For every finite dimensional semi-simple Lie group $\mathfrak{g}$, we have a loop algebra $\mathfrak{g}[t,t^{-1}]$. This loop algebra has a natural invariant inner product by taking the residue at zero of the Killing form applied to two elements (i.e. $t^k\mathfrak{g}$ and $t^{-k-1}\mathfrak{g}$ are paired by the Killing form.)

This Lie algebra actually has a Manin triple structure with respect to this inner product: the subalgebras $\mathfrak{g}[t]$ and $t^{-1}\mathfrak{g}[t^{-1}]$ are both isotropic, and non-degenerately paired by this form. This makes $\mathfrak{g}[t]$ into a Lie bialgebra, by getting the cobracket from the bracket on $t^{-1}\mathfrak{g}[t^{-1}]$.

Now, as we all know, Lie bialgebras can be quantized: in this case, the result is a quite popular Hopf algebra called the Yangian. By the usual yoga of quantization of Lie bialgebras, the dual Hopf algebra to the Yangian quantizes the universal enveloping $t^{-1}\mathfrak{g}[t^{-1}]$, so if you take a different associated graded of the Yangian, you must get the Hopf algebra of functions on the group with Lie algebra $t^{-1}\mathfrak{g}[t^{-1}]$, which is $L_<G$, the based formal loop group.

Now, all of these things also have explicit descriptions in terms of equations, and it seems as though this story must be worked out explicitly somewhere, but I've had little luck locating it. Does anyone know where? Or is this story just wrong, and that's why I can't find it?

EDIT: The comment below mostly answers this question. I would be interested if anyone out there has written something more explicit than the Etingof and Kazhdan paper, but it's the sort of thing I was looking for. If it were to be left in the form of an answer, I would probably accept it (hint, hint).

  • $\begingroup$ Thanks for the hint. I changed my comment to an answer. $\endgroup$
    – B.R.J.
    Mar 22, 2010 at 9:47
  • $\begingroup$ Well, I might misunderstand the question, but it seems to me everything is explicit, and more or less isomorphic to each other. There is "Yangian double" arxiv.org/abs/hep-th/9406194 arxiv.org/abs/q-alg/9602031 which contains both positive and negative parts of loop group quantization and all parts are explicitly described. It is the same story as for quantized loop group U_q(gl((t)) - there is no conceptual difference. As in non quantum case gl[t] gl[t^-1] - are isomorphic, more or less the same after quantization $\endgroup$ Apr 5, 2020 at 9:43

1 Answer 1


I will suggest the article Quantization of Lie bialgebras, III by Pavel Etingof and David Kazhdan. It discusses both the Yangian and the dual Yangian as examples in the context of quantization of Lie bialgebras of functions on a curve with punctures.

Also, so far as I know, no-one ever explicitly derives the relations of the Yangian from the conditions that it be a Hopf algebra quantization of the Lie bialgebra, but Drinfeld explains it in his article called Quantum Groups (Proc. ICM Berkeley, 1986). If you assume the coproduct takes a certain form on the Lie algebra (the most obvious choice given that it is a quantization of that Lie bialgebra) then impose the condition that the Hopf algebra coproduct be an algebra homomorphism then you can derive the relations of the Yangian in the first presentation term by term, using the fact that it is a homogeneous (graded) deformation.

Personally I tried to do this but I found I had to assume little things along the way, like that the RHS of equation (13) in that paper is a symmetric sum of the orthonormal basis elements of the Lie algebra. I couldn't really understand why I had to do this, but I guess it's probably obvious to people who are smarter than me. At the end you end up with an algebra that you can then prove is isomorphic to the second presentation he gives, and then get a PBW theorem based on this presentation and verify that it is a quantization of the Lie bialgebra after all (I still working on understanding this part). Then given the uniqueness, which is meant to follow from cohomological arguments described in Section 9 of that same Quantum Groups paper, you don't need to justify any assumptions you make along the way. It seems that understanding the cohomology is key.


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