Part I · The Sentence and the KernelChapter II

Blind Observers

Plate II.1The six orderings of three tokens with distinct values, joined by the two adjacent swaps. Every nontrivial relabelling moves every vertex. The figure shows a space of alternatives, not a spatial arrangement.

II.1
II.2
II.3
II.4
II.5
II.6
II.7

How can an observer who sees almost nothing still reason exactly about what it sees?

No observer sees the whole of what occurs. Most of it reaches the observer only as classes: it can tell one class of histories from another, and nothing about which member of a class took place. Chapter I met the simplest case, a sign whose origin the register cannot audit. This chapter asks what a lawful observer can say about an occurrence inside a class it cannot audit.

Picture three tokens that differ in value while the observer cannot see their order. There are six orderings. Any rule that picks one of them must break the symmetry the observer is blind to, so no lawful, covariant rule picks one. The answer the program gives is that such a rule outputs a vector over the class: one amplitude for the whole class, the same on each of its members.

The central result · The rescue theorem

Let a finite group $G$ act on a finite set $X$ of continuations, and let an observer’s records be unchanged by $G$ , so that it cannot audit which member of a $G$ -orbit occurs.

(i) A covariant law can select a single continuation only if $G$ fixes it; on an orbit of more than one member no covariant deterministic choice exists.

(ii) The covariant vectors of $\C[X]$ are exactly the functions constant on $G$ -orbits: one amplitude per auditable class, uniform over its interior.

From dimension three on, the structure of these amplitudes is forced: additive weights are Born weights (Gleason), some finite families of yes–no questions admit no noncontextual answers (Kochen–Specker), and a change of basis first carries a phase that no rephasing removes.

Status

Parts (i) and (ii) are proved; the proofs are short, and the content is in the reading. The sentence on dimension three collects classical theorems that the program’s finite checks witness on its own ray families. Together they turn the axiom “states form a complex Hilbert space with Born weights” into a statement derived from blindness modulo named premises: the rescue theorem (proved), composition over independent cells (the register’s factorization, checked), complex coefficients (for states that are Hermitian forms), Born weights (Gleason, with additivity as its hypothesis), no hidden facts (Kochen–Specker, given a rich enough family of questions) and phases (the count), together with one principle, lawful actuality: no chance and no fiat at blind steps.

The program treats the result as a theorem-candidate. It is falsifiable in the way Born’s rule is, since an established violation of the Born rule, or an identified lawless selection, would refute the principle. Blindness does not fix the relative weights of different classes; a measure must supply them, and the program has since adopted one as premise P4. That the tower’s gap is a mass is a reading.

What an observer cannot audit

Let $X$ be the set of possible continuations at an occurrence, and let a group $G$ act on $X$ . An observer is blind to $G$ when its records, and hence every law it can obey, are unchanged by $G$ . The $G$ -orbits are its auditable classes: it can tell one class from another, and nothing within a class. A law is covariant when it commutes with the action of $G$ . In the language of report maps, a feature can be recovered from a report exactly when it is constant on the report’s fibres; here the fibres are the orbits.

If $k$ tokens carry values in an alphabet and the observer is blind to which token is which, $G=S_k$ permutes positions and the auditable classes are the multisets of values. For two tokens with values in $\{1,2\}$ they are $\{11\}$ , $\{22\}$ and $\{12,21\}$ . The observer’s own record at such a step is fixed by every element of $G$ , and a covariant law must carry a fixed record to a fixed output: that one line is the whole of the no-go.

The rescue theorem

Plate II.2The only covariant vector over the six orderings is the uniform sum

\sum_\pi|\pi\rangle

at the centre: one amplitude, spread evenly over the class. For two tokens with values in

\{1,2\}

the covariant space is

\alpha|11\rangle+\beta|22\rangle+\gamma(|12\rangle+|21\rangle)

At a blind step a law has four options: select a point by rule, which covariance forbids; select a point at random, which is chance; select a point by breaking the symmetry, which is fiat; or output the class vector. The program calls the fourth option the rescue of covariant determinism, and it is the only one that keeps the law both deterministic and covariant. Its lawful freedom is exactly one amplitude per auditable class: blind positions are forced uniform, and distinguishable outcomes may be weighted.

