Wigner's classification of particles implies that the internal degrees of freedom of a particle transform under unitary representations of the subgroup of the Lorentz group that leaves its momentum invariant. In 3+1D for massless particles this group is ISO(2); we typically restrict our attention to representations that transform trivially under the translations of ISO(2) and nontrivially under rotations about the direction of propagation of the particle, which gives us its two helicity states (if one assumes parity is a symmetry of the theory).
In 2+1D the little group of a massless particle is trivial. Therefore we would naively conclude that the only degree of freedom of a massless particle is its momentum, and that there is no internal state analogous to helicity. However, in the standard classical argument for why there are two degrees of freedom in 3+1D (see Gauge theory and eliminating unphysical degrees of freedom), one finds that the redundancies of the theory lead us to subtracting two degrees of freedom from the initial total of four to give the two "physical" degrees of freedom associated to the helicity states. It doesn't appear to me naively that there is anything that would be different about the argument in 2+1D, which would lead me to conclude that EM waves in 2+1D should have 3-2 = 1 internal degrees of freedom, even though the little group analysis suggests that there are zero.
My question is -- how do these two pictures square with each other? How many degrees of freedom does the photon truly have in 2+1D?
