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I have a question regarding the Lie-Algebra of the SO(3,1), i.e. the Lorentz group. The generators are denoted by [itex] M_{\mu \nu} [/itex] with [itex] M_{\mu \nu} = - M_{\nu \mu} [/itex] and fulfill the following commutation relations

[itex] [M_{\mu \nu}, M_{\rho \sigma}] = g_{\mu \rho} M_{\nu \rho} + g_{\nu \sigma} M_{\mu \rho} - g_{\mu \sigma} M_{\nu \rho} - g_{\nu \rho} M_{\mu \sigma}[/itex],

where [itex] g_{\mu \nu} [/itex] are metric coefficients.

I can now construct from these generators the operators

[itex] \mathbf{J} = (M_{23},M_{31},M_{12}),\ \mathbf{K} = (M_{01},M_{02},M_{03})[/itex],

which fulfill the relations

[itex] [J_i,J_j] = i \epsilon_{ijk} J_k,\ [K_i,K_j]= -i \epsilon_{ijk} J_k,\ [J_i,K_j]=i \epsilon_{ijk} K_k [/itex].

In a final step, I define

[itex] \mathbf{J}^\prime = \frac{1}{2} ( \mathbf{J} + i \mathbf{K}),\ \mathbf{K}^\prime = \frac{1}{2} (\mathbf{J} - i \mathbf{K})[/itex]

with the commutation relations

[itex] [J_i^\prime,J_j^\prime] = i \epsilon_{ijk} J_k^\prime,\ [K_i^\prime,K_j^\prime]= i \epsilon_{ijk} K_k^\prime,\ [J_i^\prime, K_j^\prime]=0[/itex].

This means that I can construct generators that fulfill exactly the commutation relations of the Lie-Algebra of the SU(2). What does this mean for the group SO(3,1)? Can I deduce from this something like SU(2) x SU(2) is a representation of SO(3,1)?

Greetings, Syrius

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# Lie-Algebra of the SO(3,1)

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