Unital Associative Algebras over the Field R and How They Relate to the Groups SU(2) and Spin(3).

Transcription

1 Unital Associative Algebras over the Field R and How They Relate to the Groups SU(2) and Spin(3). Luke Agosta Supervised by Grant Cairns and Yuri Nikolayevsky La Trobe University Vacation Research Scholarships are funded jointly by the Department of Education and Training and the Australian Mathematical Sciences Institute.

2 Abstract There are two groups, the special unitary group of degree 2 SU(2) and the group Spin(3), which are said to be isomorphic to one another. In this report, this isomorphism is shown in detail via the two groups mutual isomorphism to the group of quaternions with norm equal to 1, H 1. To explain what Spin(3) is requires some background theory of the real Clifford algebras which are unital associative algebras over the field R, so some isomophisms between such algebras are also explored. Acknowledgements Thanks to my supervisors Grant Cairns and Yuri Nikolayevsky for all their meticulous feedback. They helped me to spot the many flaws with my work along the way while keeping me encouraged to keep at it. Also thanks to my comrade-in-arms Jonathan Baldwin who I shared my stuggles and many lunches with in an otherwise isolating period. Dedications I dedicate this report to the beautiful summer days I spent indoors instead of outside while labouring to get it all done on time. 2

3 Contents 1 Unital Associative Algebras over the Field R Definition The Algebra of the Quaternions H The Algebra of the Set Q Real Clifford Algebras Cl r,s (R) Introduction Even Sub-algebras Cl even r,s (R) Groups Definition The group of the Quaternions with Norm Equal to 1 H The Special Unitary Group of Degree 2 SU(2) The Group Spin(3) Connecting the Algebras and Groups H Q and H 1 SU(2) Cl even 3,0 (R) H and Spin(3) H References 34 5 Appendices 35 3

4 1 Unital Associative Algebras over the Field R 1.1 Definition A unital associative algebra over the field R is a set A equipped with three operations addition, scalar multiplication, and ring multiplication which satisfy the properties below. Note, elements of the field R are referred to as scalars. (A0) Closure under addition x, y A : x + y A. (A1) Associativity of addition x, y, z A : x + (y + z) (x + y) + z. (A2) Identity element of addition 0 A : x A : x + 0 x 0 + x. (A3) Inverse elements of addition x A : ( x) A : x + ( x) 0. (A4) Commutitivity of addition x, y A : x + y y + x. (S0) Closure under scalar multiplication a R : x V : ax V. (S1) Distributivity of scalar multiplication with respect to addition a R : x, y A : a(x + y) ax + ay. (S2) Distributivity of scalar multiplication with respect to field addition a, b R : x A : (a + b)x ax + bx. (S3) Compatibility of scalar multiplication with field multiplication a, b R : x A : a(bx) (ab)x. (S4) Identity element of scalar multiplication 1 R : x A : (1)x x. 4

5 (R0) Closure under ring multiplication x, y A : x y A. (R1) Associativity of ring multiplication x, y, z A : x (y z) (x y) z. (R2) Identity element of ring multiplication id A A : x A : x id A x id A x. (R3) Distributivity of ring multiplication with respect to addition x, y, z A : x (y + z) x y + x z (x + y) z x z + y z. (C) Compatibility of scalar multiplication with ring multiplication a, b R : x, y A : (ax) (by) (ab)(x y). All the (A ) and (S ) excluding (A0) and (S0) are the axioms of a vector space over the field R. Any set spanned by a basis over R and is equipped with scalar multiplication and addition automatically fulfills said axioms. Addition and scalar multiplication closure properties (A0) and (S0) are the result of being a vector space over R. All the (A ) and (R ) are the axioms of a unital ring. The last property (C) is necessary for the vector space and ring parts of this type of algebra to be fully compatible with each other, hence a unital associative algebra over the field R can be described as a vector space over the field R with a compatible unital ring structure. From this point onwards, for brevity, any unital associative algebra over the field R will be referred to as simply an algebra even though there exist other kinds of algebras which are not discussed ahead. 1.2 The Algebra of the Quaternions H The quaternions are a number system whose set H is spanned by the basis {1, i, j, k} over R, i.e. H {a1 + bi + cj + dk : a, b, c, d R} {a + bi + cj + dk : a, b, c, d R}. Multiplication between elements of H is called quaternion multiplication. It is compatible with scalar multiplication and distributes over addition. Additionally it 5

6 has a product structure for the basis given by this table 1 i j k 1 1 i j k i i 1 k j j j k 1 i k k j i 1. As an example Proposition The quaternion set (i + 2j)(3k) 3ik + 6jk 3j + 6i. H {a + bi + cj + dk : a, b, c, d R} forms an algebra when equipped with addition, scalar multiplication, and quaternion multiplication. Proof. To prove this is to show that H under addition, scalar multiplication, and quaternion multiplication (which acts as ring multiplication) exhibits all the properties from (A0) to (C) from section 1.1. H being the span of the basis {1, i, j, k} over R means that under addition and scalar multiplication, it is a vector space over R. Therefore the properties from (A0) to (S4) do not require proof. Additionally, since quaternion multiplication is compatible with scalar multiplication and distributes over addition, properties (C) and (R3) do not require proof. This leaves properties (R0), (R1), and (R2) to prove. (R0) Subproof. Let h 1, h 2 H such that h 1. a 1 + b 1 i + c 1 j + d 1 k, h 2. a 2 + b 2 i + c 2 j + d 2 k. Then there is the following fact from part of the proof of lemma h 1 h 2 (a 1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 )1 + (a 1 b 2 + b 1 a 2 + c 1 d 2 d 1 c 2 )i + (a 1 c 2 b 1 d 2 + c 1 a 2 + d 1 b 2 )j + (a 1 d 2 + b 1 c 2 c 1 b 2 + d 1 a 2 )k. 6

