Eq. (7): the logical operators satisfy $U(s\,x_i)U(t\,z_j)=e^{i\pi st\delta_{ij}}U(t\,z_j)U(s\,x_i)$ (printed comma read as a product)

Let $x_1,z_1,\dots,x_m,z_m\in\mathbb R^{2n}$ satisfy $\omega(x_i,x_j)=\omega(z_i,z_j)=0$ and $\omega(x_i,z_j)=\delta_{ij}$. Then for all $i,j$ and all real $s,t$,

Applied to the basis of $W^\omega/W$ (with $m=n-k$), these are the commutation relations of $n-k$ independent position and momentum shifts.

Formalization Note $U$ is the concrete realization $(U(s,t)\psi)(x)=e^{i\pi(\langle t,x\rangle+\langle s,t\rangle/2)}\psi(x+s)$ acting on all functions $\mathbb R^n\to\mathbb C$; operators are composed with $\circ$ and a scalar $c$ acts as $\psi\mapsto c\,\psi$. Commutators $[A,B]=0$ are stated as $A\circ B=B\circ A$.

Differences from the source:

• reading: U(v) = exp(iπ Σᵢ (sᵢp̂ᵢ + tᵢq̂ᵢ)) with [q̂, p̂] = 1 (2), which under no normalization of the canonical pair yields (3) as printed → the concrete realization weyl: (U(s,t)ψ)(x) = e^{iπ(⟨t,x⟩ + ⟨s,t⟩/2)} ψ(x + s); reason: (3), (7) and every example of the paper use (3); in this realization (3) and (7) hold verbatim, Z(t) = U(0,t) is the multiplication by e^{iπtx} of (11), X(s) = U(s,0) is the position shift ψ ↦ ψ(· + s), i.e. X(s)|q⟩ = |q − s⟩ (this direction is forced by (3) once Z(t) is the multiplication of (11); the p.3 example X̄(q)|000⟩ = |qqq⟩ reads the shift the other way, which is the paper's inconsistency, and no item depends on it), and the symmetric phase e^{iπ⟨s,t⟩/2} gives U(av)U(bv) = U((a+b)v)
• generalization: operators on the state space with position basis |q⟩ → operators on all functions ℝⁿ → ℂ; identities hold pointwise; reason: the position eigenvectors |q⟩ are not in L²; each U(v) is a shift followed by multiplication by a continuous phase, so it preserves L² and a.e. equality, every operator identity restricts to L²(ℝⁿ), and equality, commutation and injectivity of the U(v) are the same on L² as on all functions
• generalization: the basis x₁, z₁, …, x_{n−k}, z_{n−k} of W^ω/W constructed from a stabilizer → any vectors x₁, z₁, …, x_m, z_m ∈ ℝ²ⁿ with ω(xᵢ, xⱼ) = ω(zᵢ, zⱼ) = 0, ω(xᵢ, zⱼ) = δᵢⱼ; reason: the proof uses only these relations; the logical basis (m = n − k) satisfies them
• reading: U(s xᵢ)U(t zⱼ) = e^{iπstδᵢⱼ}U(t zⱼ), U(s xᵢ) (7), and [A, B] = 0 → U(s xᵢ) ∘ U(t zⱼ) = e^{iπstδᵢⱼ} • U(t zⱼ) ∘ U(s xᵢ); commutators as A ∘ B = B ∘ A; reason: the comma is a printing slip; with it removed (7) agrees with (3), since ω(s xᵢ, t zⱼ) = st δᵢⱼ