PHYSICAL APPLICATIONS OF GEOMETRIC ALGEBRA

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1 January 26, 1999 PHYSICAL APPLICATIONS OF GEOMETRIC ALGEBRA LECTURE 4 SUMMARY In this lecture we will build on the idea of rotations represented by rotors, and use this to explore topics in the theory of Lie groups and Lie algebras. 1. Reflections, Rotations and the rotor description. 2. Rotor groups, multivector transformations and spin-1/2 3. Lie Groups, continuous groups and the manifold structure of rotors in 3-d. 4. Bivector generators and Lie algebras. 5. Complex structures and doubling bivectors. 6. Unitary groups expressed in real geometric algebra. 1

2 PSEUDOSCALARS AND DUALITY Exterior product of n vectors in G n gives a multiple of the pseudoscalar, I. Two key properties: 1. Normalisation ji 2 j =1: Sign of I 2 depends on dimension (and signature). 2. Right-Handed. e 2 e 3 e 2 e 1 e 1 e 1^e 2 is right handed, by definition. e 1^e 2^e 3 is if looking down e 3 gives right-handed plane. Continue inductively. Product of the grade-n pseudoscalar I with grade-r multivector A r is a grade n r multivector IA r = hia r i n r Called a duality transformation. If A r is a blade, get the orthogonal complement of A r the blade formed from the space of vectors not contained in A r. 2

3 Summarise the commutative properties of I by IA r =( 1) r(n 1) A r I I always commutes with even grade. For odd grade depends on dimension of space. Important use for I: interchanging dot and wedge products. Take A r and B s, r + s» n, A r (B s I)=hA r B s Ii jr (n s)j = ha r B s Ii n (r+s) = ha r B s i r+s I = A r^b s I Have already used this in 3-d. REFLECTIONS a = a? + a k a 0 = a? a k a 0 a? a a k m Hyperplane Reflect the vector a in the (hyper)plane orthogonal to unit vector m. 3

4 Component of a parallel to m changes sign, perpendicular component is unchanged. Parallel component is the projection onto m: a k = a mm The perpendicular component is the remainder a? = a a mm =(am a m)m = a^mm Shows how the wedge product projects out component perpendicular to a vector. The reflection gives a 0 = a mm+ a^mm A remarkably neat formula! = (m a + m^a)m = mam Simple to check the desired properties. For a vector parallel to m m( m)m = mmm = m transformed to minus itself. For vector perpendicular to m m(n)m = mnm = nmm = n (n m =0) so unchanged. Also give a simple proof that lengths and angles unchanged a 0 b 0 =( mam)( mbm) =mabm 4

5 Scalar part gives a 0 b 0 = ma bm = a b, as expected. Bivector part gives a 0^b 0 = ma^bm A crucial sign change cf vectors. Origin of distinction between polar and axial vectors. ROTATIONS a 00 n m a a 0 Theorem: Successive reflections generated by two vectors m and n gives a rotation in the m^n plane. a 0 is the result of reflecting a in the plane perpendicular to m a 00 is the result of reflecting a 0 in the plane perpendicular to n. Component of a outside the plane is untouched. Simple trigonometry: angle between a and a 00 is twice angle 5

6 between m and n, so rotate through 2 in the m^n plane, where m n = cos( ). How does this look in GA? a 0 = mam a 00 = na 0 n = n( mam)n = nmamn This is beginning to look very simple! We define R = nm; ~ R = mn Note the geometric product. We can now write a rotation as a 7! Ra ~ R Incredibly, works for any grade of multivector, in any dimension, ofany signature! As seen already, R is a rotor. Now make contact with bivector approach. R is the geometric product of two unit vectors n and m, R = nm = n m + n^m = cos( ) +n^m What is the magnitude of the bivector n^m? (n^m) (n^m) = hn^m n^mi = hnm n^mi = hnm(n m mn)i =cos 2 ( ) 1= sin 2 ( ) Define a unit bivector in the m^n plane by ^B = m^n= sin( ); ^B 2 = 1 6

