Kernel of the Symmetrizing Map

[caffeinemachine]

Let $V$ be a finite dimensional vector space over a field of characteristic $0$ and let $sym:\bigotimes^k V\to \bigotimes^k V$ be the map defined as
$$
sym(\alpha)=\frac{1}{r!}\sum_{\sigma\in S_k}{^\sigma}\alpha
$$
where $S_k$ is the permutation group on $k$ letters and ${^\sigma}\alpha$ denotes the action of $\sigma$ on $\alpha$.

Since $sym^2=sym$, the kernel of $sym$ is $range(I-sym)$.

I want to show that each member of the kernel of $sym$ can be written as as a sum of the tensors of the form
$$
\beta-{^\tau}\beta\tag{1}
$$
where $\beta\in \bigotimes^k V$ is a pure tensor and $\tau$ is a transposition.

So for an example take $k=3$ and choose $u_1\otimes u_2\otimes u_3-u_2\otimes u_3\otimes u_1$ in $\ker sym$.
How do I write this in the desired form?

Thanks.

 

[jvicens]

Thank you for your question. It seems like you have already made some progress in understanding the map $sym$ and its kernel. Let me provide some further explanation and guidance to help you prove the desired result.

Firstly, let's recall that a transposition $\tau$ is a permutation that swaps two elements and leaves the rest unchanged. In the case of $k=3$, the transpositions are $\tau_1=(1,2)$, $\tau_2=(1,3)$, and $\tau_3=(2,3)$. Note that these transpositions generate the permutation group $S_3$.

Now, let's consider the pure tensor $\alpha=u_1\otimes u_2\otimes u_3-u_2\otimes u_3\otimes u_1$. We want to write this in the form of (1), with $\beta\in \bigotimes^3 V$ and $\tau$ a transposition. From the definition of $sym$, we have
$$
sym(\alpha)=\frac{1}{r!}\sum_{\sigma\in S_3}{^\sigma}\alpha=\frac{1}{3}(u_1\otimes u_2\otimes u_3-u_2\otimes u_3\otimes u_1+u_2\otimes u_1\otimes u_3-u_1\otimes u_3\otimes u_2+u_3\otimes u_1\otimes u_2-u_3\otimes u_2\otimes u_1).
$$
Note that this sum contains all possible permutations of the elements in $\alpha$. Now, let's consider the difference $\alpha-sym(\alpha)$. This is equal to
$$
\alpha-sym(\alpha)=\frac{2}{3}(u_1\otimes u_2\otimes u_3-u_2\otimes u_3\otimes u_1).
$$
We can see that this is a scalar multiple of $\alpha$, which is in the kernel of $sym$. Therefore, we can write $\alpha$ as a sum of tensors of the form (1), with $\beta=\frac{2}{3}(u_1\otimes u_2\otimes u_3-u_2\