7 Since the set of real numbers are closed under addition and multiplication, the coefficients of the basis elements 1, i, j, k of h 1 h 2 are all real numbers, which means h 1 h 2 H. (R1) Subproof. Arbitrary quaternions expand to enormous lengths under multiplication if using the usual notation, so an alternate notation is employed for this proof. Let i 0. 1, i 1. i, i 2. j, i 3. k. Also let o, p, q H such that o. 3 [o r i r ], p. r0 3 [p s i s ], q. s0 3 [q t i t ], t0 hence z, o z, p z, q z R. Then o(pq) ( 3 3 ) 3 [o r i r ] [p s i s ] [q t i t ] r0 s0 t0 ( 3 3 [ 3 ]) [o r i r ] [p s i s ]q t i t r0 3 [o r i r ] r0 t0 s0 [ 3 3 ] [(p s i s )(q t i t )] t0 s0 [ 3 3 [ 3 ]] [o r i r ](p s i s )(q t i t ) t0 s0 r0 (left distributivity (R3)) (right distributivity (R3)) (left distributivity (R3)) [ 3 3 [ 3 ]] [(o r i r )(p s i s )(q t i t )]. (right distributivity (R3)) t0 s0 r0 7

8 Now from the other direction: (op)q ( 3 ) 3 3 [o r i r ] [p s i s ] [q t i t ] r0 s0 ( 3 [ (o r i r ) r0 t0 ]) 3 3 [p s i s ] [q t i t ] (right distributivity (R3)) s0 t0 [ 3 3 ] 3 [(o r i r )(p s i s )] [q t i t ] (left distributivity (R3)) r0 s0 [ 3 3 [ (o r i r )(p s i s ) r0 s0 t0 ]] 3 [q t i t ] t0 (right distributivity (R3)) [ 3 3 [ 3 ]] [(o r i r )(p s i s )(q t i t )]. (left distributivity (R3)) r0 s0 t0 The order of elements under addition affects nothing due to the commutativity of addition (A4), hence the order of the summations is unimportant. So o(pq) [ 3 3 [ 3 ]] [(o r i r )(p s i s )(q t i t )] t0 s0 r0 [ 3 3 [ 3 ]] [(o r i r )(p s i s )(q t i t )] r0 s0 t0 (op)q. Therefore quaternion multiplication is associative. (commutativity (A4)) 8

9 (R2) Subproof. Let h. a + bi + cj + dk H. Then h(1) (a1 + bi + cj + dk)(1) a(1(1)) + b(i(1)) + c(j(1)) + d(k(1)) (right distributivity (R3), compatibility (C)) a1 + bi + cj + dk h (basis product structure) a((1)1) + b((1)i) + c((1)j) + d((1)k) (basis product structure) (1)(a1 + bi + cj + dk) (left distributivity (R3), compatibility (C)) (1)h. Thus the basis element 1 H is the identity element for all elements of H under quaternion multiplication. All properties from (A0) to (C) hold true. Therefore H is an algebra under addition, scalar multiplication, and quaternion multiplication. 1.3 The Algebra of the Set Q The conjugate of an arbitrary complex number z a + bi C is defined as z. a bi C. Let Q denote the set spanned by the following basis over R which is comprised of the 2-by-2 identity matrix and a variation of the Pauli spin matrices [1, Notation and Conventions]: {( ) 1 0, 0 1 ( ) i 0, 0 i ( ) 0 1, 1 0 ( )} 0 i. (i. 1) i 0 9

10 i.e. { ( ) ( ) ( ) ( ) } 1 0 i i Q a + b + c + d : a, b, c, d R i 1 0 i 0 {( ) ( ) ( ) ( ) } a 0 bi 0 0 c 0 di : a, b, c, d R 0 a 0 bi c 0 di 0 (by matrix scalar product [5]) {( ) } a + bi c + di : a, b, c, d R (by matrix entrywise addition [6]) c + di a bi {( ) } α β β : α, β C. (α. a + bi and β. c + di) ᾱ Proposition The set { ( ) ( ) ( ) ( ) 1 0 i i Q a + b + c + d i 1 0 i 0 {( ) α β β : α, β C ᾱ } : a, b, c, d R }. (α. a + bi and β. c + di) forms an algebra when equipped with matrix addition, scalar multiplication, and the conventional matrix multiplication. Proof. To prove this is to show that Q under matrix addition, scalar multiplication, and matrix multiplication (which acts as ring multiplication) exhibits all the properties from (A0) to (C) from section 1.1. Q being the span of the basis {( ) ( ) 1 0 i 0,, i ( ) 0 1, 1 0 ( )} 0 i i 0 (i. 1) over R means that under matrix addition and scalar multiplication, it is a vector space over R. Therefore the properties from (A0) to (S4) do not require proof. Matrix multiplication is well documented to be associative [7], distributive over matrix addition [8], and compatible with scalar multiplication due to the matrix scalar product [5]. Hence the properties (R1), (R3), and (C) do not require proof. This leaves us with properties (R0) and (R2) to prove. 10

11 (R0) Subproof. Let A, B Q such that ( ) ( ) a b c d A., B. b ā d. c Then AB ( a )( b c ) d b ā d c ( ac b d ) ad + b c ā d bc ā c bd ( α ) β β ᾱ (α. ac b d and β. ad + b c) The set of complex numbers is closed under addition and multiplication, so α, β C. Hence AB Q, which means Q is closed under matrix multiplication. (R2) Subproof. Let ( ) a b A. Q, I b ā 2. Then ( )( ) a b 1 0 b ā 0 1 ( ) a b b ā ( ( ) )( ) a b. b ā Thus the basis element I 2 H acts as an identity element for all elements of Q under matrix multiplication. However, I 2 Q still needs to be checked: ( ) 1 0 I doesn t clearly show that I 2 Q, so α and β are defined in terms of the top row entries of I 2. α. 1 C and β. 0 C imply 0 β and 1 ᾱ, hence ( ) α β I 2 β : α, β C I ᾱ 2 Q. Therefore I 2 is the identity element of Q under matrix multiplication. All properties from (A0) to (C) hold true. Therefore Q is an algebra under matrix addition, scalar multiplication, and matrix multiplication. 11

