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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?

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Assuming you mean that the parts must me alternating resp. symmetric in all pairs of indices, $A_{123} = A_{231}$ holds for both alternating and symmetric tensors. However your $e_1\otimes e_2\otimes e_3$ has $A_{123}=1$ and $A_{231}=0$. – Henning Makholm Dec 29 '11 at 23:47
Could you elaborate on this, please? What is $A_{123}$? – user20353 Dec 29 '11 at 23:51
$A_{123}$ is the coefficient of $e_1\otimes e_2\otimes e_3$ when the tensor $A$ is expanded in the basis $( e_i\otimes e_j\otimes e_k )_{i,j,k\in\{1,2,3\}}$ for $\mathbb R^3\otimes \mathbb R^3\otimes \mathbb R^3$. – Henning Makholm Dec 29 '11 at 23:52
How can I see that $A_{123}=A_{231}$ holds for both symmetric and alternating tensors? – user20353 Dec 30 '11 at 0:04
If the tensor is completely symmetric, then $A_{123}=A_{213}$ and $A_{213}=A_{231}$ by definition of symmetry. If it is completely alternating, then $A_{123}=-A_{213}$ and $A_{213}=-A_{231}$, by definition of antisymmetry. – Henning Makholm Dec 30 '11 at 1:14
up vote 10 down vote accepted

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.

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This is correct. – Qiaochu Yuan Dec 30 '11 at 0:24

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.

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