Exact scale balance and the remaining spatial supporting force

The local current action gives an exact oscillating field on the specified spatial geometry. A further calculation is needed if that geometry is to be sustained by the occupied field itself. Its size and its nonuniform deformations must both satisfy the complete variation.

The results here concern stated supporting models. They do not exclude the common characteristic size within a depth, which remains a physical working premise, or require a common size across different depths.

Exact pressure of the finite carrier

Let $\epsilon_n(R,G)$ be the smallest positive frequency of the complete component chain with diagonal entries $c(n+j)/R$ and common adjacent coupling $G$. Put $\gamma=GR/c$. Its homogeneity gives the exact identity

$$\boxed{-R\epsilon_R+G\epsilon_G=\epsilon.} \tag{1}$$

Define the radius exponent and its logarithmic gap derivative by

$$p=-\frac{R\epsilon_R}{\epsilon},\qquad p'=\frac{dp}{d\log\gamma}. \tag{2}$$

The complete inverse of the triangular chain has positive magnitudes proportional to $\gamma^{j-i}$. Let its normalized left and right Perron singular vectors assign weights $w_{ji}>0$ to these entries, with $\sum w_{ji}=1$. Differentiation of its largest singular value gives

$$p=1+\sum_{j\ge i}(j-i)w_{ji},\qquad \boxed{1<p<d,\quad p'\ge\operatorname{Var}_w(j-i)>0.} \tag{3}$$

The curvature bound follows by holding the maximizing singular vectors fixed while differentiating the logarithm of their positive bilinear sum. That fixed-vector expression is a log-sum-exp function, with second derivative equal to the stated variance. Allowing the maximizing vectors to move can only increase the curvature at their smooth maximum.

Thus the finite carrier interpolates from a radius exponent near one to the leading simplex exponent $d$. This is an exact statement about the whole finite matrix, not an assumption that its leading determinant term is already accurate.

What an additive stationary support can and cannot do

Consider the particular total energy

$$E(R,G)=\mathcal Q\epsilon_n(R,G)+U(R,G), \tag{4}$$

where $U$ is the same additive supporting potential for every occupied mode. At a joint stationary point in $R$ and $G$, equation (1) implies

$$\boxed{\mathcal Q\epsilon=R U_R-G U_G.} \tag{5}$$

If all particles were assigned the same full stationary supporting configuration $(R,G)$, the right-hand side and $U$ would be the same for each. Their complete energies in (4) would therefore coincide. This remains true even if different values of $\mathcal Q$ were allowed. The scope is the common full support state in this additive model. Common radius alone does not impose a common $G$, supporting motion, field texture, or supporting energy.

For the monomial trial $U=A R^dG^a$, joint stationarity requires

$$p=\frac{d}{d-a}. \tag{6}$$

Where a finite stationary point exists, its Hessian in $(\log R,\log G)$, divided by $\mathcal Q\epsilon$, has determinant

$$\boxed{\det\!\left(\frac{\nabla^2E}{\mathcal Q\epsilon}\right) =-(d-a+1)p'<0.} \tag{7}$$

It is a saddle. Adding a monomial supporting term and solving only one of its balance equations would miss this independent direction.

The current fixed point retains a nonzero radius force

For the homogeneous current, let $k=\epsilon_G$, $s=\sqrt{k^2+\eta^2}$, $G=f'(k)$, and $C=f(k)-kf'(k)$, with $f(k)=-g\sqrt{k^2+\eta^2}$. The selected positive branch is $\lambda=\epsilon+C$.

Hold the physical strength $g$ and $\eta$ fixed while varying the radius. The complete derivative is

$$\boxed{ -R\frac{d\lambda}{dR} =\langle\phi,H_0\phi\rangle =\epsilon-Gk=p\epsilon>0. } \tag{8}$$

The induced derivatives of $G$ and $C$ cancel. In particular,

$$\frac{dG}{d\log R} =\frac{f''G\epsilon_{GG}}{1-f''\epsilon_{GG}},\qquad \frac{\partial\lambda}{\partial g}=-s. \tag{9}$$

These relations agree with independent finite differences of the full fixed point at every maintained index. Differentiating only $\epsilon(R,G(R))$ would omit the cancellation from $C$.

