Energy momentum tensor of EM field written in symmetric form I'm reading A. Zee's book, Einstein Gravity in a Nutshell. In problem 7 of chapter IV.2, it is said that the energy momentum tensor of the electromagnetic field
\begin{align}
T^{\mu\nu}=\eta_{\lambda\sigma}F^{\mu\lambda}F^{\nu\sigma}-\frac{1}{4}\eta^{\mu\nu}F_{\sigma\rho}F^{\sigma\rho}
\end{align}
can be written in the symmetric form using the dual electromagnetic tensor defined as $\tilde{F}_{\mu\nu}=-\frac{1}{2} \epsilon_{\mu\nu\lambda\sigma}F^{\lambda\sigma}$,
\begin{align}
T^{\mu\nu}=\frac{1}{2}\eta_{\lambda\sigma}(F^{\mu\lambda}F^{\nu\sigma}+\tilde{F}^{\mu\lambda}\tilde{F}^{\nu\sigma}).
\end{align}
It is mentioned in problem 6 that the key is the identity
\begin{align}
\eta_{\lambda\sigma}(F^{\mu\lambda}F^{\nu\sigma}-\tilde{F}^{\mu\lambda}\tilde{F}^{\nu\sigma})=\frac{1}{2}\eta^{\mu\nu}F^{\rho\tau}F_{\rho\tau}
\end{align}
However, I'm stuck when proving this identity. First problem is that I can't separate a $\eta^{\mu\nu}$ from the first term in left hand side. The second problem is the second term will have something like $\epsilon^{\mu\lambda\alpha\beta}\epsilon^{\nu}_{~\lambda\gamma\eta}$ after plugging the expression of dual tensor, but I don't know how to simplify it. Any hints about proof of this identity?
 A: You can express the product of two $\epsilon$'s using the following: $$\epsilon^{\mu\lambda\alpha\beta} \epsilon_{\tau\lambda\gamma\eta}= \epsilon^{\lambda\mu\alpha\beta} \epsilon_{\lambda\tau\gamma\eta}=\epsilon^{\mu\alpha\beta} \epsilon_{\tau\gamma\eta}$$ $$=\delta^\mu_\tau(\delta^\alpha_\gamma \delta^\beta_\eta-\delta^\alpha_\eta \delta^\beta_\gamma)-\delta^\mu_\gamma(\delta^\alpha_\tau \delta^\beta_\eta-\delta^\alpha_\eta \delta^\beta_\tau)+\delta^\mu_\eta(\delta^\alpha_\tau \delta^\beta_\gamma-\delta^\alpha_\gamma \delta^\beta_\tau)$$
A: The first step is to solve the second problem, which is to find the product $\tilde{F}_{\mu\nu}\tilde{F}^{\mu\nu}$. To this, you will need to calculate the product
\begin{equation}
\epsilon^{\alpha\beta\gamma\delta}\epsilon_{\mu\nu\rho\sigma}=\left(
-1\right)  ^{s}\delta_{\mu\nu\rho\sigma}^{\alpha\beta\gamma\delta}
,\tag{1}\label{eq1}
\end{equation}
where $s$ is the number of negative eigenvalue of the metric [1] and $\delta_{\mu\nu\rho\sigma}^{\alpha\beta\gamma\delta}$ is the generalized Kronecker's delta, which we can to expresses as
\begin{equation}
\delta_{\mu\nu\rho\sigma}^{\alpha\beta\gamma\delta}=\delta_{\sigma}^{\delta
}\delta_{\mu\nu\rho}^{\alpha\beta\gamma}-\delta_{\sigma}^{\gamma}\delta
_{\mu\nu\rho}^{\alpha\beta\delta}+\delta_{\sigma}^{\beta}\delta_{\mu\nu\rho
}^{\alpha\gamma\delta}-\delta_{\sigma}^{\alpha}\delta_{\mu\nu\rho}
^{\beta\gamma\delta}.\tag{2}\label{eq2}
\end{equation}
You can find a general study of Kronecker's delta in [2].
Once knowing \eqref{eq1}, you will be able to calculate the product $\tilde{F}_{\mu\nu}\tilde{F}^{\mu\nu}$.
The first problem can be solved by means of a game of rising and falling indices, as long as you remember that 
\begin{equation}
\eta_{\mu\rho}\eta^{\mu\rho}=\delta_{\mu}^{\mu}=4\rightarrow\dfrac{1}{4}
\eta_{\mu\rho}\eta^{\mu\rho}=1.\tag{3}\label{eq3}
\end{equation}
1 Sean Carrol, Spacetime and Geometry:An Introduction to General Relativity.
2 David Lovelock and Hanna Rund, Tensor, differential forms and variational principles, section 4.2, page 109. 
