# How to prove commutation relation between charge and current in current algebra?

I am reading Gauge Theory of Elementary Particle Physics by Tapei Cheng and Lingfong Li. Proceeding equation 5.54, there is a statement which says

Then from Lorentz covariance, we can include the other components of the currents.

One would then have

$$\left[ Q^a \left( t \right) , J_{\mu}^b \left( \vec{x} , t \right) \right] = \mathrm{i} C^{a b c} J_{\mu}^c \left( \vec{x} , t \right).$$

I am having a hard time trying to figure out how would one be able to derive the equation above from Lorentz covariance and the following

$$\left[ Q^a \left( t \right) , J_0^b \left( \vec{x} , t \right) \right] = \mathrm{i} C^{a b c} J_0^c \left( \vec{x} , t \right).$$

Can someone give me a hint on this?

If your current is a Lorentz tensor, $$\Lambda^{0}_{\mu} J_{0}(\vec{x},t) = J'_{\mu}(\vec{x}',t').$$ and the charge $Q(t)$ is a Lorentz scalar: $$Q^{'}_a(t') = Q_a (t),$$ we have that $$[Q^{'a} (t') , J_{\mu}^{'b} ( \vec{x}',t')] = i C^{abc} J^{'c}_{\mu}(\vec{x},t)$$ can be written by a Lorentz transformation as $$[Q^{a} (t), \Lambda^{0}_{\mu} J_0^{b} (\vec{x} ,t) ] = \Lambda^{0}_{\mu} [Q^{a} (t), J_0^{b} (\vec{x} ,t) ] = i C^{abc} \Lambda^{0}_{\mu} J_0^{c}(\vec{x},t).$$ Therefore, it follows that if $$[Q^{a} (t), J_0^{b} (\vec{x} ,t) ] = i C^{abc} J_0^{c}(\vec{x},t).$$ is true, then $[Q^{'a} (t') , J_{\mu}^b ( \vec{x}',t')] = i C^{abc} J^{'c}_{\mu}(\vec{x},t)$ must also be true being derived by linear transformation from the previous one.