# Simplification of Index Notation Expression.

• Feb 7th 2011, 04:06 PM
jameselmore91
Simplification of Index Notation Expression.
I was asked to simplify the following expression:

$\epsilon_{ijk}\epsilon_{ijk}$

and my simplification went as follows..

$\epsilon_{ijk}\epsilon_{ijk} = \epsilon_{jki}\epsilon_{ijk} = \delta_{jj}\delta_{kk} - (\delta_{jk})^2 = (3)(3) - (3)^2 = 0$

But my book gives an answer of

$\epsilon_{ijk}\epsilon_{ijk} = 2\delta_{kk} = 2(3) = 6$

Where did I go wrong?
• Feb 7th 2011, 04:22 PM
Ackbeet
The implied summation is over all three indices, since they're all repeated. The Levi-Civita symbol's relation to the Kronecker delta is given by

$\displaystyle\sum_{i=1}^{3}\varepsilon_{ijk}\,\var epsilon_{imn}=\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km},$

where the summation is explicitly indicated. Applying this to your situation yields

$\displaystyle\sum_{j=1}^{3}\sum_{k=1}^{3}\left(\su m_{i=1}^{3}\varepsilon_{ijk}\,\varepsilon_{ijk}\ri ght)=\sum_{j=1}^{3}\sum_{k=1}^{3}\left(\delta_{jj} \delta_{kk}-\delta_{jk}\delta_{kj}\right)=\sum_{j=1}^{3}\sum_{ k=1}^{3}\delta_{jj}\delta_{kk}-\sum_{j=1}^{3}\sum_{k=1}^{3}\delta_{jk}^{2}$

$\displaystyle=\sum_{j=1}^{3}\delta_{jj}\sum_{k=1}^ {3}\delta_{kk}-\sum_{j=1}^{3}\delta_{jj}^{2}=3\cdot 3-\sum_{j=1}^{3}\delta_{jj}=9-3=6.$

I think your mistake was in evaluating

$\displaystyle\sum_{j=1}^{3}\sum_{k=1}^{3}\delta_{j k}^{2}.$

That's equal to 3, not 9.