Finite symplectic implementation by quadratic generators

Reorder interleaved coordinates by the fixed permutation $P(Q_1,P_1,\dots ,Q_r,P_r)=(Q_1,\dots ,Q_r,P_1,\dots ,P_r)$, and write $R_g:=PR$ and $W_g(w)=W(P^{T}w)$ for grouped labels. Then $J_0=PJ{P}^T$ and $S_g=PS{P}^T$ obeys $S_g^T{J}_0{S}_g=J_0$. Apply [F3] directly to the grouped Weyl system on $L^2({\mathbb{R}}^r)$ with label matrix $S_g^T$; this matrix is symplectic because $S_g^T{J}_0{S}_g=J_0$ also implies $S_g{J}_0{S}_g^T=J_0$. Since $S_g^{T}P=P{S}^T$, converting labels back gives $M_S^{\ast }W(z)M_S=M_S^{\ast }{W}_g(Pz)M_S=W_g(S_g^{T}Pz)=W(S^{T}z)\mathrm{.}$ Differentiating the bounded one-parameter Weyl identity on the common Schwartz core gives $M_S^{\ast }R{M}_S=SR$. For the rest of the proof abbreviate $R_g$ to $R$ and $S_g$ to $S$, so $R$, $K$, and $J_0$ all use grouped order.

Use exactly the endpoint definition and trajectory class in the Statement, on arbitrary compact intervals $[0,T]$ with $T>0$. Reordering by the fixed permutation of step 1.1 sends $K$ to $PK{P}^T$ and preserves entrywise integrability, reality and symmetry, so [F4] constructs and uniquely determines these evolutions in grouped order with $H_t=\frac{1}{4}{\sum }_{j,k}{K}_{jk}(t)(R_j{R}_k+R_k{R}_j)$ and no linear or scalar term. The operator acts on the common Schwartz space, and each trajectory is absolutely continuous in every seminorm and solves the equation almost everywhere. Concatenation gives products: first evolve by $U_2$ on $[0,T_2]$, then evolve by $U_1$ on $[0,T_1]$ with initial vector $U_2(T_2)f$, obtaining endpoint $U_1(T_1)U_2(T_2)$. The concatenated coefficient is integrable and symmetric on $[0,T_2+T_1]$, and the two absolutely continuous trajectories match at the join. The reversed evolution $U(T-t)U(T{)}^{-1}$ starts at $I$, ends at $U(T{)}^{-1}$, and is generated almost everywhere by $-H_{T-t}$, whose symmetric coefficient is again integrable. Thus inverses also belong to the set. By [F4], each endpoint implements the symplectic solution $\dot{S}=J_{0}K(t)S$. If two matrices had the same implementing unitary, differentiating their Weyl covariances on Schwartz space would give $(S-S^{\mathrm{\prime }})R=0$. Applied to the vacuum, $Q_k{g}_0=\sqrt{\mathrm{\hslash }\mathrm{/}2}{g}_{e_k}$ and $P_k{g}_0=i\sqrt{\mathrm{\hslash }\mathrm{/}2}{g}_{e_k}$ by [F6]; the reality of the rows forces both their position and momentum coefficients to vanish. Thus the endpoint is unique. For two lifts, $\pi (M_1{M}_2)=\pi (M_1)\pi (M_2)$ by conjugating the quadratures, so $\pi$ is a homomorphism.

If $M_1,M_2$ implement the same $S$, then [F3] makes $M_2{M}_1^{\ast }=\lambda I$ for some $\mathrm{\mid }\lambda \mathrm{\mid }=1$. For any other representation $W^{\mathrm{\prime }}$ in [F1], choose a unitary $V$ with $V{W}^{\mathrm{\prime }}(z)V^{\ast }=W(z)$ by the finite-dimensional Stone–von Neumann theorem. Then $M_S^{\mathrm{\prime }}=V^{\ast }{M}_{S}V$ is unitary and $M_S^{\mathrm{\prime }\ast }{W}^{\mathrm{\prime }}(z)M_S^{\mathrm{\prime }}=V^{\ast }{M}_S^{\ast }W(z)M_{S}V=W^{\mathrm{\prime }}(S^{T}z)\mathrm{.}$ Conjugating the phase-uniqueness statement by $V$ proves the same result in that representation.

Every generator from [F2] is a path endpoint of a homogeneous quadratic Hamiltonian. A symmetric shear is joined to the identity by scaling its symmetric block, and a partial Fourier rotation is joined by varying the oscillator angle. If $\det A>0$, polar decomposition writes $A=OP$ with $O\in SO(r)$ and $P$ positive definite. The path $P_t=(1-t)I+tP$ stays positive definite, and an orthogonal matrix in $SO(r)$ is a product of plane rotations, each joined to the identity by varying its angle. These paths connect the coordinate change to the identity. If $\det A<0$, factor one coordinate reflection from a positive-determinant change; that reflection is a $\pi$ oscillator rotation on that mode. This gives a piecewise-smooth symplectic path for each generator. Along any such path $S_t$, the matrix $X_t={\dot{S}}_t{S}_t^{-1}$ satisfies $X_t^T{J}_0+J_0{X}_t=0$; hence $K_t=-J_0{X}_t$ is symmetric and ${\dot{S}}_t=J_0{K}_t{S}_t$. The symmetrically ordered quadratic with matrix $K_t$ therefore has this classical endpoint. Thus [F2] supplies, for every $S$, a product of normalized quadratic propagators whose endpoint is $S$, proving $\pi$ is onto. The one-parameter paths also show that each such product belongs to the identity component of $\mathrm{Mp}$.

