Showing $\partial_{\mu}\tilde{F}^{\mu\nu}=0$ by the antisymmetric properties The electromagnetic dual tensor is given by
\begin{align}
\tilde{F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\delta\rho}F_{\delta\rho}
\end{align}
Here, $\epsilon^{\mu\nu\delta\rho}$ is the Levi-Civita symbol. I want to show that
$$
\partial_{\mu}\tilde{F}^{\mu\nu}=0 ,
$$
using the antisymmetric propeties of  Levi-Civita.
I have exchanged indices $\rho \to \delta$, 
\begin{align}
& \frac{1}{2}\partial_{\mu}\epsilon^{\mu\nu\rho\delta}F_{\rho\delta}\\
&=- \frac{1}{2}\partial_{\mu}\epsilon^{\mu\nu\delta\rho}F_{\rho\delta}\\
&=\frac{1}{2}\partial_{\mu}\epsilon^{\mu\nu\delta\rho}F_{\delta\rho}
\end{align}
In the last line, I've used $F_{\mu\nu}=-F_{\nu\mu}$. So, I can't use the property to prove this. In Ryder's QFT book he says that, because of the antisymmetry of $\epsilon^{\mu\nu\delta\rho}$ this will be true.
 A: In the parlance of differential forms, the identity (Abelian Bianchi identity) amounts to
$$
dF = 0.
$$
This is a trivial fact in math: exact forms ($F = dA$) are closed 
$$
dF = d^2A = 0,
$$
because of nilpotency of $d$ ($d^2 = 0$, barring Nonholonomic coordinates).
Note that the opposite may not be true: closed forms ($dF = 0$) do not necessarily imply exact forms ($F = dA$). Case in point: non-trivial de Rham cohomology. 
A: We know that 
$$
F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu .
$$
Now we get 
$$ 
\partial_\mu \epsilon^{\mu\nu\sigma\rho} F_{\sigma \rho} = 
\partial_\mu \epsilon^{\mu\nu\sigma\rho} ( \partial_\sigma A_\rho - \partial_\rho A_\sigma ). 
$$
Now, from symmetry/antisymmetry we have 
$$
\epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma A_\rho = 0
$$ 
and similar for the other term. 
To see this last step, consider
$$
\epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma 
= [\textrm{rename variables } \sigma \leftrightarrow \mu] \\  
= \epsilon^{\sigma\nu\mu\rho} \partial_\sigma \partial_\mu
= [\textrm{use property of Levi-Civita } \epsilon^{\alpha \beta} = -\epsilon^{\beta \alpha}] \\
= - \epsilon^{\mu\nu\sigma\rho} \partial_\sigma \partial_\mu  
= [\textrm{use that } \partial_\alpha \partial_\beta = \partial_\beta \partial_\alpha] \\ 
= - \epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma .  
$$
So we have 
$$ 
\epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma = - 
\epsilon^{\mu\nu\sigma\rho} \partial_\mu \partial_\sigma . 
$$
And if $a = -a$ then we must have $a = 0$. 