The program reads this as the kinematic signature of quantum theory, reached from covariance and blindness with no quantum input. It is the rescue form of a state, and the program’s transport of amplitudes by records, stipulation S6, is this form applied to the values of words. The program’s exact ranks give covariant spaces of dimension 3, 4 and 10 for two tokens over two values, three over two and three over three: the numbers of multisets.

Theorem(Rescue)

Let $G$ act on the finite set $X$ and let the observer be blind to $G$ . (i) A covariant law with values in $X$ exists exactly when $X$ has a $G$ -fixed point. (ii) The covariant vectors of $\C[X]$ form the space $\C[X]^G$ , which has a basis of orbit indicators $\mathbf 1_O$ ; its dimension is the number of auditable classes.

Proof

A covariant law $L$ satisfies $L(r)=L(gr)=g\,L(r)$ for every $g\in G$ , because the record $r$ is $G$ -fixed; so $L(r)$ is $G$ -fixed. Conversely a fixed point, or a fixed vector, can be output by a covariant law. A vector $f=\sum_xf(x)\,x$ is $G$ -fixed exactly when $f(gx)=f(x)$ for all $g$ and $x$ , that is, when $f$ is constant on each orbit. The indicators of the orbits are linearly independent and span these functions.

Three is where blindness ignites

Plate II.3The three thresholds by dimension. In dimension two each escape remains; from dimension three on, none does.

Three classical thresholds fall at the same dimension, and the program checks each on families it generates itself. In dimension three the forty-nine rays spanned by vectors with components in $\{0,\pm1,\pm\sqrt2\}$ admit no assignment of zeros and ones, while the Born weights of the density matrix $\mathrm{diag}(\tfrac12,\tfrac13,\tfrac16)$ sum to exactly one on each of their sixteen orthonormal triads. In dimension four the forty rays with components in $\{0,\pm1\}$ admit none. An explicit unitary over $\Q(i)$ has Jarlskog invariant $41472/1221025\neq0$ , so its phase cannot be removed.

In dimension two blindness can still be ignorance of a hidden fact. Assign 1 to the real rays whose angle lies in $[0^\circ,90^\circ)$ : the assignment is additive on orthogonal pairs, yet no qubit state gives it, since weight one at $0^\circ$ and at $45^\circ$ forces a Bloch vector of squared length 2. As for complex numbers, the blind classes of independent cells compose as products, so the parameter counts of independent parts must multiply; of $n(n+1)/2$ , $n^2$ and $n(2n-1)$ for real, complex and quaternionic self-adjoint matrices only $n^2$ does so for all sizes. This is the local-tomography argument of Hardy and Wootters, with its composition premise supplied by the register.

Proposition(The thresholds at three)

Let $H$ be a complex Hilbert space of finite dimension $d$ .

(i) (Gleason 1957.) If $d\ge3$ , every assignment of nonnegative weights to the rays of $H$ summing to one on each orthonormal basis has the form $P\mapsto\operatorname{Tr}(\rho P)$ for a density operator $\rho$ . For $d=2$ this fails.

(ii) (Kochen–Specker 1967.) If $d\ge3$ , there are finite families of rays admitting no assignment of 0 and 1 with exactly one 1 on each orthonormal basis. For $d=2$ such assignments exist.

(iii) (Phase count.) A $k\times k$ unitary modulo independent rephasing of rows and columns has $(k-1)^2$ real parameters: $k(k-1)/2$ angles and $(k-1)(k-2)/2$ irremovable phases. There is no phase for $k\le2$ and exactly one for $k=3$ .

Two blindnesses

Plate II.4Two alternatives the observer cannot audit, carrying transports

I

and

J

a quarter turn apart. Clause B gives the occurrence one amplitude, the uniform sum over them.

