Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 4 down vote accepted

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

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

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.