Shared Normal Geometry for Charge and Sector Frames

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.

The Common Geometric Question

The same ordinary spatial normal bundle can constrain charge representations, polarization, intermediate sector embeddings and possible contact tensors. The construction below makes that reuse explicit. It establishes geometric identities and conditional dimension criteria; it does not identify occupied particle states or a physical electric interaction.

Throughout, a normal vector is a direction in the containing spatial manifold. Spinor components and coefficient-ray spaces describe fields at a point. They are not additional physical sectors. The spatial-depth module specifies the tangent Clifford compatibility needed for each sector operator.

1. A Dimension Criterion From A Quadratic Spatial Frame

Let $N$ be an oriented Euclidean normal space of dimension $2m$, and let $S=\Delta^+$ be one complex chiral spin representation. Consider a Spin-equivariant alternating bilinear map

$$\beta:\Lambda^2S\longrightarrow N_{\mathbb C}.$$

For two commuting spinor columns, contraction with an auxiliary doublet's invariant determinant tensor makes this a quadratic spatial vector. Requiring this particular channel is a geometric premise; a different bilinear channel is a different comparison.

The all-dimension classification is

$$\dim\operatorname{Hom}_{\operatorname{Spin}(2m)} (\Lambda^2\Delta^+,N_{\mathbb C}) =\begin{cases}1,&m\equiv3\pmod4,\\0,&\text{otherwise}.\end{cases}$$

Here is a direct proof. In the exterior Clifford representation set $C=\gamma_2\gamma_4\cdots\gamma_{2m}$. Then

$$\gamma_a^T C=(-1)^m C\gamma_a,\qquad C^T=(-1)^{m(m+1)/2}C,$$

so $B(s,t)=s^TCt$ is Spin invariant and $\beta_a(s,t)=s^TC\gamma_at$ is a vector bilinear. Its same-chirality block is nonzero precisely for odd $m$, and

$$(C\gamma_a)^T=(-1)^{m(m+3)/2}C\gamma_a.$$

It is alternating for $m=3\bmod4$ and symmetric for $m=1\bmod4$. There is no second map with a different exchange sign: evaluation in one unit vector direction injects the space of vector bilinears into invariant bilinear forms of the stabilizer Spin($2m-1$). The restricted spin module is irreducible, since its even Clifford algebra is the full matrix algebra. Schur's lemma bounds that space by one. For even $m$, the central volume element already forbids a vector bilinear between equal chiralities.

Every individual chiral spinor is pure exactly for $m\le3$. For these dimensions an even spinor with scalar part is $a\exp(B/a)=a+B$, with a maximal isotropic annihilator; the remaining rays follow by continuity. For every $m\ge4$, $1+e_{1234}$ instead has annihilator dimension $m-4$. Therefore automatic chiral purity together with the alternating vector channel selects six normal directions, over all positive even dimensions. It selects a gap, not its two sector endpoints.

There is a stronger criterion that does not assume automatic purity. Require the same map to give a nonzero vector for every independent pair of columns. Fixing one column $s\ne0$ then gives

$$\ker\beta(s,\cdot)=\mathbb C s,\qquad 2^{m-1}-1\le2m.$$

Every higher map-bearing case has $m\ge7$ and violates this rank bound. At $m=3$, the invertible wedge map gives a nonzero null vector for every independent pair. Thus this universal frame prescription also selects six, with nullness following there. The universality premise matters: at higher map-bearing dimensions, the independent pure columns $1,e_{12}$ give zero output, whereas pure pairs whose annihilators intersect in one complex line give a nonzero null vector. Selecting only those pairs permits the higher dimensions again. Generic unrestricted pairs there need not even give null output; explicit pairs give metric square four in the fixed Clifford normalization.

Spin-trivial multiplicities cannot create a missing alternating normal map: their contribution factors as

$$\operatorname{Hom}_{\mathrm{Spin}}(\Lambda^2S,N_{\mathbb C}) \otimes(\Lambda^2M)^*.$$

Mixed chiralities, symmetric auxiliary tensors, selected pure rays and nonlinear purification remain alternatives with different premises. The quadratic map here acts on commuting geometric amplitudes. An expression in anticommuting field variables has a different exchange symmetry and cannot be substituted for that geometric state without specifying its expectation or other ordinary-valued reconstruction.

2. The Same Six-Normal Carrier Gives The Charge Table

Write the Hermitian normal bundle as $W$, with complex rank three, and $D=\det W$. The root-free carrier is

$$F=\mathbf1\oplus\Lambda^2W,\qquad E=F\otimes(\mathbf1\oplus D^{-1}).$$

The central rotation $W\mapsto e^{i\theta}W$ gives weights and multiplicities

$$(0,2,-3,-1),\qquad(1,3,1,3).$$

Normalizing the determinant weight three to one gives $(0,2/3,-1,-1/3)$. Both triplets have the same conjugate-fundamental SU(3) type. The determinant of $E$ is trivial, and its charge trace is zero. These are consequences of the carrier and normalization, not separate adjustments for each particle.

Locally let $L^2=D$. Then

$$E=(F\otimes L^{-1})\otimes(L\oplus L^{-1}),\qquad Q=Y\otimes1+1\otimes T_3,$$

$$Y=\operatorname{diag}(-1/2,1/6,1/6,1/6),\qquad T_3=\operatorname{diag}(1/2,-1/2).$$

The two sign ambiguities cancel, so the combined Spin-h carrier exists without a separately chosen square root when its associated bundle data match. The auxiliary doublet is not observer spatial spin. No physical hypercharge or weak interaction is identified by this notation.

Complex-linear full normal Clifford completion is $\widehat E=\Lambda^*W\otimes(1\oplus D^{-1})$, of complex rank sixteen. A connection difference commuting with all six normal Clifford vectors acts as identity on their irreducible factor. Both branch transitions then use the same determinant line $D$ and have the same reduced coefficient; their summed singlet/triplet squared norms have ratio $1:3$. A nonzero coefficient and a physical rate are still additional questions.

3. Polarization Produces The Intermediate Frame

At six, $\operatorname{Spin}(6)=SU(4)$ and $\Lambda^2\mathbb C^4=N_{\mathbb C}$. For a local four-by-two Spin-h matrix $\Phi=(\psi_1,\psi_2)$,

$$q=\psi_1\wedge\psi_2.$$

At rank two this is nonzero and null. In the invariant positive real form, $q=u+iv$ implies $u\cdot v=0$ and $\lVert u\rVert=\lVert v\rVert$. Thus an actual state gives a framed real plane. A projective state retains its oriented plane but loses a chosen phase-sensitive real line.

If the singular values satisfy $a>b>0$, the largest eigenline of $\Phi\Phi^\dagger$ gives a normal complex structure $J$ for which that plane is complex. Its framed-line stabilizer is $U(2)$. At $a=b$, the plane remains defined but a $\mathbb{CP}^1$ of choices of $J$ remains. At rank loss the frame disappears. These are distinct degeneracies.

The reconstructed $J$ need not equal the $J_0$ used to define $Q$. Writing $\Phi=[r;T]$ in its original vacuum/triplet decomposition, the old vacuum is the principal ray precisely when

$$Tr^\dagger=0,\qquad \lVert r\rVert^2>\operatorname{tr}(TT^\dagger).$$

The ordered auxiliary axis must also align to preserve the same charge endomorphism. Globally, changing $W$ while retaining the old Spin-h auxiliary bundle requires $c_1(W)^2=c_1(W_0)^2$ and $2(c_2(W)-c_2(W_0))=0$. This does not require $c_2=0$ or a trivial determinant. A splitting $W=1\oplus V$ gives explicit aligned states using its vacuum and $\Lambda^2V\otimes D^{-1}$ singlets.

4. Charge Forces A Specific Frame-Loss Locus

Every fixed fibre-charge eigenspace of $E$ occupies only one auxiliary column. It therefore has $q=0$. A charge-invariant weighted density can retain the principal ray and oriented plane without the real frame. This compares an additional eigenstate requirement; it imposes no physical superselection rule.

For the spatially lifted circle, let $G$ be its fibre generator and $K=G-(Jy)\cdot\partial_y$. A smooth state of integer total weight $M$ obeys

$$\Phi(R_\theta y)=e^{-iM\theta}U_\theta\Phi(y),\qquad q(R_\theta y)=e^{-2iM\theta}R_\theta q(y).$$

At $y=0$, nonzero $q$ would require $2M=\pm1$, impossible for integer $M$ in this periodic lift. More sharply,

$$\operatorname{ord}_{N}q\ge |M+1|+\min(|M|,|M-2|).$$

The three opposite-column products have minimum degrees $|M|+|M+1|$, $|M-2|+|M+3|$, and $|M-2|+|M+1|$; their minimum is the displayed bound. Single-coordinate monomials attain it. At $M=1$ the order is three, although vector covariance alone would allow one.

There is a global total-neutral control with first-column vacuum $\sqrt{1+b^2|y|^2}$ and second-column triplet $b\bar y\in W^*$. Its squared singular-value gap is one, but its second singular value vanishes on the base. Away from the base it yields the radial frame $y,Jy$. A spectral gap alone therefore does not give a global nonzero quadratic frame, and this example cannot supply a base sector flag through normalization at $y=0$.

5. One Complete Nested Spatial Geometry

For a metric four-dimensional spatial base $B$, a Hermitian $W=\mathbb C u\oplus V$ and unitary connection $A$, the ordinary connection metric is

$$g=\pi^*g_B+|dy+A y|^2.$$

Take $B=S^4_R$ and let $V$ be a chiral rank-two spin bundle with its chiral Levi-Civita connection. Its second Chern number is $+1$ or $-1$ according to chirality and orientation. The flag

$$S^2\subset S^3\subset S^4\subset \operatorname{Tot}(\mathbb R u)\subset \operatorname{Tot}(\mathbb C u)\subset \operatorname{Tot}(W_{\mathbb R})$$

has depths $2,3,4,5,6,10$. The first inclusions are great equators. All sectors are totally geodesic and complete. Finite length controls base motion and fibre radius; bounded-radius disk bundles over the compact base are compact, proving ambient completeness. The physical scale $R$ is free. Compact topology here does not prescribe small curled-up directions.

