How to see $\mathbf{E}\cdot\mathbf{B}$ is a total derivative? Since $\mathbf{E}\cdot\mathbf{B}$ is a Lorentz invariant of the electromagnetic fields it seems like an interesting thing to plug into a Lagrangian to see what happens. However, this ends up disappearing, and I'm told this should be obvious because it is a total derivative.
This however is not obvious to me. Is there an easy way to see that
$$ \frac{1}{2}\epsilon_{\alpha\beta\gamma\delta}F^{\alpha\beta} F^{\gamma\delta} = - \frac{4}{c}\left( \mathbf{B} \cdot \mathbf{E} \right) $$
is actually a total derivative?
I'd also appreciate if someone can show what it is the derivative of, so that I can work out the derivative to help it sink in.
 A: For completeness, here's the tensor notation version.
First rewriting:
$$ (\mathbf{E}\cdot\mathbf{B})\ \propto \ \epsilon_{\alpha\beta\gamma\delta}F^{\alpha\beta} F^{\gamma\delta} = \epsilon_{\alpha\beta\gamma\delta} \left( \partial^\alpha A^\beta - \partial^\beta A^\alpha \right)F^{\gamma\delta} = 2 \ \epsilon_{\alpha\beta\gamma\delta} \left(\partial^\alpha A^\beta \right) F^{\gamma\delta}$$
where the last step uses relabeling and the anti-symmetry of $\epsilon_{\alpha\beta\gamma\delta}$.
Similarly
$$\epsilon_{\alpha\beta\gamma\delta} \left(\partial^\alpha A^\beta \right) F^{\gamma\delta} = 2 \ \epsilon_{\alpha\beta\gamma\delta} \left(\partial^\alpha A^\beta \right) \left(\partial^\gamma A^\delta \right).$$
Now moving one of the derivatives to the front
$$\epsilon_{\alpha\beta\gamma\delta} \left(\partial^\alpha A^\beta \right) \left(\partial^\gamma A^\delta \right) = \partial^\alpha \left( \epsilon_{\alpha\beta\gamma\delta} A^\beta \left(\partial^\gamma A^\delta \right)\right) - \epsilon_{\alpha\beta\gamma\delta} A^\beta \left(\partial^\alpha \partial^\gamma A^\delta \right)$$
and note that the last term is zero because the derivatives commute and so are symmetric in the those labels, while $\epsilon_{\alpha\beta\gamma\delta}$ is anti-symmetric.
All together this gives:
$$ (\mathbf{E}\cdot\mathbf{B})\ \propto \ \partial^\alpha \left( \epsilon_{\alpha\beta\gamma\delta} A^\beta \left(\partial^\gamma A^\delta \right)\right) $$ 
A: The easiest way to examine $\mathbf{E}\cdot\mathbf{B}$ is to look at the fields in terms of the vector potential in the Weyl gauge, where $A^0=0$. In that gauge, you get:
\begin{align}
    \mathbf{E} &= - \frac{\partial \mathbf{A}}{\partial t},\ \mathrm{and} \\
    \mathbf{B} &= \nabla \times \mathbf{A}.
\end{align}
Thus you get:
\begin{align}
\mathbf{E}\cdot\mathbf{B} &= -\epsilon_{ijk}\frac{\partial A_i}{\partial t}  \left(\frac{\partial A_k}{\partial x_j}\right)
\end{align}
Now let's examine the space-time integral of this quantity:
\begin{align}
    \int \mathbf{E}\cdot\mathbf{B} \operatorname{d}x^4 & = -\int \epsilon_{ijk}\frac{\partial A_i}{\partial t}  \left(\frac{\partial A_k}{\partial x_j}\right) \operatorname{d}x^4 \\
& = \int\left[- \frac{\partial }{\partial t} \left(\epsilon_{ijk} A_i \frac{\partial A_k}{\partial x_j}\right) + \epsilon_{ijk} A_i \frac{\partial^2 A_k}{\partial x_j \partial t} \right] \operatorname{d}x^4 && \mathrm{integrate\ by\ parts\ in\ }t \\
& = \int\left[-  \frac{\partial }{\partial t} \left(\epsilon_{ijk} A_i \frac{\partial A_k}{\partial x_j}\right) + \frac{\partial }{\partial x_j} \left(\epsilon_{ijk} A_i \frac{\partial A_k}{\partial t}\right) - \epsilon_{ijk}\frac{\partial A_k}{\partial t}  \left(\frac{\partial A_i}{\partial x_j}\right) \right] \operatorname{d}x^4 && \mathrm{integrate\ by\ parts\ in\ space} \\
& = -\int \epsilon_{kji}\frac{\partial A_i}{\partial t}  \left(\frac{\partial A_k}{\partial x_j}\right) \operatorname{d}x^4 && \mathrm{drop\ surface\ terms} \\
&&& \mathrm{and\ relabel\ dummy\ variables} \\
& = \int \epsilon_{ijk}\frac{\partial A_i}{\partial t}  \left(\frac{\partial A_k}{\partial x_j}\right) \operatorname{d}x^4 && \mathrm{exchange\ indices\ of\ }\epsilon
\end{align}
Notice that we just proved that the integral equals minus itself, and therefore must vanish if the surface terms vanish.
A: First note that you can rewrite $\mathbf{E}\cdot\mathbf{B}$ as
$$
\mathbf{E}\cdot\mathbf{B}\propto F\wedge F
$$
using a field strength $2$-form $F$ where $\mathbf{E}$ and $\mathbf{B}$ are defined as
\begin{align}
F_{0i}&=E_i ,\\
F_{ij}&=\epsilon_{ijk}B_k.
\end{align}
More specifically,
\begin{align}
F\wedge F&=\frac{1}{4}F_{\mu\nu}F_{\rho\sigma}\, dx^\mu\wedge dx^\nu\wedge dx^\rho\wedge dx^\sigma \\
&=-\frac{1}{4}\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}\,\text{vol.}\\
&\propto \epsilon^{ijk}F_{0i}F_{jk}=\mathbf{E}\cdot\mathbf{B}.
\end{align}
Then it is easy to show that $\mathbf{E}\cdot\mathbf{B}$ is a total derivative using $F=dA$, i.e.,
$$
\mathbf{E}\cdot\mathbf{B}\propto F\wedge F=d(A\wedge F).
$$

As a side comment, $F\wedge F$ contains volume form but it is absent in $\mathbf{E}\cdot\mathbf{B}$. So the correct way to write is
$$
\int d^4x \,\mathbf{E}\cdot\mathbf{B}\propto\int F\wedge F.
$$
