Part II · One Graph, Four CoversChapter IX

What Space Forgets

Plate IX.1The decagon

[\lambda_1,\lambda_3]

, projected onto the plane of its periods

t_1

and

t_3

. Its four legs run around the parallelogram of periods and close. Digits are site types; gold points are sites of type 0.

IX.1
IX.2
IX.3
IX.4
IX.5
IX.6

Where do interference and holonomy live, if not in space?

Chapter VIII made space out of tallies. Walk the triangle $0\to1\to2\to0$ and then the triangle $0\to2\to3\to0$ , or walk them in the other order: both histories end at the same place, and space cannot tell them apart. What space cannot see is the difference between them: go around the first triangle, then the second, then the first backwards, then the second backwards. That loop has zero tally. In $K_4$ it is a closed walk of twelve steps, ten once a retraced letter is cancelled; in space it is a decagon that returns exactly to its start.

Such loops form the kernel of the tally map, and the chapter’s claim is that two things the program needs live there. One is the non-abelian part of every holonomy, which space cannot absorb into a momentum; a holonomy is what a rule for carrying data along the letters does to that data around a closed loop. The other is the interference between alternatives that space cannot tell apart.

The central result

The tally map $\tau\colon\pi_1(K_4)\cong F_3\to H_1(K_4;\Z)\cong\Z^3$ has kernel $[F_3,F_3]$ , the fundamental group of the $K_4$ crystal: free of infinite rank, normally generated by the commutators of the three triangles, which reduce to decagons of space.

(i) For a transport $\rho\colon F_3\to G$ along the letters, the holonomies of the closed loops of space form exactly $[\rho F_3,\rho F_3]$ . An abelian transport is flat on space and survives only as a Bloch momentum.

(ii) A report-covariant change of frame along a letter $vw$ , by a permutation of the reports, is $(v\,w)$ or $(v\,w)(c\,d)$ , with $c,d$ the other two reports. The first sends the triangles to the transpositions of the axes and every decagon to a 3-cycle, so its holonomy on space is exactly $\Z_3$ ; the second is flat. Covariance does not choose.

(iii) With the tally retained, only alternatives that share endpoint and tally, and so differ by an element of the kernel, can interfere; when their records also agree, their interference measures its holonomy.

Status

Statements (i) and (ii) are exact group theory, checked on every decagon. On the kernel graph alone, which transport is physical is not selected, and adopting either is an added choice. A larger geometry selects it: in the lift of Chapter XIII frames are compared by the parallel transport of velocity space, which realizes the first choice; the flat one is realized only by a rotation that needs extra data, the octonion table’s distinction between two kinds of cell.

Statement (iii) is exact given two things established elsewhere, the retention clause of Chapter VIII and the two-layer rule of Chapter III, and the apparatus that reads a holonomy by interference is itself an addition. That the group-theoretic kernel is the kernel of Chapter III, the place where interference lives in the two-layer world, is a reading with a named test, the kernel square of Chapter XX; it has not been run on the native dynamics. The Majorana medium on the decagons is a conditional model.

The kernel of the tally map

Plate IX.1The decagon

[\lambda_1,\lambda_3]

, projected onto the plane of its periods

t_1

and

t_3

. Its four legs run around the parallelogram of periods and close. Digits are site types; gold points are sites of type 0.

Fix report 0. The three triangles $\lambda_1=0\to1\to2\to0$ , $\lambda_2=0\to1\to3\to0$ and $\lambda_3=0\to2\to3\to0$ freely generate $\pi_1(K_4,0)$ , and their periods in space are $t_1=(1,-1,-1)$ , $t_2=(1,-1,1)$ and $t_3=(-1,-1,1)$ . The tally map is abelianization, so its kernel is the commutator subgroup: the normal closure of the commutators $[\lambda_i,\lambda_j]$ , which is the fundamental group of the crystal and is free of infinite rank. The tree of ordered histories is also the universal cover of the crystal, with deck group $[F_3,F_3]$ : above each point of space sit infinitely many ordered histories, and the kernel permutes them.

The commutator climbs by $t_1$ , then by $t_3$ , then descends by $t_1$ and by $t_3$ : its four legs run around a parallelogram of periods, and translations commute, so the loop closes. What it does not cancel is the parallelogram itself. The kernel is graded by the lower central series, and its first layer, $\Lambda^2\Z^3$ , is signed area: the vector area $\tfrac12\oint x\times dx$ of this decagon is $t_1\times t_3=(-2,0,-2)$ , and the areas $A_c$ of the six decagon classes satisfy $\sum_cA_cA_c^{\mathsf T}=16\,I$ . What space forgets first is the area a history sweeps.