The vacuum and determinant-$V$ singlets give a global unequal-singular-value polarization extracting the same $J,u,Ju$, parallel in the normal connection. Nonzero curvature and $c_2$ coexist with that frame. A Euclidean-base instanton connection gives a separate complete $\mathbb R^{10}$ control with flat lower sectors and a global unit base coorientation. Its ordinary fourth-degree Chern class is zero; an asymptotic instanton label uses additional boundary data.

These are existence controls with distinct global topologies. Neither the frame rule nor this construction selects physical particle occupations.

Minimal complex sector projectors have ranks $2,2,4,4,8,32$ and are parallel on their own sectors. Their images are invariant under every tangential Clifford vector, so each compressed symbol retains its full spatial metric. The rank-two observer projector cannot be parallel over all of round $S^4$. Nor do these minimal images contain the entire charge carrier automatically.

6. A Common Full Carrier Has A Quantifiable Cost

The real normal module $V_N=(\Lambda^*W)_{\mathbb R}$, with reporting complex structure $J_N=I\Gamma_N=-\omega_N$, has real rank sixteen, reporting-complex rank eight, and quaternionic Clifford commutant. A full ambient module retaining that original quaternionic algebra is

$$\mathcal F=V_N\otimes_{\mathbb R}U,\qquad \dim_{\mathbb R}U=8,\qquad c_N=n_a\otimes1,\quad c_B=J_N\otimes h_\mu,$$

where the $h_\mu$ generate $\mathrm{Cl}(0,4)$. The ten positive-square generators act on real rank 128. The canonical $U(3)$ normal lift fixes all base generators.

For the $U(2)$ framed-line reduction define

$$K_V=I(N_V-1)=-\frac{n_3n_4+n_5n_6}{2},\quad E_0=-K_V^2,\quad E_1=1-E_0.$$

Choose a right quaternion $r$ in the tangent-module commutant, with $r^2=-1$. Then

$$P_\pm=\frac{E_0\otimes1\pm K_V\otimes r}{2},\qquad P=E_1\otimes1+P_+$$

are orthogonal projectors of real ranks 32 and 96 respectively. They commute with the six tangential Clifford generators of $Q_6$, the original normal quaternionic algebra, charge and $U(2)$ transport. Using $P$ on each lower sector and identity on $Q_{10}$ gives a nonminimal nested projector family in the same field. It contains every original $U(2)$-restricted charge representation, but with unbalanced extra multiplicities. On this carrier, retaining only balanced copies of the original complete $E$ requires rank 128 under the simultaneous tangent Clifford, original quaternionic, charge and full complement SU(2) symmetry conditions. Reduced normal holonomy alone is a weaker condition and can permit other projections.

There is also a dimension-specific closure constraint. If a $Q_5$ projector commutes with charge, $J_N$, and $c(u)$, then

$$[G,c(u)]=c(Ju)\quad\Longrightarrow\quad[P,c(Ju)]=0.$$

It is already $Q_6$-Clifford-compatible. The minimal odd-five-volume half projection fails that retained charge condition. This is a fibre statement: the central circle still rotates a fixed real-line $Q_5$ domain out of itself, while preserving the complex-line $Q_6$ domain.

The earlier normal-commutant criterion must keep its narrower premise. On the canonical normal module without copies, a non-Abelian reporting- $J_N$-preserving commutant occurs at normal dimensions $6\bmod8$. After minimal non-Abelian full ambient completion, automatic normal-factor purity instead permits gaps two, four and six. The alternating-vector map criterion in section 1 survives Spin-trivial copies; the old commutant criterion does not.

7. Bending, Contact And Curvature Obey Shared Bounds

For a unit normal line $u$, put $\alpha=\nabla u$ and $G_{\mu\nu}=\operatorname{Re}\langle\alpha_\mu,\alpha_\nu\rangle$. The induced volume at normal line offset $t$ and the second fundamental form along the base obey

$$v(t)=\sqrt{\det(1+t^2G)},\qquad |B_{Q_5}|^2=2|\alpha|^2.$$

For rank-two polarization with singular values $a>b>0$, the same data give

$$|\nabla u|^2\le(a^{-2}+b^{-2})|D\Phi|^2,\qquad |\nabla J|_F^2\le \frac{16(a^2+b^2)}{(a^2-b^2)^2}|D\Phi|^2.$$

The constants are sharp in the full real Frobenius norm. They quantify geometric variation, not an energy without a specified action.

With the same connection preserving $J$, decompose $\alpha=iau+\eta$, where $a$ is real and $\eta\perp_{\mathbb C}u$. Put $\sigma=\det(u,\eta,\eta)/2$. Using only these first derivatives, the four-dimensional spatial base metric and orientation, the leading determinant-valued one-forms span

$$C=\iota_{a^\sharp}\sigma,\qquad S=*(a\wedge\sigma),\qquad |C|^2+|S|^2\le\frac{|\alpha|^6}{27}.$$

They can vanish independently. A zero of one proposed channel does not establish absence of every contact. The complex plane $u,Ju$ makes the linear determinant $\det(u,Ju,\alpha)$ identically zero without forcing both cubic channels to vanish.

A parallel frame kills these first-jet contacts, but residual curvature can remain. With $F$ the normal curvature, an independent cubic spatial contact is

$$\beta_\xi(X)=\sum_\mu \det\big(y,F(X,e_\mu)y,F(\xi,e_\mu)y\big).$$

The same connection metric gives

$$\Delta=\operatorname{Scal}_B-\operatorname{Scal}_{M} =\frac12\sum_{\mu<\nu}|F_{\mu\nu}y|^2,\qquad |\beta_\xi|\le|\xi||y|\Delta.$$

For isotropic chiral SU(2) curvature and $y=z_0u+w$, $w=(a,b)$, the local formula is proportional to

$$z_0 f^2\epsilon(w),\qquad \epsilon(w)=\frac{(a^2-b^2,\ i(a^2+b^2),\ -2ab)}{\sqrt2}.$$

It is null where nonzero and uses the same spinor-square polynomial as a circular three-vector polarization. Generic selfduality is insufficient: for curvature coefficients $M$, the map contains $\operatorname{cof}(M)\epsilon(w)$; universal nullness requires $\operatorname{cof}(M)^T\operatorname{cof}(M)$ to be scalar. For invertible $M$, that is precisely isotropy up to scale and rotation. The resulting axis is not a derived propagation momentum or particle helicity.

On round $S^4$, choose the smooth height gradient for $\xi$. It has two index-one zeros, forced in total by the Euler class. The contact is global and smooth but vanishes at those poles and on $Q_5,Q_6$. Nonzero curvature need not give a nonzero channel. A nonzero $c_2(V)$ also forces zeros of any chosen section of $V$; those are different zeros on a different bundle and do not count neutrino flavors.

8. Curvature Reconstruction And Transport Are Additional Choices

The six-normal frame leaves a complex rank-two complement $V$, whose traceless unitary curvature has three adjoint components. Suppose this SAME curvature is required to span one complete Hodge eigenspace of its actual base two-forms, for the induced spatial metric. A nonzero equality $*F=\pm F$ of ordinary forms requires $\dim B-2=2$. Thus this additional chiral-curvature law fixes a four-dimensional spatial base, and the six-normal rule then gives containing dimension ten. The Hodge degree condition itself is the dimension-sensitive premise; this is not a derivation from generic metric reconstruction.

There is an explicit nearby changed-law control. On an oriented three-base, three independent curvature two-forms can instead span all two-forms. With a temporary volume $\mu$, define $\iota_{X_a}\mu=F^a$ and $D=\mu(X_1,X_2,X_3)$. Then

$$h^{-1}=|D|^{-1}\sum_a X_a\otimes X_a$$

is independent of the temporary volume and adjoint frame. The actual constant connection $A_i=aT_i$ on $\mathbb R^3$ has $F^1=a^2dx^2\wedge dx^3$, cyclically, and reconstructs $h=a^2\sum_i(dx^i)^2$. Adding the same normal six-bundle gives ambient dimension nine. It obeys the broad curvature-to-metric idea but not the same-degree Hodge-chiral law. This prevents an unrestricted four-in-ten uniqueness claim.

A definite SU(2) curvature triple spans one chiral Hodge space of a unique conformal four-metric, selfdual after the appropriate orientation choice. In that selfdual basis $\Sigma_j$, write $F^a=M_{ja}\Sigma_j$. Then

$$F^a\wedge F^b=2(M^TM)_{ab}\,\mathrm{vol},\qquad U_F=\det(M)\,g\,\mathrm{vol}.$$

Here the Urbantke density convention is

$$U_F(X,Y)=\frac16\epsilon_{abc} (\iota_XF^a)\wedge(\iota_YF^b)\wedge F^c,$$

calibrated by $\Sigma_a\wedge\Sigma_b=2\delta_{ab}\,\mathrm{vol}$ and $U_\Sigma=g\,\mathrm{vol}$. Neither identity fixes a volume normalization or removes the eigenvalue ratios of $M^TM$. The isotropic subclass is an extra condition. Standard background and its Einstein specialization are developed by Fine, Krasnov and Panov, sections 2.1–2.4; the present application uses those results as geometry, not a derived physical gravity theory.

The global same-curvature Hodge identification further requires

$$w_2(TB)=c_1(V)\bmod2,\qquad c_1(V)^2-4c_2(V)=p_1(TB)\pm2e(TB).$$

These necessary characteristic-class constraints link the normal reduction to the base topology. They do not guarantee a definite connection. Perfection of the curvature Gram matrix imposes five additional pointwise conditions; it is equivalent to universal nullness of the specified cubic contact when the curvature triple is invertible. Counting both as independent evidence would count the same restriction twice.

One can instead change transport to preserve a specified $J,u$. The unique pointwise Frobenius-minimal correction starts with

$$K_J=-\tfrac12J\nabla J,\qquad F_J=\tfrac12(F-JFJ)-\tfrac14(\nabla J\wedge\nabla J),$$

and then uses $K_u=u\eta^\dagger-\eta u^\dagger-iauu^\dagger$, with $\alpha=\nabla^Ju=iau+\eta$. The residual curvature is

$$F_{\mathrm{flag}}=QF_JQ+\eta\wedge\eta^\dagger,\qquad Q=1-uu^\dagger.$$

It can be non-Abelian while the adapted first-jet contacts vanish. In a connection metric, changing $A$ this way generally changes the ambient metric. It is not a gauge rewriting of the same flat geometry. On the round parallel control, both corrections already vanish.

