Parallel Quadratic Normal Geometry and a Distinct Triplet Algebra

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 quadratic Veronese embedding

$$\nu:\mathbb{CP}^{2}\longrightarrow\mathbb{CP}^{5},\qquad [z]\longmapsto[z\odot z]$$

has a parallel, everywhere invertible complex second fundamental map $B:\operatorname{Sym}^{2}T^{1,0}\mathbb{CP}^{2}\to N^{1,0}$. Its normal bundle is genuinely $\operatorname{Sym}^{2}T^{1,0}\mathbb{CP}^{2}$, with $c_1=9H$ and $c_2=30H^2$. Its normal holonomy is the four-dimensional image $\operatorname{Sym}^{2}U(2)$, conjugate in $U(3)$ to $U(1)\times SO(3)$. The connected traceless holonomy is $SO(3)$, with Lie algebra $\mathfrak{su}(2)$, rather than the full eight-dimensional $\mathfrak{su}(3)$.

Removing the determinant line yields a globally defined rank-three bundle with a parallel symmetric scalar pairing. This extra invariant distinguishes its spin-one triplet from the fundamental triplet of full SU3. The original normal bundle's pairing is line valued; its full unitary holonomy has no untwisted scalar in $N\otimes N$.

There is no normal-section obstruction to placing this four-dimensional spatial base in a smooth 5/6/10 spatial flag. A complete ordinary ten-dimensional extension, retaining the exact Veronese metric near the base, supplies a complete occupied 2/3/4/5/6/10 flag. This extension changes the global ambient manifold from CP5 to the total normal bundle. Smooth intermediate normal lines are not parallel lines, and neither assertion proves closed global intermediate inclusions inside unchanged CP5.

1. Geometry and notation

Take a Hermitian vector space $V=\mathbb C^3$, write $B=\mathbb P(V)=\mathbb{CP}^2$, and let

$$0\longrightarrow L\longrightarrow V\otimes\mathcal O_B \longrightarrow Q\longrightarrow0$$

be the tautological sequence. Here $L=\mathcal O(-1)$, $Q$ has complex rank two, $H=c_1(L^*)$, and $\int_BH^2=1$. All complex coordinates pair ordinary real spatial directions. Thus the base has four spatial directions and its parent has ten; these are not internal or compactified dimensions. The Fubini–Study metric scale is free. Compact global topology alone does not fix a small spatial scale.

Use the Hermitian tensor metric on $\operatorname{Sym}^2V$. In orthonormal homogeneous coordinates the embedding is

$$[z_0:z_1:z_2]\longmapsto [z_0^2:\sqrt2z_0z_1:\sqrt2z_0z_2:z_1^2:\sqrt2z_1z_2:z_2^2].$$

The square-root factors matter for the standard Fubini–Study isometry statements. With the same standard normalization on source and target, $\nu^*g_{FS,5}=2g_{FS,2}$. A constant rescaling does not change the tangent connection or its holonomy.

At a line $L_p$, the ambient tangent space is

$$T_{\nu(p)}\mathbb P(\operatorname{Sym}^2V) =\operatorname{Hom}(L_p^2,\operatorname{Sym}^2V/L_p^2).$$

The tangent image comes from $L_p\odot V/L_p^2\cong L_p\otimes Q_p$. Quotienting the ambient tangent space by it therefore gives the canonical holomorphic normal-bundle identity

$$N\cong\operatorname{Hom}(L^2,\operatorname{Sym}^2Q) =L^{-2}\otimes\operatorname{Sym}^2Q =\operatorname{Sym}^2(L^{-1}\otimes Q) =\operatorname{Sym}^2T^{1,0}B.$$

Consequently a formula $N=\operatorname{Sym}^2Q$ without $L^{-2}$ misses a genuine projective line factor. In terms of the actual tangent bundle there is no additional twist.

In the affine chart $z_0=1$, the tangent coordinates are $(\sqrt2z_1,\sqrt2z_2)$ and the normal quadratic coordinates are $(z_1^2,\sqrt2z_1z_2,z_2^2)$. Their Hessian spans all three complex normals. The second fundamental map is an isomorphism at this point, and unitary homogeneity makes it an isomorphism everywhere.

