Is Aijkl a Symmetric Rank 4 Tensor? Proof Needed!

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Ressurection
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Homework Statement


Let Aijkl be a rank 4 square tensor with the following symmetries:
[tex] A_{ijkl} = -A_{jikl}, \qquad A_{ijkl} = - A_{ijlk}, \qquad A_{ijkl} + A_{iklj} + A_{iljk} = 0,[/tex]

Prove that
[tex] A_{ijkl} = A_{klij}[/tex]

Homework Equations

The Attempt at a Solution


From the first two properties I concluded that:
[tex] A_{iikl} = 0 \qquad A_{ijkk} = 0[/tex]

The last one leaded me to:
[tex] A_{ikli} = -A_{ilik} \qquad A_{ikkj} = -A_{ikjk}[/tex]

However I don't see how this last one may help me.
 
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New attempt, got further but still missing something, hope this was what you meant.
From the third property:
[tex]A_{ijkl} + A_{iklj} + A_{iljk} = 0[/tex]
[tex]A_{klij} + A_{kijl} + A_{kjli} = 0[/tex]
Therefore:
[tex]A_{ijkl} + A_{iklj} + A_{iljk} = A_{klij} + A_{kijl} + A_{kjli}[/tex]
Since the first two properties refer to switching the first pair or the last pair of indexes, I can write:
[tex]A_{ijkl} + A_{kijl} + A_{iljk} = A_{klij} + A_{kijl} + A_{kjli}[/tex]
Leading to
[tex]A_{ijkl} + A_{iljk} = A_{klij} + A_{kjli}[/tex]
However I still have one extra term on each side that I can't deal with the same way as before.
 
Ressurection said:
[tex]A_{ijkl} + A_{iljk} = A_{klij} + A_{kjli}[/tex]
However I still have one extra term on each side that I can't deal with the same way as before.

What do you get if you simply do the following renaming of the indices in this equation: ##i \leftrightarrow j##, ##k \leftrightarrow \ell##? Does it remind you of something?
 
That would result in:
[tex]A_{jilk} + A_{jkil} = A_{lkji} + A_{likj}[/tex]
The only thing it reminds me is of the third symmetry again, but if I use it I end up with a meaningless result:
[tex]A_{jlki} = A_{ljik}[/tex]
Which translates in the first two symmetries.
 
Finally got it! Thanks a lot for the help