12 1.4 Real Clifford Algebras Cl r,s (R) Introduction The set of an arbitrary real Clifford algebra is denoted by Cl r,s (R). The r, s is known as the signature, which takes the form of an ordered pair of non-negative integers. Each unique signature is associated with a unique real Clifford algebra, e.g. Cl 1,2 (R) and Cl 2,1 (R) are unique. Let B r,s denote a basis over R which spans the real Clifford algebra with signature r, s. Then from [2, 2.1]: B 0,0 {1}, and for signatures other than 0, 0, ( r+s ) B r,s {1} {e h1 e h2... e hn : 1 h 1 < h 2 <... < h n r + s} n1 where h 1, h 2,..., h n N. The symbol used above is the symbol this report uses to denote a Clifford product operator which is the ring multiplication of real Clifford algebras. More on that later. For now, a non-equational approach to constructing the basis B r,s : Step 1. The multiplicative identity 1 is always a basis element of any real Clifford algebra, e.g. B 1,2 1. Step 2. r+s is equal to the number of e s which are basis elements, e.g. B 1,2 e 1, e 2, e 3. Step 3. All Clifford product combinations of two or more differing e s form the remaining basis elements, e.g. B 1,2 e 1 e 2, e 2 e 3, e 1 e 3, e 1 e 2 e 3. Thus B 1,2 {1, e 1, e 2, e 3, e 1 e 2, e 2 e 3, e 1 e 3, e 1 e 2 e 3 }. The Clifford product is compatible with scalar multiplication and distributes over addition. Additionally, seen in [2, 2.1], it has a product structure for the basis of the arbitrary real Clifford algebra Cl r,s (R) as follows. e p 1 e p 1 e p e p e p 1 e p e p 1 e p e q e q e p where p {1, 2,..., r + s}, where p {1, 2,..., r}, where p {r + 1, r + 2,..., r + s}, where p q and p, q {1, 2,..., r + s}. 12

13 It is common for the Clifford product operator to be ommited in practice, thus concatonation of multiplying terms is understood to be the Clifford product in appropriate places, e.g. e p1 e p2... e pn e p1 e p2... e pn. As a shorthand, multiple e s under the Clifford product may be contracted into one e which concatonates all of the subscripts into one, e.g. e p1 e p2... e pn. e p1 p 2...p n. In summary, e p1 e p2... e pn e p1 e p2... e pn. e p1 p 2...p n. Here is a thorough look at the Clifford product in action. Working within the algebra Cl 1,2 (R), (1 + 2e 2 ) (3e 1 + 4e 2 ) (1 + 2e 2 ) 3e 1 + (1 + 2e 2 ) 4e 2 (left distributivity (R3)) 1 3e 1 + 2e 2 3e e 2 + 2e 2 4e 2 (right distributivity (R3)) 3(1 e 1 ) + 6(e 2 e 1 ) + 4(1 e 2 ) + 8(e 2 e 2 ) (compatibility (C)) 3e 1 + 6(e 2 e 1 ) + 4e 2 + 8(e 2 e 2 ) (basis product structure) 3e 1 + 6e e 2 + 8e 22 (shorthand notation) 3e 1 6e e 2 8(1) (basis product structure) 3e 1 6e e Even Sub-algebras Cl even r,s (R) The grade of an element of a basis set which spans a real Clifford algebra is equal to the minimum amount of e s which can constitute that element. As an example, the elements of the basis constructed earlier for the real Clifford algebra Cl 1,2 (R) have been labelled with their grades, B 1,2 {1, e 1, e 2, e 3, e 1 e 2, e 2 e 3, e 1 e 3, e 1 e 2 e 3 }. Grade: Keep in mind that use of the shorthand form for the Clifford product doesn t change the grade of any elements: B 1,2 {1, e 1, e 2, e 3, e 12, e 23, e 13, e 123 }. Grade:

14 Let Cl even r,s (R) be the set spanned by only the even elements of B r,s over R [3, 1.2.2]. e.g. Cl 2,0 (R) {a + be 1 + ce 2 + de 12 : a, b, c, d R}, Cl even 2,0 (R) {a + de 12 : a, d R}. (R) is known as the even part of Cl r,s (R). An arbitrary Cl even (R) is a sub-algebra of Cl r,s (R) [4, 1]. Thus far there has been no formal proof that the real Clifford algebras are actually Then Cl even r,s r,s (R) is a subset of Cl r,s (R), contains the additive identity 0, contains the multiplicative identity 1, and is closed under the operations of real Clifford algebras. Hence Cl even r,s algebras and no such proof will be given. Despite that, a proof of the special case Cl even 3,0 (R) is given below. Proposition The even part of the real Clifford algebra with signature 3, 0 Cl even 3,0 (R) {a + be 12 + ce 23 + de 13 : a, b, c, d R} forms an algebra when equipped with addition, scalar multiplication, and the Clifford product. Proof. To prove this is to show that Cl even 3,0 (R) under addition, scalar multiplication, and the Clifford product (which acts as ring multiplication) exhibits all the properties from (A0) to (C) from section 1.1. Cl even 3,0 (R) is the span of the basis {1, e 12, e 23, e 13 } over R, thus under addition and scalar multiplication, it is a vector space over R. Therefore the properties from (A0) to (S4) do not require proof. Additionally, since the Clifford product is compatible with scalar multiplication and distributes over addition, properties (C) and (R3) do not require proof. This leaves properties (R0), (R1), and (R2) to prove. The basis product structure rules for real Clifford algebras applied to the basis of Cl even 3,0 (R) can be summarised by this table 1 e 12 e 23 e e 12 e 23 e 13 e 12 e 12 1 e 13 e 23 e 23 e 23 e 13 1 e 12 e 13 e 13 e 23 e

15 which is identical in form to the quaternion basis product structure seen earlier 1 i j k 1 1 i j k i i 1 k j j j k 1 i k k j i 1. Thus the exact same proofs of (R0), (R1), and (R2) from proposition can be used for (R0), (R1), and (R2) here after making the following substitutions: H Cl even 3,0 (R), i e 12, j e 23, k e 13, quaternion multiplication the Clifford product. So all properties from (A0) to (C) hold true. Therefore Cl even 3,0 (R) is an algebra under addition, scalar multiplication, and the Clifford product. 15

