Section 24 4 October 2026

Conditional-rank stability with zero fibres included

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23131081

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24 Conditional-rank stability with zero fibres included

This section supplies the nonlinear step behind the radial proof. It is useful precisely because the actual clock/pointer marginal need not have a positive lower bound. Let $q=\psi(S,\cdot)$ be a radial spinor slice, $u=\psi_S$ , $z=\psi_{SS}$ , $a=\norm q$ , $\eta=q/a$ , and let $P_r$ multiply by $\one_{[0,r]}$ . Set $K=\langle\eta,P_r\eta\rangle$ and $Z_\eta=(P_r-K)\eta$ . The constant clock carrier is removed from all derivatives, but its actual drift is retained in continuity.

Lemma 24.1 (Exact weighted rank current)

On nonzero fibres, with $\kappa=\hbar/M$ ,

\rho\frac{\dd K}{\dd t}=\mathcal F[\psi] =\kappa\left\{2\Im(\eta(r)^\dagger u(r)) \Re\langle Z_\eta,u\rangle -a|\eta(r)|^2\Im\langle Z_\eta,z\rangle\right\}.

(12)

For two live conditioning coordinates $S,Z$ , the right side is the sum of the corresponding two functionals, with coefficients $\hbar/M$ and $1/(\mu_ZT_0)$ respectively.

Proof

Write $w=\int\rho\dd r$ , $A_r=\int_0^r\rho\dd y$ , and $K=A_r/w$ . Let $\delta B$ and $\delta j$ be the cumulative and full clock-current integrals after subtracting the constant carrier. Continuity and zero origin flux give

\frac{\dd K}{\dd t} =\frac{\delta J}{\rho}K_S -\frac{\partial_S\delta B-K\partial_S\delta j}{w}.

The radial current has cancelled pointwise against $K_r\dot r$ . Now $K_S=2\Re\langle q,(P_r-K)u\rangle/a^2$ and $\partial_S\delta B-K\partial_S\delta j =\kappa\Im\langle q,(P_r-K)z\rangle$ ; the $u^\dagger u$ term is real. Substitution proves (12). Applying continuity with both transverse currents gives the two summands, not an additional mixed-derivative term. Each local product sums the original spinor components. No component has been sampled as an actual sector.

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Lemma 24.2 (Stability without an exact-marginal denominator)

For normalized $\psi,G$ , write $E=\psi-G$ , and suppose the quantities

D=\norm E,\quad e_1=\norm{E_S},\quad e_2=\norm{E_{SS}},\quad g_j=\norm{\partial_S^jG},\quad Q_G^2=\int\frac{\norm{G_S(S,\cdot)}^4}{\norm{G(S,\cdot)}^2}\dd S

are finite. The ratio is zero where $G$ and its required derivatives vanish. Then

\norm{\mathcal F[\psi]-\mathcal F[G]}_1 \le\frac\hbar M (4g_1e_1+2e_1^2+14DQ_G+9Dg_2+e_2).

(13)

Here $\mathcal F[G]$ is algebraic evaluation, including whatever residual the comparison has under the actual equation.

Proof

At a fixed slice put $g=G(S,\cdot)$ , $b=\norm g$ , $\xi=g/b$ , $U=G_S$ , $V=G_{SS}$ , and $d=\norm{q-g}$ , $d_1=\norm{u-U}$ , $d_2=\norm{z-V}$ . The elementary estimates

\norm{\eta-\xi}\le2d/b,\quad \norm{Z_\eta}\le1/2, \quad\norm{Z_\eta-Z_\xi}\le3\norm{\eta-\xi}, \quad\norm{|\eta|^2-|\xi|^2}_1\le2\norm{\eta-\xi}

follow by adding and subtracting $q/b$ , by projection variance, and by $|K_\eta-K_\xi|\le2\norm{\eta-\xi}$ . Split the first term of (12) first in $u-U$ and then in normalized directions. Its derivative part is at most $4\norm U d_1+2d_1^2$ ; its direction part is at most $7\norm{\eta-\xi}\norm U^2\le14d\norm U^2/b$ . For the second term, the $z-V$ part is at most $ad_2$ . Splitting the remaining amplitude, density and projection yields $(d/2+4b\norm{\eta-\xi})\norm V\le9d\norm V$ . Integrate these inequalities and use Cauchy–Schwarz and $\int a^2\dd S=1$ . If $a=0$ , direct bounding of $\mathcal F[G]$ works with $d=b$ ; if $b=0$ and its derivatives vanish, the bound $2d_1^2+ad_2$ suffices. Thus zero fibres are included, not discarded.

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For $G=\phi(S-S_c-vt)u(S,r)$ with real $\phi$ and normalized $u$ ,

Q_G\le\norm{\phi'^2/\phi}_2+\sup_S\norm{u_S}^2, \qquad \norm{\phi_0'^2/\phi_0}_2 =\frac{4\pi^2}{\sigma^2}\sqrt{\frac3{35}}.

The last equality is a direct trigonometric integral for the unsmoothed cosine packet. Positive convolution and weighted Cauchy–Schwarz give $\phi'^2/\phi\le Z^{-1}\kappa*(\phi_0'^2/\phi_0)$ ; Young's inequality then adds only the normalization multiplier at most $1.000001$ . No actual conditional law is asserted uniform in this argument.

Lemma 24.3 (Deterministic-time CDF interface)

For every measurable radial cut $b(S)$ ,

\int w(S)|K_\psi(S,b(S))-K_G(S,b(S))|\dd S\le2\norm{\psi-G}.

Under the already transported actual-law cap $C$ , the corresponding binary-test mismatch is at most $2C\norm{\psi-G}$ per cut.

Proof

Multiply the CDF difference by $w=\int|\psi|^2$ . The exact identity is

w(K_\psi-K_G)=\int(\one_{r\le b}-K_G)(|\psi|^2-|G|^2)\dd r.

The coefficient has modulus at most one and the density difference has $L^1$ norm at most $2\norm{\psi-G}$ . Uniformity of the exact rank under its own conditional wave measure identifies this integral with the test mismatch. Domination then supplies the actual-law bound. Zero comparison fibres are charged by their exact mass. The same proof works with live conditioning pair $(S,Z)$ . It is a fixed-time result; it is not substituted at a random clock hitting time.

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