A nonzero same-field background and the joint support bound

The spectrally transparent angular construction requires a nonzero supporting background. This note gives the minimal realization of that requirement and computes its complete quadratic consequence. A retained amplitude can supply the missing metric force while leaving the reference field equation unchanged. Eliminating the amplitude always subtracts a positive Schur term from the geometric stiffness, so force cancellation alone is not stability.

This is an inverse construction. It identifies an admissible local quadratic structure and an exact test for any proposed realization in the one field. It does not derive the background, its potential, or the projected inverse angular operator from $\Psi_\infty$.

A background coupling that is spectrally transparent

Let $x$ be a real retained amplitude of the same field and let its reference value be $v\ne0$. Near one of its stable minima, write

$$U_x=\frac12\langle \xi,A\xi\rangle +\langle v+\xi,B(\Psi,g)\rangle, \qquad \xi=x-v, \qquad A>0. \tag{1}$$

The brackets allow an angularly varying amplitude; the one-component case is obtained by replacing $A$ by a positive number. Suppose

$$B(\Psi,g_*)=0 \quad\hbox{as an operator on the retained field space}. \tag{2}$$

Then the coupling energy and its field derivative vanish at the reference:

$$U_x\big|_*=0, \qquad D_\Psi U_x\big|_*=0. \tag{3}$$

The reference field, its frequency, and indeed the whole fixed-geometry linear field operator are unchanged. Nevertheless,

$$D_gU_x\big|_*=\langle v,D_gB\rangle \tag{4}$$

need not vanish. This is the nonzero-background alternative left open by the zero-background obstruction. The distinction is structural: equation (2) kills the field operator at the reference, while the background $v$ multiplies its nonzero metric derivative.

For the inverse angular channel, let the bracket in (1) be the angular $L^2$ pairing and take the angular source field schematically as

$$B(\theta)=\frac{\kappa}{v}\, \chi(\theta)\,\Psi^\dagger\mathcal T_0\Psi, \qquad \chi=\mathcal L_d^{-1}\left(\frac{\kappa}{\kappa_*}-1\right). \tag{5}$$

At the round reference $\chi=0$ as an operator coefficient and $\delta\chi=\sigma$. If $\langle\mathcal T_0\rangle=\beta b$, equation (4) contributes

$$\delta_gU_x=\kappa\int\rho\,\beta b\,\sigma\,dV, \tag{6}$$

which cancels the carrier's missing $-\kappa\int\rho\,\beta b\sigma\,dV$ channel. The ordinary local term $-\kappa\int y\,\Psi^\dagger\mathcal T_1\Psi\,dV$ cancels the separate $\mathcal L_d\sigma$ channel. Both coefficients are common operators; they are not fitted independently for each particle.

Equation (5) does not solve the origin of $\mathcal L_d^{-1}$. It shows that a nonzero retained background removes the earlier first-derivative obstruction without introducing a frequency shift.

Eliminating the background fixes the sign of its Hessian term

The stationary amplitude is

$$\xi_*=-A^{-1}B, \tag{7}$$

and completing the square gives

$$\boxed{U_{x,\mathrm{eff}} =\langle v,B\rangle-\frac12\langle B,A^{-1}B\rangle.} \tag{8}$$

Let $h$ be an angular metric perturbation and put $F=D_gB[h]|_*$. The second term in (8) contributes

$$\boxed{\delta_g^2U_{x,\mathrm{eff}}[h,h] \supset-\langle F,A^{-1}F\rangle.} \tag{9}$$

This sign is unavoidable for a stable amplitude that relaxes. Increasing the background stiffness suppresses the loss, but does not reverse it. For one harmonic, with background curvature $a>0$, background value $v$, cancelled carrier force $F_{\rm car}$, and bare geometric stiffness $G$, a linear coupling requires

$$f=-\frac{F_{\rm car}}v, \qquad \boxed{G>\frac{f^2}{a} =\frac{F_{\rm car}^2}{a v^2}.} \tag{10}$$

Thus an arbitrarily soft background destabilizes the geometry even though it cancels the first force exactly.

