Vector Parameter Forms of SU(1,1), SL(2, R) and Their Connection with SO(2,1)

Transcription

1 Vector Parameter Forms of SU(1,1), SL(, R) and Their Connection with SO(,1) VELIKO D. DONCHEV, CLEMENTINA D. MLADENOVA, IVAÏLO M. MLADENOV Geometry, Integrability and Quantization, Varna, June

2 The presentation Summary of the results The Cayley maps for the Lie algebras su(1, 1) and so(, 1) converting them into the corresponding Lie groups SU(1, 1) and SO(, 1) along their natural vector parameterizations are examined. Additionally the explicit form of the covering map SU(1, 1) SO(, 1) and its sections are presented. Finally, the vector-parameter forms of the Lie groups SU() and SU(1, 1) are compared and some of their applications are addressed.

3 This research is made within a bigger project which is about parameterizing Lie groups with small dimension and its application in physics. In [4] the Cayley maps for the Lie algebras su() and so(3) and the corresponding Lie groups SU() and SO(3, R) are examined. Parameterizations are used to describe Lie groups in an easier and more intuitive way. Let G be a finite dimensional Lie group with Lie algebra g. A vector parameterization of G is a map g G, which is diffeomorphic onto its image. Besides the exponential map, there are other alternatives to achieve parameterization. We make use of the Cayley map Cay(X) = (I + X)(I X) 1. (1.1)

4 Structure of so(, 1) The Cayley map for so(, 1) Consider the Lie algebra so(, 1) with basis P 1 = 0 0 1, P = 0 0 0, P 3 = The commutation relations of these matrices are as follows: (.1) [P 1, P ] = P 3, [P, P 3 ] = P 1, [P 3, P 1 ] = P. (.) Any C so(, 1) has a unique representation 0 c 3 c c C = c.p = c 3 0 c 1, c = (c 1, c, c 3 ). (.3) c c 1 0 The Hamilton Cayley theorem applied to C from (.3) reads as C 3 = (1 c.(ηc))c (.4)

5 where η = Structure of so(, 1) The Cayley map for so(, 1). The Cayley map applied for so(, 1) is H(c) = Cay so(,1) (C) = (I + C)(I C) 1. (.5) One checks immediately that I C is invertable if only if c.(ηc) 1 and in this case we can explicitly calculate 1 H(c) = Cay so(,1) (C) = (I + C)(I + 1 c.(ηc) C c.(ηc) C ) = I + 1 c.(ηc) C + 1 c.(ηc) C (.6) 1 c 1 c 1 c + c 3 c 1 c 3 c = c 1 c c 3 1 c c c 3 c 1 I. 1 c.(ηc) c 1 c 3 c c c 3 c c 3

6 Structure of so(, 1) The Cayley map for so(, 1) If H(c 1 ), H(c 1 ) are two SO(,1) elements represented by the vector parameters and c 1, c and c 1.(ηc 1 ) 1, c.(ηc ) 1 and 1 + c.(ηc 1 ) 0. Then H(c 3 ) = H(c )H(c 1 ), c 3 = c, c 1 SO(,1) = c + c 1 + c c 1 (.7) 1 + c.(ηc 1 ) where c c 1 := η(c c 1 ). Equation (.7) is the vector-parameter form of SO(,1) obtained by the parameterization given by the Cayley map. The same result was obtained independently by usage of pseudo-quaternions [3]. Note that in the elliptic case, i.e., 1 > c.(ηc) > 0 there exists n R,1 such that n.(ηn) = 1 and c = tanh θ n. (.8) In the hyperbolic case, i.e., c.(ηc) < 0 there exist n R,1 such that n.(ηn) = 1 and c = tan θ n.

7 Structure of su(1, 1) Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) Let us consider the Lie algebra su(1, 1) with R basis ( ) ( ) i e 1 =, e 1 0 =, e i 0 3 = The matrices E 1, E and E 3 defined by ( i 0 0 i ) (3.1) E 1 = 1 e 1, E = 1 e, E 3 = 1 e 3 (3.) also form a R basis of su(1, 1). Also [E 1, E ] = E 3, [E, E 3 ] = E 1, [E 3, E 1 ] = E. (3.3) Obviously, the map m 1 E 1 + m E + m 3 E 3 m 1 P 1 + m P + m 3 P 3 (3.4) is a linear isomorphism between the Lie algebras su(1, 1) and so(, 1).

8 Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) Let M = m.e = i m 3 m 1 + i m m 1 i m i m 3 su(1, 1). The Hamilton Cayley theorem applied to M reads as The Cayley map applied for su(1, 1) is Let us define M = m (η m )I (3.5) L(m) = Cay su(1,1) (M) = (I + M)(I M) 1. (3.6) m = 1 m 1 + m m 3 4 = 1 m (η m ) (3.7) which is exactly det (I M). Thus, Cay su(1,1) is well-defined when m 0.

