Black for Idempotency of the q-antisymmetrizer Sammy Black 2010-07-29T00:15:45Z 2010-07-29T00:15:45Z <p><em>Group Theory: Birdtracks, Lie's, and Exceptional Groups</em> by Predrag Cvitanovic has a graphical way of presenting the idempotents. His book is available online here:</p> <p><a href="http://birdtracks.eu/" rel="nofollow">http://birdtracks.eu/</a></p>