Exponentiated Weyl Relations and Regular Representations

Use finite Borel representatives as in [F2]. By [F3], the nonnegative function $\mathrm{\mid }f(x){\mathrm{\mid }}^2{\prod }_{j=n+1}^m\mathrm{\mid }\mathrm{\Omega }(x_j){\mathrm{\mid }}^2$ is product-measurable, and Tonelli gives $\mathrm{\parallel }{J}_{n,m}f{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2{\prod }_{j=n+1}^m\mathrm{\parallel }\mathrm{\Omega }{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2<\mathrm{\infty }$. For $n=0$ the first factor is $\mathrm{\mid }f{\mathrm{\mid }}^2$, and $m=n$ is the identity map. Applying the same calculation to a difference shows independence of the representative of $f$; changing a representative of $\mathrm{\Omega }$ changes the product only on finitely many null-coordinate cylinders, which are null by Tonelli and sigma-finiteness. Thus $J_{n,m}$ is a well-defined linear isometry on classes. Substitution gives $J_{m,r}{J}_{n,m}=J_{n,r}$. Computing two vectors' inner product at a common index is independent of a larger common index, since linear isometries preserve inner products by polarization. This defines a positive-definite inner product on the compatible union.

For $n\ge j$, write $x=(z,y)$ with $y=x_j$ and the other coordinates denoted by $z$. This coordinate permutation preserves Euclidean measure: on Borel rectangles both measures give the product of the side measures, so [F3] applies in this order too. Tonelli gives $\int \mathrm{\parallel }f(z,\cdot ){\mathrm{\parallel }}_2^{2}dz=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$, so almost every section lies in $L^2(\mathbb{R})$. Set $a_f(z)=\int f(z,y)\overline{\mathrm{\Phi }(y)}dy$ on those sections, and zero on the exceptional set. Cauchy--Schwarz gives $\mathrm{\mid }{a}_f(z){\mathrm{\mid }}^2\le \mathrm{\parallel }f(z,\cdot ){\mathrm{\parallel }}_2^2$. Tonelli applied to the positive and negative parts of the real and imaginary integrand makes $a_f$ measurable; its section integral is finite almost everywhere by this estimate. Define $(P_{j,n}^{\mathrm{\Phi }}f)(z,y)=a_f(z)\mathrm{\Phi }(y)$. Tonelli now gives $\mathrm{\parallel }{P}_{j,n}^{\mathrm{\Phi }}f{\mathrm{\parallel }}_2^2=\int \mathrm{\mid }{a}_f{\mathrm{\mid }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2^2$, and the same estimate for differences makes the map well-defined on classes. Since $\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_2=1$, section integration gives $a_{P_{j,n}^{\mathrm{\Phi }}f}=a_f$, hence idempotence. The products $(P_{j,n}^{\mathrm{\Phi }}f)\bar{g}$ and $f\overline{P_{j,n}^{\mathrm{\Phi }}g}$ are integrable by Cauchy--Schwarz on ${\mathbb{R}}^n$; Fubini therefore yields $⟨P_{j,n}^{\mathrm{\Phi }}f,g⟩=\int a_f\overline{a_g}=⟨f,P_{j,n}^{\mathrm{\Phi }}g⟩$. Thus $P_{j,n}^{\mathrm{\Phi }}$ is an orthogonal projection. When $n=j=1$, the $z$-space is the one-point space and the same formulas hold.

If $r\ge n\ge j$, the formula in step 1.2 acts on one of the first $n$ coordinates, while $J_{n,r}$ appends only factors $\mathrm{\Omega }$ after coordinate $n$. Substitution in the integral formula gives $P_{j,r}^{\mathrm{\Phi }}{J}_{n,r}=J_{n,r}{P}_{j,n}^{\mathrm{\Phi }}$. For a vector initially represented in $H_n$ with $n<j$, first embed it in $H_j$; the compatibility identity shows that this and every later representation give the same image. Hence the projections define a contraction $P_j^{\mathrm{\Phi }}$ on the compatible union. It extends uniquely by continuity to the Hilbert completion, and the identities of step 1.2 pass to limits, so the extension remains an orthogonal projection.

