1
$\begingroup$

Let $V$ be a finite-dimensional, real vector space and fix $k\in\mathbb N$. Recall that the symmetric group, $\mathfrak S_k$, acts linearly on $V^{\otimes k}$ by $$\sigma:v_1\otimes\dots\otimes v_k\mapsto v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(k)}.$$ According to Wikipedia, $\mathbf v\in V^{\otimes k}$ is a symmetric tensor if and only if $\sigma\mathbf v=\mathbf v$ for all $\sigma\in\mathfrak S_k$.

On the other hand, there is a linear symmetrisation map $$\mathrm{Sym}:V^{\otimes k}\to V^{\otimes k},\quad \mathrm{Sym}(\mathbf v):=\frac{1}{k!}\sum_{\sigma\in\mathfrak S_k}\sigma \mathbf v.$$


Is it the case that $\mathbf v$ is a symmetric tensor if and only if $\mathrm{Sym}(\mathbf v)=\mathbf v$? One direction I can manage, namely if $\mathbf v$ is a symmetric tensor, then $\mathrm{Sym}(\mathbf v)=\mathbf v$.

I'd appreciate any hints to prove or disprove: $$\mathrm{Sym}(\mathbf v)=\mathbf v\implies \mathbf v\text{ is a symmetric tensor}.$$

$\endgroup$
  • 1
    $\begingroup$ $\mathrm{Sym}(v)$ is always a symmetric tensor, so you immediately have that if $v$ is not a symmetric tensor, then $\mathrm{Sym}(v) \neq v$. $\endgroup$ – Joppy Apr 3 '18 at 11:26
  • $\begingroup$ Very simple - thanks. If you write a one-line answer, I will gladly accept. $\endgroup$ – Bowditch Apr 3 '18 at 11:33
1
$\begingroup$

Since $\mathrm{Sym}(v)$ is always a symmetric tensor, the contrapositive is immediate: if $v$ is not symmetric, then $\mathrm{Sym}(v) \neq v$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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