Appendix B Computation Scripts, Computations 185 to 203

Computation 185 ·  §11.3 ·  [PROOF]
W4 is not an independent gap; it follows from the slice-wise CAR limit.

Computation 181 closed the route this book had named for microcausality and moved the target to the projected Dirac propagator. That target is a corollary of the other residual listed in the same register item, and the two should be counted once. W4 asks that $\{\psi_{\alpha}(x),\bar{\psi}_{\beta}(y)\}=iS(x-y)$ vanish at spacelike separation, and $S$ is not a further object but the causal propagator $G^{+}-G^{-}$. On a globally hyperbolic spacetime a normally hyperbolic operator has unique advanced and retarded Green’s operators; a Dirac-type operator is first order but its square is normally hyperbolic, so it inherits them, with $\operatorname{supp}G^{\pm}(x,\cdot)\subseteq J^{\pm}(x)$, hence $\operatorname{supp}S(x,\cdot)\subseteq J(x)$ and CAR quantisation on a Cauchy slice gives microcausality outright [213, 214]. Three of that theorem’s four hypotheses this book already supplies: $M=\mathbb{R}\times S^{3}$ is globally hyperbolic with compact Cauchy surfaces $\{t\}\times S^{3}$, which Computation 4’s entry states while fixing the spin structure; its Cauchy problem is well posed, §4.12 deriving the signature from exactly that; and the projected field satisfies (120). The fourth is the Cauchy-slice CAR limit, which §19.9.1 lists beside W4 as the item’s other residual, so the item has one residue and it is the slice. The step the reduction leans on is measured rather than quoted. The book says the Pauli–Jordan function vanishes outside the cone “because of the mass shell”; the $1{+}1$ scalar case computed from its mode integral reproduces the exact $-\tfrac{1}{2}\,\mathrm{sgn}(t)\,\theta(t^{2}-x^{2})\,J_{0}(m\sqrt{t^{2}-x^{2}})$ to five decimals inside the cone and sits at $10^{-8}$ outside it. Change the dispersion and nothing else: a superluminal $c=2$ leaves $-1.1\times 10^{-1}$ outside the $c=1$ cone and a Schrödinger dispersion $2.7\times 10^{-2}$. The superluminal control is the informative one, having a cone of the wrong opening angle, so what is measured is the angle and not merely the presence of support. The spinor case needs no separate argument, $S$ being $(i\gamma^{\mu}\partial_{\mu}+m)\Delta$ and differentiation not enlarging support; the derivatives are measured outside the cone as central differences of the converged function, five orders below their values inside. A method note is kept because the first attempt gave the wrong answer: differentiating under the integral removes the $1/\omega$ damping and leaves a quadrature cancellation floor at $10^{-4}$ that does not improve as the cutoff rises from $K=100$ to $1600$, while the function itself converges to $10^{-8}$ over the same range. Two things are named rather than used quietly: the Cauchy slice is not supplied here and remains the item’s only residue; and the reduction needs the limit spacetime to be Lorentzian, which this book derives in §4.12 from strict hyperbolicity and from KO-dimension two, not from the Mosco theorem (§7.8), whose statement delivers $E^{\mathrm{Riem}}$ and the Laplace–Beltrami operator. A Dirichlet-form limit is an elliptic object and an elliptic operator carries no cone; reading cone support out of the Mosco limit directly would be the same elliptic-for-hyperbolic confusion Computation 181 found one level down.

 

Computation 186 ·  §11.3 ·  [SURROGATE]
The Cauchy slice is a slab, and the missing power of mass is its thickness.

Computation 185 reduced this item to one residue, the slice-wise CAR limit, and did not supply it. The construction it was missing is forced by a count this book had already made. A Cauchy slice is three-dimensional inside the four-dimensional $M$ while $\rho_{n}(D_{n})$ is finite, so $\rho_{n}^{-1}(\Sigma)$ is empty at every size, measured here; posing the relations slice-wise therefore cannot mean restricting to sites realised on $\Sigma$, because there are none. The book’s own diagnosis of the earlier error says what it does mean. That version “wrote $\delta^{(4)}(x-y)$, summing over sites realised into the four-dimensional $M$ rather than into a slice”, which is “dimensionally wrong by one power of mass”, and the missing power is a thickness: the slab $S_{n}(t)=\{a:|\tau(\rho_{n}(a))-t|\leq h_{n}/2\}$ has site density $\nu_{n}=\lambda_{n}h_{n}=h_{n}/a_{n}^{4}$ per unit $3$-volume, carrying $[\text{mass}]^{3}$ where $\lambda_{n}$ carries $[\text{mass}]^{4}$, and smearing with $\nu_{n}^{-1/2}$ lands on $\int_{\Sigma}\langle f,g\rangle\,d^{3}x$, the distributional $\delta^{(3)}$. Measured against the exact slice integral, with the old reading beside it low by exactly $1/h$. The error then splits in two and each condition kills one term: $h_{n}\to 0$ kills a bias measured as $\mathrm{O}(h^{2})$, tracking its predicted coefficient to a constant of about $1.2$ on the rows where it clears the standard error of the mean, and $\lambda_{n}h_{n}\to\infty$ kills a fluctuation measured as $\mathrm{O}(\nu_{n}^{-1/2})$, which is the CLT rate and so the Berry–Esseen rate this section says the convergence should match. Thinning the slab helps one and hurts the other, so these are two conditions rather than one stated twice, and both hold on the wide window $a_{n}^{4}\ll h_{n}\ll 1$. Balancing them picks out a preferred thickness, $h_{n}\sim a_{n}^{4/5}$ with total error $a_{n}^{8/5}$; the optimal slab is thicker than the site spacing and holds $a_{n}^{-1/5}$ sites across it. The measured RMS minimum sits within a constant of order one of that scaling and the convergence exponent at $-0.36$ against $-0.40$, both reported rather than tuned. Tamper-tested three ways, and the informative one is the slab that never shrinks: it converges perfectly well, to the slab average instead of the slice value, so the measurement distinguishes a slice from a slab rather than detecting noise. What this does not do: it is the i.i.d. surrogate, the standing of A2’s verification and A3’s corroboration (Computations 70, 83), and for the emergence-determined $\rho_{n}$ the equidistribution it leans on is A3, supplied by P3 rather than here, with P3’s time function needed besides. So the item is not closed here; it has gone from an absence to a construction resting on the condition the Mosco chain already carries. Stands in for: the emergence-determined $\rho_{n}$, by an equidistributed cloud in a unit $4$-box.

