A rank-independent current excess for same-field pair response

The cubic current excess gives a positive prepared-pair angular cost but changes the isolated source when a lower-rank current is simply inserted in a larger matrix. The following polynomial removes that algebraic defect without choosing a rank in its definition. The insertion is a conditional map of representation coefficients; the physical spatial map between nested sectors remains undetermined.

Polynomial and source

For a Hermitian Lorentz-scalar current $C$, put $T=\operatorname{Tr}C$ and $p_j=\operatorname{Tr}C^j$. Define

$$\boxed{K(C)=T p_3-p_2^2 =\sum_{i<j}\lambda_i\lambda_j(\lambda_i-\lambda_j)^2,} \tag{1}$$

where the eigenvalues include zeros. Thus $K\ge0$ on positive semidefinite currents. It can be negative on indefinite currents; for eigenvalues $2,-1$, it equals $-18$. It is nevertheless a smooth polynomial through $C=0$. Its complete matrix gradient is

$$\boxed{\nabla K=p_3 I+3T C^2-4p_2 C.} \tag{2}$$

For any occupied-rank $s$ projector $P_s$, $C=\rho P_s$ gives $K=0$ and $(\nabla K)P_s=0$, independent of the surrounding matrix rank. The gradient on the unoccupied complement is $s\rho^3 I$; thus the complete field Hessian may change, even though the occupied isolated mass source does not. The joint calculation and report check (1), (2), their full noncommuting derivatives, and all maintained representation depths 2, 3, 4, 5, 6 and 10. The largest relative gradient-derivative error in the random indefinite controls is below $2.4\times10^{-9}$.

There is an exact minimality result within unitary-invariant polynomials of this one current. Evaluate a homogeneous polynomial on $C=\rho P_s$ for every positive occupied rank $s$. Up to degree three, the invariant bases evaluate respectively as $s$, $(s^2,s)$, and $(s^3,s^2,s)$, times the appropriate power of $\rho$. Their Vandermonde matrices have full column rank, so no nonzero combination vanishes on all projector rays. At degree four the basis $(T^4,T^2p_2,p_2^2,Tp_3,p_4)$ evaluates as $(s^4,s^3,s^2,s^2,s)\rho^4$. Its nullspace is exactly the span of $Tp_3-p_2^2$. Thus (1) is the unique lowest-degree nonzero projector-transparent invariant polynomial of a single current, up to its coefficient. Since $C$ is quadratic in the field, this correction is eighth order in the field. This result does not exclude lower-order laws using the other currents, spatial derivatives or nonpolynomial functions. The batch verifies these evaluation-matrix ranks exactly using rational arithmetic.

An inverse action trial can add one coefficient $\eta K(C_k)$ to each shell's existing trace current. Since $K$ is fourth order in $C$, $\eta$ has units of inverse current cubed. The common shell weights, frequency unit, canonical norm, and separate nonlinear binding remain premises. No master equation here fixes $\eta$ or unifies it with the trace term.

Same-depth prepared pair

For two currents sharing a partner frame, let $C_k=I_r\otimes(\rho P_z+\sigma P_w)$, $s=|z^\dagger w|^2$, and $L=\rho+\sigma$. The two ray eigenvalues have product $P=\rho\sigma(1-s^k)$. Direct substitution in (1) yields

$$\boxed{K(C_k)=r^2P(L^2-4P)\ge0.} \tag{3}$$

For $s=\cos^2\theta$, its small-angle term is $r^2\rho\sigma L^2 k\theta^2$. Thus alignment is an angular minimum for every positive-order shared shell. The exact response has a useful decomposition. With $u=s^k$,

