Clebsch–Gordan decomposition for sl2

The weight multiplicities of the tensor product are $w_m=\mathrm{\#}\{(r,s):0\le r\le a,0\le s\le b,a+b-2(r+s)=m\}$ by [L1]: the sum $m$ of the two weights $a-2r$ and $b-2s$ occurs once for each such pair.

For the right-hand side ${⨁}_{i=0}^{\min (a,b)}V(a+b-2i)$ the same weight $m$ occurs in the summand $V(a+b-2i)$ exactly when $\mathrm{\mid }m\mathrm{\mid }\le a+b-2i$ and $m\equiv a+b$ modulo $2$, so its multiplicity is $w_m^{\mathrm{\prime }}=\max (0,\min (\min (a,b),\lfloor (a+b-\mathrm{\mid }m\mathrm{\mid })\mathrm{/}2\rfloor )+1)$ in that parity case and $0$ otherwise.

The counts agree, $w_m=w_m^{\mathrm{\prime }}$ for every integer $m$. If $m\equiv ̸a+b(\mathrm{m}\mathrm{o}\mathrm{d}2)$ or $\mathrm{\mid }m\mathrm{\mid }>a+b$, both counts are zero. Otherwise put $j=(a+b-m)\mathrm{/}2$. For $m\ge 0$ one has $0\le j\le (a+b)\mathrm{/}2$, and the tensor count is $c_j=\max (0,\min (j,a)-\max (0,j-b)+1)\mathrm{.}$ A direct case check for $a\le b$ and $a>b$ identifies this with the right-hand count of step 1.2. The identity for $m<0$ follows from the symmetries $w_{-m}=w_m$ and $w_{-m}^{\mathrm{\prime }}=w_m^{\mathrm{\prime }}$ obtained by reflecting the weight strings.

Both sides are direct sums of irreducibles and the left side is completely reducible by [L2]; moreover in a completely reducible ${\mathfrak{s}\mathfrak{l}}_2$-module the multiplicity $c_n$ of $V(n)$ is determined by the weight multiplicities through $w_n={\sum }_{m\ge n,m\equiv n(2)}{c}_m$, so that $c_n=w_n-{\sum }_{m>n,m\equiv n(2)}{c}_m$ is recovered by downward induction on $n$.

Applying the recovery of step 2.2 to the two modules, whose weight multiplicities agree by step 2.1, gives equal multiplicities of every $V(n)$ and hence the multiplicity-one decomposition $V(a)\otimes V(b)\cong {⨁}_{i=0}^{\min (a,b)}V(a+b-2i)$.