Is $F^{\mu\nu}F_{\mu\nu}$ equivalent to $A^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}$ for $U(1)$ gauge field lagrangian? The two seem to yield the same equation of motion is why I asked. Where of course the standard notation for exterior forms applies $dA=F$. 
We all know how the field strength tensor plays into the equations of motion. The Lagrangian density $$A^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}$$ seems to have an equation of motion (from varying the $A$) of the form:
$$\nabla^{\alpha}\nabla_{\alpha}A_{\mu}+\nabla_{\mu}\nabla_{\alpha}A^{\alpha}=0$$ Under the standard Lorenz gauge $\nabla_{\mu}A^{\mu}=0$ this becomes simply:
$$\nabla^{\alpha}\nabla_{\alpha}A_{\mu}=0$$ Now maybe this only works for an abelian field (like the Maxwell)? I'm honestly not sure. If they are equivalent can this be extended to non-abelian fields? I had thought Lagrangians were only equivalent if they're related by a total divergence term, but I'm not seeing it (maybe I'm missing it though).
EDIT
This is what I've tried, but I'm new to exterior calculus:
$$\intop_{M}\left\{ A^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}\right\} (dvol)_{M}$$
We can use the self-adjointness of the Laplace-Beltrami operator to write the above as:
$$=\intop_{M}-\langle dA,dA\rangle(dvol)_{M}$$
For some function $f$ (0-form) we have that a gauge transformation looks like:
$$A\longrightarrow A+df$$
$$=\intop_{M}-\langle d\left(A+df\right),d\left(A+df\right)\rangle(dvol)_{M}$$
$$=\intop_{M}-\left\{ \langle dA,dA\rangle+\langle d\left(df\right),dA\rangle+\langle dA,d\left(df\right)\rangle+\langle d(df),d(df)\rangle\right\} (dvol)_{M}$$
My understanding is that, $ddf=0$ which is like saying that the partial derivatives commute, or that the divergence of a curl is zero (the latter statement only being in three dimensions). Suppose all boundary terms vanish (I'm actually working where there is no boundary), this Leaves us simply with:
$$=\intop_{M}-\left\{ \langle dA,dA\rangle\right\} (dvol)_{M}$$
$$=\intop_{M}\left\{ A^{\mu}\nabla^{\alpha}\nabla_{\alpha}A_{\mu}\right\} (dvol)_{M}$$
I really don't know if that's right (I think I missed something somewhere), I'm trying to teach myself differential forms and exterior calculus. Maybe it's just wishful thinking lol!
 A: Your alternative lagrangian is not gauge invariant, since if you shift $A_\mu$ by the gradient of a scalar function, the expression clearly changes; $F_{\mu\nu}$ is antisymmetric in the indices $\mu \nu$ so basically any shift in $A$ gets cancelled. I think it's just an accident of a gauge choice that you've gotten the right EOM.
A: Your choice of Lagrangian density is equivalent, up to a total derivative, to $-(\nabla_\alpha A_\mu)^2$, while $F^2_{\alpha\mu}=2\nabla_\alpha A_\mu F^{\alpha\mu}$, so a $\nabla_\alpha A_\mu\nabla^\mu A^\alpha$ term makes them different. Even in your gauge choice, this isn't a total derivative because $A^\mu\nabla_\alpha\nabla_\mu A^\alpha=A^\mu R_{\mu\beta} A^\beta$. However, very different-looking Lagrangians can obtain the same EOMs.
A: Your expression $\intop_{M}-\langle dA,dA\rangle(dvol)_{M}$ is the usual Lagrangian for electromagnetism written in a highbrow way. You can work this out with Hodge duals and $dA=F$ if you've never seen it before. That is why it is no surprise you are showing it is gauge invariant.
Now you can go to $\intop_{M}-\langle A,d^\dagger dA\rangle(dvol)_{M}$, but this isn't the Laplacian yet. The Laplacian is $d^\dagger d+d d^\dagger $, so you need the gauge condition
$$d^\dagger A=0.$$
Again, this is just the Lorentz gauge in fancier language. So your Lagrangian does rely crucially on Lorentz gauge.
You can see all this by the way without using differential forms in flat space easily through integration by parts, by the way.
$$-\int dx^4 \partial_{[\mu}A_{\nu]}\partial^{[\mu}A^{\nu]}=-\int dx^4 \partial_{\mu}A_{\nu}\partial^{[\mu}A^{\nu]}=+\int dx^4 A_{\nu}\partial_{\mu}\partial^{[\mu}A^{\nu]}=\int dx^4 A_{\nu}\partial_{\mu}\partial^{\mu}A^{\nu}$$
where the last line I commuted the partial derivatives and used Lorentz gauge.
