- #1
gentsagree
- 93
- 1
In the context of the homomorphism between SL(2,C) and SO(3,1), I have that
\textbf{x}=\overline{\sigma}_{\mu}x^{\mu}
x^{\mu}=\frac{1}{2}tr(\sigma^{\mu}\textbf{x})
give the explicit form of the isomorphism, where \textbf{x} is a 2x2 matrix of SL(2,C) and x^{\mu} a 4-vector of SO(3,1).
Considering the linear map (the spinor map)
\textbf{x}\rightarrow\textbf{x}'=A\textbf{x}A^{\dagger}
one can show that the 4-vectors on the SO(3,1) side are also linearly related by
x'^{\mu}=\phi(A)^{\mu}_{\nu}x^{\nu}
where it is easy to show that
\phi(A)^{\mu}_{\nu}=\frac{1}{2}tr(\sigma^{\mu}A\overline{\sigma}_{\nu}A^{\dagger})
I understand all this, but I want to prove that \phi(AB)=\phi(A)\phi(B). How would I go about doing this? I tried a few things but not very successfully.
\textbf{x}=\overline{\sigma}_{\mu}x^{\mu}
x^{\mu}=\frac{1}{2}tr(\sigma^{\mu}\textbf{x})
give the explicit form of the isomorphism, where \textbf{x} is a 2x2 matrix of SL(2,C) and x^{\mu} a 4-vector of SO(3,1).
Considering the linear map (the spinor map)
\textbf{x}\rightarrow\textbf{x}'=A\textbf{x}A^{\dagger}
one can show that the 4-vectors on the SO(3,1) side are also linearly related by
x'^{\mu}=\phi(A)^{\mu}_{\nu}x^{\nu}
where it is easy to show that
\phi(A)^{\mu}_{\nu}=\frac{1}{2}tr(\sigma^{\mu}A\overline{\sigma}_{\nu}A^{\dagger})
I understand all this, but I want to prove that \phi(AB)=\phi(A)\phi(B). How would I go about doing this? I tried a few things but not very successfully.