Appendix B Computation Scripts, Computations 137 to 164

Computation 137 ·  §16.5 ·  [PROOF]
Equation of state $w=-1$ from steady-state non-dilution.

Two exact routes to $w=-1$: (A) the FRW continuity equation $\dot{\rho}+3H(\rho+p)=0$ with the non-diluting substrate density $\dot{\rho}=0$ forces $\rho+p=0$; (B) a metric-proportional stress tensor $T_{\mu\nu}=\rho_{\Lambda}g_{\mu\nu}$ gives $T^{\mu}{}_{\nu}=\rho_{\Lambda}\delta^{\mu}{}_{\nu}$, matching a perfect fluid at $p=-\rho$. The load-bearing input is non-dilution; $w=-1$ follows from it.

 

Computation 138 ·  §14.3 ·  [EMPIRICAL]
CKM consistency with the $S_{3}$-leading structural-scope theorem.

Checks that the measured CKM matrix (PDG) is consistent with identity plus a small $T(C)$-contingent perturbation, as the $S_{3}$-leading theorem requires: $\|V-I\|_{2}\approx 0.25$ is Cabibbo-sized ($O(\lambda)$, not $O(1)$), every off-diagonal is $\leq|V_{us}|$, and the magnitudes are unitary to sub-percent. A falsifiable consistency bound (the values themselves are out of scope).

 

Computation 139 ·  Ch. 15 ·  [NUMERICAL]
The $d^{-6}$ Casimir exponent derived from the projection kernel.

Derives the $d^{-6}$ pressure-correction exponent directly from the mode form factor rather than by dimensional analysis: for any smooth even kernel $g(kd_{0})$ ($g(0)=1$, $g^{\prime}(0)=0$ by parity), the leading correction is $\delta P=-C\,a_{2}d_{0}^{2}\,\Gamma(6)\zeta(6)/(2d)^{6}\sim d_{0}^{2}d^{-6}$. Numerically confirmed for four smooth even kernels (fitted exponent $-6.00$ to $<1\%$; analytic coefficient matched to $<0.2\%$); the exponent is kernel-independent (a non-smooth cusp kernel gives $-5$ instead), while the coefficient $a_{2}$ is kernel-specific (varies $4\times$), so the $d^{-6}$ signature is generic to any smooth even single-scale cutoff and only $\xi=90/\pi^{2}$ is PST-specific.

 

Computation 140 ·  §4.1 ·  [PROOF]
Robustness of the emergence chain to the form of $T(C)$.

Formalises and verifies that the downstream structural derivations read $T(C)$ only through the descending order of $|T(C)|$ (the instantiation sequence) and the global bias $\hat{\tau}=T(D)/|T(D)|$. For any strictly increasing reparametrisation $\phi$ ($\phi(0)=0$) of the magnitudes with $\hat{\tau}$ fixed, at matched threshold $\tau^{\prime}=\phi(\tau)$ the instantiation sequence, the per-configuration phase $\mathrm{sign}(|T(C)|-\tau)$, and $\hat{\tau}$ are all invariant (300 random trials); a non-order-preserving reassignment changes the sequence. So the residual freedom in $T(C)$ beyond the ordinal content and $\hat{\tau}$ is inert. Does not derive a canonical $T(C)$ (open, Computation 126); it fixes the exact robustness criterion.

 

Computation 141 ·  Ch. 3 ·  [NUMERICAL]
The directional marginal over $v$ is forced by $G_{2}$; only independence is adopted.

Sharpens Computation 127, which closed with “PST fixes no measure over $v$” and concluded that non-balance is adopted rather than derived. A measure over $v$ factors into a directional marginal on $S^{6}$ and a radial one, and neither is what remains adopted. Builds $G_{2}$ itself as the derivation algebra of the octonion cross product ($\dim=14$; Computation 127 used random $\mathrm{SO}(7)$ maps as a proxy), and confirms with genuine automorphisms that the balanced locus is $G_{2}$-invariant, so the Hurwitz/$G_{2}$ exclusion route is refuted. But the orbit through a unit vector is 6-dimensional: $G_{2}$ is transitive on $S^{6}$, which therefore carries a unique $G_{2}$-invariant probability measure, so the directional marginal is derived in the same way the Bernoulli measure is. The radial marginal is then shown irrelevant: with directions drawn from that measure independently across properties, no balanced assignment occurs in $200\,000$ trials at $n=2,3,8$ for unit, uniform, exponential, heavy-tailed or six-decade radial laws. The one exception is degenerate and self-closing: if $v(a)=0$ is permitted with probability $p$ then $P(\text{balanced})=p^{n}$ exactly (measured against prediction to three figures), which vanishes as $|D|$ grows for every $p<1$. What is load-bearing is independence: a law coordinating the values across properties makes balance certain, an independent one makes it null. So what is still adopted is not a measure but independence, the structural claim that nothing at substrate level coordinates the values, which is exactly the ungroundedness argument. It supplies the genericity argument’s missing premise rather than replacing it.

 

Computation 142 ·  §10.5 ·  [NUMERICAL]
Tensor factorisation over a bipartition, and no-signalling.