If two normalized lifts lie over the same grouped matrix $S$, the Gaussian determinant calculation in [F5] makes them differ by sign. This already holds for every $S$: the matrix $A+iB$ used there is invertible by symplecticity, including when either real block is singular. In particular it supplies the required invertible-$B$ chart argument without an imported Maslov-index parity formula. Thus every fiber has at most two elements.

The identity has two distinct lifts. The constant zero Hamiltonian gives $I$. On the first mode (and as the identity on the other modes if $r>1$), take the canonical Schwartz evolution of $H=\mathrm{\hslash }(N+1\mathrm{/}2)$ constructed in [F4]. For each Hermite vector, $e^{-it(n+1\mathrm{/}2)}\mathrm{\mid }n⟩$ solves the same Schrödinger equation with that initial vector, so uniqueness in [F4] identifies $U(t)\mathrm{\mid }n⟩=e^{-it(n+1\mathrm{/}2)}\mathrm{\mid }n⟩$. Its classical endpoint is the identity at $t=2\pi$, because its Hamilton equations are $\dot{Q}=P$, $\dot{P}=-Q$ and make one full rotation, while by [F6] $U(2\pi )\mathrm{\mid }n⟩=e^{-2\pi i(n+1\mathrm{/}2)}\mathrm{\mid }n⟩=-\mathrm{\mid }n⟩$ for every $n$. Completeness of the Hermite basis gives $U(2\pi )=-I$. For any $S$, choose a normalized lift ${\tilde{M}}_S$ supplied by step 3.1. Then ${\tilde{M}}_S$ and $-{\tilde{M}}_S$ are two distinct lifts; step 3.2 proves there are no others. The kernel is exactly $\{I,-I\}$.

Fix $S_0$ and a normalized lift $M_0$. On a matrix neighborhood $\mathcal{B}$ of $S_0$ small enough that $S{S}_0^{-1}$ lies in the local neighborhood of [F4], put $\sigma (S)=L(S{S}_0^{-1})M_0$. The parameter path $(Y,A,X)\mapsto (\tau Y,I+\tau (A-I),\tau X)$ in [F4] joins each local factor lift to $I$ smoothly, with homogeneous quadratic logarithmic generator. Thus the local factors are endpoints of the piecewise-smooth subclass of the integrable class defining this group, so $\sigma (S)$ belongs to $\mathrm{Mp}$ and has endpoint $S$. Conversely the construction of [F4] writes any integrable-coefficient propagator endpoint as a finite product of these same local lifts. Hence allowing the broader integrable temporal class in the parity lemma introduces no additional endpoint operators or phases. It depends continuously on $S$, as does its inverse, pointwise in the Schwartz topology: the explicit chirp and linear-coordinate factors have uniformly bounded seminorm estimates on compact parameter neighborhoods, and the Fourier factors are fixed continuous maps. These estimates also justify continuity when composing such a varying local factor with a varying Schwartz vector. Step 4.1 shows every lift over $S$ is exactly $\sigma (S)$ or $-\sigma (S)$.

The endpoint map is continuous for the specified operator topology, without assuming joint continuity of arbitrary Schwartz-operator composition. For each row $j$ and mode $k$, covariance and [F6] give $⟨R_{j}M{g}_0,M{g}_{e_k}⟩=\sqrt{\mathrm{\hslash }\mathrm{/}2}(S_{j,Q_k}+i{S}_{j,P_k})$, where the pairing is linear in its first argument. On the left $R_j$ is continuous on Schwartz space, its inclusion into $L^2$ is continuous, and only the two fixed input vectors $g_0,g_{e_k}$ are used. Thus every real matrix entry of $\pi (M)$ is continuous. Over $\mathcal{B}$, the scalar $\epsilon (M):=⟨M{g}_0,\sigma (\pi (M))g_0⟩$ is continuous, takes exactly the values $+1,-1$, and distinguishes the two lifts since $\mathrm{\parallel }{g}_0\mathrm{\parallel }=1$. Therefore the two inverse images ${\epsilon }^{-1}(1)$ and ${\epsilon }^{-1}(-1)$ in ${\pi }^{-1}(\mathcal{B})$ are disjoint open sheets; on each, $\pi$ and $\pm \sigma$ are continuous inverse maps onto $\mathcal{B}$. This proves the two-fold covering in the given topology.

The same charts make the set a topological group. Locally its elements are $\epsilon \sigma (S)$. Products and inverses of their explicit finite-parameter factors depend continuously on the matrix parameters and on each Schwartz vector, by the uniform seminorm bounds in step 5.1; their endpoint matrices multiply and invert continuously. The sign relative to a local section at the resulting endpoint is recovered by the continuous vacuum pairing in step 6.1. Hence multiplication and inversion are continuous in these charts, which have precisely the operator topology by step 6.1. Finally every defining normalized path is continuous in that topology by [F4], including its inverse path, so it joins its endpoint to $I$. Consequently the group is path connected. Together with step 4.1 this proves the connected double cover with kernel $\{I,-I\}$ and its defining unitary action, retaining the unrestricted phase ambiguity only for arbitrary individual Weyl implementers.