 

Computation 187 ·  §4.1 ·  [PROOF]
Only the atom (14) forces survives a generic $T$, and that is the one bounded.

Every atom this book has measured in the Born obstruction, in Computation 145, Computation 180 and Computation 183 alike, was measured on $T_{\mathrm{ex}}$, which §4.1 introduces as “a minimal example for the magnitude structure” and names as “one representative among” a family whose dimension grows as $2^{|D|}$. $T_{\mathrm{ex}}$ is integer-valued, so ties are forced on it by counting alone: at $|D|=14$ it maps $16384$ configurations onto a few dozen values. P2 makes the theory’s $T$ $V_{7}$-valued and hence $|T(C)|$ real, so asking which atoms are the theory’s and which the exemplar’s needs no new premise; it is the book’s own freedom, used. The split is exact. (14) puts $|T(C)|=0$ on exactly the configurations with no distinct pairs, which is admissibility rather than choice, so every admissible $T$ carries an atom of mass $\mu(\text{independent sets})$ and no other tie is forced: largest atom $\geq\mu(\text{independent})$, with equality for a generic one. Measured with positive real edge weights, admissible and checked here against (14) and (15) rather than asserted, the largest atom equals $\mu(\text{independent})$ exactly in every family Computation 180 tabulated, over twenty independent weightings each. A correction to Computation 180’s rate, not to its verdict: the complete graph’s atom is not $0.21$ falling as $|D|^{-0.5}$ but $9\times 10^{-4}$ falling exponentially, its independent sets being only the empty set and the singletons, so the passing families pass for a better reason than was recorded and §10.3.2.5’s sentence is corrected. An upgrade to Computation 183: its proof runs through the configurations inside a cover’s complement, which are independent sets, so the chain largest atom $\geq\mu(\text{independent})\geq 2^{-c}$ is $T$-independent at both steps and holds for every admissible $T$ rather than for the exemplar it was measured on. That also explains rather than re-measures why Computation 183’s search for a second mechanism found none: under a generic $T$ there is none, the cover bound being the whole obstruction rather than one mechanism among several. What it does not do is remove P1(iii): the surviving atom is precisely the one reducedness kills, and the star holds $1/2+2^{-|D|}$ under every admissible $T$, matched here to its closed form. Tamper-tested on the step that carries the weight: letting a third of the weights vanish breaks (14), and the atom then leaves $\mu(\text{independent})$, so the equality is a consequence of admissibility and not of merely using real numbers.

 

Computation 188 ·  §9.1 ·  [NUMERICAL]
Neither an area law nor the finiteness fixes $\Lambda$’s magnitude.

Closes a question the register carried since July 2026: whether the area-law property of $\Pi$ or the finiteness of the internal structure constrains the static valley depth $\Delta\mathcal{F}$. (i) The second edition derives no area law (Table LABEL:tab:framework-comparison), so the first leg is moot. (ii) An area law quantised in a length the theory owns would miss $A/4\ell_{P}^{2}$ by $(\ell_{P}/\ell)^{2}$: $10^{-32}$ for $\ell_{*}$, $5\times 10^{-53}$ for $d_{0}$. (iii) The energy bound such a law implies, no more vacuum energy in a region than a black hole of its size, is $3c^{4}/(8\pi G_{\ell}L^{2})$ with $G_{\ell}=\ell^{2}c^{3}/\hbar$; at the Hubble scale it is $10^{-32}$ and $5\times 10^{-53}$ of the critical density, both violated by the observed $0.7$, so such a law is false, not weak. (iv) With the emergent $G$ of (61) the bound is general relativity’s, $\rho_{\Lambda}\leq\rho_{\mathrm{crit}}$, satisfied at $0.71$ and an inequality on $j_{\tau}\,\Delta\mathcal{F}$, fixing neither factor; Computation 82’s $j_{\tau}\approx 8\times 10^{-38}$ per configuration is reproduced. (v) The finiteness sets $M_{*}^{4}$, $10^{59}\,\rho_{\Lambda}$, Mechanism A of §9.1. Negative on both legs, no fourth postulate, nothing imported; the sign, $w=-1$ and non-dilution are untouched.

 

Computation 189 ·  §8.3.6 ·  [PROOF]
The photoelectric cross-section and rate from the derived vertex.

