A smooth two-current polynomial with conditional mass and reciprocal pair response

The two-current quartic produces a simplex self frequency and a polarized pair term on a stipulated coherent constant-density spatial zero mode. At canonical norm one, its quartic energy is half its standing frequency. A nonhomogeneous current function can remove that algebraic discrepancy at a density stationary under support-volume variation. The following construction is an inverse action candidate, not a derivation of the physical particle state or master interaction.

One local function and its complete source

At retained order $k$, write $C_k=\Psi^\dagger B_0\Psi$ and $K_k=\Psi^\dagger B_2\Psi$ with the retained current maps understood. Both matrices are Hermitian Lorentz scalars in the specified doubled-field representation. Define $L_{\pm,k}=C_k\pm K_k$ and, in units of one common current scale,

$$f_\epsilon(y)=y^2[(y-1)^2+\epsilon] =y^4-2y^3+(1+\epsilon)y^2, \qquad 0<\epsilon<\frac13. \tag{1}$$

For $h_{k,d}=\binom{k+d-1}{d-1}$, the candidate action is

$$\boxed{U=g_d\sum_{k\ge0}h_{k,d}\operatorname{Tr} \left[f_\epsilon(L_{+,k})+f_\epsilon(L_{-,k})\right].} \tag{2}$$

One $f_\epsilon$ governs self and pair terms. It is a polynomial in the current, hence smooth through zero field and current cancellation. Equation (1) is nonnegative for every real eigenvalue, so (2) is nonnegative even for indefinite currents. The complete local field source, before lifting through the retained maps, is

$$\frac{\partial U}{\partial\Psi^\dagger} =g_d\sum_k h_{k,d}\left[ (B_0+B_2)\Psi f_\epsilon'(L_{+,k}) +(B_0-B_2)\Psi f_\epsilon'(L_{-,k})\right]. \tag{3}$$

The executable batch checks the full complex field derivative and actual grade-zero retained source maps at all six maintained depths. For odd $d$, it uses the doubled Clifford representation of rank $2^{(d+1)/2}$: an irreducible odd-dimensional spinor has no matrix that anticommutes with every spatial Dirac matrix, so the spatial involution requires both inequivalent Clifford blocks. The earlier rank-$2^{\lfloor d/2\rfloor}$ odd-depth matrix control did not represent this spatial algebra. The physical reason for this doubling and its canonical norm must still be established. The polynomial and equal channel coefficients remain choices in this inverse construction. The current unit, strength, shell weights and sector scales have no forward master derivation.

A selected density and exact energy-frequency identity

Assume a spatial zero mode whose upper-only partner field is pointwise $\phi=\sqrt\rho U$, with $U$ a square unitary frame and constant $\rho$. Then $L_+=2\rho P_+$ and $L_-=2\rho P_-$ on the occupied oscillator ray, where $P_++P_-=I_r$. Put $y=2\rho$ and $D_N=\sum_{k=0}^{N}h_{k,d}=\binom{N+d}{d}$. Equations (2)–(3) give

$$\omega_N=2g_d f_\epsilon'(y)D_N,\qquad U_N=g_d r f_\epsilon(y)D_N,\qquad Q=r\rho\mathcal V_d. \tag{4}$$

The displayed energy takes unit spatial volume; restoring $\mathcal V_d$ multiplies it by that volume. At fixed canonical norm, the local interaction energy per norm is proportional to $f_\epsilon(y)/y$. Its nonzero stationary densities solve $f_\epsilon(y)=y f_\epsilon'(y)$:

$$y_\pm=\frac{2\pm\sqrt{1-3\epsilon}}{3}. \tag{5}$$

$y_+$ is the local minimum and $y_-$ the intervening local maximum. At $y_+$, if the spatial kinetic eigenvalue is zero, (4) gives the exact conditional identity $E_N=\omega_N Q$. If the common frequency unit is set to one, $g_d=1/[2f_\epsilon'(y_+)]$; the same coefficient then gives every assigned $D_N$ at that depth without fitting grades separately. For the rational example $\epsilon=8/25$,

$$y_+=\frac{11}{15},\quad \rho=\frac{11}{30},\quad f_\epsilon'(y_+)=\frac{968}{3375},\quad g_d=\frac{3375}{1936},\quad \frac{y_+f_\epsilon''(y_+)}{f_\epsilon'(y_+)}=\frac34. \tag{6}$$

Within the uniform fixed-norm family, $y=2Q/(rV)$ and $E(V)=g_d rV f_\epsilon(y)D_N$. Direct differentiation gives $E'(V)=g_d rD_N(f_\epsilon-yf_\epsilon')=0$ at $y_+$ and $E''(V)=g_d rD_N y_+^2 f_\epsilon''(y_+)/V>0$. This is a local volume curvature. It is not the complete field and metric Hessian.

