Symmetric Part of Product of 2 tank 2 tensors

Let $$A_{ab}$$ be a rank 2 tensor. Let $$P_{abcd} = A_{ab}A_{cd}$$. What are the symmetric and antisymmetric parts of $$P_{abcd}$$? Is it even possible to define symmetric and antisymmetric parts? (All indices range from 1 to 3).

I was thinking to use a similar definition as matrices and say that the symmetric part is $$1/2(P + P^T)$$ where $$P^T$$ is $$P$$ transpose. The transpose of $$P_{abcd}$$ could be defined as $$A_{cd}^TA_{ab}^T = P_{dcba}$$, per the rule governing the transpose of a product of matrices, but I don’t know if that would apply here considering that$$P_{abcd}$$is not a matrix but a 4 tensor. Then the antisymmetric part could be $$1/2(P - P^T)$$. But I don’t know how to rigorously prove these are the symmetric and antisymmetric parts. I would appreciate some guidance on whether or not it is possible to define symmetric and antisymmetric parts of $$P_{abcd}$$ in this case, and if so then If my form for the symmetric and antisymmetric parts is correct and how to go about proving a different form/my form correct.

• See, for example, this question May 20, 2019 at 5:35

Symmetrization and antisymmetrization operations are well-defined for arbitrary tensors. To symmetrize $$P_{abcd}$$ one would sum over all $$24=4!$$ permutations of the symbols $$a,b,c,d$$ and divide by $$24$$. In this case $$P_{(abcd)}=\frac{1}{24}\times \Bigl[(A_{ab}+A_{ba})(A_{cd}+A_{dc})+(A_{ac}+A_{ca})(A_{bd}+A_{db})\Bigr.$$ $$+(A_{ad}+A_{da})(A_{bc}+A_{cb})+(A_{bc}+A_{cb})(A_{ad}+A_{da})$$ $$\Bigl.+(A_{bd}+A_{db})(A_{ac}+A_{ca})+(A_{cd}+A_{dc})(A_{ab}+A_{ba})\Bigr].$$ Similarly for the antisymmetrization, except now the sign of the permutation applies: $$P_{[abcd]}=\frac{1}{24}\times \Bigl[(A_{ab}-A_{ba})(A_{cd}-A_{dc})-(A_{ac}-A_{ca})(A_{bd}-A_{db})\Bigr.$$ $$+(A_{ad}-A_{da})(A_{bc}-A_{cb})+(A_{bc}-A_{cb})(A_{ad}-A_{da})$$ $$\Bigl.-(A_{bd}-A_{db})(A_{ac}-A_{ca})+(A_{cd}-A_{dc})(A_{ab}-A_{ba})\Bigr].$$