What is the Lorentz group representation for a general spin? Setup, as I understand things so far:
One way to think about where the spin of a quantum field comes from is that it is a consequence of the ways that different types of fields transform under Lorentz transformations.
The generator of a Lorentz transformation for a Dirac spinor field $\Psi$ is $S^{a b}=\frac 1 2 [\gamma^a, \gamma^b]$ (I use roman letters for the antisymmetric indices representing the rotation direction, and greek ones for the orientation of the field. Signature is $(+,-,-,-)$.)
For a vector field like $A^{\mu}$, it is $(M^{a b})^{\nu}_{\mu}=\frac 1 4 ( \eta^{a}_{\mu} \eta^{b \nu} - \eta^{b}_{\mu} \eta^{a \nu})$
For a tensor field like $g^{\mu \nu}$, it is two copies of $M$: $(M^{ab})^{\alpha}_{\mu} \otimes I+I \otimes (M^{ab})^{\beta}_{\nu}$
As one should expect, there is clearly a structure to these representations of increasing spin.
Now, although it isn't normally done, one could use the Clifford algebra relation:
$\{ \gamma^a, \gamma^b\}=2\eta^{a b}$
to express all of these generators in terms of increasingly complex products of gamma matrices.
Okay, with all that setup, my question is stated simply:


*

*Is there a general formula that one can derive that will give the $n/2$ spin representation in terms of the appropriate gamma matrix combination?

*As a particular example, what does a representation for a spin-3/2 field look like, as one might find in a supersymmetric theory?
 A: Okay, this wasn't as hard to find an answer to as I expected. However, any clarifications/ critisisms are welcome.
Weinberg basically gives the answer in section 5.6 of his QFT book:

A general tensor of rank N transforms as the direct product of N (1/2, 1/2) 
  four-vector representations. It can therefore be decomposed (by suitable 
  symmetrizations and antisymmetrizations and extracting traces) into irreducible terms (A,B) with A = N/2, N/2-1,... and В = N/2, N/2-1,... . In this way, we can construct any irreducible representation (A,B) for 
  which A + В is an integer. The spin representations, for which A + В is 
  half an odd integer, can similarly be constructed from the direct product 
  of these tensor representations and the Dirac representation $(1/2,0)\oplus(0,1/2)$.

So it does not appear that there is any simple way to unify the representations for spinor and vector generators, but one can construct the generator for arbitrary half-spins: it has as many copies of $M$ as needed, plus one copy of $S$ if it is a half-integer.
However, the converse is not true. That is, a spin-two field must transform with two copies of $M$, but doing so does not guarantee that an object is spin-two. A counterexample is the electromagnetic tensor $F$, which is certainly spin-one. The difference lies in the ability to equate the two generators as a result of the symmetry properties of the tensor, as elaborated in this answer.
Applying this to the spin 3-2 field, we expect it to have a rotation generator that looks schematically like $S \otimes I+ I \otimes M$. And indeed this is the case- the equivalent of the Dirac equation for spin 3/2 is the Rarita-Schwinger equations:
$\gamma_a \psi^{a}_\mu=0$,
$(i \gamma^\rho \partial_\rho-m)\psi^{a}_\mu=0 $
Which transforms as $\psi'^b _\nu=(\Lambda_\nu^\mu \otimes T^b_a) \psi_\mu^a$,
whose generators are the above combination of $M$ and $S$.
