The Schrödinger Weyl system is regular and irreducible

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.