= Solution
Write $a=a_1$, $b=a_2$ with $a,b\geq0$, and let $\lambda=a\omega_1+b\omega_2$ be the <highest weight>. The <positive roots> are $\alpha_1$, $\alpha_2$ and $\alpha_1+\alpha_2$, and the <Weyl vector> is $\rho=\omega_1+\omega_2$. The <Weyl dimension formula> gives
$$
d(a,b)=\prod_{\alpha>0}\frac{(\lambda+\rho,\alpha)}{(\rho,\alpha)}
=\frac{a+1}{1}\frac{b+1}{1}\frac{a+b+2}{2},\qquad
\boxed{d(a,b)=\tfrac12(a+1)(b+1)(a+b+2).}
$$
For the <quadratic Casimir operator>, use the root-length normalization $C(\lambda)=(\lambda,\lambda+2\rho)$. Substituting the <Gram matrix> in <2/a/solution> yields
$$
\boxed{C(a,b)=\frac23\bigl(a^2+ab+b^2+3a+3b\bigr).}
$$
Taking the <trace> of the defining <quadratic Casimir operator> identity gives $d(a,b)C(a,b)=\dim(\mathfrak{su}(3))c(a,b)=8c(a,b)$, where $c$ is the <Dynkin index> in the matching normalization. Therefore
$$
\boxed{c(a,b)=\frac{(a+1)(b+1)(a+b+2)}{24}
\bigl(a^2+ab+b^2+3a+3b\bigr).}
$$
In particular $c(\mathbf3)=1$ and $c(\mathbf8)=6$. The common particle-physics convention uses $C_2=C/2$ and $T=c/2$, giving $C_2(\mathbf3)=4/3$ and $T(\mathbf3)=1/2$. The subsequent discrete <gauge anomaly> congruences use the integer <Dynkin index> $c$, while the sign test for the supplied <beta function (physics)> is independent of a common rescaling of all indices.
Back to article page
