Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.