Canonical commutation relations on the component-valued core and commutation with the internal factor

[The core and the CCR] For $u=(u_l)\in {\mathcal{S}}^{(r)}$, [F1] gives $Q_{j}u,P_{k}u\in {\mathcal{S}}^{(r)}$, so both products are defined there. For each component, using the ordinary product rule for the smooth function $x_j{u}_l(x)$, $(Q_j{P}_{k}u-P_k{Q}_{j}u{)}_l=-i\mathrm{\hslash }{x}_j{\mathrm{\partial }}_k{u}_l+i\mathrm{\hslash }{\mathrm{\partial }}_k(x_j{u}_l)=i\mathrm{\hslash }{\delta }_{jk}{u}_l\mathrm{.}$ Thus $(Q_j{P}_k-P_k{Q}_j)u=i\mathrm{\hslash }{\delta }_{jk}u$ as an equality of $L^2$ classes. This proves (a).

[Linearity of the matrix action] For $f\in D(Q_j)$ and every $m$, $x_j{f}_m\in L^2$, so the components $x_j(S_{\times }f{)}_l=x_j{\sum }_m{S}_{lm}{f}_m={\sum }_m{S}_{lm}{x}_j{f}_m$ are $L^2$ classes (finite sums), and $S_{\times }f\in D(Q_j)$. For $f\in D(P_k)$, [F3] applied to the finite linear combinations ${\sum }_m{S}_{lm}{f}_m$ gives $D_k(S_{\times }f{)}_l={\sum }_m{S}_{lm}{D}_k{f}_m$, so $S_{\times }f\in D(P_k)$; the actions agree with the matrix action on each component, whence $Q_j{S}_{\times }f=S_{\times }{Q}_{j}f$ and $P_k{S}_{\times }f=S_{\times }{P}_{k}f$.

[The position spectral measure] Put $E(B):=M_{{\mathbf{1}}_{x_j^{-1}(B)}}$ on $H^{(r)}$, the componentwise multiplication by the bounded indicator. Each $E(B)$ is an orthogonal projection and $E$ is a regular PVM: regularity follows by applying [F4] to the scalar multiplier $x\mapsto x_j$ on $L^2({\mathbb{R}}^d)$ and taking the finite orthogonal sum over the $r$ components; multiplicativity and the endpoint values are pointwise, and for pairwise disjoint Borel $B_n$ with union $B$ the strong additivity follows from $\mathrm{\parallel }E(B)x-{\sum }_{n\le N}E(B_n)x{\mathrm{\parallel }}^2=\int \mathrm{\mid }{\mathbf{1}}_B-{\sum }_{n\le N}{\mathbf{1}}_{B_n}\mathrm{\mid }(x_j)\mathrm{\parallel }x{\mathrm{\parallel }}^2\to 0$ by dominated convergence. For $x\in D(Q_j)$ and $y\in H^{(r)}$, the scalar pairing measure is $E_{x,y}(B)={\int }_{x_j^{-1}(B)}⟨x(z),y(z){⟩}_{{\mathbb{C}}^r}dz$, the pushforward under $z\mapsto z_j$ of the complex measure with density $⟨x(z),y(z)⟩$. Its first absolute moment is finite because $\int \mathrm{\mid }{z}_j\mathrm{\mid }\mathrm{\mid }⟨x(z),y(z)⟩\mathrm{\mid }dz\le \mathrm{\parallel }{Q}_{j}x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ by Cauchy--Schwarz. The pushforward integral formula (first for indicator functions, then simple functions and integrable limits) therefore gives $\int \lambda d{E}_{x,y}(\lambda )=\int z_j⟨x(z),y(z)⟩dz=⟨Q_{j}x,y⟩$. Similarly, $E_x$ is the pushforward of $\mathrm{\parallel }x(z){\mathrm{\parallel }}^{2}dz$, and $\int {\lambda }^{2}d{E}_x(\lambda )=\int z_j^2\mathrm{\parallel }x(z){\mathrm{\parallel }}^{2}dz$, finite exactly when $x\in D(Q_j)$. Thus $E$ represents $Q_j$, and by uniqueness [F4] it is the spectral PVM of $Q_j$.

