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.
 A: 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:
$$
P_{[abcd]}=\frac{1}{24}\times \Bigl[(A_{ab}-A_{ba})(A_{cd}-A_{dc})-(A_{ac}-A_{ca})(A_{bd}-A_{db})\Bigr.
$$
$$
+(A_{ad}-A_{da})(A_{bc}-A_{cb})+(A_{bc}-A_{cb})(A_{ad}-A_{da})
$$
$$
\Bigl.-(A_{bd}-A_{db})(A_{ac}-A_{ca})+(A_{cd}-A_{dc})(A_{ab}-A_{ba})\Bigr].
$$
A few comments aimed at (what I guess is) the spirit of the question you are asking, and not the literal question itself.
One thing to note here is that, unlike in the rank 2 case, not all tensors are equal to the sum of their symmetrization and their antisymmetrization. The reason for this is due to group theory. Specifically, the rank 2 case is governed by the action of the symmetric group on 2 elements, which has only 2 characters: the identity (corresponding to symmetrization) and the alternating character (corresponding to antisymmetrization). But in the rank 4 case, one would need to sum over all characters of the symmetric group on 4 elements, resulting in more tensors in the symmetric/antisymmetric decomposition.
