Symmetric trace-free square of the defining orthogonal representation (${\mathrm{Sym}}_0^2({\mathbb{R}}^d)$)

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.