[Norm of the matrix action] [F5] gives the pointwise bound $\mathrm{\parallel }{S}_{\times }f(x){\mathrm{\parallel }}_{{\mathbb{C}}^r}\le \mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }f(x){\mathrm{\parallel }}_{{\mathbb{C}}^r}$, so $\mathrm{\parallel }{S}_{\times }f{\mathrm{\parallel }}_{H^{(r)}}\le \mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{(r)}}$ and $\mathrm{\parallel }{S}_{\times }\mathrm{\parallel }\le \mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. Choose a unit vector $z_0\in {\mathbb{C}}^r$ with $\mathrm{\parallel }S{z}_0\mathrm{\parallel }=\mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$, and a unit $h\in L^2({\mathbb{R}}^d;\mathbb{C})$ with $\mathrm{\parallel }h\mathrm{\parallel }=1$; then $f=h{z}_0$ (the class $x\mapsto h(x)z_0$) has $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{(r)}}=\mathrm{\parallel }h\mathrm{\parallel }\mathrm{\parallel }{z}_0\mathrm{\parallel }=1$ and $\mathrm{\parallel }{S}_{\times }f{\mathrm{\parallel }}_{H^{(r)}}=\mathrm{\parallel }h\mathrm{\parallel }\mathrm{\parallel }S{z}_0\mathrm{\parallel }=\mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$, so $\mathrm{\parallel }{S}_{\times }\mathrm{\parallel }=\mathrm{\parallel }S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$; this proves the norm statement of (b), and step 1.2 is the commutation statement.

[The momentum spectral measure] Put $E^{\mathrm{\prime }}(B):={\mathcal{F}}_2^{(r)\ast }{M}_{{\mathbf{1}}_{\{2\pi \mathrm{\hslash }{\xi }_k\in B\}}}{\mathcal{F}}_2^{(r)}$, the unitary conjugate of the multiplication PVM by the scalar multiplier $m(\xi ):=2\pi \mathrm{\hslash }{\xi }_k$; [F4] gives the scalar multiplication PVM and its regularity, which pass to the finite component sum and unitary conjugate, so $E^{\mathrm{\prime }}$ is a regular PVM. For $x\in D(P_k)$ put $g:={\mathcal{F}}_2^{(r)}x$ and for $y^{\mathrm{\prime }}\in H^{(r)}$ put $h^{\mathrm{\prime }}:={\mathcal{F}}_2^{(r)}{y}^{\mathrm{\prime }}$; then $E_{x,y^{\mathrm{\prime }}}^{\mathrm{\prime }}(B)={\int }_{m^{-1}(B)}⟨g(\xi ),h^{\mathrm{\prime }}(\xi )⟩d\xi$ is the pushforward under $m$ of the complex measure with density $⟨g(\xi ),h^{\mathrm{\prime }}(\xi )⟩$. Its first absolute moment is finite since $\int \mathrm{\mid }m(\xi )\mathrm{\mid }\mathrm{\mid }⟨g,h^{\mathrm{\prime }}⟩\mathrm{\mid }d\xi \le \mathrm{\parallel }mg{\mathrm{\parallel }}_2\mathrm{\parallel }{h}^{\mathrm{\prime }}{\mathrm{\parallel }}_2<\mathrm{\infty }$, so the pushforward integral formula (first for indicator functions, then simple functions and integrable limits) and [F2] give $\int \lambda d{E}_{x,y^{\mathrm{\prime }}}^{\mathrm{\prime }}(\lambda )=\int m(\xi )⟨g,h^{\mathrm{\prime }}⟩d\xi =⟨mg,h^{\mathrm{\prime }}⟩=⟨P_{k}x,y^{\mathrm{\prime }}⟩$. Similarly, $E_x^{\mathrm{\prime }}$ is the pushforward of $\mathrm{\parallel }g(\xi ){\mathrm{\parallel }}^{2}d\xi$, so $\int {\lambda }^{2}d{E}_x^{\mathrm{\prime }}(\lambda )=\mathrm{\parallel }mg{\mathrm{\parallel }}_2^2$, finite exactly when $x\in D(P_k)$. Thus $E^{\mathrm{\prime }}$ represents $P_k$, and by uniqueness [F4] it is the spectral PVM of $P_k$.

