Quadratic cross-depth route: a first-order $3\leftrightarrow6$ transition can generate the full $(\mathbf3,\mathbf3)$ lock after Schur elimination

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Summary: Research record, section 95: Quadratic cross-depth route: a first-order $3\leftrightarrow6$ transition can generate the full $(\mathbf3,\mathbf3)$ lock after Schur elimination.

Provenance

Preserved from interaction notebook section 95, 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.

95. Quadratic cross-depth route: a first-order $3\leftrightarrow6$ transition can generate the full $(\mathbf3,\mathbf3)$ lock after Schur elimination

Section 94 ruled out an overly direct identification of the ordinary Levi--Civita mixed connection with the desired full-rank frame-lock operator. That does not rule out the actual cross-depth block

$$D_{36}=P_3D_\infty P_6.$$

The important distinction is that a transition operator from the electron sector into a depth-three channel need not itself transform as $(\mathbf3,\mathbf3)$. The effective operator obtained after composing the transition with its adjoint can have richer representation content.

95.1 Representation content of the first-order transition

Relative to the nucleus-selected split,

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

the electron chiral four is

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

A depth-three spinor channel carried by the observed factor has representation

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

Therefore the vector space of first-order transition operators is

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

Since the $SU(2)$ doublet is pseudoreal,

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

So a bare $3\leftrightarrow6$ transition does not contain a $(\mathbf3,\mathbf3)$ representation. This agrees with the warning from Section 94 that one must not simply relabel a first-order mixed coefficient matrix as the final lock Hamiltonian.

95.2 The electron effective operator does contain $(\mathbf3,\mathbf3)$

The electron endomorphism algebra is

$$\operatorname{End}(E) \cong (\mathbf2\otimes\mathbf2,\mathbf2\otimes\mathbf2),$$

hence

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

Thus a quadratic combination such as

$$D_{63}D_{36}$$

can contain exactly the full mixed tensor sector needed by the frame lock even though $D_{36}$ itself does not.

This changes the correct target from

$$D_{36}\stackrel{?}{\sim}T_Q$$

to

$$\boxed{ D_{63}\,\mathcal R_3\,D_{36} \stackrel{?}{\sim}P_Q\text{ or }T_Q, }$$

where $\mathcal R_3$ is the appropriate depth-three resolvent/propagator in the eliminated channel.

95.3 Explicit rank-one transition already produces the exact oriented lock projector

Let

$$|S_Q\rangle$$

be the maximally entangled electron ray associated with an oriented frame map $Q$, and let

$$|n\rangle$$

be one selected normalized depth-three channel. Consider the rank-one transition

$$\boxed{ B_Q=\beta\,|n\rangle\langle S_Q|. }$$

Then

$$\boxed{ B_Q^\dagger B_Q =|\beta|^2P_Q, \qquad P_Q=|S_Q\rangle\langle S_Q|. }$$

Although $B_Q$ has rank one, the projector $P_Q$ has a full-rank $(3,3)$ correlation tensor.

At the aligned lock,

$$P_S=|S\rangle\langle S|$$

has the Pauli expansion

$$\boxed{ P_S =\frac14 \left( I-\sum_{a=1}^3\sigma_a\otimes\sigma_a \right). }$$

Its $(3,3)$ coefficient matrix is

$$\boxed{-\frac14I_3,}$$

whose determinant is

$$\boxed{-\frac1{64}\neq0.}$$

So full mixed rank does not require a full-rank transition matrix. A single coherent cross-depth channel can generate the complete oriented spin-frame correlation after composition with its adjoint.

For general orientation,

$$\boxed{ P_Q =\frac14(I-T_Q), \qquad T_Q=\sum_{a,i}Q_{ai}\sigma_a\otimes\sigma_i. }$$

This is exactly the Section-93 lock operator, now appearing in the quadratic effective electron operator rather than being incorrectly identified with the first-order Levi--Civita connection.

