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While reading "Four lectures on Poincaré gauge field theory" (available at RG) the authors present a relationship between a tetrad $e^i_{\;\gamma}$ (with Latin indices coordinates, Greek indices anholonomic) and a generator of a (local) Lorentz transformation $f_{\alpha\beta}$: $$\left[f_{\alpha\beta},e^i_{\;\gamma}\right] = \eta_{\gamma[\alpha}e^i_{\;\beta]} = \frac{1}{2}\left(\eta_{\gamma\alpha}e^i_{\;\beta} - \eta_{\gamma\beta}e^i_{\;\alpha}\right)$$ (their equation 2.7) where $\eta$ is the Minkowski metric.

Would anyone be able to help me understand this: I have been envisaging $f_{\alpha\beta}$ as acting on elements of a spin space, so would expect it to immediately commute with the tetrad, which acts on elements of the 'physical' vector space, but this appears not to be the case. I'm also not sure where the metric has arrived from: presumably the commutation relations among the Lorentz generators, but this seems a stretch. Any help would be appreciated!

Qmechanic
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AnotherShruggingPhysicist
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  • Are you asking why the vector representation of the Lorentz generators is $(f_{\alpha\beta})^{\gamma}{}{\delta}=\frac{1}{2}(\delta{\alpha}^{\gamma}\eta_{\beta\delta}-\delta_{\beta}^{\gamma}\eta_{\alpha\delta})$, or why the vielbein transforms in the vector representation of the Lorentz group? – SigmaAlpha Jul 23 '20 at 00:31
  • The former: perhaps I'm getting confused, but I was expecting $f_{\alpha\beta}$ to be an element of an arbitrary spin representation, while your expression appears to be one particular representation? – AnotherShruggingPhysicist Jul 23 '20 at 08:01
  • You are correct, your $f_{\alpha\beta}$ are the Lorentz generators in an abitrary representation. The expression I wrote above is the vector representation of $f_{\alpha\beta}$, so a more appropriate notation is $f_{\alpha\beta}^{(\text{v})}$. For an explanation of why they take that explicit form in the vector rep see, e.g., https://physics.stackexchange.com/questions/28535/lie-bracket-for-lie-algebra-of-son-m (its probably in some intro QFT textbooks too). The proof I know relies on the isomorphism between the (connected) Lorentz group and SL$(2,\mathbb{C})/\mathbb{Z}_2$. – SigmaAlpha Jul 23 '20 at 11:31
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    If you are aware of the fact that an infinitesimal Lorentz transformation takes the form $\Lambda(\omega)^{a}{}{b}=\delta^{a}{}{b}+\omega^{a}{}{b}$ where $\omega{ab}=-\omega_{ba}$, then the proof is quick from that point. – SigmaAlpha Jul 23 '20 at 11:36
  • I've seen that derivation before in a different context, but I guess I was hoping to avoid explicit representations. Or did we pick a representation when we decided to look at the vielbein's commutator? That might well explain my misunderstanding if that's the correct logic – AnotherShruggingPhysicist Jul 23 '20 at 13:44
  • The vielbein transforms in the vector representation, so the Lorentz generators take that explicit form when they act on it. Perhaps what is confusing you is the following observation: $[f_{\alpha\beta},e^i{}{\gamma}]=f{\alpha\beta}e^i{}_{\gamma}$ where the Lorentz generators act only on the vielbein (and not to the right of it). – SigmaAlpha Jul 23 '20 at 13:54
  • Thanks, this is really helping: one last question; if I had a spinor $\psi$, and wanted to evaluate $[f_{\alpha\beta},e^i_{;\gamma}]\psi$, would that be $(f_{\alpha\beta}e^i_{;\gamma})\psi - e^i_{;\gamma}(f_{\alpha\beta}\psi)$ where in the first term the generator is in the vector representation and the second term it's in the (relevant) spin representation? – AnotherShruggingPhysicist Jul 23 '20 at 15:11
  • Nope, the commutator acts on everything to the right, so: $$[f_{\alpha\beta},e^{i}{}{\gamma}]\psi=f{\alpha\beta}(e^{i}{}{\gamma})\psi+e^{i}{}{\gamma}f_{\alpha\beta}(\psi)+e^{i}{}{\gamma}\psi f{\alpha\beta}-e^{i}{}{\gamma}f{\alpha\beta}(\psi)-e^{i}{}{\gamma}\psi f{\alpha\beta}=f_{\alpha\beta}(e^{i}{}{\gamma})\psi$$ The third and last terms in the second equality are unnecessary if there is nothing after $e^{i}{}{\gamma}\psi$ in the expression you are computing. – SigmaAlpha Jul 23 '20 at 23:05
  • Thanks @SigmaAlpha, your help has been much appreciated! I think I now understand where I was going wrong (in my interpretation of $f_{\alpha\beta}$) so I'll have another think with the very helpful knowledge you've imparted – AnotherShruggingPhysicist Jul 24 '20 at 09:10

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