By Shahn Majid
Here's a self-contained creation to quantum teams as algebraic gadgets. in line with the author's lecture notes for the half III natural arithmetic direction at Cambridge college, the ebook is acceptable as a prime textual content for graduate classes in quantum teams or supplementary analyzing for contemporary classes in complicated algebra. the cloth assumes wisdom of easy and linear algebra. a few familiarity with semisimple Lie algebras may even be useful. the quantity is a primer for mathematicians however it can be invaluable for mathematical physicists.
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Additional info for A Quantum Groups Primer
Proof We adopt the convention where repeated indices are summed. Then, to verify axiom 3 of a quasitriangular structure, we need (fa(1) ®1) ®(f a(2) ®1) ®(1® ea) _ (fa ®1) ®(fb ®1) ®(1® eaeb), (fa ®1) ®(1® ea(l)) 0(10 ea(2)) = (fbfa ®1) ®(1® eb) 0(10 ea), which are easily verified by evaluating against general elements: evaluating against 0 E H* in the third factor gives both sides of the first iden- tity as 0(1) 010 (2) ®1®1. Here fa(ea, 0) _ 0 and fa 0 f b(eaeb, 0) = 8 Quantum double. Dual quasitriangular structures 46 0(1) ®0(2) Similarly on evaluating against h E H for the second identity.
The model here is the usual twist or transposition map V ® W --W 0 V for vector spaces. If we suppress 4, then the hexagon conditions are T V ®W,Z ='PV,z o T W,Z, W V,W ®z = ,VVZ o T V,W 9 Braided categories V®(W®Z) MOT 55 (V®W)®Z AY®id V®(Z®W) (V®W)®Z I 41-1 j* (V®Z)®W AF Z®(Vow) V®(W®Z) 'P (W®®V)®Z I W®/(V®Z) (W®Z)®V / id ®qY (Z®V)®W W ®(Z®V) Fig. 2. Hexagon conditions for IQ. e. transposing V ® W past Z is the same as transposing W past Z and then V past Z, and transposing V past W ® Z is the same as first transposing V past W and then V past Z.
1 - qm 1-q where we suppose (for convenience) that the q-integers [m], are nonzero for 0 < m < n. By convention, they are defined as the usual integers when q = 1. Also by convention, [, ] q = 1 when m = 0, n. If A, B are elements of an algebra obeying BA = qAB then (A+B) = n ll m=0 AmBn-m [ mI Proof We proceed by induction. m+1Bn-1-m+ E rn - 11 AmBn-m I m q m=0 m=0 IL m J q E n E qn-m n-1 Ln-11 m-1 q m=0 39 [n- -11 AmBn-m m Jq 7 q-Binomials 40 =An+Bn+ n-1(qn_m n-1 [n_i] + [m-1]q m=1 m Am Bn-m q/ Th e expression in parentheses combines to [ m ] q, as required, after an elementary computation using the identity qn-m[m]q + [n - m]q = [n]q (the addition rule for q-integers).
A Quantum Groups Primer by Shahn Majid