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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?

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

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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.

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    $\begingroup$ 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 $R = \tau c$) $\endgroup$
    – Jules Lamers
    Commented Dec 16, 2020 at 12:59
  • $\begingroup$ Is it absolutely necessary for $R$ to satisfy the YB equations? I mean can there be quasi-traingular Hopf algebras (or even quasi-triangular quasi-Hopf algebras) for which the R-matrix does not satisfy the (quasi-)YB equations? In the context of integrability, the satisfaction of the YB equations is connected with the integrability but if one does not care about integrability, can there still be an R-matrix? $\endgroup$
    – QGravity
    Commented Sep 5 at 5:16
  • $\begingroup$ Usually the definition of "$R$-matrix" is "solution of the YB equations". I suppose more generally an $R$-matrix is just a matrix describing the interaction in a physical system (like the $S$-matrix in QFT) but typically people mean "the ones leading to integrable systems/knot invariants" and for that you need the YB equations. $\endgroup$
    – Calvin McPhail-Snyder
    Commented Sep 10 at 12:50
  • $\begingroup$ Similarly, the definition of "quasi-triangular" is "has an element $R$ intertwining $\Delta$ and $\Delta^{\operatorname{op}}$ and satisfying the YB equations". $\endgroup$
    – Calvin McPhail-Snyder
    Commented Sep 10 at 12:54

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