Is the Lorentz Boost Generator Commutator Zero?

han
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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.
 
Last edited:
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han said:
[....]

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.
 
jambaugh said:
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.
 
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