Levi-Civita Product to Kronecker Delta Difference of Products: EMII Notes 2014_08_08

Summary of what's gone on before.  In the previous set of notes from the 6th, (there were no notes on the 7th), it was pointed out that the 'convenient' comment on page 11 of the notes was to cryptic.  Today's entire half hour was spent figuring out the following derivation that sprang from the convenient comment.  We want to derive:

$\epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$

Here's what to do.  First remember sum notation gives you

$\epsilon_{ijk}\epsilon_{lmk} = \epsilon_{ij1}\epsilon_{lm1} + \epsilon_{ij2}\epsilon_{lm2} +\epsilon_{ij3}\epsilon_{lm3}$

Here's the first use of the big trick for the day.  Because of the properties of the Levi-Civita symbol, $\epsilon_{ijk}$, on the indices 2 and 3 will make the first term non-zero, while only the pairs 1,3 and 1,2 will make the other two terms non-zero.  Once any of these combinations is chosen however, the other two terms will vanish.  Given that, let's get to work on the first term, ignoring the other two that evaluate to zero.

$\epsilon_{ij1}\epsilon_{lm1}$

For this expression, the only index choices that will evaluate as non-zero are shown in the table below:

Index Choices and Results

i	j	l	m	Result
2	3	3	2	-1
2	3	2	3	+1
3	2	2	3	-1
3	2	3	2	+1

At this point, we have to realize tht we have two indices that reutrn either one or negative one.  This should bring the Kronecker delta to mind.  Note that when $i=l$ and $j=m$ we get $+1$, while when $i=m$ and $j=l$ we get $-1$.   We can then right these combinations as

$\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$

So,

$\epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$

and we're done.

Handy reference that I found after doing the work:

https://en.wikipedia.org/wiki/Levi-Civita_symbol

The dudes: