Classical Color-Current Cancellation and Complete Spatial Dynamics

The zero-dielectric Gauss constraint requires the actual classical density and spatial currents to vanish. A fibre singlet is sufficient, but is not necessary. The distinction matters when interpreting inactive sectors: a zero source does not, by itself, prove that an occupied field transforms in the trivial representation.

This calculation uses the actual rank-32 Clifford carrier of the selected ten-parent, with its full normal SU3 at depths two through four, residual SU2 at five/six, and no independent normal algebra at ten. All dimensions remain ordinary spatial dimensions. Exterior factors describe field components. No particle identity, localized solution, mass-frequency or physical occupation rule is inferred from these free-mode controls.

The Current of a Positive Spatial Mode

Let the Hermitian generators commute with every retained spatial Clifford matrix:

$$[T_A,\gamma_i]=0,\qquad \{\gamma_i,\gamma_j\}=2\delta_{ij}I.$$

For a nonzero momentum $p$, put $n=p/|p|$ and $P_+=(I+\gamma\cdot n)/2$. The Clifford relation gives

$$P_+\gamma_iP_+=n_iP_+.$$

Consequently any $\psi=P_+\psi$ obeys

$$q_A=\psi^\dagger T_A\psi,\qquad j_{iA}=\psi^\dagger\gamma_iT_A\psi=n_iq_A.$$

Zero density therefore cancels every spatial current on that positive mode without discarding a spatial derivative. Its bare energy is $|p|\|\psi\|^2>0$. This is a principal-symbol result. It does not guarantee that a superposition of different source-free modes is source-free, because the currents are quadratic.

A Complex-Linear Canceled Branch at Two, Three and Four

Use $\Sigma_{10}=\Sigma_4\otimes\Lambda^*\mathbb C^3$ and an occupation basis on the normal exterior factor. Take complementary states of normal degree one and two:

$$\eta=\frac1{\sqrt2}(|100\rangle+|011\rangle),\qquad V\chi=\chi\otimes\eta.$$

The normal SU3 generators preserve exterior degree. The two terms have opposite diagonal weights and individually zero off-diagonal expectations, so

$$V^\dagger V=I_4,\qquad V^\dagger T_AV=0,\qquad V^\dagger C_3V=\frac43I_4.$$

Thus the branch has nonzero SU3 Casimir despite zero classical density. At every $d\le4$ in the maintained comparison,

$$\gamma_iV=V\gamma_i^{(4)},\qquad V^\dagger\gamma_iT_AV=0\quad(i\le d).$$

For an arbitrary spinor function $\chi(x,t)$, including interference between different positive-energy momenta, $V\chi$ has zero local color currents. The free spatial operator preserves this complex-linear rank-four branch, and its first-order kinetic form is the full nondegenerate form of $\chi$. Taking the collective connection to vanish supplies a consistent free source-free solution of the declared constrained action. Arbitrary nonflat connections or other unspecified interactions need not preserve this fixed branch.

This is not the imposed reality relation between two time-dependent conjugate amplitudes. Both components share the same independent complex envelope. Nor is it a tensor singlet: color rotations generally move the normal line, although its entire gauge orbit has the same vanishing mean source.

There is an electric-character limitation. The normal number operator $N_\perp$ has mean $3/2$ and variance $1/4$ on this line. It is not one normal central-weight eigenline. A phase generator proportional to $N_\perp$ would therefore not assign it a single electric charge. The physical electric generator and any common character must be checked separately before using this branch for an assigned particle.

A Fixed Pure Triplet Has an Exact Positive-Mode Source Floor

A positive base symbol has only two spin multiplicities. A pure state in one fixed normal triplet therefore has a reduced color density matrix $\rho_c$ of trace one and rank at most two. Fundamental completeness gives

$$\sum_Aq_A^2=\frac12\left(\operatorname{tr}\rho_c^2-\frac13\right) \ge\frac1{12}.$$

The lower bound is attained at eigenvalues $(1/2,1/2,0)$. A pure positive-mode triplet alone cannot satisfy the exact zero-source constraint at depth two. The direct-sum construction above avoids that obstruction by using the complementary exterior representation. At active three/four no zero-source constraint is imposed; their ordinary triplet sources remain allowed.

