Stress Energy Tensor in language of differential forms The motivation for this is that quantities like the electric current $J$ in maxwell's equations of motion can be expressed as a differential 3-form, so that the continuity equation can be written just as
$$dJ=0$$
Which is really nice because it can all be done without defining a metric tensor!
Now the stress-energy tensor has a similar continuity equation but is generally represented as a symmetric 2-tensor. so it can obviously not be represented as a 3 form, but can it somehow be represented potentially in the language of differential forms so that a metric tensor does not have to be defined?
 A: Good question. I suspect that the answer is no, because the (Hilbert) stress-energy tensor is defined to be
$$T_{\mu \nu} := -2 \frac{\delta \mathcal{L}}{\delta g^{\mu\nu}} + g_{\mu \nu} \mathcal{L},$$
which suggests to me that it may depend fundamentally on the metric structure of spacetime.
A: If you invite "vector bundle-valued differential forms" then you can define $$ T^\mu=T^\mu_{\ \nu}dx^\nu $$ as a "vector field-valued 1-form", then you have $$ d^\nabla\star T^\mu\sim\nabla_\nu T^{\mu\nu}. $$
However the metric is needed both for the definition of $T$ and to take the covariant exterior derivative $d^\nabla$ and to take the Hodge dual.
Note that the current $j$ in your example is naturally a vector field, as the current can be obtained by $$ j^\mu=\frac{\delta S_m}{\delta A_\mu}, $$ where $S_m$ is some "matter" Lagrangian that contains $A$ (it is the field-particle interaction Lagrangian basically). So to obtain the 1-form $j$, you need to lower it (needs the metric), and then to obtain the  current 3-form $J$, you need to take the Hodge dual $J=\star j$ (needs the metric).
Hence while the conservation equation $$ dJ=0 $$ seems metric-independent, it really isn't.
A: With the help of a Killing vector field $\xi$ one can define the current 3-form
$$J_\xi = \star\ \iota_\xi T$$
of which you can then take the exterior derivative to obtain the conservation law
$$\operatorname d J_\xi = 0.$$
Note that the metric is hidden inside both $T$ and $\xi$.
