In Sean Carroll's book he derives the two tensorial Maxwell equations from the four non-tensorial equations. I noticed that one of these equations is the Bianchi identity for the electromagnetic equations. He states that both sides of these equations "manifestly transform as tensors; therefore if they are true in one inertial frame, they must be true in any inertial frame"…" we will sometimes refer to quantities written in terms of tensors as covariant".
There is a similar question on this forum to my following question: How to tell that the electromagnetic field tensor transforms as a tensor? I am working on an assignment question that does not refer to the E&M equations. I am simply given the form:
\begin{equation} \label{eqn:equation_1}
\partial_\mu A_{\nu\rho} + \partial_\rho A_{\mu\nu} + \partial_\nu A_{\rho\mu} = 0,
\end{equation}
and that $A_{ab}$ is an antisymmetric tensor field on the spacetime $(M,g)$ with global coordinate chart $\varphi=\{x^\alpha\}$. Note that there is no reference to the Maxwell equations in this question, so I do not know the form of $A_{ab}$ other than that it is antisymmetric. Using this fact you can derive the form:
\begin{equation}
\partial_{[\mu}A_{\nu\rho]}=0,
\end{equation}
which, if I am correct, looks like the Bianchi identity for the electromagnetic field tensor. The question on my assignment is to show that the first equation holds in all coordinate charts. To me this is the same as showing that $\partial_{[\mu}A_{\nu\rho]}$ transforms like a tensor, since this would show that the equation is covariant and thus holds in all coordinate charts (or we could prove that $\partial_\mu A_{\nu\rho} + \partial_\rho A_{\mu\nu} + \partial_\nu A_{\rho\mu}$ transforms like a tensor).
So I'm wondering, how do I show that these equations transform like a tensor (without knowing anything about E&M or the form of $A_{ab}$)?
P.S. I think my question is different enough to the cited post, since they show how $A_{ab}$ is tensorial, but not how $\partial_{[\mu}A_{\nu\rho]}$ is tensorial; hopefully these are different.
Best Answer
The use of $A$ is a little weird, the usual notation is $F$. Anyway, for a $2$-form $F$ on a manifold, we can consider its exterior derivative $dF$. This is a $3$-form, and its coordinate expression is exactly what you wrote. Exterior derivatives are globally well-defined objects on any smooth manifold. So, the equation $dF=0$ is a coordinate-free statement (so the components of this $3$-form vanish in every coordinate system). The fact that $dF=0$ expresses the fact that it localy comes from a potential $A$, i.e locally, $F=dA$ (in coordinates, $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$).
Or if you don’t develop the basic machinery of exterior derivatives and the exterior algebra, then, as mentioned in the comments, this is going to be an annoying little exercise in transforming $F_{\mu\nu}$, then transforming the partial derivatives $\partial_{\rho}$, doing the product/chain rule, and carrying out an alternating summation over the indices (keeping in mind that $F_{\mu\nu}$ is anti-symmetric in its indices). When Carroll says “manifestly transforms as tensors” he’s simply delegating the tedious algebra to you.