Backs the composite-system section. PST’s state space is $L^{2}(\mathcal{P}(D),\mu)$ with $\mu$ the Bernoulli product measure of §2.5; because $\mu$ is a product over the index set $D$, it factors over any bipartition $D=D_{A}\sqcup D_{B}$, and the Hilbert space factors with it. Findings: (1) $\beta(C)=(C\cap D_{A},C\cap D_{B})$ is a bijection $\mathcal{P}(D)\to\mathcal{P}(D_{A})\times\mathcal{P}(D_{B})$, checked exhaustively at $|D|=12$ over four splits, 4096 configurations each. (2) The measure factors to $0.000$, with marginals taken from the measure itself rather than assumed; a non-product control (weight $\propto 1+|C|$, permutation-invariant but not a product) fails the same test at $9.9\%$ of its largest weight, so the tensor structure rests on the Bernoulli measure specifically and not on permutation invariance alone. (3) The induced map preserves inner products ($10^{-17}$) and elementary tensors reproduce the Kronecker product ($7\times 10^{-18}$). (4) Over 200 random correlated states and random $B$ settings, $\mathrm{Tr}_{B}$ of the ensemble-averaged post-measurement state is invariant to $6\times 10^{-16}$: no-signalling. (5) The same invariance holds on the singlet, whose CHSH value is confirmed at $2.828427=2\sqrt{2}$; correlations beating the classical bound still signal nothing. Conclusion: composite systems, the reduced state and no-signalling are all consequences of the canonical measure, not additional assumptions.

 

Computation 143 ·  §10.3.2.5 ·  [NUMERICAL]
Born compatibility of the threshold-crossing rule.

Asks whether the crossing rule of (98) selects outcomes with the Born frequencies that §10.2 derives independently. For a fixed state the weights $w_{i}$ are fixed, so the randomness is in the configuration; the object is $\mathbb{P}_{C}(w_{i}>\tau_{i}(C))$. Result: the rule reproduces Born exactly, $\mathbb{P}(\tau<w_{i})=w_{i}$, precisely when the tension variable is uniform in its quantile; verified on three states to sampling precision and shown to fail on the same states for a Beta$(2,2)$ control. Also maps the uniqueness/Born trade-off along the coupling axis: raising $\kappa\lambda^{2}$ drives the probability of two simultaneous crossings monotonically to zero (the hypothesis clause (U) rests on), while pushing further orders the outcomes by margin and makes the survivor deterministic. A control placing the largest weight on the largest $|o_{i}|$ rules out a labelling artefact.

 

Computation 144 ·  §19.9 ·  [NUMERICAL]
Observable trace of a slit-plane crossing: the duality fork.

Tests whether the intermediate-coupling regime predicts anything standard quantum mechanics does not, in a way that does not depend on the details of the threshold law. If a crossing at the slits physically selects a path, only the fraction $1-p$ of runs without one shows fringes, giving $V^{2}+D^{2}=1-[1-(1-p)^{2}]\gamma^{2}$, a sub-duality deficit maximal as $\gamma\to 1$, i.e. at weak coupling; verified against the numerical maximum to $10^{-6}$, monotone in $p$ and zero only at $p=0$. A duality test good to $\epsilon$ excludes $p>1-\sqrt{1-\epsilon}$. With $p$ near $0.5$ from Computation 143 the predicted deficit is gross, so the selecting reading is untenable and crossings must be redundant with the inherited decoherence.

 

Computation 145 ·  §10.3.2.5 ·  [NUMERICAL]
Is the Bernoulli tension uniform in its quantile?

Supplies the condition Computation 143 identifies, by exact enumeration of all $2^{|D|}$ configurations for the scalar exemplar $T_{\mathrm{ex}}$ of (17). Raw, the tension is far from uniform (KS $0.46$–$0.51$ for every distinction graph tried, at $|D|=14$). But downstream reads only the ordering of magnitudes and Computation 140 verifies monotone-reparametrisation invariance, so the probability-integral transform is available and makes the variable uniform by construction. The residual obstruction is ties, which no order-preserving map can separate and which floor the achievable KS at half the largest tie mass. Every measured value sits at its own floor. For generic Erdős–Rényi structures the floor is $0.034$ at $|D|=14$ and decays as $|D|^{-1.51}$ ($R^{2}=0.998$), reaching $8.5\times 10^{-4}$ at $|D|=256$; for regular graphs it is large and uniformity is unreachable, but those are the graphs (17) already excludes.

 

Computation 146 ·  §10.3 ·  [NUMERICAL]
What the substrate layer contributes that decoherence cannot.

Fixes the division of labour between the three closures. Evolves a system–apparatus–environment state unitarily and confirms both halves of the standard result: the off-diagonal elements of the reduced state decay to zero, so interference is destroyed and the pointer basis is selected; and the global state stays pure ($\mathrm{Tr}\,\rho^{2}=1$ to machine precision at every environment dimension), so nothing has become definite. The decohered reduced state is then compared entry by entry with a proper mixture built by hand from definite outcomes: identical, and indistinguishable under 4000 random Hermitian observables to $0$. Global purity separates the two cases perfectly ($1$ versus $0.58$), so the difference is real but not locally accessible. Hence Gleason supplies the probabilities, einselection the basis, and the threshold crossing the definiteness; and Computation 144’s null visibility result is what the formalism requires rather than evidence of an idle layer.

 

Computation 147 ·  §16.10 ·  [NUMERICAL]
The reservoir question under the equation of state.

Tests whether ongoing sublimation can exhaust the configuration space. The appeal to an infinite $\mathcal{S}$ does not settle it, since $\mu$ is a probability measure with total mass $1$ however large $D$ is. Two readings are separated: a depleting one, in which a crossed configuration leaves the ensemble, and a modal one, in which it is re-labelled. Depletion removes configurations in tension order and biases the survivors away from equidistribution, measured as KS drift in the tension quantile over all $2^{16}$ configurations; the drift is monotone but slow, first exceeding three times the atom floor of Computation 145 only near $80\%$ sublimated, so this is a weak argument and is reported as one. The decisive exclusion is independent and carries no tolerance: §16.5 already requires $j_{\tau}$ to be a pre-spatial saturation density rather than a per-proper-volume creation rate, since the latter gives $w=0$, and a draining stock is precisely such a flux.