The application Computation 166 left. (1) In the non-relativistic limit the vector vertex couples $\mathbf{A}$ to the electron’s momentum; Fermi’s golden rule on a hydrogenic $1s$ state with a plane-wave final state gives, by a symbolic angular integral, $\sigma_{B}=(2^{8}\pi/3)\,\alpha\,a_{0}^{2}\,k^{3}/(1+k^{2}a_{0}^{2})^{5}$, whose high-energy form is the Born law $(2^{8}\pi/3)\,\alpha\,a_{0}^{2}\,(I_{\mathrm{H}}/\hbar\omega)^{7/2}$. (2) The exact Coulomb-wave dipole result [215] tends to it as $k\to\infty$ (ratio $0.97$ at $ka_{0}=100$) and supplies the threshold, $6.30$ Mb for hydrogen, where the plane-wave form vanishes as $k^{3}$. (3) The axial alternative reduces to $\boldsymbol{\sigma}\cdot\mathbf{A}$, computed from the explicit Dirac matrices, a pure spin operator with zero matrix element between orthogonal eigenstates: no photoeffect at this order, so the forced form is also the needed one. (4) The rate is $\sigma$ times the photon flux: at $1\,\mathrm{W\,m^{-2}}$ and $20$ eV, $2.2$ Mb and $7\times 10^{-5}\,\mathrm{s^{-1}}$ per atom with an exponential waiting time and no accumulation, so $10^{15}$ atoms emit their first electron within $1.4\times 10^{-11}$ s, inside the Lawrence–Beams $3\times 10^{-9}$ s, where a classical field accumulating a work function through $\pi a_{0}^{2}$ needs $248$ s. (5) $\alpha$ is taken as measured (§13.3.1); the law’s form and its proportionality to $\alpha$ are the derived content.

 

Computation 190 ·  §10.3.2.5 ·  [PROOF]
Born agreement on every reduced distinction graph: the forced atom vanishes, as a theorem with a rate.

The theorem Computation 183 measured and the register carried as missing. Under $\mu=\mathrm{Bern}(\tfrac{1}{2})^{|D|}$ the forced atom is $a_{0}=\mu(C\text{ independent})$, by (14) the same for every admissible $T$ (Computation 187). (1) Lemma 1: a matching of $m$ disjoint distinctions gives $m$ independent events of probability $\tfrac{1}{4}$, none of which an independent $C$ realises, so $a_{0}\leq(3/4)^{m}$. (2) Lemma 2: for a maximal matching the unmatched properties are independent, so each has its distinction set inside the $2m$ matched ones, and reducedness (Computation 184) makes those sets distinct: $|D|-2m\leq 4^{m}$. Hence $a_{0}\leq\max\bigl{(}(3/4)^{|D|/4},(|D|/2)^{-0.21}\bigr{)}$ for every reduced $(D,\delta)$, $10^{-17}$ at $|D|=10^{80}$. (3) Both lemmas are checked exactly, $a_{0}$ enumerated over all $2^{|D|}$ configurations, on $200$ random graphs at $|D|=12$ to $16$, on the subset-graph family to $|D|=520$ by a closed form for its independent-set count, exhaustively over every reduced graph at $|D|\leq 6$ and by hill-climbing at $|D|\leq 16$; the largest reduced atom found sits well under the bound. Lemma 2 fails on the star and on $K_{k}$ with enough isolated properties, the non-reduced families of Computation 180, so the hypothesis does the work. (4) Tamper: false twins on the $c=3$ subset graph freeze the matching at $m=3$ while $|D|$ grows to $73$, break Lemma 2, and pin the atom on $2^{-c}$ to $10^{-13}$. (5) The bound is universal and not sharp: the subset-graph atom falls as $|D|^{-1}$, its $|D|^{-1.1}$ over $c=3$ to $7$ reproducing Computation 183. What remains is a condition on the law of $T$ alone, item 1’s residual.

 

Computation 191 ·  §11.3 ·  [PROOF]
The Cauchy slab on a deterministic cloud: the slab-wise CAR limit needs equidistribution with a rate, not independence.

Removes the surrogate from Computation 186. (1) For any $N$ points in the unit $4$-box the slab estimator $S_{h}=(Nh)^{-1}\sum_{\mathrm{slab}}\langle f,g\rangle$ obeys the identity $S_{h}-\int_{\Sigma}\langle f,g\rangle=\mathrm{bias}(h)+\mathrm{disc}(P,h)/h$: the bias is cloud-independent and exact ($3h^{2}/4+h^{4}/20$ on the test integrand), and the second term is the cloud’s equidistribution defect tested on the integrand times the slab indicator, which Koksma–Hlawka bounds by $C\,D_{N}/h$. So $|S_{h}-\int_{\Sigma}|\leq c\,h^{2}+C\,D_{N}/h$, the limit holds for every sequence of clouds with $D_{N}/h_{n}\to 0$, and the balanced thickness is $h_{n}\sim D_{N}^{1/3}$ with error $D_{N}^{2/3}$. (2) On the Halton and Kronecker sequences, eight disjoint blocks each, the error falls below the i.i.d. surrogate’s at every size from $2^{12}$ to $2^{20}$ and at the predicted rate, $|D_{n}|^{-0.65}$ and $|D_{n}|^{-0.48}$ against $-2/3$, where the surrogate gives $-0.39$; a jittered grid, i.i.d. below its cell scale, tracks the surrogate, so stratification at one scale is not equidistribution at every scale. (3) The discrepancy term alone falls as $|D_{n}|^{-0.92}$ and $|D_{n}|^{-0.86}$ on the two sequences against the surrogate’s $|D_{n}|^{-0.60}$, the CLT’s. (4) A smooth window in place of the sharp slab tests the cloud on a $C^{1}$ function, A3’s class, with bias $9h^{2}/7$ exactly and the same convergence. (5) Tamper: a Halton cloud with the same count but a spatial void refilled elsewhere converges to the wrong slice value, the defect flat across a sixteenfold count increase and equal to the analytic void bias, so A1 is not the hypothesis. (6) Tamper: a cloud equidistributed in space with time density $1+\epsilon\cos 2\pi t$ returns $(1-\epsilon)$ times the slice value, so the time function must be the one in which the cloud is $4$-volume equidistributed. What the item rests on is A3’s rate and that time function, nothing of its own.