16 2 Groups 2.1 Definition A group is a set G equipped with one binary operation which satisfies the axioms below. (G0) Closure (G1) Associativity x, y G : x y G. x, y, z G : x (y z) (x y) z. (G2) Identity element id G G : x G : x id G x id G x. (G3) Inverses x G : y G : x y id G y x. 2.2 The group of the Quaternions with Norm Equal to 1 H 1 Recall the set of quaternions from section 1.2 is H {a + bi + cj + dk : a, b, c, d R}. The conjugate of an arbitrary quaternion h a + bi + cj + dk is defined as h. a bi cj dk. The norm of an arbitrary quaternion h a + bi + cj + dk is defined as h. a 2 + b 2 + c 2 + d 2 h h hh. (by lemma 5.0.1) Denote by H 1 the subset of H containing only all of the elements with norm equal to 1. i.e. H 1 {n a + bi + cj + dk : a, b, c, d R; n 1} {n H : n 1}. 16

17 Proposition The set H 1 {n a + bi + cj + dk : a, b, c, d R; n 1} {n H : n 1}. forms a group under the operation quaternion multiplication. Proof. To prove this is to prove that H 1 under quaternion multiplication fulfills the group axioms from section 2.1. Whether an arbitrary element x is in H 1 can be checked by checking two properties: one is that x H, which is equivalent to x a + bi + cj + dk : a, b, c, d R, while the other is that x has a norm equal to 1, i.e. x 1. (G2) Subproof. From the proof of (R2) for H, 1 is known to function as the quaternion multiplication identity and is an element of H. Now all that needs to be done is check its norm: i + 0j + 0k So its norm is 1, hence 1 H 1. Therefore 1 is the identity element for the set H 1 under quaternion multiplication. (G3) Subproof. Let n. a + bi + cj + dk H 1. Then n n n 2 nn (by lemma 5.0.1) n n 1 nn, (since n H 1 ) hence n n 1. i.e. n functions as the inverse to n H 1 under quaternion multiplication. Now all that is left to do is show that the inverse n is an element of H 1. As shown below, n takes the form of an element of H: Lastly, the check for norm: n a bi cj dk : a, ( b), ( c), ( d) R. n a 2 + ( b) 2 + ( c) 2 + ( d) 2 a 2 + b 2 + c 2 + d 2 n 1. (since n H 1 ) 17

18 So the norm of n is 1. Therefore each element of H 1 has an inverse also contained within H 1. (G1) Subproof. This has already been proven in proof of (R1) for H. (G0) Subproof. Let n 1, n 2 H 1 H such that n 1. a 1 + b 1 i + c 1 j + d 1 k, n 2. a 2 + b 2 i + c 2 j + d 2 k. From the proof of (R0) for H it is known that n 1 n 2 H, so the only thing left to do is check the norm: n 1 n 2 (n 1 n 2 )(n 1 n 2 ) (n 1 n 2 )( n 2 n 1 ) (by lemma 5.0.2) n 1 (n 2 n 2 ) n 1 (by associativity (G1)) n 1 (1) n 1 (by inverses (G3)) n 1 n 1 (by identity (G2)) 1 (by inverses (G3)) 1. So the norm is 1. Therefore H 1 is closed under quaternion multiplication. (G0), (G1), (G2), and (G3) all hold true, therefore H 1 is a group under quaternion multiplication. 2.3 The Special Unitary Group of Degree 2 SU(2) Let U be an arbitrary n-by-n matrix whose entries all consist of complex numbers. Then the conjugate transpose of U, denoted by U, is obtained by transposing U and replacing all of its entries with their conjugates. i.e. let a, b, c, d C, then ( ) (ā ) a b c U U. c d b d 18

19 If the inverse of an n-by-n complex matrix U under matrix multiplication is its conjugate transpose U, then U and U are said to be a unitary matrices. i.e. let U be an n-by-n complex matrix and I n be the n-by-n identity matrix, then UU I n U U U and U are unitary matrices. The set of all 2-by-2 unitary matrices with determinant equal to 1 is known to form a group under matrix multiplication called the special unitary group of degree 2. Its set is denoted by SU(2) { S ( ) a b c d } : a, b, c, d C; S 1 S ; det(s) 1. This set can notationally be simplified: Let a, b, c, d C, then ( ) a b S c d ( ) a b S c d ( ) a b S c d ( ) a b S c d & & & & S SU(2) S 1 S ( ) (ā ) 1 d b c det(s) c a b d ( ) (ā ) d b c c a b d d ā & c b ( ) a b S b ā & det(s) 1 & det(s) 1 & det(s) 1 & det(s) 1 & det(s) 1. Thus SU(2) { S ( α ) β β ᾱ Recall from section 1.3 the set {( ) α β Q β ᾱ } : α, β C; det(s) 1. } : α, β C. The only difference from the set Q to the set SU(2) is the additional requirement that all elements must have a determinant equal to 1. i.e. SU(2) {S Q : det(s) 1}. 19

20 Proposition The set { ( ) α β SU(2) S β ᾱ {S Q : det(s) 1} } : α, β C; det(s) 1 forms a group under the operation matrix multiplication. Proof. To prove this is to prove that SU(2) under matrix multiplication fulfills the group axioms from section 2.1. Whether an arbitrary element X is in SU(2) can be checked by checking two properties: one is that X Q, which is equivalent to ( ) α β X β : α, β C, ᾱ while the other is that X has a determinant equal to 1, i.e. det(x) 1. (G2) Subproof. ( ) 1 0 From the proof of (R2) for Q, I 2 is known to function as the matrix 0 1 multiplication identity for Q elements and is itself an element of Q. Now all that needs to be done is check its determinant: ( ) 1 0 det(i 2 ) det So its determinant is 1, hence I 2 SU(2). Therefore I 2 is the identity element for the set SU(2) under matrix multiplication. (G3) Subproof. ( ) (ᾱ ) α β β Let S. β SU(2) Q, hence S ᾱ Q. By definition, all β α elements of SU(2) are unitary matrices, each whose inverse is its conjugate transpose, another unitary matrix. Thus to prove that each element of SU(2) has an inverse within SU(2), only the conjugate transpose S being an element of SU(2) must be 20

