Lorentz indices to label rotation irreducible representations

Question

Consider the $A_\mu\in\left(\frac{1}{2},\frac{1}{2}\right)$, the vector representation of the restricted Lorentz group. One can decompose this vector under spatial rotations as $A_\mu\in 0\oplus 1$ where $A_0$ transforms as a 3D scalar and $A_i$ transforms as a 3D vector. If say we had $\psi_a\in\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)$ a Dirac spinor, under rotations it transforms as $\psi_a=\frac{1}{2}_L\oplus\frac{1}{2}_R$ where $a=1,2$ will be the left part and $a=3,4$ will be the right part.

The question I have is when we consider higher representations, say a $\frac{3}{2}$ particle, $\psi_{\mu a}\in\left(\frac{1}{2},\frac{1}{2}\right)\otimes\left[\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)\right]$. It is prety clear that under rotations it transforms as $\psi_{\mu a}\in\left(\frac{1}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}\right)_L\oplus\left(\frac{1}{2}\oplus\frac{1}{2}\oplus\frac{3}{2}\right)_R$. However I don't see how I can distribute the indices $\mu$ and $a$ under rotations.

I am also interested in the general concept.

$\textbf{Edit: (to give more details for what my question is)}$

If we take the scalar part of the $\left(\frac{1}{2},\frac{1}{2}\right)=0\oplus1$, and multiply it by the the Dirac spinor we get $\frac{1}{2}_L\oplus\frac{1}{2}_R$. So I would say the part $\psi_{0a}$ transforms as $\frac{1}{2}_L\oplus\frac{1}{2}_R$.

What's left is $\psi_{ia}$ which transforms as $\left(\frac{1}{2}\oplus\frac{3}{2}\right)_L\oplus\left(\frac{1}{2}\oplus\frac{3}{2}\right)_R$. My question is: What part of the indices $i,a$ transforms as $\left(\frac{3}{2}\right)_L\oplus\left(\frac{3}{2}\right)_R$?

Answer 1

This is a close duplicate.

$\psi_{\mu~\alpha}$, in $$\left(\frac{1}{2},\frac{1}{2}\right)\otimes\left[\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)\right],$$ has 16 components; so you project out the spinor piece, $\gamma\cdot \psi_\alpha =0$, $ \left[\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)\right]$,which removes 4 components, leaving you with 12: $$ \left[\left({1},\frac{1}{2}\right)\oplus\left(\frac{1}{2},{1} \right)\right]. $$

However, 4 of these remaining components are gauge (local susy) degrees of freedom, $\partial_\mu \epsilon_\alpha$, that do not enter the R-S action, so they are projected out as in vector gauge theories, leaving you with just 8, $$ \left[\left(\frac{3}{2},0\right)\oplus\left(0,\frac{3}{2}\right)\right],$$ a left-handed and a right-handed quartet.

NB On-shell, one may go on: the absence of $\partial_0 \psi_{0~\alpha}$ from the action allows fixing $\psi_0=0$, reducing the 8 components to 4, and the Majorana condition reduces them to only 2 d.o.f., the extreme polarization states, analogous to photons, explained in Freedman & van Proeyen.