Locked-sector preservation theorem: invariance of the $\mathbf1\oplus\mathbf3$ split uniquely selects $q=1$

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Summary: Research record, section 84: Locked-sector preservation theorem: invariance of the $\mathbf1\oplus\mathbf3$ split uniquely selects $q=1$.

Provenance

Preserved from interaction notebook section 84, 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.

84. Locked-sector preservation theorem: invariance of the $\mathbf1\oplus\mathbf3$ split uniquely selects $q=1$

Section 83 showed that Hermiticity does not imply an observed--unobserved factor-exchange symmetry. There is, however, a cleaner representation-theoretic condition already built into the locked electron geometry.

After the diagonal frame lock,

$$(\mathbf2,\mathbf2) \downarrow_{Spin(3)_{\rm diag}} =\mathbf3\oplus\mathbf1.$$

The bound-state module interprets these as the exact three-channel $p$-like sector and one-channel $s$-like sector. If the stationary orbit generator is to act within these locked sectors rather than continuously converting the singlet into the triplet, it must preserve this decomposition.

For the full integer orbit family, that condition uniquely forces $q=1$.

84.1 General orbit family in the coupled-spin basis

Use the product-spin basis

$$|++\rangle, \quad |+-\rangle, \quad |-+\rangle, \quad |--\rangle,$$

with traceless generator

$$\widetilde G_q =\operatorname{diag} \left( q+\frac14, -q+\frac14, -\frac34, \frac14 \right).$$

Pass to the standard diagonal-spin basis

$$|1,1\rangle, \quad |1,0\rangle, \quad |1,-1\rangle, \quad |0,0\rangle,$$

where

$$|1,0\rangle =\frac{|+-\rangle+|-+\rangle}{\sqrt2},$$

and

$$|0,0\rangle =\frac{|+-\rangle-|-+\rangle}{\sqrt2}.$$

Exact transformation gives

$$\boxed{ \widetilde G_q^{\rm CG} = \begin{pmatrix} q+\frac14&0&0&0\\ 0&-\frac q2-\frac14&0&\frac{1-q}{2}\\ 0&0&\frac14&0\\ 0&\frac{1-q}{2}&0&-\frac q2-\frac14 \end{pmatrix}. }$$

The crucial matrix element is therefore

$$\boxed{ \langle1,0|\widetilde G_q|0,0\rangle =\frac{1-q}{2}. }$$

Thus every

$$q\ne1$$

mixes the locked singlet with the $m=0$ member of the triplet.

84.2 Preserving the locked irreducible sectors

Let

$$P_3 =\operatorname{diag}(1,1,1,0)$$

project onto the locked triplet and

$$P_1 =\operatorname{diag}(0,0,0,1)$$

onto the locked singlet.

The orbit generator preserves the decomposition if and only if

$$[P_3,\widetilde G_q]=0.$$

Direct calculation gives off-diagonal entries proportional only to

$$1-q,$$

so

$$\boxed{ [P_3,\widetilde G_q]=0 \quad\Longleftrightarrow\quad q=1. }$$

Equivalently, using the total locked-spin Casimir

$$J^2 =\operatorname{diag}(2,2,2,0),$$

one finds

$$\boxed{ [J^2,\widetilde G_q]=0 \quad\Longleftrightarrow\quad q=1. }$$

This does not require the orbit generator to commute with the whole diagonal $Spin(3)$. It only requires that it not mix inequivalent irreducible sectors. The beam axis may still split the triplet internally.

84.3 Why beam $SO(2)$ alone was too weak

The axial generator in this coupled basis is

$$J_z =\operatorname{diag}(1,0,-1,0).$$

For every $q$,

$$\boxed{[J_z,\widetilde G_q]=0.}$$

Both the triplet $m=0$ state and the singlet have axial quantum number zero, so beam-axis symmetry cannot distinguish or prevent their mixing.

This explains exactly why the Section-81 beam-centralizer argument failed in the physically correct basis.

The stronger invariant is the locked representation label carried by $J^2$.

84.4 Unique surviving generator

At

$$q=1,$$

the mixing vanishes and

$$\boxed{ \widetilde G_1^{\rm CG} =\operatorname{diag} \left( \frac54, -\frac34, \frac14, -\frac34 \right). }$$

This is exactly the existing V2 orbit generator, and Section 74 already showed that it can be written without any singlet--triplet discriminator as

$$\boxed{ \widetilde G_1 =-\frac34I +\frac12J_z +\frac32J_z^2. }$$

The absence of a $J^2$ term now has a sharper meaning: the selected generator acts identically on the two $m=0$ states and therefore preserves the $\mathbf3\oplus\mathbf1$ decomposition without using the representation label to shift one sector relative to the other.

Passing to $\Lambda^2\mathbf4$ then gives the already verified positive rotation rates

$$\boxed{ \frac12, \frac12, \frac32 }$$

and hence

$$\boxed{1:1:3.}$$

84.5 Status of the result

The algebraic theorem is exact:

$$\boxed{ \text{within the family }(q,-q,-1,0), \quad \text{preservation of }\mathbf3\oplus\mathbf1 \Longrightarrow q=1. }$$

This is stronger than the factor-exchange route because it does not require $O$ and $H$ to be physically interchangeable. It only asks the stationary orbit dynamics to respect the irreducible sectors produced by the already-selected diagonal frame lock.

The remaining physical question is whether that preservation law is mandatory in V2. The current electron module calls

$$\mathbf1\oplus\mathbf3$$

the exact $s\oplus p$ branching used by the chemistry construction, which strongly motivates treating the two summands as distinct locked channels. But the projected master action has not yet explicitly proved

$$[P_3,H_{\rm orb}]=0.$$

So the result should presently be stated as a representation-theoretic closure theorem conditional on channel preservation, not yet as an unconditional action theorem.

84.6 Next falsification test

The next step is to inspect every place the $\mathbf1\oplus\mathbf3$ branching is used in the bound-state, magnetic, and observer modules and ask whether any existing observable or transition law actually requires singlet/triplet preservation.

If V2 already assumes or derives that the $s$-like channel cannot be coherently mixed with the $p$-like triplet by the stationary orbit Hamiltonian, then $q=1$ is no longer free and the dimensionless $1{:}1{:}3$ spectrum is closed.

If such mixing is allowed, then this theorem remains only conditional and $q$ is still an open dynamical datum.

Related

• Research status — current proof boundaries and correction guidance.
• Reciprocity audit: Hermiticity does not force observed-unobserved exchange symmetry — preceding section in the research sequence.
• Operator-class refinement: the unwanted $q\neq1$ piece is a singlet--triplet vector intertwiner — following section in the research sequence.
• Leading-source orientation no-go: the written scalar and factorized $3\leftrightarrow6$ structures can select $s$ but cannot select the locked frame orientation — later audit limits or corrects this route.