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For the defining two-dimensional sl2 Lie algebra module with coordinates $x,y$, put $u_{ij}=x^{3-i}{y}^i\otimes x^{2-j}{y}^j$. Its three highest-weight vectors are $u_{00}$, $u_{10}-u_{01}$ and $u_{20}-2u_{11}+u_{02}$, of weights five, three and one. Lowering the last once gives $u_{30}-2u_{21}+u_{12}$, an explicit second vector of its two-dimensional irreducible summand.

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.

For generators normalized by ${\mathrm{tr}}_3(t_a{t}_b)={\delta }_{ab}/2$, the quadratic Casimir operator on the irreducible SU(3) representation of Dynkin labels $(a,b)$ has the displayed scalar value. In SU(3) highest-weight coordinates $(p,q)$ it is $p^2+q^2+p+\sqrt{3}q$.

To derive this, write the operator as $H_1^2+H_2^2+{\sum }_{i=1}^3(E_{+}^i{E}_{-}^i+E_{-}^i{E}_{+}^i)$. On a highest-weight vector, the raising operators vanish, and the three root commutators sum to $H_1+\sqrt{3}{H}_2$. Centrality and the Schur lemma extend this value to the representation. The singlet, fundamental, adjoint and decuplet values are $0,4/3,3,6$, respectively. Tensor labels and multiplet dimensions are summarized in the Particle Data Group's SU(n) multiplet guide.

Realize $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$. Contract $r$ pairs of indices, one from each symmetric tensor, with the alternating invariant tensor $ϵ$, and symmetrize the remaining indices. This is a nonzero intertwining operator into $V_{m+n-2r}$. In the bihomogeneous polynomial realization, $(z_1{w}_2-z_2{w}_1{)}^r{z}_1^{m-r}{w}_1^{n-r}$ is a highest-weight vector of weight $m+n-2r$. Its lowering string has length $m+n-2r+1$. These inequivalent irreducible summands exhaust the tensor product, since their dimensions sum to $(m+1)(n+1)$.

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.

For a highest-weight vector $v$ of weight $k$ in an irreducible finite-dimensional sl2 Lie algebra module, the basis $v,Yv,\dots ,Y^{k}v$ diagonalizes $XY$, with eigenvalue $(j+1)(k-j)$ on $Y^{j}v$. Summing those eigenvalues gives $\mathrm{tr}(XY)=k(k+1)(k+2)/6$. The formula $X{Y}^{n}v=n(k-n+1)Y^{n-1}v$ requires $v$ to be a highest-weight vector, not an arbitrary vector of the module.

For Hermitian $J_3$ and adjoint angular momentum ladder operators $J_{+}^{†}=J_{-}$ on a positive-definite inner-product space, a highest-weight vector of weight $j$ obeys $J_{+}|j,j⟩=0$. The commutation relations give $J_{+}(J_{-}{)}^{\ell }|j,j⟩=\ell (2j-\ell +1)(J_{-}{)}^{\ell -1}|j,j⟩$. Taking inner products forces $2j$ to be a nonnegative integer: otherwise the first negative factor would give a negative squared norm. The norm first vanishes at $\ell =2j+1$, so the generated unitary representation has dimension $2j+1$. Without positivity an abstract highest-weight module can remain infinite-dimensional even at an integral highest weight.