$\DeclareMathOperator{\sym}{Sym}$ Let $V$ be a finite dimensional vector space over a field of characterisitc $0$ and $\sym:\bigotimes^k V\to \bigotimes^k V$ be the map given by $$ \sym(\alpha)=\frac{1}{k!}\sum_{\sigma\in S_k}\ ^\sigma\alpha $$ where $S_k$ is the symmetric group on $k$ letters and $^\sigma\alpha$ denotes the action of $\sigma$ on $\alpha$.

What is the kernel of the map $\sym:\bigotimes^k V\to \bigotimes^k V$?

It is clear that all tensors of the form $(u\otimes v-v\otimes u)\otimes\beta$, where $\beta\in \bigotimes^{k-2}V$ are in the kernel. I suspect that these are all the members in the kernel but am unable to prove it.

Can somebody help? Thanks.

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    $\begingroup$ Your base field is of char. $0$, right? $\endgroup$ Jun 27, 2015 at 18:50
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    $\begingroup$ Yes. It's not char 2 for sure. But a proof for char 0 is also good enough for me. (Edited the question). $\endgroup$ Jun 27, 2015 at 18:52
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    $\begingroup$ Well you write $\frac{1}{k!}$ ... $\endgroup$ Jun 27, 2015 at 18:52
  • $\begingroup$ Good point. So I should have char 0. $\endgroup$ Jun 27, 2015 at 18:53

1 Answer 1


The symmetrizer $S: \bigotimes^k V \to \bigotimes^k V$ is idempotent. Hence, $\ker(S) = \mathrm{im}(\mathrm{id}-S)$. This is generated by elements of the form $\alpha-{}^\sigma \alpha$, where $\sigma$ is some permutation.

  • $\begingroup$ That's great. I just have one more question. Can each member in the kernel be written as sums of members of the form $\alpha- {^\tau}\alpha$ where $\alpha$ is a pure tensor and $\tau$ is a transposition? Thanks. $\endgroup$ Jun 27, 2015 at 19:42
  • $\begingroup$ Because of $\ker(S)=\mathrm{im}(\mathrm{id}-S)$, the kernel is generated by elements of the form $\alpha - {}^{\sigma} \alpha$, where $\sigma$ is some permutation. Now write $\sigma$ as a product of transpositions. As a warm-up, do the case that $\sigma$ is a product of two transpositions. $\endgroup$ Jun 27, 2015 at 19:46
  • $\begingroup$ So suppose for example $\sigma$ is a product of two transpositions $\tau$ and $\theta$. I need to express $\alpha-{^{\tau\theta}}\alpha$. I am unable to see how to get this into the desired form. (Okay let me try for some more time.) $\endgroup$ Jun 27, 2015 at 19:48
  • $\begingroup$ Sorry. But I don't get it. Have spent enough time on this. Can you explain the case when $\sigma$ is a product of two transpositions? Many thanks. $\endgroup$ Jun 27, 2015 at 19:55
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    $\begingroup$ Add and subtract $\alpha^{\theta}$. $\endgroup$ Jun 27, 2015 at 21:29

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