The compatible union is dense in its Hilbert completion by construction. Steps 1.1 and 2.1 prove, respectively, the well-defined incomplete tensor product and every stated local-projection property.

In coordinates from [F2], bilinearity gives $\omega ((q,p),(q^{\mathrm{\prime }},p^{\mathrm{\prime }}))=q\cdot p^{\mathrm{\prime }}-p\cdot q^{\mathrm{\prime }}$. For three vectors $v,w,u$,$\omega (v,w)+\omega (v+w,u)=\omega (v,w)+\omega (v,u)+\omega (w,u)=\omega (w,u)+\omega (v,w+u)\mathrm{.}$Thus $c(v,w)=-\frac{1}{2}\omega (v,w)$ satisfies the normalized 2-cocycle identity $c(v,w)+c(v+w,u)=c(w,u)+c(v,w+u)$. Substitution in the group law proves associativity; $\omega (v,0)=\omega (0,v)=0$ gives identity $(0,0)$, and $\omega (v,-v)=0$ gives inverse $(-v,-t)$. Addition and $\omega$ are continuous, so multiplication and inverse are continuous. The underlying finite-dimensional Euclidean space $V\times \mathbb{R}$ is locally compact and second countable.

An element $(v,t)$ commutes with every $(w,0)$ exactly when $\omega (v,w)=0$ for all $w$. Nondegeneracy in [F1] forces $v=0$, and each $(0,t)$ is central by the product formula. Hence the center is the displayed central coordinate. The projection $(v,t)\mapsto v$ is onto with kernel $\{(0,t):t\in \mathbb{R}\}$, proving exactness of the central extension in the definition.

For $q\in K$ and $(p,t)\in N$, direct multiplication gives $(q,0,0)(0,p,t)(-q,0,0)=(0,p,t-q\cdot p)$; thus ${\alpha }_q(p,t)=(p,t-q\cdot p)$. This is an automorphism because it is additive, has inverse ${\alpha }_{-q}$, and ${\alpha }_{q+q^{\mathrm{\prime }}}={\alpha }_q{\alpha }_{q^{\mathrm{\prime }}}$. Every group element has the unique factorization$(q,p,z)=(0,p,z-q\cdot p\mathrm{/}2)(q,0,0)\mathrm{.}$Therefore $(n,q)\mapsto n(q,0,0)$ is a bijection from the semidirect product of [F4] to $H(V,\omega )$; the conjugation computation makes it a homomorphism. Its coordinate formula and inverse are continuous, so it is a topological group isomorphism.

Apply [F3] to $N\cong {\mathbb{R}}^{n+1}$. Rescaling the coordinates by $2\pi$ gives every character exactly once as ${\chi }_{a,\mu }(p,t)=e^{i(a\cdot p+\mu t)}$. Since ${\alpha }_{-q}(p,t)=(p,t+q\cdot p)$, composition yields$q\cdot {\chi }_{a,\mu }={\chi }_{a+\mu q,\mu }\mathrm{.}$

The frequency parametrization in step 3.1 is a homeomorphism for the compact-open topology of [F5]. First consider ${\beta }_b(t)=e^{ibt}$ on $\mathbb{R}$. On any nonempty compact set $C\subset \mathbb{R}$, put $M={\max }_{t\in C}\mathrm{\mid }t\mathrm{\mid }$; the estimate $\mathrm{\mid }{e}^{ibt}-e^{i{b}_{0}t}\mathrm{\mid }\le \mathrm{\mid }b-b_0\mathrm{\mid }M$ proves continuity uniformly on $C$. Conversely, for any $\delta >0$, require uniform distance less than $1$ from ${\beta }_{b_0}$ on the compact interval $[-\pi \mathrm{/}\delta ,\pi \mathrm{/}\delta ]$. If $\mathrm{\mid }b-b_0\mathrm{\mid }\ge \delta$, then $t=\pi \mathrm{/}(b-b_0)$ lies in that interval and $\mathrm{\mid }{e}^{i(b-b_0)t}-1\mathrm{\mid }=2$, a contradiction. This compact-open neighborhood therefore forces $\mathrm{\mid }b-b_0\mathrm{\mid }<\delta$ and proves inverse continuity. The finite-product dual homeomorphism in [F3] now proves the full parameter map $(a,\mu )\mapsto {\chi }_{a,\mu }$ is a homeomorphism ${\mathbb{R}}^{n+1}\to \widehat{N}$; the rescaling by $2\pi$ is a Euclidean homeomorphism. Restricting to $\mu =\lambda$ gives the corresponding orbit subspace topology, including $n=0$, where the fixed-central-character orbit is a singleton.

