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I am having trouble in demonstrating that under SU(2) transformations the adjoint representation of SU(N) transforms as one spin 1, 2(N-2) spin $\frac12$ and singlets. I am trying to demonstrate it from $N \otimes \bar{N} = 1 + A$ where $A$ is the adjoint representation; and the fact that an N vector of SU(N) decomposes as one $j=\frac12$ plus two $j=0$. So basically: $(1\oplus^{N-2}0)\otimes(1\oplus^{N-2}0)= ?$. But I end up with one spin 1, $2N-3$ spin $\frac12$, and $(N-2)^2+1$ spin 0. Embarrassing. Is my equation with the question mark wrong?

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Write things out more carefully. Under the decomposition $\mathfrak{su}(N) \rightarrow \mathfrak{su}(2)$, you decompose the $\mathfrak{su}(N)$ vector representation into $$ N \to \tfrac{1}{2} \oplus 0^{\oplus(N-2)}\,. $$ Then, $$ N \otimes \bar{N} \rightarrow (\tfrac{1}{2} \oplus 0^{\oplus(N-2)})^{\otimes 2} = 1 \oplus \tfrac{1}{2}^{\oplus(2N-4)} \oplus 0^{\oplus((N-2)^2 + 1)}\,. $$ That's what you want. Don't forget that one of those singlets is the singlet in ${\rm Adj} \oplus {\rm Triv}$, since under the reduction $\mathfrak{su}(N) \rightarrow \mathfrak{su}(2)$, ${\rm Triv}$ must reduce to the spin 0 singlet representation.

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