Let $A$ be a Hopf algebra dually paired with a quasi-triangular Hopf algebra $B$. If $x$ is some fixed element of $A$, then we can define a linear map $$ P_x: A \to \mathbb{C} $$ by setting $$ P_x:a \mapsto \langle R,x \otimes a \rangle. $$

Let us take the case $A = SL_q(2)$, $B = U_q({\mathfrak sl}_2)$, and let $R$ be the standard universal $R$-matrix for $U_q({\mathfrak sl}_2)$, for which $$ \langle R, u^i_m \otimes u^j_n \rangle = R^{ij}_{mn} = q^{-\frac{1}{2}}.(q^{\delta_{ij}}\delta_{im}\delta_{jn} + (q-q^{-1})\theta (i-j)\delta_{in}\delta_{jm}), $$ where $\theta$ is the Heaviside symbol. If we take $x=u^k_l$, then $$ P_{u^k_l}(a) = \langle R, u^k_l \otimes a \rangle. $$ Now since $ab = qba$, we should have $$ P_{u^k_l}(u^1_1u^1_2) = q P_{u^k_l}(u^1_2u^1_1), \qquad \qquad \text{ for all } \quad k,l = 1,2. $$ However, $$ P_{u^2_1}(u^1_1u^1_2) = \langle R, u^2_1 \otimes u^1_1u^1_2 \rangle = \sum_{z=1}^2 \langle R,u^2_z \otimes u^1_1 \rangle \langle R, u^z_1 \otimes u^1_2 \rangle = \sum_{z=1}^2 R^{21}_{z1}R^{z1}_{12}. $$ From the formula for $R^{ij}_{mn}$, we get that $$ P(u^1_1u^1_2) = \sum_{z=1}^2 R^{21}_{z1}R^{z1}_{12} = R^{21}_{11}R^{11}_{12} + R^{21}_{21}R^{21}_{12} = 0.0 + q^{-\frac{1}{2}}.1.q^{-\frac{1}{2}}.(q-q^{-1}) = q^{-1}(q-q^{-1}). $$

On the other hand, we have $$ qP_{u^2_1}(u^1_2u^1_1) = \langle R,u^2_1 \otimes u^1_2u^1_1 \rangle = \sum_{z=1}^2 \langle R,u^2_z \otimes u^1_2 \rangle \langle R,u^z_1 \otimes u^1_1 \rangle =\sum_{z=1}^2R^{21}_{z2}R^{z1}_{11}. $$ From the formula for $R^{ij}_{mn}$, we now get that $$ qP_{u^2_1}(u^1_2u^1_1) = q\sum_{z=1}^2R^{21}_{z2}R^{z1}_{11} = qR^{21}_{12}R^{11}_{11} + qR^{21}_{22}R^{21}_{11} = q.q^{-\frac{1}{2}}.(q-q^{-1}).q^{-\frac{1}{2}}.q + q.0.0 = q(q-q^{-1}). $$

Thus, the two results are not equal, but instead differ by a factor of $q^2$. A similar problem arises for the action of $P_{u^2_1}$ on $bd - qdb$. We get $$ P_{u^2_1}(u^1_2u^2_2) = q^{-1}(q-q^{-1}), $$ whereas $$ qP_{u^2_1}(u^2_2u^1_2) = q(q-q^{-1}). $$

I've checked and rechecked everything very carefully but can't seem to spot my error. Can anyone see what is going wrong here?

  • $\begingroup$ The duality between SL_q(2) and u_q(sl_2) is worked out in Kassel's book. Have you tried comparing your R-matrix and other definitions with the ones there? He has ba=qab and db=qbd, so your q is his q^{-1}. $\endgroup$
    – M T
    Aug 14, 2010 at 16:03
  • $\begingroup$ The actual pairing and the form of the R-matrix in $U_q(sl_2)$ isn't so important, what matters is just the formula for $R^{ij}_{rs}$. $\endgroup$ Aug 14, 2010 at 16:37

5 Answers 5


Dear John,

I tried to follow your computation until the first place where I couldn't understand a step. This comes at:

However, $$ P_{u^2_1}(u^1_1u^1_2) = \langle R, u^2_1 \otimes u^1_1u^1_2 \rangle = \sum_z \langle R,u^2_z \otimes u^1_1 \rangle \langle R, u^z_1 \otimes u^1_2 \rangle, $$

Rather than the RHS, I would expect $$<(\operatorname{id}\otimes \Delta)(R), u^2_1 \otimes u^1_1 \otimes u^1_2> =<R_{13}R_{12}, u^2_1\otimes u^1_1\otimes u^1_2> =\sum_z<R,u^2_z\otimes u^1_2><R,u^z_1\otimes u^1_1> $$

which seems different than what you wrote. It seems you have used the opposite comultiplication in your computations so that where I wrote $R_{13}R_{12}$ above, you instead had $R_{12}R_{13}$. I hope this helps. I am aware that pairing of Hopf algebras sometimes requires matching multiplication of $H$ with opposite co-multiplication of $H^*$. However, you seem to be working from Klymik and Schmudgen's text, which doesn't not use opposite co-product in the definition of dual pairing of Hopf algebras.

