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In the defining representation $V={\mathbb{C}}^{2n}$, the basis vectors $v_i$ and $v_{\bar{i}}$ have weights ${\epsilon }_i$ and $-{\epsilon }_i$, respectively, each with multiplicity one. Thus its formal character isThe vector $v_1$ is killed by every positive root space, since all positive-root matrices are upper triangular, and is a highest-weight vector. Any nonzero invariant subspace splits into weight spaces and therefore contains one of the natural weight vectors, since those weight spaces are one-dimensional. The root matrices connect that vector to all the others, so this representation is irreducible for $n\ge 2$. In terms of the fundamental weights of Dn, its highest weight is ${\omega }_1$ for $n\ge 3$ and ${\omega }_1+{\omega }_2$ for $n=2$.

The flip of the two tensor factors is equivariant, henceUse the invariant symmetric bilinear form $Q(u,v)=u^{T}Jv$. Contraction $C(u\otimes v)=Q(u,v)$ is an equivariant map to the trivial Lie algebra representation. Its inverse-form tensoris invariant and satisfies $C(\tau )=2n$. ConsequentlyThe traceless summand contains the highest-weight vector $v_1\otimes v_1$ of weight $2{\epsilon }_1$. The Weyl complete reducibility theorem states that every finite-dimensional complex semisimple Lie algebra representation is a direct sum of irreducible representations, so the submodule generated by this highest vector is $L(2{\epsilon }_1)$. The Weyl dimension formula verifies that it is the entire traceless space: only positive roots involving ${\epsilon }_1$ change their factors, andThis equals $\dim {\mathrm{Sym}}^{2}V-1=n(2n+1)-1$. Thus the traceless symmetric square of the defining even orthogonal representation is irreducible, including $n=2$.

For the other summand, the exterior square of the defining orthogonal representation has the equivariant identificationThe image is skew relative to $Q$, and in a basis making $Q$ the identity the images are the independent elementary skew matrices. The map is therefore an isomorphism onto the Adjoint representation of a Lie algebra. For $n\ge 3$, the connected Dn Dynkin diagram gives a simple Lie algebra (with $D_3=A_3$); its adjoint module is irreducible because its invariant subspaces are exactly its ideals. Its highest weight is ${\epsilon }_1+{\epsilon }_2$, and $v_1\wedge v_2$ is a highest vector. The tensor-square decomposition of the defining even orthogonal representation is consequentlyThe three dimensions are $(n+1)(2n-1)$, $n(2n-1)$ and one, adding to $4n^2$. The highest weights are $2{\omega }_1$, ${\omega }_2$ and zero for $n\ge 4$; for $n=3$ the middle weight is ${\omega }_2+{\omega }_3$.

For $n=2$, the Chiral decomposition of the complexified so4 Lie algebra splits the adjoint module into its two three-dimensional ideals. ThusThe weights are $2{\omega }_1+2{\omega }_2$, $2{\omega }_1$, $2{\omega }_2$ and zero. Equivalently, labelling highest weights by the two sl2 Lie algebra factors, this is $(2,2)\oplus (2,0)\oplus (0,2)\oplus (0,0)$. If $n=1$ is allowed, the abelian algebra has $V=L({\epsilon }_1)\oplus L(-{\epsilon }_1)$ and $V\otimes V=L(2{\epsilon }_1)\oplus 2L(0)\oplus L(-2{\epsilon }_1)$; there is then no single highest weight for an irreducible natural module.

For $V={\mathbb{C}}^{2n}$, $n\ge 3$, the symmetric square splits into an invariant line and its irreducible traceless part of highest weight $2{\epsilon }_1$. The exterior square of the defining orthogonal representation is the adjoint module of highest weight ${\epsilon }_1+{\epsilon }_2$. For $n=2$, the exterior square splits instead into highest weights ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_1+{\epsilon }_2$. This exception reflects that so4 Lie algebra is semisimple but not simple.