Symmetric tensor decomposition in higher tensor powers

Solution 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 $\otimes^nV=Symm^nV\oplus W$, and $W$ can always be described explicitly as the kernel of the symmetrization map $\otimes^nV\to Symm^nV$. 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 $V\otimes(Symm^2 V\oplus\Lambda^2V)\cong (V\otimes Symm^2V)\oplus (V\otimes\Lambda^2V)$. Now $V\otimes Symm^2V=Symm^3V\oplus W_1$, 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 $Symm^3V$ or $\Lambda^3V$. (Once irreducibility has been proved, this simply follows from dimensions for example.) So the decomposition has the form $\otimes^3 V=Symm^3V\oplus (W\oplus W)\oplus\Lambda^3V$, and $W$ is a "new" irreducible representation. As mentioned above, the general case can be described in terms of Young diagrams via Young symmetriyers.