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Generators for Lorentz group represented in spinor space are $M_{\mu\nu} = \frac12 \sigma_{\mu\nu}$ where sigma matrices are defined to be: $$\sigma_{\mu\nu} = \frac i2 [\gamma_{\mu},\gamma_{\nu}]$$ and is obviously anti-symmetric.
But usually when deriving the form of generators, before knowing the above (namely that $M_{\mu\nu}$ is proportional to commutator of gamma matrices), we assume the solution form, knowing that $M_{\mu\nu}$ is an anti-symmetric tensor and just verify the assumption.

Question: How do we know that $M_{\mu\nu}$ is anti-symmetric?

I would suppose it comes from the way we define tensor $M_{\mu\nu}$ and the relation to $\omega^{\mu\nu}$. $\omega^{\mu\nu}$ is anti-symmetric, so it would be silly and impractical to define $M_{\mu\nu}$ anything else then anti-symmetric. Is this again just physics notation thing and a consequence of us not wanting to write $$S(\Lambda) = exp(-\frac i2 \omega^{\mu\nu}M_{\mu\nu})$$ with single index on parameters and generators, but instead using two? (Which probably later enables easier use of generators as observables or something like that...)

Qmechanic
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lojle
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1 Answers1

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$SO(n, m)$ is an example of a Lie group. Their infinitesimal approximation or linearisation is called a Lie algebra. This is always an antisymmetric algebra. Their elements are called generators of the Lie group.

Mozibur Ullah
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