Symmetric square ($S^{2}V$)

The symmetric square is the quotient of $V\otimes V$ by $v\otimes w-w\otimes v$, and records symmetric tensors of rank two.

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 $V={\mathbb{C}}^{2n}$, $n\ge 3$, 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.

Over every field, including characteristic two, there is a natural isomorphismThe middle summand maps $v\otimes w$ to the mixed product $vw$. Monomial bases prove bijectivity, and the Leibniz rule makes the map equivariant for Lie algebra representations.

For the defining real representation of $SO(d)$, the symmetric square splits as a scalar trace plus symmetric trace-free tensors: ${\mathrm{Sym}}^2({\mathbb{R}}^d)=\mathbb{R}\oplus {\mathrm{Sym}}_0^2({\mathbb{R}}^d)$. The trace projection is $S\mapsto S-(\mathrm{tr}S)I/d$, so the second summand has dimension $d(d+1)/2-1$.

For $d\ge 3$ the trace-free summand is an irreducible real group representation. To see this, diagonalize a nonzero symmetric trace-free matrix in a nonzero invariant subspace. Two diagonal entries differ. Infinitesimal rotation in their plane yields a nonzero symmetric off-diagonal matrix; rotations carry it to every coordinate pair, and a rotation through $\pi /4$ produces a difference of two diagonal entries. These off-diagonal matrices and diagonal differences span all symmetric trace-free matrices, so the invariant subspace is the whole summand. For $d=24$, its dimension is 299, the third-level nontrivial multiplet of the lowest levels of a fully transverse ND bosonic string.