What is the asymptotic charge for a two-form theory in Lorenz gauge? I'm trying to derive the generating charge of the asymptotic symmetries for a two-form field in Lorenz gauge at future null infinity.
I'm working in retarded Bondi coordinates $(u,r,x^A, x^B)$.
First of all, I don't understand how they calculate their charges.
My ansatz would be to start from
$$Q = \int_{S(u)} (*k)^{AB} dS_{AB} =  \int_{S(u)} r^2 k^{ur} d\Gamma$$
where I'm integrating over a two-sphere at constant $u$ at large $r$.
$k^{ab} = H^{cab} \epsilon_a$ is the Noether two-form.
The fall-off conditions are described in the papers.
Then,
$$k^{ur} = g^{AB} (\partial_u B_{rB}^{(-1)} + \partial_B B_{ur}^{-1)}) \epsilon_A.$$
But I don't get to the charges in the papers if I insert this into the charge.
Where is my mistake or how do the papers calculate their charges?
 A: Covariant Phase Space Prescription
The most straightforward way to compute the surface charge associated to an asymptotic symmetry is the covariant phase space prescription which I will outline in this answer. Details can be found in https://arxiv.org/abs/1801.07064 and https://arxiv.org/abs/2009.14334. The basic idea goes back to standard Hamiltonian mechanics. If $(\Gamma,\Omega)$ is your phase space, where $\Omega$ is the sympletic form, a vector field on phase space $X\in \sec T\Gamma$ generates a canonical transformation when $$\Omega(X,\cdot) = -\mathbf{d}H_X(\cdot )\tag{1}$$
where $\mathbf{d}$ is the phase space exterior derivative and $H_X\in C^\infty(\Gamma)$ is the associated Hamiltonian charge. Equivalently, we can write this as $$\Omega(X,Y)=-Y(H_X).\tag{2}$$
In the covariant phase space formalism, we start with the space of allowed field configurations. The phase space will be the subspace of solutions to the equations of motion modulo trivial gauge transformations. The important parts are: (1) field variations are vector fields on the phase space and (2) the sympletic form can be constructed following an algorithm which I'll outline. You write a general variation of the Lagrangian form $L$ as $$\delta L=E[\Phi]\delta \Phi+d\Theta[\Phi,\delta \Phi].\tag{3}$$
We call $\Theta$ the pre-sympletic potential density. Knowing $\Theta[\Phi,\delta\Phi]$ you define $$\omega[\Phi,\delta_1\Phi,\delta_2\Phi]=\delta_1\Theta[\Phi,\delta_2\Phi]-\delta_2\Theta[\Phi,\delta_1\Phi]\tag{4},$$
which is known as pre-sympletic density. The pre-sympletic form is then $$\Omega[\Phi,\delta_1\Phi,\delta_2\Phi]=\int_\Sigma \omega[\Phi,\delta_1\Phi,\delta_2\Phi],\tag{5}$$
where $\Sigma$ is a Cauchy slice on spacetime. This is called pre-sympletic form because it might be degenerate, not qualifying as a sympletic form. In this case a gauge-fixing is needed. So the idea is: construct (5), then take $\delta_1\Phi$ to be a gauge transformation and $\delta_2\Phi$ to be a generic variation. If the final result turns out to be $-\delta Q$ you have identified the charge. Moreover if for some subset of the gauge transformations you get $Q=0$ the sympletic form will turn out to be degenerate. This identifies the trivial/small gauge transformations that must be gauge fixed to define the phase space properly.
Two-form theory
So let us consider ${\cal H}=d{\cal B}$ for a two-form field ${\cal B}$. The Lagrangian form is $$L=-\dfrac{1}{2}{\cal H}\wedge \star{\cal H}.\tag{6}$$
We study $\delta L$. It is $$\delta L=-{\delta \cal B}\wedge d\star{\cal H}-d(\delta{\cal B}\wedge \star{\cal H}).\tag{7}$$
Therefore we have $\Theta = -\delta{\cal B}\wedge \star{\cal H}$ and $\omega = \delta_1 {\cal B}\wedge \star \delta_2{\cal H}-{\delta_2}{\cal B}\wedge \star{\delta_1{\cal H}}$. Now suppose that $\delta_1{\cal B}=\delta_\alpha{\cal B}=\alpha$ with closed inexact $\alpha$, i.e., $d\alpha=0$ with $\alpha\neq d\beta$. This is a gauge transformation. In particular since ${\cal H}=d{\cal B}$ we have $$\delta{\cal H}=d\delta{\cal B}=d^2\alpha=0\tag{8}$$. Let $\delta_2{\cal B}=\delta{\cal B}$ a generic variation. We find $$\omega[{\cal B},\delta_\alpha{\cal B},\delta{\cal B}] = \alpha \wedge \star{\delta}d{\cal B}=\delta(\alpha \wedge \star d{\cal B})\tag{9}.$$
Therefore the sympletic form is $$\Omega[{\cal B},\delta_\alpha{\cal B},\delta{\cal B}]=\delta \int_{\Sigma}\alpha \wedge \star {\cal H}\tag{10}.$$
We already identify the charge as an integral over the Cauchy slice $\Sigma$. We now want to write it as a surface charge at $\partial\Sigma$, and to do so we use the duality described in the first paper you mention. Indeed, let us define a scalar $\phi$ such that ${\cal H}=\star d\phi$. Then $\star {\cal H}=\star^2 d\phi=d\phi$. Therefore
$$\Omega[{\cal B},\delta_\alpha{\cal B},\delta{\cal B}]=\delta \int_{\Sigma}\alpha \wedge d\phi\tag{11}.$$
Finally since $\alpha$ is closed, invoking Liebnitz rule we have that $\alpha \wedge d\phi = d(\phi\alpha)$. As a result we obtain a surface charge
$$\Omega[{\cal B},\delta_\alpha{\cal B},\delta{\cal B}]=-\delta Q,\quad Q=-\int_{\partial \Sigma} \phi\alpha\tag{12}.$$
Now recall that $\alpha$ is a two form on spacetime and you will pull it back to $\partial \Sigma$ to carry out this integral. Since $\partial \Sigma$ is two-dimensional, if $\alpha$ is not proportional to the volume form of $\partial \Sigma$ the result will be zero. In that case you have a trivial gauge symmetry which must eventually be gauge fixed. If $\alpha = f\epsilon_{\partial \Sigma}$ is proportional to it then the charge will be the integral of $f\phi$ over $\partial \Sigma$. Working with future asymptotics in flat spacetime $\partial \Sigma = {\cal I}^+_-$ which is a sphere.
