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Math Help - Simplification of Index Notation Expression.

  1. #1
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    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?
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  2. #2
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    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.
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