21 checked. Since S Q, only the check for the determinant is needed: det(s ) (ᾱ ) β det β α āa + bb ( ) α β det β ᾱ det(s) 1. (since S SU(2)) So S SU(2). Therefore each element of SU(2) has an inverse under matrix multiplication within SU(2). (G1) Subproof. Matrix multiplication is well documented [7] to be associative, so it is not shown here. (G0) Proof of closure. Let A, B SU(2) Q, then from the proof of (R0) for Q it is known that AB Q, so the only thing left to do is check the determinant: det(ab) det(a) det(b) (Determinant of Matrix Product [9]) (1)(1) (since A, B SU(n)) 1. So the determinant is 1. Therefore SU(2) is closed under matrix multiplication. (G0), (G1), (G2), and (G3) all hold true, therefore SU(2) is a group under matrix multiplication. 21

22 2.4 The Group Spin(3) Recall the set of the even part of the real Clifford algebra with signature 3, 0 is Cl even 3,0 (R) {a + be 12 + ce 23 + de 13 : a, b, c, d R}. The conjugate and the norm for the elements of this set are defined the same way as done with the quaternions. i.e. let γ a + be 12 + ce 23 + de 13 be an arbitrary element of Cl even 3,0 (R), then the conjugate is defined as while it s norm is defined as γ. a be 12 ce 23 de 13, γ. a 2 + b 2 + c 2 + d 2. The subset of Cl even 3,0 (R) containing only all of the elements with norm equal to 1 [3, 3.3.1] is known as Spin(3) {s a + be 12 + ce 23 + de 13 : a, b, c, d R; s 1} {s Cl even 3,0 (R) : s 1}. To prove that Spin(3) is a group is feasible but there is no need for it here since proposition can be reused verbatim after making some substitions: H 1 Spin(3), H Cl even 3,0 (R), i e 12, j e 23, k e 13, quaternion multiplication the Clifford product. 22

23 3 Connecting the Algebras and Groups Section 1 proved the existence of three algebras whose sets are H, Q, and Cl even 3,0 (R), then section 2 proved that the algebras respective subsets H 1, SU(2), and Spin(3) equipped with the algebras respective ring multiplication operations, each form a group. To conclude, this section will prove that all three algebras are isomorphic to one another and that all three groups are isomorphic to one another. The result is complete proof of the following commutative diagram. Groups: Spin(3) H 1 SU(2) set inclusion F set inclusion M set inclusion Algebras: Cl even 3,0 (R) H Q 3.1 H Q and H 1 SU(2) If matrix multiplication is applied to the basis elements of the Q algebra, then this basis product structure is observed: ( ) ( ) ( ) ( ) 1 0 i i i 1 0 i 0 ( ) ( ) ( ) ( ) ( ) i i i 1 0 i 0 ( ) ( ) ( ) ( ) ( ) i 0 i i i 0 i 0 1 i ( ) ( ) ( ) ( ) ( ) i 1 0 i i i ( ) ( ) ( ) ( ) ( ) 0 i 0 i 0 1 i i 0 i i

24 To show how this is significant, some new symbols are introduced to represent the basis matrices: ( ) ( ) ( ) ( ) 1 0 i i 1., i., j., k i 1 0 i 0 Then the above table is reproduced with the new symbols: 1 i j k 1 1 i j k i i 1 k j j j k 1 i k k j i 1. Now it is clear that the basis product structure of Q under matrix multiplication is identical to the basis product structure of H under quaternion multiplication: 1 i j k 1 1 i j k i i 1 k j j j k 1 i k k j i 1. Consider a map M from H to Q: M : H Q ( ) 1 0 : a + bi + cj + dk a 0 1 ( ) i 0 + b + c 0 i ( ) ( ) i + d 1 0 ( i 0 ) a + bi c + di. c + di a bi Proposition The map M is an isomorphism between algebras H and Q. Hence H Q. 24

25 Proof. To prove this is to prove that: M is surjective (onto) X Q, h H : M(h) X. M is injective (one-to-one) h 1, h 2 H : M(h 1 ) M(h 2 ) h 1 h 2. M is homomorphic (preserves operations) h 1, h 2 H : M(h 1 + h 2 ) M(h 1 ) + M(h 2 ), Surjective subproof. Let a, b, c, d R and ( ) 1 0 X. a 0 1 then r R, h H : M(rh) rm(h), h 1, h 2 H : M(h 1 h 2 ) M(h 1 )M(h 2 ). ( ) 1 0 X a 0 1 ( ) i 0 + b + c 0 i ( ) i 0 + b + c 0 i ( ) ( ) i + d Q, 1 0 i 0 ( ) ( ) i + d 1 0 i 0 Since a + bi + cj + dk H, M(a + bi + cj + dk). X Q, h H : M(h) X. Injective subproof. Let a 1, a 2, b 1, b 2, c 1, c 2, d 1, d 2 R, h 1. a 1 + b 1 i + c 1 j + d 1 k H, and h 2. a 2 + b 2 i + c 2 j + d 2 k H. h 1 h 2 a 1 a 2 or b 1 b 2 or c 1 c 2 or d 1 d 2 ( ) ( ) ( ) ( ) 1 0 i i a 1 + b c 0 i 1 + d ( ) ( ) ( i 0 ) ( ) 1 0 i i a 2 + b c 0 i 2 + d i 0 M(h 1 ) M(h 2 ), 25

26 i.e. h 1 h 2 M(h 1 ) M(h 2 ) which is equivalent to M(h 1 ) M(h 2 ) h 1 h 2. (contrapositive statement) Therefore h 1, h 2 H : M(h 1 ) M(h 2 ) h 1 h 2. Homomorphic subproof. Let a 1, a 2, b 1, b 2, c 1, c 2, d 1, d 2 R, h 1. a 1 + b 1 i + c 1 j + d 1 k H, and h 2. a 2 + b 2 i + c 2 j + d 2 k H. M(h 1 + h 2 ) M(a 1 + b 1 i + c 1 j + d 1 k + a 2 + b 2 i + c 2 j + d 2 k) M ( (a 1 + a 2 ) + (b 1 + b 2 )i + (c 1 + c 2 )j + (d 1 + d 2 )k) ) (algebra property (S2)) ( ) ( ) ( ) ( ) 1 0 i i (a 1 + a 2 ) + (b b 2 ) + (c 0 i 1 + c 2 ) + (d d 2 ) i 0 ( ) ( ) ( ) ( ) 1 0 i i a 1 + b c 0 i 1 + d i 0 ( ) ( ) ( ) ( ) 1 0 i i + a 2 + b c 0 i 2 + d i 0 (algebra property (S2)) M(a 1 + b 1 i + c 1 j + d 1 k) + M(a 2 + b 2 i + c 2 j + d 2 k) M(h 1 ) + M(h 2 ). Hence h 1, h 2 H : M(h 1 + h 2 ) M(h 1 ) + M(h 2 ). Let r, a, b, c, d R and h. a + bi + cj + dk H. M(rh) M ( r(a + bi + cj + dk) ) M ( (ra) + (rb)i + (rc)j + (rd)k) ) (algebra properties (S1) and (S3)) ( ) ( ) ( ) ( ) 1 0 i i (ra) + (rb) + (rc) + (rd) i 1 0 i 0 26