The complete joint Hessian

Field and metric perturbations can mix. Let $p$ denote a physical field perturbation after phase and other collective zero modes have been removed. Write the uncoupled field Hessian as $P$, its field--metric block as $M$, the geometric block including $vD_g^2B$ as $G$, and the mixed second derivative of $B$ as $N$. The full quadratic operator on $(p,h,\xi)$ is

$$\mathbb H= \begin{pmatrix} P&M+vN&0\\ (M+vN)^*&G&F^*\\ 0&F&A \end{pmatrix}. \tag{11}$$

Assuming $P>0$ and $A>0$ on the projected physical subspaces, two successive Schur complements give the necessary and sufficient quadratic condition

$$\boxed{ G-F^*A^{-1}F -(M+vN)^*P^{-1}(M+vN)>0.} \tag{12}$$

Equation (12) is the complete local test. Checking only (10) can miss an instability caused by field relaxation. Conversely, the negative term in (9) is not by itself a no-go theorem: sufficiently positive retained geometry and field blocks can satisfy (12).

Collective position, orientation, phase, and radius directions must be projected or constrained according to their actual symmetries before using an inverse in (12). A zero eigenvalue associated with a true symmetry is not a negative mode, but it cannot be treated as a positive denominator.

Angular scaling and the lowest-harmonic bottleneck

For the inverse channel in (5), the force per normalized density harmonic is independent of the Jacobi eigenvalue:

$$|F_\ell|\propto|\beta b|, \qquad \ell\ge2. \tag{13}$$

If the retained geometric stiffness is proportional to

$$\mathcal L_d(\ell) =-1+\frac{\ell(\ell+d-1)}d, \tag{14}$$

then the normalized requirement $|F_\ell|^2/[a\mathcal L_d(\ell)]$ is largest at $\ell=2$ and decreases strictly thereafter. The inverse channel therefore has no high-angular-degree instability forced by its Schur term. Its most demanding positive harmonic is the quadrupolar one.

This conclusion depends on separating the two force channels. If a single relaxing background is made to carry the local-curvature term as well, then $|F_\ell|\sim\mathcal L_d(\ell)$, and a background with constant $A$ produces a penalty growing as $\ell^4$, faster than a second-order geometric stiffness. Such a combined realization needs angular stiffness in $A$ or a different nonrelaxing structure. Keeping the inverse and local channels distinct is therefore dynamically meaningful.

A causal background can satisfy the static bound

Give the background fluctuation the quadratic action

$$S_\xi=\frac12\int dt\left[ Z\|\dot\xi\|^2-\frac1{R^2} \langle\xi,(a_0+a_1\mathcal L_d)\xi\rangle\right], \qquad Z,a_0,a_1>0. \tag{15}$$

On the positive angular subspace,

$$\Omega_\ell^2=\frac{a_0+a_1\mathcal L_d(\ell)}{ZR^2}. \tag{16}$$

Its high-degree propagation speed is $c_\xi=\sqrt{a_1/(Zd)}$ in units where the containing-field limiting speed is one. The common sufficient choice $a_1/Z\le1$ is subluminal at every maintained depth. If the bare geometry has $G_\ell=g\mathcal L_d(\ell)$, the background part of the static Schur condition becomes

$$\boxed{ g>\frac{|F_\ell|^2} {\mathcal L_d(\ell)[a_0+a_1\mathcal L_d(\ell)]}.} \tag{17}$$

For the inverse channel's harmonic-independent force, the right side is strictly largest at $\ell=2$. Hence causal propagation and positive static stiffness are algebraically compatible under one common kinetic normalization. The required coefficient must still be obtained from the physical action. Choosing it after inspecting each particle would be a fit, and equation (17) does not include the field-relaxation term in (12).

A conventional local way to obtain a nonzero real background is a double-well potential $V(x)=\lambda(x^2-v^2)^2/4$, for which the radial curvature at either minimum is $2\lambda v^2>0$. Its minimum energy can be normalized to zero, so a constant background need not carry an infinite vacuum offset. This example proves local realizability of the assumed quadratic neighborhood; it does not establish that this potential belongs to the IDWT master action.