The founding sentence has no clause for many histories or for many rules; both enter through blindness. The branchial clause B says that an occurrence the observer cannot audit is one occurrence over every history consistent with its records, and that the observer’s amplitude for it is the uniform sum over those histories. The rulial clause Ru says that the observer’s rule, its vantage and its choice of representatives, is itself a record and a coordinate: the laws are covariant under changes of rule, re-anchoring is a move like any other, and freedom from a frame is uniformity over the rule.

The two clauses are one blindness at two levels. A choice that is a rule for a register, which rod stands for an axis when memory identifies two words, is a history for a composite built of registers: which member of the crowd realized the move. Composition turns rulial blindness into branchial blindness, and this is where the branchial and rulial levels of the program’s hierarchy first meet. Clause Ru has a sharp exact form in the forcing theorem: an observer uniform over which report it stands on recovers nothing of the qubit but its trace. To see anything it must stand somewhere; a vantage is the price of sight.

Proposition(A blind decoder sees only the trace)

No decoder of the selection theorem is equivariant for the unmarked actions of the tetrahedral group on the comparison algebra and on the reported qubit. There are four decoders $\mathcal E_i$ indexed by a retained report mark, with $\mathcal E_{g(i)}\beta_g=\alpha_g\mathcal E_i$ . On an element $(a,B,C)$ of $\C\oplus M_3\oplus M_2$ their uniform average is $\tfrac14\sum_i\mathcal E_i(a,B,C)=\tau(C)\,I$ , the normalized trace of the $M_2$ block, which does not recover the returning qubit.

Proof

The Klein four-group fixes the three perfect matchings of the reports. It acts trivially on the two-dimensional returning sector, which is spanned by contrasts of matchings, and nontrivially on the tetrahedral qubit, so no surjective decoder intertwines the two actions. Fixing a mark leaves a stabilizer of order three, whose actions on the two qubit algebras are conjugate; transporting one intertwiner around the four marks gives the family. Its average is equivariant, so its image is fixed by the Klein group and is a scalar, fixed by the trace.

The price of a blind sum

Plate II.5The blind sum

B=\tfrac12(I+J)

is the midpoint of the chord: an eighth of a turn, of length

\sqrt2/2

. Applied twice it gives

B^2=J/2

, the quarter turn at half length; the complementary port

\tfrac12(I-J)

carries the rest of the norm.

Clause B sums amplitudes, not states. For a finite group of unrecorded transports $U_h$ it gives $\frac1{|G|}\sum_hU_h$ acting on amplitudes, not the trace-preserving twirl, and the two are not interchangeable. Nor is it completion, the identification of histories an observer cannot distinguish: completion says which histories an observer identifies, and clause B says what amplitude an identified class carries. A blind sum is not a projection: two of them compose to half of the observer’s complex structure, not to themselves.

The program’s tower, its model of a register’s memory extended to composites built from many registers, spends this price in a pattern that separates its two kinds of composite. For a composite of odd radius, matter in the program’s reading, the transport of one coarse move factorizes as $T=\tfrac14\,\sigma_x\gamma_{x_2}\gamma_x\sigma_{x_2}$ , with $\sigma_x=\gamma_x+\gamma_{\bar x}$ and $\sigma_x^2=-2$ : a blind sum at the destination, the bare hop, and a blind sum at the source. Each costs $\sqrt2/2$ , so matter’s transport has modulus $\tfrac12$ . For even radius, radiation in the same reading, both sums are trivial and radiation pays nothing.

Proposition(A blind sum is a scaled eighth-turn)

Let two alternatives the observer cannot audit carry real transports $I$ and $J$ , with $J^2=-I$ and $J^\top=-J$ . Their blind sum $B=\tfrac12(I+J)$ satisfies $B^\top B=\tfrac12I$ , $B^2=\tfrac12J$ and $B=\tfrac{\sqrt2}{2}\,e^{\pi J/4}$ . The complementary combination $\tfrac12(I-J)$ carries the other half of the norm. If the two transports are equal, the blind sum is $I$ and costs nothing.