9. A Parallel Real Frame Constrains Curvature Equality

Let the same connection preserve the actual unit frame $u$, so its curvature on $W=\mathbb C u\oplus V$ is $F=0\oplus G$. Define the positive curvature response

$$H=\sum_{i<j}F_{ij}^{\dagger}F_{ij},\qquad H_0=H-\frac{\operatorname{tr}H}{3}I.$$

If the eigenvalues of $H$ are $0,\lambda_1,\lambda_2$, then

$$\operatorname{tr}H_0^2 =\frac{(\operatorname{tr}H)^2}{6} +\frac{(\lambda_1-\lambda_2)^2}{2}.$$

The bound is sharp at $H=kP_V$. Every traceless $SU(2)$ curvature has this form after summing its squared components, because each Pauli generator combination squares to a scalar. Saturation therefore does not imply the stronger perfect-curvature condition used above. This is also the response entering the connection metric: $y^\dagger Hy=2(\operatorname{Scal}_B-\operatorname{Scal}_M)$.

The actual charge triplets require their own calculation. Their curvature representations are

$$F_{\Lambda^2W}=(\operatorname{tr}G)\oplus G,\qquad F_{\Lambda^2W\otimes D^{-1}} =0\oplus(G-\operatorname{tr}G\,I).$$

The second triplet is $W^*$ and retains a parallel singlet. A scalar response on this entire triplet must therefore be zero. Positivity of the squared-curvature sum and faithfulness of the dual representation then force $F=0$. Thus both triplets can have isotropic nonzero curvature response only after changing the parallel-real-frame premise.

There is an explicit such change. On a flat four-dimensional spatial base choose real closed two-forms

$$f=dx^1\wedge dx^2,\qquad g=-\frac12f+\frac{\sqrt3}{2}dx^3\wedge dx^4.$$

All of $f,g,f+g$ have norm one. A diagonal trace-free curvature $i\,\operatorname{diag}(f,g,-f-g)$ gives $H=I$ on $W$, $\Lambda^2W$ and $W^*$. Its connection preserves the complex line but rotates the phase of $u$, bending its real five-sector. By contrast, $i\,\operatorname{diag}(0,f,g)$ preserves $u$ and gives isotropic response on $\Lambda^2W$ alone. These smooth Abelian controls separate the real frame, complex plane and complete triplet conditions.

Equal curvature response is distinct from the common algebraic transition coefficient of Section 2. The latter survives this comparison. No physical color breaking, mass splitting or force rate follows from $H$ alone.

10. Actual Spatial Rotations And Common Charge-Preserving Sections

In the perfect-curvature controls, compatible spin-lift data identify $V$ with a chiral base spin doublet. A selected base coorientation gives a locked $\operatorname{Spin}(3)=SU(2)$ action

$$(v,t,z,w)\longmapsto(R(q)v,t,z,qw).$$

Here $v$ is a base three-vector, $t$ its fourth coordinate, $z$ the coordinate of $\mathbb C u$, and $w$ the four real coordinates of $V$. These are all ordinary spatial coordinates. The central element $q=-1$ fixes the base but sends $w$ to $-w$. Consequently this actual ambient action does not descend to $SO(3)$: a $2\pi$ base rotation ends at a nontrivial normal inversion, and $4\pi$ closes it.

The Euclidean instanton and round chiral-spin controls have actual lifted isometries with this action. Generic curvature-frame covariance alone does not establish an isometry. The quadratic contact

$$p(w)=\left(-\frac{i}{2}(a^2-b^2), \frac12(a^2+b^2),iab\right),\qquad w=(a,b),$$

is even under $w\mapsto-w$ and transforms as an $SO(3)$ vector. An output channel can descend even though the full spatial action does not.

At a fixed zero-section point, the complex ambient Dirac carrier branches under the locked action as

$$\Delta_{10}|_{SU(2)}=4[0]\oplus4[1]\oplus8[1/2],$$

where $[j]$ has dimension $2j+1$. This differs from rotating only three tangent directions while fixing the normals. For an actual section, orbital pullback matters too. If $N_V$ counts normal exterior factors and $\ell_V$ is its degree in $w,\bar w$, the central sign is $(-1)^{1+N_V+\ell_V}$.

Each charge triplet restricts to a normal singlet plus doublet. Tensoring a base doublet gives $[1/2]\oplus[0]\oplus[1]$, so the fibre triplet is not already three spin-one-half copies. The same two ordinary coordinate maps provide charge-preserving spin-one-half section subspaces in both charged doublet branches:

$$H_1(z,w)=\bar z\,w,\qquad H_2(z,w)=z\,\mathcal J\bar w,\qquad \mathcal J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$

They obey $H_i(z,qw)=qH_i(z,w)$ and are invariant under the charge circle $(z,w)\mapsto(e^{i\theta}z,e^{i\theta}w)$. Under section transport their normal factor is invariant, leaving the base doublet. Thus the original lifted charge is retained without a separate shift for each branch. Constant base spinors in the flat instanton control realize actual doublet section spaces. On a curved base the normal construction alone does not supply such a global base-section space. This is an additional choice of sections.

There is a complete polynomial classification for this input:

$$H=P(|z|^2,|w|^2)H_1+Q(|z|^2,|w|^2)H_2.$$

Indeed $\operatorname{Sym}^p\mathbf2\otimes \operatorname{Sym}^r\mathbf2$ contains a doublet exactly when $|p-r|=1$, with multiplicity one. Charge invariance fixes the remaining $z$-weight to $-1$ or $+1$. The module is free of rank two over $\mathbb C[|z|^2,|w|^2]$, with Hilbert series $2t^2/(1-t^2)^2$. These are counts of available polynomial maps, not flavors or particles.

Every continuous map with these symmetries vanishes when $w=0$ or $z=0$. At the first locus the entire $SU(2)$ fixes the input; at the second, a charge rotation of $w$ can be realized by its $SU(2)$ stabilizer circle, which fixes no nonzero output doublet. The zeros therefore extend beyond the polynomial examples. Algebraic channels using those values inherit them, while derivatives such as $dH_1=d\bar z\,w$ at $z=0$ can remain nonzero.

Writing $K=[H_1\ H_2]$ gives

$$K^\dagger K=|z|^2|w|^2I,\qquad \det K=|z|^2|w|^2.$$

One common normalization gives an $SU(2)$ normal frame off both zero sets. At fixed positive $z$, its restriction to the unit $w$ sphere identifies $S^3$ with $SU(2)$ and cannot extend over the four-ball. This does not trivialize the original normal bundle over the base. The same pair also obeys

$$p(H_1)=\bar z^2p(w),\qquad p(H_2)=z^2\overline{p(w)}.$$

Their nonzero circular axes are opposite. Charge and spin compatibility therefore leave both handed outputs available. Neither axis is yet a derived propagation momentum. Combining the normal singlet with these two maps gives three equal-charge doublet section copies, but does not derive a full color $SU(3)$ action or select occupied states.

The same connection differentiates both maps:

$$DH_1=d\bar z\,w+\bar z\,Dw,\qquad DH_2=\mathcal J\overline{DH_1},\qquad |DH_1|^2+|DH_2|^2=4|w|^2+8|z|^2.$$

The norm sums an orthonormal real vertical frame of the connection metric. The combined differential has real rank two on $z=0,w\ne0$, rank four on $w=0,z\ne0$, and rank zero at their intersection. Moreover, $D^2K=FK$: off the axes these two maps detect the entire normal curvature. Vanishing horizontal first derivatives of tautological coordinates do not make a globally parallel frame, because the horizontal distribution has curvature.

A local color action on these specified three copies meets an additional obstruction. Their evaluation matrix is $T=\operatorname{diag}(1,K)$. If a pointwise fibre action implemented constant $SU(3)$ coefficient mixing, it would satisfy $B_gT=Tg$ and preserve $\ker T$. On either zero locus that kernel is a proper two-plane, impossible for the defining irreducible $SU(3)$ action. An abstract action on coefficients or a different action moving spatial points is outside this obstruction.

11. Global Round Sections Link Chirality To A Two-Sector

The round example admits an explicit global realization of the base doublets needed above. On $S^4_R\subset\mathbb R^5$ take the Hermitian Clifford matrices

$$\Gamma_i=\sigma_i\otimes\sigma_3\quad(i=1,2,3),\qquad \Gamma_4=I\otimes\sigma_1,\quad\Gamma_5=I\otimes\sigma_2.$$

Put $N=x^a\Gamma_a/R$ and $P_\chi=(I+\chi N)/2$, for $\chi=\pm1$. The images $E_\chi$ are the intrinsic chiral spin bundles: at a pole they are the two Weyl representations of the $Spin(4)$ isotropy, and the equivariant associated-bundle identification extends them over $Spin(5)/Spin(4)$. Their sum is the trivial complex Dirac bundle, with curved spin connection

$$\nabla=d+\frac12N\,dN.$$

The actual observer $Spin(3)$ rotates $x^1,x^2,x^3$ and fixes $x^4,x^5$. The fixed-plane Clifford volume $K=i\Gamma_4\Gamma_5$ commutes with that group. Each eigenspace $U_\kappa$ is a constant doublet, and $s_u(x)=P_\chi(x)u$, $u\in U_\kappa$, is a global doublet space of sections under the lifted spatial action.

Its evaluation Gram has eigenvalues and determinant

$$\lambda_\pm=\frac{1\pm r_3/R}{2},\qquad \det(T^\dagger T)=\frac{(x^4)^2+(x^5)^2}{4R^2},\qquad r_3^2=(x^1)^2+(x^2)^2+(x^3)^2.$$

Thus evaluation has rank two off the great $Q_2=S^2$ defined by $x^4=x^5=0$, and rank one on it. A nonzero constant $u$ has unit Bloch vector $a_i=u^\dagger\Gamma_i u/|u|^2$, and its chiral section has precisely one zero, at $x=-\chi R(a_1,a_2,a_3,0,0)$. At that zero $ds_u(X)=\chi\Gamma(X)u/(2R)$ is an invertible real map of four-dimensional spaces, proving transversality. With outward orientation and $\Gamma_1\cdots\Gamma_5=-I$, the Chern number is $c_2(E_\chi)=-\chi$.