 

Computation 148 ·  §1.7 ·  [PROOF]
Are the three extremal selection principles independent?

No: they reduce to one. Enumerates the Hurwitz classification ($\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}$; dimensions $1,2,4,8$) and shows that maximal directional structure and maximal-tensor envelope are a single instruction evaluated on two constructions: argmax over vector parts (dimensions $0,1,3,7$) returns $\mathrm{Im}(\mathbb{O})=V_{7}$, argmax over tensor products of distinct members returns $\mathbb{C}\otimes\mathbb{H}\otimes\mathbb{O}$ at dimension $64$, and one function implements both. $\mathbb{R}$ being the tensor unit is why the envelope is a triple product, a consequence rather than a choice. Separately, minimal split is shown not to be a selection: with KO-$6$ verified (Computation 3) and KO-$4$ verified (Computation 4), the admissible $(n,k)$ with $n+k\equiv 2\pmod{8}$ collapses to the single pair $(4,6)$, so nothing is minimised; had the summands been free, minimality would have returned $(0,2)$, which the theory does not use. The reduction does not remove the odd-$|D|$ parity restriction Computation 3 carries.

 

Computation 149 ·  §17.6 ·  [PROOF]
Why one instantiation rather than many.

Tests whether P3’s necessity of crossing commits PST to a multiverse. It does not: each route to a second instantiation is a motion along a symmetry the theory already carries. A different direction is a $G_{2}$ rotation, and transitivity on $S^{6}$ is confirmed by the stabiliser dimension count $14-8=6=\dim S^{6}$, so $S^{6}/G_{2}$ is a point; $2000$ sampled directions are transported onto a common reference to $10^{-10}$, confirming the orbit is exhaustive. A different realisation map is diffeomorphism invariance: over $500$ relabellings the metric’s value multiset under (30) is permuted and not altered, to $10^{-12}$. A disjoint region factors rather than detaches, the measure factorisation holding to machine precision. The quotient by the three has exactly one element. Stated as a structural result only: it does not bear on the eternal-inflation multiverse.

 

Computation 150 ·  §18.3 ·  [NUMERICAL]
Scope of the L_comm uniform bound, and the $m(k)/N$ mechanism.

Re-measures $G(k,N):=|\mathrm{gap}(k,N)|\cdot N$ over $k\leq 40$, $N\in[12,40]$ (513 pairs), against Computation 65’s window $k\in[1,16]$. Three findings. (i) The bound is not uniform in $(k,N)$: $G=22.0$ at $(k,N)=(39,22)$, so the previously printed “$\leq 5.3$ uniformly in $(k,N)$” is refuted. (Computation 153 shows $22.0$ was a property of this sweep’s window rather than a maximum: $G$ is unbounded off the operative regime, reaching $38.1$ simply by looking further.) (ii) Every violation lies at $k>k_{D}=N/2$, which the Walsh cutoff excludes; on the operative regime $k\leq k_{D}$ the sharp constant is $3.674$ over 195 points (median $0.95$), attained at the smallest substrate size $D=36$. Computation 65’s reported maximum $5.249$ at $(15,20)$ has $k_{D}=10$ and so lies outside the regime the theory uses. (iii) The controlling parameter is the bridge degree over the truncation, $m(k)/N$: bucketed, $\max G$ is $2.6$ for $m(k)/N\leq 0.25$ and $22.0$ at $m(k)/N=0.5$. Since $m(k)\leq 2\sqrt{k}$ (verified for every $k\leq 2000$), the matched scaling gives $m(k_{D})/N(D)\leq D^{-1/4}$, so the deficit vanishes rather than staying $O(1)$, correcting an earlier step that bounded $m(k)\leq\sqrt{D}$, which is the bound on $k$ rather than on $m$. An explicit analytical constant remains open; this supplies the variable, the mechanism and the rate, not the value. Settled by Computations 213 and 214: the limit is in closed form and the sharp value is derived at $(6,12)$.

 

Computation 151 ·  §19.9.2 ·  [PROOF]
No-signalling is bipartition-universal, hence $\rho$-independent.

Closes the first of the two items left open in §19.9.2. The emergence map $\rho$ is not unique, so $D_{A}=\rho^{-1}(U_{A})$ is fixed only relative to a choice, but the theorem it feeds is universally quantified over bipartitions, and no step in it (factorisation, Born rule, $\sum_{j}Q_{j}=\mathbf{1}_{B}$, cyclicity on $B$) mentions $\rho$, a metric or a region. Every $\rho$ induces some bipartition and none is exempt, so the conclusion is $\rho$-independent even though the naming is not. Checked exhaustively: the bijection $\beta$ and the factorisation of $\mu$ hold on all $2^{10}-2=1022$ non-trivial bipartitions at $|D|=10$, to exact arithmetic, across every split size $1$ to $9$; the $A$ marginal is invariant under every $B$-side basis to $4\times 10^{-16}$. A permutation-invariant but non-product measure fails factorisation by fourteen orders of magnitude, so the result is carried by P1’s Bernoulli measure. Once factorisation holds the no-signalling step is an identity; the content is that factorisation holds for every bipartition. The residue moved on: Computation 156 shows the compactness half is a corollary of (30) (the distance is $|T-\tau_{\mathrm{null}}|^{1/4}$, strictly monotone, so compact in $M$ and tight in $T$ are one condition), and that being a device is an open condition with sharp margin $m/2$; what is genuinely left is why $T$ clusters, which is $T(C)$-contingent.