For the positive action at fixed phase charge, its standing-wave energy is $\mathcal Q\lambda$. Equation (8) is therefore a real remaining radius force. The current law has closed the spatial field equation on the given geometry, while leaving the size as an input.

A radial minimum from a stated curvature law

It is useful to test a single bounded geometric coupling rather than assigning separate particle radii. As an explicit inverse trial, choose

$$g(R)=\frac{g_0}{1+(R/R_0)^4}. \tag{10}$$

Equivalently, with curvature scale $\kappa=1/R$, this is $g(\kappa)=g_0\kappa^4/(\kappa^4+\kappa_0^4)$, $\kappa_0=1/R_0$. This function and its scales are inputs; they are not derived by writing them in curvature notation.

First examine the saturated current branch, $\eta=0$, where the occupied current is strictly negative. Then $G=g$, $C=0$, and $\lambda=\epsilon$. Set $q_g=-d\log g/d\log R$. The radius equation becomes

$$\frac{d\log\epsilon}{d\log R}=-p+q_g(p-1)=0. \tag{11}$$

At a stationary point its logarithmic second derivative is

$$\frac{d^2\log\epsilon}{d(\log R)^2} =q_g'(p-1)-(q_g-1)^2p'. \tag{12}$$

For (10), the tested roots are radial minima. In the leading large-gap limit their common radius within a depth is

$$\boxed{\frac{R_*}{R_0} =\left(\frac{d}{3d-4}\right)^{1/4}.} \tag{13}$$

The exact finite-chain radii retain mode dependence. At the illustrative $g_0R_0/c=1000$, the depth-two controls give radii from about $1.0118R_0$ to $1.0185R_0$; the depth-three controls range from $0.880114R_0$ to $0.880238R_0$. Consequently, even the radial test does not establish an exactly common finite radius.

The complete nonuniform metric variation

The decisive additional test is a conformal deformation of the actual spatial metric, $g_{ab}\mapsto e^{2\sigma(x)}g_{ab}$. Work in units $c=R_0=1$, and transport the spinor canonically so its $L^2$ inner product is fixed. The transported Dirac operator is

$$\widetilde D_\sigma=e^{-\sigma/2}D e^{-\sigma/2}. \tag{14}$$

For the positive Laplacian $\widehat\Delta=R^2\Delta$, the curvature scale becomes

$$\kappa_\sigma=\kappa e^{-\sigma} \sqrt{1+\frac2d\widehat\Delta\sigma -\frac{d-2}{d}|\widehat\nabla\sigma|^2}, \qquad \delta\kappa=\kappa\left(-\sigma+\frac{\widehat\Delta\sigma}{d}\right). \tag{15}$$

The stated small variations keep the square root real. Define component expectations

$$b_n=\langle\phi,\chi_1\phi\rangle,\qquad a_n=\left\langle\phi, \chi_1\operatorname{diag}(j-d/2)\phi\right\rangle, \qquad \beta=n+d/2. \tag{16}$$

For a normalized spatial eigenfield with density $\rho(x)$, put $C_n=a_n-sg_\kappa$. Its full first variation is

$$\delta E_1=\kappa\int\rho\left[ -\beta b_n\sigma+C_n\left(-\sigma+ \frac{\widehat\Delta\sigma}{d}\right)\right]dV. \tag{17}$$

At the radial stationary point, $C_n=-\beta b_n$. What remains is

$$\boxed{\delta E_1= \frac{\kappa C_n}{d}\int\sigma\widehat\Delta\rho\,dV.} \tag{18}$$

For the highest-weight density $\rho=q^n/M(n)$, choose the smooth variation $\sigma=q-\langle q\rangle_\rho$. Then

$$\left\langle\widehat\Delta q\right\rangle_\rho =\frac{2(d-1)n}{n+(d+1)/2}>0. \tag{19}$$

Since $b_n>0$, equations (18)--(19) exhibit a nonzero direction that lowers the energy. The radial minimum is not stationary against this ordinary spatial deformation. For a positive standing branch, the positive action gives the corresponding force at fixed phase charge; it does not remove (18).

This argument applies to the stated highest-weight family and this curvature trial. It does not exclude other geometries, other local interactions, constant-density eigensections, or supporting motion.

