# Tensor Algebra

mikeeey
Can some one write for me the Symmetric part of a third order tensor (as a tensor form)

Thanks .

## Answers and Replies

Homework Helper
What do you mean by "symmetric" for a third order tensor? A second order tensor would be represented (in a given coordinate system) by a 3 by 3 matrix (or 4 by 4 if you are counting time) and it would be symmetric if and only if $A_{ij}= A_{ji}$. But a third order tensor would be represented by a "3 by 3 by 3" array, $A_{ijk}$. And then we can have several different kinds of "symmetry":
$A_{ijk}= A_{jik}$, or $A_{ijk}= A_{ikj}$, or $A_{ijk}= A_{kji}$. You could even have a kind of symmetry by "rotating" the indices: $A_{ijk}= A_{kij}= A_{jki}$.

mikeeey
Ok i will explain .
$$T_{[abc]} = \frac{1}{6} \big( T_{abc} -T_{acb} + T_{bca} -T_{bac} + T_{cab} -T_{cba} \big)$$
this is the anti-symmetric part of a third order tensor, i want to write me the symmetric part of a third order tensor

mikeeey
$T_{[abc]} = \frac{1}{6} \big( T_{abc} -T_{acb} + T_{bca} -T_{bac} + T_{cab} -T_{cba} \big)$

lpetrich
The symmetric part is like the antisymmetric part, but with all signs +. Its symmetry should be easy to verify.

mikeeey
Thanks you very much . Lperrich.