[Commutation with the spectral measures] The multiplier $M_{{\mathbf{1}}_{x_j^{-1}(B)}}$ commutes with $S_{\times }$ because both are multiplication by functions on the finite-dimensional values: pointwise, ${\mathbf{1}}_{x_j^{-1}(B)}(x)Sz=S({\mathbf{1}}_{x_j^{-1}(B)}(x)z)$ for every $z\in {\mathbb{C}}^r$. For the momentum case, $S_{\times }{\mathcal{F}}_2^{(r)}={\mathcal{F}}_2^{(r)}{S}_{\times }$ because ${\mathcal{F}}_2^{(r)}$ acts on components and matrix multiplication acts on the finite index, so ${\mathcal{F}}_2^{(r)}\left(S_{\times }g\right)=({\sum }_m{S}_{lm}{\mathcal{F}}_2{g}_m{)}_l=S_{\times }{\mathcal{F}}_2^{(r)}g$; and $S_{\times }$ commutes with the scalar multiplier by the same pointwise argument. Hence $S_{\times }$ commutes with both $E_{Q_j}(B)$ and $E_{P_k}(B)$ by steps 1.3 and 2.2. For unitary $V\in U(r)$ and $f\in H^{(r)}$, $V_{\times }$ is unitary (it is matrix multiplication by a unitary pointwise) and commutes with the PVMs, so $\mathrm{\parallel }E(B)V_{\times }f\mathrm{\parallel }=\mathrm{\parallel }{V}_{\times }E(B)f\mathrm{\parallel }=\mathrm{\parallel }E(B)f\mathrm{\parallel }$, which is the last assertion of (c).

[Amplification] Let $u\in H$ and $z={\sum }_l{z}_l{e}_l\in {\mathbb{C}}^r$, and put $w:=\mathrm{\Theta }(\kappa (u\otimes z))$; by Unitary identifications of the component-valued wave-function space clause (b), $w={\sum }_l{z}_l{\iota }_l(u)$, where ${\iota }_l(u)\in H^{(r)}$ denotes the vector with $u$ in slot $l$ and $0$ elsewhere. Then ${\mathrm{\Theta }}^{-1}{S}_{\times }\mathrm{\Theta }\kappa (u\otimes z)={\mathrm{\Theta }}^{-1}\left(S_{\times }w\right)$, and $S_{\times }w={\sum }_l{z}_l{S}_{\times }{\iota }_l(u)={\sum }_{l,m}{z}_l{S}_{ml}{\iota }_m(u)$; applying ${\mathrm{\Theta }}^{-1}$ with clause (b) gives ${\sum }_{l,m}{z}_l{S}_{ml}\kappa (u\otimes e_m)=\kappa (u\otimes Sz)$, where the last step expands $Sz={\sum }_m\left({\sum }_l{S}_{ml}{z}_l\right)e_m$; by [F5] this equals $(I\otimes S)\kappa (u\otimes z)$. Both sides are bounded operators (Bounded tensor maps: extension, norm estimate, adjoint and commutation clause (a) and step 2.1) that agree on the elementary tensors, whose span is dense in $H\widehat{\otimes }{\mathbb{C}}^r$ (Hilbert tensor construction: inner product, completion and product bases clause (c)); hence ${\mathrm{\Theta }}^{-1}{S}_{\times }\mathrm{\Theta }=I\otimes S$, which is (d).