 

Computation 192 ·  §11.3 ·  [PROOF]
The time function the slab is cut in: isotropy forces the foliation, its level sets are the Cauchy surfaces, and the slab’s normalisation is the proper $3$-volume.

Supplies what Computation 191 left: the time function, built from what the book already has. P3 fixes $\hat{\tau}$ globally; Chapter 5 maps it to $\Phi(\hat{\tau})=\partial_{t}$ along the $\mathbb{R}$ factor; the Cosmological Principle makes the geometry Friedmann–Robertson–Walker; A3 equidistributes the cloud against $a(t)^{3}\,dt\,d\Omega_{3}$. (1) The foliation is forced. A timelike field is tangent to a foliation iff $u\wedge du=0$, iff its twist vanishes; the twist is a spatial vector, and isotropy admits no invariant spatial vector, so it is zero. Symbolically on (220), $u=dt$ has $u\wedge du=0$ for every $a(t)$, while the rotating tamper $u=dt+\Omega r^{2}d\phi$ gives $2\Omega r\,dr\wedge d\phi\wedge dt$: no foliation, no slab. (2) The level sets are compact Cauchy surfaces: $dt/d\lambda>0$ along every future-directed causal curve, checked on two hundred null geodesics on $a=e^{Ht}$ and on the matter-plus-instantiation solution, each crossing each of six level sets exactly once. (3) The normalisation is the proper $3$-volume. A3 makes the cloud’s time marginal proportional to $a^{3}$, so the uniform normalisation of Computation 191 would return $a(t_{0})^{3}/\langle a^{3}\rangle$ times the slice value, from $0.2$ to $2.3$ across the range tested, while $\nu=\lambda\,a(t_{0})^{3}V_{3}\,h$ lands on it at every slice to $0.1\%$. (4) The one choice left carries no content: relabelling the leaves by a monotone $f$ moves the uniform estimator by $1/f^{\prime}(t_{0})$ and the covariant one not at all. With Computation 185 and Computation 191, the slab-wise CAR limit, and W4 with it, is a theorem given A1 to A4 with A3 supplied by P3: the standing of equicoercivity A7, and the item leaves the structural front on those terms.

 

Computation 193 ·  §19.9.1 ·  [PROOF]
One condition, two faces: the Born condition on $T$’s contingent content and non-balance are both the non-degeneracy of that content.

The contingent content of an admissible $T$ is a point $(t,v)$: a magnitude $t$, any order-preserving function on $2^{D}$ vanishing exactly on the independent sets (a cone of dimension $2^{|D|}-i(G)$), and the directions $v\in(S^{6})^{|D|}$. (1) Magnitude face. The law of $|T|$ under $\mu$ has an atom off zero iff two non-independent configurations tie; the tie set is a finite union of $\binom{2^{|D|}-i(G)}{2}$ hyperplanes, closed of codimension one. On generic points of the cone (increments continuous, two hundred draws on three graphs at $|D|=6,8,10$) the largest atom is $i(G)/2^{|D|}$ exactly and there is no unforced atom; $T_{\mathrm{ex}}$ has unforced atoms, taking at most $|E|+1$ values. (2) Direction face. The balance set is closed of codimension seven: for $v$ uniform on $(S^{6})^{8}$, $\Pr(|\sum v|<r)$ scales as $r^{6.7}$ over $r\in[0.45,1]$ on four million draws, and $\mu_{\pm}$ balances with $\binom{|D|}{|D|/2}2^{-|D|}$, zero at odd $|D|$. (3) Independence: in the table (generic or $T_{\mathrm{ex}}$) $\times$ (continuous or balanced $v$), unforced atoms follow the first factor only and balance the second only. (4) The distance to degeneracy, $\min\bigl{(}\min_{C\neq C^{\prime}}|t_{C}-t_{C^{\prime}}|/\sqrt{2},\ |\sum v|/\sqrt{|D|}\bigr{)}$, is positive on every generic draw and zero on every degenerate one. So both faces are the same kind of statement, open and dense, null under continuous content, positive under discrete content, fixed by the theory on neither; the register carries one condition with two faces, the magnitude face being item 1’s residual, and the count is unchanged.

 

Computation 194 ·  §9.5 ·  [PROOF]
The unit of orbital angular momentum: $L_{z}=nQ$ exactly, the quantum of $Q$ is an action the classical theory does not have, and the only action scale is the normalisation of the gradient term.

(1) From (81), $xT_{y0}-yT_{x0}=\partial_{\phi}\theta\cdot\Pi$ with $\Pi=2r_{0}^{2}\,\partial_{t}\theta$, and for every axisymmetric $\theta=n\phi+\chi(\rho,z,t)$, $\partial_{\phi}\theta=n$ identically, so $L_{z}=nQ$ with $Q=\int\Pi\,d^{3}x$ the Noether charge: an identity, no far-field assumption, checked symbolically. (2) $Q$ has the dimension of action (checked on $(M,L,T)$ with $r_{0}^{2}$ an energy per length as (90) has it); the canonical commutator makes $Q\in\hbar\mathbb{Z}$ and the Bohr–Sommerfeld cycle makes $2\pi Q\in h\mathbb{Z}$, each with the unit as input; classically $Q$ is continuous. (3) $r_{0}^{2}d_{0}^{2}/c$ is the unique monomial in $(r_{0}^{2},d_{0},c)$ with the dimension of action (exponents searched in $[-3,3]$), so a derived unit $\hbar=\kappa\,r_{0}^{2}d_{0}^{2}/c$ is $r_{0}^{2}=\hbar c/(\kappa d_{0}^{2})$, the absolute Goldstone stiffness, which is what the normalisation of P3’s gradient term sets through $G=6\pi d_{0}^{2}/(c\,r_{0}^{2})$ (74): item 4’s residual in another variable. Illustration, not prediction, and corrected by Computation 197: $m_{0}c^{2}=(4\pi^{3}/3\kappa)\,\hbar c/d_{0}=3.5$ keV$/\kappa$ at $d_{0}=2.3$ nm was printed with $r_{0}$ left free, while the book fixes $r_{0}=v$, which makes $\kappa=1.2\times 10^{-19}$ there and the mass $2.9\times 10^{13}$ GeV. (4) On the regulated vortex $\theta=n(\phi-\omega t)$, $L_{z}=nQ$ at every outer radius while both grow as $R^{3}$ and the energy per unit length as $\ln(R/d_{0})$: the global vortex, a property of the class and not a gap; $\pi_{2}(S^{1})=0$ keeps the lump of (92) a mass scale and not a vortex. The item has nothing of its own and folds into item 4, as that item’s closing clause already says.

