Multiplying two tensors of the Levi-Civita type How to multiply two epsilons with one another?
We know $$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho'\sigma'}=-2(\delta^{\rho}_{\rho'}\delta^{\sigma}_{\sigma'}-\delta^\rho_{\sigma'}\delta^{\sigma}_{\rho'})$$
So if we had $$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\rho\sigma\mu\nu}$$ then this is always zero? I am confused if this is the case.
I think it is zero because it appears the first expression vanishes when we contract $\rho$ with $\rho'$ and $\sigma$ with $\sigma'$. Yet, the second expression would seem to be nonzero as it is just $(\pm 1)^2$ summed over the distinct rearrangements of $1,2,3,4$.
 A: The answer is that the second contracted pair of deltas amounts to a single delta contracted which gives $4$ whereas the first amounts to the product of two contracted deltas hence $(4)(4)=16$. Thus, $-2(16-4)=-24$. I'm not sure about the source of the minus just yet.
I'll add some detail as tensor arithmetic seems unclear to some users. First, the consider: (which uses repeated index implied summation on $\mu, \nu$ on the l.h.s.)
$$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho'\sigma'}=-2(\delta^{\rho}_{\rho'}\delta^{\sigma}_{\sigma'}-\delta^\rho_{\sigma'}\delta^{\sigma}_{\rho'}) \ \ \ \star $$
Observe, $\delta^\rho_{\sigma}\delta^{\sigma}_{\rho'} = \delta^{\rho}_{\rho'}$ as is easily verified by thinking about cases. Take $\star$ and contract $\rho$ with $\rho'$ and $\sigma$ with $\sigma'$, keeping $\rho$ and $\sigma$ as the dummy indices of summation, we obtain:
$$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho\sigma}=-2(\delta^{\rho}_{\rho}\delta^{\sigma}_{\sigma}-\delta^\rho_{\rho}) \ \ \ \star \star $$
But, $\delta^\rho_{\rho} = 1+1+1+1 = 4 = \delta^{\sigma}_{\sigma}$. Hence, following $\star \star$
$$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho\sigma} = -2(16-4) = -24. $$
Notice $$\epsilon_{\mu\nu\rho'\sigma'} = \epsilon_{\rho'\sigma'\mu\nu}$$ hence $\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho\sigma}$ ought to be the same as $\epsilon^{\rho\sigma\mu\nu}\epsilon_{\rho\sigma \mu\nu}$. To calculate the completely contracted levi-civita symbol with itself we note $\epsilon_{1234} = 1$ and then all other values follow from complete antisymmetry of $\epsilon_{ijkl}$ thus we directly calculate
$$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho\sigma}= 24.$$
As there are $24$ ways to rearrange $1,2,3,4$ into a list of $4$ numbers without repeats.
My conclusion, modulo context or some other Minkowski metric convention, the original identity is off by a sign. We should modify $\star$ to:
$$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho'\sigma'}=2(\delta^{\rho}_{\rho'}\delta^{\sigma}_{\sigma'}-\delta^\rho_{\sigma'}\delta^{\sigma}_{\rho'}). $$
