Trivial representation in Clebsch-Gordan decomposition My professor defined the Clebsch-Gordan series as the direct sum decomposition of the tensor product of two representations of the Lie group SU(2):
$$
D_{j_1} \otimes D_{j_2} = D_{j_1+j_2} \oplus D_{j_1+j_2-1} \oplus ... \oplus D_{|j_1-j_2|}
$$
I think it's an interesting fact that the trivial representation $D_0$ will be in the tensor product if and only if $j_1=j_2$.
Is there an intuitive explanation for that? What common property of all systems with $j_1=j_2$ leads to the trivial representation being part of the tensor product?
This question could be generalized to a similar fact about $D_{1/2}$ and $D_1$.
 A: Generically, given a representation $V$ of a group, the tensor representation $V\otimes V$ will decompose into the symmetric and antisymmetric parts
$$ V\otimes V = \Lambda^2 V \oplus S^2 V$$
and in the case of the rotation group (or its universal cover), the symmetric 2-tensors have a certain invariant under rotations - their trace! So when $j_1 = j_2$, the tensor representation is really the 2-tensors (think: matrices) on the $D_{j_1}$ space, and the trace part of the matrices forms the trivial subrepresentation.
If $j_1 \neq j_2$, the tensor representation is not simply such sqaure matrices on the original space, so we can't speak of a trace, and we don't have a trivial subrepresentation.
A: Yes.  To obtain a scalar (or the trivial representation) one must multiply a representation $D^{\lambda}$ by its conjugate $D^{\lambda^*}$: the weights in the conjugate representation $\lambda^*$ are the negative of those in $\lambda$, and the scalar occurs when combining conjugate functions, so,as to produce a final function with weight $0$. 
In $SU(2)$ all representations are self conjugate, so $\lambda^*$ is equivalent to $\lambda$, and indeed the weights in $\lambda$ runs from $-\lambda$ to $\lambda$, so that any weight and its negative are in the representation.  This means that, to get a scalar, you need to have $\lambda_1=\lambda_2$.
In other groups, v.g. $SU(3)$, not all irreps are self-conjugate.  To get a scalar one must then combine two properly conjugate irreps, such as $(1,0)$ and $(0,1)$.  Indeed $(1,0)\otimes (0,1)=(0,0)\oplus (1,1)$ etc. 