Here $g_d$ denotes the coefficient after choosing the common frequency unit. A physical field normalization, spatial volume and relation among depths must determine it; setting it algebraically does not derive those quantities. For $Q=1$, unit volume would require $r\rho=1$, which is false at ranks other than the matching value. Thus the unit-volume calculation alone is not a one-particle normalization across depths. With one universal coefficient and current unit, the selected $y_+$ would give the same frequency unit at every depth. The maintained observed families instead require different units: the approximate W-sector unit is $27.44$ MeV and the electron-sector unit is $27.53$ eV, about a millionfold apart. The required depth dependence of $g_d$ or its geometric projection is an unsolved part of the law; inserting separate fitted $g_d$ values would not derive it. The six-to-ten charged-lepton unit relation is an additional simultaneous constraint.

There is also an exact support limitation. On an unbounded flat carrier, a smooth field of fixed norm can spread over volume $V$, making its density $y\propto V^{-1}$. Equation (1) gives $U\sim\mathrm{const}/V\to0$, while a massless first-order spatial kinetic expectation can also tend to zero with increasing size. The finite-density $y_+$ branch is only a local interaction-energy minimum; this action without a derived spatial support cannot make it the global isolated particle minimum.

Reciprocal prepared-pair energy

For equal-density coherent upper fields with matching spatial involutions, let $s=|z^\dagger w|^{2k}$ be their ray overlap at order $k$. Their two nonzero current eigenvalues are $y(1\pm\sqrt s)$ in each occupied spatial channel. Subtracting the isolated energies and using (5) yields

$$\boxed{V_{AB,k}^{\rm match} =2g_d r y^3s[(3+s)y-2]h_{k,d}\ge0.} \tag{7}$$

The inequality holds for $y=y_+>2/3$ and $0\le s\le1$. The expression is symmetric and its complete source varies both fields. For opposite spatial involutions, $L_+$ and $L_-$ place the two fields in orthogonal current channels and the prepared cross term vanishes exactly, even when their oscillator rays overlap. At (6), equation (7) per partner rank is $77/60$ for $s=1$ and $693/4000$ for $s=3/10$. The actual retained grade-zero calculation uses $s=1$, independently of the full coherent-ray angle, and checks both reciprocal derivatives.

This is a polarization-sensitive pair energy, not a measured vertex or interaction rate. Matching gives a positive cross cost under this preparation. The spatial orientations, phase-bearing photon states, encounter trajectories and nested cross-depth maps remain unselected.

Frozen stability boundary

At the selected density, the normalized density curvature simplifies to

$$q_\epsilon=\frac{y_+f_\epsilon''(y_+)}{f_\epsilon'(y_+)} =\frac{3y_+-2}{1-y_+}. \tag{8}$$

Thus $q_\epsilon\le1$ exactly when $\epsilon\ge5/16$. Complete finite real-linear uniform-field tests at depths two, four and six, with one or two oscillator columns and momenta $0.01$ through $100$ in seven samples, find high-momentum growth below this boundary. At $\epsilon=0.31$, the maximum sampled imaginary frequency is $0.28135062$; at $0.25$ it is $1.41389553$. At $\epsilon=5/16,0.315,0.32,0.33$, no sampled imaginary frequency exceeds $2.2\times10^{-12}$. This numerical result is consistent with (8); it is not an all-momentum stability theorem. The odd-depth checks in the batch establish Clifford algebra and retained maps in the doubled representation, not odd-dimensional spatial spectra. Neither complete spatial stability, positive collective inertia nor the physical photon response follows from a uniform frozen test.

