# Homomorphism SL(2,C) with restricted Lorentz

• I
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?

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

$$\sum_{\mu}Tr(G\sigma_{\mu})Tr(\overline{\sigma}^{\mu}H) = 2 Tr(GH)$$

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

ChrisVer
Gold Member
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...

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ChrisVer
Gold Member
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.