Ivancevic_Applied-Diff-Geom

Applied Bundle Geometry 749good coordinate near that point, a D ≈ c 0 (u − u 0 ), with some constant c 0 .Using τ D = dh D /da D , we learn thata(u) = −h D (u) ≈ a 0 + i π a D ln a D ≈ a 0 + i π c 0(u − u 0 ) ln(u − u 0 ),for some constant a 0 = a(u = u 0 ). This constant a 0 cannot be zero becauseif it had been zero, all the electrically charged particles would have beenmassless at u = u 0 and the computation using light monopoles only wouldnot be valid.Now we can read off the monodromy. When u circles around u 0 ,soln(u − u 0 ) → ln(u − u 0 ) + 2πi,then one has a D → a D , a → a − 2a D . (4.240)This effect is a sort–of dual of the monodromy at infinity. Near infinity,the monopole gains electric charge, and near u = u 0 , the electron gainsmagnetic charge. (4.240) can be represented by the 2 × 2 monodromymatrix( ) 1 0M 1 = ST 2 S −1 = . (4.241)−2 14.14.7.4 The Third SingularityWith our assumption that there are only three singularities (counting u =∞) and with two of the three monodromies determined in (4.239) and(4.241), we can now determine the third monodromy, which we call M −1 .With all of the monodromies taken in the counter clockwise direction, themonodromies must obey M 1 M −1 = M ∞ , and from this we get( ) −1 2M −1 = (T S)T 2 (T S) −1 = .−2 3The matrix M −1 is conjugate to M 1 . Actually,( ) −1 1If A = T M 1 = , (4.242)−2 1Then M −1 = AM 1 A −1 . (4.243)Hence, M −1 can arise from a massless particle, just like M 1 . (4.243) wouldalso hold if Ais replaced by AM r 1 for any integer r.What kind of particle should become massless to generate this singularity?If one arranges the charges as a row vector q = (n m , n e ), then