When $\mu =\lambda \mathrm{\ne }0$, the map $q\mapsto a+\lambda q$ is a bijection of ${\mathbb{R}}^n$ with itself. It is continuous with continuous inverse $a\mapsto (a-a_0)\mathrm{/}\lambda$, so the orbit through $(a_0,\lambda )$ is all of ${\widehat{N}}_{\lambda }$ and its stabilizer is $\{0\}$. This proves the free transitive orbit and stabilizer claims.

Steps 1.1–5.1 establish the group, topological, central, semidirect, and dual-orbit assertions, including the zero-dimensional case $n=0$ where the Lagrangian subgroup $K$ is trivial.

Bilinearity gives $m_{\lambda }(v,w)m_{\lambda }(v+w,z)=m_{\lambda }(w,z)m_{\lambda }(v,w+z)$, so the two parenthesizations of ${\delta }_v{\delta }_w{\delta }_z$ agree. Also $m_{\lambda }(0,v)=m_{\lambda }(v,0)=1$, so ${\delta }_0$ is a two-sided unit. Since $\omega (v,-v)=0$, ${\delta }_v^{\ast }{\delta }_v={\delta }_v{\delta }_v^{\ast }={\delta }_0$. Finally, $\overline{m_{\lambda }(v,w)}=m_{\lambda }(-w,-v)$, which gives $({\delta }_v{\delta }_w{)}^{\ast }={\delta }_w^{\ast }{\delta }_v^{\ast }$; by bilinearity and conjugate-linearity these identities extend to all finite sums.

For any positive unital $\phi$, the form $B_{\phi }(a,b)=\phi (b^{\ast }a)$ is positive semidefinite. Positivity of $B_{\phi }(a+tb,a+tb)$ for every complex $t$ gives $\mathrm{\mid }{B}_{\phi }(a,b){\mathrm{\mid }}^2\le B_{\phi }(a,a)B_{\phi }(b,b)$: if $B_{\phi }(b,b)>0$, minimize the quadratic in $t$; if it is zero, varying the phase and magnitude of $t$ forces $B_{\phi }(a,b)=0$. Therefore its null space $N_{\phi }=\{a:B_{\phi }(a,a)=0\}$ is orthogonal to every vector and the quotient has the inner product specified in [F1].

Write $a={\sum }_v{a}_v{\delta }_v$ with finite support and define ${\phi }_0(a)=a_0$. The coefficient of ${\delta }_0$ in $a^{\ast }a$ is ${\sum }_v\overline{a_v}{a}_v{m}_{\lambda }(-v,v)={\sum }_v\mathrm{\mid }{a}_v{\mathrm{\mid }}^2$, because $\omega (v,-v)=0$. Thus ${\phi }_0(a^{\ast }a)\ge 0$ and ${\phi }_0({\delta }_0)=1$; the positive unital functional set in [F1] is nonempty.

