TITLE: Why does the electromagnetic tensor $F$ satisfy $**F = -F$?
QUESTION [5 upvotes]: I don't understand how tensors work, can someone please explain? 
In particular, in the context of Electromagnetism, the dual of the field tensor $F$ is $$(*F)^{\mu\nu}:={1\over 2}\epsilon^{\mu\nu\rho\sigma}F_{\rho \sigma}$$
I am wondering what happens if we take the dual twice. It is given that $$**F=-F$$ But I don't see it trivially. 
What I think (probably wrong):
I suppose $$(**F)^{ab}={1\over 4} \epsilon^{ab\mu\nu}\epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma}$$
Actually I am very confused as to whether to put the indices upstairs or downstairs.For example, I don't know whether to write the above or $$(**F)^{ab}={1\over 4} \epsilon^{ab\mu\nu}\epsilon_{\mu\nu}.^{\rho\sigma}F_{\rho\sigma}$$
Or in fact other variants, providing that indices to be summed over are such that one's up and one's down.
Help would be appreciated. Thank you.

REPLY [4 votes]: $\def\a{\alpha}
\def\b{\beta}
\def\e{\epsilon}
\def\d{\delta}
\def\m{\mu}
\def\n{\nu}
\def\r{\rho}
\def\s{\sigma}$We have 
$(*F)^{\m\n} = \frac{1}{2}\e^{\m\n\r\s} F_{\r\s}$,
so 
$(*F)_{\m\n} = \frac{1}{2}\e_{\m\n\r\s} F^{\r\s}.$
Indices are raised and lowered with the Minkowski metric $\eta_{\m\n}$ using the fact that
$$\eta^{\m\n}\eta_{\n\r} = \d^\m_\r.$$
Thus, 
$$\begin{eqnarray*}
(**F)^{\m\n} &=& \frac{1}{4} \e^{\m\n\r\s}\e_{\r\s\a\b} F^{\a\b} \\
&=& \frac{1}{4} \e^{\r\s\m\n}\e_{\r\s\a\b} F^{\a\b}
\end{eqnarray*}$$
as you've written. 
To evaluate this we need a result about the Levi-Civita symbol, 
$$\e^{a_1\cdots a_k a_{k+1}\cdots a_d}\e_{a_1\cdots a_k b_{k+1}\cdots b_d} =
(-1)^s k!(d-k)!\ \d^{[a_{k+1}}_{b_{k+1}}\cdots \d^{a_d]}_{b_d},$$
where $(-1)^s = \textrm{sign}\,\det\eta$ and where indices in square brackets are antisymmetrized. 
In particular, for $d=4$, $k=2$, and $(-1)^s = -1$ we have 
$$\e^{\r\s\m\n}\e_{\r\s\a\b} = -4 \d_\a^{[\m} \d_\b^{\n]}.$$
Thus, 
$$\begin{eqnarray*}
(**F)^{\m\n} &=& -\d_\a^{[\m} \d_\b^{\n]} F^{\a\b} \\
&=& -\frac{1}{2}(\d_\a^\m \d_\b^\n - \d_\a^\n \d_\b^\m) F^{\a\b} \\
&=& - \frac{1}{2}(F^{\m\n} - F^{\n\m}) \\
&=& -F^{\m\n},
\end{eqnarray*}$$
where we have used the fact that $F$ is antisymmetric. 
It is possible to write all this in the language of differential forms in which case we are verifying that the inverse of the Hodge star operation $*$ on 2-forms is $-*$.