A Pure Spatial Spinor Pencil Links the Four-Plane, Normal Frame and Charge Carrier

In the four-plus-six spatial reference split, the base has four ordinary spatial dimensions and its perpendicular complement has six. Time is a separate coordinate in the infinite-dimensional physical spacetime. This four-plus-six split concerns ten spatial directions in the matter domain.

Two compatible polarizations of a spatial spinor give a common route to several previously separate geometric constructions. Their pure pencil selects a real four-plane. Applying the earlier universal alternating frame rule to its complement then selects six normal directions and a ten-dimensional parent. Fixing one polarization and the other's ray leaves the determinant-trivial block-unitary group whose exterior character gives the familiar one-family charge weights.

These are conditional geometric consequences. The pencil and universal normal-frame prescription are explicit premises; a positive potential below makes the first a possible energetic condition. Neither premise is an observed particle property, and the ambient spinor carrier does not assign particles to depths. The orthogonal normal six-plane must also remain distinct from the nested physical six-dimensional sector.

The ten-dimensional pure-pair stabilizer is established mathematics: Krasnov, Geometry of Spin(10) symmetry breaking, theorem A and sections VII A–C. The present calculation compares other parent dimensions, connects that construction to the existing normal-frame criterion, supplies an alternative positive polarization potential, and tests compatibility with the physical sector flag. It does not adopt an independent Higgs field, mediator particles, or the source paper's proposed gauge-field dynamics.

1. The Pencil Premise Selects a Four-Plane

Let the parent ordinary spatial dimension be $2m$, with Hermitian Clifford matrices $\gamma_i$. A positive-chiral spinor is represented in $\Sigma^+=\Lambda^{\mathrm{even}}\mathbb C^m$. A spinor is pure when its complex Clifford annihilator has dimension $m$.

Take two independent orthogonal spinors $u,v$ and impose the common condition

$$\alpha u+\beta v\quad\text{is pure for every }(\alpha,\beta)\ne(0,0).$$

This is a projective line lying in the pure-spinor variety, rather than an assumption that every parent spinor is pure. After a spatial Spin rotation and a harmless normalization, set $u=1$. Orthogonality removes the scalar part of $v$. The pure graph expansion near $u$ is

$$1+t v=\exp B(t) =1+B(t)+\frac12B(t)\wedge B(t)+\cdots.$$

Its degree-two part gives $B(t)=t v_2$. Comparing the higher grades at every $t$ forces $v_4=v_6=\cdots=0$ and $v_2\wedge v_2=0$. A nonzero two-form with this property is decomposable. A unitary change of the ordinary spatial complex frame therefore gives

$$u=1,\qquad v=e_1\wedge e_2.$$

For a normalized pure spinor define the real skew matrix

$$(J_\psi)_{ij}=-i\psi^\dagger\gamma_i\gamma_j\psi\quad(i\ne j), \qquad (J_\psi)_{ii}=0.$$

The sign is opposite to one earlier normal-alignment convention; all products and ranks below are unaffected by reversing both signs. Each $J_\psi$ is an orthogonal complex structure on the actual spatial tangent space. The two endpoint structures commute. They have opposite signs on the first two complex planes and equal signs on the remaining ones. Thus

$$\boxed{P_B=\frac{I+J_uJ_v}{2},\quad \operatorname{rank}_{\mathbb R}P_B=4, \qquad P_N=\frac{I-J_uJ_v}{2},\quad\operatorname{rank}_{\mathbb R}P_N=2m-4.}$$

The projectors transform covariantly when the spatial field pair rotates. This split is selected by the field polarization; it is not an inherent decomposition of every particle or a change in the kind of spatial dimension.

The condition is restrictive. The individual endpoint $e_1\wedge\cdots\wedge e_{2j}$ is pure, but for $j\ge2$ its generic sum with $1$ has annihilator dimension $m-2j$, not $m$. Merely choosing two pure spinors does not select four directions. At parent depth two the chiral rank is one and no independent pair exists. At every even parent depth at least four, a pure pencil does exist.

2. Combining the Four-Plane with the Normal Frame Rule

The universal normal-frame criterion asks for an alternating Spin-equivariant vector bilinear

$$\beta:\Lambda^2\Delta^+_{2k}\longrightarrow N_{\mathbb C}$$

that is nonzero for every independent pair of its normal-spinor columns. The alternating channel exists only for $k\equiv3\pmod4$. Universal nonvanishing also requires $2^{k-1}-1\le2k$. Together these conditions select $k=3$ over all positive even normal dimensions.

The pencil has already given a four-plane and complement $2k=2m-4$. Applying that same normal rule to the complement gives

$$\boxed{\dim B=4,\qquad \dim N=6,\qquad \dim\text{ parent}=10.}$$

Higher map-bearing complements fail the universal rank bound, so this conditional conclusion extends beyond the displayed scan. Selecting only some admissible normal-spinor pairs would weaken the premise and permit higher dimensions again.

