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Lie Algebras of Generalized Quaternion Groups 1 Introduction 2

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Lie Algebras of Generalized Quaternion Groups 1 Introduction 2 Lie Algebras of Generalized Quaternion Groups Samantha Clapp Advisor: Dr. Brandon Samples Abstract Every finite group has an associated Lie algebra. Its Lie algebra can be viewed as a subspace of the group algebra with certain bracket conditions imposed on the elements. If one calculates the character table for a finite group, the structure of its associated Lie algebra can be described. In this work, we consider the family of generalized quaternion groups and describe its associated Lie algebra structure completely. 1 Introduction The Lie algebra of a group is a useful tool because it is a vector space where linear algebra is available. It is interesting to consider the Lie algebra structure associated to a specific group or family of groups. A Lie algebra is simple if its dimension is at least two and it only has f0g and itself as ideals. Some examples of simple algebras are the classical Lie algebras: sl(n), sp(n) and o(n) as well as the five exceptional finite dimensional simple Lie algebras. A direct sum of simple lie algebras is called a semi-simple Lie algebra. Therefore, it is also interesting to consider if the Lie algebra structure associated with a particular group is simple or semi-simple. In fact, the Lie algebra structure of a finite group is well known and given by a theorem of Cohen and Taylor [1]. In this theorem, they specifically describe the Lie algebra structure using character theory. That is, the associated Lie algebra structure of a finite group can be described if one calculates the character table for the finite group. In particular, the quaternions are a way to generalize complex numbers to higher dimensions. The quaternion group has many scientific applications; for example, aerospace and robotics engineers use them to model the positions and orientations of planes and robots. In this paper, we will begin with basic definitions and examples of Lie algebras of quaternions and then describe the Lie algebra structure of the generalized quaternion group completely. 2 Preliminaries In order to describe the Lie algebra structure associated with the generalized quaternion group, we must first consider the definition of a Lie algebra. Definition 2.1. Let L be a vector space over a field F . Then L is a Lie algebra if it possesses a bracket operation [ ; ]: L × L ! L that satisfies the following axioms: i. The bracket operation is bilinear. ii. [x; x] = 0 for all x 2 L. iii. [x; [y; z]] + [y; [x; z]] + [z; [x; y]] = 0 (x; y; z 2 L): 1 Note that (i.) and (ii.) imply anticommutativity, [x; y] = −[y; x], and (iii.) is called the Jacobi identity. With the definition of a Lie algebra, we can consider a few examples. The first is the set of all x1 3 3 vectors in R under the bracket operation of the cross product. Let x; y; z 2 R with x = x2 ; x3 y1 z1 x2y3 − x3y2 y = y2 , and z = z2 . The cross product