Finite graph vertex Haar averaging is an orthogonal projection

For each dual factor, choose an orthonormal basis of $V_e$ and use its dual-basis inner product; this is basis-independent because a change between orthonormal bases preserves the Euclidean coefficient norm. The matrix of the dual action is the transpose of the inverse unitary matrix, hence is unitary. Tensor products of orthonormal bases are an orthonormal basis for the displayed tensor inner product, so the diagonal action is unitary by [F3]–[F4]. The graph has finitely many incident edges and each $V_e$ is finite-dimensional, so this orthonormal basis of $W_v$ is finite. Its coordinate map is an isometry to some ${\mathbb{C}}^N$, which is complete by coordinatewise completeness; hence $W_v$ is a complex Hilbert space. Its matrix entries are finite sums and products of continuous entries of the edge representations, so [F7] makes $R_v$ strongly continuous.

For fixed $w\in W_v$, define $T_w:\mathbb{C}\to W_v$ by $T_w(z)=zw$, with the trivial representation on $\mathbb{C}$. It is linear and bounded by [F8], with $\mathrm{\parallel }{T}_w\mathrm{\parallel }=\mathrm{\parallel }w\mathrm{\parallel }$. The intertwining equation $T_w(z)=R_v(g)T_w(z)$ for all $z$ is equivalent to $w\in W_v^{SU(2)}$.

By [F1], $SU(2)$ is compact Hausdorff; by [F2] and [F5], its normalized Haar probability exists under the stated full AC assumption. The hypotheses of [F9]–[F10] now hold for $H=\mathbb{C}$, $J=W_v$, and $T=T_w$, using [F6], [F7], and steps 1.1 and 2.1. Since the trivial representation acts as the identity, the weak Haar average of $T_w$ is $T_{P_{v}w}$, where the finite-dimensional vector integral is defined coordinatewise. The range clause of [F10] therefore gives $P_v^2=P_v$, $\mathrm{\parallel }{P}_v\mathrm{\parallel }\le 1$, and $\mathrm{ran}{P}_v=W_v^{SU(2)}$.

If $x\in \mathrm{ran}{P}_v$ and $y\in \ker P_v$, then $P_v(x+ty)=x$ for every complex scalar $t$. Contractivity gives $\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }x+ty\mathrm{\parallel }$. Squaring and taking real $t$ of both signs forces $\mathrm{Re}⟨x,y⟩=0$; taking $t$ purely imaginary of both signs forces $\mathrm{Im}⟨x,y⟩=0$. Hence range and kernel are orthogonal. Since an idempotent has $W_v=\mathrm{ran}{P}_v\oplus \ker P_v$, the kernel is the orthogonal complement of the range, and [F11] identifies $P_v$ as the orthogonal projection onto its invariant range.

Equip $W={⨂}_v{W}_v$ with the tensor-product inner product, defined on elementary tensors by $⟨{⨂}_v{x}_v,{⨂}_v{y}_v⟩={\prod }_v⟨x_v,y_v⟩$ and extended sesquilinearly. The graph has finitely many vertices and every $W_v$ has a finite orthonormal basis, so the corresponding tensor basis is finite and $W$ is finite-dimensional. Each vertex operator acts as $P_v$ on one factor and the identity on the others. Operators on distinct tensor factors commute, so their product is an idempotent whose range is ${⨂}_v\mathrm{ran}{P}_v$, the tensors invariant under every vertex action. Tensor products of orthogonal projections are orthogonal projections for this inner product, so the product projects orthogonally onto that joint invariant subspace. If the graph has no vertices, the empty product is the identity on $\mathbb{C}$, consistent with the statement.