This is an ambient/base/normal result. It does not demonstrate electric-charge preference at four particle depths, and it does not replace the all-sector color-coupling comparison. The odd sectors three and five still require their actual embeddings and tangent operators. Borrowing a neighboring even carrier as a shadow does not give an intrinsic four-plane inside three dimensions or identify a six-dimensional complement at physical depth five.

The ambient pair itself has the factorization $u=1_B\otimes1_N$, $v=e_{12,B}\otimes1_N$. Its normal spinors are proportional. Their alternating normal bilinear is therefore zero. The new pair selects the base and its complement, but does not also generate the separate normal framed line needed for the intermediate sector flag.

3. A Positive Potential Can Enforce the Polarization Premise

For an unnormalized chiral spinor $\psi$, let $n=\|\psi\|^2$ and define the raw matrix $M_{ij}=-i\psi^\dagger\gamma_i\gamma_j\psi$. The homogeneous quartic defect

$$\mathcal D_m(\psi)=2m\,n^2-\|M\|_F^2$$

is nonnegative and vanishes exactly on pure spinors and zero.

To see this, first normalize $\psi$. For any orthogonal unit real vectors $a,b$, the Hermitian operator $-i\gamma(a)\gamma(b)$ has eigenvalues $\pm1$. Hence $|a^TMb|\le1$, every skew singular value of $M$ is at most one, and $\|M\|_F^2\le2m$. Equality forces every singular value to be one. In a corresponding orthogonal plane frame, the spinor is a simultaneous extremal eigenvector of the commuting plane operators. Their common eigenspace is a pure-spinor line. Rescaling restores the displayed quartic formula.

For two mode polarizations of the same field, take

$$\begin{aligned} V(u,v)={}&\kappa\left[\mathcal D_m(u)+\mathcal D_m(v) +\frac14\sum_{\zeta\in\{1,i,-1,-i\}}\mathcal D_m(u+\zeta v)\right] +\eta|u^\dagger v|^2\\ &+\lambda_u(\|u\|^2-a^2)^2+\lambda_v(\|v\|^2-b^2)^2, \end{aligned}$$

with positive coefficients and unspecified nonzero amplitude scales $a,b$. Every term is positive and Spin invariant. The four phase samples are the exact phase average: the quartic defect has no phase Fourier order beyond two. Separate phase changes of the two polarizations preserve that average.

The zero set consists of orthogonal fixed-norm pure pencils. For completeness, once the endpoints are pure, nonnegativity makes their first derivatives of $\mathcal D_m$ vanish. The restricted quartic can then contain only $|\alpha\beta|^2 A+2\operatorname{Re}(\alpha^2\bar\beta^2 B)$. A zero phase average gives $A=0$; positivity at every phase then forces $B=0$. Thus every linear combination is pure, rather than only the four sampled combinations.

This is an inverse construction of a possible mode-polarization energy. It does not add two fundamental fields: the two columns would have to be extracted as independent spatial modes or correlations of the one field. Deriving that extraction and its complete kinetic terms remains necessary. Nor does the potential select parent ten by itself; it has zero minima at every even parent at least four.

At the unit-norm control $a=b=1$ and shared $\kappa=\eta=\lambda_u=\lambda_v=1$, the exact real Hessian has the following spectrum. These are polarization-energy curvatures under a stated coordinate normalization, not particle masses or frequencies.

For $m\ge3$, the zero count is $m^2+3m-6$, exactly the dimension of the fixed-pair Spin orbit. At $m=2$ an additional common phase gives the fourth zero. There are no extra negative or accidental flat directions in these checked preparations. Positive canonical mode kinetics would make the nonzero curvatures restoring directions; the actual field kinetics and dimensional amplitude factors must be supplied before calculating physical response times.

4. The Stabilizer and Its Charge Character

Write $k=m-2$. Fixing $u$ exactly gives $SU(m)$. Fixing the ray of the decomposable two-form $v$ preserves $\mathbb C^2\oplus\mathbb C^k$. Their joint group is

$$S(U(2)\times U(k)),\qquad \dim=k^2+3\quad(k\ge1).$$

The exact real Clifford nullspace calculation gives dimensions $4,7,12,19$ at parents $6,8,10,12$. At parent ten the Lie algebra is $\mathfrak{su}(2)\oplus\mathfrak{su}(3)\oplus\mathfrak u(1)$. Fixing both spinors exactly removes the relative circle and reduces these dimensions by one. Fixing both rays adds a circle and increases them by one. Keeping a phase-sensitive datum for one spinor is therefore material to the result; ray geometry alone gives a larger group.

