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The Clebsch-Gordan decomposition for SU2 iswhere $j$ increases by one and every summand occurs once. The tensor product of Lie algebra representations is antihermitian for the product inner product, so an invariant subspace has an invariant orthogonal complement; this gives complete reducibility. To see the particular summands, suppose $j_1\ge j_2$. The product weight diagram has one state of weight $j_1+j_2$, two of weight $j_1+j_2-1$, and increasing multiplicities down to the plateau determined by $j_1-j_2$. Subtracting the weight string of $V_{j_1+j_2}$ leaves a single highest state of weight $j_1+j_2-1$. Repeating highest-weight character subtraction yields exactly the displayed sequence, down to $j_1-j_2$. Its dimensions exhaust the tensor product becauseThe case $j_2\ge j_1$ is identical with the factors exchanged.

Let $C_m^j(m_1,m_2)=⟨m_1:m_2|j,m⟩$ be the Clebsch-Gordan coefficients. Since the total $J_3$ is $J_3^{(1)}\otimes 1+1\otimes J_3^{(2)}$, orthogonality of its eigenspaces gives the selection rule $C_m^j(m_1,m_2)=0$ unless $m_1+m_2=m$. Normalize the coupled states by lowering from a unit highest-weight vector, as in part (i); a common phase for an entire summand is immaterial. ThenTake the inner product with $⟨m_1:m_2|$. On the other hand, the total ladder operator is $J_{\mp }^{(1)}\otimes 1+1\otimes J_{\mp }^{(2)}$, and $J_{\mp }^{†}=J_{\pm }$. Its action on the product bra is thereforeEquating the two evaluations and multiplying by $\sqrt{2}$ proves both signs of the ladder recurrence for Clebsch-Gordan coefficients:Terms with an index outside its allowed weight range are zero. Merely specifying eigenvectors of $J_3$ would allow unrelated phases for different $m$; the coherent positive-ladder convention above is what makes these recurrences hold with the displayed positive square roots.