Finiteness of Maxwell gauge field symplectic form? The symplectic form for a Maxwell $U(1)$ gauge field is
$$
\omega = \int_\Sigma d \Sigma^\mu \delta F_{\mu \nu}  \wedge \delta A^\nu
$$
where $\wedge$ acts on field space. (In this notation, $\delta$ is like the exterior derivative $\mathrm{d}$ but for field variations instead of spacetime variations.) Here, our phase space is solutions to the Maxwell e.o.m.. Remember that the symplectic form will take in two "tangent vectors" in phase space, i.e. two field variations $\delta A$ which are "based" at a point, and output a number. In particular, the 'point' we're basing this at is an on-shell solution $A$, and the variations $A+\delta A$ must also be on-shell at the linearized level. For instance, unwrapping the notation a bit, we have
$$
\omega[A; \delta_1 A, \delta_2 A] = \int_\Sigma d \Sigma^\mu \big( \delta_1 F_{\mu \nu}  \delta_2 A^\nu - \delta_2 F_{\mu \nu}  \delta_1 A^\nu \big).
$$
Here $\Sigma$ is a Cauchy slice.
My question concerns the finiteness of the symplectic form $\omega$ given some reasionable boundary conditions. If we take a simple Cauchy slice $t = 0$, then the integrand will roughly be
$$
\sim \int d^2\Omega r^2 dr F_{t i} A^i
$$
where $i=1,2,3$ are the spatial directions. (Here I dropped the variations $\delta$ because all I care about is the fall offs.)
Now, $F_{ti}$ is the electric field, and if we assume that there is no radiation at infinitely far distances, then we can assume it falls off as $F_{ti} \sim r^{-2}$. Furthermore, let's assume that $A^i$ takes the form of the Leonard-Wiechert potential, in which case it will fall off as $A^i \sim r^{-1}$. Our $r$ integral is then
$$
\int dr r^2 \frac{1}{r^2} \frac{1}{r}
$$
which seems to diverge as $\log(r)$ for large radius. Isn't this a problem? Shouldn't the symplectic form be finite for these very reasonable fall off conditions?
 A: Yes, the symplectic form should be finite in order to be mathematically well-defined. The main idea is to impose appropriate asymptotic fall-off conditions at spatial infinity as OP already writes. The solution involves (i) imposing parity conditions and (ii) adding boundary terms to the symplectic form, see Ref. 1 for details.
References:

*

*M. Henneaux & C. Troessaert, Asymptotic symmetries of electromagnetism at spatial infinity, arXiv:1803.10194
; section 3.

A: I accepted Qmechanic's answer for pointing me to the right answer, but will give more details here.
The retarded vector potential in Lorenz gauge $(\partial_\mu A^\mu = 0)$ for a charge of constant velocity $\vec{\beta}$ which passes through the origin is
\begin{equation}
    A_\mu^{\rm ret}(t, \vec{r}) = \frac{e}{4 \pi \epsilon_0} \frac{(-1, \vec{\beta})}{\sqrt{ (\vec{r} - \vec{\beta} t)^2 + (\vec{r} \cdot \vec{\beta} )^2 - \beta^2 r^2 }}.
\end{equation}
Notice that the components of the gauge field fall off as $A_\mu \sim r^{-1}$.
Let's now look at the symplectic form on a $t = \mathrm{const}$ slice.
\begin{equation}\label{omega_large_r}
    \omega = \int_0^\infty r^2 dr \; \int_{S^2} d^2 \Omega \left( \delta_1 F^{t i} \delta_2 A_i - \delta_2 F^{t i} \delta_1 A_i  \right).
\end{equation}
If $F \sim r^{-2}$, and $A \sim r^{-1}$, then the $r$ integral is $\int \frac{dr}{r}$ which diverges logarithmically.
However, let's take a closer look at our vector potential $A_\mu^{\rm ret}$. Notice that if we hold $t$ fixed and send $|\vec{r}| \to \infty$, then it is even under sending $\vec{r} \mapsto - \vec{r}$.
\begin{equation}\label{Aantipodal}
    A_\mu^{\rm ret}(t, \vec{r}) = A_\mu^{\rm ret}(t, - \vec{r}) \hspace{0.5 cm} (\text{for large }r).
\end{equation}
This fact, that the vector potential takes the same value at antipodal points at spatial infinity, gives us another key boundary condition we need to adopt in order to make $\omega$ well defined. Let's now figure out what antipodal condition $F_{ti}$ satisfies. From the equation $A_\mu^{\rm ret}(-t, \vec{r}) = A_\mu^{\rm ret}(t, -\vec{r})$, we can see that $F_{ti} = \partial_t A_i - \partial_i A_t$ satisfies
\begin{equation}\label{Fantipodal}
    F_{ti}(t, -\vec{r}) = -F_{ti}(t, - \vec{r}) \hspace{0.5 cm} (\text{for large }r).
\end{equation}
So while $A_\mu$ is even, $F_{\mu \nu}$ is odd. These are our final boundary conditions. Using them, we can see that the $\int d^2 \Omega$ integral in the expression for $\omega$ at large $r$ vanishes identically.
