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Quantum Group U_q (f (k)) to achieve equivalent

Author: PanYan
Tutor: LiLiBin
School: Yangzhou University
Course: Basic mathematics
Keywords: Quantum groups invertible operator equitable presentation
CLC: O152.5
Type: Master's thesis
Year: 2008
Downloads: 27
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Abstract


Quantum group developing in the mid of the eighties is one of the importa -nt branches of algebra. And the theory of quantum groups has been widely studied during the last twenty years. The aim of this dissertation is to study the quantum algebra Uq（ f （ k ）） which has a presentation with generators x±1 , y ,z and relations: We call x±1 , y ,z the equitable generator and define x = xm,then we prove that x±1 , x±1 , y ,z are invertible in finite dimensional Uq（ f （ k ））-modules. Then we display a linear operatorΩthat acts on finite dimensional Uq（ f （ k ））-module -s, and satisfiesΩ-1 xΩ= y ,Ω-1 yΩ= z ,Ω-1zΩ= x.At last, we obtain the explicit action ofΩon the simple Uq（ f （ k ））-module V （ n ,α）.Concretely, in the first part, we introduce the background of Uq（ g ）, and especially introduce the equitable presentation for Uq（ sl （2））. Moreover, we lead to the target of this dissertation: quantum group Uq（ f （ k ）） and its equitable presentation.In the second part, we collect some important results of quantum group Uq（ f （ k ））. The main results are following: the quantized enveloping algebra Uq（ f （ k ）） over C is generated by four generators k±1 , e , f associated with t -he relations kk -1 = k -1k = 1,ke = q2 ek , kf = q-2fk , ef -fe = f （ k）; Uq（ f （ k ）） ad -mits a Hopf algebra structure （lemma 2.2）; the center Z （Uq（ f （ k ））） generated by analog of the Casimir element C qm is a subalgebra of Uq（ f （ k ））, especially (Z（Uq（f（k））） = C[Cqm]; Uq（ f （ k ）） is a noetherian domain with a basis [eifjks]i、j∈N , s∈Z};each finite dimensional Uq（ f （ k ））-module is semi-simple; and so on.In the third part, we mainly discuss the equitable presentation for Uq（ f （ k ））.The main results are the following.Theorem 3.1 The algebra Uq（ f （ k ）） is isomorphic to the unital associativ -e C -algebra with generators x±1 , y ,z and the following relations:Definition 3.3 By the equitable presentation for Uq（ f （ k ）） we mean the presentation given in Theorem 3.1 We call x±1 , y ,z the equitable generators.In the fourth part, we introduce an infinite dimensional Uq（ f （ k ））-module in order to show that y and z are not invertible in Uq（ f （ k ））.The main result is following:Theorem 4.2 LetΓy be a Uq（ f （ k ））-module given in Lemma 4.1. Then the following （i）-（iii） hold（i） yu00 = 0, where the vector u00 is from Lemma 4.1;（ii） y is not invertible onΓy;（iii） y is not invertible in Uq（ f （ k ））.We also can obtain the similar result for z .In the fifth part, we show that y and z are invertible on each finite dimensional Uq（ f （ k ））-module. According to the theory of representations we only need to consider that y and z are invertible on each finite dimensional simple Uq（ f （ k ））-module. In this section we firstly introduce the notation of the simple Uq（ f （ k ））-module V （ n ,α）, where n∈N,αis the primitive 2m-th root of unity. Then we obtain the action of the equitable generators that act on V （ n,α）, and we lead to the main results:Theorem 5.4 Let V （ n ,α） be a finite dimensional simple Uq（ f （ k ））-mod -ule. The following （i）,（ii） hold（i） x is semi-simple with eigenvaluesαqn ,αqn-2,... ,αq-n; each of x , y ,z is semi-simple with eigenvaluesαm qmn ,αm qm （n-2）, ,αm q-mn;（ii） Each of x , x , y ,z is invertible.Theorem 5.5 On each finite dimensional Uq（ f （ k ））-module the actions of y and z are invertible and diagonalizable. Let y-1 (resp. z-1 ) denote the linear operator that acts on each finite dimensional Uq（ f （ k ））-module as the inverse of y （resp. z ）. In the sixth part, we define some elements n x , n y , n z of U q（ f （ k ）） and show that these are nilpotent on each finite dimensional U q（ f （ k ））-module. We then recall the q -exponential function expq m and derive a number of equ -ations involving expqm（nx）,expqm（ny）,expqm（nz） . Using these equations we will show that on finite dimensional U q（ f （ k ））-modules the operators y-1 , z-1 satisfy:In the last part, we display a linear operatorΩthat acts on finite dimensi -onal U q（ f （ k ））-module , and show that it satisfiesΩ-1 xΩ= y ,Ω-1yΩ= z,Ω-1zΩ= x.And then we get the explicit action ofΩon the module V （ n ,α）.Theorem 7.10 For an integer n≥0 andαlet u 0 , u1 , , u n denote the basis for V （ n ,α）. Then for 0≤j≤n,we have:

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