TITLE: $e_1\otimes e_2 \otimes e_3$ cannot be written as a sum of an alternating tensor and a symmetric tensor
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: The relevant context in which to put this result is in the representation theory of finite groups, although you don't need much of it. Let $V = \text{span}(e_1, e_2, e_3)$ be a $3$-dimensional real vector space. The tensor cube $V^{\otimes 3} = V \otimes V \otimes V$ is a $27$-dimensional vector space with a natural action of the symmetric group $S_3$. The space of symmetric tensors is the subspace spanned by copies of the trivial representation of $S_3$ (which sends everything to the identity), whereas the space of alternating tensors is the subspace spanned by copies of the sign representation of $S_3$ (which sends transpositions to $-1$ and the identity and $3$-cycles to $+1$). 
The tensor $e_1 \otimes e_2 \otimes e_3$, on the other hand, generates the regular representation of $S_3$, which contains a $2$-dimensional irreducible representation and in particular cannot be a direct sum of trivial and sign representations. Another way to see this, which is the argument in Ben Blum-Smith's answer, is that the subgroup $A_3 \subset S_3$ acts nontrivially on the regular representation of $S_3$ but trivially in both the trivial and sign representations.