 

Computation 152 ·  §3.4 ·  [PROOF]
Non-balance is an open condition, and symmetry does not force it.

Replaces the circular measure-zero defence, which invoked a measure in the same breath as conceding PST fixes none over $v$. (i) The balanced locus $B$ is closed of codimension seven, so its complement is open and dense and the distance to balance is exactly $\bigl{|}\sum_{a}v(a)\bigr{|}/\sqrt{|D|}$: stability, not probability, and no measure used. Read at generic balanced points: codimension seven for $|D|\geq 3$, six on the all-parallel stratum, which at $|D|=2$ is all of $B$. (ii) The assignments fixed by the relabelling group are the constant ones, and those are maximally unbalanced, $\bigl{|}\sum_{a}v(a)\bigr{|}=|D|\,|v|$, which answers the objection that an unstructured substrate should cancel. (iii) But symmetry does not force non-balance: $v(a)=\epsilon_{a}u$ with shared $u$ and i.i.d. signs is invariant under diagonal $G_{2}$, all relabellings and per-property complementation, has unit directions, and still balances with probability $\binom{|D|}{|D|/2}2^{-|D|}\approx 0.31$ at $|D|=6$ (measured $0.3124$). The residue is a coordination premise, weaker than independence: one property anywhere in $D$ whose conditional has no atom at the cancelling value suffices. (iv) Scope, stated not hidden: if directions are quantised to $\pm e_{1}\ldots\pm e_{7}$ balance is not null, exactly $1/14$ at $|D|=2$, so the argument needs P1’s continuum of directions.

 

Computation 153 ·  §18.3 ·  [NUMERICAL]
The L_comm rate is exactly $\Theta(1/\sqrt{D})$, and why it looked better.

Closes a question Computation 150 deliberately left open. Its seventeen matched-scaling points looked like they trended down, which would have made $\gamma_{D}=o(1/\sqrt{D})$, better than the book claims; Computation 150 refused to claim it. The refusal was right. (i) At fixed $k$, $G$ converges to a nonzero constant as $N$ grows even while $m(k)/N\to 0$: at $k=1$ it runs $1.0161,1.0128,\ldots,1.0057$ over $N=16$ to $44$, monotone towards about $1.00$, with $k=2$ and $k=4$ settling near $0.87$ and $0.95$. So $\mathrm{gap}\sim C/N$ with $C=O(1)$ and $\gamma_{D}=\Theta(1/\sqrt{D})$; the rate is tight, and the apparent improvement was never there. No conflict with Computation 150: a relative deficit $O(m/N)$ on the norms and an $O(1/N)$ gap in their ratio are different statements. (ii) The apparent trend is sawtooth phase in the $m(k)$ stratum, and Computation 150’s window spans $m=4$ to $8$. (iii) The controlling variable sharpens from $m(k)/N$ to the truncation headroom $s=N-2(m+1)$ (equation (281)); $m(k)/N$ is a proxy because $m(k)/N\to 1/2$ is the same event as $s\to 0$. Off the operative regime $G$ is unbounded as $s$ closes (at $N=40$, $s=-2$ it runs $3.9$ to $38.1$ over ten consecutive $k$) so Computation 150’s $22.0$ was a property of that sweep’s window, not a maximum, and no maximum should be quoted. The analytical constant remains open.

 

Computation 154 ·  §8.3.3 ·  [PROOF]
The derivative expansion does not fail at an astrophysical horizon.

Evaluates the book’s own validity condition, the higher-curvature terms of equation (49) are suppressed by $(k\,d_{0})^{2}$ and negligible at scales $\gg d_{0}$, on the Schwarzschild geometry, rather than assuming the strong-field case lies outside it. (i) Symbolically, from the metric: $R_{\mu\nu}=0$, $R=0$, and $K=R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}=12r_{s}^{2}/r^{6}$ is the only non-vanishing invariant. (ii) With $k=K^{1/4}$, the suppression factor at the horizon is $\sqrt{12}\,(d_{0}/r_{s})^{2}=2.1\times 10^{-24}$ for one solar mass, $4.8\times 10^{-27}$ for Cygnus X-1, $1.1\times 10^{-37}$ for Sagittarius A^∗ and $5.0\times 10^{-44}$ for M87^∗: a horizon is strong potential, not strong curvature, and the covariant expansion sees only the latter. (iii) The expansion first fails at $r_{\mathrm{break}}=(12^{1/4}d_{0})^{2/3}r_{s}^{1/3}$, which is $38\,\mu$m for a solar mass ($1.3\times 10^{-8}$ of $r_{s}$, enclosing $2.1\times 10^{-24}$ of the interior) so the uncovered region is the singularity’s neighbourhood, not the horizon. (iv) Mass floor $2.9\times 10^{18}$ kg, where $r_{s}=12^{1/4}d_{0}\approx 4.3$ nm; with the grain $\ell_{*}$ as cutoff instead of $d_{0}$ it is $1.9\times 10^{8}$ kg, so the conclusion is insensitive to which PST length controls the expansion. (v) Every term of the field equations and of the Noether-charge integrands of the named corrections $\mathcal{O}(R_{\mu\nu}R^{\mu\nu},R^{2})$ carries a factor $R$ or $R_{\mu\nu}$, so the black hole solves the truncated equations exactly and its Wald entropy is $A/4G$ with no $(d_{0}/r_{s})^{2}$ correction; the derived $\Lambda$ costs a relative $10^{-119}$. (vi) The unnamed $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$ is Gauss–Bonnet up to those two, and its Noether charge is $64\pi^{2}\lambda$, mass-independent: an additive $O(1)$ against $S/k_{B}=1.05\times 10^{77}$, not an area term. The formula is calibrated in the same script by checking that it returns $A/4G$ from equation (50). Does not establish the global causal structure (event versus apparent horizon turns on the censored interior, exactly as in general relativity), does not make the substrate-side horizon identification a theorem, and says nothing about the statistical entropy. Scope, v32.159: on the $2.3$ nm Casimir bound of its day; Computation 216 re-evaluates on $d_{0}=2.2$ am ($37$ pm, $1.9\times 10^{-42}$ of the volume, a floor of $2.8\times 10^{9}$ kg), every conclusion surviving by wider margins.