Example(A commutator, by hand)

$[\lambda_1,\lambda_3]=\lambda_1\lambda_3\lambda_1^{-1}\lambda_3^{-1}$ is the walk $0\to1\to2\to0\to2\to3\to0\to2\to1\to0\to3\to2\to0$ . The steps $2\to0\to2$ retrace a letter; cancelling them leaves the closed walk $0\to1\to2\to3\to0\to2\to1\to0\to3\to2\to0$ of length ten. With the chord rule its addresses are $(0;0,0,0)$ , $(1;0,0,0)$ , $(2;1,0,0)$ , $(3;1,0,1)$ , $(0;1,0,1)$ , $(2;1,0,1)$ , $(1;0,0,1)$ , $(0;0,0,1)$ , $(3;0,0,1)$ , $(2;0,0,0)$ and back to $(0;0,0,0)$ : ten distinct sites, a decagon of the crystal, with tally $(1,0,0)+(0,0,1)-(1,0,0)-(0,0,1)=0$ . The crystal has no shorter cycles; thirty oriented decagons start at each site, and up to translation, reversal and change of starting point they fall into six classes, which the report group permutes transitively.

What a holonomy shows to space

Plate IX.2The triangles

\lambda_1

and

\lambda_3

K_4

. A transport assigns a group element to each oriented letter; the loops of space see only the commutators of the triangles’ holonomies.

A transport along the letters assigns to each oriented letter $e_{ab}$ an element $g_{ab}$ of a group $G$ acting on some fiber, with $g_{ba}=g_{ab}^{-1}$ ; multiplying along a closed walk at 0 gives its holonomy, a homomorphism $\rho\colon\pi_1(K_4,0)\to G$ . The same assignment lifts to the crystal, each of its edges carrying the element of its letter, and the holonomy of a closed loop of space is $\rho$ of its image in $\pi_1(K_4)$ , which lies in $[F_3,F_3]$ .

The register’s own transport is an instance. Writing $\gamma_0,\dots,\gamma_5$ for the Clifford units of the letters 01,23,02,13,03,12, the holonomies of the three triangles are the three-letter blades $H_1=-\gamma_0\gamma_2\gamma_5$ , $H_2=\gamma_0\gamma_3\gamma_4$ and $H_3=-\gamma_1\gamma_2\gamma_4$ ; any two share exactly one unit, so they commute, and every closed loop of space has holonomy $I_8$ , as an independent census of all 120 rooted decagons confirms. What space does with an abelian holonomy is remember it as momentum: a transport with values in $U(1)$ factors through the tally, a point of the crystal’s Brillouin torus. Up to gauge, a connection on the crystal is a homomorphism from its fundamental group, the kernel, so every flux of every gauge field on space is a function on what space forgets.

Proposition(Space sees only commutators)

For every homomorphism $\rho\colon F\to G$ , $\rho([F,F])=[\rho(F),\rho(F)]$ . Hence the holonomies of the closed loops of space, for a transport lifted from $K_4$ , form exactly the commutator subgroup of its holonomy group. If that group is abelian, every closed loop of space has trivial holonomy.

Proof

$\rho([x,y])=[\rho x,\rho y]$ , and every commutator of elements of $\rho(F)$ has this form. Both subgroups are generated by their commutators.

The order-three holonomy

Plate IX.3The transposition transport: on the axes

\rho(\lambda_1)=(A\,B)

and

\rho(\lambda_3)=(B\,C)

, and their commutator is the 3-cycle

(A\,C\,B)

. The Klein-four transport is flat.

Report relativity allows few transports. Along the letter from $v$ to $w$ take a permutation $T_{wv}$ of the four reports that takes $v$ to $w$ , reverses under reversal, and is covariant. The permutations fixing the ordered pair $(v,w)$ are the identity and the transposition $(c\,d)$ of the other two reports, so exactly two choices remain: $r$ , with $T_{wv}=(v\,w)$ , and $d$ , with $T_{wv}=(v\,w)(c\,d)$ . The report group acts on the axes $A=\{01,23\}$ , $B=\{02,13\}$ and $C=\{03,12\}$ through $S_4\to S_3$ , whose kernel is the Klein four-group, so $d$ does nothing to the axes: it is a translation, a displacement, while $r$ fixes the zero displacement and permutes the three steps. The sign of the holonomy of a closed walk is the parity of its length, which factors through the tally: the part of the holonomy that space keeps is its sign, and the part it forgets is the 3-cycle.

