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

In the Hopf algebra $SL_q(N)$, it can be shown, using direct calculations, that $S(u^1_i)u^j_1 = q^{-1}u^j_1S(u^1_i)$. Can anyone see a more elegant way of establishing this?

Moreover, does anyone know of a similar relation in the more general case of $S(u^1_r)u^i_j$?

Edit (referneces): By $SL_q(N)$ I mean the quantized coordinate algebra (not the quantized enveloping algebra). I am using the conventions of Klimyk and Schmudgen, Chpt 4 for the N=2 case, or Chpt 9 for the general case.

share|cite|improve this question
Could you add a reference for this model of quantum groups? It is not exactly the same as the usual $U_q(sl(N))$. And your question depends on precise conventions. – Greg Kuperberg Oct 15 '10 at 18:15
Sorry for the confusion. It's not the quantum enveloping algebra, it's the quantised coordinate algebra. I've added the references. – John McCarthy Oct 15 '10 at 19:53
up vote 2 down vote accepted

This is a reasonably known result. That $S(u^1_i)u^j_1 = q^{-1}u^j_1S(u^1_i)$, was originally proven (to the best of my knowdledge) in FRT's '89 paper "Quantum Groups and Lie Algebras" - the paper is in Russian though. The only English write up of the proof that I known is in Theorem 1 of Vainermann and Podkolzin's '99 paper on Quantum Stiefel Manifolds. It gives a general comm rel for $[S(u^i_j),u^r_s]$, for the general $N$ case, using just the $R$-matrix construction of the $SU_q(N)$. I am sure there are other versions around somewhere though.

share|cite|improve this answer

For the first question, I would use the dual pairing with $U_q(\mathfrak{sl}_N)$. The $u_i^j$'s are defined to be matrix coefficients of the vector representation of $U_q(\mathfrak{sl}_N)$ with respect to some distinguished basis, usually a basis of weight vectors. There are unfortunately a lot of different conventions in use. My standard reference is Klimyk and Schmudgen. See, for example, Theorem 19 of Chapter 9 of their book. It states:

There is a unique dual pairing $( , )$ of Hopf algebras between $U_q^{ext}(\mathfrak{sl}_N)$ and $\mathcal{O}(SL_q(N))$ such that $(f, u^k_l) = t_{kl}(f)$ for all $f \in U_q^{ext} (\mathfrak{sl}_{N})$.

Here $((t_{kl}(f))$ is the matrix for $f$ in the vector representation. OK, this theorem is a little bogus in the sense that it is more of a definition. But the point is that $\mathcal{O}(SL_q(N))$ is generated by the matrix coefficients of all finite-dimensional irreducible representations of $U_q^{ext} (\mathfrak{sl}_{N})$, and these separate points of $U_q^{ext} (\mathfrak{sl}_{N})$, so the pairing is nondegenerate.

So, to show that your two guys are equal, just show that they pair the same way with $U_q^{ext} (\mathfrak{sl}_{N})$. Since it is a pairing of Hopf algebras, you just need to check on the generators $E_i, F_i, K_\lambda$. This just requires you to have a handle on the vector representation. In my opinion this is much cleaner than doing the calculations directly.

share|cite|improve this answer
Great, this is just what I was looking for. Just one thing though, I don'e see why the fact that its a Hopf algebra pairing implies that I only need to check it on the generators. How do I get the value of $< x,E^2> = <x_{(1)},E> < x_{(2)}, E >$ from the pairings of $x$ with the generators? – John McCarthy Oct 12 '10 at 21:48
... take the example, for $N=2$, of $<b^2,K>=<b^2,E>=<b^2,F>=0$, while of course $b^2 \neq 0$. – John McCarthy Oct 13 '10 at 0:28
Yeah, you're right. Hmmm... not quite sure how to resolve that. I guess the generators aren't enough. So perhaps this doesn't work after all. My bad. – MTS Oct 13 '10 at 20:34

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.