TITLE: Symmetric decompositions of $SU(2)$ representations.
QUESTION [8 upvotes]: Let us consider the representation theory of $SU(2)$. There is a unique irreducible representation of dimension $n$ for each $n \ge 1$, which we will denote $\mathbf{n}$, with the defining $2$-dimensional representation $\mathbf{2}$ being called the fundamental representation. It is well known that all the irreducible representations can be built up from tensor powers of the fundamental representation. Moreover, from the Schur-Weyl duality, we have ways of classifying the symmetries of the tensor representations.
For example, the tensor power $\mathbf{2}\otimes \mathbf{2}$ decomposes as
$$\mathbf{2}\otimes \mathbf{2} = \mathbf{1} \oplus \mathbf{3},$$
where $\mathbf{1} \simeq\Lambda^2(\mathbf{2})$ is the space of alternating tensors over $\mathbf{2}$ and where $\mathbf{3} \simeq S^2(\mathbf{2})$ is the space of symmetric tensors over $\mathbf{2}$. 
Likewise, for higher tensor powers of the fundamental representation, we can decompose the space into irreducible representations composed of tensors of mixed symmetry, corresponding to different Young tableaux. 
As far as I'm aware, the techniques that I know of only allow us to do this sort of decomposition for tensor powers of the fundamental rep. Is there anyway to generalize this to other irreducible reps? For example, we know that
$$\mathbf{3} \otimes \mathbf{3} = \mathbf{1} \oplus \mathbf{3} \oplus \mathbf{5}.$$
Do these irreducible subspaces correspond to symmetric/antisymmetric tensors over $\mathbf{3}$? If so, how can we find the relevant decomposition? Are there ways of classifying the symmetric/anti-symmetric/mixed-symmetric subspaces of $\mathbf{n}^{\otimes k}$ in general?
I am looking for references relating to the above questions (although if the answer is short and simple, feel free to just answer the question). I don't know very much representation theory, so I don't know if this is a well known problem with a solution or if it's intractable. If there are solutions for $SU(N)$ in general, I would be interested to know as well.

REPLY [2 votes]: Good question! The answer,  known as the Cayley-Sylvester formula, has been figured out in the 1850's. First let me switch notations from $\mathbf{n}$ to $S^{n-1}\mathbf{2}$, for the $n$-dimensional irreducible representation of $SU(2)$ realized as the $(n-1)$-th symmetric power of the fundamental two-dimensional one. In general, one has
$$
S^m(S^n\mathbf{2})=\oplus_{w=0}^{\lfloor\frac{mn}{2}\rfloor} M_{m,n,w}\ S^{mn-2w}\mathbf{2}
$$
where the multiplicities $M_{m,n,w}$ are given by
$$
M_{m,n,w}=[m,n,w]-[m,n,w-1]\ .
$$
Here, $[m,n,w]$ denotes the number of integer partitions of weight $w$ with at most $m$ parts of length at most $n$. By convention $[m,n,w]=0$ if $w<0$.
For the totally antisymmetric part, one can reduce to the previous situation via the isomorphism
$$
\wedge^m S^n\mathbf{2}=S^m(S^{n-m+1}\mathbf{2})\ .
$$
For more on this Wronskian isomorphism, see my article with Chipalkatti page 46.
For example, the symmetric part of $\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}$, i.e., $S^{3}S^2\mathbf{2}$ can be computed as follows. One has $[3,2,0]=1$ (the empty partition), $[3,2,1]=1$ (the partition (1)), $[3,2,1]=2$ (the partitions (2) and (1,1)), and $[3,2,3]=2$ (the partitions (2,1) and (1,1,1)). Therefore $M_{3,2,0}=1$,
$M_{3,2,1}=0$, $M_{3,2,2}=1$ and $M_{3,2,0}=0$. So
$$
S^{3}\mathbf{3}=\mathbf{7}\oplus\mathbf{3}\ .
$$
The formula for the decomposition of $S^m(S^n\mathbf{2})$ was discovered by Cayley in a letter from around 1854 starting by "Dear Sylvester, Eureka." However he left a substantial gap in the derivation and the first complete proof is due to Sylvester in an 1878 article. This is also the first proof of unimodality of $q$-binomial coefficients.
A nice account of the history around the Cayley-Sylvester formula is in the article "The rise of Cayley's invariant theory (1841–1862)" by T. Crilly.
I don't know if one has similar nice formulas for mixed symmetry. There is a formula for $s_{\lambda}(1,q,q^2,\ldots)$ due to Stanley, but I am not sure it is easy to extract the multiplicities from it.
The formula (Stanley's hook-content formula) is a $q$-analogue of the hook length formula and says
$$
s_{\lambda}(1,q,q^2,\ldots,q^{m-1})=q^{\sum_i (i-1)\lambda_i}
\prod_{\square\in\lambda}\frac{1-q^{m+c(\square)}}{1-q^{h(\square)}}
$$ 
where $h(\square)$ is the hook and $c(\square)$ is the content.
In the $SU(N)$ case, some examples are known but in the general this a major open problem in representation theory called the plethysm problem.