$Alt(T)=0$ if $T$ is a symmetric tensor

Question

Question is to prove that $Alt(T)=0$ if $T$ is a symmetric tensor.

We have $$Alt(T)=\sum_{\sigma}sgn(\sigma)T^{\sigma}$$ As $T$ is symmetric we have $T^{\sigma}=T$ for all $\sigma$. So, we have $$Alt(T)=\sum_{\sigma}sgn(\sigma)T=T\sum_{\sigma}sgn(\sigma)$$ As no of odd permutations is same as that of no of odd permutations we have $\sum_{\sigma}sgn(\sigma)=0$ thus, $Alt(G)=0$.

Now, another question is, does the converse hold?

Suppose $Alt(T)=0$ do we have that $T$ is a symmetric tensor?

I see that this is true for $2$ tensor. Suppose $T$ is a $2$ tensor then,

$0=Alt(T)=sgn(1)T+sgn(1~2)T=T-T^{\sigma}$ implies that $T=T^{\sigma}$. Thus, $T$ is a symmetric tensor.

I am sure this is true for all $k$ tensors but i am unable to prove.

Do i have to go by contradiction or what... I am not able to guess some path..

Please provide some hints..

Answer 1

There is nice characterisation of tensors which $Alt(T)=0.$

Thm. Let $V$ be a vector space over $\mathbb{R}$ (without any additional assumptions). Consider subspace $N^n(V)$ of $\otimes^nV$ generated by elements $v_1\otimes\dots\otimes v_n$ such that $v_i=v_j$ for at least one pair $i\neq j.$ We have that $$\ker(Alt)=N^n(V).$$ Equivalently. For every $T\in\otimes^nV$ $$Alt(T)=0\iff T\in N^n(V).$$

It requires some work, but you can find it in Greub's Multilinear Algebra.

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EDIT Explicit example

Consider space $V=(\mathbb{R}^2)^*.$ Then $3-$tensors in $V$ can be treated as multilinear maps $$T:\mathbb{R}^2\times\mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}.$$ Set $T$ by formula $$T((a,b),(c,d),(e,f))=ace+bde-acf-bdf.$$ Multilinear: easy to verify.

Symmetric: Nope $$T((1,1),(1,1),(1,0))=2\neq 0=T((1,0),(1,1),(1,1))$$

$Alt(T)=0.$ Yes. See first that $$T((a,b),(c,d),(e,f))=T((c,d),(a,b),(e,f))$$ (this is the main idea, compare with the thm. in quotation above). In fact

$$T((a,b),(c,d),(e,f))=ace+bde-acf-bdf=cae+dbe-caf-dbf=T((c,d),(a,b),(e,f)).$$ Now $Atl(T)$ has $6$ elements. To permutation $\sigma=id$ corresponds $\gamma=(12),$ such that $$T^\sigma=T^\gamma\hspace{5pt}\text{and}\hspace{5pt}sgn(\sigma)=-sgn(\gamma).$$ Similarly to $\sigma=(13)$ you have $\gamma=(123)$ and to $\sigma=(23)$ you have $\gamma=(132).$ Hence the whole sum $Alt(T)$ cancels out.