Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

what are the differences between classical Yang-Baxter Equation and quantum Yang-Baxter Equation? Thank you very much.

share|cite|improve this question
up vote 3 down vote accepted

This is basically a comment to the answer of Mariano Suárez-Alvar, but it's too long for a comment. I want to say what is the connection of classical and quantum YBE and to correct a little omition.

A solution of Yang-Baxter equation is called an R-matrix. In the quantum YBE, $R$ is an element of $H\otimes H$, where $H=U_h\mathfrak{g}$ is a quantum group, i.e. a deformation of the enveloping algebra $U\mathfrak{g}$. ($R$ is supposed to make the category of $H$-modules to a braided monoidal category. In particular it should give a representation of the braid group $B_n$ on the $n$-fold tensor power of any module $V$; quantum YBE is just the relevant relation in $B_3$.)

Classical YBE is a 1st order approximation of the quantum YBE. It is an equation for an element $r\in\mathfrak{g}\otimes\mathfrak{g}$, it says $[r_{12}, r_{13}] + [r_{12}, r_{23}] + [r_{13}, r_{23}] = 0$. If $r$ is skew-symmetric then it's the equation written by Mariano. If not (and in the interesting cases it is not skew-symmetric) it can be written as $[r_{skew},r_{skew}]=\phi$ where $r_{skew}$ is the skew-symmetric part of $r$ and $\phi\in\wedge^3\mathfrak{g}$ is obtained from the symmetric part and from the structure constants of $\mathfrak{g}$.

By a theorem of Etingof and Kazhdan (building upon Drinfeld's results), any clqssical R-matrix can be extended to a quantum one.

edit: typo

share|cite|improve this answer
Thank you very much. – user Mar 25 '11 at 23:49
@Jianrong: It's nice that you're being polite, but the standard way of saying "thank you" on this site is to upvote the answer. (Click on the "up" arrow next to the answer.) – Hans Lundmark Mar 26 '11 at 9:08
@Hans, thank you very much. Now I know. – user Mar 26 '11 at 15:01
@Jianrong: You're welcome! :) – Hans Lundmark Mar 26 '11 at 17:32

Well, they are different equations...

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 $$R_{12} \ R_{13} \ R_{23} = R_{23} \ R_{13} \ R_{12},$$ where the notation is explained, for example, in Wikipedia.

share|cite|improve this answer
thank you very much. – user Mar 25 '11 at 23:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.