A particle-sourced positive multiplet fails transparency

The previous construction leaves $\mathcal L_d^{-1}$ explicit. It is natural to ask whether local positive modes can generate that inverse. Let their static energy instead contain

$$U_z=\frac12\langle z,\mathcal L_dz\rangle -\langle z,S(\Psi)\rangle, \tag{18}$$

on the projected $\ell\ge2$ space. Here $z$ may be a multiplet and $\mathcal L_d$ may include any positive component matrix. Elimination gives

$$z_*=\mathcal L_d^{-1}S, \qquad U_{z,\mathrm{eff}}=-\frac12 \langle S,\mathcal L_d^{-1}S\rangle. \tag{19}$$

For the required quadratic particle source, $S(a\Psi)=|a|^2S(\Psi)$. The field-amplitude direction therefore obeys

$$D_\Psi U_{z,\mathrm{eff}}[\Psi] =-2\langle S,\mathcal L_d^{-1}S\rangle<0 \quad\hbox{when }S\ne0. \tag{20}$$

The positivity is strict because $\mathcal L_d>0$ on the projected space. At fixed canonical norm, the same nonzero component appears as a shift of the Lagrange multiplier, hence of the standing frequency. Thus every unbiased stable quadratic construction in which the particle density is the sole source of the inverse background fails spectral transparency; adding more positive response components does not cancel the sign.

This narrows the viable interpretation of (1). Its nonzero background must be present independently of the occupied particle source, or a constraint, indefinite pairing, an independently derived bias satisfying the full orthogonality condition, or nonlinear structure must remove the positive amplitude overlap. An independently present background can multiply the vanishing operator in (5), but then the origin of the inverse coefficient still needs a local derivation. A particle-sourced positive quadratic multiplet does not provide both ingredients.

The radius and the conformal kernel

The inverse construction in (2) is used on the centered even harmonics of degree at least two, where $\mathcal L_d$ is positive. Two scalar sectors lie outside it and require separate treatment.

On the radius, $\mathcal L_d(0)=-1$. The inverse exists there, but with the opposite sign: $\chi=-y$. A uniform conformal variation $\sigma_0$ rescales the radius and every standing frequency as $\omega\propto1/R$, so $\delta y=-\sigma_0$ and the radius frequency shifts at first order. The $\ell=0$ channel is therefore not spectrally transparent, unlike the $\ell\ge2$ inverse. The causal background curvature on the radius is $A(0)=a_0-a_1$, so positive radius dynamics requires the additional condition $a_0>a_1$. This is strictly stronger than the positivity used on $\ell\ge2$ and is not implied by it: the two-parameter family $a_0,a_1>0$ that softens the angular Schur penalty can still give non-positive radius curvature when $a_0\le a_1$, and the radius Schur penalty $F_0^2/(a_0-a_1)$ diverges as $a_0\to a_1^+$.

On degree one, $\mathcal L_d(1)=0$ exactly. The operator is singular, its kernel is the degree-one scalar harmonics with multiplicity $d+1$, and no inverse exists. These are the conformal collective directions; they must be projected out or carried as collective coordinates rather than absorbed by an inverse.

The three scalar sectors are therefore disjoint: the one-dimensional radius, the $d+1$-dimensional kernel, and the positive inverse on $\ell\ge2$.

Scope across the maintained depths

The identities use only positivity of $\mathcal L_d(\ell)$ for $\ell\ge2$ and apply at depths $2,3,4,5,6,10$. The separate depth-two form coefficients change $\mathcal T_0$ and $\mathcal T_1$, but not the background or Schur argument. This includes the maintained massive depth-two indices; it does not construct a photon state or derive physical transverse polarization.

The result also does not repair the finite-chain mass-unit drift or the failed shared electron--tau unit. It preserves the existing reference frequencies deliberately. A complete candidate still needs a common background potential, the projected inverse as an actual response of $\Psi_\infty$, a physical origin for the radius curvature $a_0-a_1$, and verification of (12) using the physical field and geometry Hessians rather than symbolic positive blocks.

Verification

nonzero_support_background_checks.py verifies the exact force cancellation, eliminated energy, Schur determinant, stable and unstable sides of the joint bound, the causal dispersion bound, the positive particle-sourced overlap, and the lowest-harmonic monotonicity for every maintained index at depths $2,3,4,5,6,10$. It also verifies the two scalar sectors outside the inverse: the opposite-sign radius response, the $a_0>a_1$ positivity condition with its degenerate and wrong-sign controls, the diverging radius Schur penalty, and the $d+1$-dimensional degree-one kernel. The rational component values are algebraic controls rather than measured physical coefficients. The selected 4,218-check report records the normal Python run; an optimized run passes the same conditions.