Representation of the Gamma matrices and the metric

Question

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.

Answer 1

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.