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I Homomorphism SL(2,C) with restricted Lorentz

  1. Apr 18, 2016 #1
    We have that

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

    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?
     
  2. jcsd
  3. Apr 18, 2016 #2
    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

    [tex]
    \sum_{\mu}Tr(G\sigma_{\mu})Tr(\overline{\sigma}^{\mu}H) = 2 Tr(GH)
    [/tex]

    While I still don't know the proof of this, this is the correct answer.
     
  4. Apr 18, 2016 #3

    ChrisVer

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    A map of the form:
    [itex] \phi : A \rightarrow B[/itex]
    [itex] a \rightarrow b=\phi(a)[/itex]
    is roughly speaking an homomorphism if for elements [itex]x, y \in A[/itex] the [itex] \phi(x) \cdot \phi(y) = \phi(x*y)[/itex]
    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:
    [itex] G_{ai} \sigma_{ia}^\mu \bar{\sigma}^\mu_{jb} H_{bj}[/itex]
    and then using seperately the [itex]\sigma^0, \sigma^k[/itex]'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...
     
    Last edited: Apr 18, 2016
  5. Apr 18, 2016 #4

    ChrisVer

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    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 o0))...

    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.
     
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