2. Why the second fundamental map is parallel

There is a short symmetry proof. At $[1:0:0]$, the unitary map $\operatorname{diag}(1,-1,-1)$ of $V$ preserves the Veronese image. Its induced projective isometry acts by minus the identity on the tangent directions $L\odot Q$, and by plus the identity on the normal directions $\operatorname{Sym}^2Q$.

The covariant derivative $\nabla B$ has three tangent arguments and one normal output. Invariance under this isometry makes it equal to its negative at the fixed point, so it vanishes there. Homogeneity gives $\nabla B=0$ on all of CP2. This establishes parallelity for the actual induced tangent and normal connections, rather than inferring it from constant quadratic coordinate coefficients.

The primary normal-holonomy calculation of Console and Di Scala, Parallel submanifolds of complex projective space and their normal holonomy independently supports this example: Table 1 gives the Veronese normal action as the isotropy action of $Sp(n)/U(n)$, and Proposition 2.4 identifies normal holonomy with the effective slice action. For $n=2$, that action is the symmetric-square representation used here. Their theorem concerns the Fubini–Study geometry; it does not assert full U3 for this parallel example.

3. Characteristic classes and determinant lines

The Euler sequence gives

$$c(TB)=1+3H+3H^2.$$

For formal tangent Chern roots $x,y$, the normal roots are $2x,x+y,2y$. Therefore

$$c_1(N)=3(x+y)=9H,$$ $$c_2(N)=2(x+y)^2+4xy=30H^2, \qquad \det N=(\det TB)^3=\mathcal O(9).$$

An independent ambient check uses $\nu^*\mathcal O_{CP5}(1)=\mathcal O_B(2)$:

$$c(N)=\frac{(1+2H)^6}{(1+H)^3} =1+9H+30H^2\quad\text{in }\mathbb Z[H]/(H^3).$$

Let $D=\det TB=\mathcal O(3)$. Unlike an arbitrary determinant cube-root choice, this geometry supplies the particular root $D^3=\det N$ directly. The determinant-free normal bundle is

$$W=N\otimes D^{-1} =\operatorname{Sym}^2TB\otimes(\det TB)^{-1}, \qquad c_1(W)=0,\quad c_2(W)=3H^2.$$

As a bundle with connection,

$$W\cong\mathfrak{su}(TB)\otimes_{\mathbb R}\mathbb C.$$

Indeed the algebraic rank-two identity $\operatorname{Sym}^2E\otimes(\det E)^{-1}\cong\operatorname{End}_0E$ identifies the compact real form with $\mathfrak{su}(E)$. It follows that

$$p_1(\mathfrak{su}(TB))=c_1(TB)^2-4c_2(TB)=-3H^2, \qquad w_2(\mathfrak{su}(TB))=H\bmod2.$$

Thus its SO3 frame bundle does not lift to a global SU2 principal bundle on CP2. The phrase “SU2 spin one” describes the representation of the covering group and the Lie algebra correctly; its effective connected group on this bundle is SO3. This global qualification does not invalidate the three-dimensional representation.

4. Actual holonomy and the additional pair invariant

The Fubini–Study tangent holonomy of CP2 is U2. For example, its complex curvature endomorphisms, up to an overall nonzero scale, are $E_{ij}+\delta_{ij}I$; these span the complexification $\mathfrak{gl}(2,\mathbb C)$. Parallelity and invertibility of $B$ identify normal parallel transport with the symmetric square of tangent parallel transport. Consequently

$$\operatorname{Hol}(N)=\operatorname{Sym}^2U(2) \cong U(2)/\{\pm I\}\cong U(1)\times SO(3).$$

CP2 is simply connected, so the full and restricted normal holonomy groups coincide here. The last group is embedded as scalar phases times the complexified real standard SO3 action. A central tangent phase $e^{i\theta}I_2$ acts on $N$ by $e^{2i\theta}I_3$. The connected commutator group is SO3, and after removing the determinant connection the holonomy on $W$ is precisely SO3 inside SU3. The normal algebra has dimension four and its traceless algebra dimension three. No commutator closure of these three generators produces the five missing SU3 directions.

