# Definition of the (anti-)symmetric part of a tensor

Consider the following tensor: $$A_{ij} = B_{k,i}C_{k,j}$$ which can be decomposed into its symmetric and antisymmetric parts: $$A_{ij} = A_{ij}^s + A_{ij}^a$$. I'm all confused as to how to express these parts: if I use transpose of the tensor, I get \begin{align} A_{ij}^s = \tfrac{1}{2} (A_{ij} + A_{ji}) = \tfrac{1}{2}(B_{k,i}C_{k,j} + C_{j,k}B_{i,k}) \\ A_{ij}^a = \tfrac{1}{2} (A_{ij} - A_{ji}) = \tfrac{1}{2}(B_{k,i}C_{k,j} - C_{j,k}B_{i,k}) \end{align} but they are not symmetric/antisymmetric upon interchange of $$i,j$$. The more sensible decomposition would be using (anti-)symmetrization \begin{align} A_{ij}^s = B_{k,(i|}C_{k,|j)} = \tfrac{1}{2}(B_{k,i}C_{k,j} + B_{k,j}C_{k,i}) \\ A_{ij}^a = B_{k,[i|}C_{k,|j]} = \tfrac{1}{2}(B_{k,i}C_{k,j} - B_{k,j}C_{k,i}) \end{align} which does the job, but it's different from the standard transposition method.

• $A_{ji} = B_{kj}C_{ki}$.
– kipf
Jun 13, 2023 at 8:53
• "The more sensible decomposition would be...". Indeed: what you wrote in the first version of $A^s_{ij}$ is not the symmetrization because $C_{k,j}$ and $B_{i,k}$ are numbers that commute. Jun 13, 2023 at 9:03