One calibrated displacement still leaves full receiver ambiguity
A known nonzero incoming probe displacement breaks R26’s sign involution but still permits distinct receiver states with exactly the same ten final apparatus coordinates and the same two initial energies. A Borsuk–Ulam argument produces actual equal-record pairs on a compact preparation sphere. The resulting positive risk concerns the full receiver state in fixed component units; it does not establish a positive canonical x–P error product.
R27, 2026-09-12. Retain R25’s smooth finite-duration apparatus, receiver energy E>0, positive clock speed, fixed masses, pulse supports and cutoff margins. Require initial H_s=E, H_app=H_0=M_c v_0^2/2, and q_1=c with c nonzero and |c|/L_1<b/4. Here L_1 is the fixed position unit and b is the dimensionless apparatus box half-width about R25’s nominal preparation. Choose b sufficiently small for its uniform flow bounds. Every point satisfying these constraints inside the box is admitted. All ten final apparatus coordinates at known T are observed exactly. Source coverage and proof review are recorded in B58 and the review.
1. A nonempty preparation chart with eleven free coordinates
The apparatus energy is purely kinetic before the pulses. Set q_1=c and solve its positive clock-momentum branch exactly:
\[p_s=\sqrt{2M_c\left(H_0-\sum_{j=1}^4\frac{\pi_j^2}{2M_j}\right)}. \tag{1}\]
The remaining apparatus coordinates q_2,q_3,q_4,pi_1,…,pi_4,s are eight independent local coordinates near zero probe phase and s=s_0. At their centre, (1) gives p_s=M_c v_0. Continuity supplies a fixed neighbourhood with positive radicand and every apparatus coordinate inside half the box. This is interior relative to the two imposed apparatus constraints.
For the receiver use three independent coordinates x,P,Q near zero, solving its energy for the upper y branch:
\[y=\frac gd x+ \sqrt{\frac{2E-k_xx^2-P^2/\mu-Q^2/\nu}{d}}, \qquad k_x=a-g^2/d>0. \tag{2}\]
Equations (1)–(2) give a smooth injective chart Psi from an open neighbourhood of zero in R^11 into the admissible initial states. Scale each of its eleven free coordinates by its fixed physical component unit (and centre s at s_0). Choose r>0 such that the closed parameter ball of radius r lies in this neighbourhood with strict energy radicands and the b/2 apparatus margin. The radius r depends on E,H_0,b,c and the component units, but not on lambda. Initial pulses are absent throughout this chart. Reduce the common positive coupling upper bound to retain R25’s trajectory margins.
2. A topological theorem yields exact common records
Write G_lambda(z,eta) for the complete final apparatus record, in its ten fixed component units. On the parameter sphere |u|_2=r, define
\[f_\lambda(u)=G_\lambda(\Psi(u))\in\mathbb R^{10}. \tag{3}\]
This is a continuous map S^10 to R^10. The Borsuk–Ulam theorem gives some u_lambda on that sphere with
\[f_\lambda(u_\lambda)=f_\lambda(-u_\lambda). \tag{4}\]
The two initial states are distinct because Psi is injective. Both have exactly q_1=c, H_s=E and H_app=H_0 by construction, and positive box margins. The equality in (4) is exact, not an infinitesimal kernel or a dimension count. Antipodes here are chart parameters: they do not reverse the physical receiver or the calibrated displacement. No Hamiltonian sign symmetry is used.
3. The receiver states differ by a uniform amount
The equal-record pair cannot differ only in apparatus preparation. In fixed Euclidean component norms, precondition G_lambda by the invertible free shear S and put F_lambda=S^{-1}G_lambda. R25’s smooth-flow argument, with constants chosen in these norms on the same convex receiver ball and apparatus box, gives
\[\|D_\eta F_\lambda-I\|\le C\lambda\le\tfrac12, \qquad \|D_zF_\lambda\|\le L\lambda. \tag{5}\]
For two states with equal G_lambda, integrate the first derivative along the apparatus segment at fixed receiver state, and the second along the receiver segment at fixed apparatus state. This yields
\[\tfrac12\|\eta_+-\eta_-\|_2 \le L\lambda\|z_+-z_-\|_2. \tag{6}\]
The segments are used only to estimate the unconstrained smooth map on its convex domain; they need not stay on the two energy shells or calibration surface. In particular, equal receiver states would force equal apparatus states, contradicting injectivity of the chart.
There is also a quantitative bound. The eleven chart coordinates are an orthogonal coordinate projection of the fourteen dimensionless initial coordinates, up to constant centring. Consequently
\[2r=\|u_\lambda-(-u_\lambda)\|_2 \le\sqrt{\|z_+-z_-\|_2^2+\|\eta_+-\eta_-\|_2^2},\]
and (6) gives
\[\|z_+-z_-\|_2\ge \frac{2r}{\sqrt{1+4L^2\lambda^2}}. \tag{7}\]
For deterministic estimation of all four initial receiver coordinates, let R_z be the infimum over estimators of the worst-case Euclidean error in these fixed component units on the constrained preparation class. The triangle inequality at the common record proves
\[R_z\ge\frac{r}{\sqrt{1+4L^2\lambda^2}}>0. \tag{8}\]
The constants and r are fixed as lambda decreases. Thus liminf R_z>=r in this weak-coupling limit. This dimensionless full-state risk need not be in x or P: the unresolved difference may lie in the internal coordinates y,Q. It gives neither an action-valued lower bound nor a universal scale.
4. What calibration changes and the next test
The involution of R26 sends q_1=c to -c and leaves the admitted class when c is nonzero. Equations (1)–(8) replace it by a topological obstruction to full receiver recovery. They use a fixed positive preparation neighbourhood, exact initial energies and exact final records; they require no pulse-rank assumption. Borsuk–Ulam is the established imported theorem; the constrained chart and quantitative receiver bound are consequences for this apparatus.
Next R28: determine whether an equal-record pair under this same calibrated two-energy preparation can be forced to differ in both x and P. Seek an actual canonical pair with margins or prove canonical recovery despite hidden internal coordinates. The full-state norm bound alone settles neither.