 

Computation 195 ·  §19.9.1 ·  [PROOF]
Maximum entropy supplies non-degeneracy, if P3’s typicality reaches the content of $T$.

The argument Computation 193 named as the one that would close the condition, and what it needs. The Gibbs variational principle: relative to the invariant reference measure (the uniform product on $(S^{6})^{|D|}$, Lebesgue measure on the admissible cone with a scale constraint), $D(P\,\|\,\nu)\geq 0$ with equality only at $\nu$, and the maximum-entropy law under moment constraints is an exponential family with a density; so every closed nowhere-dense set is null under maximum entropy, the tie set and the balance set among them. (1) Directions: the von Mises–Fisher family of concentration $\kappa$ about $\pm u$, $u$ uniform and the signs fair (the signs Computation 178 forces), lies inside the symmetry package; its entropy, exact from the Bessel functions, decreases strictly in $\kappa$ from $\log(16\pi^{3}/15)$ at the uniform product; exact balance has probability zero at every finite $\kappa$ (a power law of slope six to seven at small radius on a million draws) and the binomial mass $\binom{|D|}{|D|/2}2^{-|D|}$ only at $\kappa=\infty$, which is $\mu_{\pm}$, of infinite relative entropy. (2) Magnitudes: exponential increments along inclusion, the maximum-entropy increments under a mean constraint, carry only the forced atom on 200 draws per graph; a lattice of $L$ increment levels ties, with mass $0.27$ at $L=1$ ($T_{\mathrm{ex}}$) falling to $0.01$ at $L=256$. (3) The reference measure is load-bearing: on the fourteen octonion units, Computation 152’s quantised scope, the maximum-entropy law is uniform on a finite set and balances with probability $1/14$ at $|D|=2$ and positively at every even $|D|$. (4) Both faces hold on every one of 500 joint maximum-entropy draws. What the argument needs: that P3’s clause, which selects the realised configuration consistent with the tension data, reach the content of $T$ that is its input. A reading of P3, not a new postulate; under it the condition has A3’s standing, and the register carries it as what that reading would discharge.

 

Computation 196 ·  §7.5 ·  [PROOF]
The truncation spacing $a_{*}$ is comoving, and under the conformal identification it is the thermal wavelength of the state’s radiation.

(1) On the Friedmann geometry with the curvature $k$ symbolic, the divergence of a homogeneous isotropic traceless tensor is $\dot{\rho}+4H\rho$ exactly, the spatial components zero; $\rho=Ca^{-4}$ solves it and $\rho=\mathrm{const}$, which is a fixed physical $a_{*}$, leaves $4H\rho$; dust gives $3H\rho$, so the fourth power is the tracelessness; a perfect fluid is traceless iff $w=1/3$. (2) Under $\Pi^{0}{}_{0}=\rho_{r}$: $\rho_{\gamma}(2.7255~{}\mathrm{K})=4.17\times 10^{-14}$ J m^-3, neutrino multiplier $1.69$, $a_{*}=(\hbar c/\rho_{r})^{1/4}=0.818$ mm, equal to $(15/\pi^{2}f)^{1/4}=0.974$ times $\hbar c/kT_{\mathrm{CMB}}$ and scaling as $1/T$; $a_{*}=d_{0}$ at $kT=83$ eV and $a_{*}=0.41\,\ell_{*}$ at $kT=M_{*}$. (3) Tamper: a fixed $a_{*}$ fails conservation at the relative rate $4H_{0}$; a general $w$ gives $3(1+w)H\rho$. The magnitude of $\Pi_{\mu\nu}$ is supplied by the state and not owed by the theory, which was the first structural item’s only remaining gap. Scope, from Computation 201: the $83$ eV crossing is at the Casimir bound; at the derived $d_{0}$ it is $kT\approx 40$ GeV.

 

Computation 197 ·  §19.9.1 ·  [PROOF]
The normalisation item with $r_{0}=v$: the derived-unit coefficient is $1.2\times 10^{-19}$ at the Casimir bound, and the dependency graph of $K$ has one free node, $d_{0}$.

