Kalb-Ramond current fall-offs at future null infinity I can couple the electromagnetic field to a current generated by the complex scalar field for example:
$S=- \int d^4x \frac{1}{4} F_{\mu\nu} F^{\mu\nu} + A_\mu J^\mu$
with $J_\mu = i(\partial_\mu \phi^* \phi - \phi^* \partial_\mu \phi)$.
From the large-$r$ falloffs for the scalar field $\phi \in O(r^{-1})$ I can get the large-$r$ falloffs for the current by plugging the asymptotic expansion of the scalar field into the equation for the current.
In this example I get  $J_u \in \mathcal{O}(r^{-2})$ , $J_r \in \mathcal{O}(r^{-4})$ and $J_A \in \mathcal{O}(r^{-2})$ (in retarded Bondi coordinates $(u,r,A,B)$).
If we are looking at the two-form Kalb-Ramond theory with a coupled current
$S = - \int \frac{1}{6} H_{\mu\nu\rho} H^{\mu\nu\rho} + J^{\mu\nu} B_{\mu\nu}$, where $H=dB$ is the 3-form field strenght and $B$ the 2-form gauge field, how can I obtain the fall-offs for the coupled current at large-$r$?
I know it is antisymmetric, $\partial_\mu J^{\mu\nu}=0$ and the equations of motion are $\partial_\mu H^{\mu\nu\rho} = - J^{\nu\rho}$.
I also know the form of the current
$J^{\mu\nu} = q \int \delta^{(n)} (x-z) dz^\mu \wedge dz^\nu$
(p-form electrodynamics, Henneaux)
Now I need to calculate similar fall-off conditions for the different $J$ components like in electrodynamics above (e.g. $J_{uA} \in O(r^{n})$).
How do I do this?
 A: This will not be an answer, as I have some counter-questions regarding your post! However, hopefully it will lead to you answering my counter-questions (presumably in the comments) and me replying back/contributing, clearing, in this way, some things in your mind regarding the behavior of various components' of $J$ in the asymptotic /large $r$ limit!
First of all, you state that you expand the scalar field $\phi(x)$ near the asymptotic regions of space, obtaining, thus, the asymptotic expression for $\phi(x)$! From reading Strominger's "Lectures on the Infrared structure of Gravity and Gauge Theory" (https://arxiv.org/abs/1703.05448), I understand that the large $r$ behavior of the field at hand (i.e. the photon field) is determined by requiring that the charge or the energy flux of a Cauchy slice being finite (see discussion above Eq. (2.10.2) in Strominger's lecture notes). To be honest, I do not know which is more fundamental (demanding that some flux be finite or transforming the fields in retarded/advanced Bondi coordinates and expanding around the large $r$ limit). I also do not know if choosing one of the two above-mentioned approaches plays an essential role. My gut tells me that it shouldn't matter, but I still feel that Strominger's demand should not be neglected, so I leave it there for you to think about...
Second, after having a quick look at a paper you cite at a different question (https://arxiv.org/abs/1810.05634), I can see the exact formula for the three-form field strength components
$$H_{\mu\nu\rho}=\partial_{\mu}B_{\nu\rho}+
\partial_{\rho}B_{\mu\nu}+
\partial_{\nu}B_{\rho\mu}$$
with $B_{\mu\nu}$ being an antisymmetric rank-two tensor gauge field. Hence, following the same approach you seem to describe, the latter should admit an expansion of the form
$$B_{\mu\nu}(x)=e\int\frac{d^3\vec{q}}{(2\pi)^3}\frac{1}{2\omega_q}
\Big[a_{\mu\nu}(\vec{q})e^{iq\cdot x}+
a_{\mu\nu}^{\dagger}(\vec{q})e^{-iq\cdot x}\Big]$$
where $a_{\mu\nu}^{\dagger}(\vec{q}),\ a_{\mu\nu}(\vec{q})$ are the creation/annihilation operators for the respective field multiplied with some polarization vectors and summed over the potential polarization states (you can find the exact form of the latter operators yourself). The latter field, can be expressed in terms of the Bondi retarded coordinates
$$B_{\mu\nu}(x)=\frac{e}{8\pi^2}\int_0^{\infty}d\omega_q\omega_q
\int_0^{\pi}d\theta\sin\theta
\Big[a_{\mu\nu}(\vec{q})e^{-i\omega_q u-i\omega_qr(1-\cos\theta)}+
a_{\mu\nu}^{\dagger}(\vec{q})e^{+i\omega_q u+i\omega_qr(1-\cos\theta)}\Big]$$
and the limit of large $r$ can be taken such that
$$B_{\mu\nu}(x)=-\frac{ie}{8\pi^2r}\int_0^{\infty}d\omega_q
\Big[a_{\mu\nu}(\omega_q\vec{x})e^{-i\omega_q u}-
a_{\mu\nu}^{\dagger}(\omega_q\vec{x})e^{+i\omega_q u}\Big]$$
Then, the expression for the various components of the transformed antisymmetric rank-two tensor will be given by general coordinate transformations, as
$$B_{a\beta}(x')=\frac{\partial x^{\mu}}{\partial x^{'a}}
\frac{\partial x^{\nu}}{\partial x^{'\beta}} B_{\mu\nu}(x)$$
where with $x'$ I denote the retarded Bondi coordinates, whereas with $x$ I denote the Minkowski ones (expressed in terms of Bondi coordinates, as we have done above). For instance, it can already be seen (I think and if I am not mistaken) that
$$B_{uu}=B_{tt}=\mathcal{O}(r^{-1})$$
and you can plug in this and the remaining components you find in the expression for the current (from the equations of motion) and determine the current dependence on $r$...
This should give you a valid method for determining the various components of the large $r$ expansion of the $B$ field. Sorry for not being more concrete, but I haven't done the calculation myself. I reply though, in hopes of transferring the way of thinking of other similar calculations (just like some of the ones performed in Strominger's lecture notes) to your case. I hope this helps. If there are any questions, please do not hesitate to leave a comment.
