Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 10 / 3

A unitary family with the required Weyl relations isTranslations preserve the $L^2$ norm and the multiplier has modulus one. Multiplication of two such operators leaves the phase $x_1{y}_2-y_1{x}_2$ after extracting $W(x_1+x_2,y_1+y_2)$, proving the relation. Translation and modulation are strongly continuous, first on compactly supported smooth functions by direct estimates, then on all of $L^2$ by density and unitarity.

If a bounded operator $A$ commutes with all $W$, it commutes with translations and every modulation $e^{2iyt}$. The spectral theorem for the multiplication coordinate therefore makes it commute with all bounded multiplication operators and interval projections. One can see its form directly: on a finite interval $I$, set $h_I=A{1}_I$. For bounded $\varphi$ supported in $I$, commutation gives $A\varphi =\varphi h_I$. Testing characteristic functions shows $|h_I|\le \Vert A\Vert$ almost everywhere. On increasing intervals these functions agree on overlaps and yield a single $h\in L^{\infty }$ with $A=M_h$.

Commutation with translations says $h(t+x)=h(t)$ almost everywhere for every $x$. Convolving with a smooth approximate identity produces smooth functions invariant under every translation, hence constants. Passing to the local $L^1$ limit makes $h$ constant almost everywhere. Therefore $A$ is a scalar operator, and the unitary family acts irreducibly.

Let $\tilde{W}$ be the given regular irreducible family. Write $a=(x,y)$ and $a\wedge b=x_1{y}_2-y_1{x}_2$. Define the strongly integrated operatorThe Gaussian is integrable and the operators are unitary, so the integral defines a bounded operator on each vector. The Weyl relation and $\tilde{W}(a{)}^{\ast }=\tilde{W}(-a)$ give $P^{\ast }=P$. For its square, the coefficient at $\tilde{W}(c)$ isby the two-dimensional Gaussian integral. Thus $P^2=P$. The same Gaussian integration, with an inserted Weyl operator, givesThis is the Gaussian projection in a Weyl representation.

It is nonzero. Otherwise all its conjugates would vanish; the Weyl relation makes their matrix coefficients the Fourier transforms of $e^{-|a{|}^2/2}(\tilde{W}(a)\xi ,\eta )$, with the dual variable $2b\wedge a$. Uniqueness of the Fourier transform of an integrable function would make each such function zero. Strong continuity then makes its value at zero vanish, contradicting $(\xi ,\xi )>0$ for a nonzero vector.

Choose a unit vector $v\in PH$. Irreducibility makes the span of $\tilde{W}(a)v$ dense. If $w\in PH$ is perpendicular to $v$, the boxed identity makes it perpendicular to this entire span, so $w=0$. Hence $P$ has rank one, and its identity gives $(\tilde{W}(a)v,v)=e^{-|a{|}^2/2}$.

In the concrete representation of part (a), the Gaussian $v_0(t)=(2/\pi {)}^{1/4}{e}^{-t^2}$ has precisely this expectation. With the linear-in-first inner-product convention, both orbit families have Gram matrixTherefore the map $\sum c_{j}W(a_j)v_0\mapsto \sum c_j\tilde{W}(a_j)v$ is well-defined and isometric. Its source and target spans are dense, so it extends to a unitary operator intertwiner. This proves the unitary equivalence, the Stone-von Neumann theorem in the normalization of this question, rather than invoking that theorem as the proof.

The real matrix $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with determinant one preserves $u\wedge v$. Hence $W_g(u)=W(gu)$ obeys the same Weyl relations. It is strongly continuous and irreducible because $g$ is invertible and its arguments range over the whole plane. Part (b) gives a unitary $V$ intertwining $W$ and $W_g$, so $VW(u)V^{\ast }=W(gu)$. If $V_1,V_2$ both work, $V_2^{\ast }{V}_1$ commutes with all $W$ and is scalar by part (a); its scalar has modulus one. Thus the implementer is unique up to phase.

For the displayed quarter-turn, choose the scaled inverse Fourier transformThe formula starts on Schwartz functions and extends unitarily to $L^2$ by Plancherel and a change of scale. It conjugates $T_{x}f(t)=f(t+x)$ to multiplication by $e^{-2ixs}$, and multiplication by $e^{2iyt}$ to $T_y$. Keeping the Weyl phase therefore gives $VW(x,y)V^{\ast }=W(y,-x)$, as required. With the opposite Fourier-kernel sign one obtains the opposite quarter-turn, so that convention must be checked.