# Homework Help: Tensors, Levi-Civita symbol

1. Sep 11, 2010

### fluidistic

1. The problem statement, all variables and given/known data
If $$\epsilon _{ijjk}$$ is the Levi-Civita symbol:
1)Demonstrate that $$\sum _{i} \epsilon _{ijk} \epsilon _{ilm}=\delta _{jl} \delta _{km} -\delta _{jm} \delta _{kl}$$.
2)Calculate $$\sum _{ij} \epsilon _{ijk} \epsilon _{ijl}$$.
3)Given the matrix M, calculate $$\sum _{ijk} \sum _{lmn} \epsilon _{ijk} \epsilon _{lmn} M_{il} M_{jm} M_{kn}$$.

2. Relevant equations

Maybe some properties on tensors but I'm not sure.

3. The attempt at a solution
I'm trying to start with 1) first. I'm so new with tensors that I don't understand well what I have to do.
I know that Levi-Civita symbol is either worth -1, 0 or 1 though I didn't understand what is an even and odd permutation of ijk. And I know Kronecker's delta which is worth either 0 or 1, depending if i=j or not.
So starting with $$\sum _{i} \epsilon _{ijk} \epsilon _{ilm}$$, can I assume i to go from 1 to 3? And what about j and k? Fixed constants which are either 1, 2 or 3?

2. Sep 12, 2010

### vela

Staff Emeritus
(132) would be an odd permutation of (123) because you can obtain it using an odd number of transpositions (swap of a pair), e.g. swap 2 and 3. (312) on the other hand would be an even permutation of (123) because obtaining it requires an even number of transpositions, e.g. swap 1 and 3 to get (321) and then swap 1 and 2 to get (312).

http://en.wikipedia.org/wiki/Parity_of_a_permutation
Yes, the index i runs from 1 to 3; j, k, l, and m are fixed. There aren't too many combinations of j, k, l, and m to check since you can easily see that most are 0.

Last edited: Sep 12, 2010
3. Sep 12, 2010

### fluidistic

Thank you very much. Now I understand and I still find that there are lots of choices (4^4 I think) even though most of them are worth 0, I must show all. Part 1) solved.
By the way I had seen the wikipedia article but didn't understand. Your words were much clearer to me. Now I'll try part 2.

Edit: I'm a bit confused on the notation of the sum. There are 9 cases. I take the first as $$k=l=1$$. Is $$\sum _{ij} \epsilon _{ijk} \epsilon _{ijl}$$ worth $$\epsilon _{111}\epsilon _{111}+\epsilon _{211}\epsilon _{211}+\epsilon _{311}\epsilon _{311}+\epsilon _{121}\epsilon _{121}+\epsilon _{131}\epsilon _{131}+\epsilon _{231}\epsilon _{231}+\epsilon _{321}\epsilon _{321}=0$$?

Last edited: Sep 12, 2010
4. Sep 12, 2010

### vela

Staff Emeritus
The only non-vanishing terms are the last two, and they're both equal to 1, so the sum is equal to 2.

5. Sep 12, 2010

### fluidistic

Indeed you're right, I made a mistake in the last term of the sum, I considered it as -1 instead of 1. My question was about if my sum was right since I don't recall having used a double subindex in sums before.
Now I'll try to tackle part 3).
Nevermind, I see it's only a huge arithmetic "mess". I don't see any trick but to calculate the whole sum...
Thanks for all, and you've taught me what was an even and odd permutation. :)

Last edited: Sep 12, 2010