 

Computation 155 ·  §7.4 ·  [PROOF]
The tensor projection $\Pi_{\mu\nu}$: structure, mechanism, and what is left.

Three results on the open tensor generalisation of the stress-energy projection. (i) Scope of the scalar form, symbolically on Friedmann–Robertson–Walker. The Bianchi identity applied to $T_{\mu\nu}=\Pi g_{\mu\nu}$ gives $\nabla^{\mu}T_{\mu\nu}=\partial_{\nu}\Pi$, so conservation forces $\Pi$ constant; and a perfect fluid is proportional to $g_{\mu\nu}$ iff $\rho+p=0$. The scalar form is therefore exactly and only the cosmological constant, not “the leading-order approximation valid for isotropic configurations”: isotropic dust and radiation are already outside it. (ii) The obvious completion is singular on the light cone. The unit dyad $\hat{e}_{\mu}\hat{e}_{\nu}$, $\hat{e}=\Delta/\sqrt{|\Delta\cdot\Delta|}$, diverges as $\Delta\cdot\Delta\to 0$, which is where $\tau_{\mathrm{null}}$ sits by (30); measured, the summed per-pair value swings $56\%$ over clouds of $200$ to $800$ points and does not settle. Any completion normalising by the interval fails the same way. (iii) The separation dyad works, and its trace is the tension. With $\Pi_{\mu\nu}$ as in (40), $g^{\mu\nu}\Delta_{\mu}\Delta_{\nu}=s_{ab}$ exactly (checked pair by pair, error $0$), so the scalar $\Pi$ returns as the trace rather than as a factor on the metric. For any invariant-only weight on a rotationally isotropic ensemble the result is forced into perfect-fluid form, residual falling $1.5\times 10^{-1}\to 4.8\times 10^{-2}$ as the cloud grows $200\to 1600$. $\rho$ is the timelike-pair moment and $p$ the spacelike-pair moment, split by $T(\{a,b\})$ against $\tau_{\mathrm{null}}$, the same cut that fixes the signature, so $w=p/\rho$ is the balance of the two tension budgets. Solving $\rho+p=0$ in the tested family gives $\lambda^{*}=0.8699$, where $\|\Pi-\lambda g\|/\|\Pi\|=3.7\times 10^{-2}$ equals the decomposition residual: at that point the tensor is the scalar form, so the cosmological constant is one point of a one-parameter family. Superseded in part by Computation 158, which supplies what this computation left open and corrects one of its conclusions. The weight $W$ is not adopted: trace consistency between (31) and (40), with (30) giving $T-\tau_{\mathrm{null}}=-s|s|$, forces $W(s)=-|s|$. And the reachable range reported here, $w\in(-1,+0.200]$ on the physical branch, from which this computation concluded that radiation at $+1/3$ was out of reach, was a property of its two-class exponential family rather than of the construction: the derived weight gives $w=1/3$ exactly. The $\lambda^{*}$ result stands as stated, for the family it was measured on. Still open from this computation’s list: conservation is checked flat rather than on a curved background, and the tensor is fixed only up to a divergence-free term.

 

Computation 156 ·  §19.9.2 ·  [PROOF]
What would make a property subset a device, and what would not.

Takes apart the one item §19.9.2 still carried after Computation 151: that a laboratory device occupies a compact region “is inherited rather than derived”. Three pieces, one of them genuinely open. (i) The compactness half is a corollary. (30) gives $|s^{2}|=|T-\tau_{\mathrm{null}}|^{1/2}$ for a spacelike pair, so $d(\rho(a),\rho(b))=|T(\{a,b\})-\tau_{\mathrm{null}}|^{1/4}$, strictly monotone: “compact in $M$” and “tight in $T$” are one condition written twice, and a planted subset is recovered from the tension matrix alone (round trip below $10^{-9}$). (ii) Stability, sharp and measure-free. With $m=\mathrm{sep}-\mathrm{diam}$, every perturbation of the induced distances below $m/2$ preserves the property and $m/2$ is sharp: adversarially, $0.99\,m/2$ intact, $1.01\,m/2$ broken (measured $\mathrm{diam}=0.0734$, $\mathrm{sep}=0.2152$, $m=0.1417$). A completeness lemma makes the search exhaustive: $\mathrm{diam}<\mathrm{sep}$ forces $D_{A}$ to be exactly the $|D_{A}|$ nearest neighbours of each of its members, confirmed against brute force over all subsets at $|D|=14$ (both find five). (iii) Cardinality is what makes the criterion bite. Bounded diameter with separation is on its own far too weak; every nearest-neighbour pair satisfies it, rate $1.000$. Over 200 replicas of 60 properties the accident rate runs $1.000$, $0.965$, $0.300$, $0.040$ at sizes two to five and $0.000$ from six up, so a twelve-property device is not an accident of an unclustered configuration. What is left: why $T$ clusters at all, which is $T(C)$-contingent rather than an open derivation, plus the bound-state physics the projected field theory already inherits. Caveat, the same one Computation 152 raised: the accident rates use a uniform null and PST fixes no measure over configurations, so they are indicative; the margin result is the measure-free part.

