Quantum Group U_q (f (k)) Is Equivalent To Achieve

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 U_q( 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 U_q( f ( k ))-modules. Then we display a linear operatorΩthat acts on finite dimensional U_q( 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 U_q( f ( k ))-module V ( n ,α).Concretely, in the first part, we introduce the background of U_q( g ), and especially introduce the equitable presentation for U_q( sl (2)). Moreover, we lead to the target of this dissertation: quantum group U_q( f ( k )) and its equitable presentation.In the second part, we collect some important results of quantum group U_q( f ( k )). The main results are following: the quantized enveloping algebra U_q( 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); U_q( f ( k )) ad -mits a Hopf algebra structure (lemma 2.2); the center Z (U_q( f ( k ))) generated by analog of the Casimir element C qm is a subalgebra of U_q( f ( k )), especially (Z(U_q(f(k))) = C[C_q^m]; U_q( f ( k )) is a noetherian domain with a basis [eⁱf^jk^s]_i,j∈N , s∈Z};each finite dimensional U_q( f ( k ))-module is semi-simple; and so on.In the third part, we mainly discuss the equitable presentation for U_q( f ( k )).The main results are the following.Theorem 3.1 The algebra U_q( 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 U_q( 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 U_q( f ( k ))-module in order to show that y and z are not invertible in U_q( f ( k )).The main result is following:Theorem 4.2 LetΓ_y be a U_q( 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 onΓ_y;(iii) y is not invertible in U_q( 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 U_q( 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 U_q( f ( k ))-module. In this section we firstly introduce the notation of the simple U_q( 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 U_q( f ( k ))-mod -ule. The following (i),(ii) hold(i) x is semi-simple with eigenvaluesαqⁿ ,αq^n-2,... ,αq^-n; each of x , y ,z is semi-simple with eigenvaluesα^m q^mn ,α^m q^{m (n-2)}, ,α^m q^-mn;(ii) Each of x , x , y ,z is invertible.Theorem 5.5 On each finite dimensional U_q( 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 U_q( 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 exp_q^m(n_x),exp_q^m(n_y),exp_q^m(n_z) . 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:...