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I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}_q({\mathfrak sl}_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}_N$, and let $R$ be a universal R-matrix for ${\cal U}_q({\mathfrak sl}_N)$ . If we denote the usual dual pairing of ${\cal U}_q({\mathfrak sl_N})$ with $SU_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$R^{ir}_{js} = \langle R, u^i_j \otimes u^r_s \rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (r-s)\delta_{is}\delta_{jr}.$$ i-r)\delta_{is}\delta_{jr}.$$A natural question to ask is whether such a formula exists for (R^{-1})_{js}^{ir}=\langle R^{-1}, u^i_j \otimes u^r_s \rangle. An obvious guess would be to take the inverse of the matrix [R_{js}^{ir}]_{i,r,j,s}. That is, to guess that$$[(R^{-1})_{js}^{ir}] _{i,r,j,s}$$is equal to$$([R_{js}^{ir}]_{i,r,j,s})^{-1}.$$This guess is confirmed by the fact that$$ \delta_{ij}\delta_{rs} = \langle R R^{-1},u^i_j \otimes u^r_s\rangle = \sum_{k,l} \langle R,u^i_k \otimes u^r_l \rangle \langle R^{-1},u^k_j \otimes u^l_s \rangle= \sum_{k,l} R_{kl}^{ir} (R^{-1})^{kl}_{js}. $$The matrix inverse is easy to calculate and gives us the formula$$\langle R, u^i_j \otimes u^r_s \rangle = q^{-\delta_{ir}}\delta_{ij}\delta_{rs} - (q-q^{-1})\theta (r-s)\delta_{is}\delta_{jr}. i-r)\delta_{is}\delta_{jr}. $$Now let's try and test this result: As is very well known (S \otimes id)R = R^{-1}. Thus,$$\langle R^{-1}, u^i_j \otimes u^r_s \rangle = \langle R, S(u^i_j) \otimes u^r_s \rangle.$$In the case of N=2, i=j=r=s=1, we have S(u^1_1) = u^2_2, and so,$$ (R^{-1})^{11}_{11}=\langle R^{-1},u^1_1 \otimes u^1_1 \rangle = \langle R,S(u^1_1) \otimes u^1_1 \rangle = \langle R,u^2_2 \otimes u^1_1 \rangle = R^{21}_{21} = 1. $$But the formula above gives us that (R^{-1})^{11}_{11} is equal to$$ q^{-\delta_{11}}\delta_{11}\delta_{11} - (q-q^{-1})\theta (1-1)\delta_{11}\delta_{11} = q^{-1},$$Moreover, performing the analogous calculations for the other possible values of i,j,r,s we get two different matrices: Using the general formula we get$$ \left( \begin{array} {cccc} q^{-1} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -\lambda & 1 & 0 \\ 0 & 0 & 0 & q^{-1} \\ \end{array} \right); $$and using the equality (S \otimesid)(R) = R^{-1} we get$$ \left( \begin{array} {cccc} 1 & 0 & 0 & -\lambda \\ 0 & q & 0 & 0 \\ 0 & 0 & q & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right), $$where \lambda = (q-q^{-1}). I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text 5 added 749 characters in body I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let {\cal U}_q({\mathfrak sl}_N) denote the quantised enveloping algebra of {\mathfrak sl}_N, and let R be a universal R-matrix for {\cal U}_q({\mathfrak sl}_N) . If we denote the usual dual pairing of {\cal U}_q({\mathfrak sl_N}) with SU_q(N) by \langle \cdot , \cdot \rangle, then it is well known that$$R^{ir}_{js} = \langle R, u^i_j \otimes u^r_s \rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (r-s)\delta_{is}\delta_{jr}.$$A natural question to ask is whether such a formula exists for (R^{-1})_{js}^{ir}=\langle R^{-1}, u^i_j \otimes u^r_s \rangle. An obvious guess would be to take the inverse of the matrix [R_{js}^{ir}]_{i,r,j,s}. That is, to guess that$$[(R^{-1})_{js}^{ir}] _{i,r,j,s}$$is equal to$$([R_{js}^{ir}]_{i,r,j,s})^{-1}.