Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:57:30Z http://mathoverflow.net/feeds/question/35057 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35057/map-constructed-from-the-coquasitriangular-structure-of-slq2-which-appears-not Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations John McCarthy 2010-08-09T22:35:05Z 2010-08-18T13:01:25Z <p>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 $$P_x:a \mapsto \langle R,x \otimes a \rangle.$$</p> <p>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 <code>$$\langle R, u^i_m \otimes u^j_n \rangle = R^{ij}_{mn} = q^{-\frac{1}{2}}.(q^{\delta_{ij}}\delta_{im}\delta_{jn} + (q-q^{-1})\theta (i-j)\delta_{in}\delta_{jm}),$$</code> 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, <code>$$P_{u^2_1}(u^1_1u^1_2) = \langle R, u^2_1 \otimes u^1_1u^1_2 \rangle = \sum_{z=1}^2 \langle R,u^2_z \otimes u^1_1 \rangle \langle R, u^z_1 \otimes u^1_2 \rangle = \sum_{z=1}^2 R^{21}_{z1}R^{z1}_{12}.$$</code> From the formula for $R^{ij}_{mn}$, we get that <code>$$P(u^1_1u^1_2) = \sum_{z=1}^2 R^{21}_{z1}R^{z1}_{12} = R^{21}_{11}R^{11}_{12} + R^{21}_{21}R^{21}_{12} = 0.0 + q^{-\frac{1}{2}}.1.q^{-\frac{1}{2}}.(q-q^{-1}) = q^{-1}(q-q^{-1}).$$</code></p> <p>On the other hand, we have <code>$$qP_{u^2_1}(u^1_2u^1_1) = \langle R,u^2_1 \otimes u^1_2u^1_1 \rangle = \sum_{z=1}^2 \langle R,u^2_z \otimes u^1_2 \rangle \langle R,u^z_1 \otimes u^1_1 \rangle =\sum_{z=1}^2R^{21}_{z2}R^{z1}_{11}.$$</code> From the formula for $R^{ij}_{mn}$, we now get that <code>$$qP_{u^2_1}(u^1_2u^1_1) = q\sum_{z=1}^2R^{21}_{z2}R^{z1}_{11} = qR^{21}_{12}R^{11}_{11} + qR^{21}_{22}R^{21}_{11} = q.q^{-\frac{1}{2}}.(q-q^{-1}).q^{-\frac{1}{2}}.q + q.0.0 = q(q-q^{-1}).$$</code></p> <p>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 $$qP_{u^2_1}(u^2_2u^1_2) = q(q-q^{-1}).$$</p> <p>I've checked and rechecked everything very carefully but can't seem to spot my error. Can anyone see what is going wrong here?</p> http://mathoverflow.net/questions/35057/map-constructed-from-the-coquasitriangular-structure-of-slq2-which-appears-not/35111#35111 Answer by David Jordan for Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations David Jordan 2010-08-10T12:44:15Z 2010-08-18T13:01:25Z <p>Dear John,</p> <p>I tried to follow your computation until the first place where I couldn't understand a step. This comes at:</p> <blockquote> <p>However, $$P_{u^2_1}(u^1_1u^1_2) = \langle R, u^2_1 \otimes u^1_1u^1_2 \rangle = \sum_z \langle R,u^2_z \otimes u^1_1 \rangle \langle R, u^z_1 \otimes u^1_2 \rangle,$$</p> </blockquote> <p>Rather than the RHS, I would expect <code>$$&lt;(\operatorname{id}\otimes \Delta)(R), u^2_1 \otimes u^1_1 \otimes u^1_2&gt; =&lt;R_{13}R_{12}, u^2_1\otimes u^1_1\otimes u^1_2&gt; =\sum_z&lt;R,u^2_z\otimes u^1_2&gt;&lt;R,u^z_1\otimes u^1_1&gt;$$</code></p> <p>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.</p> <p>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.</p> <p>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 $q^2$</p> <p>thanks, -david</p> http://mathoverflow.net/questions/35057/map-constructed-from-the-coquasitriangular-structure-of-slq2-which-appears-not/35781#35781 Answer by DamienC for Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations DamienC 2010-08-16T17:44:53Z 2010-08-16T17:51:57Z <p>First of all I believe a factor $q^{-1/2}$ is missing in your definition of the coefficients of the universal R-matrix. </p> <p>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 $ab=qba$. </p> <p>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> <p><code>$P_c(ab) = R^{21}_{11} R^{11}_{12} + R^{21}_{21} R^{21}_{12} = 0$</code></p> <p>and </p> <p><code>$P_c(ba) = R^{21}_{12} R^{11}_{11} + R^{21}_{22} R^{21}_{11} = q(q-q^{-1})$</code></p> <p>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). </p> <p>Now if I compute following Kassel's definition of R-matrix coefficients I find : </p> <p><code>$P_c(ab) = R^{21}_{11} R^{11}_{12} + R^{21}_{21} R^{21}_{12} = 0$</code></p> <p>and </p> <p><code>$P_c(ba) = R^{21}_{12} R^{11}_{11} + R^{21}_{22} R^{21}_{11} = 0$</code></p> <p>By the way, even following uniquely your definitions I can't see how you get (on line 16) the following: </p> <p><code>$P(u^1_1u^1_2) = \quad \sum_z R^{21}_{z1}R^{zi}_{1z} \quad = \quad R^{21}_{12}R^{21}_{12} \quad = \quad q^{-1}(q-q^{-1})$</code></p> <p>First of all there is a typo, the second term should be <code>$\sum_z R^{21}_{z1}R^{z1}_{12}$</code>. Then there seems to be two errors: </p> <ol> <li>how can you find <code>$R^{21}_{12}R^{21}_{12}$</code> ?</li> <li>I can't see how <code>$R^{21}_{12}R^{21}_{12}=q(q-q^{-1})$</code>. </li> </ol> http://mathoverflow.net/questions/35057/map-constructed-from-the-coquasitriangular-structure-of-slq2-which-appears-not/35806#35806 Answer by John McCarthy for Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations John McCarthy 2010-08-16T22:54:21Z 2010-08-16T22:54:21Z <p>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.</p> <p>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.</p> <p>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).</p> <p>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.</p> <p>With regard to the $R^{ij}_{mn}$ calculations, I just use the fact that the only non-zero entries are <code>$$R^{11}_{11} = R^{22}_{22} = q^{\frac{1}{2}}, \qquad R^{12}_{12}=R^{21}_{21}=q^{-\frac{1}{2}}, \qquad R^{21}_{12} = q^{-\frac{1}{2}}(q-q^{-1}).$$</code> (But I think it was my typos that were causing the confusion here.) </p> http://mathoverflow.net/questions/35057/map-constructed-from-the-coquasitriangular-structure-of-slq2-which-appears-not/35816#35816 Answer by Abtan Massini for Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations Abtan Massini 2010-08-17T00:27:18Z 2010-08-17T00:27:18Z <p>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 <a href="http://mathoverflow.net/questions/35814/establishing-the-co-quasi-triangular-structure-of-frt-algebras" rel="nofollow">here</a>. Hopefully someone can find an answer to both questions.</p> http://mathoverflow.net/questions/35057/map-constructed-from-the-coquasitriangular-structure-of-slq2-which-appears-not/35957#35957 Answer by DamienC for Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations DamienC 2010-08-18T12:08:22Z 2010-08-18T12:08:22Z <p>I think I now see where is the problem in your computation. </p> <p>a) First of all let me recall the problem. </p> <p>You find $r(c\otimes ab)=q^{-1}r(c\otimes ba)$, while you would like to find $r(c\otimes ab)=qr(c\otimes ba)$. </p> <p>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), </p> <p><code>$$r(ab\otimes c)=r(u_1^1u_2^1\otimes u_1^2)=\sum_zr(u_1^1\otimes u_z^2)r(u_2^1\otimes u_1^z)$$</code> <code>$$=R_{1z}^{12}R^{1z}_{12}=R_{12}^{12}R^{12}_{12}=q^{-1}(q-q^{-1}).$$</code></p> <p>On the other hand <code>$$r(ba\otimes c)=r(u_2^1u_1^1\otimes u_1^2)=\sum_zr(u_2^1\otimes u_z^2)r(u_1^1\otimes u_1^z)$$</code> <code>$$=R_{2z}^{12}R^{1z}_{11}=R_{21}^{12}R^{11}_{11}=q-q^{-1}.$$</code></p> <p>Then we find $r(ab\otimes c)=q^{-1}r(ba\otimes c)$ while we would hope to have $r(ab\otimes c)=qr(ba\otimes c)$. </p> <p>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). </p> <p>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 $\Delta(c)=\Delta(u_1^2)=\sum_zu_1^z\otimes u_z^2=a\otimes c+c\otimes d$. </p> <p>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. </p> <p>I hope this answers yor question. </p>