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edited May 22 2010 at 20:23 |
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);
$$
and using the equality $(S \otimes$id$)(R) = R^{-1}$ we get
$$
\left(
\begin{array} {cccc}
1 & 0 & 0 & -\lambda \\
0 & q & 0 & 0 \\
0 & 0 & q & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right),
$$
where $\lambda = (q-q^{-1})$.
I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text
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edited May 22 2010 at 19:11 |
I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}_q({\mathfrak sl}_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}_N$, and let $R$ be a universal R-matrix for ${\cal U}_q({\mathfrak sl}_N)$ . If we denote the usual dual pairing of ${\cal U}_q({\mathfrak sl_N})$ with
$SU_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$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);
$$
and using the equality $(S \otimes$id$)(R) = R^{-1}$ we get
$$
\left(
\begin{array} {cccc}
1 & 0 & 0 & -\lambda \\
0 & q & 0 & 0 \\
0 & 0 & q & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right),
$$
where $\lambda = (q-q^{-1})$.
I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text
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edited May 22 2010 at 18:55 |
I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}_q({\mathfrak sl}_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}_N$, and let $R$ be a universal R-matrix for ${\cal U}_q({\mathfrak sl}_N)$ . If we denote the usual dual pairing of ${\cal U}_q({\mathfrak sl_N})$ with
$SU_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$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},$$ I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text
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edited May 20 2010 at 22:39 |
The Inverse of a Universal R-Matrix for Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2)
I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}_q({\mathfrak sl}_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}_N$, and let $R$ be a universal R-matrix for ${\cal U}_q({\mathfrak sl}_N)$ . If we denote the usual dual pairing of ${\cal U}_q({\mathfrak sl_N})$ with
$SU_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$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 \otimes 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 \otimes S(u^1_1) \rangle = \langle R,u^1_1 \otimes u^2_2 \rangle = R^{12}_{12} = 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},$$ I can't see why these two results don't agree and am guessing I have made some basic beginners beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text
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edited May 20 2010 at 21:11 |
The Inverse Universal R-Matrix for Quantum Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2)
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asked May 20 2010 at 21:04 |
The Inverse Universal R-Matrix for Quantum Enveloping Algebra of sl2 and the Dual Pairing with SUq(2)
I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}_q({\mathfrak sl}_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}_N$, and let $R$ be a universal R-matrix for ${\cal U}_q({\mathfrak sl}_N)$ . If we denote the usual dual pairing of ${\cal U}_q({\mathfrak sl_N})$ with
$SU_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$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 \otimes 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 \otimes S(u^1_1) \rangle = \langle R,u^1_1 \otimes u^2_2 \rangle = R^{12}_{12} = 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 beginners mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text
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