most recent 30 from http://mathoverflow.net 2013-05-25T08:18:48Z http://mathoverflow.net/feeds/question/86254 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86254/alternative-definition-of-the-quantum-determinant Mihail Matrix 2012-01-20T20:57:00Z 2012-01-21T15:27:54Z

<p>Let $M_q(n)$ be the standard quantum matrices (over the complex numbers) with generators $u^i_j$ for $i,j = 1, \ldots ,N$, and reations $$ u^i_ju^k_j = qu^k_ju^i_j, \text{ for } i &lt; k, ~~~~~~~ u^i_ju^k_j = q^{-1}u^k_ju^i_j, \text{ for } i > k, $$ and so on .... The determinant element $\mathrm{det}$ is defined by $$ \mathrm{det} = \sum_{\pi \in S_n} (-q)^{\mathrm{l}(\pi)}u^1_{\pi(1)} \cdots u^N_{\pi(N)}, $$ where $S_N$ is the group of permutations on $N$ objects, and $\mathrm{l}(\pi)$ is the length of $\pi$. Can anyone find a slick way to show that one can also define $\mathrm{det}$ by $$ \mathrm{det} = \sum_{\pi \in S_n} (-q)^{-\mathrm{l}(\pi)}u^N_{\pi(N)} \cdots u^1_{\pi(1)}? $$ I am almost sure it is true (it works for the lower orders) but I can't seem to be able to prove it.</p>

http://mathoverflow.net/questions/86254/alternative-definition-of-the-quantum-determinant/86295#86295 Alexander Chervov 2012-01-21T10:37:59Z 2012-01-21T15:27:54Z

<p>The good approach to prove anything about (q)-determinats was proposed by Y. I. Manin. It is via (q)-Grassman algebra.</p> <p>If I am understanding yours question correctly, then answer can be obtained on the following route.</p> <p>Consider (q)-Grassman variables $\psi_i \psi_j = -q \psi_j \psi_i , ~ i &lt; j $ and $\psi_i^2=0$. Consider matrix $U$ with elements $u_{ij}$</p> <p>Notation: consider variables $\psi_i^U= \sum_k \psi_k U_{ki}$,</p> <p>Standard Lemma (Manin): $det_q(M) \prod_i \psi_i = \prod_i \psi_i^U$ - this holds for any matrix "U" - do not need not satisfy quantum group relations - no relation at all is necessary.</p> <p>KEY OBSERVATION (Manin): If $U$ satisfy relations of quantum group, then $\psi_i^U$ will q-commute again !!! i.e. $\psi_i^U \psi_j^U= -q \psi_j^U\psi_i^U, ~(\psi_i^U)^2=0 $. (Actually you need only "half" of the relation of quantum group for this lemma to be true. We proposed to call such "half"-quantum matrices "q-Manin" matrices see <a href="http://arxiv.org/abs/0901.0235" rel="nofollow">http://arxiv.org/abs/0901.0235</a>).</p> <p>Now the question you are asking about become rather obvious. Just consider the product $\psi_1^U\psi_2^U...\psi_n^U$ in the opposite order $=(-q)^{n(n-1)/2} \psi_n^U\psi_{n-1}^U...\psi_1^U$ and also pay attention that variables in the opposite order become $q^{-1}$-commuting. So treating all these power of $q$ correctly we should arrive to yours formula, if I am not mistaking.</p> <p>If you write me e-mail al. mysurname gmail dot com I can send you some some materials about q-Manin matrices where we discuss things like that...</p> <p>For q=1 - these q-Manin matrices are NOT commutative - but all theorems of linear algebra can be extended to them in the form precisely like standard commutative. See <a href="http://arxiv.org/abs/0901.0235" rel="nofollow">http://arxiv.org/abs/0901.0235</a> Algebraic properties of Manin matrices 1 A. Chervov, G. Falqui, V. Rubtsov </p>