- #1
gentsagree
- 96
- 1
In the context of the homomorphism between SL(2,C) and SO(3,1), I have that
[tex]\textbf{x}=\overline{\sigma}_{\mu}x^{\mu}[/tex]
[tex]x^{\mu}=\frac{1}{2}tr(\sigma^{\mu}\textbf{x})[/tex]
give the explicit form of the isomorphism, where [itex]\textbf{x}[/itex] is a 2x2 matrix of SL(2,C) and [itex]x^{\mu}[/itex] a 4-vector of SO(3,1).
Considering the linear map (the spinor map)
[tex]\textbf{x}\rightarrow\textbf{x}'=A\textbf{x}A^{\dagger}[/tex]
one can show that the 4-vectors on the SO(3,1) side are also linearly related by
[tex]x'^{\mu}=\phi(A)^{\mu}_{\nu}x^{\nu}[/tex]
where it is easy to show that
[tex]\phi(A)^{\mu}_{\nu}=\frac{1}{2}tr(\sigma^{\mu}A\overline{\sigma}_{\nu}A^{\dagger})[/tex]
I understand all this, but I want to prove that [itex]\phi(AB)=\phi(A)\phi(B)[/itex]. How would I go about doing this? I tried a few things but not very successfully.
[tex]\textbf{x}=\overline{\sigma}_{\mu}x^{\mu}[/tex]
[tex]x^{\mu}=\frac{1}{2}tr(\sigma^{\mu}\textbf{x})[/tex]
give the explicit form of the isomorphism, where [itex]\textbf{x}[/itex] is a 2x2 matrix of SL(2,C) and [itex]x^{\mu}[/itex] a 4-vector of SO(3,1).
Considering the linear map (the spinor map)
[tex]\textbf{x}\rightarrow\textbf{x}'=A\textbf{x}A^{\dagger}[/tex]
one can show that the 4-vectors on the SO(3,1) side are also linearly related by
[tex]x'^{\mu}=\phi(A)^{\mu}_{\nu}x^{\nu}[/tex]
where it is easy to show that
[tex]\phi(A)^{\mu}_{\nu}=\frac{1}{2}tr(\sigma^{\mu}A\overline{\sigma}_{\nu}A^{\dagger})[/tex]
I understand all this, but I want to prove that [itex]\phi(AB)=\phi(A)\phi(B)[/itex]. How would I go about doing this? I tried a few things but not very successfully.