7 NB Using the correct right-handed orientation for ^B, as defined as angle between m and n in positive sense from m to n. Nowhave R =cos( ) ^B sin( ) Familiar? it is the polar decomposition of a complex number back again. Unit imaginary replaced by the unit bivector ^B. Write R =expf ^B g (Exponential defined by power series this always converges for multivectors). But our formula was for a rotation through 2. For rotation through, need the half angle formula R =expf ^B =2g which gives a 7! e ^B =2 a e ^B =2 for a positive rotation through in the ^B plane. In GA think of rotations taking place in a plane not around an axis. 7

8 THE ROTOR GROUP Form composite of two rotations Define a 7! R 2 (R 1 a R ~ 1 ) R ~ 2 = R 2 R 1 a R ~ 1 ~R 2 R = R 2 R 1 Still have a 7! Ra ~ R. R is a geometric products of an even number of unit vectors, R = kl mn This defines a rotor. The reversed rotor is ~R = nm lk so still have normalisation condition R ~ R = kl mnnm lk =1= ~ RR In non-euclidean spaces require this. Now decompose general multivector into sum of blades. Write each as product of orthogonal vectors. Take A r = a 1 a 2 a r 8

9 Rotate each vector to a 0 = Ra ~ i i R. Get A 0 = r a0 1a 0 2 a 0 = Ra r 1 RRa ~ ~ 2 R Ra ~ r R = Ra 1 a 2 a ~ r R = RA ~ r R Recover the same law as for vectors! All multivectors transform same way when component vectors are rotated. Spin-1/2 Have rotors R and R = exp( ^B =2). Product rotor is R 0 = R R = e ^B =2 R: Now increase from 0 through to 2ß. =2ßis identity operation. But R transforms to R 7! R 0 = e ^Bß R = (cos ß ^B sin ß)R = R: Rotors change sign under 360 o rotations. Just like fermions in quantum theory! But no quantum mechanics here. Can see effect with coupled rotations. LIE GROUPS Rotors form an infinite-dimensional continuous group a Lie group. Not a vector space, actually a manifold. See this in 3-d. Write R = x 0 + x 1 Ie 1 + x 2 Ie 2 + x 3 Ie 3 9

10 Then R ~ R = x x1 2 + x2 2 + x3 2 =1 Defines a unit vector in 4-d. Group manifold is a 3-sphere. Not same as rotation group. Rotations are formed by a 7! Ra ~ R,soR and R give same rotation. Rotation group manifold is 3-sphere with opposite points identified. Attitude of a rigid body described by a rotor, so configuration space for rigid body dynamics is a 3-sphere. Important for 1. Finding best fit rotation. 2. Extrapolating between two rotations. 3. Lagrangian treatment and conjugate momenta. 4. Quantum rigid rotor. Abstract idea. Lie group = Manifold + product z = ffi(x; y). BIVECTOR GENERATORS Question: can any rotor be written as the exponential of a bivector? Define a one-parameter Abelian subgroup R( ) R( + μ) =R( )R(μ) Must have R(0) = 1. Look at vector a( ) =Ra 0 ~R 10

11 Differentiate with respect to to get a 0 ( ) =R 0 a 0 ~ R + Ra 0 ~ R 0 = R 0 ~ Ra ar 0 ~ R Have used familiar result (R R) ~ 0 =0=R 0 R ~ + R R ~ 0 R 0 R ~ could have grades 2, 6, 10 etc. But commutator product with any vector is a vector, so R 0 R ~ is a bivector only, R 0 ( ) = 1 2 B( )R( ) But have R 0 ( + μ) = 1 B( + μ)r( + μ) 2 = 1 B( + μ)r( )R(μ) 2 = [R( )R(μ)] 0 = 1 2 B( )R( )R(μ) So B is constant along this curve. Integrate to get R( ) = e B=2 Any rotor on this curve is exponential of a bivector. Manifold idea ) true for all near identity. Euclidean space: All rotors = bivector exponential Mixed Signature: R( ) =± e B=2 Converse result: exponential of a bivector = rotor. Form a( ) = e B=2 a 0 e B=2 : 11