Left multiplication by ${\delta }_v$ is an isometry on this quotient: $B_{\phi }({\delta }_{v}a,{\delta }_{v}a)=\phi (a^{\ast }{\delta }_v^{\ast }{\delta }_{v}a)=\phi (a^{\ast }a)=B_{\phi }(a,a)$. Hence it preserves the null space. For a finite sum $a={\sum }_v{a}_v{\delta }_v$, the triangle inequality gives $\mathrm{\parallel }{L}_a[b]\mathrm{\parallel }\le ({\sum }_v\mathrm{\mid }{a}_v\mathrm{\mid })\mathrm{\parallel }[b]\mathrm{\parallel }$. Left multiplication therefore extends to a bounded operator on the Hilbert completion of the quotient, and its adjoint is $L_{a^{\ast }}$ because $B_{\phi }(ab,c)=B_{\phi }(b,a^{\ast }c)$. Associativity from step 1.1 makes ${\pi }_{\phi }(a):=L_a$ a unital $\ast$-representation.

The supremum in [F1] is over a set of functionals (a subset of the function set ${\mathbb{C}}^{A_0}$), is nonempty by step 2.1, and is finite since step 2.2 gives $\mathrm{\parallel }{\pi }_{\phi }(a)\mathrm{\parallel }\le {\sum }_v\mathrm{\mid }{a}_v\mathrm{\mid }$ for every $\phi$. Each GNS Hilbert space is nonzero because $\mathrm{\parallel }[{\delta }_0]{\mathrm{\parallel }}^2=\phi ({\delta }_0)=1$, so [F3] applies to every ${\pi }_{\phi }$. Since each ${\pi }_{\phi }$ is a $\ast$-representation, $\mathrm{\parallel }{a}^{\ast }a{\mathrm{\parallel }}_u={\sup }_{\phi }\mathrm{\parallel }{\pi }_{\phi }(a{)}^{\ast }{\pi }_{\phi }(a)\mathrm{\parallel }={\sup }_{\phi }\mathrm{\parallel }{\pi }_{\phi }(a){\mathrm{\parallel }}^2=\mathrm{\parallel }a{\mathrm{\parallel }}_u^2$. Submultiplicativity and $\mathrm{\parallel }{a}^{\ast }{\mathrm{\parallel }}_u=\mathrm{\parallel }a{\mathrm{\parallel }}_u$ follow by the same operator-norm calculation. Thus $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_u$ is a nonzero C*-seminorm; its kernel is a $\ast$-ideal, and the induced norm on $A_0\mathrm{/}\ker \mathrm{\parallel }\cdot {\mathrm{\parallel }}_u$ is a C*-norm.

Let $\rho :A_0\to \mathcal{B}(\mathcal{K})$ be any unital $\ast$-representation. If $\mathcal{K}=\{0\}$, the unique zero representation extends; assume otherwise and choose a unit vector $\xi \in \mathcal{K}$. The vector functional ${\phi }_{\xi }(a)=⟨\rho (a)\xi ,\xi ⟩$ is positive and unital. The map $[a]\mapsto \rho (a)\xi$ preserves the inner product on the GNS quotient, extends to a unitary from its completion onto the cyclic subspace $\overline{\rho (A_0)\xi }$, and intertwines ${\pi }_{{\phi }_{\xi }}$ with the restriction of $\rho$. Therefore $\mathrm{\parallel }\rho (a)\xi \mathrm{\parallel }\le \mathrm{\parallel }{\pi }_{{\phi }_{\xi }}(a)\mathrm{\parallel }\le \mathrm{\parallel }a{\mathrm{\parallel }}_u$; taking the supremum over unit vectors gives $\mathrm{\parallel }\rho (a)\mathrm{\parallel }\le \mathrm{\parallel }a{\mathrm{\parallel }}_u$.

The quotient completion in [F1] is a C*-algebra: multiplication and $\ast$ extend continuously by submultiplicativity and isometry, and the C*-identity passes to limits. Step 4.1 makes every unital $\ast$-representation contractive for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_u$, so it extends uniquely to the completion; conversely each representation of the completion restricts to one of $A_0$. The state ${\phi }_0$ from step 2.1 sends ${\delta }_0$ to a unit vector in its GNS space, so $\mathrm{\parallel }{\delta }_0{\mathrm{\parallel }}_u=1$ and the completion is nonzero. This proves the universal property and all claims.

