Projector-curvature candidate: nonzero geometry beyond the pure-gauge frame map

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 61: Projector-curvature candidate: nonzero geometry beyond the pure-gauge frame map.

Provenance

Preserved from interaction notebook section 61, 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.

61. Projector-curvature candidate: nonzero geometry beyond the pure-gauge frame map

Section 60 showed that differentiating the scattering-selected frame itself,

$$A=dQ_{\rm scat}Q_{\rm scat}^{-1},$$

produces a Maurer--Cartan connection with zero curvature. A different V2-native object is available: the sector projector.

Let

$$P^2=P=P^T$$

be a smooth rank-three projector selecting the observed three-plane inside a six-dimensional space. Differentiating $P^2=P$ gives

$$\boxed{P(dP)P=0.}$$

Thus $dP$ is off-diagonal between the selected and complementary subspaces. Nevertheless, two different projector derivatives need not commute. The natural curvature of the induced projector connection is

$$\boxed{F_P=P(dP\wedge dP)P.}$$

Unlike $dQ\,Q^{-1}$, this can be nonzero.

61.1 Exact base-point test

Take

$$P_0=\begin{pmatrix}I_3&0\\0&0\end{pmatrix}$$

and two pure observed--unobserved generators $X,Y$. At the base point,

$$P_x=[X,P_0],\qquad P_y=[Y,P_0].$$

Choosing $X$ to mix observed direction $1$ with unobserved direction $1$ and $Y$ to mix observed direction $2$ with the same unobserved direction gives

$$P_0P_xP_0=P_0P_yP_0=0,$$

while

$$F_{xy}=P_0[P_x,P_y]P_0$$

has observed block, up to orientation convention,

$$F_{xy}^{(O)}=\pm \begin{pmatrix} 0&-1&0\\ 1&0&0\\ 0&0&0 \end{pmatrix}.$$

The sign depends only on the ordering convention for the two tangent directions. The invariant result is

$$\boxed{-\frac12\operatorname{tr}[(F_{xy}^{(O)})^2]=1.}$$

Thus projector geometry supplies genuine nonzero curvature with fixed dimensionless algebraic normalization for unit tangent motions.

61.2 Full unit-tangent basis

The tangent space of the rank-three Grassmannian at $P_0$ has nine elementary mixed directions labelled by

$$(a,i),\qquad a=1,2,3,\quad i=1,2,3,$$

where $a$ labels the observed direction and $i$ the unobserved direction. With the normalization

$$\frac12\operatorname{tr}(dP_{ai}^2)=1,$$

the observed curvature between two basis tangents is nonzero exactly when they share the same unobserved index and have different observed indices:

$$\boxed{i=j,\ a\ne b.}$$

Every such nonzero curvature has the same invariant norm

$$\boxed{-\frac12\operatorname{tr}(F_{ai,bi}^2)=1.}$$

So the unit projector geometry contains a universal curvature quantum at the algebraic level.

61.3 Unobserved-channel multiplicity

For a fixed observed rotation plane $(ab)$ there are three unobserved channels $i=1,2,3$. If their curvature matrices enter with equal phase and are added before taking the norm, their amplitudes add to three and

$$\boxed{\|F_{ab}^{\rm coh}\|^2=9.}$$

If instead the three unobserved channels are orthogonal/incoherent in the physical quadratic form, their norms add to

$$\boxed{\sum_i\|F_{ab,i}\|^2=3.}$$

This is an important new fork. The projector algebra fixes the individual unit curvature to one and fixes the multiplicity to three, but does not by itself decide whether the physical interaction contracts those channels coherently or orthogonally.

61.4 Why this evades the Maurer--Cartan no-go

The full moving frame may have a flat Maurer--Cartan connection while the induced connection on a moving subbundle has nonzero curvature. The off-diagonal pieces of $dP$ encode how the selected subspace bends inside the ambient space, and their commutator returns a tangential rotation:

$$\boxed{\text{off-diagonal }dP\times\text{off-diagonal }dP\longrightarrow\text{tangential curvature}.}$$

Therefore the pure-gauge result of section 60 does not rule out curvature generated by sector projector geometry.

61.5 Remaining normalization issue

If projector derivatives are rescaled,

$$P_x\mapsto aP_x,\qquad P_y\mapsto bP_y,$$

then

$$F_{xy}\mapsto abF_{xy},$$

and

$$\boxed{\|F_{xy}\|^2\mapsto a^2b^2\|F_{xy}\|^2.}$$

Thus the Grassmann algebra fixes the curvature generated by unit normalized projector motions, but not the physical magnitude of those motions.

61.6 Strongest current closure candidate

The most V2-native chain is now

$$\boxed{ \text{scattering data} \to Q_{\rm scat} \to P_{\rm scat} \to dP_{\rm scat} \to P(dP\wedge dP)P \to \text{lock curvature}. }$$

This candidate survives the present tests. The unresolved master equation must fix both

$$\boxed{\|\nabla P\|}$$

and the physical contraction law among the three unobserved curvature channels. If V2 derives those two pieces, the remaining algebraic normalization is no longer arbitrary.

Related

• Research status — current proof boundaries and correction guidance.
• Maurer--Cartan test: the scattering frame alone cannot generate its own restoring curvature — preceding section in the research sequence.
• Canonical projector geometry: the rank-three Grassmannian has isotropic Ricci eigenvalue four — following section in the research sequence.