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I've come across a reference in a paper to the

Hecke relation for the universal R-matrix of a quasi-triangular Hopf algebra.

I've looked around, standard references, online etc, but can't seem to find what this Hecke relation is. Can anyone point me in the right direction, or even better, tell me what it is?

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    $\begingroup$ I am uncomfortable with this question as phrased, because the universal R-matrix for a quasi-triangular Hopf algebra does not satisfy a Hecke relation. The universal R-matrix is a unit in H\otimes H, and in the case of U_q(sl_n), U_q(gl_n), and a few exotic examples, its evaluation on the tensor square of a certain representation of H satisfies a Hecke relation. However, even in the case of U_q(sl_n) and U_q(gl_n), the universal R-matrix itself doesn't satisfy a Hecke relation. $\endgroup$
    – David Jordan
    Jun 6, 2012 at 16:49

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For the Drinfeld-Jimbo quantum universal enveloping algebras, see Proposition 24 of Chapter 8 in the book Quantum Groups and Their Representations, by Klimyk and Schmudgen. This relation is just in the type A situation, for $\mathfrak{gl}_n$ or $\mathfrak{sl}_n$. The relation they get is $$ (\hat{R} - q)(\hat{R} + q^{-1}) = 0, $$ where $R$ is the R-matrix for the vector representation of $U_q(\mathfrak{g})$, $\hat{R} = \tau \circ R$, where $\tau$ is the tensor flip, and all the conventions are those of Klimyk and Schmudgen.

In other situations, the map $\hat{R}$ has more than just two eigenvalues, so the spectral decomposition is more complicated, and the characteristic polynomial (i.e. the Hecke relation) is more complicated.

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  • $\begingroup$ I should say also that the spectral decompositions for the R-matrices for the other quantized enveloping algebras are there as well in Propositions 25 and 26. $\endgroup$
    – MTS
    May 22, 2010 at 13:39
  • $\begingroup$ I've done some calculations and this equality seems to imply that $R^2 - (q-q^{-1})R - 1$ is also true. But I don't see it anywhere in the book. I'm worried that this means I've made a mistake somewhere. My work was as follows: I showed directly that $\hat{R}^{-1} = \hat{R^{-1}}$. Then I multiplied out the relation to get $\hat{R}^2 -(q-q^{-1})\hat{R} - 1 = 0$ and multiplied both sides by $\hat{R^{-1}}$ to get $\hat{R} -(q-q^{-1}) - \hat{R^{-1}} = 0$. I operated on both sides of this equality with $\tau$ to get $R -(q-q^{-1}) - R^{-1}=0$. Finally, I multiplied both sides by $R$. Is this right? $\endgroup$
    – Dyke Acland
    May 22, 2010 at 17:08

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