= Symmetric trace-free square of the defining orthogonal representation
{title2=$\operatorname{Sym}_0^2(\mathbb R^d)$}
= Symmetric traceless square
{synonym}
For the defining real representation of $SO(d)$, the <symmetric square> splits as a scalar trace plus symmetric trace-free tensors: $\operatorname{Sym}^2(\mathbb R^d)=\mathbb R\oplus\operatorname{Sym}_0^2(\mathbb R^d)$. The trace projection is $S\mapsto S-(\operatorname{tr}S)I/d$, so the second summand has dimension $d(d+1)/2-1$.
For $d\geq3$ 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>.
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