9 Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) In this case we have (I M) 1 = 1 m (I + M) and thus we can explicitly calculate L(m) = Cay su(1,1) (M) = 1 (I + M) = 1 (I + M + M ) = m I m m m (3.8) = m I + m.e = 1 ( ) m + im 3 m 1 + im. m m m m 1 im m im 3 (3.9) Now direct calculation shows ( that ) det (L(m)) = 1 and also 1 0 L(m) ηl(m) = η, η =. Thus L(m) SU(1, 1). 0 1

10 Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) To invert the Cayley map Cay su(1,1) : su(1, 1) SU(1, 1) let us consider an arbitrary SU(1, 1) matrix ( ) α1 + iα L(α 1, α, β 1, β ) = β 1 + iβ, α1 + α β1 β = 1. β 1 iβ α 1 iα L ICay su(1,1) if and only if there exist m R 3 : m 0 such that Cay su(1,1) (A(m)) = L. This is only possible if α 1 1 and in this case the inversion is Finally, the Cayley acts (m 1, m, m 3 ) = 1 + α 1 (β 1, β, α ). (3.10) Cay su(1,1) : {m.e su(1, 1); m 0} {L(α 1, α, β 1, β ) SU(1, 1); α 1 1}.

11 Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) Theorem Let M, A su(1, 1) M = m.e, m = (m 1, m, m 3 ), A = a.e, a = (a 1, a, a 3 ) be such that m 0, a 0 and (a.(ηa))(m.(ηm)) + 8a.(ηm) (3.11) Let L(m) = Cay su(1,1) (M), W(a) = Cay su(1,1) (A). Then, if L = W.L is the composition of the images in SU(1, 1) then L = Cay su(1,1) (Ã) where à = m.e and m = (1 + m (η m ))a + (1 + a (η a ))m + a m 1 + a (η m ) + ( a (η a ))( m (η m (3.1) ))

12 Idea for proof 1 Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) Let us calculate ( a W(a).L(m) = I + ) ( m a.e I + ) m.e a m m a (??) = ( a)( m ) + a m I + ( m)a + ( a )m + a m E. a m a m Now, the condition for existence of m is ( a )( m ) + a m a m (1 a )(1 m ) + a (η m ) and after simplification we obtain that it is equivalent to (3.11).

13 Idea for proof Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) Thus, m exists and is such that m = ( a)( m ) + a.(ηm) m a m (3.14) m = ( m)a + ( a )m + a m m a m (3.15) From (3.14) we immediately find m = a m 1 + a (η m ) + ( a (η a ))( m (η m )) 0 (3.16) and thus after some algebraic simplifications the composition law in vector-parameter form follows.

14 Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) It is well know fact that SU(1, 1) is isomorphic as a group to the real special linear group of order, i.e., SL(, R) by the map ϕ : SL(, R) SU(1, 1), (( )) ( ( )) (3.17) 1 ( ) ( ) a b 1 1 i a b 1 1 i ϕ =. c d i 1 c d i 1 Now, to automatically obtain the vector-parameter form we need to invert ϕ and calculate ϕ 1 (L(m)) where m R 3, m 0. A straight forward calculation shows that ψ(m) := ϕ 1 (L(m)) = M(m) = 1 m We have ( m + m m 1 + m 3 m 1 m 3 m m tr L(m) = tr M(m) = 4 m m, m 0. (3.18) ).

15 Structure of su(1, 1) The Cayley map for su(1, 1) Composition law in SU(1,1) Vector-parameter form of SL(, R) In a straight forward manner we obtain that if m 0 then hyperbolic if m < 1 m.(ηm) > 0 M(m) is elliptic if m > 1 m.(ηm) < 0 parabolic if m = 1 m.(ηm) = 0. (3.19) Corollary In the terms of Theorem 1, for the composition vector m we have hyperbolic if ξζ > 0 M( m) is elliptic if ξζ < 0 parabolic if ξ = 0 ξ = ξ(a, m) = (a + m).(η(a + m)), ζ = ζ(a, m) = (4 + (a.m)) (a m).(η(a m)).

16 Consider the homomorphism map [1] ϕ ( : SU(1, ) 1) SO(, 1) α β which sends the SU(1, 1) matrix L = where β α α = α 1 + iα, β = β 1 + iβ and αα ββ = 1 into 1 (β + β α α i ) (α + β α β ) i(αβ αβ) ϕ(l) = i (β β α + α 1 ) (α + β + α + β ) αβ + αβ. i(αβ αβ) αβ + αβ αα + ββ The homomorphism map ϕ is a double cover with ker ϕ = {±I} and SU(1, 1)/Z = SO(, 1). We are interested in the vector-parameter form of ϕ.

