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: $$\mathrm{Sym}(\mathbf v)=\mathbf v\implies \mathbf v\text{ is a symmetric tensor}.$$

  • 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, 2018 at 11:26
  • $\begingroup$ Very simple - thanks. If you write a one-line answer, I will gladly accept. $\endgroup$
    – Bowditch
    Apr 3, 2018 at 11:33

1 Answer 1


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


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