$$K=r^2\rho\sigma(\rho-\sigma)^2(1-u) +4r^2(\rho\sigma)^2u(1-u). \tag{3a}$$

For equal densities, only the second term remains. Since $\|[P_z^k,P_w^k]\|_{\rm HS}^2=2u(1-u)$, its prepared-pair energy is $2r^2\rho^4\|[P_z^k,P_w^k]\|_{\rm HS}^2$. The current therefore penalizes noncommuting ray projectors in this restricted preparation. Orthogonal equal-density rays also have $K=0$, whereas unequal orthogonal rays have $K=r^2\rho\sigma(\rho-\sigma)^2$. Therefore this algebraic energy alone does not guarantee that two freely evolving patterns align; it can have another stationary configuration and a barrier. For equal densities the barrier occurs at $\cos^{2k}\theta=1/2$ and has height $r^2\rho^4$ per shell. On the alignment side of this angle the energy slope tends to restore alignment; on the orthogonal side it tends toward orthogonality. If two ray angles are allowed to respond in one plane, their derivatives of this shared energy are equal and opposite. Their actual angular kinetic terms and spatial coupling have not been derived. Spatial overlap, phases in the full fields, and reciprocal shape motion must be included before interpreting a rate or force.

Conditional cross-depth insertion

Suppose a rank-$s$ lower current $\sigma P_s$ is zero padded in a rank-$r$ parent. Equation (2) gives zero source on the occupied lower subspace for every $s\le r$, unlike the cubic excess. For a nested commuting prepared pair $C=\rho P_r+\sigma P_s$, the exact cross term is

$$\boxed{K(C)=s(r-s)\rho(\rho+\sigma)\sigma^2.} \tag{4}$$

The batch checks the isolated identity and (4) for all fifteen ordered depth pairs. The largest absolute formula error in those double-precision controls is below $2.3\times10^{-13}$. This is an algebraic condition on a specified representation embedding. It is not a derivation of how independent physical particle patterns in nested spatial sectors couple. In particular, (4) begins quadratically in the lower amplitude and has no orientation dependence for rotations entirely inside a filled parent projector. With equal local amplitudes, the rank factor in (4) is 12 for a depth-two rank-2 current inside depth six's rank 8, but 16 for a depth-five rank-4 current there. The rank-only comparison therefore does not yield weak neutrino engagement. Different spatial support, occupied modes or a derived geometric map would have to carry any such suppression.

Frozen partner threshold and physical limit

At a unit-density coherent rank-$r$ current, the traceless partner-split Hessian of $\operatorname{Tr}C+\eta K(C)$ matches the earlier trace-current metric with $q=2\eta r$. At local density $\rho$, $q_{\rm loc}=2\eta r\rho^3$. Its frozen high-momentum partner block therefore requires

$$\boxed{2\eta r\rho(x)^3<1} \tag{5}$$

for the bounded real-frequency side of that block. This is a local necessary condition, not a whole-core stability theorem. Complete finite real-linear controls at depths 2, 4 and 6, with one or two oscillator columns, core gaps 0 and 1, and momenta 0.1, 1 and 10, found no resolved imaginary frequency above $9\times10^{-15}$ at one quarter and three quarters of the threshold. At 1.25 times the threshold, the largest imaginary frequency was about 0.484666 in particle-frequency units at each tested depth, approaching the analytic high-momentum limit 0.5.

The polynomial is a stronger conditional coupling lead because the same expression preserves isolated occupied sources across projector ranks and gives a nonzero angular pair cost. It is not the physical coupling law or mass operator. The relative strength, the complete cross-depth spatial embedding, permitted polarization states, and encounter dynamics are unselected. The inherited upper-only localized core still has the constrained energetic saddle: its coherent energy is flat under norm-preserving dilation while the lower-spinor mixed source is nonzero. In fact, any correction that vanishes on every coherent current $\rho P$ is identically zero along this full upper-only dilation, so it contributes no scale-scale curvature. Its occupied first derivative also vanishes when it is projector-transparent. No strength of (1) can cure that particular energetic saddle without changing the occupied core or another part of the action. A complete same-field law must also supply spatial support and a positive full response in one normalization.