What can we say about the dimensionality of the irreducible representations of $SU(3)$

Question

I've read about $SU(2)$ group in my physics course. The irreducible representations of $SU(2)$ are $(2j+1)$ dimensional where $j=0,1/2,1,3/2,...$

What can we say about the dimensionality of the irreducible representations of $SU(3)$?

Answer 1

The dimensions of finite dimensional irreducible representations of $SU(3)$ are given by Weyl's dimension formula. The root system of $SU(3)$ is of type $A_2$, so the positive roots are $\alpha_1,\alpha_2,\alpha_1+\alpha_2$, the half-sum of positive roots it $\rho=\lambda_1+\lambda_2$ where $\lambda_i$ are the fundamental dominant weights determined by duality: $(\lambda_i,\alpha_j)=\delta_{ij}$.

When we look at the irreducible representation $V(\lambda)$ with highest weight $\lambda=m_1\lambda_1+m_2\lambda_2$, $m_1,m_2$ non-negative integers, Weyl's formula gives $$ \dim V(\lambda)=\frac{(\lambda+\rho,\alpha_1)(\lambda+\rho,\alpha_2)(\lambda+\rho,\alpha_1+\alpha_2)}{(\rho,\alpha_1)(\rho,\alpha_2)(\rho,\alpha_1+\alpha_2)}=\frac{(m_1+1)(m_2+1)(m_1+m_2+2)}{2}. $$ So

• $V(\lambda_1)$ and $V(\lambda_2)$ are 3-dimensional. One is the defining rep, the other its dual. Depends on how you select the positive roots which is which.
• $V(\lambda_1+\lambda_2)$ is 8-dimensional. This is the adjoint representation.
• $V(2\lambda_1)$ and $V(2\lambda_2)$ are both 6-dimensional. They are the symmetric squares of the 3-dimensional reps.
• $V(3\lambda_1)$ and $V(3\lambda_2)$ are 10-dimensional. They are the symmetric third powers of the 3-dimensional reps, and IIRC played a role in the development of particle physics (as did the adjoint rep).
• In general $V(m_1\lambda_1+m_2\lambda_2)$ and $V(m_2\lambda_1+m_1\lambda_2)$ are each others duals.