The full Dirac section $(P_+u,P_-u)$ is everywhere nonzero and has norm $|u|$. It is not parallel. Tensoring this actual global doublet with either $H_i$ completes the charge-preserving rotation construction on the same round geometry without adding a base zero. The chiral version retains its extra zero. These are section representations, not finite-norm particle solutions on the noncompact total space.

The relationship with $Q_2$ requires the stated $K$ splitting. Observer symmetry alone permits any multiplicity line in $\mathbb C^2$. A line with Bloch vector $m$ gives Gram

$$\frac12\left[ \left(1+\chi\frac{m_1x^4+m_2x^5}{R}\right)I +\chi m_3\frac{x^i\sigma_i}{R}\right].$$

Its defect is generally a displaced two-sphere; at $m_3=0$ it collapses to a rank-zero point. The fixed-plane volume selects the great sphere. Nor is $Q_3$ unique: all equators $\cos\alpha\,x^4+\sin\alpha\,x^5=0$ contain this $Q_2$ and are equally observer-invariant. The fixed-plane rotation moves them through an $\mathbb{RP}^1$ family.

Finally these section spaces do not replace the Clifford projectors. Individual tangent Clifford vectors exchange the two chiral bundles, and the constant $U_\kappa$ projector fails to commute with all four tangent directions. The full Dirac bundle retains the four-dimensional symbol; a two-dimensional invariant space of sections does not establish a rank-two Clifford subbundle over the four-dimensional spatial base.

12. A Nonspin Base Tests The Global Determinant Data

The common geometry does not require a spin base or a trivial normal determinant. Take $B=\mathbb{CP}^2$ with its complex orientation and Fubini–Study metric, $V=(T^{1,0}B)^*$, and $W=\mathbb C u\oplus V$. With $\int_Bh^2=1$,

$$c_1(V)=-3h,\quad c_2(V)=3h^2,\quad p_1(TB)=e(TB)=3h^2,\quad D=\det W=K_B.$$

The base and normal bundles separately have $w_2=h\bmod2$, while $TB\oplus W_{\mathbb R}$ is spin. Its complex determinant is trivial. Neither separate determinant has a square root. The same endpoint identity holds:

$$c_1(V)^2-4c_2(V)=-3h^2=p_1(TB)-2e(TB).$$

The parallel identification $\mathfrak{su}(V)=\Lambda^2_-T^*B$ comes from primitive $(1,1)$ forms. Fubini–Study curvature is Einstein with $W^-=0$, making its traceless normal curvature perfect and anti-self-dual. The determinant curvature $F_D=i\rho$ is retained, with $\rho$ the self-dual Ricci form. The curvature decomposition and metric normalization are described in Gursky and LeBrun, sections 1–2. Their standard normalization gives scalar curvature $24$ and volume $\pi^2/2$; the physical scale here remains free.

A small embedded geodesic $S^3$ in $B$ and an equatorial $S^2$ supply $Q_3,Q_2$, followed by $B$, $B\times\mathbb R u$, $B\times\mathbb C u$ and $\operatorname{Tot}(W_{\mathbb R})$. Every induced metric is complete. The lower spheres generally bend. This $S^2$ is null-homologous in $B$.

The global vacuum and $\Lambda^2V\otimes D^{-1}$ identity sections again give an aligned gapped normal polarization. The quaternionic commutant is now the algebra bundle with skew part $\mathbb R\oplus D_{\mathbb R}$, not three globally fixed matrices. Using the tangent $Spin^c$ determinant $D^{-1}$ cancels the normal determinant in the complex rank-32 ambient spin carrier. A corresponding real-rank-eight tangent Clifford factor, with its retained scalar complex structure, globalizes the real-128 and $P_{96}$ construction. Its parallelness along $Q_6$ survives; the minimal moving lower-volume projectors need not be parallel on the bent $Q_3,Q_2$.

There is an exact description of the additional components. If $S_{10}$ is the irreducible real-rank-32 ambient Clifford bundle, then

$$\mathcal F=S_{10}\otimes_{\mathbb R} (\mathbb R^2\oplus D_{\mathbb R}),\qquad \nabla_{\mathrm{mult}}=d\oplus\nabla_D.$$

The complexified multiplicity has characters $1,1,D,D^{-1}$; the paired quaternionic real structures combine them into the stated real bundle. Equivalently, paired determinant half-lifts act on a quaternion $x=z+w j$ by $x\mapsto a x a$, $a=e^{i\theta/2}$, rotating $z$ by $e^{i\theta}$ and fixing $w j$. Both half-lift signs cancel. This transition calculation proves the splitting, beyond matching characteristic classes. A full-ambient Clifford submodule retaining the entire original quaternion algebra needs all four multiplicity dimensions. The lower-sector $P_{96}$ is not a full ten-direction Clifford submodule.

The retained determinant curvature provides another shared test. Write $F_V=F_0+aI$, with $F_0$ anti-self-dual and $a$ self-dual. The quadratic central contribution to the determinant cubic vanishes, and the cross term vanishes because opposite Hodge skew endomorphisms commute. Thus $\beta_F=\beta_{F_0}$, although the scalar curvature response retains the positive central contribution. For $F_0=f\Sigma^-_aT_a$, $a=it\omega$,

$$H_V=\left(\frac32f^2+2t^2\right)I.$$

At the standard Fubini–Study values $f=2,t=3$, this is $24I$. The normalized contact fraction is

$$\frac{|\beta_\nu|^2}{|y|^2 (\operatorname{Scal}_B-\operatorname{Scal}_M)^2} =\frac1{18}\frac{|z|^2}{|z|^2+|w|^2},$$

for unit $\nu$ and nonzero $w$. The same contact can therefore remain unchanged while the metric's total curvature response increases.

Finally the old quaternionic conjugation has the global target $j:V\to V\otimes D^{-1}$. Consequently $H_1=\bar z w$ and $H_2=zj(w)$ belong to different bundles here. The two original charge triplets split as $D\oplus V$ and $1\oplus V D^{-1}$. Their relative twist survives multiplication by a common base $Spin^c$ factor; a nowhere-zero determinant section that identified them does not exist. Thus the round two-compensator construction cannot be transferred unchanged. The full charge carrier remains global, while particular section identifications acquire additional restrictions. This is global representation data; it does not establish a scalar conserved charge of the full ambient kinetic operator.

13. The Same Covariants Under Spatial Differentiation

The two charge-preserving maps also have a complete angular classification. Set $\sigma=|z|^2$, $\rho=|w|^2$ and $r^2=\sigma+\rho$ in the ordinary six-dimensional normal space. For either $H_a$,

$$\Delta_N(FH_a)=4\left( \sigma F_{\sigma\sigma}+\rho F_{\rho\rho} +2F_\sigma+3F_\rho\right)H_a.$$

For a scalar homogeneous polynomial of degree $n$, harmonicity reduces to the coefficient recurrence

$$(j+1)(j+2)a_{j+1}+(n-j)(n-j+2)a_j=0.$$

It has exactly one solution after setting $a_0=1$:

$$F_n=\sum_{j=0}^n \frac{(-1)^j}{j+1}\binom nj\binom{n+2}j \sigma^j\rho^{n-j} =\frac{(-1)^n}{n+1}r^{2n} P_n^{(2,1)}\!\left(\frac{\sigma-\rho}{r^2}\right).$$

Thus precisely two equivariant harmonic maps occur at each ordinary polynomial degree $L=2n+2$: $F_nH_1$ and $F_nH_2$. Their angular eigenvalues and individual norms on the unit normal sphere are

$$-\Delta_{S^5}(F_nH_a)=4(n+1)(n+3)F_nH_a,\qquad \int_{S^5}|F_nH_a|^2\,d\Omega =\frac{\pi^3}{(n+1)(n+2)(n+3)}.$$

The sphere here separates angular variables in ordinary normal space; it imposes neither a spatial boundary nor a new sector topology. Writing $t=\sigma/r^2$ gives the positive Jacobi weight $t(1-t)^2$. Orthogonality and completeness in that weighted space give the complete angular closure with these symmetries. The full polynomial module has Hilbert series $2q^2/(1-q^2)^2$, whereas its harmonic part has $2q^2/(1-q^2)$. The counts concern available covariant maps.

At each fixed $n$, both columns share $n$ simple interior angular nodes in addition to the forced axes. For example $F_1=\rho-\frac32\sigma$ vanishes at $\sigma/\rho=2/3$. The Jacobi three-term recurrence has a nonzero backward coefficient: a common root of consecutive polynomials would descend to the nonzero constant polynomial. Therefore adjacent degrees have no common interior node. Retaining both degrees as separate channels removes their common interior zeros, although the symmetry-forced axes remain. A particular superposition can still have its own zeros. The quadratic circular output scales as $F_n^2$, so crossing a sign-changing angular node does not reverse its axis. Neither angular eigenvalues nor equal norms choose one handedness.

The Clifford derivative is a separate, stricter calculation. Use the normal exterior basis $e_0,e_1,e_2$, with $e_0$ along the framed complex line, and embed the maps as

$$h_1=\bar z_0(z_1e_{01}+z_2e_{02}),\qquad h_2=z_0(-\bar z_2e_{01}+\bar z_1e_{02}).$$

With creation operators $a_j^\dagger$ and their adjoints $a_j$,

$$D_N=2\sum_{j=0}^2 \left(a_j^\dagger\partial_{\bar z_j}+a_j\partial_{z_j}\right), \qquad D_N^2=\Delta_N.$$

Direct differentiation gives

$$g_1=D_Nh_1=-4\bar z_0e_0,\qquad g_2=D_Nh_2=2(-\bar z_2e_1+\bar z_1e_2-2z_0e_{012}), \qquad D_Ng_1=D_Ng_2=0.$$

The original three even-normal copies $e_{12},h_1,h_2$ therefore need two opposite-normal-chirality descendants to close. Their five-dimensional complex polynomial span is invariant under $D_N$. All five have the same total circle weight under

$$\mathcal M_b=N-\sum_j (z_j\partial_{z_j}-\bar z_j\partial_{\bar z_j})-3b: \qquad \mathcal M_b h=(2-3b)h,\quad b=0,1.$$

The normalized weights remain $2/3$ and $-1/3$. This is a finite polynomial representation, not a finite spatial norm or occupied spectrum.

