Let $V$ be a vector space over a field $k$ of characteristic $0$ (not assumed to be algebraically closed, if it makes any difference). Consider the sequence of tensor powers $$V^{\otimes 2} = V\otimes V, \quad V^{\otimes 3} = V\otimes V\otimes V,\dots $$ It's well-known that there is a "braiding" representation of $S_n$ on $V^{\otimes n}$ given by permuting the tensor factors; i.e., $\sigma\in S_n$ acts linearly and on basis vectors via $$\sigma(v_{i_1}\otimes\dots\otimes v_{i_n}) = v_{i_{\sigma(1)}}\otimes\dots\otimes v_{i_{\sigma(n)}}.$$ The symmetric tensors are defined as the subspace $\text{Sym}^n (V)\subseteq V^{\otimes n}$ on which $S_n$ acts trivially.
Now for $n=2$ there is a very nice decomposition: namely $S_2=C_2$ is cyclic of order $2$, and the nontrivial automorphism $\epsilon$ of $V\otimes V$ has order $2$. Therefore its eigenvalues are $\pm 1$; $\text{Sym}^2 (V)$ is the $+1$ eigenspace and $\Lambda^2 (V)$ (the alternating square) is the $-1$ eigenspace. This decomposes $V\otimes V$ as a direct sum: $$V\otimes V = \text{Sym}^2 (V)\oplus \Lambda^2 (V).$$
My questions are:
- Does a generalisation of this decomposition hold in the higher tensor powers? i.e. Can we write $V^{\otimes n} = \text{Sym}^n (V)\oplus W$ for some explicit direct sum $W$?
- As a specialisation, assume that $\Lambda^2 (V) \cong k$ is one-dimensional, so that $\dim V = 2$, say with basis vectors $v_1$ and $v_2$. In this case I have tried working out the decomposition of the higher tensor powers into symmetric tensors plus "alternating" parts. For example, $V^{\otimes 3} = \text{Sym}^3 (V)\oplus W$ for a $4$-dimensional space $W$ on which $S_3$ acts nontrivially. I calculated that $$W = (V\otimes\Lambda^2 (V)) \oplus (\Lambda^2 (V)\otimes V)\cong V\oplus V.$$ I want to convince myself (and make precise) that in the decomposition $V^{\otimes n} = \text{Sym}^n (V)\oplus W$ for larger $n$, the direct summands appearing in $W$ are "all things we've seen before" (e.g. for $n=3$ there were just two copies of $V$, rather than some horrible "higher-alternating-square" thing), and so the only genuinely new things (representations of the symmetric group?) occurring in this sequence are the symmetric tensors $\text{Sym}^n (V)$. Can anyone explain how this holds in general?