is defined as the following x × y = x3y1 − x1y3 . y3 z3 x1y2 − x2y1 x2y3 − x3y2 3 Note that x and y are general vectors and x × y = x3y1 − x1y3 2 R ; therefore, the set of x1y2 − x2y1 vectors in R3 is closed under the cross product. Next we will consider the three axioms. i. First, let a; b 2 R. Then we have the calculation for bilinearity as follows, 2 3 2 3 (ax2 + by2)z3 − (ax3 + by3)z2 ax2z3 + by2z3 − ax3z2 − by3z2 [ax + by; z] = 4(ax3 + by3)z1 − (ax1 + by1)z35 = 4ax3z1 + by3z1 − ax1z3 − by1z35 (ax1 + by1)z2 − (ax2 + by2)z1 ax1z2 + by1z2 − ax2z1 − by2z1 2 3 a(x2z3 − x3z2) + b(y2z3 − y3z2) = 4a(x3z1 − x1z3) + b(y3z1 − y1z3)5 = a[x; z] + b[y; z]; a(x1z2 − x2z1) + b(y1z2 − y2z1) 2 3 2 3 x2(ay3 + bz3) − x3(ay2 + bz2) x2ay3 + x2bz3 − x3ay2 − x3bz2 and [x; ay + bz] = 4x3(ay1 + bz1) − x1(ay3 + bz3)5 = 4x3ay1 + x3bz1 − x1ay3 − x1bz35 x1(ay2 + bz2) − x2(ay1 + bz1) x1ay2 + x1bz2 − x2ay1 − x2bz1 2 3 (ax2y3 − x3y2) + b(x2z3 − x3z2) = 4a(x3y1 − x1y3) + b(x3z1 − x1z3)5 = a[x; y] + b[x; z]: a(x1y2 − x2y1) + b(x1z2 − x2z1) Therefore, the cross product satisfies bilinearity. ii. Next, consider an element bracketed with itself: 2 3 2 3 x2x3 − x3x2 0 [x; x] = 4x3x1 − x1x35 = 405 : x1x2 − x2x1 0 So we have that the second axiom is satisfied. iii. Consider the Jacobi identity: 2 2 33 2 2 33 2 2 33 y2z3 − y3z2 z2x3 − z3x2 x2y3 − x3y2 [x; [y; z]] + [y; [z; x]] + [z[x; y]] = 4x; 4y3z1 − y1z355 + 4y; 4z3x1 − z1x355 + 4z; 4x3y1 − x1y355 y1z2 − y2z1 z1x2 − z2x1 x1y2 − x2y1 2 3 2 3 x2(y2z3 − y3z2) − x3(y2z3 − y3z2) y2(z2x3 − z3x2) − y3(z2x3 − z3x2) = 4x3(y3z1 − y1z3) − x1(y3z1 − y1z3)5 + 4y3(z3x1 − z1x3) − y1(z3x1 − z1x3)5 x1(y1z2 − y2z1) − x2(y1z2 − y2z1) y1(z1x2 − z2x1) − y2(z1x2 − z2x1) 2 3 2 3 z2(x2y3 − x3y2) − z3(x2y3 − x3y2) 0 + 4z3(x3y1 − x1y3) − z1(x3y1 − x1y3)5 = 4 0 5 z1(x1y2 − x2y1) − z2(x1y2 − x2y1) 0: Consequently, the cross product satisfies the Jacobi identity. 2 Since we have shown that all of the properties for a Lie algebra are satisfied by the cross product, then we have that the vectors of R3 with the bracket operation of the cross product is indeed an example of a Lie algebra. Next, consider the set of all 2 × 2 matrices of trace zero with entries in C with the bracket operation being ab − ba is a special linear algebra known as sl(2). Consider the matrices a a b b c c a = 1 2 ; b = 1 2 ; and c = 1 2 , such that the entries are in . Note a; b; c 2 sl(2) a3 −a1 b3 −b1 c3 −c1 C and a a b b b b a a ab − ba = 1 2 1 2 − 1 2 1 2 a3 −a1 b3 −b1 b3 −b1 a3 −a1 a b + a b a b − a b b a + b a b a − b a = 1 1 2 3 1 2 2 3 − 1 1 2 3 1 2 2 1 a3b1 − a1b3 a3b2 + a1b1 b3a1 − b1a3 b3a2 + b1a1 a b + a b − b a − b a a b − a b − b a + b a = 1 1 2 3 1 1 2 3 1 2 2 3 1 2 2 1 2 sl(2): a3b1 − a1b3 − b3a1 + b1a3 a3b2 + a1b1 − b3a2 − b1a1 Thus, we see that sl(2) is closed under the bracket operation. Now consider the three axioms for a Lie algebra. i. First, let m; n 2 C. The bilinearity calculation is as follows, ma + nb ma + nb c c [ma + nb; c] = 1 1 2 2 ; 1 2 ma3 + nb3 −ma1 − nb1 c3 −c1 ma c + nb c + ma c + nb c ma c + nb c − ma c + nb c = 1 1 1 1 2 3 2 3 1 2 1 2 2 1 2 1 