The Inverse of a Universal R-Matrix for Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:29:21Z http://mathoverflow.net/feeds/question/25423 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25423/the-inverse-of-a-universal-r-matrix-for-quantized-universal-enveloping-algebra-of The Inverse of a Universal R-Matrix for Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2) Abtan Massini 2010-05-20T21:04:26Z 2010-05-22T23:37:59Z <p>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 $$R^{ir}_{js} = \langle R, u^i_j \otimes u^r_s \rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}.$$ 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 $[R_{js}^{ir}]_{i,r,j,s}$. </p> <p>That is, to guess that $$[(R^{-1})_{js}^{ir}] _{i,r,j,s}$$ </p> <p>is equal to $$([R_{js}^{ir}]_{i,r,j,s})^{-1}.$$ This guess is confirmed by the fact that</p> <p>$$\delta_{ij}\delta_{rs} = \langle R R^{-1},u^i_j \otimes u^r_s\rangle = \sum_{k,l} \langle R,u^i_k \otimes u^r_l \rangle \langle R^{-1},u^k_j \otimes u^l_s \rangle= \sum_{k,l} R_{kl}^{ir} (R^{-1})^{kl}_{js}.$$ The matrix inverse is easy to calculate and gives us the formula $$\langle R, u^i_j \otimes u^r_s \rangle = q^{-\delta_{ir}}\delta_{ij}\delta_{rs} - (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}.$$</p> <p>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, $$(R^{-1})^{11}_{11}=\langle R^{-1},u^1_1 \otimes u^1_1 \rangle = \langle R,S(u^1_1) \otimes u^1_1 \rangle = \langle R,u^2_2 \otimes u^1_1 \rangle = R^{21}_{21} = 1.$$ </p> <p>But the formula above gives us that $(R^{-1})^{11}_{11}$ is equal to</p> <p>$$q^{-\delta_{11}}\delta_{11}\delta_{11} - (q-q^{-1})\theta (1-1)\delta_{11}\delta_{11} = q^{-1},$$ </p> <p>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 $$\left( \begin{array} {cccc} q^{-1} &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; -\lambda &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; q^{-1} \\ \end{array} \right);$$ and using the equality $(S \otimes$id$)(R) = R^{-1}$ we get $$\left( \begin{array} {cccc} 1 &amp; 0 &amp; 0 &amp; -\lambda \\ 0 &amp; q &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; q &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{array} \right),$$ where $\lambda = (q-q^{-1})$.</p> <p>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</p> http://mathoverflow.net/questions/25423/the-inverse-of-a-universal-r-matrix-for-quantized-universal-enveloping-algebra-of/25604#25604 Answer by David Jordan for The Inverse of a Universal R-Matrix for Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2) David Jordan 2010-05-22T21:38:49Z 2010-05-22T23:37:59Z <p>I think the only issue here is a harmless error in your calculation and that there is a normalization of the $R$-matrix for $U_q(sl_N)$ by a factor of $q^{1/2}$ which you have omitted (See 8.4.2 of Klymik Schmudgen).</p> <p>First, I get <code>$(R^{-1})^{21}_{12} = -q^{-1} R^{21}_{12}$</code>, because $\langle(S\otimes id)(R),a^2_1\otimes a^1_2\rangle = \langle R,S(a^2_1)\otimes a^1_2 \rangle = -q^{-1} \langle R,a^2_1\otimes a^1_2\rangle$, using that $S(b)=-q^{-1} b$ from Proposition 4.1.2.3 of Klymik Schmudgen. So the actual matrix you should get should just be $q^{-1}$ times what you had expected to get.</p> <p>Now the factor of q^{-1} here is because you had multiplied the actual universal R matrix by $q^{1/2}$ and so $(\lambda A)^{-1} = \lambda^{-1} A^{-1}$, so there's a factor of $\lambda^2$ as a discrepancy.</p> <p>I hope this helps! </p>