= Solution
Choose a <Cartan subalgebra>, <simple roots> $\alpha_1,\ldots,\alpha_r$, and <root vectors> $e_i,f_i$ normalized by $[e_i,f_i]=h_i=\alpha_i^\vee$. Let
$$
\rho^\vee=\tfrac12\sum_{\alpha\in R^+}\alpha^\vee,\qquad h=2\rho^\vee=\sum_i c_i h_i,
\qquad e=\sum_i e_i,\qquad f=\sum_i c_i f_i.
$$
Here the <coroots> lie in the <Cartan subalgebra>, and $\alpha_i(\rho^\vee)=1$. This last identity is the <Weyl vector> identity applied to the <dual root system>. For distinct <simple roots>, $\alpha_i-\alpha_j$ is not a root, hence $[e_i,f_j]=0$. Consequently
$$
[e,f]=\sum_i c_i h_i=h,\qquad [h,e]=2e,\qquad[h,f]=-2f.
$$
These are exactly the <sl2 Lie algebra> relations. For nonzero $\mathfrak g$ the resulting homomorphism from $\mathfrak{sl}_2$ is nonzero and therefore injective, since $\mathfrak{sl}_2$ is simple. Its image is a <Principal sl2 subalgebra>. Equivalently its raising element is a <principal nilpotent element>, a nilpotent element with <Lie algebra centralizer> of <dimension> equal to the rank. The construction works for a <semisimple Lie algebra> with several simple factors by taking the triple in every factor, giving one diagonally embedded $\mathfrak{sl}_2$. This is the <construction of a principal sl2 triple>.
For a representation $V$, restrict to this subalgebra and take its <formal character>. If $V_\mu$ is the <weight space> for the original <Cartan subalgebra>, the resulting <q-character of a highest-weight representation>, or the analogous q-character for any finite-dimensional representation, is
$$
\boxed{\operatorname{ch}_q V=\operatorname{tr}_V(q^{2\rho^\vee})=\sum_\mu\dim(V_\mu)\,q^{\mu(2\rho^\vee)}.}
$$
It is independent of conjugating the chosen principal triple. In particular a root $\beta=\sum_i k_i\alpha_i$ has grading $\beta(h)=2\sum_i k_i$. Some conventions use $q^{h/2}$ instead; that convention divides all exponents here by two. The normalization above agrees with the ordinary <sl2 Lie algebra> character in the preceding part.
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