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Any quantized universal enveloping algebra (in fact, any toplogically quasi-triangular Hopf algebra) has an (in its completion) an element u called the Drinfeld element which gives an isomorphism from a representation to its double dual.

However, most people prefer to use a different pivotal structure on the category of representations of the quantized universal enveloping algebra, where u is replaced by g=v-1u. Several obvious references don't seem to have a formula for this element, even though my dim recollection is that it is very simple. Is there anywhere where this is written down properly?

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  • $\begingroup$ I was going to ask what quasi-triangular means, but it turns out there's a wiki page en.wikipedia.org/wiki/Quasitriangular_Hopf_algebra, $\endgroup$
    – David Zureick-Brown
    Commented Oct 12, 2009 at 17:58
  • $\begingroup$ Yeah, though I wouldn't worry too much about the general theory. My question is very specifically about quantized universal enveloping algebras. The exact problem is that books like Chari and Pressley work this stuff out for general quasi-triangular Hopf algebras before defining quantized enveloping algebras, which makes the particular formulas in that case hard to find. $\endgroup$
    – Ben Webster ♦
    Commented Oct 12, 2009 at 18:04
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    $\begingroup$ So u doesn't actually give a pivotal structure because it isn't grouplike. In particular the functor from V->V** given by u is not a tensor functor and hence doesn't give an actual "pivotal structure." $\endgroup$
    – Noah Snyder
    Commented Oct 12, 2009 at 20:34

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Up to sign conventions it's just K_\rho, where \rho is the Weyl vector (half the sum of the positive roots). It's easy to check that this does satisfy the basic properties of g, namely that it is grouplike and that conjugation by it acts by S^2. Thus it gives a pivotal structure. (You still need to check that it actually is the g that comes from the usual ribbon element, which is a bit more delicate.)

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  • $\begingroup$ That's what I thought, but when I couldn't find a formula in Chari and Pressley (or your paper with Peter), I started to wonder. $\endgroup$
    – Ben Webster ♦
    Commented Oct 12, 2009 at 20:34
  • $\begingroup$ I apologize. I also assumed it would be in the paper with Peter and looked it up to try to find a reference and was sad when it wasn't there. $\endgroup$
    – Noah Snyder
    Commented Oct 12, 2009 at 20:57
  • $\begingroup$ @NoahSnyder Shouldn't it be twice the Weyl vector? $S^2(E_i)=K_i E_i K_i^{-1}=q^{(\alpha_i,\alpha_i)} E_i = q^{(2\rho,\alpha_i)} E_i =K_{2\rho} E_i K_{2\rho}^{-1}$ $\endgroup$
    – Bipolar Minds
    Commented Apr 5, 2017 at 20:46
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It looks like your question was already answered, but I highly recommend the book by Klymik and Schüdgen for this sort of question. In particular, they spell out the formula for a ribbon element in a QUE's quite explicitly as I recall. Both an advantage and a drawback of that book are that they do everything very explicitly with formulas.

I could be wrong (don't have a text with me) but I thought that the formula was a slight modification of what Noah wrote: namely one takes as a first guess u=\mu(R_21 R_12), i.e. you take the two components of the squared R-matrix and multiply them. This is almost a ribbon element; it satifies a coproduct relation similar to the ribbon element, and more precisely, uS(u) is the square of the actual ribbon element. Thus, u has to be corrected by the factor Noah mentioned, e^{-h\rho}. In the end, I believe the ribbon element is e^{-h rho}u.

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  • $\begingroup$ I think I misunderstood the question, so I answered about the element u which Ben mentioned originally, instead of g=e^{-h\rho} which Noah already explained. My bad. $\endgroup$
    – David Jordan
    Commented Oct 23, 2009 at 15:03
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Could this be related to Drinfeld associator?

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  • $\begingroup$ Only incredibly distantly. I think it's much easier than that. Remember, Drinfeld (justifiably) has his name on a lot of not particularly related things. $\endgroup$
    – Ben Webster ♦
    Commented Oct 12, 2009 at 17:45
  • $\begingroup$ Yeah, they're not related beyond the fact that they both have to do with Drinfeld-Jimbo quantum groups. $\endgroup$
    – Noah Snyder
    Commented Oct 12, 2009 at 20:28

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