I haven't checked the details to see if the above resolves your issue. Perhaps this is still not your source of confusion, but it confused me when I first read it in your post.

Looking again at what you wrote, this means that the two computations you did for $P_c(ab)$ and $P_c(ba)$ are thus switched, so that you are multiplying $P_c(ab)$ by $q$ instead of $P_c(ba)$, as you thought. Multiplying instead of dividing by $q$ gives the discrepancy of $q^2$

thanks, -david

  • $\begingroup$ That's it!!! I proved that $\langle R, ab \otimes c \rangle = \langle R,a \otimes c_{(1)} \rangle \langle b \otimes c_{(2)} \rangle$ and stupidly assumed that $\langle R, a \otimes bc \rangle = \langle R,a_{(1)} \otimes b \rangle \langle a_{(2)} \otimes c \rangle$. Thank very much, I've been pulling my hair out for a week now ... what a relief! $\endgroup$ Aug 18, 2010 at 16:49
  • $\begingroup$ I am wondering. Are David and me saying the same thing ? (I am refering to my second answer) $\endgroup$
    – DamienC
    Aug 18, 2010 at 17:35
  • $\begingroup$ No I don't think so. The problem lies not with the definition of the R-matrix $R^{ij}_{rs}$ but with how the it operates on $H \otimes H$. Changing the R-matrix wouldn't resolve this. Changing the coproduct might solve the specific problem you dealt with, but will throw up new ones. Thanks alot for all your work though, unlucky you didn't get the bounty. $\endgroup$ Aug 18, 2010 at 18:14
  • $\begingroup$ Damien, yes looking at your second answer, I think you are correct (perhaps John can be convinced to re-assign his bounty =] ). Basically (as you know), the quantum group is not a well-defined object but a torsor over a rather annoying group of arbitrary choices, such as swapping $\Delta$ for $\Delta^op$, $R$ for $R^{-1}$, etc. It seems that each author chooses a slightly different convention, just to keep things interesting. I pointed out that John used opposite coproduct when simplifying the pairing, while you pointed out that one can take opposite co-product to begin with (a la Kassel). $\endgroup$ Aug 18, 2010 at 18:16
  • $\begingroup$ @John: oops, our comments crossed paths. Anyways, I'm glad to help. $\endgroup$ Aug 18, 2010 at 18:37

First of all I believe a factor $q^{-1/2}$ is missing in your definition of the coefficients of the universal R-matrix.

Then, as far as I remember, the commutation relations of $\mathcal O_q(SL_2)$ are $ba = qab$, $db = qbd$, $ca = qac$, $dc = qcd$, $bc = cb$, $da-ad=(q-q^{-1})bc$, and $ad-q^{-1}bc=1$. So we do NOT have the relation $ab=qba$.

Finally, according to your convention it seems that you have $a=u_1^1$, $b=u_2^1$ $c=u_1^2$ $d=u_2^2$ (it seems that Kassel has a different convention for indices, but his R-matrix coefficients are also organized in a different way, so...). So let me compute $P_c(ab)$ and $P_c(ba)$ following your notation.

$P_c(ab) = R^{21}_{11} R^{11}_{12} + R^{21}_{21} R^{21}_{12} = 0$


$P_c(ba) = R^{21}_{12} R^{11}_{11} + R^{21}_{22} R^{21}_{11} = q(q-q^{-1})$

Then I believe the definition of the coefficients you gave is wrong (also I can't really follow your computations: there are a few typos, and also errors - or it might be that I did not understand what is going on).

Now if I compute following Kassel's definition of R-matrix coefficients I find :

$P_c(ab) = R^{21}_{11} R^{11}_{12} + R^{21}_{21} R^{21}_{12} = 0$


$P_c(ba) = R^{21}_{12} R^{11}_{11} + R^{21}_{22} R^{21}_{11} = 0$

By the way, even following uniquely your definitions I can't see how you get (on line 16) the following:

$P(u^1_1u^1_2) = \quad \sum_z R^{21}_{z1}R^{zi}_{1z} \quad = \quad R^{21}_{12}R^{21}_{12} \quad = \quad q^{-1}(q-q^{-1})$

First of all there is a typo, the second term should be $\sum_z R^{21}_{z1}R^{z1}_{12}$. Then there seems to be two errors:

  1. how can you find $R^{21}_{12}R^{21}_{12}$ ?
  2. I can't see how $R^{21}_{12}R^{21}_{12}=q(q-q^{-1})$.
  • $\begingroup$ Please see my answer by way of a comment. $\endgroup$ Aug 16, 2010 at 23:46

I tried to find a resolution of this problem by looking at it in the greater generality of FRT-algebras. However, I also ran into an apparent contradiction. I have posted my calculations as a new question here. Hopefully someone can find an answer to both questions.