Fixed-norm energy saddle and a third-current control

There is a decisive missing direction in the apparent stable window. Let the full-rank upper field be $\sqrt\rho I_r$ and choose a unit vector $v$ in the $\beta=+1$ spinor subspace. Add a lower field $t vv^\dagger$, while multiplying the upper field by $a_t=\sqrt{1-t^2/(r\rho)}$. This keeps the total canonical norm $Q=r\rho$ exactly fixed. With $P_v=vv^\dagger$, the two currents have spectra equivalent to

$$L_+=y a_t^2P_+, \qquad L_-=y a_t^2P_- -2t^2P_v. \tag{9}$$

Thus one shell has the exact energy

$$U(t)=g_d\{r f_\epsilon(y-2t^2/r)+f_\epsilon(-2t^2)\} =U(0)-2g_df_\epsilon'(y)t^2+O(t^4). \tag{10}$$

At the unit-frequency normalization in (6), the coefficient is $-1$. The batch checks the fixed norm and approaches that coefficient at all six depths; at $t=0.001$ the computed coefficients lie between $-0.999991$ and $-0.999990$. If the upper spatial field is a zero mode of an off-diagonal Dirac kinetic operator, its kinetic cross term with an arbitrary lower perturbation vanishes by self-adjointness. The lower-only quadratic kinetic term also vanishes for that off-diagonal operator. Consequently the same negative direction persists for that spatial zero-mode realization. Real frozen frequencies do not remove this constrained energy saddle.

The existing third scalar current $J=\Psi^\dagger B_1\Psi$ suggests one simple repair. Add $g_da\operatorname{Tr}J^2$ to (2). It vanishes with zero first source on the isolated upper field, and does not change the prepared upper-field pair formula (7). In the direction (9), its energy curvature changes the coefficient to $-1+4g_d\rho a$. Energetic repair requires

$$\boxed{a>\frac{f_\epsilon'(y)}{y}.} \tag{11}$$

The complete frozen tests expose the countervailing response. At $\epsilon=8/25$ the threshold is $a=0.391111\ldots$; the frozen uniform field already has high-momentum growth about $0.5$ at that value. At $\epsilon=0.33$ the threshold is $a=0.42$: $a=0.41$ retains a negative energy direction, while $a=0.43$ has positive curvature in that direction but sampled growth $0.153653$ at intermediate momentum and $0.077152$ at momentum $1000$. Depths two, four and six give the same values within numerical precision. The tested simple $J^2$ completion therefore has no interval with both positive curvature in this direction and no resolved frozen growth. This is a bounded result for that completion, not a theorem excluding all same-field actions.

As a diagnostic, equal squares of $J_x=\Psi^\dagger (\sigma_x\otimes\beta)\Psi$ and $J_y=\Psi^\dagger(\sigma_y\otimes\beta)\Psi$ have a tuned coefficient $a=2f_\epsilon'(y)/y$. At $\epsilon=0.33$ this is $a=0.84$. The complete local rotating Hessian is nonnegative in the tested rank-two, four and eight blocks, and sampled frozen frequencies are real; moving the coefficient by $0.01$ restores a negative Hessian direction or growth. This apparent completion is not Lorentz scalar in the action's boost representation. For boost generator $G=\sigma_x\otimes\alpha_1$, $\{G,\sigma_y\otimes\beta\}\ne0$, whereas the corresponding anticommutators vanish for $B_0,B_1,B_2$. A direct boost changes $J_y$ at every tested depth. The symmetric result therefore identifies a response that would be needed, not an admissible physical term in this same-field action.

There is a different fourth bilinear, $J_0=\Psi^\dagger (\sigma_y\otimes I)\Psi$, which is scalar under these boosts. Testing its square alongside $J_x^2$ leaves the minimum rotating Hessian at $-2$ in particle-frequency units for every tested coefficient. The remaining negative direction has a direct algebraic explanation that applies to a much larger class than (2).

Restricted no-go for algebraic scalar currents

Let $F(C,K,J_x,J_0)$ be any differentiable local function of these four Hermitian Lorentz-scalar currents. Suppose its coherent upper-only square-unitary branch $\Psi_0=(\sqrt\rho I_r,0)^\mathsf T$ has a positive standing self frequency $\omega$ and the canonical norm $Q=r\rho$.

