Quasitriangular Hopf algebra

（重定向自Quasitriangular）

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

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

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, $(\epsilon \otimes 1) R = (1 \otimes \epsilon) R = 1 \in H$ ; moreover $R^{-1} = (S \otimes 1)(R)$ , $R = (1 \otimes S)(R^{-1})$ , and $(S \otimes S)(R) = R$ . One may further show that the antipode S must be a linear isomorphism, and thus S is an automorphism. In fact, S is given by conjugating by an invertible element: $S^2(x)= u x u^{-1}$ where $u := m (S \otimes 1)R^{21}$ (cf. Ribbon Hopf algebras).

单词	Quasitriangular
释义	Quasitriangular 英语百科 Quasitriangular Hopf algebra （重定向自Quasitriangular） In mathematics, a Hopf algebra, H, is quasitriangular if there exists an invertible element, R, of $H \otimes H$ such that where $R_{12} = \phi_{12}(R)$ , $R_{13} = \phi_{13}(R)$ , and $R_{23} = \phi_{23}(R)$ , where $\phi_{12} : H \otimes H \to H \otimes H \otimes H$ , $\phi_{13} : H \otimes H \to H \otimes H \otimes H$ , and $\phi_{23} : H \otimes H \to H \otimes H \otimes H$ , are algebra morphisms determined by 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, $(\epsilon \otimes 1) R = (1 \otimes \epsilon) R = 1 \in H$ ; moreover $R^{-1} = (S \otimes 1)(R)$ , $R = (1 \otimes S)(R^{-1})$ , and $(S \otimes S)(R) = R$ . One may further show that the antipode S must be a linear isomorphism, and thus S is an automorphism. In fact, S is given by conjugating by an invertible element: $S^2(x)= u x u^{-1}$ where $u := m (S \otimes 1)R^{21}$ (cf. Ribbon Hopf algebras).
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