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I would like to know why it is interesting to define the quasi-triangular structure on a Hop algebra. I understand that the pseudo-co-commutative (the existence of an intertwining operator between the co-product and the opposite co-product) together with the quasi-triangular property implies the Yang-Baxter equation.

  • What does the quasi-triangular property means?

  • In physics I the YBE leads to the notion of integrable systems. However what role does in play in mathematics?

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The point is that not only quasi-triangular structures give rise to braidings on the monoidal category $(H\text{-Mod},-\otimes-)$, but that they are essentially the same thing: if you write out what is means to put a braiding on $(H\text{-Mod},\otimes)$ - a natural question, I hope? - you end up with a quasi-triangular structure:

Namely, suppose a natural family of isomorphisms $\{\text{br}_{X,Y}: X\otimes Y\to Y\otimes X\}_{X,Y\in H\text{-Mod}}$ is given, providing $(H\text{-Mod},\otimes)$ with a braiding. Then, let $X,Y\in H\text{-Mod}$ and $x\in X$, $y\in Y$ be arbitrary, and consider the morphisms $c_x: H\to X$ and $c_y: H\to Y$ given by $c_x(1) := x$ and $c_y(1) := y$. Then, by naturality of $\text{br}_{-,-}$, $$\text{br}_{X,Y}(x\otimes y) = \text{br}_{X,Y}((c_x\otimes c_y)(1\otimes 1)) = (c_y\otimes c_x)(\text{br}_{H,H}(1\otimes 1)),$$ so we see that $\text{br}_{-,-}$ is determined by $R := \text{br}_{H,H}(1\otimes 1) \in H\otimes H$. You can then check (let me know if you need details) that $\text{br}_{-,-}$ being a braiding translates into $R$ being a quasi-triangular structure on $H$.

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  • $\begingroup$ @Anne O'Nyme: Do you have any questions? $\endgroup$ – Hanno Apr 21 '15 at 4:41

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