Spin of a gauge field I was wondering what is the simplest way to understand the reduction of the Wigner's little group from $SO(d-1)$ to $SO(d-2)$ when one considers massive and massless fields respectively (in a $d$ dimensional spacetime). I would like a way to see this in the equations of motion. For example a spin one matter field denoted by $C_{\mu}$ can be described by $SO(d-1)$ little group. Let's now identify $C_\mu$ and $C_{\mu} +\partial_{\mu} \epsilon $ ($\epsilon$ the gauge parameter) as physically indistinguishable. How does that reduce the little group to $SO(d-2)$.
 A: I assume you are familiar with Wigner's classification in d=4, as you might be implying. The m→0 limit is best appreciated on the Poincaré sphere, but let us skip that here to count particle states. 
So, reviewing Wigner, for a massive state, we can Lorentz-transform the momentum to the rest frame, (m,0,0,0) so the little group is SO(3) and its vector rep is a triplet, so spin 1, and in d-dimensions the vector of SO(d-1), the spin classification being pointless. 
For vanishing mass, there is no rest frame: the best we can do is (p,0,0,p), cf. parabolic Lorentz transformations , and further, 
for instance, so the little group is SO(2), (actually, E(2)), and the invariant segues to the helicity. The eqns of motion are the same as before, except the absence of the mass term allows gauge invariance: with the emergence of gauge invariance, the massive triplet vector loses its gauge state, and reduces to a massless doublet. For d you have the vector of SO(d-2), again, its massive brother minus one (gauge) state.
Note how this works for the symmetric traceless tensor rep; you have (d-1)d/2 -1=(d-2)(d+1)/2 components for non-vanishing mass, so a spin two quintet for d=4. And (d-2)(d-1)/2-1 = d(d-3)/2  for vanishing mass, so a massless graviton doublet for d=4.  
