Cross-depth channel audit: the canonical factorized normal-spinor contraction does not lock the frames

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 97: Cross-depth channel audit: the canonical factorized normal-spinor contraction does not lock the frames.

Provenance

Preserved from interaction notebook section 97, 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.

97. Cross-depth channel audit: the canonical factorized normal-spinor contraction does not lock the frames

Sections 95--96 established that a second-order cross-depth process can in principle produce both the positive coarse lock and the full oriented singlet projector. The remaining danger is circularity: the explicit rank-one transition

$$B_Q=\beta|n\rangle\langle S_Q|$$

already contains the desired oriented singlet. The next test is therefore whether the canonical $3\subset6$ spinor factorization naturally supplies that channel. It does not.

97.1 The direct normal-polarization contraction is factorized

Relative to the nucleus-selected split,

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

the electron chiral carrier is

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

A depth-three output spinor carried by the observed factor is

$$N=(\mathbf2,\mathbf1).$$

Selecting a normalized unobserved/normal spinor $|\chi\rangle\in\mathbf2_H$ gives the most direct contraction

$$\boxed{ B_\chi=I_O\otimes\langle\chi|. }$$

Consequently

$$\boxed{ B_\chi^\dagger B_\chi =I_O\otimes|\chi\rangle\langle\chi|. }$$

This is a rank-two product projector. Its Pauli expansion contains only

$$(\mathbf1,\mathbf1)\oplus(\mathbf1,\mathbf3),$$

and its entire mixed correlation matrix vanishes:

$$\boxed{ \frac14\operatorname{tr} \left[B_\chi^\dagger B_\chi (\sigma_a\otimes\sigma_i) \right]=0 \qquad\forall a,i. }$$

So the canonical factorized restriction has no $(\mathbf3,\mathbf3)$ lock content at all for a fixed normal polarization.

97.2 Eliminating a complete unobserved doublet gives only a scalar shift

For an orthonormal unobserved basis $|\uparrow\rangle,|\downarrow\rangle$,

$$\boxed{ B_\uparrow^\dagger B_\uparrow +B_\downarrow^\dagger B_\downarrow =I_4. }$$

Therefore a complete or unpolarized normal-channel elimination produces

$$\Delta H_{\rm fact}\propto-I_4,$$

which cannot distinguish the diagonal-spin decomposition

$$\mathbf1\oplus\mathbf3.$$

This gives a clean no-go:

$$\boxed{ \text{the bare statement that the depth-six spinor carries a two-state normal polarization does not generate the frame lock.} }$$

A nonfactorizing correlation between the observed and unobserved spin factors is required.

97.3 Once a relative frame is supplied, two reduced doublet channels are allowed

If the nuclear/scattering background supplies a relative frame $Q$, the product symmetry can be compared through the corresponding diagonal subgroup

$$Spin(3)_{\rm diag}.$$

Then

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

while the depth-three carrier is a doublet

$$N=\mathbf2.$$

A transition involving a background/nuclear doublet can therefore pass through either

$$\mathbf2\otimes\mathbf1\to\mathbf2$$

or

$$\mathbf2\otimes\mathbf3\to\mathbf2.$$

Because

$$\mathbf2\otimes\mathbf1=\mathbf2,$$

and

$$\mathbf2\otimes\mathbf3=\mathbf2\oplus\mathbf4,$$

there are exactly two reduced doublet channels: one through the electron singlet and one through the electron triplet.

Therefore

$$\boxed{ \text{diagonal rotational covariance alone does not force singlet dominance.} }$$

The triplet channel is symmetry-allowed.

97.4 Exact Clebsch--Gordan normalization gives a quantitative test

Let $P_S$ and $P_T=I-P_S$ denote the electron singlet and triplet projectors. For a normalized singlet reduced map, summing over a complete depth-three doublet gives

$$\boxed{ \sum_m B_{S,m}^\dagger B_{S,m}=2P_S. }$$

For the normalized Clebsch--Gordan projection

$$\mathbf2\otimes\mathbf3\to\mathbf2,$$

direct exact matrix calculation gives

$$\boxed{ \sum_m B_{T,m}^\dagger B_{T,m}=\frac23P_T. }$$

Let the two reduced matrix elements be $a$ and $b$. If the eliminated channel has a common positive gap $\Delta>0$, the rotationally averaged second-order shift is

$$\boxed{ \Delta H_e =-\frac{2|a|^2}{\Delta}P_S -\frac{2|b|^2}{3\Delta}P_T. }$$

The singlet is therefore the lower branch exactly when

$$\boxed{ |a|^2>\frac{|b|^2}{3}. }$$

Equality gives no singlet--triplet splitting; reversing the inequality favors the triplet and kills this lock route.

This is now a sharp falsification criterion for the projected master operator.

97.5 The canonical factorized contraction lies exactly on the no-splitting boundary

For the direct product contractions,

$$\sum_\chi B_\chi^\dagger B_\chi=I_4=P_S+P_T.$$

Comparing with the reduced-channel normalization above gives

$$2|a|^2=1, \qquad \frac{2|b|^2}{3}=1,$$

hence

$$\boxed{|b|^2=3|a|^2.}$$

That is precisely the equality condition

$$|a|^2=\frac{|b|^2}{3}.$$

So the naive factorized $3\subset6$ restriction is not merely too weak to orient the lock. It sits exactly at zero singlet--triplet preference.

A successful interaction must depart from factorization in a measurable way.

97.6 Representation content of the bare transition confirms the same conclusion

Before choosing a relative frame,

$$\operatorname{Hom}(E,N) \cong (\mathbf2,\mathbf1)\otimes(\mathbf2,\mathbf2)^*.$$

Using pseudoreality of the $SU(2)$ doublet,

$$\boxed{ \operatorname{Hom}((\mathbf2,\mathbf2),(\mathbf2,\mathbf1)) =(\mathbf1,\mathbf2)\oplus(\mathbf3,\mathbf2). }$$

There is no bare $(\mathbf3,\mathbf3)$ transition representation. The full mixed lock only appears after a quadratic electron endomorphism is formed:

$$\operatorname{End}(\mathbf2,\mathbf2) =(\mathbf1,\mathbf1) \oplus(\mathbf3,\mathbf1) \oplus(\mathbf1,\mathbf3) \oplus(\mathbf3,\mathbf3).$$

This independently supports the second-order route of Sections 95--96 and rules out returning to the first-order shortcut corrected in Section 94.

97.7 Updated proof obligation

The remaining calculation is now unusually concrete. In the actual nuclear/scattering background one must extract the two reduced amplitudes

$$\boxed{a\quad\text{and}\quad b}$$

from

$$P_3D_\infty P_6$$

or the corresponding master interaction and test

$$\boxed{|a|^2>\frac{|b|^2}{3}.}$$

In addition, the nonfactorizing term must be covariantly tied to the physically supplied relative frame, for example $Q_{\rm scat}$, so that the favored singlet is

$$|S_{Q_{\rm scat}}\rangle.$$

If both conditions hold, then the positive-gap Schur mechanism supplies the oriented low-energy lock and Sections 89--92 give

$$\boxed{ q=1\Longrightarrow1:1:3. }$$

If the explicit cross-depth calculation yields the factorized boundary relation $|b|^2=3|a|^2$, or a triplet-dominant value, this dynamical derivation of the $1:1:3$ generator fails.

Related

• Research status — current proof boundaries and correction guidance.
• Leading-source orientation no-go: the written scalar and factorized $3\leftrightarrow6$ structures can select $s$ but cannot select the locked frame orientation — following section in the research sequence.