A common core with graph-differentiable orbits integrates Heisenberg commutators

Fix $a,b$ and $\xi \in D$, abbreviate $U=U_a$ and $c=c_{ab}$, and set $F(s)=U(s)A_{b}U(-s)\xi$. Every product is defined: $U(-s)\xi \in D$, $A_{b}D\subset D\subset D(A_a)$, and $A_{a}D\subset D\subset D(A_b)$. The graph-orbit hypothesis gives the norm derivative of $\eta (s)=A_{b}U(-s)\xi$ as ${\eta }^{\mathrm{\prime }}(s)=-i{A}_b{A}_{a}U(-s)\xi$. The difference quotient for $U(s)\eta (s)$ splits into $U(s+h)(\eta (s+h)-\eta (s))\mathrm{/}h$ and $(U(s+h)-U(s))\eta (s)\mathrm{/}h$. Strong continuity and norm one of $U(s+h)$ handle the first term; [F1] applies to the fixed vector $\eta (s)\in D(A_a)$ in the second. Consequently $F^{\mathrm{\prime }}(s)=iU(s)(A_a{A}_b-A_b{A}_a)U(-s)\xi =-c\xi$. The curve $G(s)=F(s)+cs\xi$ is norm differentiable on $\mathbb{R}$ with derivative zero, hence is continuous there; its real-variable norm derivative is also its Fréchet derivative, since the remainder divided by $\mathrm{\mid }h\mathrm{\mid }$ tends to zero. For any $s$, apply [F5] to $G$ on the closed interval with endpoints $0,s$ (the case $s=0$ is immediate). It gives $G(s)=G(0)=A_b\xi$, so $F(s)=A_b\xi -cs\xi$.

For fixed $s$, the closed operator $B=U(s)A_{b}U(-s)$ has domain $U(s)D(A_b)$ and is self-adjoint: substituting $U(-s)x,U(-s)y$ in the adjoint-domain condition gives $B^{\ast }=U(s)A_b^{\ast }U(-s)=B$. Moreover $D$ is a core for $B$, because $U(s)$ transports graph approximations from the core $D$ of $A_b$ and $U(s)D=D$. The operator $A_b-csI$ is closed on $D(A_b)$: convergence of $x_j$ and $(A_b-csI)x_j$ implies convergence of $A_b{x}_j$ and closedness of $A_b$ applies. The two graph norms of $A_b$ and $A_b-csI$ are equivalent by the triangle inequality, so $D$ is a core for this shifted operator as well. Step 1.1 identifies their restrictions to $D$; taking their graph closures yields the exact operator equality $B=A_b-csI$, including equality of domains. In particular $U(s)D(A_b)=D(A_b)$ has now been proved, rather than assumed.

Fix $s$ and consider $K(t)=U_a(s)U_b(t)U_a(-s)$ and $L(t)=e^{-i{c}_{ab}st}{U}_b(t)$. Both are strongly continuous unitary groups by their formulas and [F1]. For $K$, taking difference quotients after unitary conjugation shows its exact derivative domain is $U_a(s)D(A_b)$, with derivative $iB$. For $L$, differentiating the scalar factor shows that every vector of $D(A_b)$ has derivative $i(A_b-c_{ab}sI)$. Conversely, if the derivative of $L(t)x$ at zero exists, writing $U_b(t)x=e^{i{c}_{ab}st}L(t)x$ proves the derivative of $U_b(t)x$ exists, so $x\in D(A_b)$ by [F1]. Thus the exact generators agree by step 2.1; [F2] gives $K=L$. Multiplying on the right by $U_a(s)$ proves the claimed pairwise Weyl relation on all of $H$. This is the functional-calculus exponentiation step; no power series of unbounded operators was used.

All factors defining $W(f)$ are unitary, and $W(0)=I$. Their finite product is strongly continuous jointly in $f$: subtract the products at two points and telescope, using norm one for the factors and strong continuity on each fixed vector in the finitely many resulting terms. To multiply $W(f)$ and $W(g)$, move each $U_a(g_a)$ left past $U_b(f_b)$ for $b>a$. Step 3.1 contributes $\exp (i{c}_{ab}{f}_b{g}_a)$ for each such pair; same-index factors combine by the group law. The scalar exponent after dividing by the defining scalar of $W(f+g)$ is $\frac{i}{2}{\sum }_{a<b}{c}_{ab}(f_a{f}_b+g_a{g}_b-(f_a+g_a)(f_b+g_b))+i{\sum }_{a<b}{c}_{ab}{f}_b{g}_a=-\frac{i}{2}{\sum }_{a<b}{c}_{ab}(f_a{g}_b-f_b{g}_a)=-i\sigma (f,g)\mathrm{/}2$. This proves the full displayed Weyl relation. Since $\sigma (f,-f)=0$, it gives $W(f)W(-f)=W(-f)W(f)=I$, so unitarity yields $W(f{)}^{\ast }=W(-f)$.

In the stated block-matrix case the commutators on $D$ are $[Q_j,Q_k]=[P_j,P_k]=0$ and $[Q_j,P_k]=i\kappa {\delta }_{jk}I$. Hence $[(i\mathrm{/}\kappa )P_j,Q_k]={\delta }_{jk}I$ when $\kappa \mathrm{\ne }0$. Domain invariance makes these restrictions endomorphisms of $D$. Since $D$ is a complex vector space, pointwise addition and scalar multiplication and composition make ${\mathrm{End}}_{\mathbb{C}}(D)$ a unital associative $\mathbb{C}$-algebra: composition is associative, the identity map is its unit, and $\mathbb{C}$-bilinearity follows by evaluating on vectors. Regard $E_n$ as the span of the generators $x_j,p_j$ and send them to $Q_j{\mathrm{\mid }}_D$ and $(i\mathrm{/}\kappa )P_j{\mathrm{\mid }}_D$. By [F6] this extends to a unital algebra map from $T(E_n)$. The displayed commutators show that this map kills each defining relator in [F4]; its kernel is a two-sided ideal because it is an algebra homomorphism, so it contains the generated ideal by [F9]. Then [F7] factors it through $A_n(\mathbb{C})$; write $\rho$ for this factor map. The factor is $\mathbb{C}$-linear because the quotient map is surjective, and evaluation $a\cdot \xi :=\rho (a)\xi$ on $D$ gives a unital left module by [F8]. Step 4.1 gives its eligible exponentiated realization, with the exact phase and finite-mode scope claimed. The proof uses Choice through [F1]–[F2]'s spectral/Stone construction and Countable Choice in the self-adjoint/graph supplier closure [F3]; it makes no assertion that the common-core and graph-orbit assumptions follow from algebraic relations alone.