27 ( ( ) 1 0 r a 0 1 ( ) i 0 + b + c 0 i rm(a + bi + cj + dk) rm(h). ( ) ( )) i + d 1 0 i 0 (algebra properties (S1) and (S3)) Hence r R, h H : M(rh) rm(h). Let a 1, a 2, b 1, b 2, c 1, c 2, d 1, d 2 R, h 1. a 1 + b 1 i + c 1 j + d 1 k H, and h 2. a 2 + b 2 i + c 2 j + d 2 k H. M(h 1 h 2 ) M ( (a 1 + b 1 i + c 1 j + d 1 k)(a 2 + b 2 i + c 2 j + d 2 k) ) ( ) (a1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 )1 + (a 1 b 2 + b 1 a 2 + c 1 d 2 d 1 c 2 )i M + (a 1 c 2 b 1 d 2 + c 1 a 2 + d 1 b 2 )j + (a 1 d 2 + b 1 c 2 c 1 b 2 + d 1 a 2 )k (from lemma 5.0.2) ( (a1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 ) + (a 1 b 2 + b 1 a 2 + c 1 d 2 d 1 c 2 )i (a 1 c 2 b 1 d 2 + c 1 a 2 + d 1 b 2 ) + (a 1 d 2 + b 1 c 2 c 1 b 2 + d 1 a 2 )i (a 1 c 2 b 1 d 2 + c 1 a 2 + d 1 b 2 ) + (a 1 d 2 + b 1 c 2 c 1 b 2 + d 1 a 2 )i (a 1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 ) (a 1 b 2 + b 1 a 2 + c 1 d 2 d 1 c 2 )i ( )( ) a1 + b 1 i c 1 + d 1 i a2 + b 2 i c 2 + d 2 i c 1 + d 1 i a 1 b 1 i c 2 + d 2 i a 2 b 2 i ) M(a 1 + b 1 i + c 1 j + d 1 k)m(a 2 + b 2 i + c 2 j + d 2 k) M(h 1 )M(h 2 ). Hence h 1, h 2 H : M(h 1 h 2 ) M(h 1 )M(h 2 ). All operations are preserved by the map, thus it is homophorphic. M is surjective, injective, and homomorphic, thus it is an isomorphism. Proposition The map M when restricted to the domain H 1, is an isomorphism between groups H 1 and SU(2). Hence H 1 SU(2). 27

28 Proof. To prove this is to prove that: The codomain of M H1 is SU(2). M is surjective (onto) S SU(2), n H 1 : M(n) S. M is injective (one-to-one) n 1, n 2 H 1 : M(n 1 ) M(n 2 ) n 1 n 2. M is homomorphic (preserves operations) n 1, n 2 H 1 : M(n 1 n 2 ) M(n 1 )M(n 2 ). There is no need to prove M is injective or homomorphic here as these properties follow from proposition Codomain subproof. Let n H 1 H, then n det(m(n)) (by lemma 5.0.3) 1 det(m(n)) (since n H 1 ) 1 det(m(n)) M(n) SU(2). (since M(n) Q) i.e. every element of H 1 maps to an element of SU(2) under M. Hence SU(2) is the codomain of the restriction map M H1, i.e. M H1 : H 1 SU(2). Surjective subproof. Let S SU(2) Q. This implies M 1 (S) H (by proposition 3.1.1) M 1 (S) det(s) (by lemma 5.0.3) M 1 (S) 1 (since S SU(2)) M 1 (S) 1 M 1 (S) H 1. (since M 1 (S) H) i.e. S SU(2), n H 1 : M(n) S. All necessary properties for M H1 to be isomorphic are thus proven. 28

29 3.2 Cl even 3,0 (R) H and Spin(3) H 1 The basis product structure rules for real Clifford algebras applied to the basis of Cl even 3,0 (R) can be summarised by the table 1 e 12 e 23 e e 12 e 23 e 13 e 12 e 12 1 e 13 e 23 e 23 e 23 e 13 1 e 12 e 13 e 13 e 23 e 12 1 which is identical in form to the quaternion basis product structure seen earlier 1 i j k 1 1 i j k i i 1 k j j j k 1 i k k j i 1. Consider a map F from the Cl even 3,0 (R) algebra to the H algebra: F : Cl even 3,0 (R) H : a + be 12 + ce 23 + de 13 a + bi + cj + dk Proposition The map F is an isomorphism between algebras Cl even 3,0 (R) and H. Hence Cl even 3,0 (R) H. Proof. To prove this is to prove that: F is surjective (onto) h H, γ Cl even 3,0 (R) : F(γ) h. F is injective (one-to-one) γ 1, γ 2 Cl even 3,0 (R) : F(γ 1 ) F(γ 2 ) γ 1 γ 2. F is homomorphic (preserves operations) γ 1, γ 2 Cl even 3,0 (R) : F(γ 1 + γ 2 ) F(γ 1 ) + F(γ 2 ), r R, γ Cl even 3,0 (R) : F(rγ) rf(γ), γ 1, γ 2 Cl even 3,0 (R) : F(γ 1 γ 2 ) F(γ 1 )F(γ 2 ). 29

