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Let $R^{+}$ be the positive roots, $W$ the Weyl group, $\ell (w)$ its Coxeter length, $\rho$ the half-sum of positive roots, and ${\alpha }^{\vee }$ a coroot. For a dominant integral highest weight $\lambda$, the Weyl character formula isTaking the value at the identity gives the Weyl dimension formula

For the q-character convention relevant to the Principal sl2 subalgebra, set $h_{\mathrm{pr}}=2{\rho }^{\vee }$, so ${\alpha }_i(h_{\mathrm{pr}})=2$ for every simple root, and defineThe q-character formula is the principal specialization of the Weyl character formula:

For the defining three-dimensional special linear Lie algebra representation $V$, contraction $(uv)\otimes f\mapsto f(u)v+f(v)u$ maps $(S^{2}V)\otimes V^{\ast }$ onto $V$. The kernel contains the highest-weight vector $e_1^2\otimes e_3^{\ast }$ of Dynkin labels $(2,1)$. The Weyl dimension formula gives its irreducible module dimension $15$, the entire kernel dimension; the Weyl complete reducibility theorem supplies the displayed splitting.

Symmetrize the $p$ upper and $q$ lower indices separately, and remove every upper-lower trace. The resulting representation has highest weight $p{\omega }_1+q{\omega }_2$. Its dimension isThe subtraction is valid because contraction is onto: in the polynomial realization its adjoint, for the monomial-factorial inner product, is multiplication by the nonzero polynomial ${\sum }_i{x}_i{y}_i$, an injective map. The resulting kernel contains the highest-weight constituent and has precisely its Weyl dimension formula dimension, hence is irreducible. This highest-weight constituent is not always the largest-dimensional constituent of the full mixed tensor.

For $n\ge 2$, contraction with the invariant symmetric bilinear form splits off the trivial line in the symmetric square of the defining special orthogonal group representation. Its null space is the traceless symmetric square. It contains a highest-weight vector of weight $2e_1$, and the Weyl dimension formula gives $(n+1)(2n-1)$, exactly its dimension. Complete reducibility of compact-group representations therefore makes that null space irreducible. The entire symmetric square is reducible because of the invariant line.

Split the two dual factors into antisymmetric and symmetric tensors. The invariant Levi-Civita symbol identifies the antisymmetric pair with $3$, giving $6\oplus \overline{3}$ after multiplying by the remaining triplet. On the symmetric part, contraction $T^a{}_{ba}$ maps onto $\overline{3}$; the equivariant splitting is $T^a{}_{bc}=({\delta }_b^a{v}_c+{\delta }_c^a{v}_b)/4$. Its trace-free kernel has dimension $18-3=15$ and highest weight with Dynkin labels $(1,2)$. The Weyl dimension formula gives dimension $15$, so this is exactly the irreducible kernel. It is conjugate to the other fifteen-dimensional module with labels $(2,1)$; dimension alone does not distinguish the two.

In the symmetric power ${\mathrm{Sym}}^m{\mathbb{C}}^n$ of the defining unitary group representation, diagonal torus weights are indexed by nonnegative integer tuples summing to $m$, with one monomial per weight space. Torus averaging extracts a monomial from any nonzero invariant complex subspace. The complexified Lie algebra acts by $E_{ij}=x_i{\partial }_{x_j}$ on the symmetric algebra. Whenever $a_j>0$, the displayed action moves one unit of degree from $j$ to $i$ with nonzero coefficient. Finite transfers connect every degree-$m$ monomial, so the subspace is the entire representation. The case $m=0$ is the one-dimensional trivial representation. This proof needs neither the Weyl character formula nor the Weyl dimension formula.

When ${\alpha }_1$ is the short simple root, ${\omega }_1$ is the spin fundamental weight and ${\omega }_2$ is the vector fundamental weight. The Weyl dimension formula gives