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In the classical equation, one looks for $R\in\Lambda^2\mathfrak g$ such that $$[R,R]=0,$$ where the bracket is Schouten's bracker in $\Lambda^\bullet\mathfrak g$, the exterior algebra on a Lie algebra $\mathfrak g$. In the quantum one (in its non-parametric form...), one looks for endomorphisms $R:V\otimes V\to V\otimes V$ of tensor squares of vector spaces $V$ such that $$R_{12} \ R_{13} \ R_{23} = R_{23} \ R_{13} \ R_{12},$$

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What is the difference between the Yang--Baxter equation and the quantum Yang--Baxter equation?

For a vector space $V$ and a linear operator $R:V \otimes V \to V \otimes V$, we say that $R$ satisfies the Yang--Baxter equation if $$(R\otimes id)(id\otimes R)(R\otimes id) = (id\otimes R)(R\otimes …
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