(1) $\hbar c/(r_{0}^{2}d_{0}^{2})$ is dimensionless on $(M,L,T)$ and $6\pi d_{0}^{2}/(Gv^{2})$ carries $[M]^{-2}$, as the note after (74) says. (2) With $r_{0}=v$: $\kappa=\hbar c/(v^{2}d_{0}^{2})=1.21\times 10^{-19}$ at $d_{0}=2.3$ nm; $\kappa=1$ at $d_{0}=\hbar c/v=0.80$ am, $2.9\times 10^{9}$ below the bound; (92) gives $m_{0}=2.9\times 10^{13}$ GeV at the bound and $(4\pi^{3}/3)v=10.2$ TeV at $\kappa=1$, linear in $d_{0}$ ($1$ TeV at $d_{0}=7.9\times 10^{-20}$ m); Computation 194’s $3.55$ keV$/\kappa$ divided by that $\kappa$ is the same $2.9\times 10^{13}$ GeV, so the illustration is corrected and the arithmetic is not. (3) The graph: $v$, $m_{h}$, $G$ measured; $\varepsilon=m_{h}^{2}/4$ and $b=m_{h}^{2}/(8v^{2})=0.032$ fixed by them; $c=6\pi d_{0}^{2}/(Gv^{2})=6.3\times 10^{48}$ GeV^-2 at the bound; $K$, $c$, $\kappa$ and $m_{0}$ hang on $d_{0}$ and on nothing else. (4) Tamper: with $r_{0}$ free the coefficient is unconstrained, which is the reading Computation 194 took. The item stays open as an assumption; what closes it is a prediction of $d_{0}$. Scope, from Computation 201: the graph omits (205), which ties $K$ to $d_{0}$; with it the canonical reading fixes $d_{0}=2.2$ am and $\kappa=0.13$, order one.

 

Computation 198 ·  §12.2 ·  [NUMERICAL]
Stability of the regulated vacuum at two loops: the running quartic is positive on the whole range the theory contains.

Computation 123’s two-loop running copied and completed with the standard two-loop $\beta_{\lambda}$ (GUT-normalised $g_{1}$). (1) Validation: $\lambda(M_{*})=0.09145$ at one loop and $0.09234$ without the $\lambda$ piece, Computation 123’s numbers to five decimals; the two-loop $\lambda$ terms move it by $-0.00006$ to $0.09228$, within the $0.0007$ of the literature’s $0.0927$; the quartic crosses zero at $10^{9.7}$ GeV, inside the known window; $0.4\sigma$ from $\mathrm{e}^{-1}/4$. (2) On $[m_{t},M_{*}]$, $\lambda$ decreases monotonically from $0.126$ to $0.092$ with its minimum at $M_{*}$, so $V=\lambda(\phi)\phi^{4}/4$ is positive at every field value the regulated theory contains; the instability scale is $6.5$ decades above $M_{*}$; $\lambda(M_{*})=0$ would need $m_{t}=224$ GeV; over the $3\sigma$ box of $(m_{t},\alpha_{s},m_{h})$ it stays in $[0.088,0.096]$. (3) The mass parameter’s running is multiplicative ($m^{2}(M_{*})/m^{2}(m_{t})=1.06$), so the sign driving symmetry breaking is the same at every scale. (4) Tamper: without $-6y_{t}^{4}$ the quartic never crosses zero below $10^{19}$ GeV, so the check is sensitive to the terms it carries. Stability beyond leading loop order is established in the RG-improved sense; the item is the construction at all orders. Forward reference: the SM-side value quoted here is the anchor of its day; Computation 202 restates it on current inputs at three loops, $0.0930\pm 0.0008$ on the 2026 Review, $1.4\sigma$ from $\mathrm{e}^{-1}/4$, which is what the book states from v32.153 (v32.137 printed $1.6\sigma$ on the 2024 Review).

 

Computation 199 ·  §7.5 ·  [PROOF]
The conformal part of the pair sum reaches exactly the traceless tensors obeying the null energy condition, so the identification has no obstruction.

Removes the obstruction Computation 164 found. (1) Term by term, $\Delta_{\mu}\Delta_{\nu}-\tfrac{1}{4}s\,g_{\mu\nu}$ is traceless identically (symbolic); the conformal part of the sum is the sum of these; the trace of the raw sum is $\sum|s|\,s$, minus Computation 158’s summed $T-\tau_{\mathrm{null}}$, the balance scalar. A timelike unit bond gives $\rho=3/4$, $p=1/4$; a spacelike one gives pressure $3/4$ along itself and tension $1/4$ across. (2) Explicit clouds: $96$ points on concentric regular polygons in the $x$-$y$ plane give $\mathrm{diag}(u,u,u,-u)$ after the projection with deviation $3\times 10^{-16}$ (magnetic, field along $z$), the $y$-$z$ plane gives the electric tensor along $x$; one $x$-bond plus one $y$-bond is $\tfrac{1}{2}\mathrm{diag}(1,1,1,-1)$ exactly. In Computation 158’s balanced regime the trace fraction runs $2.6\times 10^{-2}$, $2.9\times 10^{-3}$, $2.4\times 10^{-4}$, $7.4\times 10^{-5}$ as the window doubles, and the projection changes the sum by a quarter of that, so Computations 158 to 161 are untouched; Computation 164’s six shapes all obey the null energy condition after projection. (3) Inclusion, symbolic: for null $k=(1,\mathbf{n})$, $G(k,k)=(\Delta\cdot k)^{2}\geq 0$. (4) Equality, numerical: random positive semidefinite metric-traceless $Y$ decompose into $200$ sampled null dyads with zero residual (the dual cone); random traceless tensors obeying the condition with margin decompose into $1600$ sampled generators with zero residual, violators keep a residual above $0.5$ times their violation, a null dyad being the certificate; the magnetic and electric tensors are reached at $10^{-3}$ and $7\times 10^{-5}$ while the zero-energy anisotropic stress and negative-energy radiation are not. (5) Pataki: a spectrahedron with two equations has rank-one extreme points only, so the dual’s extreme rays are null dyads; the rank-two case by hand, $200$ draws each equal to $c_{1}k_{1}k_{1}+c_{2}k_{2}k_{2}$ on the two null directions of its Lorentzian plane, residual $10^{-13}$, coefficients positive. (6) Tamper: a signed weight $W=-s$ reaches the zero-energy anisotropic stress (residual $0$ against $1$) and gives a cloud strung along the time axis a negative energy density. The identification is of conformal parts; an unbalanced sum carries a trace it does not use.

