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:

  1. 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 $W$?
  2. 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?
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    $\begingroup$ I don't have time right now to write a detailed answer, but look up "Schur functor". $\endgroup$ Commented Nov 23, 2016 at 15:51

1 Answer 1


This is a big topic and I'll just give a short and very incomplete reply. To get mor information, you should look up "Schur functor" as suggested by @JeremyRickard or "Young symmetrizer".

The answer to your question 1. is definitely positive, so we always have $\otimes^nV=\operatorname{Sym}^nV\oplus W$, and $W$ can always be described explicitly as the kernel of the symmetrization map $\otimes^nV\to \operatorname{Sym}^nV$. This is a special case of the statement that any representation of a finite group is completely reducible, so any invariant subspace has an invariant complement.

For question 2., things get much more difficult in higher dimensions, even if you just look at $\otimes^3V$ for $\dim(V)\geq 3$. You can simply decompose this as $V\otimes(\operatorname{Sym}^2 V\oplus\Lambda^2V)\cong (V\otimes \operatorname{Sym}^2V)\oplus (V\otimes\Lambda^2V)$. Now $V\otimes \operatorname{Sym}^2V=\operatorname{Sym}^3V\oplus W_1$, where as above $W_1$ can be identified with the kernel of the symmetrization map. Likewise $V\otimes\Lambda^2V=\Lambda^3V\oplus W_2$ where $W_2$ is the kernel of the alternation map. Now it turns out that $W_1$ and $W_2$ are isomorphic representations of $S_3$ but they are not isomorphic to $\operatorname{Sym}^3V$ or $\Lambda^3V$. (Once irreducibility has been proved, this simply follows from dimensions for example.) So the decomposition has the form $\otimes^3 V=\operatorname{Sym}^3V\oplus (W\oplus W)\oplus\Lambda^3V$, and $W$ is a "new" irreducible representation. As mentioned above, the general case can be described in terms of Young diagrams via Young symmetriyers.


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