# Rank 4 Tensor Symmetry Proof

## Homework Statement

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

Prove that
$$A_{ijkl} = A_{klij}$$

## The Attempt at a Solution

From the first two properties I concluded that:
$$A_{iikl} = 0 \qquad A_{ijkk} = 0$$

The last one leaded me to:
$$A_{ikli} = -A_{ilik} \qquad A_{ikkj} = -A_{ikjk}$$

However I don't see how this last one may help me.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
I suggest not trying to make any contractions and instead just apply the given symmetries, including the last one involving three tensor components.

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

Orodruin
Staff Emeritus
Homework Helper
Gold Member
$$A_{ijkl} + A_{iljk} = A_{klij} + A_{kjli}$$
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:
$$A_{jilk} + A_{jkil} = A_{lkji} + A_{likj}$$
The only thing it reminds me is of the third symmetry again, but if I use it I end up with a meaningless result:
$$A_{jlki} = A_{ljik}$$
Which translates in the first two symmetries.

Orodruin
Staff Emeritus