Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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
2  
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
add comment

2 Answers 2

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.

share|improve this answer
1  
This is correct. –  Qiaochu Yuan Dec 30 '11 at 0:24
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.