QUANTUM MECHANICS AND NON-ABELIAN THETA FUNCTIONS ...

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40RĂZVAN GELCA AND ALEJANDRO URIBEFor that we need to make sure that v k+1 and v k−1 are nonzero, and we alsoneed to understand the action of Y on them.Set Y v k+1 = αv k + v k+2 and Y v k−1 = βv k + v k−2 , where Xv k±2 =v k±2 . It might be possible that the scalars α and β are zero. Thevectors v k+2 , v k−2 might as well be zero; if they are not zero, then they areeigenvectors of X, and their respective eigenvalues are as specified becauseof the above argument.Using again the relations satisfied by the three generators of ˜RT t (T 2 ×[0,1]) we write2cos (k±2)πr2cos π r Y XY − (Y 2 X + XY 2 ) = 2cos 2π r X,which impliescos π rThis is equivalent to((k + 2)πcos + cos kπ r rthat is(k + 1)πcos α + cos π r rsin(k − 1)πcos β − cos kπ r r(α + β)= cos 2π r cos kπ r − cos kπ r .) ((k − 2)π(α − 1) + cos + cos kπ r r(k + 1)π(α − 1) + sinrFor further use, we write this as(k − 1)π(β − 1) = 0.r(t 4k+4 − 1)(α − 1) + (t 4k − t 4 )(β − 1) = 0.)(β − 1) = 0Reasoning similarly with the last of the three relations in ˜RT t (T 2 ×[0,1])we obtain the equality(t −6 + t 2 − 2t −2 (2k + 1)π)(α + β) + 4cos α + 4cos π r r α(2k − 1)π+4cos β + 4cos π r r β − (2k + 2)π2t2 cos α − 2t 2 αr−2t 2 (2k − 2)πcos β − 2t 2 β − 2t −2 cos 2kπrr α − 2t−2 cos 2kπr β−2t −2 α − 2t −2 β = 2(t 6 + t −6 − t 2 − t −2 ) − 2(t 6 + t −2 − 2t 2 )−(t 6 + t −2 − 2t 2 )(t 4k + t −4k ).
QUANTUM MECHANICS AND GENERALIZED THETA FUNCTIONS 41Using the fact that t = cos kπ2r+ isinkπ2rwe can transform this into(t −6 + t 2 − 2t −2 (2k + 1)π)((α − 1) + (β − 1)) + 4cos α + 4cos π r r α(2k − 1)π+4cos β + 4cos π (2k + 3)π (2k + 1)πβ − cos α − cos αr r rr(2k + 3)π (2k + 1)π−isin α + isin α − 2cos π rr r α − 2isin π r α(2k − 1)π (2k − 3)π (2k − 1)π (2k − 3)π− cos β − cos β − isin β − isin βrrrr−2cos π r β − 2isin π (2k − 1)π (2k + 1)πβ − cos α − cos αr rr(2k − 1)π (2k + 1)π (2k − 1)π (2k + 1)π−isin α + isin α − cos β − cos βrrrr(2k − 1)π (2k + 1)π−isin β + isin β − 2cos π rr r α + 2isin π r α − 2cos π r β+2isin π r β = −(t6 + t −2 − 2t 2 )(t 4k + t −4k ).After cancellations we obtain()(t −6 + t 2 − 2t −2 (2k + 3)π (2k + 3)π)((α − 1) + (β − 1)) − cos + isin αrr() ()(2k − 3)π (2k − 3)π (2k + 1)π (2k + 1)π− cos − isin β + 2 cos + isin αrrrr() ()(2k − 1)π (2k − 1)π (2k − 1)π (2k − 1)π+2 cos − isin β − cos + isin αrrrr()(2k + 1)π (2k + 1)π− cos − isin βrr= −t 4k+6 + t −4k+6 + t 4k−2 + t −4k−2 − 2t 4k+2 − 2t −4k+2 .This is the same as(t 4k+6 + t 4k−2 + 2t −2 − 2t 4k+2 − t −6 − t 2 )(α − 1)+(t −4k+6 + t −4k−2 + 2t −2 − 2t −4k+2 − t −6 − t 2 )(β − 1) = 0.Dividing through by t −6 + t 2 − 2t −2 we obtain(t 4k+4 − 1)(α − 1) + (t −4k+4 − 1)(β − 1) = 0.Combining this with the relation obtained before, we obtain the system oftwo equations(t 4k+4 − 1)u + (t 4k − t 4 )v = 0(t 4k+4 − 1)u + (t −4k+4 − 1)v = 0in the unknowns u = α − 1 and v = β − 1. Recall that t = e iπ2r .The coefficient of v in one of the equations is equal to zero if and only ifk = 1, in which case we are forced to have β = 0, because zero is not aneigenvalue of X. The coefficient of u in one of the equations is equal to zero
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