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1.2. Homomorphism: SL(2,C) ~ L 0 Let Φ be a map between groups G 1 & G 2, i.e., Φ: G 1 → G 2. Φ is a homomorphism if Example: G 1 = (Z,+), G 2 = C 2, Φ(n)

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Presentation on theme: "1.2. Homomorphism: SL(2,C) ~ L 0 Let Φ be a map between groups G 1 & G 2, i.e., Φ: G 1 → G 2. Φ is a homomorphism if Example: G 1 = (Z,+), G 2 = C 2, Φ(n)"— Presentation transcript:

1 1.2. Homomorphism: SL(2,C) ~ L 0 Let Φ be a map between groups G 1 & G 2, i.e., Φ: G 1 → G 2. Φ is a homomorphism if Example: G 1 = (Z,+), G 2 = C 2, Φ(n) = (-1) n  n  Z. Example: G 1 = (Z,+), G 2 = C k, Φ(n) = exp(2πi n/k)  n  Z. → Multiplication in C k ~ addition modulo k in Z

2 Let M be the Minkowski space, i.e., R 4 with the Lorentz metric →(c = 1) A Lorentz transformation B preserves the Lorentz metric: The set of all B forms the Lorentz group L. Rewrite x as a 2×2 self-adjoint (Hermitian) matrix: Note: all eigenvalues λ j of a Hermitian matrix must be real.

3 Pauli matrices The action of a 2×2 matrix A on X is defined as (Hermitian) A  SL(2,C) → → Φ is a homomorphism

4 Let C be a Lorentz transformation. By definition Let Metric tensor  x x →

5 Let A  SU(2) → A -1 = A + →A I A + = I Let→ Let C = Φ(A) be the corresponding Lorentz transformation: i.e., C is a spatial rotation → The subspace orthogonal to e 0 is For all → O(3) is the subgroup of L defined by  Φ: SU(2) → O(3) Φ: SL(2,C) → L 0

6 Example 1 → Φ(U Θ ) is a rotation of angle 2Θ about the x 3 axis. Θ = π corresponds to 2π rotation in 3-space.

7 Example 2 Φ(V Θ ) = rotation of angle 2α about x 2 -axis

8 Example 3 Φ(M r ) = Boost along z-axis Set

9 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. 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 i.e., Let a t be any continuous curve of non-zero complex numbers with a 0 = 1 and a 1 = a. Likewise for b t with b 0 = 0 and b 1 = b.

10 A Lorentz transformation C satisfies→ 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 x 0. Elements with detC = +1 form a subgroup L 0 called the proper Lorentz group. A change of the sign of x 0 requires reflection so that detC = -1 → L 0 preserves the forward & backward light cones. To be shown: Φ[ SL(2,C) ] = L 0 Φ[ SU(2) ] = SO(3)

11 Lemma Every B  L 0 can be written as R 1 & R 2 are spatial rotations. L u z a boost along the z-axis. Proof: Let S be a rotation that aligns x with the z-axis: Letwith → (boost along z ) Φ[ SL(2,C) ] = L 0

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