Tensor product - Direct Sum equality for SU(2) (I initially posted this on Physics SE, but haven't had helpful responses so I'm reposting here.)
I'm working on Problem 3A of Georgi's Lie Algebra's in Particle Physics. I'm trying to show that, $$\{j\} \otimes \{s\} = \sum\limits_{\oplus\ell =|s-j|}^{s+j}\{\ell\}$$ where $\{k\}$ is the spin-$k$ representation of $SU(2)$. I've seen an explicit version of this in an answer to this post (https://physics.stackexchange.com/questions/231497/decomposition-of-tensor-product-space-into-direct-sum) using characters.
My current thinking is to use the property $$J_\alpha^{j\otimes s} = [J_\alpha^j]_{ab}\delta_{cd}+\delta_{ab}[J_\alpha^s]_{cd}.$$
But I'm generally just confused about how I'm supposed to use the highest weight decomposition for this and any tips would be very much appreciated.
 A: The spin-$j$ irrep $V_j$ (where $j$ is an integer or half-integer) has dimension $2j+1$. It has a basis consisting of vectors with weights $-2j,\cdots,2j$ (each weight is the last $+2$). The complex lie algebra called $\mathfrak{su}(2)$ (not to be confused with the real lie algebra...) has three generators: (i) a diagonal operator whose eigenvectors are the weight vectors and whose eigenvalues are the weights; (ii) a raising operator which turns one basis weight vector into the next one (the one with the next highest weight), or $0$ if that's not possible; and (iii) a lowering operator which turns one basis weight vector into the previous one (the one with the next lowest weight), or $0$ if that's not possible. The spin is half of the highest weight. Given an arbitrary rep, to decompose it into irreps, find its highest weight vectors (i.e. vectors annihilated by the raising operator which are eigenvectors of the diagonal operator). You should be able to find one per irrep (up to linear combinations for irreps of the same type) and determine the spin of the irreps from the weights themselves.
If $u\otimes v$ is the element of a tensor product of two spin reps and we apply an operator $Z$ to it from the lie algebra, we get $(Zu)\otimes v+u\otimes(Zv)$ (by definition - it's supposed to mimic the product rule because of how lie algebras are related to Lie groups). Thus, if $u$ and $v$ are weight vectors, this splits the tensor of two weight vectors into a sum of a pair of such tensors. If we form an array of pure tensors of weight vectors, that means the raising operator turns any entry of the array into the sum of the entries to the right and below it.
Let's consider an array for a spin-$\frac{3}{2}$ and spin-$2$ rep:
$$ \begin{array}{c|ccccc}
   & -4 & -2 & 0 & +2 & +4 \\ \hline
-3 & \cdot & \cdot & \cdot & \cdot & \color{magenta}- \\
-1 & \cdot & \cdot & \cdot & \color{magenta}+ & \color{limegreen}+ \\
+1 & \cdot & \cdot & \color{magenta}- & \color{limegreen}- & \color{cyan}- \\
+3 & \cdot & \color{magenta}+ & \color{limegreen}+ & \color{cyan}+ & \color{darkorange}+
\end{array} $$
The rows are labelled with the weights of the spin-$\frac{3}{2}$ rep and the columns are labelled with the weights of the spin-$2$ rep, in increasing order. The coefficients along the colored diagonals tell us how to form linear combinations which are highest weight vectors of the tensor products (i.e. tensors annihilated by the raising operator). For example, if we use $a_\lambda$ to denote a weight-$\lambda$ vector from $V_{3/2}$ and $b_\mu$ to denote a weight-$\mu$ vector from $V_2$, then the magenta diagonal corresponds to the linear combination
$$ a_{+3}\otimes b_{-2}-a_{+1}\otimes b_0+a_{-1}\otimes b_{+2}-a_{-3}\otimes b_{+4} $$
which we can check has weight $+1$ (the sum of the weights of any pure tensor in the combination). I leave it as an exercise to show these are all the highest weight vectors (start with "suppose a highest weight vector has this pure tensor in it" then go "well then it must also have this other one, but with opposite coefficient" and so on). We can see in general then for $V_r\otimes V_s$ (with $s\ge r$) with these diagonals at the corner of the array we get one irrep of spin $t$ from $t=s-r$ all the way to $t=s+r$, increasing by $1$ between weights:
$$ \quad V_r\otimes V_s = \bigoplus_{s-r}^{s+r} V_t \quad (s\ge r) $$
