Master Dissertation

〈y, y〉 = det σ(y) = det A det σ(x) det A ∗ = det σ(x) = 〈x, x〉.
Clearly the parallelogram identity holds for the Lorentz metric 2 , thus
〈x, y〉 =
= det σ(x) + det σ(y) − det σ(x) − det σ(y)
〈x + y, x + y〉 − 〈x, x〉 − 〈y, y〉
2
.
2
From this we see that 〈ΛAx, ΛAy〉 = 〈x, y〉, thus it is a Lorentz
transformation.
Since
fixme: make sure the following is correct
Hence
ΛB(x) = y ⇔ σ(y) = Bσ(x)B ∗ and ΛA(y) = z ⇔ σ(z) = Aσ(y)A ∗ ,
ΛAB(x) = z ⇔ ABσ(x)B ∗ A ∗ = Aσ(y) = σ(z).
ΛAB = ΛAΛB.
In this way we can construct a representation Λ : A ↦→ ΛA of SL(2, C), the
group of all complex 2 × 2 matrices with determinant 1, onto L ↑
+ . We call
SL(2, C) the spinor representation. The vectors in the representation space
C2 are called spinors.
Note that the kernel of the representation Λ is
ker Λ = Λ −1 ({I4}) = {A ∈ SL(2, C)|AHA ∗ for all H ∈ H(2)}.
Letting H = I2 we see that A ∈ ker Λ must be unitary and A ∈ ker Λ if and
only if AH = HA for all H ∈ H(2). Hence A = −I2 or A = I2 and
L ↑
+ ∼ = SL(2)/{I2, −I2}.
Hence Λ(A) = Λ(−A) and SL(2, C) is a two-valued representation of L ↑
+ .
2 The proof is exactly the same as the usual one for inner product spaces.
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