Depth-two forms require their own metric transport

The bosonic form parent does not use the spinor weight in (14). Let $P_{\rm even}$ and $P_{\rm odd}$ project form parity. Its exact canonical conformal transport on the two-sphere is

$$\widetilde D_{{\mathrm f},\sigma} =E_\sigma D_{\mathrm f}E_\sigma,\qquad E_\sigma=e^{-\sigma}P_{\rm even}+P_{\rm odd}. \tag{20}$$

For the electric family in the form construction, write its densities as $\rho_e$ and $\rho_o$, and let $\rho=\rho_e+\rho_o$. The complete transported generating energy is

$$\begin{aligned} E_1(\sigma)={}&2n\kappa b_n\int e^{-\sigma}\rho_e\,dV\\ &+\int\kappa_\sigma \left[a_{0,n}\rho+\frac{b_n}{2}(\rho_e-\rho_o)\right]dV -s\int g(\kappa_\sigma)\rho\,dV, \end{aligned} \tag{21}$$

where $a_{0,n}=\langle\chi_1\operatorname{diag}(j-1/2)\rangle$. Direct variation of (21) agrees with independent spatial quadrature and also leaves a nonzero force at the radial minimum. The conclusion is therefore not an artifact of assigning a spinor transformation law to a bosonic field.

Independent angular stresses constrain local curvature support

The nonuniform force is not specific to the bounded example (10). Keep the round reference geometry, the highest-weight spatial state, and a coupling depending only on the local scalar curvature scale $g(\kappa)$. Allow an additional rotationally invariant functional of the geometry. At a round reference metric its first variation is isotropic, so it contributes only to the constant angular channel. This includes local scalar-curvature supporting terms; it excludes support with its own nonuniform field pattern or motion.

For spinors, (17) gives the nonconstant force density

$$\kappa\left[-\beta b_n\rho+ C_n\left(-\rho+\frac{\widehat\Delta\rho}{d}\right)\right]. \tag{22}$$

Changing $g$ or its slope changes $C_n$, but the same value must balance every angular component. Set $b=(d-1)/2$. Two zero-mean axisymmetric scalar harmonics are

$$H_1(q)=q-\frac1{b+1},\qquad H_2(q)=q^2-\frac{4q}{b+3}+\frac2{(b+2)(b+3)}, \tag{23}$$

with Laplacian eigenvalues $2(d+1)$ and $4(d+3)$. For $\rho=q^n/M(n)$, their exact density moments are

$$\begin{aligned} \langle H_1\rangle_\rho &=\frac{bn}{(b+1)(n+b+1)},\\ \langle H_2\rangle_\rho &=\frac{b(b+1)n(n-1)} {(b+2)(b+3)(n+b+1)(n+b+2)}. \end{aligned} \tag{24}$$

Both are nonzero for $n\ge2$. Stationarity in these two independent directions would require

$$-\beta b_n+C_n\frac{d+2}{d}=0,\qquad -\beta b_n+C_n\frac{3(d+4)}d=0. \tag{25}$$

The equations are incompatible when $b_n\ne0$, as in the retained positive branches. No isotropic first variation of an additional geometric functional changes either equation. The spinor $n=1$ control has no degree-four density component and is outside this two-harmonic obstruction; the earlier explicit radial test still applies to it under its narrower assumptions.

The depth-two form family has a separate constraint. On the unit reference sphere, its even and odd densities obey

$$\rho_o=r\rho_e-t\widehat\Delta\rho_e, \qquad r=\frac n{n+1},\qquad t=\frac1{2(n+1)^2}. \tag{26}$$

This follows from $\widehat\Delta|Y_n|^2=2n(n+1)|Y_n|^2-2|\widehat\nabla Y_n|^2$ and the actual form weights. Put $C_{0,n}=a_{0,n}-sg_\kappa$. From (21), the coefficient of an angular component of $\rho_e$ with eigenvalue $\Lambda$ is

$$\boxed{ P_n(\Lambda)=-2n b_n+ \left(-1+\frac\Lambda2\right) \left[C_{0,n}(1+r-t\Lambda) +\frac{b_n}{2}(1-r+t\Lambda)\right]. } \tag{27}$$

For $n\ge3$, the density contains at least the three nonconstant degrees two, four and six. A quadratic polynomial vanishing at all three eigenvalues must vanish identically. Its quadratic coefficient would require $C_{0,n}=b_n/2$; after that substitution,

$$P_n(\Lambda)=b_n\left[\frac\Lambda2-(2n+1)\right], \tag{28}$$

which is not identically zero. Thus the maintained form controls also cannot be sustained by this class of scalar local curvature laws and isotropic supporting variations.