95.4 Schur/Feshbach elimination fixes the favorable sign conditionally on a positive gap

Consider a block Hamiltonian/operator of the schematic form

$$\mathcal H = \begin{pmatrix} H_e & B_Q^\dagger\\ B_Q & H_N \end{pmatrix}.$$

If the relevant eliminated depth-three channel has positive gap

$$H_N=\Delta I, \qquad \Delta>0,$$

then the static Schur complement on the electron sector is

$$H_{\rm eff} =H_e-B_Q^\dagger H_N^{-1}B_Q.$$

Therefore

$$\boxed{ \Delta H_e =-\frac{|\beta|^2}{\Delta}P_Q. }$$

Using

$$P_Q=\frac14(I-T_Q),$$

this becomes

$$\boxed{ \Delta H_e =\frac{|\beta|^2}{4\Delta}T_Q -\frac{|\beta|^2}{4\Delta}I. }$$

The last term is a common energy shift. The physically relevant mixed coefficient is therefore

$$\boxed{ \kappa_{\rm eff}=\frac{|\beta|^2}{4\Delta}>0. }$$

Consequently the singlet-oriented ray $|S_Q\rangle$ is lowered relative to the triplet by

$$\boxed{ \Delta E=\frac{|\beta|^2}{\Delta}. }$$

This is important: if the lock is generated by virtual mixing with a positive-gap depth-three channel, the sign required in Section 93 is not arbitrary. It is the ordinary negative-semidefinite second-order level shift from eliminating that channel.

95.5 Relation to the scattering-selected orientation

The remaining orientation problem may now be phrased at the level of the transition amplitude rather than the Levi--Civita connection.

One needs the cross-depth matrix element to select

$$\langle S_{Q_{\rm scat}}|$$

or, more generally, a coherent family whose dominant singular vector is the electron lock ray associated with the scattering/nuclear frame.

Then

$$D_{36} \longrightarrow B_{Q_{\rm scat}}^\dagger B_{Q_{\rm scat}} \longrightarrow P_{Q_{\rm scat}} \longrightarrow T_{Q_{\rm scat}},$$

and the positive-gap Schur sign selects that ray as the low-energy electron configuration.

Sections 90--92 then apply directly:

$$P_{Q_{\rm scat}} \text{ stationary under }G_q \Longrightarrow q=1 \Longrightarrow 1:1:3.$$

95.6 What is now proved and what remains

The following statements are exact:

1. the bare transition space is $$(\mathbf1,\mathbf2)\oplus(\mathbf3,\mathbf2),$$ not $(\mathbf3,\mathbf3)$;
2. quadratic electron endomorphisms contain a $(\mathbf3,\mathbf3)$ sector;
3. a rank-one transition $B_Q=\beta|n\rangle\langle S_Q|$ gives exactly $$B_Q^\dagger B_Q=|\beta|^2P_Q;$$
4. $P_Q$ has the full mixed correlation tensor required for frame lock;
5. eliminating a positive-gap channel produces the sign that lowers the selected singlet ray.

The unresolved physical step is now highly specific:

$$\boxed{ \text{does }P_3D_\infty P_6\text{ actually have a dominant singular channel whose electron right-singular vector is }|S_{Q_{\rm scat}}\rangle? }$$

Equivalently, one must derive the singular-vector structure of the cross-depth transition operator from the master interaction and the nuclear/scattering background.

This is a substantially narrower problem than deriving an arbitrary full-rank mixed connection. The full $(3,3)$ lock can emerge automatically at second order from a much smaller first-order transition space.

Related

• Research status — current proof boundaries and correction guidance.
• Correction to Section 93: the Levi--Civita mixed connection contracts to mean curvature and does not by itself supply the nine-component lock operator — preceding section in the research sequence.
• Cross-depth channel audit: the canonical factorized normal-spinor contraction does not lock the frames — later audit limits or corrects this route.