Do local Lorentz rotations (see below definition) actually transform the Dirac Gamma matrices? If so, how can they collude with coordinate transformations to make the Gamma matrices $\gamma_a$ invariant?

Conceptually speaking, Gamma matrices $\gamma_a$ are relics of the vierbein/tetrad 1-form $e$ of curved spacetime: $$ e = \sum_{a, \mu} e^{a}_{\mu}\gamma_a dx^{\mu} $$ and metric is $$g_{\mu\nu} = \sum_{a, b}e^{a}_{\mu}e^{b}_{\nu}\eta_{ab}$$

General relativity has two independent symmetries, namely:

  • Local Lorentz symmetry: which is about transformation at the same space-time point and the "gauge symmetry" is the local Lorentz symmetry. For example, a fermion $\psi$ would transform as $\psi \rightarrow R^{1/2}\psi$ and the Dirac Gamma matrices (and hence the vierbein/tetrad $e$) would transform as $$\gamma_a \rightarrow R^{1/2}(x)\gamma_a R^{-1/2}(x)$$ while the underlying coordinate $x_{\mu}$ stays put.

  • Diffeomorphism symmetry, which does involve space-time coordinate transformation $x_\mu \to x'_\mu$. Since the vierbein/tetrad $e$ is a 1-form dependent on $dx^{\mu}$, it transforms accordingly.

In special relativity, the metric is fixed to Minkowskian metric, or in terms of the vierbein/tetrad: $$ e = \sum_{a, \mu} \delta^{a}_{\mu}\gamma_a dx^{\mu} = \sum_{\mu}\gamma_{\mu} dx^{\mu} $$

which effectively breaks both the Local Lorentz symmetry and the diffeomorphism symmetry.

However, there is a residual symmetry, i.e. the global Lorentz symmetry, which combines the global portion of the local Lorentz symmetry($\gamma_{\mu} \rightarrow R^{1/2}\gamma_\mu R^{-1/2}$) and the global rotation portion of the diffeomorphism symmetry ($dx^{\mu} \rightarrow \sum_{\nu}\Lambda^\mu_\nu dx^\nu$), so that $e$ (and hence metric $g_{\mu\nu}$) is invariant $$ e = \sum_{\mu}\gamma_{\mu} dx^{\mu} \rightarrow \sum_{\mu,\nu}(R^{1/2}\gamma_\mu R^{-1/2}) (\Lambda^\mu_\nu dx^\nu) = \sum_{\nu}\gamma_{\nu} dx^{\nu} = e $$ or equivalently the combined transformation (forgetting about $dx^{\mu}$ now) is $$ \gamma_{\mu} \rightarrow \sum_{\mu}R^{1/2}\gamma_\mu R^{-1/2}\Lambda^\mu_\nu = \gamma_\nu $$ Since $\mu$ and $\nu$ are dummy indices, we regard the Gamma matrices (thus the Pauli matrices $\sigma_i$ as well) as Lorentz invariant as long as the dummy indices are properly contracted in the Lagrangian.

  • $\begingroup$ The gamma matrices have nothing to do with tetrads. Your first formula in particular disagrees with what is usually meant by "gamma matrices". $\endgroup$ – alexarvanitakis Jun 1 at 16:30
  • $\begingroup$ The gamma matrices are square roots of the metric as Dirac first stumbled upon. And tetrads are also square roots of the metric. You will see the light if you try connecting the dots... $\endgroup$ – MadMax Jun 3 at 20:28
  • $\begingroup$ Somebody tell my students... $\endgroup$ – alexarvanitakis Jun 4 at 16:31
  • $\begingroup$ It's widely known in the field for many years when Riemann-Cartan geometries are concerned. See for example arxiv.org/abs/gr-qc/9403026. $\endgroup$ – MadMax Jun 4 at 19:16
  • $\begingroup$ No. The formula (2) there, which is the one currently confusing you, defines Clifford-algebra-valued forms. So in their (2), upon picking a representation for the Clifford algebra, we find $\gamma_a$ carries two additional spinor indices (omitted in the notation of that paper). So that $\gamma_a$ is a Gamma matrix in the conventional sense. Gamma matrices are elements of the Clifford algebra --- nothing to do with tetrads a priori, although of course they can be simultaneously relevant. $\endgroup$ – alexarvanitakis Jun 4 at 23:20

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