By Shahn Majid

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**Extra resources for A Quantum Groups Primer**

**Sample text**

For the quasicocommutativity axiom 2 on DE, we need R(E ® g + 1® E) = (E ®1 + g,& E)R. But R9(1®E) = (g ®E)R9 by the gE = g2Eg relations and a change of variable in the sum for R,. Similarly for E ®1 since Rg is 38 6 Roots of unity. uq(sl2) symmetric. Hence (E®1+g®E)Rg =Rg(E®g-1+1®E). In view of this, we require eq 2 (B + C) = (C + D)eq 2, where A=(q-q-1)E®F,B=E®g,C=1®E,D=E®g-1. 2 in the next lecture for elements of any algebra obeying AB = g2BA, AD = q-2DA, [C, A] = B - D. In our case, these hold in the tensor product algebra via the algebra relations.

32 5 Quasitriangular structures Proof We write T R'") ® R'(2) as an explicit notation for a second copy of R, with the prime used to distinguish it from the first copy. Then Adh((S (Sid)(Q)) = h(1)(SQ('))Sh(2) ® h(3) Q(2) Sh(4) = h(,)(SR('))(SR' 2 )Sh(2)® h(3) R'«'R(2)Sh(4, h(1)(SR"))Sh(3)(S)Z/(2)) ®R'("h(2)R(2)Sh(4) = = h(1)Sh(Z)(SR('))(SR'(2)) ®R'(')R(2)h(3) Sh(4) = e(h)(S(9 id)(Q), using that S is an antialgebra map, and axiom 2 of a quasitriangular structure for the third and fourth equalities.

R(h ®1) = e(h) = R(1(9 h), dh E H. 2. The Yang-Baxter equation holds in the form E R(h(u 09(1))R(h(2) 0 fa>)R(g(2) ® = E R(g(1) f(2)) ®f(2))R(h(2) ®9(2)) for all h, g, f E H. 3. In the Hopf algebra case R(Sh ®g) = R-1(h ®9), R-1 (h ®Sg) = R(h ®g), R(Sh (9 Sg) = R(h ®g), Vh, g E H. Proof For part 2, expand R(h ® f (l)g(1))R(g(2) 0 f (2)) in two ways, either using the quasicommutativity axiom first and then the bicharacter axiom, or the bicharacter axiom directly. 6 Let (H, R) be a dual quasitriangular Hopf algebra.