Communications in Mathematical Physics - Volume 202 by A. Jaffe (Chief Editor)

By A. Jaffe (Chief Editor)

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Asaeda, U. Haagerup It is too hard to obtain the above gauge matrix by calculating all the elements by multiplication of matrices. Note that  00100   0 0 0 1 0   G   = 1 0 0 0 0    e e 0 1 0 0 0   00001   ˜ d u G 2,l  G  d  = u  G 2,l  1 G ∼  00100  0   1  0  0  0  0   = 1  0  0   =  u e 5 4,r e , 1   0 0 1 0  G  0 0 0 0  e 1 0 0 0  0001  0100   t d˜ 0 0 1 0   u G 2,l  0 0 0 0  u  1 0 0 0  0001 u   00100 d t G 2,l  0   1   0 1  0   0 0 1 0  G  0 0 0 0  1 0 0 0  0001 u e e t 5 4,r 1  d t G 2,l u  G   t d˜ G 2,l 1 u e t 5 4,r e , 1 t d d d = u G = u G .

Asaeda, U. Haagerup Table 8. ) 1 β2 1 β2 β 2 β−1 β 2 −2 β 2 −1 2β1 β 2 +1 dd˜ 1 ff fh f d˜ l1 l2 m1 m2 n1 n2 p1 p2 q1 q2 r1 r2 √ 3β 2 −1 2β−1 −1 2 −1 β−1 f b˜ hf hb˜ 1 ˜h ˜ h ˜ d˜ h ˜∗ h˜ ˜ dd ˜ df ˜ d˜h d˜d˜ ˜ bf ˜ bh ˜ ∗˜ h ge 1 √ df fd ce 1 1 s1 s2 β2 β−1 t1 t2 1 β 2 −1 u1 u2 β2 β−1 1 The broken edges correspond to the trivial connection 1. We will now compute √ the connection αα−1,which ˜ is determined only up to vertical gauges. As in the (5+ 13)/2case, we assume that 1 × 1 gauge transform unitaries corresponding to single vertical edges which connect different vertices in the graph G1 G1t ∪ G3 G3t are 1.

4, 4) and check that the square sums are equal to 1 respectively, and also calculate the (5,2)-entry is equal to 0. Then the (3,2)-entry is determined by using the fact that the square sum of the entries in the second column is equal to 1, and by the sign of the entry obtained by the numerical calculation of the product of matrices by Mathematica. Then we have (3,3) and (3,4)-entries by the orthogonal relation of the second column and the third and fourth. Now we know that the matrix is block diagonal, and we have the rest (5,5) and (5,6)-entries by unitarity and signs obtained by Mathematica.

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