Normalize the determinant character on the complex two-plane to one. Its trace-free diagonal generator has weights $1/2$ on each of the two directions and $-1/k$ on each normal direction. For normal exterior degree $p$ and two-plane degree $q$,

$$Y=-\frac pk+\frac q2,\qquad p+q\text{ even}.$$

Only $q=1$ is an $SU(2)$ doublet. With $T_3=\pm1/2$ there and zero on the two singlets, $Q=Y+T_3$. At $k=3$ the complete positive-chiral rank-sixteen carrier gives

This is standard one-family representation bookkeeping, including conjugate representations. It does not solve IDWT's physical distinction between particles and antiparticles. The earlier eight-component normal charge carrier and this sixteen-component ambient chiral carrier are different bundles; their matching charge values do not identify their field dynamics or occupations.

At arbitrary $k$, the fundamental doublet instead has

$$Q_u=1-\frac1k,\qquad Q_d=-\frac1k.$$

The minimal family with a colorless lepton doublet and fixed lepton charge has the usual trace-compatible values $(k+1)/(2k)$ and $(1-k)/(2k)$. These agree with the geometric character only at $k=3$. More explicitly, for odd $k\ge3$ select just the normal grades $0,1,k-1,k$ with their displayed two-plane partners. This gives $4(k+1)$ components. Its geometric traces are

$$\boxed{\operatorname{Tr}_{\rm minimal}Y^3=-\frac34(k-3),\qquad \operatorname{Tr}_{\rm weak\ doublets}(T_3^2Y)=\frac14(k-3).}$$

At three, $\Lambda^2\mathbb C^3$ is already the conjugate fundamental and the six blocks exhaust the full carrier. At higher odd rank, additional exterior representations are present. For example $k=5$ has 64 full components versus 24 minimal components, leaving 40. The full carrier's displayed cubic and weak mixed traces vanish for every $k\ge2$; those extra representations cancel the minimal subset's traces. A colorless weak doublet is absent from this positive-chiral carrier when $k$ is even. These controls prevent an assertion that anomaly cancellation alone excludes every higher parent.

The traces are exact representation diagnostics. Interpreting them as anomaly constraints requires additional chiral quantum gauge dynamics, which has not been derived for the classical spatial field. Likewise the normalization of $Y$ and the use of $Q=Y+T_3$ are explicit character choices, not an observed electric-force derivation.

5. The Orthogonal Split and the Nested Sectors Must Coexist

In a flat existence control, choose one ordinary spatial flag

$$Q_2\subset Q_3\subset Q_4\subset Q_5\subset Q_6\subset Q_{10}, \qquad Q_d=\operatorname{span}_{\mathbb R}(e_1,\ldots,e_d).$$

Let the pencil's four-plane be $Q_4$. Its normal complement is then $\operatorname{span}(e_5,\ldots,e_{10})$, whereas the physical $Q_6$ is $\operatorname{span}(e_1,\ldots,e_6)$. They are different six-planes. Specifying a normal complex line $\operatorname{span}(e_5,e_6)$ gives $Q_6=Q_4\oplus\mathbb C e_5$, and its real axis gives $Q_5$. The pencil does not supply this extra line or the lower two- and three-sector flag.

There is a further force-interpretation constraint. If the normal $SU(3)$ acts as literal ambient rotations and must preserve that physical $Q_6$, it reduces to the line stabilizer $U(2)$, of dimension four. Preserving the real-line $Q_5$ reduces it to $SU(2)$, of dimension three. The exact ordinary rotation generators verify both reductions. Thus the full observed color symmetry cannot simply be called the group of ambient rotations preserving every fixed sector. Independent equivalent-mode transport, a moving compatible embedding, or another common field mechanism would need explicit construction.

Global topology also remains a separate problem. The round-four-sphere chiral spin bundle in the earlier curved existence control has nonzero second Chern number. It cannot contain two everywhere independent global sections as a trivial rank-two frame. A local pencil or a suitably twisted pencil subbundle is a weaker datum, and must be checked with the chosen global geometry. The existing projective tangent splitting obstruction is not removed by a pointwise spinor calculation.

Verification

The single checker and report construct actual spatial Clifford annihilators and bilinear rotation structures through parent depth twelve; solve exact real stabilizer equations; rotate the complete pair covariantly; compare non-pencils; compute the positive potential's exact Hessians; compare complete exterior charge characters through complex normal rank ten; and enforce one nested flag. The calculation passes 3,900 checks across 22 cases. Executed code hashes are preserved. No particle simulation, physical amplitude fit or new particle assignment is used.

Independent Observer Spin Is a Further Test

The full normal-color calculation verifies that the six-normal triplet commutes with observer spin and every three/four-dimensional tangent Clifford vector. It also finds that the standard ambient exterior charge generator fails to commute with two observer-spin generators. In an ordinary spatial interpretation, the familiar charge table therefore cannot directly be identified with measured electric charge. A separate normal charge operator with an auxiliary Clifford-volume branch is spin independent and reproduces the earlier weights locally, but its extension beyond the three-sector and its total orbital/domain action remain to derive. The known pure-pair stabilizer theorem and the conditional four-dimensional spatial base/six-normal dimension criterion are unaffected.