To see the extra invariant explicitly, use the basis $(u^2,uv,v^2)$ for $\operatorname{Sym}^2\mathbb C^2$. The infinitesimal action of $A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ is

$$\rho(A)=\begin{pmatrix}2a&b&0\\2c&a+d&2b\\0&c&2d\end{pmatrix}.$$

For $q=a_0u^2+b_0uv+c_0v^2$, the discriminant $b_0^2-4a_0c_0$ has character $(\det g)^2$. Its polarized form has matrix

$$K=\begin{pmatrix}0&0&-2\\0&1&0\\-2&0&0\end{pmatrix}, \qquad \rho(A)^TK+K\rho(A)=2\operatorname{tr}(A)K.$$

Geometrically this is a nondegenerate parallel symmetric pairing

$$b:\operatorname{Sym}^2N\longrightarrow D^2=\mathcal O(6).$$

It is not a parallel scalar-valued pairing on the untwisted normal bundle. After tensoring both inputs by $D^{-1}$, it becomes a parallel scalar-valued pairing on $W$. Equivalently the traceless normal connection preserves a complex symmetric quadratic form as well as its Hermitian metric and volume.

The distinction in tensor products is exact:

$$\mathbf3_{\mathrm{spin1}}\otimes\mathbf3_{\mathrm{spin1}} =\mathbf1_s\oplus\mathbf3_a\oplus\mathbf5_s,$$ $$\mathbf3_{SU3}\otimes\mathbf3_{SU3} =\mathbf6_s\oplus\overline{\mathbf3}_a.$$

Only the first has a singlet. Including the full U1 normal phase again removes the untwisted scalar singlet. This observation distinguishes bilinear pair contractions from the Hermitian singlet in $\mathbf3\otimes\overline{\mathbf3}$, which exists in both cases, and from determinant contractions, which also occur in both determinant-free groups.

5. Smooth intermediate sectors and a complete occupied flag

A tubular flag inside the fixed CP5 geometry

The complex normal rank is three and the real base dimension four. A smooth section of $N$ transverse to its zero section would have zeros of codimension six. Since six exceeds four, such a section has no zeros. This supplies a nowhere-zero smooth normal section $s$. Normalize it using the normal metric.

Its real span $\ell_{\mathbb R}=\mathbb R s$ and complex span $\ell_{\mathbb C}=\mathbb C s$ are globally trivial smooth subbundles. A tubular neighborhood of the compact Veronese base embeds a sufficiently small normal disk bundle in CP5. Restricting it to these two subbundles gives global neighborhoods of the entire base with real dimensions five and six:

$$B\subset\operatorname{Disk}(\ell_{\mathbb R}) \subset\operatorname{Disk}(\ell_{\mathbb C}) \subset\operatorname{Disk}(N_{\mathbb R})\subset CP5.$$

These are smooth inclusions. The complex line subbundle need not be holomorphic, and its tubular image is not asserted to be a complex submanifold. There is no contradiction with $c_1(N)=9H$ and $c_2(N)=30H^2$: a trivial complex line splits off smoothly, leaving a rank-two bundle with those two nonzero classes. Topological existence of a section neither trivializes the full normal bundle nor supplies a parallel section.

In fact neither the real line nor the complex line is preserved by the normal connection. The irreducible SO3 triplet, together with the central phase in the original bundle, permits no parallel proper normal line. Imposing a parallel line as an additional requirement would exclude this example. A mere smooth nested flag does not impose that requirement.

Open finite-radius tubular sectors generally have incomplete induced metrics. Their existence does not solve a request for complete intermediate sectors inside the unchanged compact parent. No closed global 5/6 inclusion in unchanged CP5 is claimed here.

Complete ambient extension preserving the Veronese neighborhood

There is nevertheless a concrete complete ordinary spatial realization. Set

$$X=\operatorname{Tot}(N_{\mathbb R})$$

and pull the CP5 metric back by the normal tubular exponential map on a small disk bundle; denote this metric by $g_V$. Choose the complete connection metric

$$g_0=\pi^*g_B+|dy+A_Ny|^2$$

on all of $X$, where $A_N$ is the Veronese normal connection. It is complete: finite length bounds both base distance and fibre norm, and bounded-radius disk bundles over compact $B$ are compact.

