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.
Best Answer
This is a question that touches on many issues. On the one hand, things are indeed easier to deal with in the language of the complexification. For $SL(2,\mathbb{C})$, there is a specific element $H$ in the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$, which acts diagonalizably on any finite dimensional representation. The eigenvalues are called weights, and in the case of an irreducible representation, they are all different. Specifically, the $n$-dimensional irreducible representation $\mathbf{n}$ admits a basis of weight vectors of weight $-n+1,-n+3,\dots,n-3,n-1$. Now in a tensor product, the tensor product of two eigenvectors for $H$ is an eigenvector with eigenvalue the sum of the eigenvalues of the two factors. There is a general result saying that the weight vector with highest possible weight will always generate an irreducible subrepresentation, and there there is an invariant complement to this subrepresentation. This allows you to compute decompositions in an elementary way.
Take the example of $\mathbf{3} \otimes \mathbf{3}$. In $\mathbf{3}$ you have weights $-2$, $0$, and $2$, so in the tensor product, the possible weights are $-4$, $-2$, $0$, $2$, and $4$ and the dimensions of the eigenspaces are $1$, $2$, $3$, $2$, and $1$. The weight vector of weight $4$ generates a subrepresentation $\mathbf{5}$, which has one weight vector for each of the listed weights. So in the complement, you get weights $-2$, $0$, and $2$ with dimensions $1$, $2$, and $1$. The highest of these gives you a subrepresentation $\mathbf{1}$ again including one of each of the weights, so there is just one copy of $\mathbf{0}$ left. To describe the result in terms of symmetry, you observe that the weight vector of the maximal weight $4$ is the tensor product of a highest weight vector with itself, so this sits in $S^2\mathbf{3}$. Either counting dimensions or by direct analysis of the weights you can see that $S^2\mathbf{3} \cong \mathbf{5} \oplus \mathbf{1}$ and $\Lambda^2\mathbf{3} \cong \mathbf{3}$.
This works similarly for general tensor products $\mathbf{n} \otimes \mathbf{m}$ for $n \geq m$. This is isomorphic to $\mathbf{n+m-1} \oplus \dots \oplus \mathbf{n-m+1}$ with dimension decreasing by two in each step (so there are always $m$ summands). For $n=m$, things look similar as in the special case, $S^2\mathbf{n} = \mathbf{2n-1} \oplus \mathbf{2n-5} \oplus \dots$ and $\Lambda^2\mathbf{n} = \mathbf{2n-3} \oplus \mathbf{2n-7} \oplus \dots$ with dimensions going down in steps of $4$.
You can always construct higher tensor powers step by step, say $\mathbf{2}\otimes \mathbf{2} \otimes \mathbf{2} \cong (\mathbf{3} \oplus \mathbf{1})\otimes \mathbf{3} \cong (\mathbf{3} \otimes \mathbf{2}) \oplus \mathbf{2}$ and then the first summand splits as $\mathbf{4} \oplus \mathbf{2}$. However, this is just one possible way to “decompose according to symmetry”, since one has distinguished the first two factors. In fact, the canonical way to decompose higher tensor products is just into so-called isotypical components. Here this would be $\mathbf{2} \otimes \mathbf{2} \otimes \mathbf{2} \cong \mathbf{4} \oplus W$, where $W$ is an invariant subspace isomorphic to a direct sum of two copies of $\mathbf{2}$. However, there are various possible realizations for this isomorphism, none of which is canonical. the systematic way to deal with this is simultaneously decomposing as a representation of $\mathfrak{sl}(2,\mathbb{C})$ and the permutation group $\mathfrak{S}_3$. This indeed leads towards Young diagrams, which are also needed to deal with $SL(n,\mathbb{C})$ and hence with $SU(n)$. If you are looking for literature in that direction, I would recommend the book of Fulton and Harris on representation theory.