The same choice appears on the fiber. A report-covariant $\mathrm{Spin}(6)$ connection built from the letters alone has decagon holonomy $H$ with $H^3=I$ , trace $-4$ and eigenvalues $e^{\pm2\pi i/3}$ , each four times; an equally covariant flat connection has holonomy $+I_8$ around every decagon; and deforming the first by a rotation within the destination spin changes the trace continuously, $-4$ at angle 0 and $-2$ at angle $\pi/2$ . The covariant, reversible connections form seven components. On the kernel graph alone the class is a choice. Across clocks the flat option disappears, because $\PSL(2,7)$ is simple and admits no retraction onto $S_3$ ; within one clock the Klein four-group supplies exactly such a retraction, which is why $d$ exists.

Proposition(Triangles to transpositions, decagons to 3-cycles)

Under $r$ the triangles go to the three transpositions of the two reports they do not start from: $\lambda_1\mapsto(1\,2)$ , $\lambda_2\mapsto(1\,3)$ , $\lambda_3\mapsto(2\,3)$ , which act on the axes as $(A\,B)$ , $(A\,C)$ , $(B\,C)$ . The holonomy group is all of $S_3$ , the image of the kernel is exactly $A_3\cong\Z_3$ , the commutator of any two distinct triangles is a 3-cycle, and every one of the thirty rooted decagons at a site carries a 3-cycle. Under $d$ every loop of $K_4$ , and so of space, has trivial holonomy.

Proof

For $\lambda_1$ the frame changes are $(0\,1)$ , then $(1\,2)$ , then $(2\,0)$ ; their product sends $0\mapsto1\mapsto2\mapsto0$ , $1\mapsto0\mapsto0\mapsto2$ , $2\mapsto2\mapsto1\mapsto1$ and fixes 3, so it is $(1\,2)$ , which fixes the axis $C$ and exchanges $A$ and $B$ . The other triangles are the same computation. The three transpositions generate $S_3$ , so the kernel goes onto $[S_3,S_3]=A_3$ . For two involutions the commutator is the square of their product: $(A\,B)(B\,C)=(A\,B\,C)$ , whose square $(A\,C\,B)$ has order three. The decagon statement is a finite check. For $d$ the frame changes lie in the Klein four-group, and around any triangle they multiply to the identity.

The lift chooses

Plate IX.4One clock’s tetrahedron in the lift: four cusps, the reports, and six edges, the letters. Around a face the null-rotation transport is a half-turn, around a corner it is trivial, and around a decagon it is a third-turn.

In the lift of Chapter XIII the eight ends are the eight sky points, the directions from which light can arrive, and each letter is a rest frame on an edge joining two of them. One clock’s four reports are the four ends of a tetrahedron of the lift, half of that clock’s cube, and its six letters are the tetrahedron’s six edges. Two letters that share a report share a light direction, and there is a canonical way to compare their frames: the Lorentz transformation that fixes the shared light direction and turns nothing about it, a null rotation. It is the parallel transport of velocity space, and on one clock’s letters its holonomy has exactly the structure of $r$ : each triangle carries a half-turn, the Gauss–Bonnet rotation of an ideal triangle, each corner is trivial, the holonomy group at a letter is $S_3$ , and all thirty rooted decagons carry order three, a third-turn about the base letter’s line of sight.

On spinors the same transport is Thomas precession: the holonomy group on one clock’s letters is the binary dihedral group of order 12, faces carry trace zero and square $-I$ , and every decagon has order exactly three. The flat option is realized too, by turning the clock’s own cube about the shared light direction, but that needs the octonion table to tell the cubes of clocks from the cubes of lines. For the comparison of frames, and of spin, the choice is the null rotation, so the order-three holonomy is not an added choice: it is the curvature of velocity space, reduced to the kernel graph. Read modulo $\sqrt{-3}$ , every decagon’s holonomy becomes the identity, and the third-turn survives only as the next digit, which is always the letter at which the decagon is based, read as one of the thirteen lines of $\mathfrak{sl}_2(\F_3)$ . Read at the prime three, what space forgets is written in letters.

A binary flux on the decagons

Plate IX.5The same decagon carrying a conserved

\Z_2

flux, the product of the link variables around it. Of the eight translation-invariant flux patterns, zero flux has the lowest energy on every cell computed.