For a smooth radial cutoff $\chi$ equal to one near the zero section and zero before leaving the tubular chart, define

$$g=\chi g_V+(1-\chi)g_0.$$

The interpolation occurs on a compact disk bundle, so $g\ge c g_0$ for one $c>0$. Thus $g$ is complete. Near $B$ it is exactly the pulled-back CP5 metric. In particular the second fundamental form, its parallelity and the normal connection along the four-dimensional spatial base are unchanged. Using the unmodified connection metric alone would make the zero section totally geodesic; the interpolation retaining $g_V$ near the base is essential.

The entire subbundles $\operatorname{Tot}(\ell_{\mathbb R})$ and $\operatorname{Tot}(\ell_{\mathbb C})$ are closed embedded submanifolds of this complete ambient. Their induced metrics are complete: an intrinsic Cauchy sequence is ambient Cauchy, its ambient limit remains in the closed submanifold, and local smooth coordinates give convergence in the induced metric.

Choose a small smooth $S^3$ in an ordinary coordinate chart of $B$, and an equatorial $S^2$ inside that sphere. Their compact induced metrics are complete. Then one common spatial geometry contains the complete occupied flag

$$S^2\subset S^3\subset\mathbb{CP}^2 \subset\operatorname{Tot}(\ell_{\mathbb R}) \subset\operatorname{Tot}(\ell_{\mathbb C}) \subset\operatorname{Tot}(N_{\mathbb R}),$$

with real dimensions $2,3,4,5,6,10$. The lower spheres generally bend and the chosen upper lines generally rotate under normal transport. Neither fact obstructs smooth nesting. The construction does not independently choose incompatible geometries at different depths.

This ambient is globally $\operatorname{Tot}(N_{\mathbb R})$, rather than CP5 or Euclidean ten-space. It demonstrates a complete allowed curved-parent flag, not a complete extension inside the original CP5. It also supplies no claim about the color holonomy of every lower sector or its physical coupling; those require their own specified carrier and restriction rule.

Why the older projective obstruction is different

A real five-dimensional intermediate manifold containing the whole CP2 would require a real normal line subbundle; over simply connected CP2 such a line is trivial and would provide a forbidden nowhere-zero normal section.

The Veronese normal bundle has real rank six, so that rank-two Euler obstruction does not apply. Likewise the chosen lower $S^2$ is a small null-homologous sphere in a coordinate chart, not the projective $CP1\subset CP2$. Requiring a global real-three intermediate around that projective CP1 would reintroduce its own nonzero normal Euler obstruction. The full flag above does not make that additional choice.

6. What the comparison teaches about color selection

An invertible quadratic normal map and three complex normal directions do not by themselves select full SU3. The parallel curved example has a smaller holonomy and an additional symmetric pair invariant. The maintained flat-ambient graph instead has nonparallel bending and full U3 normal holonomy. Parallelity therefore separates two materially different geometric realizations of the same pointwise rank match.

If full SU3 is the desired traceless normal algebra, maintaining a parallel isomorphism $\operatorname{Sym}^2T\to N$ at complex tangent rank two prevents it in any ambient geometry: tangent holonomy has dimension at most four and its symmetric-square image remains that small. The canonical curved example realizes this restriction exactly. Full SU3 requires relaxing that parallel identification or changing the proposed carrier/connection; a smooth nested flag alone selects neither possibility.

This is useful geometric discrimination, not a derivation of higher-sector force absence, an observed coupling strength or quark occupation. CP2's compact topology, its complex structure, the Fubini–Study metric and the choice of quadratic Veronese embedding remain conditional geometric premises.

Reproduction

The exact checker and report preserve 55 exact checks of the Lie homomorphism, four/three algebra dimensions, central and determinant characters, discriminant covariance, tensor singlets and characteristic classes. The tubular existence and completeness arguments are given above; a finite matrix check does not replace them. The original independent derivation, primary paper bytes and input hashes are preserved separately.