A question on Andrew Strominger's lecture (I now use the same conventions)
(I think the notations are clear enough if you are familiar with differential geometry. Further, I tagged this post as homework-and-excercises. What is the problem with this post? Why not simply leave it to others who may be interested if you do not like it?)
I am currently reading Andrew Strominger's lectures "Lectures on the infrared structure of gravity and gauge theory".
On page 9, it wrotes:

Maxwell's theory of electromagnetism is described by the action
  $$S=-\frac{1}{4e^2}\int d^4x\sqrt{-g}F_{\mu\nu}F^{\mu\nu}+S_M,$$
  where $F=dA$.
  The equation of motion is 
  $$d*F=e^2*j  \implies \nabla^\mu F_{\mu\nu}=e^2 j_\nu,\tag{1}$$
  where $*$ is the Hodge dual and the charge current is 
  $$j^\nu=-\frac{\delta S_M}{\delta A_\nu}.$$

What I can not obtain is the symbol $\implies$ in Eq.(1). My proof of it is as following.
According to the definition of Hodge star,
$$(*F)_{\mu\nu}=\frac{1}{2!}F^{\rho\sigma}\varepsilon_{\rho\sigma\mu\nu},$$
where $\varepsilon_{\mu\nu\rho\sigma}$ is the volume element
$$\varepsilon_{\mu\nu\rho\sigma}=\sqrt{|g|}(dx^1)_\mu\wedge(dx^2)_\nu\wedge(dx^3)_\rho\wedge(dx^4)_\sigma.$$
According to the definition of exterior derivative,
$$(d*F)_{\gamma\mu\nu}=(2+1)\frac{1}{2!}\nabla_{[\gamma}F^{\rho\sigma}\varepsilon_{|\rho\sigma|\mu\nu]}.$$
Taking the Hodge star again on the above equation, we have
$$(*d*F)_\delta=\frac{1}{3!}\times\frac{3}{2}\nabla^{[\gamma}F_{\rho\sigma}\varepsilon^{|\rho\sigma|\mu\nu]}\varepsilon_{\gamma\mu\nu\delta}
=\frac{1}{4}\nabla^{[\gamma}F_{\rho\sigma}\varepsilon^{|\rho\sigma|\mu\nu]}\varepsilon_{\gamma\mu\nu\delta}
=\frac{1}{4}\nabla^{\gamma}F_{\rho\sigma}\varepsilon^{\rho\sigma\mu\nu}\varepsilon_{\gamma\mu\nu\delta}.\tag{2}$$
Now using the properties for volume elements:
$$\varepsilon^{\rho\sigma\mu\nu}\varepsilon_{\gamma\mu\nu\delta}=\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{\mu\nu\gamma\delta}=(-1)^s 2!(4-2)!\delta^{[\rho}_{\ \ \gamma}\delta^{\sigma]}_{\ \ \delta},$$
where $s$ is the number of minus signs in the diagonalized metric tensor ($s$=1 or 3 depending on (-, +,+,+) or(+,-,-,-)). Thus 
$$(*d*F)_\delta=-\nabla^\rho F_{\rho\delta}$$
On the other hand, we have 
$$**j_\delta=(-1)^{s+1(4-1)}j_\delta=j_\delta.$$
Therefore, the first equation in Eq.(1) implies
$$\nabla^\rho F_{\rho\delta}=-e^2 j_\delta.$$
My derivation arrives at a result which differs from Eq.(1) by a sign. Where is going wrong in the above derivation (since Andrew uses the second equation in Eq.(1) in several papers, I therefore expect something wrong in my derivation rather than a typo in his lecture notes)? 
 A: After comparing with Prahar's answer, I see the problem in my derivation. We used different definitions for the Hodge star. In my derivation, it is (take $*j$ as an example)
$$(*j)_{\mu\nu\rho}=j^\sigma\varepsilon_{\sigma\mu\nu\rho}$$
while in Prahar's answer, he used (see Eq.(1) in the link)
$$(*j)_{\mu\nu\rho}=j^\sigma\varepsilon_{\mu\nu\rho\sigma}$$ which differs by minus sigh from here.
If I had used Andi's definition, there will be no difference in most of the equations but Eq.(2) in the post will be
$$(*d*F)_\delta=\frac{1}{3!}\times\frac{3}{2}\nabla^{[\gamma}F_{\rho\sigma}\varepsilon^{\mu\nu]\rho\sigma}\varepsilon_{\delta\gamma\mu\nu}
=-\frac{1}{4}\nabla^{[\gamma}F_{\rho\sigma}\varepsilon^{|\rho\sigma|\mu\nu]}\varepsilon_{\gamma\mu\nu\delta}
=-\frac{1}{4}\nabla^{\gamma}F_{\rho\sigma}\varepsilon^{\rho\sigma\mu\nu}\varepsilon_{\gamma\mu\nu\delta}.$$
However, I would like to comment that, actually, the definition used here is more standard since it appears in Nakahara, John Baez and the text book of GR in my hand. I point it our for convenience of other readers. 
