Minimal self-adjoint unobserved-visible lock with exactly zero relaxed backreaction

Physical premise (IDWT): Physical spacetime has infinitely many ordinary spatial dimensions plus time. Matter occupies up to ten spatial dimensions; all remaining spatial directions are unbound for every particle. Depth counts a pattern's bound spatial involvement within that space. “Observed” and “unobserved” describe an observer's access, not different kinds of dimension. Older projective and nested-geometry calculations below retain their additional hypotheses; they do not establish a separate physical manifold or compactified direction. See spatial meaning and mathematical scope.

Summary: Research record, section 44: Minimal self-adjoint unobserved-visible lock with exactly zero relaxed backreaction.

Provenance

Preserved from interaction notebook section 44, recorded 2026-09-05 and corrected 2026-09-06. Original sequence contents. This is an exploratory research record, not an independently established V2 result. Read the related later audits before using an intermediate conclusion.

44. Minimal self-adjoint unobserved-visible lock with exactly zero relaxed backreaction

Choose the local observer-relative coordinate grouping

$$V_6=O\oplus H\cong\mathbb R^3\oplus\mathbb R^3$$

within six ordinary spatial dimensions of the same kind. This is not an inherent electron split. Additionally assume an orthogonal link

$$Q:H\to O,\qquad Q\in SO(3).$$

For relative coordinates $u\in O$ and $h\in H$, consider

$$\boxed{ V_{\rm lock}(u,h)=\frac{\kappa}{2}\|h-Q^Tu\|^2,\qquad \kappa>0. }$$

Its Hessian is

$$\boxed{ K_{\rm lock}=\kappa \begin{pmatrix} I&-Q\\ -Q^T&I \end{pmatrix}. }$$

Because $Q^TQ=QQ^T=I$, this operator is self-adjoint and positive semidefinite. The explicit symbolic check gives

$$\operatorname{rank}K_{\rm lock}=3,$$

and

$$\boxed{ \operatorname{spec}K_{\rm lock}=\{0,0,0,2\kappa,2\kappa,2\kappa\}. }$$

The zero eigenspace is exactly

$$\boxed{h=Q^Tu.}$$

Thus the interaction does not trap common motion; it penalizes only visible-unobserved disagreement.

44.1 Exact Schur cancellation

With

$$A=\kappa I,\quad B=-\kappa Q,\quad C=-\kappa Q^T,\quad D=\kappa I,$$

the relaxed visible correction is

$$A-BD^{-1}C.$$

Since

$$BD^{-1}C =(-\kappa Q)(\kappa^{-1}I)(-\kappa Q^T) =\kappa QQ^T =\kappa I,$$

one obtains the exact identity

$$\boxed{A-BD^{-1}C=0.}$$

Therefore if

$$K_6= \begin{pmatrix} K_{\rm obs}+\kappa I&-\kappa Q\\ -\kappa Q^T&\kappa I \end{pmatrix},$$

then unobserved relaxation gives

$$\boxed{K_{\rm eff}=K_{\rm obs}.}$$

Strong six-dimensional locking and an exactly unchanged reduced static law are therefore mathematically compatible.

44.2 Dynamical relative mode

With equal unit kinetic normalization and

$$r=h-Q^Tu,$$

the equations give

$$\boxed{\ddot r+2\kappa r=0,}$$

so

$$\boxed{\omega_{\rm lock}^2=2\kappa.}$$

The collective locked coordinate is a zero mode at this order.

44.3 Status

This structure is now algebraically verified. It is not yet derived from the V2 master action. The remaining obligations are to derive the difference-norm term, determine $\kappa$, and explain how a free electron pair obtains $Q$ without borrowing the nucleus-selected frame.

Related

• Research status — current proof boundaries and correction guidance.
• Implication for visible-unobserved collision correlation — preceding section in the research sequence.
• Canonical free-pair operator from two electron complex structures — following section in the research sequence.