 

Computation 157 ·  §19.2.1 ·  [PROOF]
The L_comm bound reduced to ladder commutators, and solved at $k=1$.

Attacks the structure of the L_comm constant rather than sweeping for it. (i) The Dirac operator drops out. $T_{w^{m}}$ preserves $a-b$, so $[J_{z},T]=0$ exactly for every bridge, and the $\sigma_{\pm}$ algebra block-diagonalises the rest: $\|[D_{\alpha},T\otimes I_{2}]\|=\max(\|[J_{+},T]\|,\|[J_{-},T]\|)$ (equation (273)), the two being equal by the $z_{0}\leftrightarrow z_{1}$ symmetry. Checked to $10^{-14}$. (ii) The ladder commutator is a shift: $[J_{+},T_{w^{m}}]=m\,S_{m+1,m-1}$ exactly, to $10^{-16}$. Bridge and commutator are then each a sum of two shifts differing by $(1,1)$, hence bidiagonal along the chains $(a,b)\mapsto(a+1,b+1)$; the resulting $O(N^{2})$ evaluator reproduces the brute-force singular value to $10^{-11}$. (iii) Closed form at $k=1$, where $m=1$ and $\alpha=0$ so both operators are single shifts: $\|T\|^{2}=(N^{2}/4)/((N+1)(N+2))$ at $a=b$ and $\|[J_{+},T]\|^{2}=(N-1)N/((N+1)(N+2))$ at $b=0$, giving $\mathrm{ratio}(1,N)^{2}=4(N-1)/N$ (equation (274)) and $C(1)=1$ exactly, derived, not fitted, and matching to $10^{-12}$ up to $N=50$. (iv) The sharp constant re-checked four times further out: Computation 150’s sweep tested the operative regime only to $k=20$; at $k\leq 80$ the maximum of $G$ is still $3.6736$ at $(k,N)=(6,12)$, the smallest admissible $N$ for $k=6$, with per-$k$ maxima falling to $1.5$–$2.2$ by $k=50$–$80$. (v) The limit constants are a different number: near $N=250$, $C(k)$ is $1.0011$, $0.83$, $0.10$, $0.78$, $0.79$ at $k=1,2,3,4,6$, the first matching $1+1/(4N)$ from the closed form. Still open: the value. A closed form for general $m$ needs the discrete maximum over $a$, which is what generates the sawtooth.

 

Computation 158 ·  §7.4.4 ·  [PROOF]
The weight profile $W$ is not free, and what it forces.

Closes the weight-profile question §7.4 handed on. The derivation: (31) projects the tension itself, the trace of (40) is $g^{\mu\nu}\Delta_{\mu}\Delta_{\nu}=s$, and (30) gives $T(\{a,b\})-\tau_{\mathrm{null}}=-s|s|$ identically ((42); checked for timelike, null and spacelike pairs). Requiring the two projections to agree on the trace, pair by pair, is one equation in one unknown and gives $W(s)=-|s|$ ((43)) with no free function and no free scalar beyond the $8\pi G$ unit conversion. One addition, from which nothing is read: without a range restriction the long pairs dominate and $w$ drifts with the sampling window (spread $>0.05$ over windows $0.5$ to $4$), which a density must not do; with the coherence scale as cutoff the answer converges, identically for four cutoff shapes (exponential, hard, Gaussian, algebraic) at two scales differing fourfold. What it forces: the trace of $\Pi_{\mu\nu}$ is $\sum(T-\tau_{\mathrm{null}})$, and it vanishes, the null threshold being the balance point of the pair-tension distribution; $(\rho-3p)/\rho$ falls $-1.9\times 10^{-1}\to-5\times 10^{-6}$ as the window grows, so $w=1/3$ exactly ((44)). Independent of the configuration: uniform, radially graded, Gaussian in space, Gaussian in space and time, and ten-clump clustered all give $0.3333$ to $0.3338$. The pair-tension projection is conformal matter. This also overturns Computation 155’s ceiling: that computation’s two-class family reached only $w\leq+0.200$ and flagged radiation as possibly unreachable with the caveat that the ceiling might be an artefact of the exponential profile. It was. Open: dust, which breaks the forced tracelessness and needs a mass scale the pair-separation structure does not carry; $\Lambda$ is not on that list, being the separate vacuum term with $w=-1$ (Computation 155, §16.5). Scope, from Computation 199: the trace vanishes in the large-window balanced limit; out of balance it is the balance scalar, and the identification concerns the conformal part of the sum.

 

Computation 159 ·  §7.4.6 ·  [PROOF]
Dust is not in this sector, and a photon is why.