Proof

$B^\top B=\tfrac14(I-J)(I+J)=\tfrac14(I-J^2)=\tfrac12I$ , and $B^2=\tfrac14(I+2J+J^2)=\tfrac12J$ . Since $J^2=-I$ , $e^{\pi J/4}=\cos\tfrac\pi4\,I+\sin\tfrac\pi4\,J=\tfrac{\sqrt2}{2}(I+J)$ .

Mass as the cost of blindness

Plate II.6Matter pays

\sqrt2/2

at each of its two blind sums, so its transport has modulus

\tfrac12

; radiation’s two transports agree, and its blind sums cost nothing.

The price is observable inside the model. Every composite carries an internal clock operator, a rotation generator of its internal frame rather than a timekeeper. Matter’s clock runs at level $\sqrt2$ when both histories are blind, 2 when either one is audited, and $2\sqrt2$ when both are, while radiation’s runs at $2\sqrt2$ under every audit. Each audited history restores one factor of $\sqrt2$ .

These blind sums are couplings inside a generator, and a modulus below one is a smaller hop, not lost probability. The same moduli used as whole occurrences would need their complementary outcomes kept.

Reading(Mass is the cost of blindness) a reading

With the two blind sums switched off, the matter composite is massless at the natural mobility $\rho=1$ : its memory moves and its re-anchoring moves balance on a forty-eight-dimensional kernel, as radiation’s do on a hundred and forty-four. The blind sums attenuate the moves and break the balance. What is left is the class constant 0.021387, the smallest eigenvalue in modulus of a real scalar walk whose hops carry the factor $\tfrac12$ . These statements are exact in the tower model. That the gap is a mass is a reading, and the value $\tfrac12$ is not special: the gap as a function of the hop modulus is largest, 0.0315, near 0.4214. The value also belongs to the tower’s flat transports and moves when the lift’s curved transport replaces them. The test of the reading is a bridge from the model’s cycle to a physical time.

What blindness cannot decide

Plate II.7Two lawful worlds at the first memory-bearing depth: equal weight per move, and re-anchoring weighted twice. Both are uniform on every blind class.

The rescue theorem fixes the form of a state within each auditable class. It says nothing about the relative weights of different classes, and those are where the program’s dynamics lives. The generator has three kinds of move, which the records tell apart: a hop changes one letter of the word and keeps the anchor; a memory rewrite changes two letters and keeps the anchor; a re-anchoring changes the anchor and keeps the word. Applied to the generator, the rescue theorem leaves exactly three free amplitudes, one per kind. Maximal ignorance adds the two equalities between kinds and gives natural mobility $\rho=1$ ; it is an added clause.

Read globally, clause B would sum uniformly over every hidden history compatible with the observer’s log, and there is no such measure: a hidden partner can append its current letter any number of times without changing the log, so equal weights cannot be normalized, and finite replacements disagree, one-sixth against one-seventh for reception against an own continuation. So clause B is used only locally, over finite auditable classes at an occurrence. Blindness does select where a record fails to distinguish: with a relational vantage record and a common-intensity premise, uniform weight forces $\rho=1$ , and a receipt for a change of clock that reveals nothing about the source anchor forces equal weights on the two admissible laws. Both selections are conditional.

Example(Two lawful worlds)

At the first memory-bearing depth there are 144 hops, 144 memory rewrites and 108 re-anchorings. Equal weight per move gives the three kinds total weights $\tfrac4{11},\tfrac4{11},\tfrac3{11}$ . Doubling the weight of re-anchoring gives $\tfrac27,\tfrac27,\tfrac37$ . Both laws are uniform on every blind class, use the same moves and make at most one move per cycle; blindness cannot choose between them.

The rescue theorem gives the form of a state, and from dimension three on that form is quantum. What remains free is the weight between classes, which a measure must supply. The later parts of the volume use each piece: the branchial tree of Chapter VII is clause B’s home, and rulial relativity, Chapter XI, is clause Ru’s. The next step is nearer. Chapter III asks where blind alternatives live, and finds them in what the record does not keep, where they can interfere.

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