# Levi-Civita symbol and its effect on anti-symmetric rank two tensors

Jason Bennett
Homework Statement:
below
Relevant Equations:
below
I am trying to understand the following:

$$\epsilon^{mni} \epsilon^{pqj} (S^{mq}\delta^{np} - S^{nq}\delta^{mp} + S^{np}\delta^{mq} - S^{mp}\delta^{nq}) = -\epsilon^{mni} \epsilon^{pqj}S^{nq}\delta^{mp}$$

Where S^{ij} are Lorentz algebra elements in the Clifford algebra/gamma matrices representation.

The pattern I recognize is that, the only term to remain is the one where the two indices of both the delta and the Lorentz algebra matrix are in the same "slot" of the Levi-Civita symbol. Notably, the m and the p of the delta are both in the first "slot" of the L-C symbol, and the n and q are both in the second "slot" of the L-C symbol.

Can someone help me by pointing out which property of the L-C symbols I ought to be using?

Some additional points that may be on the right track

- anti-symmetric times symmetric = 0
- L-C = anti-sym,
- delta = sym (?), and
- the Lorentz matrices = anti-sym

Homework Helper
Gold Member
Homework Statement:: below
Relevant Equations:: below

I am trying to understand the following:

$$\epsilon^{mni} \epsilon^{pqj} (S^{mq}\delta^{np} - S^{nq}\delta^{mp} + S^{np}\delta^{mq} - S^{mp}\delta^{nq}) = -\epsilon^{mni} \epsilon^{pqj}S^{nq}\delta^{mp}$$

Where S^{ij} are Lorentz algebra elements in the Clifford algebra/gamma matrices representation.

The pattern I recognize is that, the only term to remain is the one where the two indices of both the delta and the Lorentz algebra matrix are in the same "slot" of the Levi-Civita symbol. Notably, the m and the p of the delta are both in the first "slot" of the L-C symbol, and the n and q are both in the second "slot" of the L-C symbol.

Can someone help me by pointing out which property of the L-C symbols I ought to be using?

Some additional points that may be on the right track

- anti-symmetric times symmetric = 0
- L-C = anti-sym,
- delta = sym (?), and
- the Lorentz matrices = anti-sym
It seems to me that the right hand side of the equation is missing a factor of 4.
All you have to do is to relabel some of the dummy indices so that all the terms are the same as in the expression on the right side. You will only need to use the antisymmetry of the LC symbol and of S.
For example, in the first term, you simply have to switch the indices ##m## and ## n ## and then you will have to use ##\epsilon^{nmi} = - \epsilon^{mni}##.

Last edited:
• Jason Bennett
Jason Bennett
It seems to me that the right hand side of the equation is missing a factor of 4.
All you have to do is to relabel some of the dummy indices so that all the terms are the same as in the expression on the right side. You will only need to use the antisymmetry of the LC symbol and of S.
For example, in the first term, you simply have to switch the indices ##m## and ## n ## and then you will have to use ##\epsilon^{nmi} = - \epsilon^{mni}##.

UGH! you're absolutely right! Thank you so much! Completely forgot that these are dummy indices (I really need to obey upper/lower even in Euclidean because I forgot about summed over indices all the time if i right things as all upper.) Cheers!

• nrqed