# 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 $$(R\otimes id)(id\otimes R)(R\otimes id) = (id\otimes R)(R\otimes id)(id\otimes R).$$ If instead $$R$$ satisfies $$R_{12}R_{13}R_{23} = R_{23}R_{13}R_{12}$$ we say that $$R$$ satisfies the quantum Yang--Baxter equation.

So what is the difference between the Yang--Baxter equation and the quantum Yang--Baxter equation? I guess that YBE came first and then came QYBE, but I don't see what is quantum about QYBE. What different properties do both have, and why should one consider them two versions of the same thing?

• this is answered in math.stackexchange.com/q/29054/87355 Dec 15, 2020 at 21:19
• Thanks for the link! However, I would like to keep the question open, in the hope that some other answers arise. Dec 15, 2020 at 21:32
• So in fact the question is not what the title suggests; 'quantum' is just often omitted Dec 16, 2020 at 12:55

Your two equations are equivalent, and are both versions of the quantum YBE. (The question from the comments does a good job of answering your classical versus quantum question.) Write the first as $$(c \otimes id)(id \otimes c)(c \otimes id) = (id \otimes c)(c \otimes id)(c \otimes id)$$ and the second as $$R_{12}R_{13}R_{23} = R_{23}R_{13}R_{12}.$$ Let $$\tau : V \otimes V \to V \otimes V$$ be the permutation map $$\tau(v_1 \otimes v_2) = v_2 \otimes v_1$$. Then a solution $$R$$ to the second gives a solution $$c = \tau R$$ to the first, and vice-versa. Whether or not to include the permutation in the definition of $$R$$-matrix is just a convention.
In the quantum groups literature, $$V$$ is a module for a quasitriangular Hopf algebra $$H$$ and $$R$$ can be given by the action of the universal $$R$$-matrix $$\mathcal{R} \in H \otimes H$$. Since $$\tau$$ will not be given by the action of something in $$H \otimes H$$, it's natural to leave it out. (I think the YBE for $$R$$ is also more natural in the spin chain context, but I know less about that.)
However, in quantum topology we focus more on $$c = \tau R$$ (because it gives braid group representations) and less on $$R$$, so some authors eliminate the distinction and call $$c$$ an $$R$$-matrix. I prefer to call $$c$$ a braiding and $$R$$ an $$R$$-matrix, but this convention is not universal.
• In quantum integrability we often denote your $c$ by $\check{R}$, call it (when speaking) the 'R-check matrix', and might say it obeys the 'braid-like form' of the YBE. (Let me also point out that $\bar{R} \coloneqq \tau R \tau = c \tau$ might differ from $R = \tau c$) Dec 16, 2020 at 12:59