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Take, for instance, the $R$ matrix,

\begin{equation} R(u)=\begin{pmatrix}u+1 & 0 & 0 & 0\\0 & u & 1 & 0\\0 & 1 & u & 0\\0 & 0 & 0 & u+1\end{pmatrix}, \end{equation} wich satisfies the Yang-Baxter equation \begin{equation} R^{12}(u-v)R^{13}(u)R^{23}(v)=R^{23}(v)R^{13}(u)R^{12}(u-v). \end{equation} This $R$ matrix is associated to the $sl(2)$ Lie algebra, but how can I prove that? Actually I have a very complicated $R$ matrix: I'm looking for a general guide to find the Lie algebra associated to a given $R$ matrix.

In other question posted here, (Solutions of the Quantum Yang-Baxter Equation) I found the comment: "The general theory (due to Jimbo) is that each irreducible finite dimensional representation of the quantised enveloping algebra of a Kac-Moody algebra (not of finite type) gives a trigonometric R-matrix." I think I'm looking for the converse of this: given a $R$-matrix satisfying the Yang-Baxter equation, what is the Lie algebra associated to it?

Thank you in advance.

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  • $\begingroup$ From a solution of the TB equation you can construct an algebra, using the FRT construction. But that algebra willin general not come from a Lie algebra in any reasonable way.. $\endgroup$ Aug 7, 2016 at 22:46

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To the best of my knowledge this is a very hard problem and the answer to this question is, unfortunately, open.

A famous example that illustrates this in the context of quantum integrability comes from the one-dimensional Hubbard model in condensed-matter physics. Its (quite complicated) $R$-matrix was known since the late eighties, yet the corresponding quantum group was only found in the last decade, "by accident" in a very different context. Namely: the $R$-matrix of the Hubbard model came out of the study of the quantum group associated to a Lie (super)algebra, and the choice of the latter was motivated by string theory (more precisely: the AdS/CFT correspondence). For a bit more about this, and further references, see for example (the introduction of) arXiv:1509.06205.

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