Beam-axis spinor decomposition: the orbit generator requires a bilinear lock term

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 72: Beam-axis spinor decomposition: the orbit generator requires a bilinear lock term.

Provenance

Preserved from interaction notebook section 72, 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.

72. Beam-axis spinor decomposition: the orbit generator requires a bilinear lock term

Section 71 showed that a beam axis explains the twofold transverse degeneracy but not the longitudinal factor three. The spinor branching gives a sharper test.

Under the nucleus/scattering-induced split,

$$Spin(6)\supset Spin(3)_O\times Spin(3)_H,$$

the electron homogeneous-coordinate representation is

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

If the scattering state selects one axis in each three-plane, each spinor doublet acquires an axial Pauli generator. In the product basis

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

the natural diagonal operator basis is

$$I, \qquad \sigma_z\otimes I, \qquad I\otimes\sigma_z, \qquad \sigma_z\otimes\sigma_z.$$

The desired traceless orbit generator is

$$\widetilde G =\frac14\operatorname{diag}(5,-3,-3,1).$$

72.1 Correct exact decomposition

Direct exact decomposition gives

$$\boxed{ \widetilde G =\frac14\,\sigma_z\otimes I +\frac14\,I\otimes\sigma_z +\frac34\,\sigma_z\otimes\sigma_z }$$

for the displayed product-state ordering.

This corrects the first write of this section, which incorrectly stated coefficients $1/2,1/2,1/4$. The symbolic calculation itself returned the correct coefficients $1/4,1/4,3/4$; the prose inserted afterward was wrong and must not be used.

The decomposition can be checked state by state:

$$\begin{array}{c|c} |++\rangle & \frac14+\frac14+\frac34=\frac54,\\ |+-\rangle & \frac14-\frac14-\frac34=-\frac34,\\ |-+\rangle & -\frac14+\frac14-\frac34=-\frac34,\\ |--\rangle & -\frac14-\frac14+\frac34=\frac14. \end{array}$$

72.2 Axial one-body generators are insufficient

Trying to represent $\widetilde G$ using only

$$I, \qquad \sigma_z\otimes I, \qquad I\otimes\sigma_z$$

has no solution.

Therefore the beam-selected axes in the observed and unobserved spin frames do not by themselves generate the required orbit spectrum. A genuine two-frame bilinear is necessary:

$$\boxed{\sigma_z\otimes\sigma_z.}$$

The missing ingredient is therefore an interaction between the two axial spin-frame labels, not merely another preferred direction.

72.3 Observed--unobserved exchange symmetry fixes the bilinear magnitude

The ordering of the four product states is conventional, so all permutations of the four target eigenvalues were tested. If no physical relation between the two spin factors is imposed, the Pauli-basis coefficients depend on that relabelling.

However, the locked construction treats the observed and unobserved $Spin(3)$ factors symmetrically before the external scattering axis distinguishes a common direction. Requiring the one-body axial coefficients to respect exchange symmetry,

$$b=c$$

in

$$\widetilde G =b\,\sigma_z\otimes I +c\,I\otimes\sigma_z +d\,\sigma_z\otimes\sigma_z,$$

leaves only exchange-symmetric relabellings whose bilinear coefficient has invariant magnitude

$$\boxed{|d|=\frac34.}$$

Thus, within the observed--unobserved exchange-symmetric identification, the required two-frame interaction strength relative to the chosen primitive generator normalization is not $1/4$ but $3/4$.

72.4 Relation to the adopted depth-six coupling

The adopted V2 value is

$$g_{66}=\frac14.$$

The corrected exact decomposition therefore gives the numerical relation

$$\boxed{|d|=\frac34=3g_{66}.}$$

This is a different and more constrained observation than the erroneous earlier claim $d=g_{66}$.

It is potentially interesting because the factor three is independently present in the three-dimensional frame geometry and in the $1{:}1{:}3$ vector-representation spectrum. But at present

$$\boxed{d=3g_{66}}$$

is only a candidate structural identification. No current V2 theorem says that the two-frame axial bilinear must carry three copies of the depth-six scalar coupling.

72.5 Why the factor three appears in the vector rates

With the correct spinor generator

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

the fundamental weights are exactly

$$\left(\frac54,-\frac34,-\frac34,\frac14\right).$$

Passing to the real vector representation

$$\Lambda^2\mathbf4$$

adds these weights pairwise and gives positive rotation-plane rates

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

up to signs and ordering. Hence the $1{:}1{:}3$ spectrum is reproduced exactly.

72.6 Revised binary proof target

The strongest remaining operator-level test is now:

Does the depth-six projected interaction, in the beam-axis-adapted $(\mathbf2,\mathbf2)$ basis and with observed--unobserved exchange symmetry, contain a traceless block proportional to

$$\boxed{ \frac14\sigma_z\otimes I +\frac14I\otimes\sigma_z +\frac34\sigma_z\otimes\sigma_z? }$$

Equivalently, if the one-body axial normalization is tied to $g_{66}=1/4$, does the same projected geometry force the bilinear coefficient to be

$$\boxed{3g_{66}?}$$

If yes, the orbit generator follows with no additional dimensionless parameter. If the projected master interaction gives another coefficient or no exchange-symmetric bilinear, this closure fails.

Related

• Research status — current proof boundaries and correction guidance.
• Axial mixed tensors do not determine the orbit-generator spectrum — preceding section in the research sequence.
• Three-axis contraction test: summing three depth-six channels does not yield the required axial bilinear — 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.