Quasitriangular Hopf algebra

In mathematics, a Hopf algebra, H, is quasitriangular if there exists an invertible element, R, of $H\otimes H$ such that

• $R\ \Delta (x)=(T\circ \Delta )(x)\ R$ for all $x\in H$, where $\Delta$ is the coproduct on H, and the linear map $T:H\otimes H\to H\otimes H$ is given by $T(x\otimes y)=y\otimes x$,

• $(\Delta \otimes 1)(R)=R_{13}\ R_{23}$,

• $(1\otimes \Delta )(R)=R_{13}\ R_{12}$,

where $R_{12}=\phi _{12}(R)$, $R_{13}=\phi _{13}(R)$, and $R_{12}=\phi _{12}(R)$, where $\phi _{12}:H\otimes H\to H\otimes H$, $\phi _{13}:H\otimes H\to H\otimes H$, and $\phi _{23}:H\otimes H\to H\otimes H$, are algebra morphisns determined by

$\phi _{12}(a\otimes b)=a\otimes b\otimes 1,$

$\phi _{13}(a\otimes b)=a\otimes 1\otimes b,$

$\phi _{23}(a\otimes b)=1\otimes a\otimes b.$

R is called the R-matrix.

As a consequence of the properties of quasitriangularity, the R-matrix, R, is a solution of the Yang-Baxter equation (and so a module V of H can be used to determine quasi-invariants of braids, knots and links). Also as a consequence of the properties of quasitriangularity, $R^{-1}=(S\otimes 1)(R)$, and $R=(1\otimes S)(R^{-1})$, and so $(S\otimes S)(R)=R$.

It is possible to contruct a quasitriangular Hopf algebra from a Hopf algebra and its dual, using the Drinfrl'd quantum double construction.

Twisting

The property of being a quasi-triangular Hopf algebra is preserved by twisting via an invertible element $F=\sum _{i}f^{i}\otimes f_{i}\in {\mathcal {A\otimes A}}$ such that $(\varepsilon \otimes id)F=(id\otimes \varepsilon )F=1$ and satisfying the cocycle condition

$(F\otimes 1)\circ (\Delta \otimes id)F=(1\otimes F)\circ (id\otimes \Delta )F$

Furthermore, $u=\sum _{i}f^{i}S(f_{i})$ is invertible and the twisted antipode is given by $S'(a)=uS(a)u^{-1}$, with the twisted comultiplication, R-matrix and co-unit change according to those defined for the quasi-triangular Quasi-Hopf algebra. Such a twist is known as an admissible (or Drinfeld) twist.