Physical $(3,1)\oplus(1,3)$ audit: beam symmetry leaves $q$ free; frame exchange would select $q=1$

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 82: Physical $(3,1)\oplus(1,3)$ audit: beam symmetry leaves $q$ free; frame exchange would select $q=1$.

Provenance

Preserved from interaction notebook section 82, 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.

82. Physical $(3,1)\oplus(1,3)$ audit: beam symmetry leaves $q$ free; frame exchange would select $q=1$

Section 81 distinguishes physical beam-axis symmetry from a convenient rotation-plane repartition. Here the physical branching is computed explicitly.

The $Spin(3)_O\times Spin(3)_H$ branching fixes which split the matrix must use.

82.1 The physical six-vector split

For the locked-spin product basis

$$\mathbf4=(\mathbf2,\mathbf2),$$

the real/vector representation branches as

$$\boxed{ \Lambda^2(\mathbf2,\mathbf2) =(\mathbf3,\mathbf1)\oplus(\mathbf1,\mathbf3). }$$

This is the physical

$$V_6=O\oplus H$$

split into observed and unobserved three-frames.

Use the spinor basis

$$e_1=|++\rangle, \quad e_2=|+-\rangle, \quad e_3=|-+\rangle, \quad e_4=|--\rangle,$$

and wedge basis

$$e_{12},e_{13},e_{14},e_{23},e_{24},e_{34}.$$

An orthonormal Clebsch--Gordan basis adapted to $(\mathbf3,\mathbf1)\oplus(\mathbf1,\mathbf3)$ is

$$\begin{aligned} O_+&=e_{12}, & O_0&=\frac{e_{14}-e_{23}}{\sqrt2}, & O_-&=e_{34},\\ H_+&=e_{13}, & H_0&=\frac{e_{14}+e_{23}}{\sqrt2}, & H_-&=e_{24}. \end{aligned}$$

82.2 The full $q$ family in the physical split

For projective weights

$$(q,-q,-1,0),$$

the traceless spinor generator has weights

$$\left(q+\frac14,-q+\frac14,-\frac34,\frac14\right).$$

Its $\Lambda^2$ weights in the wedge basis are

$$\frac12, \quad q-\frac12, \quad q+\frac12, \quad -q-\frac12, \quad \frac12-q, \quad -\frac12.$$

Transforming exactly to the physical $O\oplus H$ basis gives

$$\boxed{ G_6^{O\oplus H}(q)= \begin{pmatrix} \frac12&0&0&0&0&0\\ 0&0&0&0&q+\frac12&0\\ 0&0&-\frac12&0&0&0\\ 0&0&0&q-\frac12&0&0\\ 0&q+\frac12&0&0&0&0\\ 0&0&0&0&0&\frac12-q \end{pmatrix} }$$

in the complex weight basis displayed above.

The interpretation is immediate:

• the observed triplet has an axial pair within its representation with weights $\pm1/2$;
• the unobserved triplet has an axial pair within its representation with weights $\pm(q-1/2)$;
• the two zero-weight axial states $O_0$ and $H_0$ mix with coefficient $q+1/2$.

Most importantly, the physical observed--unobserved mixed block has

$$\boxed{\operatorname{rank}M_{OH}=1}$$

for generic $q$, including $q=1$.

A full-rank mixed $3\leftrightarrow3$ block for this orbit generator is basis-split-dependent. It is true only after repartitioning the three independent rotation planes by taking one axis from each plane; that repartition is not the physical $(\mathbf3,\mathbf1)\oplus(\mathbf1,\mathbf3)$ frame split.

So the orbit generator cannot replace the independently constructed full-rank scattering frame link $Q_{\rm scat}$.

82.3 The physical beam generator commutes for every $q$

The diagonal beam-axis generator on the spinor product is

$$J_z^{\rm diag} =\frac12(\sigma_z\otimes I+I\otimes\sigma_z) =\operatorname{diag}(1,0,0,-1).$$

Both $J_z^{\rm diag}$ and $G_q$ are diagonal in the same spinor basis, hence

$$\boxed{[J_z^{\rm diag},G_q]=0}$$

for every $q$.

Functoriality of $\Lambda^2$ then gives

$$\boxed{ [\Lambda^2J_z^{\rm diag},\Lambda^2G_q]=0 }$$

for every $q$ as well.

In the physical $O\oplus H$ basis,

$$\Lambda^2J_z^{\rm diag} =\operatorname{diag}(1,0,-1,1,0,-1),$$

and the exact symbolic commutator still vanishes identically.

Therefore

$$\boxed{ \text{beam-axis }SO(2)\text{ symmetry does not select }q=1. }$$

This is the beam-symmetry result stated in Section 81.

82.4 A different symmetry would select $q=1$: observed--unobserved frame exchange

There is nevertheless an exact conditional selection rule. Exchange of the two spin-$1/2$ factors acts as

$$|+-\rangle\leftrightarrow|-+\rangle.$$

Let $S_{OH}$ denote this factor-swap operator. Direct calculation gives

$$[S_{OH},G_q]=0$$

if and only if

$$-q=-1,$$

hence

$$\boxed{q=1.}$$

Equivalently, in the Pauli-product decomposition,

$$\widetilde G_q =\frac14\sigma_z\otimes I +\frac{2q-1}{4}I\otimes\sigma_z +\frac{2q+1}{4}\sigma_z\otimes\sigma_z,$$

and exchange symmetry forces equality of the two one-frame coefficients:

$$\frac14=\frac{2q-1}{4} \quad\Longrightarrow\quad \boxed{q=1.}$$

At $q=1$ this becomes the corrected section-72 result

$$\widetilde G =\frac14\sigma_z\otimes I +\frac14I\otimes\sigma_z +\frac34\sigma_z\otimes\sigma_z.$$

82.5 Status of the exchange route

The mathematics is exact, but the physical premise must not be smuggled in. The labels $O$ and $H$ are not automatically interchangeable merely because they have equal dimension.

The scattering-frame functional

$$\|\widehat O-Q\widehat H\|_F^2$$

is reciprocal under exchange accompanied by

$$O\leftrightarrow H, \qquad Q\leftrightarrow Q^T,$$

which makes exchange symmetry a plausible property of the lock interaction. What has not yet been proved is that the stationary Hamiltonian circle acting on the electron spinor must inherit that reciprocity.

Thus the corrected status is

$$\boxed{ \text{beam }SO(2)\text{ leaves }q\text{ free,} \qquad O\leftrightarrow H\text{ reciprocity would force }q=1. }$$

82.6 Revised proof target

The next decisive test is whether the scattering-selected lock operator or the projected cross-depth block

$$P_3D_\infty P_6$$

is exchange-self-adjoint in precisely the sense required to induce

$$S_{OH}G S_{OH}^{-1}=G.$$

If that follows from Hermiticity/reciprocity of the electron--nucleus contact rather than being imposed as a new symmetry, then $q=1$ and hence the $1{:}1{:}3$ spectrum are genuinely selected. If not, the integer $q$ remains an undetermined orbit datum.

Related

• Research status — current proof boundaries and correction guidance.
• Beam-axis symmetry leaves the integral charge free — preceding section in the research sequence.
• Reciprocity audit: Hermiticity does not force observed-unobserved exchange symmetry — 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.