# Levi-Civita Identity help

1. Aug 29, 2011

### cozmo

Can somebody show me how

$\epsilon_{mni}a_{n}(\epsilon_{ijk}b_j c_{k})$

Turns in to

$\epsilon_{imn}\epsilon_{ijk}a_{n}b_j c_{k}$

Something about the first $\epsilon$ I'm not seeing here when the terms are moved around.

Last edited: Aug 29, 2011
2. Aug 29, 2011

### vela

Staff Emeritus
The tensor changes sign when you transpose two indices, right? A cyclic permutation is an even number of transpositions, so $\epsilon_{mni} = \epsilon_{nim} = \epsilon_{imn}$.

3. Aug 30, 2011

### cozmo

When $\epsilon_{ijk}$ and $a_{n}$ change places the $\epsilon_{mni}$ changes to a cyclic permutation that is still positive and $\epsilon_{mni} =\epsilon_{imn}=\epsilon_{nim}$ but each one of these will give a different final answer.

I don't see how $\epsilon_{mni}$ turns to $\epsilon_{imn}$ when $\epsilon_{ijk}$ doesn't change.

This is from a problem proving the A X (B X C) = (A*C)B-(A*B)C identity.

4. Aug 30, 2011

### vela

Staff Emeritus
You can swap $a_n$ and $\epsilon_{ijk}$ because real numbers commute. Swapping them has nothing to do with reordering the indices of $\epsilon_{mni}$.

$\epsilon_{mni} = \epsilon_{imn}$ for all i, m, and n, so you can simply replace $\epsilon_{mni}$ with $\epsilon_{imn}$ in the summation. There's no relabeling of indices going on if that's what you think is happening.