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Open Journal of Applied & Theoretical Mathematics (OJATM) Vol. 2, No. 4, December 2016, pp. 583~586 ISSN: 2455-7102 UNIVERSAL COVERING GROUPS OF MATRIX LIE GROUPS N. V. RAMANA MURTY G. M. VICTOR EMMANUEL & CH. BADARI NARAYANA Dept. of Mathematics, Andhra Loyola College, Vijayawada – 520008 Dept. of Mathematics, The Hindu College, Machilipatnam [email protected] [email protected] [email protected] Article Info Keyword: Universal Covering group, Fundamental group, Matrix Lie group ABSTRACT The concept of Universal Cover of a topological space has a significant role in the subject of Algebraic topology. Therefore, this paper makes an attempt to study Universal Covering Groups of Matrix Lie Groups. Mainly, it has been concentrated on the Matrix groups SO 3 and U n . It is found that Universal Cover of the matrix group SO 3 is SU 2 and Universal cover of U n is R SU n . Throughout this paper, all Matrix Lie groups are considered to be connected, since Universal Cover exists only for connected Lie groups. The Universal Cover of a connected topological space X is a simply connected topological space X together with a continuous map P : X X satisfying the property that each x X has a neighborhood U such that P 1 U is a disjoint union of open sets V such that P maps each V homeomorphically onto U . The map P is known as projection map. In case of Lie groups G , the universal cover G is again Lie group with the projection map P : G G is a smooth homomorphism and group G is known as a universal covering group of G . It is also given a relation between a universal covering group and the Fundamental group of a matrix Lie group. Copyright © 2015 Open Journal of Applied & Theoretical Mathematics (OJATM) All rights reserved. Journal homepage: http://ojal.us/ojatm/ 583 Open Journal of Applied & Theoretical Mathematics (OJATM) Vol. 2, No. 4, December 2016, pp. 583~586 ISSN: 2455-7102 INTRODUCTION It is well known that the set of all n n invertible matrices over real numbers field or complex numbers field forms a group with respect to matrix multiplication, and they are denoted by GL n; and GL n; respectively. A sequence Am of complex matrices in the set of all n n matrices is said to be convergent to matrix A if each element of Am converges to the corresponding element of A , as m . Any subgroup G of GL n; is called a matrix Lie group if any sequence Am of matrices in G converges to some matrix A , then either A G, or A is not invertible. An n-rowed square matrix A is known as orthogonal if AAT I n AT A, where AT is the transpose of the matrix A , I n is the unit matrix of order n. It is clear that A 1, where A stands for determinant of the matrix A . It is easy to see that the set of all n n orthogonal matrices forms a group with respect to matrix multiplication, and it is denoted by O n . If we consider the set of all n n orthogonal matrices with determinant one only, it also forms a group with respect to matrix multiplication known as the Special orthogonal group, and it is denoted by SO n . It is obvious that SO n O n GL n; GL n; . An n2 rowed square matrix with complex entries A is known as unitary matrix if AA* I n A* A, where A* is the transposed conjugate of A. Therefore, determinant of a unitary matrix is unit modulus. The set of all n n unitary matrices forms a group with respect to matrix multiplication, and it is denoted by U n . The set of all unitary matrices with determinant one also forms a group with respect to the same operation, called Special unitary group, and it is denoted by SU n . It is clear that SU n U n GL n; . The Lie algebra of a matrix Lie group G is denoted by ɡ and it is defined as the set of all matrices A such that ekA is in G for all real numbers k . Main Results The importance of universal covering groups lies in obtaining Lie group homomorphism for a matrix Lie group G. If a matrix Lie group G is simply connected, then it is possible to obtain a Lie group homomorphism by exponentiation of its Lie algebra homomorphism. This is not possible if the matrix Lie group G is not simply connected. Therefore, it is necessary to find a group H such that it is simply connected and it has same Lie algebra as that of group G . Such a group is known as the universal covering group of G. Definition 1: The Universal covering group of a connected Lie group G is a simply connected Lie group X together with a Lie group homomorphism : X G such that the associated Lie algebra homomorphism : ɦ → ɡ is a Lie algebra isomorphism, where ɦ, ɡ are Lie algebras of the Lie groups X and G respectively. The homomorphism is called Projection map or Covering homomorphism. The Universal covering group of a group G is denoted by G, . Journal homepage: http://ojal.us/ojatm/ 584 Open Journal of Applied & Theoretical Mathematics (OJATM) Vol. 2, No. 4, December 2016, pp. 583~586 ISSN: 2455-7102 Theorem 2: For any connected Lie group, a universal covering group exists and it is unique up to isomorphism. Proof: See [1] Towards the relation of universal covering group and the Fundamental group we have Theorem 3: If G, is the universal covering group of a group G , then Ker 1 G , where 1 G is the Fundamental group of G. Proof: It is clear that the kernel of any homomorphism is a closed normal subgroup. By definition 1, is isomorphism, and hence kernel of must be discrete. Therefore, it is a subgroup of the centre of group G. In view of this, it is easy to establish an isomorphism between the kernel of and the Fundamental group 1 G of G. Theorem 4: If : G G is a universal covering map, then the automorphism group A G, is isomorphic to the fundamental group 1 G of G. Proof: See [1]. Universal covering group of SO 3 : It is known that the set of all 2x2 complex matrices which are self-adjoint and have trace zero forms a 3dimensional real vector space. Now, consider the Special unitary group SU 2 . We see that, for each X in SU 2 , we can define a linear map X on the above vector space such that X A XAX 1. 3 It is easy to see that the linear map is an orthogonal transformation on the vector space. Therefore, the map X X is a homomorphism from SU 2 into O 3 . We see that is continuous and hence it is a Lie group homomorphism. Since SU 2 is simply connected, is continuous, I I , which has the value of determinant is one, must map SU 2 into SO 3 . Therefore, the universal covering group of SO 3 is SU 2 and is the covering homomorphism. Also, kernel of is {I , I }. Therefore, the fundamental group 1 SO 3 of SO(3) is isomorphic to the group 2 , where 2 is the group of addition modulo 2. (by theorem 3) Universal covering group of U n : Since the Lie algebra of U n is the space of all n-rowed square complex matrices X such that X* X, the associated Lie algebra homomorphism of the covering homomorphism : SU n U n , defined by , X ei X , is an isomorphism, and the group is simply connected, it is easy to see that SU n is a universal cover of U n . Journal homepage: http://ojal.us/ojatm/ 585 SU n Open Journal of Applied & Theoretical Mathematics (OJATM) Vol. 2, No. 4, December 2016, pp. 583~586 ISSN: 2455-7102 In fact, it is tedious to compute directly the Universal Covering groups of matrix Lie groups. It is better to find them via fundamental groups. References: [1] Satya Deo, Algebraic Topology, Hindusthan Book Agency, New Delhi, 2003 [2] Allen Hatcher, Algebraic Topology, Cambridge University Press, Cambridge, 2003 [3] Joseph J. Rotman, An Introduction to Algebraic Topology, Springer, 2004 [4] David S. Dummit, Richard M. Foote, Abstract Algebra, John Wiley & Sons, New York, 1999 Journal homepage: http://ojal.us/ojatm/ 586