How to compute the dimension of an $(m_1, m_2)$ irreducible representation of $\mathfrak{sl}(3,\mathbb{C})$?

Question

We know that finite dimensional irreducible (complex) representations of the Lie algebra $\mathfrak{sl}(3,\mathbb{C})$ are indexed by their highest weight $(m_1, m_2)$ (both non negative integers). Is there an abstract way to compute the dimension of such representation, and if yes, how? I, of course, don't want to make all the generators act on the highest weight vector and see which ones are redundant or zero.

Thanks.

Answer 1

Denote the fundamental dominant weights by $\omega_1,\omega_2$. Abbreviate the $\mathfrak{sl}_3(\mathbb{C})$-module $V(m_1ω_1+ m_2ω_2)$ by $V(m_1,m_2)$. Then Weyl’s dimension formula shows $$ \dim V(m_1,m_2)= \frac{1}{2}(m_1 +1)(m_2 +1)(m_1 +m_2 +2). $$ For a reference, see Humphrey's book on Lie algebras and representation theory.