The best-known physics on what space forgets is Kitaev’s. In his honeycomb model each site carries four Majorana operators; three pair across the three bonds into link variables that commute with the Hamiltonian, and their products around hexagons are conserved $\Z_2$ fluxes. The honeycomb is itself a maximal abelian cover, and its hexagons are commutator loops. Yao and Lee gave each site six Majoranas, three for the bonds and three itinerant. The six letters supply those six Majoranas without choice: at a site of type $v$ the three letters of its star pair across the bonds and their complements are itinerant, and on the physical sector $D_i=+1$ , which keeps one of the fiber’s two chiral quartets, each bond becomes $-J\,\tau_i^a\tau_j^a(\sigma_i\cdot\sigma_j)$ , the Yao–Lee model on the crystal.

The fluxes live on the decagons, and so on the kernel. The six decagon classes obey four relations over $\mathrm{GF}(2)$ of rank three, which leaves exactly eight translation-invariant flux patterns, and uniform $\pi$ flux through every decagon is inconsistent. Among all eight, with all eight boundary twists, on cells of 32, 256 and 864 sites, zero flux has the lowest energy, in agreement with Hermanns and Trebst’s numerical finding for the Kitaev model on the same lattice; no Lieb-type theorem is available there. The binary flux is a conserved state variable with eigenvalues $\pm1$ , while the order-three holonomy is part of a transport law, with eigenvalues 1 and $e^{\pm2\pi i/3}$ , and no choice of $\Z_2$ link signs reproduces it. The medium’s site algebras, quartic coupling, projection and couplings are added.

Interference on the kernel

Plate IX.6Two routes recombine at one address, one carrying a decagon’s holonomy

H_C

. With the fiber maximally mixed,

p(+)=\tfrac12+\tfrac1{16}\mathrm{Re}\operatorname{Tr}H_C

: 1 for the flat connection,

\tfrac14

for the order-three one, and

\tfrac12

whenever a record names the route.

Retain the tally and start from a sharp address. Two alternatives that end with different tallies end at different addresses, orthogonal states of the retained address, so their cross term vanishes; two that end at the same address differ by a closed walk of zero tally, an element of the kernel. So with the tally retained, interference happens only across the kernel, and not across all of it: the two-layer rule multiplies the cross term of two alternatives by the overlap of what they leave in the permanent record. Go out and back along 01 and then along 02, or along 02 first: both tallies are zero, and if the two orders leave the same permanent records the alternatives remain fully coherent, while if the records name the order the coherence is exactly zero.

When both agree, the interference reads the kernel. Let two routes from the same start recombine at the same address, one carrying the holonomy $H_C$ of a decagon $C$ relative to the other, and let the fiber be maximally mixed. The probability of the symmetric outcome is $p(+)=\tfrac12+\tfrac1{16}\mathrm{Re}\operatorname{Tr}H_C$ , which depends only on the conjugacy class of $H_C$ , so any observer with the same apparatus reads the same value. The apparatus is an addition, and the native records do not measure a flux: every effect that a complete record can register is a multiple of the identity on the fiber.

Example(Reading a holonomy by interference)

For the flat connection $\operatorname{Tr}H_C=8$ and $p(+)=1$ : the routes recombine perfectly. For the order-three connection $\operatorname{Tr}H_C=-4$ and $p(+)=\tfrac14$ . At angle $\pi/2$ in its deformation, $\operatorname{Tr}H_C=-2$ and $p(+)=\tfrac38$ . If the permanent record notes which route was taken, the interference term is multiplied by zero and $p(+)=\tfrac12$ for every connection. The loop is invisible to space and to the tally, and visible to interference exactly when nothing else has recorded it.

The covers of Part II now fit in one sequence: the tree keeps the order of a history of report changes, the crystal keeps its tally, and $K_4$ keeps its current frame, with deck groups $[F_3,F_3]$ and $\Z^3$ . Two threads leave from here. The first layer of the kernel is signed area, which a $U(1)$ flux on the crystal would couple to, a candidate for light that the volume examines later. And the holonomy of the transposition transport acts on the three axes, which at the level of slots are the fiber’s three colour lines; what that holonomy does physically is open. Everything in Part II has happened at one clock. Part III turns to the twenty-eight observers of the seven clocks, the pairs of points of a finite celestial sphere, among which the flat choice disappears.

Concepts: incarnationcontinuum
Objects: the seven pointsthe sky