 

Computation 200 ·  §12.2 ·  [PROOF]
The interacting vacuum of the regulated scalar sector exists nonperturbatively; what remains is the gauge and fermion sector, the cutoff Standard Model’s own.

Three classical theorems applied to a theory that never removes its cutoff. Finite volume: with finitely many modes the Hamiltonian is a Schrödinger operator on $L^{2}(\mathbb{R}^{N})$ with a quartic potential bounded below, essentially self-adjoint with compact resolvent, and its semigroup is positivity improving, so the ground state is unique and strictly positive. Infinite volume: the Euclidean lattice theory at spacing $\ell_{*}$ has ferromagnetic couplings and an even single-site weight, so with Dirichlet data Griffiths’ second inequality makes every correlation non-decreasing in the volume, and the quartic bounds it; the limit exists in any dimension, the dimension entering only the continuum limit the theory does not take. Back to a Hamiltonian: the nearest-neighbour action is reflection positive, so the transfer matrix is a positive operator and $-\ell_{*}^{-1}\log T$ is bounded below with the Euclidean vacuum as its ground state. Checks: (1) one and two sites by finite differences, every ground state nodeless and even, the double well’s gap closing as it deepens ($0.35$ to $3\times 10^{-5}$) and as a site is added ($0.0097$ to $5\times 10^{-5}$); (2) the one-site Trotter transfer matrix positive to round-off with its gap converging to the Hamiltonian’s, $0.3344$, $0.3427$, $0.3451$, $0.3457$ against $0.3458$ as the step halves; (3) four-dimensional lattice Metropolis with Dirichlet data at the book’s couplings in units of $\ell_{*}$, $\lambda=0.092$ and $\mu^{2}=-(m_{h}\ell_{*})^{2}/2$, the central nearest-neighbour correlation $0.0264$, $0.0271$, $0.0295$ at $L=4,6,8$ and $\langle\phi^{2}\rangle$ bounded at $0.15$, and at a massive test point flat within errors; (4) tampers: an antiferromagnetic coupling leaves Griffiths’ ordering and gives negative correlations, and a negative quartic has no ground state, the lowest eigenvalue running off as the grid widens, so $\lambda>0$ (Computation 198) is what the construction needs. Not closed and said so: the gauge and fermion sector below $M_{*}$, whose infinite-volume limit at weak coupling and whose chiral fermions on a regulator are the cutoff Standard Model’s own open problems, inherited with it.

 

Computation 201 ·  §15 ·  [PROOF]
The gradient normalisation and the coherence length are one unknown: under the canonical reading, $d_{0}=\sqrt{2}\,\hbar c/m_{h}$, and the nanometre expectation does not follow from the equations.

(205) with $|a_{\scriptscriptstyle LG}|=m_{h}^{2}/4$ leaves one free quantity, $c_{\scriptscriptstyle LG}$, which is the open item. (1) Symbolic: $V^{\prime\prime}(r_{0})=4\varepsilon$ for $V=-\varepsilon r^{2}+br^{4}$ at $r_{0}^{2}=\varepsilon/(2b)$, independent of $b$; the Ginzburg–Landau length $\sqrt{c/(2|a|)}$ is $\sqrt{2}$ times the radial correlation length $\sqrt{c/V^{\prime\prime}}$; at $c_{\scriptscriptstyle LG}=1$, $d_{0}m_{h}=\sqrt{2}$ in the neutral frame and $1$ in the doublet frame. (2) Numbers: $d_{0}=2.23$ am (neutral), $1.58$ am (doublet); $d_{0}/\ell_{*}=14.5$ and $10.26=M_{*}/m_{h}$; $d_{0}/\ell_{P}=10^{17}$; the bound exceeds $d_{0}$ by $10^{9}$; the Casimir correction $\xi(d_{0}/d)^{2}$ is $1.1\times 10^{-19}$, $1.8\times 10^{-20}$, $1.8\times 10^{-21}$ at $20$, $50$, $160$ nm against $0.12$, $0.019$, $0.0019$ at the bound and a precision of $1.9\times 10^{-3}$; Computation 197’s $\kappa=m_{h}^{2}/(2v^{2})=0.129$; (92) gives $28$ TeV; Computation 196’s $a_{*}$ passes through $d_{0}$ at $kT=36$ GeV. (3) A nanometre $d_{0}$ needs $c_{\scriptscriptstyle LG}=1.1\times 10^{18}$, which through Computation 176’s linearity moves $M_{*}$ by $10^{9}$; even $c^{2}=\mathrm{e}$ is excluded at $121\sigma$. (4) The BCS analogy quantified, $d_{0}/\ell_{*}=\sqrt{2}\,M_{*}/m_{h}$ with $M_{*}/m_{h}=4\pi\sqrt{2/3}$ the book’s own; tamper, $|a_{\scriptscriptstyle LG}|$ left free: $2.3$ nm needs $|a_{\scriptscriptstyle LG}|$ at $10^{-18}$ of its value, a radial mode of $10^{-7}$ GeV. The $0.4\sigma$ agreement is not spent: reading A is the standing premise and $d_{0}$ follows from it. Forward reference: the SM-side value quoted here is the anchor of its day; Computation 202 restates it on current inputs at three loops, $0.0930\pm 0.0008$ on the 2026 Review, $1.4\sigma$ from $\mathrm{e}^{-1}/4$, which is what the book states from v32.153 (v32.137 printed $1.6\sigma$ on the 2024 Review).

 

Computation 202 ·  §19.7 ·  [NUMERICAL]
The Higgs self-coupling test at three loops on current inputs: the third loop is worth nothing at $M_{*}$, the 2013 anchor is not reproducible, and the honest agreement is $1.6\sigma$ on the 2024 Review and $1.4\sigma$ on the 2026 one.

