Symmetric Subspace Dimension And Haar Moment

For each weak composition ${\alpha }_1+\cdots +{\alpha }_d=k$ with ${\alpha }_j\ge 0$, sum and normalize the elementary tensors in which index $j$ occurs ${\alpha }_j$ times. Different compositions have disjoint orthonormal tensor supports, and permutation invariance forces coefficients to be constant on each such orbit, so these vectors form a basis of ${\mathrm{Sym}}^k({\mathbb{C}}^d)$ by [F1]. Weak compositions correspond bijectively to placements of $d-1$ separators among $k+d-1$ positions (the numbers of stars before, between, and after separators recover the parts). There are $\left(\genfrac{}{}{0px}{}{k+d-1}{d-1}\right)=N_k$ such placements. For $k=0$ this gives the single empty tensor.

The pushforward measure $\mu$ is invariant under every $V\in U(d)$: left invariance of Haar gives $VU$ the same distribution as $U$, and $(VU)e_1=V(U{e}_1)$. Define the finite-dimensional operator integral $M=\int \mathrm{\mid }u⟩⟨u{\mathrm{\mid }}^{\otimes k}d\mu (u)$. By [F5], every integrand is a rank-one projection onto a symmetric tensor and has trace one, so $M={\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m}}M{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m}}$ by [F1]. Invariance gives $V^{\otimes k}M(V^{\otimes k}{)}^{\ast }=M$.

The coordinate integral in step 1.2 is self-adjoint and nonnegative: each integrand is a rank-one orthogonal projection by [F5], so $M^{\ast }=M$ and $⟨Mv,v⟩=\int \mathrm{\mid }⟨v,u^{\otimes k}⟩{\mathrm{\mid }}^{2}d\mu (u)\ge 0$. Since $M={\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m}}M{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m}}$, its restriction to the symmetric subspace is self-adjoint and nonnegative. By the spectral theorem [F4], that restriction has an orthogonal decomposition into eigenspaces. Each eigenspace is invariant under $U(d)$ because $M$ commutes with the representation by step 1.2; irreducibility [F2] therefore permits only one eigenspace, so $M=c{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m}}$ for some $c\ge 0$. Taking traces gives $1=\mathrm{tr}M=c\mathrm{tr}{\mathrm{\Pi }}_{\mathrm{s}\mathrm{y}\mathrm{m}}=c{N}_k$, since every integrand has trace one by [F5] (also for $k=0$). Thus $c=1\mathrm{/}{N}_k$, proving the moment identity.