Properties of tensor $\gamma^\mu\gamma^\nu\gamma^\lambda$

Question

As the title suggests, I am trying to derive some properties of the tensor $\Gamma^{\mu\nu\lambda}=\gamma^\mu\gamma^\nu\gamma^\lambda$. My motivation is that $\bar{\psi}\Gamma^{\mu\nu\lambda}\psi$ should not create any more Dirac bilinears, meaning that it should be reducible to the existing one $1$, $\gamma^\mu$, $\gamma^\mu\gamma^\nu$.

Using anticommutation relation of gamma-matrices I managed to prove that

$\Gamma^{\mu\nu\lambda}=2\eta^{\mu\nu}\gamma^\lambda-\Gamma^{\nu\mu\lambda}$

And similar identities. Also, I derived that

$tr\,\Gamma^{\mu\nu\lambda}=0$

$\Gamma^{\mu\,\nu}_{\,\mu}=D\gamma^{\nu}$

How do I reduce the tensor and why doesn't it create more Dirac bilinears?

Answer 1

To analyze $\Gamma^{\mu \nu \rho}$, it is useful to think about different cases. When two of the three indices are equal, it should reduce to a single gamma matrix. When the three indices are different, it should reduce to a completely antisymmetric tensor. Thus, it is natural to expect something of the form $$ \Gamma^{\mu \nu \rho} \sim \eta \gamma + \epsilon^{\mu \nu \rho \sigma} M_\sigma $$ Going in this direction and checking explicitly some cases, you'll find $$ \Gamma^{\mu \nu \rho} = \eta^{\mu \nu}\gamma^\rho + \eta^{\nu \rho} \gamma^\mu - \eta^{\mu \rho} \gamma^\nu - i \epsilon^{\sigma \mu \nu \rho} \gamma_\sigma \gamma^5 $$ as mentioned in Wikipedia.