## The relation $S(u^1_i)u^j_1 = q^{-1}u^j_1S(u^1_i)$

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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.

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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 2010 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 2010 at 19:53

## 2 Answers

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.

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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.

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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 2010 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 2010 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 2010 at 20:34
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