Counting degrees of freedom without fixing the gauge?

Question

In electrodynamics, the current-current interaction in the momentum space is described by

$$p^2 A_\mu J^\mu = J_\mu J^\mu \, ,$$

where $J$ denotes an arbitrary external current. Since photon-mediated interactions respond only to the transverse part of the current, we can write

$$ J_\mu J^\mu = J^\mu \Pi_{\mu\nu} J^\nu \, $$

where we define $\Pi_{\mu\nu}$ as a projection operator that projects an on-shell current to its transverse part,

$$ \Pi_{\mu\nu} = \eta_{\mu\nu} - p_\mu \bar{p}_\nu - p_\nu \bar{p}_\mu \, ,$$

where $p$ denotes the exchanged on-shell momentum ($p^2=0$) and $\bar{p}$ denotes some vector such that $\bar{p}^2 = 0$ and $p \cdot \bar{p} = 1$.

Now, obviously,

$$\eta_{\mu\nu} \Pi^{\mu\nu} = D - 2 \, , $$

in $D$-dimensional spacetime. The claim is that the trace of $\Pi$ counts the number of degrees of freedom (polarizations) that take part in the current-current interaction, without fixing the gauge.

This isn't so obvious to me, could someone give it a shot and try to explain it?

Answer 1

I think I understand it, so I'm posting this as an answer. It makes sense to me, but correct me if I'm wrong.

We can view $$ J_\mu J^\mu = J^\mu \Pi_{\mu\nu} J^\nu $$ as an operator equation, suppressing the indices, $$ \Pi J = J. $$ This is an eigenvalue problem, with a fully degenerate set of eigensolutions with eigenvalues equal to one.

Since the trace of a linear operator equals the sum of its eigenvalues, if all eigenvalues are one, it simply counts the number of eigenvectors spanning the solution space, which is $D-2$ in this case. Furthermore, the solution space is the space of vectors transverse to $p^\mu$, so it makes physical sense that its dimension is the number of degrees of freedom in the current-current interaction.