Understanding the Properties of Levi-Civita Symbol in Tensor Calculus

[fluidistic]

Homework Statement

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}.

Homework Equations

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

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?

 

[vela]

Homework Statement

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}.

Homework Equations

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

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.

(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

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?

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.

 

[fluidistic]

(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.

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?

 

[vela]

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?

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

 

[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. :)