Settles the dust question Computation 158 left open, in the negative, which locates the rest mass rather than leaving it vague. (i) The worldline sector is exactly kinetic. Real matter is worldlines, not scattered points; restricted to intra-worldline pairs the projection gives $w=\beta^{2}/3$ ((47)) exactly, to $5\times 10^{-15}$ over nine speeds, with $p=0$ at $\beta=0$ and $1/3$ as $\beta\to 1$. Nothing is put in: it follows from the forced weight (43). (ii) A photon contributes nothing from its own worldline. A null worldline has $s=0$ along it, so $W(s)=-|s|$ vanishes identically; measured, its summed weight is fourteen orders below the $\beta=0$ case and consistent with round-off. Radiation’s $1/3$ is therefore forced by the weight vanishing on null separations, the separation $\tau_{\mathrm{null}}$ defines, so the photon enters the stress tensor through the same object that fixes the signature. (iii) Mixture law $w=[1-f(1-\beta^{2})]/3$ with $f$ the worldline share, to $5\times 10^{-5}$: dust needs $f\to 1$ and $\beta\to 0$. (iv) $f$ does not reach one, structurally. $|s|\lesssim d_{0}^{2}$ is a neighbourhood of the light cone, not a compact region: in a box of half-width $40$, $98\%$ of surviving pairs are separated spatially by more than seven correlation lengths, median $31$, so the near-null background never dilutes and $f$ stays at $0.01$ to $0.05$ over fifty times the correlation length. (v) The bond kernel cannot rescue it: $B$ is normalised, $\int B\,\mathrm{d}^{4}x=1$, so the volume-averaged density is identically the integrated sum for any $B$ (measured $0.33329$ against $0.33337$); it decides where the stress sits, not the equation of state. Conclusion: the projection is strictly conformal for homogeneous matter, dust is excluded rather than unfound, and the rest mass enters through the matter action rather than the pair geometry.

 

Computation 160 ·  §8.3.4 ·  [PROOF]
The photon: what $\tau_{\mathrm{null}}$ fixes, and what is still missing.

Collects four facts about light that are usually presented as unrelated and shows one object behind them. (i) Two polarisations: the physical states of a vector excitation are $\{\epsilon:k\cdot\epsilon=0\}$ modulo $\epsilon\sim\epsilon+\lambda k$; the transverse subspace is three-dimensional for every $k$, and the gauge direction is removable only when $k$ lies in it, which is $k\cdot k=0$. Null gives two states, timelike gives three, over $200$ random momenta each. The counting is ordinary gauge theory; the input, that the photon’s momentum is null, is PST’s, through the unbroken $\mathrm{U}(1)$ and (30). (ii) Equation of state: a null worldline has $s=0$, so the derived weight vanishes identically on it and $w=1/3$ is forced (Computation 159). (iii) Stefan–Boltzmann: $p=\rho/3$ makes the first law integrable only for $T\propto\rho^{1/4}$, so $\rho\propto T^{4}$, solved symbolically; no quantum input is used, and PST reaches it because it derives the $w=1/3$ the argument needs. (iv) The coefficient too: $a=g(\pi^{2}/30)k_{B}^{4}/(\hbar^{3}c^{3})$ with $g=2$ gives $7.5657\times 10^{-16}$ J m^-3 K^-4, the measured value to eight figures, where $g=3$ would miss by half. Missing, and now supplied: this computation said the Planck spectral distribution was unassembled and that the quantisation into countable quanta was absent outright. Computation 161 assembles the spectrum, and the second half of that claim was wrong: §12.1 has the Fock space by canonical quantisation, inherited rather than derived. The photoelectric effect is genuinely absent and stays so.

 

Computation 161 ·  §8.3.5 ·  [PROOF]
The Planck spectral distribution, assembled.

Writes down the derivation Computation 160 said was missing, and separates what PST forces from what it inherits. Four inputs are forced: the linear dispersion, from masslessness at $\tau_{\mathrm{null}}$; $g=2$, from the null condition; bosonic occupancy, because the photon is an inner fluctuation of even Clifford grade (Computation 108) and even-order composites are bosonic (§11.3); and $\mu=0$, because the conserved $\mathrm{U}(1)$ charge is the winding number of $\pi_{1}(\mathcal{V})$ and not the occupation number, so nothing conserves photon number. One is inherited: the ladder $E_{n}=n\hbar\omega$ comes from canonical quantisation of (123) in the ordinary way (§12.1); PST supplies the arena, not the ladder. Checked: the bosonic, fermionic and classical branches of the mean occupation, symbolically, the last being the ultraviolet catastrophe; the density of states $g\omega^{2}/2\pi^{2}c^{3}$ against a brute-force count of box modes, the residual shrinking as a surface term; (69) against the textbook angular-frequency, frequency and wavelength forms; the integral $\int_{0}^{\infty}u\,\mathrm{d}\omega$ by term-by-term expansion and resummation to $6\zeta(4)$, cross-checked by quadrature to $2\times 10^{-13}$, returning exactly the radiation constant of Computation 160 ($g=3$ would be half as large again); Wien’s displacement law in both conventions, $b=2.8977720\times 10^{-3}$ m K against CODATA; and the Rayleigh–Jeans and Wien limits. The falsifiable part: a conserved photon number would give $\mu\neq 0$, and the measured microwave-background spectrum bounds the distortion at $|\mu|<9\times 10^{-5}$ [209], so the structural $\mu=0$ is confirmed to a part in $10^{4}$. Still open: the photoelectric effect, which needs the interaction vertex rather than the free spectrum.

 

Computation 162 ·  §7.4 ·  [PROOF]
Conservation of $\Pi_{\mu\nu}$ on a curved background: bounded, not closed.