ma3c1 + nb3c1 + ma1c3 + nb1c3 ma3c2 + nb3c2 + ma1c1 + nb1c1 c ma + c nb + c ma + nc b c ma + c nb − c ma − c nb − 1 1 1 1 2 3 2 3 1 2 1 2 2 1 2 1 c3ma1 + c3nb1 − c1ma3 + c1nb3 c3ma2 + c3nb2 + c1ma1 + c1nb1 m(a c + a c ) + n(b c + b c ) m(a c − a c ) + n(b c + b c ) = 1 1 2 3 1 1 2 3 1 2 2 1 1 2 2 1 m(a3c1 + a1c3) + n(b3c1 + b1c3) m(a3c2 + a1c1) + n(b3c2 + b1c1) m(c a + c a ) + n(c b + c b ) m(c a − c a + n(c b 1 − c b ) − 1 1 2 3 1 1 2 3 1 2 2 ) 1 2 2 1 m(c3a1 − c1ma3) + n(c3b1 + c1b3) m(c3a2 + c1a1) + n(c3b2 + c1b1) = m[a; c] + n[b; c]; a a mb + nc mb + nc and [a; mb + nc] = 1 2 ; 1 1 2 2 a3 −a1 mb3 + nc3 −mb1 − nc1 a mb + a nc + a mb + na c a mb + a nc − a mb − a nc = 1 1 1 1 2 3 2 3 1 2 1 2 2 1 2 1 a3ma1 + a3nc1 − a1mb3 + a1nc3 a3mb2 + a3nc2 + a1mb1 + a1nc1 mb a + nc a + mb a + nc a mb a + nc a − mb a + nc a − 1 1 1 1 2 3 2 3 1 2 1 2 2 1 2 1 mb3a1 + nc3a1 + mb1a3 + nc1a3 mb3a2 + nc3a2 + mb1a1 + nc1a1 m(a b + a b ) + n(a c + a c ) m(a b − a b ) + n(a c − a c ) = 1 1 2 3 1 1 2 3 1 2 2 1 1 2 2 1 m(a3a1 − a1b3) + n(a3c1 + a1c3) m(a3b2 + a1b1) + n(a3c2 + a1c1) m(b a + b a ) + n(c a + c a ) m(b a − b a ) + n(c a + c a ) − 1 1 2 3 1 1 2 3 1 2 2 1 1 2 2 1 = m[a; b] + n[a; c]: m(b3a1 + b1a3) + n(c3a1 + c1a3) m(b3a2 + b1a1) + n(c3a2 + c1a1) ii. Next, consider an element bracketed with itself: a a a a a a a a 0 0 [a; a] = 1 2 1 2 − 1 2 1 2 = : a3 −a1 a3 −a1 a3 −a1 a3 −a1 0 0 3 iii. Consider the Jacobi identity calculation: b b c c c c b b [a; [b; c]] + [b; [c; a]] + [c; [a; b]] = a; 1 2 1 2 − 1 2 1 2 b3 −b1 c3 −c1 c3 −c1 b3 −b1 c c a a a a c c + b; 1 2 1 2 − 1 2 1 2 c3 −c1 a3 −a1 a3 −a1 c3 −c1 a a b b b b a a + c; 1 2 1 2 − 1 2 1 2 a3 −a1 b3 −b1 b3 −b1 a3 −a1 b c − c b 2b c − 2b c c a − a c 2c a − 2c a = a; 2 3 2 3 1 2 2 1 + b; 2 3 2 3 1 2 2 1 2b3c1 − 2b1c3 b3c2 − c3b2 2c3a1 − 2c1a3 c3a2 − a3c2 a b − b a 2a b − 2a b + c; 2 3 2 3 1 2 2 1 2a3b1 − 2a1b3 a3b2 − b3a2 a a b c − c b 2b c − 2b c b c − c b 2b c − 2b c a a = 1 2 2 3 2 3 1 2 2 1 − 2 3 2 3 1 2 2 1 1 2 a3 −a1 2b3c1 − 2b1c3 b3c2 − c3b2 2b3c1 − 2b1c3 b3c2 − c3b2 a3 −a1 b b c a − a c 2c a − 2c a c a − a c 2c a − 2c a b b + 1 2 2 3 2 3 1 2 2 1 − 2 3 2 3 1 2 2 1 1 2 b3 −b1 2c3a1 − 2c1a3 c3a2 − a3c2 2c3a1 − 2c1a3 c3a2 − a3c2 b3 −b1 c c a b − b a 2a b − 2a b c c a b − b a 2a b − 2a b + 1 2 2 3 2 3 1 2 2 1 − 1 2 2 3 2 3 1 2 2 1 c3 −c1 2a3b1 − 2a1b3 a3b2 − b3a2 c3 −c1 2a3b1 − 2a1b3 a3b2 − b3a2 2a b c − 2a b c − 2a b c + 2a b c 4a b c − 4a b c + 2a b c − 2a c b = 2 3 1 2 1 3 3 1 2 3 2 1 1 1 2 1 2 1 2 3 2 2 3 2 2a3b2c3 − 2a3c2b3 − 4a1b3c1 + 4a1b1c3 2a3b1c2 − 2a3b2c1 − 2a2b3c1 + 2a2b1c3 2b c a − 2b c a − 2b c a + 2b c a 4b c a − 4b c a + 2b c a − 2b a c + 2 3 1 2 1 3 3 1 2 3 2 1 1 1 2 1 2 1 2 3 2 2 3 2 2b3c2a3 − 2b3a2c3 − 4b1c3a1 + 4b1c1a3 2b3c1a2 − 2b3c2a1 − 2b2c3a1 + 2b2c1a3 2c a b − 2c a b − 2c a b + 2c a b 4c a b − 4c a b + 2c a b − 2c b a + 2 3 1 2 1 3 3 1 2 3 2 1 1 1 2 1 2 1 2 3 2 2 3 2 2c3a2b3 − 2c3b2a3 − 4c1a3b1 + 4c1a1b3 2c3a1b2 − 2c3a2b1 − 2c2a3b1 + 2c2a1b3 0 0 = : 0 0 Thus, we have shown that sl(2) satisfies all of the conditions for a Lie algebra.
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