MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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 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!

share|cite|improve this answer
Great! Thanks a lot for that! – Abtan Massini May 22 '10 at 23:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.