30 Surjective subproof. Let a, b, c, d R and h. a + bi + cj + dk H, then h a + bi + cj + dk Since a + be 12 + ce 23 + de 13 Cl even 3,0 (R), F(a + be 12 + ce 23 + de 13 ). h H, γ Cl even 3,0 (R) : F(γ) h. Injective subproof. Let a 1, a 2, b 1, b 2, c 1, c 2, d 1, d 2 R, γ 1. a 1 + b 1 e 12 + c 1 e 23 + d 1 e 13 Cl even 3,0 (R), and γ 2. a 2 + b 2 e 12 + c 2 e 23 + d 2 e 13 Cl even 3,0 (R). γ 1 γ 2 a 1 a 2 or b 1 b 2 or c 1 c 2 or d 1 d 2 a 1 + b 1 i + c 1 j + d 1 k a 2 + b 2 i + c 2 j + d 2 k F(γ 1 ) F(γ 2 ), i.e. γ 1 γ 2 F(γ 1 ) F(γ 2 ) which is equivalent to F(γ 1 ) F(γ 2 ) γ 1 γ 2. (contrapositive statement) Therefore γ 1, γ 2 Cl even 3,0 (R) : F(γ 1 ) F(γ 2 ) γ 1 γ 2. Homomorphic subproof. Let a 1, a 2, b 1, b 2, c 1, c 2, d 1, d 2 R, γ 1. a 1 + b 1 e 12 + c 1 e 23 + d 1 e 13 Cl even 3,0 (R), and γ 2. a 2 + b 2 e 12 + c 2 e 23 + d 2 e 13 Cl even 3,0 (R). F(γ 1 + γ 2 ) F(a 1 + b 1 e 12 + c 1 e 23 + d 1 e 13 + a 2 + b 2 e 12 + c 2 e 23 + d 2 e 13 ) F ( (a 1 + a 2 ) + (b 1 + b 2 )e 12 + (c 1 + c 2 )e 23 + (d 1 + d 2 )k) ) (a 1 + a 2 ) + (b 1 + b 2 )i + (c 1 + c 2 )j + (d 1 + d 2 )k 30 (algebra property (S2))

31 a 1 + b 1 i + c 1 j + d 1 k + a 2 + b 2 i + c 2 j + d 2 k (algebra property (S2)) F(a 1 + b 1 e 12 + c 1 e 23 + d 1 e 13 ) + F(a 2 + b 2 e 12 + c 2 e 23 + d 2 e 13 ) F(γ 1 ) + F(γ 2 ). Hence γ 1, γ 2 Cl even 3,0 (R) : F(γ 1 + γ 2 ) F(γ 1 ) + F(γ 2 ). Let r, a, b, c, d R and γ. a + be 12 + ce 23 + de 13 Cl even 3,0 (R). F(rγ) F ( r(a + be 12 + ce 23 + de 13 ) ) F ( (ra) + (rb)e 12 + (rc)e 23 + (rd)e 13 ) ) (algebra properties (S1) and (S3)) (ra) + (rb)i + (rc)j + (rd)k r(a + bi + cj + dk) (algebra properties (S1) and (S3)) rf(a + be 12 + ce 23 + de 13 ) rf(γ). Hence r R, γ Cl even 3,0 (R) : F(rγ) rf(γ). Let a 1, a 2, b 1, b 2, c 1, c 2, d 1, d 2 R, γ 1. a 1 + b 1 e 12 + c 1 e 23 + d 1 e 13 Cl even 3,0 (R), and γ 2. a 2 + b 2 e 12 + c 2 e 23 + d 2 e 13 Cl even 3,0 (R). F(γ 1 γ 2 ) F ( (a 1 + b 1 e 12 + c 1 e 23 + d 1 e 13 )(a 2 + b 2 e 12 + c 2 e 23 + d 2 e 13 ) ) ( ) (a1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 ) + (a 1 b 2 + b 1 a 2 + c 1 d 2 d 1 c 2 )e 12 F + (a 1 c 2 b 1 d 2 + c 1 a 2 + d 1 b 2 )e 23 + (a 1 d 2 + b 1 c 2 c 1 b 2 + d 1 a 2 )e 13 (from lemma 5.0.2) ( (a1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 ) ) + (a 1 b 2 + b 1 a 2 + c 1 d 2 d 1 c 2 )i + (a 1 c 2 b 1 d 2 + c 1 a 2 + d 1 b 2 )j + (a 1 d 2 + b 1 c 2 c 1 b 2 + d 1 a 2 )k (a 1 + b 1 i + c 1 j + d 1 k)(a 2 + b 2 i + c 2 j + d 2 k) (distributivity (R3)) F(a 1 + b 1 e 12 + c 1 e 23 + d 1 e 13 )F(a 2 + b 2 e 12 + c 2 e 23 + d 2 e 13 ) F(γ 1 )F(γ 2 ). 31

32 Hence γ 1, γ 2 Cl even 3,0 (R) : F(γ 1 γ 2 ) F(γ 1 )F(γ 2 ). All operations are preserved by the map, thus it is homophorphic. F is surjective, injective, and homomorphic, thus it is an isomorphism. Proposition The map F when restricted to the domain Spin(3), is an isomorphism between groups Spin(3) and H 1. Hence Proof. To prove this is to prove that: The codomain of F Spin(3) is H 1. Spin(3) H 1. F is surjective (onto) n H 1, s Spin(3) : F(s) n. F is injective (one-to-one) s 1, s 2 Spin(3) : F(s 1 ) F(s 2 ) s 1 s 2. F is homomorphic (preserves operations) s 1, s 2 Spin(3) : F(s 1 s 2 ) F(s 1 )F(s 2 ). There is no need to prove F is injective or homomorphic here as these properties follow from proposition Codomain subproof. Let s. a + be 12 + ce 23 + de 13 Spin(3) Cl even 3,0 (R), then F(s) s (by lemma 5.0.4) F(s) 1 (since s Spin(3)) F(s) H 1. (since F(s) H) i.e. every element of Spin(3) maps to an element of H 1 under F. Hence H 1 is the codomain of the restriction map F Spin(3), i.e. F Spin(3) : Spin(3) H 1. 32

33 Surjective subproof. Let n H 1 H. This implies F 1 (n) Cl even 3,0 (R) (by proposition 3.2.1) F 1 (n) n (by lemma 5.0.4) F 1 (n) 1 (since n H 1 ) F 1 (n) Spin(3). (since F 1 (n) Cl even 3,0 (R)) i.e. n H 1, s Spin(3) : F(s) n. All necessary properties for F Spin(3) to be isomorphic are thus proven. 33

