Let $(e_1,e_2,e_3)$ be the standard dual basis for $(\mathbb{R}^3)^\ast$. How can I show that $e_1\otimes e_2 \otimes e_3$ cannot be written as a sum of an alternating (or antisymmetric) tensor and a symmetric tensor?
Tensors – $e_1\otimes e_2 \otimes e_3$ as Sum of Alternating and Symmetric Tensors
tensor-productstensors
Best Answer
This is a very tentative answer; I am new to the world of tensor products. HOWEVER:
It seems to me that any 3-tensor that is the sum of an alternating and a symmetric tensor should have a value that is fixed under action of $A_3\subset S_3$ on the input. In other words, given any triple of vectors $(v_1,v_2,v_3)$ with each $v_i\in\mathbb{R}^3$, a symmetric tensor evaluates to the same thing on $(v_1,v_2,v_3)$ as on any permuted triple, while an alternating tensor should evaluate to the same thing on $(v_2,v_3,v_1)$ and $(v_3,v_1,v_2)$ (and the opposite on the others). Thus a tensor that is a sum of an alternating and a symmetric tensor should still evaluate to the same thing on $(v_1,v_2,v_3)$ as on $(v_2,v_3,v_1)$ and $(v_3,v_1,v_2)$, if not on the other permuations.
But, taking $v_1,v_2,v_3$ to be the standard basis for $\mathbb{R}^3$, your tensor evaluates to $1$ on $(v_1,v_2,v_3)$ but to $0$ on any other permutation of the $v$'s. This seems to me to imply that it can't be the sum of an alternating and a symmetric tensor.
I submit all this with great tentativeness and would appreciate a fact-check from any of you who know more about this.