At even depth in the irreducible complex spatial Clifford representation, these four exhaust the Hermitian bilinear scalar insertions for the stipulated boosts. Indeed, decompose an insertion as $I\otimes A_0+\sigma_x\otimes A_x+ \sigma_y\otimes A_y+\sigma_z\otimes A_z$. Anticommutation with every boost generator $\sigma_x\otimes\alpha_i$ requires $A_0,A_x$ to anticommute with every $\alpha_i$, hence each is proportional to $\beta$; $A_y,A_z$ commute with every $\alpha_i$, hence each is proportional to $I$. The odd-depth doubled representation can have additional commutant structure, so the theorem below is stated for these four currents there.

Choose one Hermitian spatial Clifford matrix $\alpha_1$ that anticommutes with $\beta$. The fixed-norm variation

$$\Psi_t=\begin{pmatrix} \sqrt{1-t^2/(r\rho)}\sqrt\rho I_r\\ t\alpha_1/\sqrt r \end{pmatrix} \tag{12}$$

has, exactly,

$$C_t=\left(\rho-\frac{2t^2}{r}\right)I_r, \quad K_t=\left(\rho-\frac{2t^2}{r}\right)\beta, \quad J_{x,t}=J_{0,t}=0. \tag{13}$$

Thus every action in this restricted class follows its original coherent current ray toward smaller density. On that ray the standing source implies $dU/d\rho=r\omega$. Consequently

$$\boxed{U(\Psi_t)-U(\Psi_0)=-2\omega t^2+O(t^4)<0.} \tag{14}$$

The batch checks (12)–(14) at depths two, three, four, five, six and ten, both with the original polynomial and after adding tested $J_x^2$, $J_0^2$ and $\operatorname{Tr}(C^2-K^2)$ coefficients. The computed quadratic coefficients approach $-2$ throughout. If the upper field is an exact spatial zero mode of an off-diagonal Dirac operator, that quadratic kinetic energy also vanishes by self-adjointness. The result obstructs a constrained energy minimum for that upper-only zero-mode class. It does not prove exponential growth, exclude occupied lower components or rule out derivative or Lorentz-tensor interactions with a different complete response.

The same direction has a broader Lorentz explanation. Let $G=\sigma_x\otimes\alpha_1$ be a Hermitian boost generator, $G^2=I$, and set $S_\theta=e^{\theta G/2}$. For an upper-only field, $\langle\Psi_0,G\Psi_0\rangle=0$, so $Q(S_\theta\Psi_0)=Q_0\cosh\theta$. The normalized field

$$\widehat\Psi_\theta=(\cosh\theta)^{-1/2}S_\theta\Psi_0 \tag{15}$$

has exactly the original canonical norm. Let $U$ be any differentiable algebraic Lorentz-scalar potential made only from the field, with no derivatives or fixed Lorentz-breaking background. Pointwise boost invariance gives $U(\widehat\Psi_\theta)= U((\cosh\theta)^{-1/2}\Psi_0)$. If its occupied stationary source is $\omega\Psi_0$ with $\omega>0$, amplitude variation gives $dU(\lambda\Psi_0)/d\lambda|_{\lambda=1}=2\omega Q_0$. Since $(\cosh\theta)^{-1/2}=1-\theta^2/4+O(\theta^4)$,

$$\boxed{U(\widehat\Psi_\theta)-U(\Psi_0) =-\frac12\omega Q_0\theta^2+O(\theta^4)<0.} \tag{16}$$

This is the same coefficient as (14) after expressing the lower-field norm in terms of $\theta$. For an upper zero mode of a self-adjoint off-diagonal spatial Dirac operator, the kinetic cross term with the induced lower field vanishes, and a lower-only field has no quadratic kinetic expectation. A smooth local boost in a spatial patch suffices when a global tangent frame is unavailable. Equation (16) therefore extends the fixed-geometry constrained-energy obstruction to every algebraic Lorentz-scalar field potential in this upper-only zero-mode class. It does not apply to a state with occupied lower components, a derivative-dependent interaction or a nonzero spatial kinetic balance, and it is not a dynamical growth theorem.

The polynomial repairs the quartic branch's algebraic $E/\omega Q$ factor on a selected constant-density zero mode, and it supplies a reciprocal polarized pair response. Equations (12)–(16) show why this does not establish a stable one-law mass operator. A successful action must change the occupied upper/lower structure or supply a response outside this upper-only algebraic class whose complete spatial dynamics remains sound. Canonical norm, spatial size, master selection, photon structure and cross-sector coupling remain unresolved.

The calculation report records the tested values.