12 Differentiate to get Taylor series: a 0 = e B=2 a B e B=2 a 00 = e B=2 (a B) B e B=2 ; etc. NB a B always a vector, so preserves grade. Get e B=2 a e B=2 = a + a B + 1 (a B) B + 2! LIE ALGEBRAS AND BIVECTORS Bivectors are generators of the Lie group, by exponentiation. These generate a Lie algebra. Expresses fact that rotations do not commute. Form compound rotation: Ra ~ R = ~ R 2 ~ R 1 (R 2 R 1 a ~ R 1 ~ R 2 )R 1 R 2 : The resulting rotor is R = e B=2 = e B 2=2 B1=2 B2=2 B1=2 e e e Expanding exponentials we find (exercise) B = B 1 B 2 + higher order terms The Baker-Campbell-Hausdorff formula. Abstract idea. Lie algebra is a linear space (tangent space at the identity element of the group manifold) with a Lie bracket. This is Antisymmetric + Closed + Jacobi identity. 12

13 For bivectors, Lie bracket = commutator product, A B. Gives a third bivector. Jacobi identity is (A B) C +(C A) B +(B C) A =0 Proof Expand out into geometric products. Nothing special about grades. True for any 3 multivectors. One consequence: (a^b) B =(a B)^b (b B)^a NB proves closure. Another view: Basis set of bivectors fb i g. Can write B j B k = C i jkb i C i jk are the structure constants. Compact encoding of properties of a Lie group. Can always construct a matrix rep of Lie algebra from structure constants COMPLEX STRUCTURES GA in 2-d gives complex numbers. What about complex vectors? Natural idea is work in 2n-d space. Introduce basis fe i ;f i g f i f j = e i e j = ffi ij e i f j =0; 8i; j: Introduce complex structure through doubling bivector J = e 1 f 1 + e 2 f e n f n = e i^f i 13

14 Sum of n commuting blades, each an imaginary in own plane. J satisfies J f i =(e j ^f j ) f i = e j ffi ij = e i J e i =(e j ^f j ) e i = f i J maps from one half of vector space to other. Follows that J (J e i )= J f i = e i ; J (J f i )=J e i = f i ; Hence J (J a) =(a J) J = a 8a Phase rotation becomes a rotation in J plane. Expand e Jffi=2 a e Jffi=2 = a + ffia J + ffi2 2! =(1 ffi2 2! (a J) J + + )a +(ffi ffi3 3 )a J =cosffia+sinffia J HERMITIAN INNER PRODUCT Complex vectors Z and W : Z i = x i + iy i ; and W i = u i + iv i Hermitian inner product is hw jzi = W Λ i Z i = u i x i + v i y i + i(u i y i v i x i ) 14

15 Want analog in 2n-d space. Introduce vectors x = x i e i + y i f i ; and w = u i e i + v i f i : Real part of hw jzi is x w. Imaginary part is w e i x f i w f i x e i =(w e i x x e i w) f i =[(x^w) e i ] f i =(x^w) (e i ^f i )=(x^w) J: NB Antisymmetric bilinear form. Can now write hajbi = a b i (a^b) J Maps from 2n-d space onto complex numbers. UNITARY GROUPS Unitary group is invariance group of Hermitian inner product. Must leave inner product and skew term invariant. Build from rotations. Require that (a 0^b 0 ) J =(a^b) J with a 0 = Ra R, ~ b 0 = Rb R. ~ Find (a 0^b0 ) J = ha 0 b 0 Ji = hra RRb ~ RJi ~ = hab RJRi ~ =(a^b) ( RJR) ~ Must hold for all a and b, so ~RJR = J 15

16 Unitary group U(n) is subgroup of rotor group which leaves J invariant. Get complex groups as sub-groups of real rotation groups! Unusual approach, but has a number of advantages. With R =exp( B=2) must have B J =0: Get bivector form of Lie algebra of the unitary group, u(n). Use Jacobi identity to prove [(a J)^(b J)] J = (a J)^b +(b J)^a = (a^b) J: Follows that [a^b +(a J)^(b J)] J =0: Work through all combinations of fe i ;f i g. Write down the following Lie algebra basis for u(n): E ij = e i e j + f i f j (i<j=1:::n) F ij = e i f j f i e j (i<j=1:::n) J i = e i f i : Can establish closure (exercise). Algebra contains J, commutes with all other elements, gives global phase term. Removing this gives special unitary group, SU(n). 16