The two equal-norm angular maps nevertheless have different Clifford derivatives. If $N_n=\|F_nh_a\|_{S^5}^2$, their exact derivative Gram is

$$\operatorname{Gram}\big(D_N(F_nh_1),D_N(F_nh_2)\big) =8(n+2)N_n \begin{pmatrix}n+2&0\\0&n+3\end{pmatrix}.$$

For example the normalized squared derivatives are $32$ and $48$ at $n=0$. To obtain this identity, put $a=F_\sigma$, $b=F_\rho$. The pointwise squared derivatives are $4[\sigma^2\rho a^2+\sigma(\rho b+2F)^2]$ and $4[\rho(\sigma a+F)^2+\sigma(\rho b+2F)^2]$; their cross term is zero. On the sphere, integration uses $2\pi^3(1-t)\,dt$. The Jacobi equation in Sturm form,

$$\big(t^2(1-t)^3g_n'\big)' +n(n+4)t(1-t)^2g_n=0,$$

reduces both integrals by parts to the stated Gram. Its positive eigenvalues exclude a nonzero monogenic combination inside the restricted two-doublet target. The comparison concerns the specified homogeneous extension and Clifford embedding. It is not the intrinsic spherical Dirac operator or a physical kinetic preference.

Completing the target gives a constructive alternative. For a harmonic spinor $h$ of degree $L$, the Clifford anticommutator yields

$$M=h-\frac{c(y)D_Nh}{2L+4},\qquad D_NM=0.$$

At degree two this gives

$$M_1=\frac12(h_1+\bar z_0^2\,1),\qquad M_2=\frac14\big(h_2+(2\sigma-\rho)e_{12}\big).$$

Both retain total circle weight two. The first includes a vacuum component with compensating orbital weight. The second stays entirely inside the original fibre-number-two triplet and has

$$|M_2|^2=\frac{4\sigma^2-3\sigma\rho+\rho^2}{16}>0 \quad\text{for }\sigma+\rho>0.$$

The singlet here is the $SU(2)$ singlet $e_{12}$ within that triplet. The symmetry-forced axes of a map into its doublet complement are therefore not forced zeros of a complete triplet section. A constant $e_{12}$ already avoids those axes; this construction additionally retains homogeneous degree two and solves the normal monogenic equation.

There is an all-degree classification behind this example. Every $SU(2)$-equivariant even polynomial spinor of total weight two is uniquely

$$\Psi=\bar z^2 A(\sigma,\rho)\,1+B(\sigma,\rho)e_{12} +C(\sigma,\rho)h_1+T(\sigma,\rho)h_2.$$

The normal Dirac equation reduces to

$$\sigma A_\sigma+2A=\rho C_\rho+2C,\qquad A_\rho=-C_\sigma, \qquad B_\rho=-\sigma T_\sigma-T,\qquad B_\sigma=\rho T_\rho+2T.$$

Compatibility forces both $C$ and $T$ to satisfy the same Jacobi radial equation as $F_n$. The first pair then determines $A$ uniquely; the second determines $B$ up to the separate degree-zero constant. Consequently at each ordinary degree $2k\ge2$ the polynomial, harmonic and monogenic dimensions are respectively $4k+1$, $4$ and $2$. Imposing the additional fibre eigencondition $N\Psi=2\Psi$ leaves exactly one monogenic copy. Degree zero has the constant $e_{12}$ in every comparison.

For $k=n+1$, that single fibre-number-two copy is

$$M_{2,n}=\frac{n+1}{2(n+2)}F_nh_2+B_ne_{12},\qquad B_n=\frac{(2\sigma-\rho)F_n+ \sigma\rho(F_{n,\rho}-F_{n,\sigma})}{2(n+2)}.$$

Equivalently its scalar coefficient is the binomial harmonic

$$B_n=-\frac1{2(n+2)} \sum_{j=0}^{n+1}(-1)^j \binom{n+1}j\binom{n+2}j \sigma^j\rho^{n+1-j}.$$

Every individual $M_{2,n}$ is nonzero off the origin. At either axis, $B_n$ is nonzero. At an interior root of $F_n$, its derivative difference is nonzero by simplicity of the Jacobi root, so $B_n$ again survives. The orthogonal doublet and singlet components therefore cannot vanish together. This is a property of each homogeneous branch; arbitrary sums of degrees can acquire cancellation zeros.

These statements use the $SU(2)$ determinant framing. With general $U(2)$ gluing, the second completion naturally takes values in $\Lambda^2W\otimes D^{-1}=1\oplus VD^{-1}$ as an ordinary bundle decomposition, and the whole normal Clifford carrier must have that common twist. Under the maintained circle rotating all three complex directions, $D=\det W$ has weight three: this twisted branch has total weight $2-3=-1$, hence normalized charge $-1/3$. Its displayed trivial line also has that nonzero circle weight. Identifying $D$ with $\det V$ by the chosen $u$ is an ordinary bundle identification, not an identification of their circle actions: the framed line itself has circle weight one. The $\mathbb{CP}^2$ control therefore cannot borrow both untwisted branches or erase their relative character.

The completed maps also allow a precise topology comparison. In the fixed frame, their normalized coefficient vectors lie in the sphere $S^4\subset\mathbb C^2\oplus\mathbb R$, because $B_n$ is real. They are invariant under common coordinate phase and descend from the ordinary angular $S^5$ to a map $f_n:\mathbb{CP}^2\to S^4$. These are an angular quotient and a coefficient target; neither is being identified with a physical sector. With complex orientation on the quotient and boundary orientation on the target, the degree is

$$\deg f_n=\frac{1+(-1)^n}{2}.$$

To see the cancellation of extra radial excursions, use the affine coordinate $v=w/z$ and write the normalized map as $(h(r)j(v/r),b(r))$, where $h^2+b^2=1$. Its endpoints are $b(0)=(-1)^n$ and $b(\infty)=-1$. The quaternionic map $j$ preserves real orientation, and the target volume primitive is $K(b)=b-b^3/3$. Thus the degree is $[K((-1)^n)-K(-1)]/[K(1)-K(-1)]$. The hyperplane $z=0$ collapses to the south pole; at $n=0$ the affine chart maps diffeomorphically onto the complementary target chart. This does not make $\mathbb{CP}^2$ diffeomorphic to $S^4$.

Allowing the scalar coefficient to be complex enlarges the normalized target to $S^5$. Every resulting map $\mathbb{CP}^2\to S^5$ is nullhomotopic, because the source has CW dimension four and the target is four-connected. The original-domain $S^5\to S^5$ composite is therefore also nullhomotopic. The parity distinction requires the restricted real target; it is not a protected invariant of the complete complex carrier or a particle-flavor count.

14. Curvature Decides Which Polynomial Closure Survives

For the actual connection metric, the spatial Dirac operator contains more than the separate base and normal derivatives. With positive-square Hermitian Clifford matrices and no overall factor $i$, the ordered spin lift is

$$\Omega_X=-\frac14\sum_{B,C} \langle\nabla_Xe_B,e_C\rangle c(e_B)c(e_C).$$

At a base normal frame set $f_{ija}=\langle F_{ij}y,E_a\rangle$. The mixed Koszul coefficients are $\langle\nabla_{H_i}H_j,E_a\rangle=-f_{ija}/2$ and $\langle\nabla_{E_a}H_i,H_j\rangle=f_{ija}/2$. Their horizontal spin contraction contributes one half of the unordered curvature sum, and the vertical contraction subtracts one quarter. Hence the actual operator is

$$D_M=D_{\rm split}+C_F,\qquad C_F=\frac14\sum_{i<j}\gamma_i\gamma_jc_N(F_{ij}y).$$

Here $c_N(v)=\Gamma_B\otimes n(v)$ includes base chirality. For pure-Hodge $SU(2)$ normal curvature, $C_F$ kills the opposite base Weyl factor. On the flat four-dimensional spatial base with the BPST normal connection, that factor has constant parallel spinors $s$. The natural covariants $h_a$ and $g_a$ are horizontally parallel: the connection's rotation of their values cancels its rotation of their position arguments. Thus, on this same curved ambient metric,

$$D_M(s\otimes h_a)=\chi\,s\otimes g_a,\qquad D_M(s\otimes g_a)=0,\qquad \Gamma_Bs=\chi s.$$

This proves actual ambient closure, beyond a flat-normal symbol check. It requires the flat base and its parallel opposite-Weyl sections. The round base's global doublet sections are not parallel; its base Dirac term changes Weyl chirality, so the same proof does not transfer.

The other Weyl factor behaves differently. For perfect curvature $F=f\Sigma_a\otimes T_a$, $T_a=-i\sigma_a/2$, the three real vectors $T_aw$ are orthogonal with length $|w|/2$. The curvature block reduces to the Pauli coupling

$$\sum_a\sigma_a\otimes\sigma_a=2\,\mathrm{Swap}-I,$$

whose eigenvalues are $1$ and $-3$, with multiplicities three and one. Consequently $C_F$ has singular values $|f||w|/4$ and $3|f||w|/4$ on the active rank-sixteen complex block, with multiplicities twelve and four. It is invertible there when $fw\ne0$. The curvature term raises normal polynomial degree by one, while $D_{\rm split}$ preserves or lowers it. An active highest-degree term therefore acquires unbounded degree under repeated application: it cannot have a finite polynomial derivative span.

The $\mathbb{CP}^2$ control supplies an exact contrast using the same operator. Its determinant curvature $it\omega I$ acts on the complementary Weyl block, with

$$C_{\det}^2=-\frac{t^2|w|^2}{4}I$$

there. For $f,t\ne0$ and $w\ne0$, the full $C_F$ is invertible on all thirty-two complex components. No nonzero finite-degree normal polynomial section has a finite bounded-degree derivative span in that control. At $f=2,t=3,|w|=1$, the eigenvalues of $C_F^*C_F$ are $1/4$ with multiplicity twelve and $9/4$ with multiplicity twenty. Yet the selected cubic still has $\beta_F=\beta_{F_0}$. The same determinant geometry is invisible to that contact and decisive for this Clifford closure.

