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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 $$R^{ir}_{js} = \langle R, u^i_j \otimes u^r_s \rangle = q^{\delta_{ir}}\delta_{ij}\delta_{rs} + (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}.$$ 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 $$\langle R, u^i_j \otimes u^r_s \rangle = q^{-\delta_{ir}}\delta_{ij}\delta_{rs} - (q-q^{-1})\theta (i-r)\delta_{is}\delta_{jr}. $$

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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According to Klimyk and Schmudgen's book, your formula for the $R$-matrix is incorrect. They have $\theta(i-r)$ where you have $\theta(r-s)$. That term is nonzero only when $s=i$, so your formula is the same as $\theta(r-i)$. I don't know if this anyway gives the same result, but that's all I can come up with. –  MTS May 22 '10 at 19:46
    
Thanks for pointing that out, it's now correct. Unfortunately it was just a typo with no affect on the calculations. –  Abtan Massini May 22 '10 at 20:26
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1 Answer 1

up vote 8 down vote accepted

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).

First, I get $(R^{-1})^{21}_{12} = -q^{-1} R^{21}_{12}$, 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.

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.

I hope this helps!

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Great! Thanks a lot for that! –  Abtan Massini May 22 '10 at 23:14
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