If $n=0$, $L^2({\mathbb{R}}^0)=\mathbb{C}$ with unit mass and $S_{\lambda }(0,0,t)=e^{i\lambda t}I$, which is a strongly continuous irreducible representation with the asserted central character. Assume $n\ge 1$. Write $T_{q}f(x)=f(x-q)$ and $M_{p}f(x)=e^{i\lambda p\cdot x}f(x)$. These are unitaries because translations preserve Lebesgue measure and the multiplier has modulus one. The formula is $S_{\lambda }(q,p,t)=e^{i\lambda (t-q\cdot p\mathrm{/}2)}{M}_p{T}_q$.

Suppose a nonzero proper closed subspace $M\subset L^2({\mathbb{R}}^n)$ is invariant under every $W_{\lambda }(q,p)$. To obtain $g\mathrm{\ne }0$ in $M^{\perp }$, choose $x\mathrm{\notin }M$ and set $d={\inf }_{m\in M}\mathrm{\parallel }x-m\mathrm{\parallel }$. Countable Choice selects $m_k\in M$ with $\mathrm{\parallel }x-m_k{\mathrm{\parallel }}^2<d^2+1\mathrm{/}(k+1)$. If $d=0$, then $m_k\to x$, contradicting closedness; hence $d>0$. Since $(m_k+m_l)\mathrm{/}2\in M$, the parallelogram identity gives $\mathrm{\parallel }{m}_k-m_l{\mathrm{\parallel }}^2\le 2\mathrm{/}(k+1)+2\mathrm{/}(l+1)$, so $m_k\to m\in M$ and $\mathrm{\parallel }x-m\mathrm{\parallel }=d$. Minimality, applied to $m\pm th$ and $m\pm tih$ for real $t$ and $h\in M$, gives $g:=x-m\perp M$; also $g\mathrm{\ne }0$ since $d>0$. Choose $0\mathrm{\ne }f\in M$. Invariance under the unitaries and their inverses makes $M^{\perp }$ invariant too. For every $q,p$, orthogonality and the formula for $S_{\lambda }$ give$0=⟨W_{\lambda }(q,p)f,g⟩=e^{-i\lambda p\cdot q\mathrm{/}2}\int f(x-q)\overline{g(x)}{e}^{i\lambda p\cdot x}dx\mathrm{.}$For fixed $q$, $h_q(x)=f(x-q)\overline{g(x)}$ lies in $L^1$ by [F7]. As $p$ ranges over ${\mathbb{R}}^n$ and $\lambda \mathrm{\ne }0$, the last integral is its Fourier transform at every frequency (up to the fixed sign and $2\pi$ normalization). It vanishes everywhere, so [F5] gives $h_q=0$ almost everywhere for each $q$.

For $q,p,q^{\mathrm{\prime }},p^{\mathrm{\prime }}$, applying the formula twice gives$S_{\lambda }(q,p,t)S_{\lambda }(q^{\mathrm{\prime }},p^{\mathrm{\prime }},s)f(x)=e^{i\lambda (t+s+(p+p^{\mathrm{\prime }})\cdot x-p\cdot q\mathrm{/}2-p^{\mathrm{\prime }}\cdot q^{\mathrm{\prime }}\mathrm{/}2-p^{\mathrm{\prime }}\cdot q)}f(x-q-q^{\mathrm{\prime }})\mathrm{.}$On the other hand, the phase in $S_{\lambda }(q+q^{\mathrm{\prime }},p+p^{\mathrm{\prime }},t+s-\omega ((q,p),(q^{\mathrm{\prime }},p^{\mathrm{\prime }}))\mathrm{/}2)$ is the same, since $\omega ((q,p),(q^{\mathrm{\prime }},p^{\mathrm{\prime }}))=q\cdot p^{\mathrm{\prime }}-p\cdot q^{\mathrm{\prime }}$. Thus $S_{\lambda }(g)S_{\lambda }(h)=S_{\lambda }(gh)$ for all $g,h\in H$; the identity and inverse laws follow from [F1], so the unitary operators form a group representation.