17 Consider the SU(1, 1) matrix L(m) = Cay su(1,1) (m). We have α = m m + i m 3 m, β = m 1 m + i m m (4.1) Substitution of (4.1) in (4.1) leads to the matrix ϕ(m) = H(m) i.e., ( ) + I. m m m 3 m 1 m + m 3 ( m) m 1 m 3 + m ( m) m 1 m m 3 ( m) m 1 m 3 m m 3 + m 1 ( m) m 1 m 3 + m ( m) m m 3 + m 1 ( m) m 1 + m This vector parameter form will allow us to obtain the connection between the vector parameter in SO(,1) and its cover SU(1,1).

18 Theorem Let L(m) is an SU(1, 1) element, represented by the vector-parameter m such that m (η m ) 1. Then in SO(, 1) this element is represented by H(c) where ηm c = 1 + m (η m ) (4.) On the other hand, if c represents the SO(, 1) element H(c), then the in SU(1, 1) the elements represented by the vector-parameters m + (c) = 1 c.(ηc) c.(ηc) correspond to H(c). ηc, m (c) = + 1 c.(ηc) ηc c.(ηc)

19 Moreover, the following relations hold m +.(ηm ) = 4, (m +.(ηm + ))(m.(ηm )) = 16 (4.3) 4 m + = m.(ηm ) m 4, m = m +.(ηm + ) m +. (4.4)

20 Idea for proof 1 Let us equate the following expressions from H(c) (cf. (??)) and H(m) (cf. (4.)) tr H(c) = tr H(m) (4.5) H(c) 1, H(c),1 = H(m) 1, H(m),1 H(c) 1,3 + H(c) 3,1 = H(m) 1,3 + H(m) 3,1, (4.6) H(c),3 + H(c) 3, = H(m),3 + H(m) 3,. Equation (4.5) is equivalent to 4 (3 c.(ηc)) 3 = 1 c.(ηc) m.(ηm) + 3 (4.7) m from where after some algebraic manipulation 1 c.(ηc) = ( m) (4.8) m

21 Idea for proof Equation (4.6) reads as where as (??) reads as 4 1 c.(ηc) c 3 = 4 m d 3 (4.9) m 4 1 c.(ηc) c = 4 m d, m 4 1 c.(ηc) c 1 = 4 m d 1 (4.10). m It is obvious that from (4.9) and (4.10) 1 1 c.(ηc) c = m ηm (4.11) m and thus substituting (4.8) into (4.11) we obtain the result (4.).

22 Idea for proof 3 Let us find the vector-parameter form of the sections of the homomorphism i.e., invert (4.). From it we have obtain ηm ηm c.(ηc) = 1 + m (η m η ) 1 + m (η m = ) m.(ηm) ( 1 + m.(ηm) 4 This leads to the quadratic equation for x = m.(ηm) ) (4.1) c.(ηc)x + 8(c.(ηc) )x + 16c.(ηc) = 0. (4.13) x Note that the expression 1 c.(ηc) = 1 (1 + x and straight ) 4 forward calculus shows that 1 c.(ηc) 0 with 1 c.(ηc) = 0 if only if x = 4. Thus, equation (4.13) has two real roots m +.(ηm + ) = 4 c.(ηc) ± 1 c.(ηc) (4.14) c.(ηc)

23 Idea for proof 4 Direct calculation shows that 1 + m ± (η m ± ) = 1 c.(ηc) c.(ηc) (4.15) and thus the two sections of (4.) are m + (c) = 1 c.(ηc) ηc, c.(ηc) m (c) = + 1 c.(ηc) ηc. (4.1 c.(ηc) Note that the properties (4.3) follow directly from (4.14) and (4.16).

24 Remark Note that if we assume the axis-rapidity representation of SO(,1) elliptic vector-parameter c = tanh θ n, n.(ηn) = 1, θ R (5.1) then for m + (c) and m (c) we obtain m (c) = tanh θ 4 ηn, m +(c) = 1 tanh θ 4 ηn = coth θ 4 ηn.(5.) If we assume the representation of SO(,1) hyperbolic vector-parameter c = tan θ n, n.(ηn) = 1, θ [0, π) (5.3) then for m + (c) and m (c) we obtain m (c) = tan θ ηn, m + (c) = 1 ηn = cot θ ηn.(5.4)

25 1 Basu D., Introduction to Classical and Modern Analysis and Their Application to Group Representation Theory, World Scientific, Singapore 009. Brezov D., Mladenova C. and Mladenov I., Vector Decompositions of Rotations, J. Geom. Symmetry Phys. 8 (01) Brezov D., Mladenova C. and Mladenov I., Vector Parameters in Classical Hyperbolic Geometry, J. Geom. Symmetry Phys. 30 (013) Donchev, V., Mladenova K. and Mladenov I. Vector Parameter Form of the SU SO(3, R) Map, accepted for publication in Ann. Univ. Sofia (015). 5 Fedorov F., The Lorentz Group (in Russian), Nauka, Moscow Serre J., Lie Algebras and Lie Groups, nd Edn, Springer, Berlin 006.

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