I Homomorphism SL(2,C) with restricted Lorentz

[gentsagree]

We have that

\Lambda^{\mu}_{\nu} = \frac{1}{2}Tr(\overline{\sigma}^{\mu}A\sigma^{\nu}A^{\dagger})

I would like to make sense of the statement that this is a homomorphism because the correspondence above is preserved under multiplication.

Can someone clarify how I could see this?

 

[gentsagree]

Nevermind. While the tool needed was clear from the beginning, i.e. an identity about products of traces, I was completely oblivious to the existence of one. It turns out that

<br /> \sum_{\mu}Tr(G\sigma_{\mu})Tr(\overline{\sigma}^{\mu}H) = 2 Tr(GH)<br />

While I still don't know the proof of this, this is the correct answer.

 

[ChrisVer]

A map of the form:
\phi : A \rightarrow B
a \rightarrow b=\phi(a)
is roughly speaking an homomorphism if for elements x, y \in A the \phi(x) \cdot \phi(y) = \phi(x*y)
and that's why the comment about multiplication.

As for the proof of your last equation in post2, if I recall well you can prove it better by writting the traces with summed indices:
G_{ai} \sigma_{ia}^\mu \bar{\sigma}^\mu_{jb} H_{bj}
and then using seperately the \sigma^0, \sigma^k's and use their completeness relation (generalization of the : https://en.wikipedia.org/wiki/Pauli_matrices#Completeness_relation)... but I don't really remember the complete proof...

 

[ChrisVer]

yup it's straightforward (5-6 single-expression lines depending on how rigorous you want to be) with the completeness relation I gave you (obviously I just tried it )...

So I guess it would be a better "exercise" for you to prove the completeness relation for the pauli matrices that is given in the wiki article I sent you.