$$This guess is confirmed by the fact that$$ \delta_{ij}\delta_{rs} = \langle R R^{-1},u^i_j \otimes u^r_s\rangle = \sum_{k,l} \langle R,u^i_k \otimes u^r_l \rangle \langle R^{-1},u^k_j \otimes u^l_s \rangle= \sum_{k,l} R_{kl}^{ir} (R^{-1})^{kl}_{js}. $$The matrix inverse is easy to calculate and gives us the formula$$\langle R, u^i_j \otimes u^r_s \rangle = q^{-\delta_{ir}}\delta_{ij}\delta_{rs} - (q-q^{-1})\theta (r-s)\delta_{is}\delta_{jr}. $$Now let's try and test this result: As is very well known (S \otimes id)R = R^{-1}. Thus,$$\langle R^{-1}, u^i_j \otimes u^r_s \rangle = \langle R, S(u^i_j) \otimes u^r_s \rangle.$$In the case of N=2, i=j=r=s=1, we have S(u^1_1) = u^2_2, and so,$$ (R^{-1})^{11}_{11}=\langle R^{-1},u^1_1 \otimes u^1_1 \rangle = \langle R,S(u^1_1) \otimes u^1_1 \rangle = \langle R,u^2_2 \otimes u^1_1 \rangle = R^{21}_{21} = 1. $$But the formula above gives us that (R^{-1})^{11}_{11} is equal to$$ q^{-\delta_{11}}\delta_{11}\delta_{11} - (q-q^{-1})\theta (1-1)\delta_{11}\delta_{11} = q^{-1},$$Moreover, performing the analogous calculations for the other possible values of i,j,r,s we get two different matrices: Using the general formula we get$$ \left( \begin{array} {cccc} q^{-1} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -\lambda & 1 & 0 \\ 0 & 0 & 0 & q^{-1} \\ \end{array} \right); $$and using the equality (S \otimesid)(R) = R^{-1} we get$$ \left( \begin{array} {cccc} 1 & 0 & 0 & -\lambda \\ 0 & q & 0 & 0 \\ 0 & 0 & q & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right), $$where \lambda = (q-q^{-1}). I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text 4 edited body I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let {\cal U}_q({\mathfrak sl}_N) denote the quantised enveloping algebra of {\mathfrak sl}_N, and let R be a universal R-matrix for {\cal U}_q({\mathfrak sl}_N) . If we denote the usual dual pairing of {\cal U}_q({\mathfrak sl_N}) with SU_q(N) by \langle \cdot , \cdot \rangle, then it is well known that$$R^{ir}_{js} = \langle R, u^i_j \otimes u^r_s \rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (r-s)\delta_{is}\delta_{jr}.$$A natural question to ask is whether such a formula exists for (R^{-1})_{js}^{ir}=\langle R^{-1}, u^i_j \otimes u^r_s \rangle. An obvious guess would be to take the inverse of the matrix [R_{js}^{ir}]_{i,r,j,s}. That is, to guess that$$[(R^{-1})_{js}^{ir}] _{i,r,j,s}$$is equal to$$([R_{js}^{ir}]_{i,r,j,s})^{-1}.$$This guess is confirmed by the fact that$$ \delta_{ij}\delta_{rs} = \langle R R^{-1},u^i_j \otimes u^r_s\rangle = \sum_{k,l} \langle R,u^i_k \otimes u^r_l \rangle \langle R^{-1},u^k_j \otimes u^l_s \rangle= \sum_{k,l} R_{kl}^{ir} (R^{-1})^{kl}_{js}. $$The matrix inverse is easy to calculate and gives us the formula$$\langle R, u^i_j \otimes u^r_s \rangle = q^{-\delta_{ir}}\delta_{ij}\delta_{rs} - (q-q^{-1})\theta (r-s)\delta_{is}\delta_{jr}. $$Now let's try and test this result: As is very well known (S \otimes id)R = R^{-1}. Thus,$$\langle R^{-1}, u^i_j \otimes u^r_s \rangle = \langle R, u^i_j S(u^i_j) \otimes S(u^r_s) u^r_s \rangle.$$In the case of N=2, i=j=r=s=1, we have S(u^1_1) = u^2_2, and so,$$ (R^{-1})^{11}_{11}=\langle R^{-1},u^1_1 \otimes u^1_1 \rangle = \langle R,u^1_1 R,S(u^1_1) \otimes S(u^1_1) u^1_1 \rangle = \langle R,u^1_1 R,u^2_2 \otimes u^2_2 u^1_1 \rangle = R^{12}_{12R^{21}_{21} = 1. $$But the formula above gives us that (R^{-1})^{11}_{11} is equal to$$ q^{-\delta_{11}}\delta_{11}\delta_{11} - (q-q^{-1})\theta (1-1)\delta_{11}\delta_{11} = q^{-1}, I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text