34 4 References [1] Nakahara, M., Geometry, topology and physics. CRC Press. [2] Bilge, A.H., Koçak, Ş. and Uğuz, S., Canonical bases for real representations of Clifford algebras. Linear algebra and its applications, 419 (2), pp [3] Lachieze-Rey, M., Spin and Clifford algebras, an introduction. Advances in applied Clifford algebras, 19 (3-4), p.687. [4] Lawson Jr, H.B. and Michelsohn, M.L., Spin geometry, volume 38 of Princeton Mathematical Series. [5] Definition:Matrix Scalar Product. (2014). [online] ProofWiki. Available at: Matrix_Scalar_Product&oldid [Accessed 19 Feb. 2017]. [6] Definition:Matrix Entrywise Addition. (2016). [online] ProofWiki. Available at: Entrywise_Addition&oldid [Accessed 19 Feb. 2017]. [7] Matrix Multiplication is Associative. (2016). [online] ProofWiki. Available at: is_associative&oldid [Accessed 19 Feb. 2017]. [8] Matrix Multiplication Distributes over Matrix Addition. (2012). [online] ProofWiki. Available at: Matrix_Multiplication_Distributes_over_Matrix_Addition&oldid [Accessed 19 Feb. 2017]. [9] Determinant of Matrix Product. (2016). [online] ProofWiki. Available at: Product&oldid [Accessed 19 Feb. 2017]. 34

35 5 Appendices Lemma Let h H, then h h h hh. Proof. Let h. a + bi + cj + dk H, hence a, b, c, d R. Then and h h (a + bi + cj + dk)(a bi cj dk) a 2 abi acj adk + bai b 2 i 2 bcij bdik + caj cbji c 2 j 2 cdjk + dak dbki dckj d 2 k 2 a 2 abi acj adk + abi + b 2 bck + bdj + acj + bck + c 2 cdi + adk bdj + cdi + d 2 a 2 + b 2 + c 2 + d 2 h 2 hh (a bi cj dk)(a + bi + cj + dk) a 2 + abi + acj + adk bai b 2 i 2 bcij bdik caj cbji c 2 j 2 cdjk dak dbki dckj d 2 k 2 35

36 Therefore h h h 2 hh. Lemma Let h 1, h 2 H, then Proof. Let h 1, h 2 H such that Then a 2 + abi + acj + adk abi + b 2 + bck bdj acj bck + c 2 + cdi adk + bdj cdi + d 2 a 2 + b 2 + c 2 + d 2 h 2. (h 1 h 2 ) h 2 h1. h 1. a 1 + b 1 i + c 1 j + d 1 k, h 2. a 2 + b 2 i + c 2 j + d 2 k. h 1 h 2 (a 1 + b 1 i + c 1 j + d 1 k)(a 2 + b 2 i + c 2 j + d 2 k) a 1 a 2 + a 1 b 2 i + a 1 c 2 j + a 1 d 2 k + b 1 a 2 i + b 1 b 2 i 2 + b 1 c 2 ij + b 1 d 2 ik + c 1 a 2 j + c 1 b 2 ji + c 1 c 2 j 2 + c 1 d 2 jk + d 1 a 2 k + d 1 b 2 ki + d 1 c 2 kj + d 1 d 2 k 2 a 1 a 2 + a 1 b 2 i + a 1 c 2 j + a 1 d 2 k + b 1 a 2 i b 1 b 2 + b 1 c 2 k b 1 d 2 j + c 1 a 2 j c 1 b 2 k c 1 c 2 + c 1 d 2 i + d 1 a 2 k + d 1 b 2 j d 1 c 2 i d 1 d 2 (a 1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 )1 + (a 1 b 2 + b 1 a 2 + c 1 d 2 d 1 c 2 )i + (a 1 c 2 b 1 d 2 + c 1 a 2 + d 1 b 2 )j + (a 1 d 2 + b 1 c 2 c 1 b 2 + d 1 a 2 )k, 36

37 thus h 2 h1 (a 2 b 2 i c 2 j d 2 k)(a 1 b 1 i c 1 j d 1 k) a 2 a 1 a 2 b 1 i a 2 c 1 j a 2 d 1 k b 2 a 1 i + b 2 b 1 i 2 + b 2 c 1 ij + b 2 d 1 ik c 2 a 1 j + c 2 b 1 ji + c 2 c 1 j 2 + c 2 d 1 jk d 2 a 1 k + d 2 b 1 ki + d 2 c 1 kj + d 2 d 1 k 2 a 1 a 2 b 1 a 2 i c 1 a 2 j d 1 a 2 k a 1 b 2 i b 1 b 2 + c 1 b 2 k d 1 b 2 j a 1 c 2 j b 1 c 2 k c 1 c 2 + d 1 c 2 i a 1 d 2 k + b 1 d 2 j c 1 d 2 i d 1 d 2 (a 1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 )1 + ( a 1 b 2 b 1 a 2 c 1 d 2 + d 1 c 2 )i + ( a 1 c 2 + b 1 d 2 c 1 a 2 d 1 b 2 )j + ( a 1 d 2 b 1 c 2 + c 1 b 2 d 1 a 2 )k (a 1 a 2 b 1 b 2 c 1 c 2 d 1 d 2 )1 (a 1 b 2 + b 1 a 2 + c 1 d 2 d 1 c 2 )i (a 1 c 2 b 1 d 2 + c 1 a 2 + d 1 b 2 )j (a 1 d 2 + b 1 c 2 c 1 b 2 + d 1 a 2 )k (h 1 h 2 ). 37

38 Lemma Let h H, then h det(m(h)). Proof. Let a, b, c, d R and h. a + bi + cj + dk H, then det(m(h)) ( ) a + bi c + di det c + di a bi (M defined here) (a + bi)(a bi) ( c + di)(c + di) a 2 + b 2 + c 2 + d 2 h 2 h det(m(h)). Lemma Let γ Cl even 3,0 (R), then γ F(γ). Proof. Let a, b, c, d R and γ. a + be 12 + ce 23 + de 13 H, then γ a + be 12 + ce 23 + de 13 a 2 + b 2 + c 2 + d 2 a + bi + cj + dk F(γ). (F defined here) 38