1.2. Homomorphism: SL(2,C) ~ L0

Let Φ be a map between groups G1 & G2 , i.e., Φ: G1 → G2.Φ is a homomorphism if ab a b 1,a b G

Example: G1 = (Z,+), G2 = C2, Φ(n) = (-1)n n Z. 1 1 1a b a ba b a b

Example: G1 = (Z,+), G2 = Ck, Φ(n) = exp(2πi n/k) n Z. 2 ( ) / 2 / 2 /i a b k ia k ib ka b e e e a b

→ Multiplication in Ck ~ addition modulo k in Z

Let M be the Minkowski space, i.e., R4 with the Lorentz metric

2 2 2 2 20 1 2 3x x x x x 0 1 2 3

Tx x x xx → (c = 1)

A Lorentz transformation B preserves the Lorentz metric: 2 2B x x M x

The set of all B forms the Lorentz group L.Rewrite x as a 2×2 self-adjoint (Hermitian) matrix:

0 3 1 2

1 2 0 3

x x x ixX

x ix x x

X X

Note: all eigenvalues λj of a Hermitian matrix must be real.0 1 22TrX x 0 1 2

12

x d d

22 2 2 20 3 1 2 1 2detX x x x x x

3 1 212

x d d

0 3 1 20 1 1 2 2 3 3

1 2 0 3

x x x ixX x e x x x

x i x x x

1 00 1

e

1

0 11 0

2

00i

i

3

1 00 1

Pauli matrices

The action of a 2×2 matrix A on X is defined as

X A X AX A

AX A A X A (Hermitian

)AX A

det detA X AX A 2det detA Xdet det detA X A

A SL(2,C) →

2det detA X X x

AB X ABX AB ABX B A A BX B A

A B X A B X

→ Φ is a homomorphism A A

Let C be a Lorentz transformation. By definition

2 2C x x

1 2 3

1 11 12 13

2 21 22 23

3 31 32 33

a v v vw R R R

Cw R R Rw R R R

TaR

vw

Let0

1 0

2

3

xx xxx

xr

2 20

Tx x r r 00

1 T xx

I

0r

r0TGx x 1 T

GI

00

Metric tensor

2 TC C G Cx x x T TC GCx x TGx x x

→ TC GC G1T T T

T

a aR I R

w 0 vv 0 w

T T

T

a aR R

w vv w

2 T T T

T T T

a a Ra R R R

w w v wv w vv

2 1Ta w w Ta Rv w

T TR R I vv

Let A SU(2)

→ A-1 = A+ → A I A+ = I

Let

0 1 0 0 0 Te 1 0~

0 1I

0 0A A I A e e 0e→

Let C = Φ(A) be the corresponding Lorentz transformation: 0 0C e e

i.e., C is a spatial rotation

→1 1T aa

R

v0 w 0w

1 T

CR

00

TR R I

The subspace orthogonal to e0 is

0 0, T e r

01 T

CR

0x

r0For all

0x e 0

R r

0 e

→ O(3) is the subgroup of L defined by

0 0C e e

Φ: SU(2) → O(3) Φ: SL(2,C) → L0

Example 1

02

0

i

i

eU SU

e

Ux x 0 3 1 2

1 2 0 3

0 00 0

i i

i i

x x x ixe ex ix x xe e

20 3 1 2

21 2 0 3

i

i

x x e x ixe x i x x x

→ Φ(UΘ) is a rotation of angle 2Θ about the x3 axis.

0 3 1 2 1 2

1 2 1 2 0 3

cos2 sin 2 sin 2 cos2cos2 sin2 sin 2 cos2

x x x x i x xx x i x x x x

~U XU

0 1 2 1 2 3cos2 sin 2 sin 2 cos2 Tx x x x x x x

Θ = π corresponds to 2π rotation in 3-space.

Example 2

cos sin2

sin cosV SU

1 cos sinsin cos

V V V

Vx x

0 3 1 2

1 2 0 3

cos sin cos sinsin cos sin cos

x x x ixx i x x x

~V XV

0 1 3 1 3 2

1 3 2 0 1 3

sin 2 cos2 cos2 sin 2cos2 sin 2 sin 2 cos2x x x x x i xx x i x x x x

0 1 3 2 1 3cos2 sin 2 sin 2 cos2 Tx x x x x x x

1

2

3

cos2 0 sin 20 1 0sin 2 0 cos

xxx

Φ(VΘ ) = rotation of angle 2α about x2-axis

Example 3

1

02

0r

rM SU

r

rMx x

0 3 1 21 1

1 2 0 3

0 00 0

x x x i xr rx i x x xr r

~ r rM XM

20 3 1 2

21 2 0 3

r x x x ixx i x r x x

2 2 2 2 2 2 2 20 3 1 2 0 3

1 1 1 12 2 2 2

T

r r x r r x x x r r x r r x

x

0

1

2

3

cosh2 0 0 sinh20 1 0 00 0 1 0

sinh2 0 0 cosh2

xt txxxt t

Φ(Mr ) = Boost along z-axis

1

00r r

rM M

r

0 3 1 2 0 3cosh2 sinh 2 sinh 2 cosh 2 Tx t x t x x x t x t x

1cosh2

t tt e e Set tr e

2ztL x

2

00

iz

i

eR

e

2

cos sinsin cos

yR

2

00

tztt

eL

e

Every A SL(2,C) is continuously joined to I by a curve of matrices in SL(2,C)

Every matrix is conjugate (similar) to an upper triangular matrix.

1

0a b

A B Bc

It is similar to a diagonal matrix if the number of its independent eigenvectors is equal to its dimension.For a 2×2 matrix A

If A SL(2,C), then

acdet0a bc

1 det A

110

a bA B B

a

i.e.,

Let at be any continuous curve of non-zero complex numbers with a0 = 1 and a1 = a. Likewise for bt with b0 = 0 and b1 = b.

110

t tt

t

a bA B B

a

TC GC GA Lorentz transformation C satisfies

→ det 1C

Since elements with detC = -1 cannot be joined continuously to those with detC = +1, the former is outside the range of Φ.

A pure boost cannot change the sign of x0.

Elements with detC = +1 form a subgroup L0 called the proper Lorentz group.

A change of the sign of x0 requires reflection so that detC = -1→ L0 preserves the forward & backward light cones.

To be shown:• Φ[ SL(2,C) ] = L0

• Φ[ SU(2) ] = SO(3)

LemmaEvery B L0 can be written as

1 2zuB R L R R1 & R2 are spatial

rotations. Luz a boost

along the z-axis.Proof: 0

0

1 xB B

e0 x

220 1x x

Let S be a rotation that aligns x with the z-axis:

00

3

xSB

ex e 0 0 0

Tx x 0

0

0~

0x

Xx

xx

Let 1

00r

rM

r

with

120 0r x x

x x

0 ~r r rM SB M X M e

0 0x x x x

1 00 1

~ 1 0 0 0 T

2

0

20

0

0

r x

r x

x

x

0 0rM SB e e → 2rM SB R 112rB S M R

(boost along z )

Φ[ SL(2,C) ] = L0