Symmetric Functions And Infinite Dimensional Algebras [phd Thesis]

Symmetric Functions and In nite Dimensional Algebras Timothy Howard Baker, B.Sc. (Hons.) A thesis submitted in ful lment of the requirements for the degree of Doctor of Philosophy at the University of Tasmania. September, 1994 Declaration Except as stated herein this thesis contains no material which has been accepted for the award of any other degree or diploma in any University. To the best of my knowledge and belief, this thesis contains no material previously published or written by another person, except where due reference is made in the text of the thesis. Timothy H. Baker ii Acknowledgements Firstly, I'd like to thank my supervisor Dr. Peter Jarvis for putting up with me and providing continuous encouragement during the course of this work. I'd also like to thank the head of the theory group Professor Bob Delbourgo, for creating a great environment in which to work. Special thanks also go to Dr. Ming Yung for lengthy discussions on symmetric functions, and taking a continued interest in my work. Now to the story-tellers: Dr. Roland Warner whose breadth and depth of knowledge about subjects in and out of physics is astonishing Dr. Dong-Sheng Liu who has the ability to crack people up, any place, any time Dr. Ioannis Tsohantjis who can brighten up the gloomiest of days with his numeracy skills and his iridescent smile Dr. Dirk Kreimer, the cultural and professional beacon in my life, who has rekindled the deutschophile in me. I sincerely thank Julio Herrera Coronado and my friends at spanish classes for providing me with the social highlights of my week. 0 p+j +1 dpj = ((qt qq))j ((qqp+j t qq))j : (2.70) j j It turns out that by applying the formula (2.69) to the functions appearing in (2.67), we can reexpress them in terms of the functions Q(n m) (q t) in the form p 2 X Q(2p) (q t2) = tp;j ((tt qq))2j Q(p+j p;j)(q t) (2.71) 2j j =0 p 2 X Q(2p;1) (q t2) = tp;j ((tt qq))2j;1 Q(p+j;1 p;j)(q t): (2.72) 2j ;1 j =1 25 Let us prove this for (2.72). From (2.67) we write Q(2p;1) (q t2 ) = = = where p X (tp;k + tp;1+k ) Q(p+k;1)(q t)Q(p;k)(q t) k=1 p X 0p;k 1 X (tp;k + tp;1+k ) @ d2l k;1A Q(p+k;1+l p;k;l)(q t) k=1 p X j =1 l=0 tp;j fj Q(p+j;1 p;j)(q t) fj = jX ;1 (tn + t2j;1;n) d2nj;1;2n: n=0 Although it is not apparent, we can rewrite this as a terminating basic hypergeometric series of the type 2 1, which is de ned by 1 (a q )n (b q )n X ( a b c q z ) = zn: 2 1 ( q q ) ( c q ) n n n=0 Indeed, 2;2j ;2 (2.73) fj = 2 1(t q1;2j q2;2j t;1 q q) = t2j;1 ((qq2;2j tt;1 qq))2j;1 2j ;1 where we have used the q-Vandermonde summation formula ;n n (c=a q )n : (2.74) 2 1 (a q c q q ) = a (c q)n From (2.73) the result (2.72) follows, and the proof of (2.71) is similar. The question then arises as to whether the function Q (q tk ) can be expanded in the form X Q (q tk ) = a (q t k) Q (q t) (2.75) where the functions a (q t k) 2 Z(q t), the eld of quotients of integer-valued polynomials in q and t. For the case of j j = 3 we have explicitly k ; qtk )(1 ; q 2 tk ) Q(3) (q tk ) = (1(1;;t t)(1 )(1 ; qt)(1 ; q2t) Q(3) (q t) k k )(1 ; qtk ) (1 ; tk )(t ; tk )(t2 ; tk ) Q 3 (q t) + (1 ;(1t ;)(tt);2 (1t ; Q ( q t ) + (21) qt2 ) (1 ; t)(1 ; t2)(1 ; t3) (1 ) k 2 2 )(1 ; qt2k ) Q(21) (q tk ) = (1(1;;tt))2 (1(1;;qtq 2t)(1 ; q2tk ) Q(21) (q t) k 2 tk )(1 ; qt)(1 + t + qt + tk + qtk + qtk+1) Q 3 (q t) + (1 ; t ) (t ; (1 ) (1 ; t)(1 ; t2)(1 ; t3 )(1 ; q2tk ) k )(1 ; t2k )(1 ; t3k )(1 ; qt)(1 ; qt2 ) Q(13 )(q tk ) = (1(1;;tt)(1 ; t2)(1 ; t3 )(1 ; qtk )(1 ; qt2k ) Q(13 ) (q t): 26 The