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This question concerns the Dirac equation and the $4\times4$ $\gamma$-matrices. The task is to prove that a similarity transformation of the standard $\gamma$-matrix conserves the commutation relation

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

where $2g^{\mu\nu}$ is the metric tensor $\text{diag}(1,-1,-1,-1)$, and the similarity transformation is defined as

$$ \tilde{\gamma}^\mu = S \gamma^\mu S^\dagger, $$

and $S$ is a unitary matrix. I will write down the start of my proof to show where I stop. First of all, we can use that $S$ is unitary and show that $\gamma^\mu = S^\dagger\tilde{\gamma}^\mu S$, and insert this into the commutator. This leaves us, again using that $SS^\dagger = I$, with

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

which again gives us

$$ \{\tilde{\gamma}^\mu,\tilde{\gamma}^\nu\} = 2Sg^{\mu\nu}S^\dagger. $$

In order for the proof to hold, it requires that $g^{\mu\nu}$ and $S$ commute so that

$$ 2Sg^{\mu\nu}S^\dagger = 2g^{\mu\nu}SS^\dagger = 2g^{\mu\nu}. $$

So my question is: Do all unitary matrices commute with the metric tensor $g^{\mu\nu}$? If yes, how can I show this easily?

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    $\begingroup$ Hint to the question (v2): The commutation relation has an (implicit) identity matrix on the rhs: $\{\gamma^\mu,\gamma^\nu\} ~=~ 2g^{\mu\nu}~{\bf1}_{4\times 4}$. $\endgroup$ – Qmechanic May 2 '14 at 14:59
  • $\begingroup$ I don't see it, even with the hint.. $\endgroup$ – camzor00 May 2 '14 at 15:33
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    $\begingroup$ If you fix $\mu$ and $\nu$ $g^{\mu\nu}$ is a number, so it commutes with every matrix. $\endgroup$ – Valter Moretti May 2 '14 at 17:30

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