I think I now see where is the problem in your computation.

a) First of all let me recall the problem.

You find $r(c\otimes ab)=q^{-1}r(c\otimes ba)$, while you would like to find $r(c\otimes ab)=qr(c\otimes ba)$.

b) Let me now compare $r(ab\otimes c)$ with $r(ba\otimes c)$ and see if you end up with the same problem. On the one hand (using the same computation rule as yours),

$$ r(ab\otimes c)=r(u_1^1u_2^1\otimes u_1^2)=\sum_zr(u_1^1\otimes u_z^2)r(u_2^1\otimes u_1^z) $$ $$ =R_{1z}^{12}R^{1z}_{12}=R_{12}^{12}R^{12}_{12}=q^{-1}(q-q^{-1}). $$

On the other hand $$ r(ba\otimes c)=r(u_2^1u_1^1\otimes u_1^2)=\sum_zr(u_2^1\otimes u_z^2)r(u_1^1\otimes u_1^z) $$ $$ =R_{2z}^{12}R^{1z}_{11}=R_{21}^{12}R^{11}_{11}=q-q^{-1}. $$

Then we find $r(ab\otimes c)=q^{-1}r(ba\otimes c)$ while we would hope to have $r(ab\otimes c)=qr(ba\otimes c)$.

c) The problem might come from the definition of the $R$-matrix (it may be that somewhere $R$ and $\hat{R}:=R\tau$ have been mixed).

But the problem might also come from a mistake in the way the coproduct is written. Namely, according to what your wrote $\Delta(c)=\Delta(u_1^2)=\sum_zu^2_z\otimes u^z_1=c\otimes a+d\otimes c$; while I am used to $\Delta(c)=\Delta(u_1^2)=\sum_zu_1^z\otimes u_z^2=a\otimes c+c\otimes d$.

Now doing again the computation with this second definition of the coproduct I find: $r(c\otimes ab)=q(q-q^{-1})=qr(c\otimes ba)$... which is precisely what you were expecting.

I hope this answers yor question.


I'm going to put my comment to Damien's answer as an answer since there's not enough room to place it as a comment. Firstly, thank you for pointing the typos. I have corrected them and apologise for not checking what I had written thoroughly enough at the start.

With regard to the normalisation factor $q^{\frac{-1}{2}}$, I tacitly dropped it because it cancels out for the calculation I'm interested in. However, you're right, it should be included in my definition and I've changed it.

With regard to the commutation relations of $SL_q(2)$ there are two conventions: one is as I have written, with, for example, $ab=qba$, and another has $ab=q^{-1}ba$, as you have written. Both algebras are of course isomorphic. I have taken my conventions from Klimyk and Schmuedgen, both for the relations (Chapter 4) and for the definition of $R^{ij}_{nm}$ (Chapter 9).

I don't have Kassel's book at hand, so I can't really comment at the moment on his conventions. I will try to have a look tomorrow though.

With regard to the $R^{ij}_{mn}$ calculations, I just use the fact that the only non-zero entries are $$ R^{11}_{11} = R^{22}_{22} = q^{\frac{1}{2}}, \qquad R^{12}_{12}=R^{21}_{21}=q^{-\frac{1}{2}}, \qquad R^{21}_{12} = q^{-\frac{1}{2}}(q-q^{-1}). $$ (But I think it was my typos that were causing the confusion here.)

  • $\begingroup$ Thank you. I agree that with these coefficients and this presentation there is something wrong. nevertheless, if you take these coefficients and the other presentation (with ($ba=qab$), it seems to work. $\endgroup$
    – DamienC
    Aug 17, 2010 at 8:21
  • $\begingroup$ Well actually it doesn't seem to work with Kassel's approach either: Taking all our conventions from Page 195 (along with the usual coproduct $\Delta(b) = a\otimes b + b \otimes d$) we see that $r(b \otimes (ac)) = r(a \otimes a)r(b \otimes c) + r(b \otimes a)r(d \otimes c) = \lambda q(q-q^{-1}) + 0 = q^{\frac{1}{2}}(q-q^{-1}).$ Whereas $q^{-1}r(b \otimes (ca)) = q^{-1}r(a \otimes c)r(b \otimes a)+ q^{-1}r(b \otimes c)r(d \otimes a) = 0 + q^{-1}\lambda (q-q^{-1}) = q^{-\frac{3}{2}}(q-q^{-1}).$ (As well I'm baffled by the $\Delta(a)= a \otimes a + b \otimes d$ on the following page.) $\endgroup$ Aug 18, 2010 at 4:47
  • $\begingroup$ About $\Delta(a)$ on page 196 of Kassel's book, I guess it is a typo. $\Delta(a)=a\otimes a+b\otimes c$. $\endgroup$
    – DamienC
    Aug 18, 2010 at 9:44

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