# Section 24: Conditional-rank stability with zero fibres included
## 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, $u=\psi_S$, $z=\psi_{SS}$, $a={\left\lVert{q}\right\rVert}$, $\eta=q/a$, and let $P_r$ multiply by ${\mathbf1}_{[0,r]}$. 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 $\kappa=\hbar/M$,
$$
\rho\frac{{\,\mathrm d} K}{{\,\mathrm d} t}=\mathcal F[\psi]
=\kappa\left\{2\Im(\eta(r)^\dagger u(r))
\Re\langle Z_\eta,u\rangle
-a|\eta(r)|^2\Im\langle Z_\eta,z\rangle\right\}.
$$
Equation (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 $w=\int\rho{\,\mathrm d} r$, $A_r=\int_0^r\rho{\,\mathrm d} y$, 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
$$
\frac{{\,\mathrm d} K}{{\,\mathrm d} t}
=\frac{\delta J}{\rho}K_S
-\frac{\partial_S\delta B-K\partial_S\delta j}{w}.
$$
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)](/quantum-measurement/research/nonequilibrium-records/conditional-rank-stability-with-zero-fibres-included#eq:rankF). 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
$$
D={\left\lVert{E}\right\rVert},\quad e_1={\left\lVert{E_S}\right\rVert},\quad e_2={\left\lVert{E_{SS}}\right\rVert},\quad
g_j={\left\lVert{\partial_S^jG}\right\rVert},\quad
Q_G^2=\int\frac{{\left\lVert{G_S(S,\cdot)}\right\rVert}^4}{{\left\lVert{G(S,\cdot)}\right\rVert}^2}{\,\mathrm d} S
$$
are finite. The ratio is zero where $G$ and its required derivatives vanish. Then
$$
{\left\lVert{\mathcal F[\psi]-\mathcal F[G]}\right\rVert}_1
\le\frac\hbar M
(4g_1e_1+2e_1^2+14DQ_G+9Dg_2+e_2).
$$
Equation (13).
Here $\mathcal F[G]$ is algebraic evaluation, including whatever residual the comparison has under the actual equation.
**Proof.**
At a fixed slice put $g=G(S,\cdot)$, $b={\left\lVert{g}\right\rVert}$, $\xi=g/b$, $U=G_S$, $V=G_{SS}$, and $d={\left\lVert{q-g}\right\rVert}$, $d_1={\left\lVert{u-U}\right\rVert}$, $d_2={\left\lVert{z-V}\right\rVert}$. The elementary estimates
$$
{\left\lVert{\eta-\xi}\right\rVert}\le2d/b,\quad {\left\lVert{Z_\eta}\right\rVert}\le1/2,
\quad{\left\lVert{Z_\eta-Z_\xi}\right\rVert}\le3{\left\lVert{\eta-\xi}\right\rVert},
\quad{\left\lVert{|\eta|^2-|\xi|^2}\right\rVert}_1\le2{\left\lVert{\eta-\xi}\right\rVert}
$$
follow by adding and subtracting $q/b$, by projection variance, and by $|K_\eta-K_\xi|\le2{\left\lVert{\eta-\xi}\right\rVert}$. Split the first term of [(12)](/quantum-measurement/research/nonequilibrium-records/conditional-rank-stability-with-zero-fibres-included#eq:rankF) first in $u-U$ and then in normalized directions. Its derivative part is at most $4{\left\lVert{U}\right\rVert} d_1+2d_1^2$; its direction part is at most $7{\left\lVert{\eta-\xi}\right\rVert}{\left\lVert{U}\right\rVert}^2\le14d{\left\lVert{U}\right\rVert}^2/b$. For the second term, the $z-V$ part is at most $ad_2$. Splitting the remaining amplitude, density and projection yields $(d/2+4b{\left\lVert{\eta-\xi}\right\rVert}){\left\lVert{V}\right\rVert}\le9d{\left\lVert{V}\right\rVert}$. Integrate these inequalities and use Cauchy–Schwarz and $\int a^2{\,\mathrm d} S=1$. 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 $u$,
$$
Q_G\le{\left\lVert{\phi'^2/\phi}\right\rVert}_2+\sup_S{\left\lVert{u_S}\right\rVert}^2,
\qquad
{\left\lVert{\phi_0'^2/\phi_0}\right\rVert}_2
=\frac{4\pi^2}{\sigma^2}\sqrt{\frac3{35}}.
$$
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 $b(S)$,
$$
\int w(S)|K_\psi(S,b(S))-K_G(S,b(S))|{\,\mathrm d} S\le2{\left\lVert{\psi-G}\right\rVert}.
$$
Under the already transported actual-law cap $C$, the corresponding binary-test mismatch is at most $2C{\left\lVert{\psi-G}\right\rVert}$ per cut.
**Proof.**
Multiply the CDF difference by $w=\int|\psi|^2$. The exact identity is
$$
w(K_\psi-K_G)=\int({\mathbf1}_{r\le b}-K_G)(|\psi|^2-|G|^2){\,\mathrm d} r.
$$
The coefficient has modulus at most one and the density difference has $L^1$ norm at most $2{\left\lVert{\psi-G}\right\rVert}$. 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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