These are restrictions on geometric operators and polynomial sections. The polynomial sections have divergent integrals over the unbounded normal fibres; no particle state, physical helicity or mass is inferred. The connection's ordinary normal circle is an isometry and has a commuting lifted orbital-plus-spin action. This differs from treating the scalar fibre charge alone as conserved. Locally put $E=\sum_j(z_j\partial_{z_j}-\bar z_j\partial_{\bar z_j})$. Since $F$ commutes with the normal complex structure,

$$[N,c_N(F_{ij}y)]=[E,c_N(F_{ij}y)],\qquad [N-E,C_F]=0.$$

The same total generator commutes with the split terms. A constant scalar offset preserves the result; a branch-dependent offset in a larger carrier additionally requires a parallel commuting multiplicity action. The natural spin lift can be a double cover of the geometric circle until its scalar determinant factor is specified. Ambient circle symmetry also does not preserve one fixed real-line $Q_5$ domain.

15. Uniform Higher-Normal-Dimension Controls

For every complex normal rank $m$ retain exactly the same prescription

$$F_m=\Lambda^{\rm even}W,\qquad E_m=F_m\otimes(\mathbf1\oplus D^{-1}),\qquad D=\det W.$$

An even exterior degree $k$ has multiplicity $\binom mk$ and the two charges $k/m$ and $k/m-1$. The denominator follows the same determinant normalization at every rank. For $m\ge2$,

$$\det F_m=D^{2^{m-2}},\qquad \det E_m=1, \qquad \operatorname{tr}Q=0.$$

Thus determinant cancellation is not a six-normal discriminator. Under the full $SU(m)$, odd $m\ge3$ gives two singlets in $E_m$ and $(m-1)/2$ distinct nontrivial exterior types per determinant branch. Even $m\ge2$ gives four singlets and $m/2-1$ such types. Consequently the additional carrier premise of two singlets and one nontrivial type per branch selects $m=3$, whose nontrivial rank is then $\binom32=3$. The triplet rank was not separately prescribed. This is a simplicity condition on the specified carrier, not an observed selection law.

For completeness the character and its odd moments are

$$\chi_m(t)=\frac{1+t^{-m}}2\big((1+t)^m+(1-t)^m\big), \qquad \operatorname{tr}Q^a= \begin{cases} 0,&m\text{ even},\\ -m!\,\left\{\!\begin{smallmatrix}a\\m\end{smallmatrix}\!\right\}/m^a, &m\text{ odd}, \end{cases} \quad a\text{ odd}.$$

The braces are Stirling numbers of the second kind; the identity follows by a finite difference of $k^a$. In particular the cubic trace is $-2/9$ at $m=3$ and zero at every other $m\ge2$. This is a carrier invariant, not a gauge-anomaly calculation: no spacetime chirality or quantized matter content has been assigned here.

Write $W=\mathbb C u\oplus V$. A quaternionic antiunitary $j^2=-1$ on $V$ requires its complex rank to be even. It therefore permits every odd $m\ge3$ after a suitable holonomy reduction. If it must commute with the full residual $SU(m-1)$, its center gives

$$j\zeta=\bar\zeta j=\zeta j,\qquad \zeta=e^{2\pi i/(m-1)}.$$

For a nonzero quaternionic complement this forces $m-1=2$. Full $U(m-1)$ equivariance would instead fail for every nonzero complement; the determinant-valued formulation retains that distinction.

Higher-rank complete controls remain available. For $m=2r+1$ take $V=(S^+)^{\oplus r}$ over round $S^4$, with its direct-sum chiral connection and the same connection metric. The ambient dimension is $6+4r$. The parallel line and first spin-bundle summand realize

$$S^2\subset S^3\subset S^4\subset\operatorname{Tot}(\mathbb R u) \subset\operatorname{Tot}(\mathbb C u) \subset\operatorname{Tot}(\mathbb C u\oplus S^+).$$

All are complete and totally geodesic: connection-invariant subbundles are fixed sets of isometric complement reflections, and bounded-radius disk bundles over the compact base are compact. Thus the entire $2,3,4,5,6,10$ flag coexists in these larger ordinary spatial ambients. They retain a quaternionic complement, nonzero curvature and determinant cancellation while changing the full charge representation.

For example, at $m=5$ the $Sp(2)$ branching is

In general the preserved symplectic bivector supplies one invariant $\Omega^l$ at every even $V$ degree $2l$, $0\le l\le r$. The two branches give $2(r+1)$ singlet lines at charges $2l/(2r+1)$ and $2l/(2r+1)-1$. These extra channels cannot be silently discarded while claiming the same complete carrier. These controls do not satisfy the earlier universal alternating-frame premise.

16. Charge Compatibility Forces Complex Saturation

Use the canonical complex exterior Clifford module $\Lambda^*W$ and its number operator $N$. With positive Clifford generators,

$$[N,c(v)]=-i c(Jv).$$

If an orthogonal projector $P$ commutes with $N$ and every $c(v)$ for $v$ in a real tangent subspace $L$, the Jacobi identity forces commutation with every $c(Jv)$ as well. Put

$$U=L+JL,\qquad k=\dim_{\mathbb R}L,\qquad 2h=\dim_{\mathbb R}(L\cap JL),\qquad r=\dim_{\mathbb C}U=k-h.$$

The creation and annihilation operators for $U$ generate $\operatorname{End}(\Lambda^*U)$. Hence

$$P=1_{\Lambda^*U}\otimes P_0,\qquad [P_0,N_{U^\perp}]=0.$$

Every nonzero image has complex rank divisible by $2^r$. Its charge character has the form

$$(1+t)^r\sum_q p_qt^q,\qquad p_q\in\mathbb Z_{\ge0}.$$

The minimum is $2^r$; it has one residual occupation vector. Without additional multiplicity, the simultaneous commutant has complex dimension

$$\sum_q\binom{m-r}{q}^2=\binom{2(m-r)}{m-r}.$$

These are canonical exterior-complex ranks, not the reporting-complex counts of the real quaternionic carrier in section 6. A parallel multiplicity charge offset changes the residual operator but not the active factor $2^r$.

At $m=3$, real normal ranks $k=0,1,2,3,4,5,6$ permit complex saturation ranks respectively $\{0\},\{1\},\{1,2\},\{2,3\},\{2,3\},\{3\},\{3\}$. A real two-plane that is a complex line has minimum active rank two; a noncomplex real two-plane needs rank four. The real line in $Q_5$ and its complex-line $Q_6$ partner have the same saturation, so this rule does not separate those particle assignments.

There is also a spatial-domain obstruction. The normal circle fixes the zero section $B$. An invariant smooth submanifold containing $B$ must have a $J$-invariant normal tangent there, of even real rank. Therefore no $Q_5$ containing the four-dimensional zero section is invariant under that circle, even if bent. The complex-line $Q_6$ is invariant. Odd-dimensional invariant submanifolds away from the fixed locus are not excluded. This distinguishes domain invariance from projector invariance.

Finally $[N,c(p)]^\dagger[N,c(p)]=|p|^2$ for real normal $p$. Scalar fibre number alone cannot commute with a kinetic symbol retaining a nonzero normal momentum. The existing orbital-plus-fibre generator can cancel that commutator. No physical charge-to-depth assignment is selected by this compatibility theorem.

17. Completed Monogenic Branches Feed The Same Frame Map

For rank three the alternating map has a root-free unitary expression

$$\Lambda^2(\mathbf1\oplus\Lambda^2W)\otimes D^{-1} \cong W\oplus W^*=(W_{\mathbb R})_{\mathbb C}.$$

Write its two inputs as $(a,\alpha)$ and $(b,\beta)D^{-1}$. In a unit determinant frame put $A=(\alpha_{12},-\alpha_{02},\alpha_{01})$ and $B=(\beta_{12},-\beta_{02},\beta_{01})$. Then

$$v=A\times B,\qquad \xi=aB-bA,\qquad \xi(v)=0.$$

Its Hermitian norm is the exterior Gram determinant. Consequently independent inputs produce a nonzero complex null vector and an ordinary real normal frame. The determinant factors make this statement global.

Take the completed $M_{1,m}$ and $M_{2,n}D^{-1}$ of section 13, and set

$$c=F_m/2,\qquad d=\frac{n+1}{2(n+2)}F_n,\qquad a=\bar z^2A_m,\qquad B_0=B_n.$$

The same map gives

$$v=(cd\sigma\rho,cB_0\bar zw_1,cB_0\bar zw_2),\qquad \xi=a(B_0,-dz\bar w_1,-dz\bar w_2),$$

$$\|v\|^2+\|\xi\|^2 =(\sigma^2A_m^2+c^2\sigma\rho) (B_n^2+d^2\sigma\rho).$$

The factors are the squared norms of the two orthogonal completed inputs. Thus the frame vanishes exactly on $z=0$, at every pair of grades; all interior polynomial nodes and the former $w=0$ zero are removed. At $w=0,z\ne0$ its nonzero covector is $aB_nu^*$, selecting the real plane underlying $\mathbb C u$. A constant untwisted vacuum paired with $M_{2,n}D^{-1}$ instead gives a frame everywhere off the normal origin.

Charge accounting remains explicit. The first completed pair has total integer weights $2$ and $-1$, producing a normal-vector channel of weight $1$; it is not a single fixed-charge state. The constant-vacuum pair produces weight $-1$. These examples connect the same charge carrier, polynomial completion and normal-frame tensor without selecting particle occupation or physical handedness.