By [F3], $q\mapsto T_q$ is strongly continuous. For fixed $f\in L^2$,$\mathrm{\parallel }{M}_{p}f-M_{p_0}f{\mathrm{\parallel }}_2^2={\int }_{{\mathbb{R}}^n}\mathrm{\mid }{e}^{i\lambda (p-p_0)\cdot x}-1{\mathrm{\mid }}^2\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx⟶0$as $p\to p_0$: the integrand converges pointwise to zero and is bounded by $4\mathrm{\mid }f{\mathrm{\mid }}^2$, so [F4] applies. Scalar multiplication by $e^{i\lambda t}$ is continuous. A finite product of strongly continuous unitaries is jointly strongly continuous, by inserting and subtracting one factor at a time and using that each factor has norm one. Therefore $(q,p,t)\mapsto S_{\lambda }(q,p,t)$ is strongly continuous.

By [F8], choose finite-valued Borel representatives $\tilde{f}$ and $\tilde{g}$ of $f$ and $g$, and put $G(x,q)=\mathrm{\mid }\tilde{g}(x){\mathrm{\mid }}^2\mathrm{\mid }\tilde{f}(x-q){\mathrm{\mid }}^2$. For each fixed $q$, the representative discrepancies, including the translated discrepancy for $f(x-q)$, are null sets, so step 1.2 still gives ${\int }_{{\mathbb{R}}^n}G(x,q)dx=0$. The integrand is product-measurable by [F8], so Tonelli [F6] gives ${\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}G(x,q)dxdq=0$. Reversing the order and using the measure-preserving change $q\mapsto x-q$ from [F9] gives instead${\int }_{{\mathbb{R}}^n}\mathrm{\mid }\tilde{g}(x){\mathrm{\mid }}^2({\int }_{{\mathbb{R}}^n}\mathrm{\mid }\tilde{f}(x-q){\mathrm{\mid }}^{2}dq)dx=\mathrm{\parallel }g{\mathrm{\parallel }}_2^2\mathrm{\parallel }f{\mathrm{\parallel }}_2^2>0,$ a contradiction. Hence no such $M$ exists and the Weyl system is irreducible.

Set $W_{\lambda }(q,p):=S_{\lambda }(q,p,0)$. Step 2.1 gives the Weyl relation with phase $e^{-i\lambda \omega \mathrm{/}2}$ by [F1]. The full strong continuity just proved implies regularity on every line. Also $S_{\lambda }(0,0,t)=e^{i\lambda t}I$, so the central character is as stated.

Steps 1.1–3.1 prove the unitary representation, central character, strong continuity and Weyl relation; step 2.3 proves irreducibility. These are exactly the asserted properties.

By [F7], $\mathrm{\parallel }{\mathrm{\Omega }}_a{\mathrm{\parallel }}_2^2=(a\mathrm{/}\pi {)}^{1\mathrm{/}2}\int e^{-a{x}^2}dx=1$. For fixed $f\in L^2(\mathbb{R})$, the map $(q,p)\mapsto e^{-a{q}^2\mathrm{/}4-p^2\mathrm{/}(4a)}W(q,p)f$ is strongly measurable by [F1] and has integrable norm. Therefore [F5] defines a bounded operator $A_a$ by the displayed weak integral.

Suppose $g\in L^2(\mathbb{R})$ is orthogonal to $W(0,p){\mathrm{\Omega }}_a$ for every $p$. Then $h(x)={\mathrm{\Omega }}_a(x)\overline{g(x)}$ belongs to $L^1$ by [F8], and $\widehat{h}(\xi )=0$ for every $\xi$, because $\lambda =1$ in the modulation and $p$ ranges over all real numbers. By [F6], $h=0$ almost everywhere; ${\mathrm{\Omega }}_a(x)>0$ everywhere, so $g=0$. Hence the modulation orbit, and therefore the full Weyl orbit, has dense linear span.