When dealing with the fundamental and adjoint representations, it is more useful to use tensor methods. I will below explain the methods for SU(2) and SO(3).
SU(2): Here the fundamental representation (i.e., the smallest non-trivial space on which the elements of SU(2) act) is spanned by 2-component vector X_{a}, a = 1,2. We often write [n] for the n-dimensional (irreducible) representation space spanned by objects having n independent components. Now, let us consider the tensor product X_{a}Y_{b} which represents [2] \otimes [2]. We write
<br />
X_{a}Y_{b} = \frac{1}{2}(X_{a}Y_{b} + X_{b}Y_{a}) + \frac{1}{2}(X_{a}Y_{b} - X_{b}Y_{a}). \ \ (1)<br />
Since the index a and b take values in the set {1,2}, the first object on the right-hand-side has only 3 independent elements; \{X_{1}Y_{1}, (1/2)(X_{1}Y_{2}+X_{2}Y_{1}),X_{2}Y_{2}\}. We think of it either as 3-component vector V_{i}, i = 1,2,3 or as 2 \times 2 symmetric matrix G_{ab}=G_{ba}. Both span the same representation space [3] of SU(2). The 2nd object in Eq(1) has only one independent element; \{(1/2)(X_{1}Y_{2}-X_{2}Y_{1})\} = (1/2)\epsilon^{ab}X_{a}Y_{b}. Thus, it belongs to the trivial (invariant) representation space [1]; notice that it is proportional to the invariant SU(2) “metric”. So, we can rewrite Eq(1) as
<br />
X_{a}Y_{b} = G_{ab} + (1/2) \epsilon_{ab} \epsilon^{cd} X_{c}Y_{d}.<br />
This shows that
[2]\otimes [2] = [3] \oplus [1].
Next, let us decompose the tensor product [3] \otimes [2]. For this we need to break the tensor G_{ab}X_{c} into symmetric and anti-symmetric parts. So, we write
<br />
G_{ab}X_{c} = \frac{1}{3}(G_{ab}X_{c} + G_{ca}X_{b} + G_{bc}X_{a}) + \frac{1}{3}(G_{ab}X_{c}-G_{bc}X_{a}) + \frac{1}{3}(G_{ab}X_{c}-G_{ca}X_{b}). \ \ (2)<br />
So what do we have here? The 1st object on the right-hand-side is a rank-3 totally symmetric tensor T_{(abc)} in 2-dimension. Such tensor has only 4 independent components; \{T_{(111)},T_{(112)},T_{(122)},T_{(222)}\}. We can think of this as a vector spanning 4-dimensional vector space [4].
The 2nd + 3rd objects on the RHS of Eq(2) can be written as
X_{abc}\equiv \epsilon_{ac}Z_{b} + \epsilon_{cb}Z_{a},
where
Z_{a} = G_{ab}\epsilon^{bc}X_{c}.
Now, using the identity
\epsilon_{ab}Z_{c} + \epsilon_{ca}Z_{b} + \epsilon_{bc}Z_{a} = 0,
we finally find
G_{ab}X_{c} = T_{(abc)} + \epsilon_{ab}Z_{c},
which proves
[3] \otimes [2] = [4] \oplus [2].
[SO(3)]: I will give you the result of decomposing the tensor product [3] \otimes [3] and leave you to fill in the details to conclude that
[3] \otimes [3] = [5] \oplus [3] \oplus [1],
follows from
U^{i}V^{j} = \frac{1}{2}(g^{ij} - \frac{1}{3}\delta^{ij}g) + \frac{1}{2}A^{ij} + \frac{1}{6}\delta^{ij}g,
where g^{ij} = g^{ji}, A^{ij} = -A^{ji} and g = \delta_{ij}g^{ij}.
Question for you: Why didn’t we subtract a trace from the symmetric SU(2) tensor G_{ab}?
Sam