This identifies the missing kind of supporting response for the stated states: it must carry the independent angular stresses, or the spatial state and geometry must change. A patterned or moving component of the same field remains a possible source. The calculation does not exclude all common-size states, all geometries, or all same-field interactions.

A constant-density exception at the maintained depth-three index

There is an explicit spatial state outside the nonuniform obstruction. On the unit sphere, choose a constant ambient spinor $\zeta$ and set

$$U(x)=\Gamma_x\zeta,\qquad \|\zeta\|=1. \tag{29}$$

For odd $d$, choose the opposite ambient chirality for $\zeta$ so that $U$ lies in the chosen physical spinor bundle. The ambient notation adds no spatial coordinate. Direct differentiation gives

$$\boxed{\|U(x)\|^2=1,\qquad BU=-\frac{d+1}{2R}U,\qquad DU=-\frac d{2R}U.} \tag{30}$$

Thus this is the $n=1$ negative spatial branch of the same parent. Its component chain has diagonal entries $-d/R,\ldots,-1/R$, and exactly the same positive singular frequency as the positive $n=1$ chain. The spatial spectral sign is not assigned a particle--antiparticle meaning. The maintained down-sector index is $d=3,n=1$; $n=1$ at the other tested depths is a mathematical control, and the depth-two spinor here is not the bosonic form realization.

This Killing spinor has an isotropic Dirac metric stress. In a unit orthonormal frame, direct differentiation verifies $\nabla_aU=\tfrac12\mathrm{cl}(e_a)U$ and

$$\frac12\operatorname{Re}\langle U, \mathrm{cl}(e_a)\nabla_bU+ \mathrm{cl}(e_b)\nabla_aU\rangle=-\frac12\delta_{ab}. \tag{31}$$

Restore $\kappa=1/R$, use canonical field transport, and let $h_{ab}=\delta g_{ab}$ be an arbitrary metric variation in the orthonormal frame. The integrated divergence terms in $\delta\kappa$ vanish against the constant density. With the component expectations $b_n,a_n$ of (16), and $C_n=a_n-sg_\kappa$, the complete first variation is

$$\boxed{\delta E_1= \frac{\kappa}{2d}\left(\frac d2 b_n-C_n\right) \langle\operatorname{tr}h\rangle.} \tag{32}$$

The average uses normalized physical volume. A uniform variation in (32) is exactly the radius equation. Consequently a radial stationary point now makes the first variation vanish for every metric direction, including the nonuniform and traceless ones. This conclusion uses constant density and the actual Killing stress, not an angular average of the earlier nonuniform field.

For the specified bounded law (10), the saturated negative-current branch has the previously computed nondegenerate radial minimum. The direct component calculation and the full tensor stress agree with (32). This supplies a stationary metric source at the maintained depth-three $n=1$ index under that inverse law.

For sufficiently small positive $\eta$, the radial minimum continues by the implicit-function theorem while the current remains strictly negative. The Killing stress stays isotropic, so radius balance continues to cancel the full first variation. No quantitative interval of $\eta$ or full metric stability is claimed here.

This does not select the function $g(\kappa)$, supply its dynamical metric origin, establish the second variation against arbitrary metric deformations, or support the higher assigned modes at a common radius. It is a concrete stationary exception within the tested class, and shows why the spatial polarization state must remain part of the support calculation.

Verification

support_scale_checks.py passes 608 exact and numerical scale-balance checks. current_radius_checks.py passes 112 checks of the fully self-consistent radius force and the stated curvature trial. current_metric_force_checks.py passes 89 checks of the actual spinor and form metric variations, including independent quadrature. The retained particle indices are unchanged; the two bottom indices remain component controls rather than two separate bottom particles. current_local_support_constraint_checks.py passes 87 checks of the independent angular harmonics, their exact density moments, and the separate form force polynomial. The broader support constraint is restricted to the class and states specified above. killing_current_support_checks.py passes 224 checks of the actual constant-density spinor, physical chirality, complete isotropic Dirac stress, spectral sign equivalence, and cancellation of the first metric source at the radial minimum. Only its depth-three index is a maintained particle control.