# Weinberg's proof that Levi-Civita Symbol is a tensor

In Gravitation and Cosmology, S.Weinberg states the following:

$$\Lambda_{\epsilon}^{\alpha}\Lambda_{\zeta}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda} \propto \epsilon^{\alpha \beta \gamma \delta}$$

and the argument is that the left-hand side must be odd under a single permutation of the indices. I don't see why this is true. Say I interchange $\alpha\leftrightarrow \beta$:

$$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}$$

I don't see why the above expression must satisfy

$$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}=-\Lambda_{\epsilon}^{\alpha}\Lambda_{\zeta}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}$$

Any hint will be appreciated.

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First of all, by interchanging $\Lambda_{\epsilon}^{\beta}$ with $\Lambda_{\zeta}^{\alpha}$ nothing changes:

$$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}=\Lambda_{\zeta}^{\alpha}\Lambda_{\epsilon}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}$$

Now switch $\epsilon$ with $\zeta$ only in the Levi-Civita symbol:

$$\Lambda_{\zeta}^{\alpha}\Lambda_{\epsilon}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda} = -\Lambda_{\zeta}^{\alpha}\Lambda_{\epsilon}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\zeta \epsilon \kappa \lambda}$$

and rename all the $\epsilon$ with $\zeta$ and all the $\zeta$ with $\epsilon$ (again nothing changes):

$$-\Lambda_{\zeta}^{\alpha}\Lambda_{\epsilon}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\zeta \epsilon \kappa \lambda}\to-\Lambda_{\epsilon}^{\alpha}\Lambda_{\zeta}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}$$

So from (1),(2) and (3) you get

$$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda} = -\Lambda_{\epsilon}^{\alpha}\Lambda_{\zeta}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}$$

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Use the anti-symmetry of $\epsilon$ to switch the indices $\epsilon$ and $\zeta$. Then relabel your dummy indices.

EDIT: Allow me to expand on this answer on Bebop's behalf.

You have two steps. In the first step, you use the fact that $\epsilon$ is anti-symmetric. That means you'll get a minus-sign if you exchange $\zeta$ and $\epsilon$. In the second step, you use the fact that you're ultimately summing over those indices. That means that $\zeta$ and $\epsilon$ are so-called dummy variables. Their names don't mean anything outside the sum, and so we can rename them any way we want. The idea here is basically that $\sum_n a_n = \sum_m a_m$. This means that you can just as well flip $\zeta$ and $\epsilon$ again, but this time "everywhere" in the expression (in $\epsilon$ and $\Lambda$), and this time without a sign change.

This technique (flipping first using the symmetry or antisymmetry and then a second time using the fact that we just have dummy variables) comes up quite a lot in physics.

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Hi, thanks. I'm sorry but I still don't get it.. when I switch $\epsilon$ and $\zeta$ in the Levi-Civita symbol I add a minus sign, and the indices in the Lorentz transformation should be switched too. Then how I relabel the dummy indices? switching back $\zeta$ to $\epsilon$? – Jorge Mar 19 '13 at 21:52
@Nivalth: you definitely don't flip the indices in the lorentz transformations--the point of the einstein summation is that once you've reduced the problem to there, everything is just sets of numbers. – Jerry Schirmer Mar 19 '13 at 22:25