3 forgot (-) in def of q-Grassman; added 3 characters in body

The good approach to prove anything about (q)-determinats was proposed by Y. I. Manin. It is via (q)-Grassman algebra.

If I am understanding yours question correctly, then answer can be obtained on the following route.

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

Notation: consider variables $\psi_i^U= \sum_k \psi_k U_{ki}$,

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.

KEY OBSERVATION (Manin): If $U$ satisfy relations of quantum group, then $\psi_i^U$ will q-commute again !!! i.e. $\psi_i^U \psi_j^U= q \psi_j^U\psi_i^U$. (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 http://arxiv.org/abs/0901.0235).

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.

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...

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 http://arxiv.org/abs/0901.0235 Algebraic properties of Manin matrices 1 A. Chervov, G. Falqui, V. Rubtsov

1

The good approach to prove anything about (q)-determinats was proposed by Y. I. Manin. It is via (q)-Grassman algebra.

If I am understanding yours question correctly, then answer can be obtained on the following route.

Consider (q)-Grassman variables $\psi_i \psi_j = q \psi_i \psi_j$. Consider matrix $U$ with elements $u_{ij}$

Notation: consider variables $\psi_i^U= \sum_k \psi_k U_{ki}$,

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.

KEY OBSERVATION (Manin): If $U$ satisfy relations of quantum group, then $\psi_i^U$ will q-commute again !!! i.e. $\psi_i^U \psi_j^U= q \psi_j^U\psi_i^U$. (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 http://arxiv.org/abs/0901.0235).

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.

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...

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 http://arxiv.org/abs/0901.0235 Algebraic properties of Manin matrices 1 A. Chervov, G. Falqui, V. Rubtsov