Residual SU2 at Five and Six

The residual fundamental has four positive spatial spin multiplicities. Choose orthonormal vectors $u_1,u_2$ in the simultaneous positive spatial-symbol and $T_z=1/2$ subspace. With $T_-=T_x-iT_y$, define

$$\psi_+=\frac1{\sqrt2}(u_1+T_-u_2).$$

Orthogonality makes the off-diagonal color means vanish, and equal weights cancel $T_z$. Thus

$$q_A=j_{iA}=0,\qquad \langle C_2\rangle=\frac34,\qquad \|\psi_+\|^2=1.$$

A common independent complex amplitude retains a positive kinetic norm. The construction works for arbitrary nonzero momentum directions at both physical depths. It proves existence of source-free positive modes in a nonsinglet carrier, rather than a complete finite-energy localized branch.

Two independently chosen such modes generally acquire nonzero cross-current terms when coherently superposed. The executable checks exhibit this directly. A one-direction transport envelope can retain the cancellation, but gives no localization in the transverse directions. A full spatial wave packet needs further conditions.

There is also a precise fixed-subspace obstruction. On the colored residual carrier, the complex algebra generated by the five spatial Clifford matrices is $M_4(\mathbb C)\oplus M_4(\mathbb C)$; with six matrices it is $M_8(\mathbb C)$. Any nonzero complex-linear invariant subspace under this complete algebra contains a spin block times a nonzero pure fundamental color vector. Such a color vector obeys

$$\sum_{a=1}^3q_a^2=\frac14\|\eta\|^4>0.$$

Hence there is no nonzero fixed colored complex-linear subspace on which all density compressions vanish while preserving the complete spatial Clifford action. This does not exclude a nonlinear source-free solution set, a momentum-dependent constrained field, or the actual residual fibre-singlet branches already proved to preserve all spatial directions.

Antiunitary Pairing Repeats the Symplectic Obstruction

On the eight-component six-dimensional spatial spinor, take $C=BK$ with complex conjugation $K$ and

$$B=\gamma_1\gamma_3\gamma_5,\qquad C^2=-I,\qquad C\gamma_i=\gamma_iC\quad(i\le6).$$

The real-linear relation $\Psi=(f,Cf)$ in the residual fundamental cancels all density and spatial-current expectations at five/six. Its norm is positive, $\|\Psi\|^2=2\|f\|^2$. Nevertheless its first-order symplectic form vanishes:

$$\operatorname{Im}\langle(\delta_1f,C\delta_1f),(\delta_2f,C\delta_2f)\rangle=0.$$

Conjugating the full spatial wave reverses momentum and temporal phase. The positive bare energy of one wave and the negative bare energy of its constrained conjugate cancel. This reality relation therefore does not supply nondegenerate positive first-order particle dynamics. The independent common-amplitude positive mode above has different constraints and avoids that particular cancellation; it does not solve the packet problem.

All Occupied Sectors and Physical Interpretation

The symmetric Hodge rule still distinguishes the occupied quark depths in the declared action. The more careful conclusion about inactive depths is absence of a permitted classical source, not proof of a unique fibre-singlet occupation. Classical expectation cancellation, quantum Gauss invariance, electric charge and a physical strong response remain different questions.

The checker and report preserve 328 current and kinetic checks across 32 cases, plus 192 checks in the exact imported geometry dependency. They include every occupied depth, full generator and spatial matrices, mixed-momentum interference, exact source floors, positive-mode cancellation, complete Clifford algebra ranks and symplectic controls. No particle simulation or fitted sector coefficient is used.