Lorentz boost generator commutator

[han]

Homework Statement

Show that $[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=0$ where $M_{\mu\nu}$ is a Lorentz boost generator

Relevant Equations

The commutation relation of $M_{\mu\nu}$ is given: $$[M_{\rho \sigma},M_{\alpha\beta}]=i(g_{\rho\beta}M_{\sigma\alpha}+g_{\sigma\alpha}M_{\rho\beta}-g_{\rho\alpha}M_{\sigma\beta}-g_{\sigma\beta}M_{\rho\alpha}).$$

Using above formula, I could calculate the given commutator.
$$
[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\rho\sigma},M_{\alpha\beta}]M_{\mu \nu})
$$
(because $\epsilon^{\mu\nu\rho\sigma}=\epsilon^{\rho\sigma\mu\nu}$, $(\mu\nu)\leftrightarrow(\rho\sigma)$ preserves the result in the 2nd term)
$$
=i\epsilon^{\mu\nu\rho\sigma}(g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})+g_{\sigma\alpha}(M_{\mu\nu}M_{\rho\beta}+M_{\rho\beta}M_{\mu\nu})-g_{\rho\alpha}(M_{\mu\nu}M_{\sigma\beta}+M_{\sigma\beta}M_{\mu\nu})-g_{\sigma\beta}(M_{\mu\nu}M_{\rho\alpha}+M_{\rho\alpha}M_{\mu\nu}))
$$

And my calculation stuck here. I could not find any clue that the terms in above formula cancel each other.

I personally checked that for a specific example like taking $\alpha=1, \beta=2$, the commutator is indeed zero.

It feels like any sign in the 3rd or 4th term is miscalculated and symmetricity in $\rho$ and $\sigma$ combines with the antisymmetric tensor and give the result zero, but I could not find where did I make a mistake on the signs.

Additionally, the term $\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma}$ is not explicity zero for example on spinor, where $M_{\mu \nu}=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]$, you can check that the given expression is proportional to $\gamma^5$.

Edit: I found out by directly calculating that $$
\epsilon^{\mu\nu\rho\sigma}g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})$$
term itself is already zero. Again for example when $\alpha=1,\beta=2$ case,
$$
\begin{align}
\epsilon^{\mu\nu\rho\sigma}g_{\rho 2}(M_{\mu\nu}M_{\sigma 1}+M_{\sigma 1}M_{\mu\nu})\\ \nonumber
&=\epsilon^{\mu\nu 23}g_{22}(M_{\mu\nu}M_{31}+M_{31}M_{\mu\nu})+\epsilon^{\mu\nu 20}g_{22}(M_{\mu\nu}M_{01}+M_{01}M_{\mu\nu})\\ \nonumber
&=\epsilon^{0123}g_{22}(M_{01}M_{31}+M_{31}M_{01})+\epsilon^{1320}g_{22}(M_{13}M_{01}+M_{01}M_{13})\\ \nonumber
&=-(M_{01}M_{31}+M_{31}M_{01})+(M_{31}M_{01}+M_{01}M_{31})=0 \nonumber
\end{align}
$$
(Using $g_{00}=+1, g_{11}=g_{22}=g_{33}=-1$ convention)
So it's enough to show that the form $\epsilon^{\mu\nu\rho\sigma}g_{\rho\beta}(M_{\mu\nu}M_{\sigma\alpha}+M_{\sigma\alpha}M_{\mu\nu})$ is zero. But I have no clue how to show this formula is zero with algebraic steps, like switching indicies and cancel the terms out.

 

[jambaugh]

[....]

Using above formula, I could calculate the given commutator.
$$
[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\rho\sigma},M_{\alpha\beta}]M_{\mu \nu})
$$

I believe this should read...
$$
[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\mu\nu},M_{\alpha\beta}]M_{\rho \sigma})
$$
Lie brackets obey a Leibniz rule: [AB,C] = A[B,C]+[A,C]B.
In detail: [AB,C]=ABC-CAB = ABC-ACB+ACB-CAB.

 

[han]

I believe this should read...
$$
[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=i\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\mu\nu},M_{\alpha\beta}]M_{\rho \sigma})
$$
Lie brackets obey a Leibniz rule: [AB,C] = A[B,C]+[A,C]B.
In detail: [AB,C]=ABC-CAB = ABC-ACB+ACB-CAB.

You can check that
$$
[\epsilon^{\mu\nu\rho\sigma} M_{\mu \nu}M_{\rho\sigma},M_{\alpha\beta}]=\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\mu\nu},M_{\alpha\beta}]M_{\rho \sigma})=\epsilon^{\mu\nu\rho\sigma}(M_{\mu \nu}[M_{\rho\sigma},M_{\alpha\beta}]+[M_{\rho\sigma},M_{\alpha\beta}]M_{\mu \nu})
$$

Because
$$
\epsilon^{\mu\nu\rho\sigma}[M_{\mu\nu},M_{\alpha\beta}]M_{\rho \sigma})=\epsilon^{\rho\sigma\mu\nu}[M_{\rho \sigma},M_{\alpha\beta}]M_{\mu\nu})=\epsilon^{\mu\nu\rho\sigma}[M_{\rho \sigma},M_{\alpha\beta}]M_{\mu\nu})
$$
$\mu\nu\rho\sigma$ and $\rho\sigma\mu\nu$ both are even permutations.