
\noindent
{\bf Problem (\ref{Levi_Civita}): Four-dimensional Levi-Civita tensor.}
\smallskip

\noindent (1) Using the Einstein summation convention of the left-hand
side, but not on the right-hand side, of the following equation we
have
$$ g_{\alpha_1\alpha_2} g_{\beta_1\beta_2} g_{\gamma_1\gamma_2} 
   g_{\delta_1\delta_2} \epsilon^{\alpha_2\beta_2\gamma_2\delta_2}
   = g_{\alpha_1\alpha_1} g_{\beta_1\beta_1} g_{\gamma_1\gamma_1} 
   g_{\delta_1\delta_1} \epsilon^{\alpha_1\beta_1\gamma_1\delta_1}\,.$$
The right-hand side is zero unless $\alpha_1\beta_1\gamma_1\delta_1$ are
a permutation $\pi_0\pi_1\pi_2\pi_3$ the numbers $0,1,2,3$ and in that 
case
$$ \epsilon_{\pi_0\pi_1\pi_2\pi_3} = g_{\pi_0\pi_0} g_{\pi_1\pi_1} 
   g_{\pi_1\pi_2} g_{\pi_1\pi_3} \epsilon^{\pi_0\pi_1\pi_2\pi_3} = 
   (-1)^3\,\epsilon^{\pi_0\pi_1\pi_2\pi_3}\,, $$
which proves $\epsilon_{\alpha\beta\gamma\delta}=-
\epsilon^{\alpha\beta\gamma\delta}$.
\smallskip

\noindent (2) We have $\epsilon_{\alpha\beta\gamma_1\delta_1}
\epsilon^{\alpha\beta\gamma_2\delta_2}=0$ unless either $\gamma_1=
\gamma_2$, $\delta_1=\delta_2$ or $\gamma_1=\delta_2$, $\delta_1=
\gamma_2$ or (otherwise one of the already used numbers will be 
repeated by $\alpha$ or $\beta$ of the sums). Therefore,
$$\epsilon_{\alpha\beta\gamma_1\delta_1} 
  \epsilon^{\alpha\beta\gamma_2\delta_2} = 
  a\,\delta_{\gamma_1}^{\ \gamma_2}\,\delta_{\delta_1}^{\ \delta_2}+
  b\,\delta_{\gamma_1}^{\ \delta_2}\,\delta_{\delta_1}^{\ \gamma_2}$$
holds. With no summation in the permutation indices $\pi_0$ to $\pi_3$
the constants follow from
\begin{eqnarray} \nonumber
 \epsilon_{\alpha\beta\pi_2\pi_3}\epsilon^{\alpha\beta\pi_2\pi_3} =
 \epsilon_{\pi_0\pi_1\pi_2\pi_3}\epsilon^{\pi_0\pi_1\pi_2\pi_3} +
 \epsilon_{\pi_1\pi_0\pi_2\pi_3}\epsilon^{\pi_1\pi_0\pi_2\pi_3} =
 -2 = a\,, \\ \nonumber
 \epsilon_{\alpha\beta\pi_2\pi_3}\epsilon^{\alpha\beta\pi_3\pi_2} =
 \epsilon_{\pi_0\pi_1\pi_2\pi_3}\epsilon^{\pi_0\pi_1\pi_3\pi_2} +
 \epsilon_{\pi_1\pi_0\pi_2\pi_3}\epsilon^{\pi_1\pi_0\pi_3\pi_2} =
 +2 = b\,. 
\end{eqnarray}
\smallskip

\noindent (3) We have $\epsilon_{\alpha\beta_1\gamma_1\delta_1}
\epsilon^{\alpha\beta_2\gamma_2\delta_2}=0$ unless $\alpha_2\beta_2
\gamma_2$ is a permutation of $\alpha_1\beta_1\gamma_1$. There are
six such permutations, so that the results is a sum of the form
\begin{eqnarray} \nonumber
   \epsilon_{\alpha\beta_1\gamma_1\delta_1} 
   \epsilon^{\alpha\beta_2\gamma_2\delta_2}\, &=&
    a_1\,\delta_{\beta_1}^{~\beta_2}\delta_{\gamma_1}^{~\gamma_2}
         \delta_{\delta_1}^{~\delta_2}
  + a_2\,\delta_{\beta_1}^{~\gamma_2}\delta_{\gamma1}^{~\delta_2}
         \delta_{\delta_1}^{~\beta_2}
  + a_3\,\delta_{\beta_1}^{~\delta_2}\delta_{\gamma_1}^{~\beta_2}
         \delta_{\delta_1}^{~\gamma_2} \\ \nonumber &+&
    b_1\,\delta_{\beta_1}^{~\beta_2}\delta_{\gamma_1}^{~\delta_2}
         \delta_{\delta_1}^{~\gamma_2}
  + b_2\,\delta_{\beta_1}^{~\delta_2}\delta_{\gamma_1}^{~\gamma_2}
         \delta_{\delta_1}^{~\beta_2}
  + b_3\,\delta_{\beta_1}^{~\gamma_2}\delta_{\gamma_1}^{~\beta_2}
         \delta_{\delta_1}^{~\delta_2} \,,
\end{eqnarray}
where it follows from (no summation)
$\epsilon_{\pi_0\pi_1\pi_2\pi_3}\epsilon^{\pi_0\pi_1\pi_2\pi_3}=-1$
that $a_i=-1$ and $b_i=1$ holds for $i=1,2,3$.
\bigskip % \vfil\eject