18. Exact Angular Contact And A One-Field Differential Channel

On the unit angular $S^5$ put $t=\sigma$, $\rho=1-t$ and $f_n(t)=F_n(t,1-t)$. Its invariant measure gives

$$\int_{S^5}p(t)\,d\Omega =2\pi^3\int_0^1(1-t)p(t)\,dt.$$

Define

$$a_n=\frac{2f_n-(1-t)f_n'}{2(n+2)},\qquad b_n=\frac{(3t-1)f_n-t(1-t)f_n'}{2(n+2)},\qquad h_n=\frac{n+1}{2(n+2)}.$$

For $U_n=M_{1,n}$, $V_n=M_{2,n}$ and a neutral scalar insertion $q(t)$, the cross branch pointwise product is zero, while the two diagonal kernels are

$$d^1_{nm}=\frac{t(1-t)}4f_nf_m+t^2a_na_m,\qquad d^2_{nm}=h_nh_m t(1-t)f_nf_m+b_nb_m.$$

If $qd^a_{nm}=\sum_j c_jt^j$, the exact angular coefficient divided by $\pi^3$ is $\sum_j2c_j/((j+1)(j+2))$. In particular

$$\|U_n\|^2=\frac{\pi^3}{2(n+1)(n+2)(n+3)},\qquad \|V_n\|^2=\frac{\pi^3}{2(n+2)^2(n+3)}.$$

Different grades are orthogonal. The insertion $q=t$ is tridiagonal, with normalized diagonal expectations

$$\langle t\rangle_{U_n}=\frac12,\qquad \langle t\rangle_{V_n}=\frac12-\frac1{2(n+2)(2n+5)}.$$

The projected doublet instead has expectation $1/2-3/(2(2n+3)(2n+5))$. Completion therefore changes angular response, not only pointwise zeros. For the neutral scalar harmonic $g_r=(-1)^rP_r^{(1,0)}(2t-1)$ the necessary nonzero ranges are

$$|n-m|\le r\le n+m+1\quad(U),\qquad |n-m|\le r\le n+m+2\quad(V).$$

The first upper zero follows because the top coefficient in $d^1_{nm}$ cancels. The rational integral decides further zeros inside these ranges.

The complete charge branches also admit a determinant-balanced cubic:

$$\mathcal V(T_0,T_1,T_2)=\det(T_0,T_1,T_2),\qquad T_0\in\Lambda^2W,\quad T_1,T_2\in\Lambda^2W\otimes D^{-1}.$$

Here $\det(\Lambda^2W)=D^2$ cancels the two twists; the integer circle weights are $2-1-1=0$. The tensor uses triplet components, while the underlying field carrier remains complete. In basis $(e_{01},e_{02},e_{12})$, $\det(H_1,H_2,e_{12})=\sigma\rho$. The constant twisted singlet $C=e_{12}$ therefore gives

$$\int_{S^5}\mathcal V(U_m,V_n,C)\,d\Omega =\delta_{mn}\frac{\pi^3}{4(n+2)^2(n+3)}.$$

The original carrier already contained this singlet; restricting to doublets hid its available contraction. For three positive grades set

$$\frac{C_{mnl}}{\pi^3} =\int_0^1 t(1-t)^2 f_m \big(h_nf_nb_l-h_lf_lb_n\big)\,dt.$$

It obeys $C_{mnl}=-C_{mln}$ and vanishes unless $\max(0,|n-l|-1)\le m\le n+l+1$. This follows from the ordinary harmonic triangle for degrees $2(m+1),2(n+1),2(l+1)$. For example $C_{001}/\pi^3=1/360$, $C_{101}/\pi^3=-1/2240$, and $C_{201}/\pi^3=1/4032$. Three $V$ towers alone have pointwise rank at most two and zero determinant at every grade. A nonvanishing completed section does not make every contact channel nonzero.

For reproducibility, define finite power coefficients

$$\alpha_{ni}=(-1)^i\binom ni\frac{(n+4)_i}{(i+1)!},\qquad \beta_{ni}=(-1)^{i+1}\binom{n+1}i \frac{(n+3)_i}{2(n+2)i!}.$$

They vanish outside $0\le i\le n$ and $0\le i\le n+1$ respectively; $(a)_i$ is the rising factorial. Then the integral is the exact finite sum

$$\frac{C_{mnl}}{\pi^3} =\sum_{i,j,s}\alpha_{mi} \big(h_n\alpha_{nj}\beta_{ls}-h_l\alpha_{lj}\beta_{ns}\big) \frac{2}{(i+j+s+2)(i+j+s+3)(i+j+s+4)}.$$

For normal-valued components, the same commuting field in both twisted slots makes the algebraic cubic identically zero, even after a mode decomposition. This also holds within one fixed base polarization. A specified geometric derivative supplies a nonzero one-field alternative. Let $Y=y^a\partial_{y^a}$ be the tautological radial normal vector and set

$$\mathcal J_Y(T_0,T_1)=\det(T_0,T_1,\nabla_YT_1).$$

For the actual connection metric, the vertical spin correction contracts to $\langle F_{ij}y,y\rangle=0$. Thus $\nabla_Y$ is the Euler derivative in the adapted spin frame; the normal exterior projector is parallel in this direction. The tensor retains unitary covariance and total circle neutrality. If $T_0=A U_m$ and $T_1=\sum_n B_nV_n$,

$$\int_{S^5}\mathcal J_Y\,d\Omega =A\sum_{n<l}2(l-n)B_nB_l C_{mnl}.$$

For $m=0$ and twisted grades $0,1$ the coefficient is $\pi^3/180$. A single homogeneous twisted grade still gives zero. A common radial factor $f$ in $T_1$ contributes $f^2$; its derivative cancels by antisymmetry. A radial factor $g$ in $T_0$ contributes another $g$. The angular selection rules therefore survive arbitrary common radial factors, while the radial integral and a physical action remain open. This is an available differential tensor of one field, not an assigned interaction strength, rate or synchronized particle phase.

An additional uniform complex phase is a separate symmetry: $\Psi\mapsto e^{i\alpha}\Psi$ sends $\mathcal J_Y\mapsto e^{3i\alpha}\mathcal J_Y$. Its real part would not preserve that symmetry. The scalar $|\mathcal J_Y|^2$ preserves both the uniform phase and the geometric charge, retaining the same channel zeros but changing the interaction order. Choosing either expression in a physical action is additional input, not a consequence of charge neutrality alone.

For the complete spatial Dirac field there is a further distinction. If the components retain a tangent-spin factor $S_B$, the normal contraction leaves

$$(\mathcal J_Y)_{ABC} =\det(T_{0,A},T_{1,B},\nabla_YT_{1,C}) \quad\text{in }S_B^{\otimes3}.$$

It is not a full spatial-spin scalar: the central element $-1$ acts oddly on three spinors. The auxiliary doublet used to construct $E$ does not replace those tangent indices. The invariant Hermitian norm $\|\mathcal J_Y\|^2_{S_B^{\otimes3}}$ contracts them and preserves both the normal charge and a uniform complex phase. For a factorized unit base polarization and the lowest mixed normal grades its angular integral is

$$\int_{S^5}\|\mathcal J_Y\|^2\,d\Omega =\frac{\pi^3}{24192},\qquad \mathcal J_Y(t)=\frac{t(1-t)(5t^2-3t+1)}{24}.$$

The earlier $\pi^3/180$ integrates the normal tensor component, not this full-spin scalar. For several base polarizations the algebraic tensor $\det(T_{0,A},T_{1,B},T_{1,C})$ can already be nonzero when $B\ne C$; it is antisymmetric in those indices. Thus a derivative is needed for the stated single-polarization example, not for every tensor-valued field. The sixth-order norm is an available invariant geometric density; no physical action or coefficient is selected.

Existence of a neutral cubic alone is not dimension-specific. For every odd $m\ge3$, $\Lambda^2W$ contracts with two copies of $\Lambda^{m-1}W\otimes D^{-1}\cong W^*$, at charges $2/m,-1/m,-1/m$. Even a same-exterior-degree version reappears at $m=3s$ for odd $s$, using degree $2s$: three complementary $s$-forms wedge to the determinant. Rank nine supplies an explicit comparison.

The complete number of these alternating fibre tensors can nevertheless be derived. With $F_m=\Lambda^{\rm even}W$, define the precise class

$$N_m=\dim_{\mathbb C}\operatorname{Hom}_{U(m)} \big(F_m\otimes\Lambda^2(F_m\otimes D^{-1}),\mathbb C\big).$$

For an exterior block $(k_1,k_2,k_3)$, central balance requires $k_1+k_2+k_3=2m$. The complementary degrees $l_i=m-k_i$ sum to $m$. Their dual forms wedge to $D^{-1}$, cancelling the remaining $D$. The coordinate torus forces each coordinate index to occur exactly once; permutations relate all ordered partitions. Thus the wedge functional is the unique invariant for every eligible ordered block. Unequal last-two degrees give one channel in the exterior square; equal degrees survive precisely when the complementary degree is odd. Counting these blocks gives

$$N_{2r}=N_{2r+1}=\left\lfloor\frac{(r+1)^2}{4}\right\rfloor.$$

For odd $m=2r+1$, write $l_i=2a_i+1$, so $a_1+a_2+a_3=r-1$. Unordered $a_2,a_3$ give $\sum_{s=0}^{r-1}(\lfloor s/2\rfloor+1)$, the displayed result. For even $m=2r$, the $l_i=2a_i$ instead require $a_2<a_3$, giving the same count. Hence $m=1,2,3,4,5,6,7,8,9$ have respectively $0,1,1,2,2,4,4,6,6$ channels. Within odd rank at least three, a single tensor of this class selects $m=3$; without that premise $m=2$ also survives. At rank two it is a cross-degree scalar pairing, not the rank-three triplet determinant. This counts the specified fibre tensors, not derivative operators or all possible interaction actions.

19. Round Curvature Further Constrains Polynomial Closure

On the exact round model write the actual ambient operator as $\mathcal D=A+P+C$, where $A$ is the horizontal Dirac operator, $P=\Gamma_BD_N$, and $C$ is section 14's curvature term. Put $k=R^{-2}$. Call the base Weyl block annihilated by $C$ dark. For every homogeneous normal degree, including arbitrary smooth base tensor coefficients, its induced twisting curvature has the opposite Hodge type. The twisted Lichnerowicz identity therefore gives

$$\|Au\|_B^2=\|\nabla u\|_B^2+3k\|u\|_B^2 \quad\text{on the dark block}.$$

No nonzero invariant polynomial space can remain wholly in that block. More generally, if a $\mathcal D$-invariant space has uniform normal degree at most $m$, its top coefficient must be dark because $C$ is invertible on the active block for $w\ne0$. Its derivative's degree-$m$ coefficient is active and must vanish by the same condition. Repeating on the degree filtration proves

$$\mathcal D^{m+1}=0$$

on that space. Thus a nonzero bounded-degree invariant space would contain a nonzero harmonic polynomial.

At degrees zero and one none exists. At degree one the coupled top equations for dark $u_1$ and active constant $v_0$ imply $A^2v_0+PCv_0=0$. The actual coefficient $1/4$ in $C$ cancels the horizontal normal-spin curvature exactly, leaving $(\nabla^*\nabla+3k)v_0=0$.

