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Suppose I have the Gamma matrices given by $\gamma ^\mu$. Under some unitary transformation $U$ I can consider $\tilde{\gamma^{\mu}} = U\gamma ^\mu U^\dagger$. Since I have:

$$ \{ \gamma^\mu, \gamma^\nu \} = 2g^{\mu \nu}\textbf{I} $$

I can compute $\{\tilde{\gamma^{\mu}},\tilde{\gamma^{\nu}}\} $ and show that

$$\{\tilde{\gamma^{\mu}},\tilde{\gamma^{\nu}}\} = U \{ \gamma^\mu, \gamma^\nu \} U^{\dagger} = 2U g^{\mu \nu} U^{\dagger} $$

My question is the following. If I consider $\tilde{\gamma^{\mu}}$ to be a different representation of the gamma matrices, do I take

$$\tilde{g^{\mu}} = U g^{\mu \nu} U^{\dagger}$$ as a different representation of the metric, or as a transformed version of the metric or at least what happens to the metric?

In my QFT class, my instructor said one can take

$$ U g^{\mu \nu} U^{\dagger} = g^{\mu \nu} UU^{\dagger} = g^{\mu \nu}$$

in flat space time when $g^{\mu \nu}$ is just the Minkowski metric. But even in this case in non-Cartesian coordinates, this metric depends on coordinates so I'm not sure how to understand the transformation of the metric.

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  • $\begingroup$ You need to consider the gamma matrices in an orthonormal (not necessarily coordinate) basis. $\endgroup$
    – octonion
    Oct 30, 2021 at 11:01

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The Gamma matrices are on $4$-dimensional space, but this is not the $4$-dimensional spacetime; it's a different space entirely. (For $n$-dimensional spacetime, the Gamma matrices would be on a space of dimension $2^{\lfloor n/2\rfloor}$, if memory serves.) This distinction addresses all your concerns.

Let's use Roman indices for this unphysical space. Write $\bar{\gamma}^\mu:=U\gamma^\mu$ so $\{\gamma^\mu,\,\gamma^\nu\}_{ab}=2g^{\mu\nu}\delta_{ab}$ and$$\{\bar{\gamma}^\mu,\,\bar{\gamma}^\nu\}_{ab}=U_{ac}U_{de}(\gamma^\mu_{cd}\gamma^\nu_{eb}+\gamma^\nu_{cd}\gamma^\mu_{eb}),$$which for suitably chosen $U$ will be $2\bar{g}^{\mu\nu}\delta_{ab}$ (say).

Your instructor's point would be more carefully written as $Ug^{\mu\nu}\mathbf{I}U^\dagger=g^{\mu\nu}\mathbf{I}$, or even more explicitly $U_{ac}g^{\mu\nu}\delta_{cd}U^\dagger_{db}=g^{\mu\nu}\delta_{ab}$. As you can see, the $g^{\mu\nu}$ factor is just a spectator here; this idea works for any unitary $U$ on the Roman space.

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  • $\begingroup$ Do the Roman indices $a,b$ refer to explicit matrix elements of the gamma matrices in some representation? So that would mean my unitary transformation affects the matrix elements (which don't necessarily depend on coordinates) but since the metric is a tensor in spacetime, that is unaffected? $\endgroup$
    – newtothis
    Oct 30, 2021 at 11:37
  • $\begingroup$ @newtothis That's right, yes. $\endgroup$
    – J.G.
    Oct 30, 2021 at 12:28

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