The complete three-loop Standard Model $\beta$ functions and the next-to-next-to-leading-order matching conditions at the top mass, both from [51], coded and validated by reproducing that paper’s Planck-scale couplings to four decimals. (1) The third loop moves $\lambda_{\mathrm{SM}}(M_{*})$ by $+0.00007$, a tenth of the error, so the running was converged at two loops. (2) The anchor Computation 123 used, $0.0927\pm 0.0007$, is not reproduced: the source’s formulas on the source’s 2013 inputs give $0.0924\pm 0.0011$ at every loop order, a $0.4\sigma$ agreement, and the $\pm 0.0007$ counted $m_{t}$ at $\pm 0.5$ GeV where 2013 had $\pm 0.76$. (3) On PDG 2024 inputs [216] ($m_{t}=172.57\pm 0.29$, $m_{h}=125.20\pm 0.11$, $\alpha_{s}=0.1180\pm 0.0009$, $M_{W}=80.3692\pm 0.0133$) the value is $0.0932\pm 0.0006$ from the quoted errors, $2.2\sigma$, and $0.0932\pm 0.0008$ once the $0.5$ GeV ambiguity between the measured and the pole top mass is counted, $1.6\sigma$ and $1.4\%$; the cross-section pole mass $172.4\pm 0.7$ gives $0.0934\pm 0.0009$, also $1.6\sigma$. (4) Exact agreement needs $m_{t}=173.7$ GeV, $4\sigma$ above the direct measurement. (5) Re-solving Computation 176’s equation for $c$ on current inputs gives $0.9927\pm 0.0044$, the same $1.6\sigma$, with $\sqrt{\mathrm{e}}$ excluded at $149\sigma$. Tampers: the three-loop QCD piece of the $y_{t}$ matching is worth nine times the third loop of the running, so the precision lives in the matching; Computation 123’s error set is identified. (6) On the 2026 Review [233] ($m_{t}=172.60\pm 0.27$, $m_{h}=125.13\pm 0.11$, $\alpha_{s}$ unchanged, $M_{W}=80.3625\pm 0.0077$) the Higgs mass’s $0.07$ GeV does most of the work: $0.0930\pm 0.0008$ with the pole-mass ambiguity counted, $1.4\sigma$ and $1.2\%$ ($1.9\sigma$ on the quoted errors alone), exact agreement at $m_{t}=173.6$ GeV, and $c=0.9938\pm 0.0044$. The book printed $1.6\sigma$ from v32.137 and prints $1.4\sigma$ from v32.153, on the cover too. (7) Run back down, $\lambda_{\mathrm{SM}}(M_{*})=\mathrm{e}^{-1}/4$ gives $\lambda_{\mathrm{SM}}(m_{t})=0.1251$ against the $0.1260$ the matching extracts from the observed $(m_{h},v)$, so the prediction returns $v=247.1$ GeV from $m_{h}$ alone, $0.4\%$ above $246.22$; the earlier “$0.01\%$” applied the tree-level relation to the running coupling at $\mu=v$ (which gives $256$ GeV even on the data) and is withdrawn.

 

Computation 203 ·  §19.2.1 ·  [PROOF]
The single-monomial L_comm problem closed for every $m$, and the sharp constant certified in exact arithmetic.

Supplies the closed form Computation 157 left as the remaining work. (1) Along a fixed degree the commutator weight obeys the exact ratio test $f(a+1)/f(a)=(a+m+2)\,b/((a+1)(b+m-1))$, at least one exactly while $(m+1)b\geq(m-1)(a+1)$, and both shift weights increase with the degree, so on degree $\leq N$ the maxima sit at the top degree $s=N-2m$: $\|T_{w^{m}}\|^{2}$ at $a=b=s/2$ and $\|[J_{+},T_{w^{m}}]\|^{2}$ at $a^{*}=\lfloor((m+1)s-(m-1))/(2m)\rfloor+1$; the ceiling is the sawtooth. Verified in exact rational arithmetic against every lattice point for $m\leq 8$ and $N\leq 2m+60$. (2) The finite-$N$ single-monomial ratio is therefore an explicit factorial ratio for every $m$, generalising (274); its limit is Lemma 5’s closed form to $10^{-6}$ at $N=4000$, and the deficit $(\mathrm{ratio}_{\mathrm{cf}}-\mathrm{ratio}_{m}(N))N$ tends to $1$ at $m=1$ and to $0$ for $m\geq 2$, the $1/N$ term cancelling, with an $O(1/N^{2})$ sawtooth left. (3) The two-monomial bridge’s $1/N$ constant does not cancel: $0.85$, $0.17$, $0.82$, $0.81$ at $k=2,3,4,6$ measured to $N=480$; extremal chains identified; a harmonic (semiclassical) correction to the chain symbol does not reproduce it. (4) Each chain norm is the top eigenvalue of a nonnegative tridiagonal matrix; Collatz–Wielandt bounds with a rational test vector enclose it exactly. Over $k\leq 20$ at the ten smallest admissible $N$ every $G(k,N)$ is certified below $3.6736$; at $(6,12)$ the enclosure is narrower than $10^{-7}$ and reads $[3.67356,3.67356]$; the widest enclosure, at $k=18$, still certifies $G<1.4$. Caveat: $\alpha(k)$ is an algebraic root taken at fifteen digits. Tampers: one step off $a^{*}$ or one degree down lowers the weight. What remains: an analytic bound uniform in $N$ for the two-monomial bridge. Settled by Computation 213 and Computation 214: the bridge’s $1/N$ term is a limit in closed form, and the sharp value is derived at its corner.