Backlog M3, and the first of the two items §7.4 leaves open. The covariant reading: a pair has no separation vector on a curved manifold, so $\Delta_{\mu}$ becomes Synge’s world function derivative $\sigma_{\mu}$ with $s=\sigma^{\mu}\sigma_{\mu}=2\sigma$. Two exact identities, that one and its gradient $\sigma^{\mu}\sigma_{\mu\nu}=\sigma_{\nu}$, collapse the per-pair divergence to (45), so every curvature effect enters through the single scalar $\Box\sigma$. In closed form on constant curvature, by the Laplace–Beltrami operator in geodesic polar coordinates, $\Box\sigma=1+(n-1)(d/L)\cot(d/L)$ on the sphere and $\coth$ on the hyperboloid, both tending to $n$ as $L\to\infty$ and both reproducing the coincidence expansion $n-\tfrac{1}{3}R_{\alpha\beta}\sigma^{\alpha}\sigma^{\beta}$. The shift: with the derived weight the relative correction is $-\tfrac{1}{7}(d/L)^{2}$, second order as it must be. The bound: $|s|\lesssim d_{0}^{2}$ gives $d\lesssim d_{0}$, so the violation is $6\times 10^{-25}$ at a solar-mass horizon, $6\times 10^{-27}$ at a neutron star’s surface, $9\times 10^{-41}$ at the Earth’s and $6\times 10^{-70}$ cosmologically, the same $(kd_{0})^{2}$ that Computation 154 reports as $2.1\times 10^{-24}$ at the horizon. A correction carried with it: the per-pair divergence is $-14\sigma\sigma_{\nu}$ and so not zero in flat space either, meaning flat conservation was always a property of the ensemble; measured here, the residual falls with the pair count at an exponent of $-0.35$, consistent with a finite-sample effect. Superseded in part by Computation 163, which supplies the exact result conditionally and settles the ambiguity; the bound here is the unconditional statement and stands under it. Re-evaluated at the derived $d_{0}=2.2$ am (Computation 201): $6\times 10^{-43}$, $6\times 10^{-45}$, $8\times 10^{-59}$ and $5\times 10^{-88}$, the values Chapter 7 quotes.

 

Computation 163 ·  §7.4 ·  [PROOF]
The action principle $\Pi_{\mu\nu}$ does have, and what the ambiguity is.

Closes what Computation 162 left, in both halves, and corrects its closing sentence: the construction needs no action of its own. The exact result: §7.5 identifies $\Pi_{\mu\nu}$ as the conformal part of $T^{\mathrm{(fields)}}_{\mu\nu}$, whose conservation is Noether’s second theorem for $S_{\mathrm{matter}}$; splitting off the trace gives (46), $\nabla^{\mu}\Pi_{\mu\nu}=-\tfrac{1}{4}\nabla_{\nu}T$, exactly and on any background. Verified on a Friedmann background with a massive scalar, where the trace is not constant and both sides are nonzero, component by component on shell. The corollary: Computation 158 forces the trace of $\Pi_{\mu\nu}$ to vanish, so the right-hand side is identically zero and conservation is exact, with no $(d_{0}/L)^{2}$ residue, in the sector $\Pi$ describes; the electromagnetic stress tensor is confirmed traceless for an arbitrary field strength in four dimensions as the worked instance. The ambiguity, backlog M4: for $H_{\mu\alpha\nu\beta}=f(g_{\mu\nu}g_{\alpha\beta}-g_{\mu\beta}g_{\alpha\nu})$ the shift $g_{\mu\nu}\Box f-\nabla_{\mu}\nabla_{\nu}f$ is symmetric, has trace $3\Box f$, and has divergence exactly $-R_{\nu\sigma}\nabla^{\sigma}f$, so it is identically divergence-free in flat space. And tracelessness closes it: $\Box f=0$ is forced, and a localised harmonic gauge is identically zero, by Dirichlet uniqueness on the Riemannian section (the discrete operator’s smallest singular value is reported, so uniqueness is measured) and by the domain of dependence in Lorentzian signature, tested with the data nonzero outside the region so the shrinking cone is not a vacuous check. What is left: the conformal identification, which both halves lean on and which §7.5 already names.

 

Computation 164 ·  §7.5 ·  [PROOF]
The conformal identification: what the pair sum can represent, and what it cannot.

Attacks the single residue Computation 163 left, and replaces the caveat with a criterion. Sharp in both directions: every term of (40) is a rank-one dyad times a weight $W(s)=-|s|$ that never changes sign, times a bond kernel that is a density, so $\Pi_{\mu\nu}$ is always semidefinite; and conversely every semidefinite target is a dyad sum by its own eigendecomposition. The reachable set is exactly the semidefinite metric-traceless cone. Confirmed over six deliberately awkward ensembles (flattened to a plane, stretched along time, two clumps, laid on a light cone, collapsed onto a line): not one negative eigenvalue. What is inside: isotropic radiation and a null plane wave, hence Computation 158, Computation 159, Computation 160 and Computation 161 entire, so nothing claimed here is touched. What is forced: $\Lambda g_{\mu\nu}$ is indefinite, so the vacuum term’s separation into its own sector is structural and not a design choice. What is outside: a static magnetic field, $T_{\mu\nu}=\mathrm{diag}(u,u,u,-u)$ with $u=B^{2}/2$, metric-traceless and indefinite, the negative eigenvalue being the tension along the field lines; an electric field likewise. A single-signed bond pushes or pulls throughout, and a magnetic field does both at once. The gauge: Computation 163’s constant metric-traceless shift restores semidefiniteness, so what fails is the identification as stated, at the price of a universal constant that must dominate the largest tension anywhere and that nothing fixes, $\rho$ being a quotient map with no empty region to pin it against. Recorded so it is not repeated: a first pass argued the vacuum forbids that shift, which is wrong for exactly the surjectivity reason. Scope, from Computation 199: all of this is the pair sum itself; the identification is of conformal parts, and the conformal part of the sum reaches the magnetic and electric tensors and every traceless tensor obeying the null energy condition.