The quadratic case can also be excluded without restricting base coefficients. Its active linear coefficient $v_1$ must satisfy

$$(\nabla^*\nabla+3k)v_1=K v_1,\qquad K=-\frac12\sum_{i<j}b_ib_jL_{F_{ij}}-CP.$$

This is a parallel endomorphism on the complete complex-rank-96 coefficient bundle. The round normalization is $R^{\rm spin}_{ij}=k b_ib_j/2$, so $\sum_{i<j}b_ib_jR^{\rm spin}_{ij}=-3k$. The exact Clifford matrix gives

$$\operatorname{spec}(K/k)= \left\{0,-\frac12,-\frac32,\frac32,\frac92, \frac{1-\sqrt{10}}2,\frac{1+\sqrt{10}}2\right\}.$$

For clarity the algebra certificate consists of two copies of each of the following degree-24 characteristic polynomials:

$$\frac{\lambda^8(2\lambda-9)(2\lambda-3)^3 (2\lambda+1)^9(2\lambda+3)^3}{65536}, \qquad \frac{\lambda^8(2\lambda+1)^8 (4\lambda^2-4\lambda-9)^4}{65536}.$$

All coefficient isotropy types are $(j,0)$ under $Spin(4)$, or $(0,j)$ after orientation reversal. The homogeneous-bundle rough Laplacian is the $Spin(5)$ Casimir minus the fibre Casimir:

$$C_5(\lambda)=\lambda_1(\lambda_1+3) +\lambda_2(\lambda_2+1),\qquad C_4(j,0)=2j(j+1).$$

Weights $\lambda_1,\lambda_2$ are both integral or both half-integral, with the parity of $j$ fixed by the common central character. In the half-integral case the two Casimirs are integers plus $5/2$ and $3/2$ respectively; their difference is an integer. Thus every rough Laplacian eigenvalue is a nonnegative integer multiple of $k$. Parallelness of $K$ lets the necessary equation be checked on each base eigenspace. Its left eigenvalue is an integer at least $3k$; the only right eigenvalue above $3k$ is the noninteger $9k/2$. The spectra are disjoint, so $v_1=0$, then $u_2=0$ by the dark identity.

The Casimir construction is standard homogeneous-bundle mathematics; see Goodman and Wallach, Symmetry, Representations, and Invariants. The normalization, parity calculation and exact coefficient matrix above are the application to this connection metric. No finite base spectral cutoff enters the proof.

Hence the flat polynomial closure does not transfer to round curvature, even allowing coupled chiralities through normal degree two. At higher degrees the orbital and $CP$ terms remain; no general no-go is claimed. Polynomial sections over the noncompact normal fibres are not localized particles or a physical mass spectrum.

20. A Quartic Identity And An Exact Cubic Sum Rule

Retaining the tangent-spin factor of the original charge carrier also supplies lower-field-order scalars preserving spatial-spin covariance and uniform phase. Let $B$ be its base-spin factor of complex rank $r$ and write one field's two branch components as $X\in B\otimes F$ and $Y\in B\otimes F\otimes D^{-1}$. These are the even charge components inside the full normal Clifford completion. Their use in a geometric tensor does not assert that the even projection is preserved by Clifford differentiation. Their Hermitian coherence is

$$K=\langle X,Y\rangle\in D^{-1},\qquad I_K=|K|^2.$$

The vacuum and triplet contributions take values in the same determinant line, so $K=K_s+K_t$ and

$$|K|^2=|K_s|^2+|K_t|^2+2\operatorname{Re}(\bar K_sK_t).$$

This neutral quartic contraction retains their relative interference without an extra interaction field. It selects no action coefficient.

In a unit base frame let $x_A,y_B$ be the normal components and retain both base indices of the frame tensor $T_{AB}=\beta(x_A,y_B)$. Define

$$Q=\|T\|^2,\quad N_0=\sum_A\|x_A\|^2,\quad N_1=\sum_B\|y_B\|^2,\quad M_{AB}=\langle x_A,y_B\rangle.$$

The ordinary exterior Gram identity, summed over the two base indices, gives

$$Q=N_0N_1-\|M\|_{\rm HS}^2, \qquad K=\operatorname{tr}M,$$

$$\boxed{Q+\frac{|K|^2}{r} +\left\|M-\frac Kr I_r\right\|_{\rm HS}^2=N_0N_1}.$$

All three terms are global invariant quartic scalars. Equivalently $\|M\|_{\rm HS}^2=\operatorname{tr}(\rho_0\rho_1)$ with $\rho_0=\sum_Ax_Ax_A^\dagger$ and $\rho_1=\sum_By_By_B^\dagger$. These are normal reduced density endomorphisms, without assigning quantum probabilities or physical amplitudes. The aggregate $Q$ need not choose one ordinary normal plane.

For factorized branches, a common base-spin ray recovers the simpler $Q+|K|^2=N_0N_1$. This is sufficient, not necessary; orthogonal normal supports also give $M=K=0$. The simpler identity fails for general correlated components: $x=y=(e_0,e_1)$ gives $(N_0N_1,|K|^2,Q)=(4,4,2)$. Orthogonal base polarizations with the same normal column instead give $K=Q=0$. Thus vanishing coherent contact alone does not establish a nonzero normal frame. For identical normalized arrays, the exact specialization is

$$Q=1-\operatorname{tr}(\rho_N^2),\qquad 0\le Q\le1-\frac1{\min(r,4)}.$$

The upper bound follows from the squared Schmidt weights and is attained by equal weights. It is $1/2$ for a rank-two chiral base factor and $3/4$ for a rank-four Dirac factor. These are field-component ranks, not physical spatial dimensions.

There is a direct sum rule connecting this quartic frame scalar to section 18's cubic angular coefficients. In the determinant-framed normal model use $V_n=M_{2,n}$ in one branch and $V_l$ in the other, with a common unit base polarization. Put

$$W_{nl}=h_nf_nb_l-h_lf_lb_n.$$

Then $Q_{nl}(t)=t(1-t)W_{nl}(t)^2$. The $f_k$ form an orthogonal polynomial basis for the weight $t(1-t)^2$, and

$$J_k=\int_{S^5}\sigma\rho F_k^2\,d\Omega =\frac{\pi^3}{(k+1)(k+2)(k+3)},\qquad C_{knl}=\frac12\int_{S^5}\sigma\rho F_kW_{nl}\,d\Omega.$$

Expanding the finite polynomial $W_{nl}$ in this basis and applying orthogonality therefore proves

$$\frac1{\pi^3}\int_{S^5}Q_{nl}\,d\Omega =4\sum_k(k+1)(k+2)(k+3) \left|\frac{C_{knl}}{\pi^3}\right|^2 =24\sum_k\binom{k+3}{3} \left|\frac{C_{knl}}{\pi^3}\right|^2.$$

The finite sum uses the same harmonic triangle as the cubic vertex. For $(n,l)=(0,1)$ the coefficients already displayed in section 18 give the exact frame integral $53\pi^3/241920$. The binomial factor is the inverse Jacobi norm written combinatorially; it does not identify $k$ with a particle index or select a depth-three physical sector.

Different grades have zero integrated bilinear overlap, but their quartic coherence remains positive: at $t=0$, $d^2_{nl}(0)=1/(4(n+2)(l+2))\ne0$. Continuity gives a positive integral of its squared modulus. Thus angular orthogonality, a quartic contact zero and a frame zero are distinct tests. The same polynomial basis, determinant line and complete field carrier constrain all three.

The normal charge, rotation and oriented-volume note adds an exact character comparison and a geometric reading of the charge weights. The rank-three even carrier alone has a nonzero cubic trace among $m\ge2$; its full Clifford completion cancels that trace. Every rank-three chiral polarization determines an orthogonal normal complex structure, whereas an explicit fixed-charge state at every higher rank can fail that property. Definite-grade charge weights measure relative normal rotation alignment with the existing determinant-branch shift. Correlated frame tensors also show that complete oriented volume is not determined by their scalar norm. These are conditional consequences of the same normal construction; they do not select physical particle charge or a new interaction coefficient.

Volume Contraction and Pair Response

The determinant-contact calculation reuses the exterior Gram identity at the level of a complete color-mode carrier. A rank-three volume-squared term, with one isotropic mode slot contracted, gives $\|a\|^2\|b\|^2-|a^\dagger b|^2$ and hence the two-slot antisymmetrizer. At general rank the full three-form norm gives the same expression multiplied by $k-2$. This supplies a concrete interaction interpretation and a higher-rank control for the invariant contractions used here, while retaining the distinction between the normal spinor carrier and a physical color multiplicity space.

21. Interpretation And Reproduction

These results provide a conditional normal-dimension discriminator and explicit common geometric realizations. They also constrain how charge, polarization and sector projections can coexist. They do not establish the previously proposed preference at four of the six assigned particle depths. The sector-case paper places this evidence beside the separately scoped mass comparisons.

The single exact checker and recorded results verify representation signs, projectors, covariant maps, sharp-bound witnesses, curvature and competing examples. The all-dimension and global conclusions also depend on the proofs above; a check count is not a count of independent empirical predictions. No particle evolution is required for these tests.

A Pure Pencil Can Supply the Four-Base Premise

The pure spatial spinor pencil derives a real four-plane from two orthogonal pure polarizations whose entire span is pure. Applying section 1's universal alternating normal-vector rule to its complement selects parent depth ten. The fixed/projective pair stabilizer gives the determinant-trivial two-plus-three complex split and its known charge character. A positive mode-polarization potential realizes the pencil as a zero-energy condition, without selecting ten on its own. Its normal six-plane is distinct from the nested physical depth-six sector, and the pair shares one normal vacuum rather than supplying the additional framed normal line. Full higher-rank carriers and fixed-sector rotation controls remain explicit.

An Explicit Induced U3 Graph and Its Controls

The nested normal-curvature note gives a second common spatial realization, using one invertible graph map in ordinary ten-space. The complex second fundamental count selects a four-dimensional spatial base with six normals under complete quadratic normal generation, and actual curvature generates U3 there and on its three-sector pullback. Exact higher-jet controls show that nesting alone permits larger normal holonomy at depth two. A parallel quadratic tangent-to-normal identification would reduce the holonomy and lose full U3, so its dynamics cannot be inferred from the count alone.