# Kernel of the Symmetrizing Map $Sym:\bigotimes^k V\to \bigotimes^k V$

$\DeclareMathOperator{\sym}{Sym}$ 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.

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

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.
• 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. Jun 27, 2015 at 19:42
• 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. Jun 27, 2015 at 19:46
• 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.) Jun 27, 2015 at 19:48
• 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. Jun 27, 2015 at 19:55
• Add and subtract $\alpha^{\theta}$. Jun 27, 2015 at 21:29