Part II · One Graph, Four CoversChapter VII

The Branchial Tree

Plate VII.1The tree of reduced histories from report 0 to depth three; each vertex shows the report reached. Two histories that differ once stay different.

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

What does a world keep when it keeps the order of everything that happened?

Chapter VI set time aside as a count of occurrences. The covers of the kernel graph remain, and this chapter takes the largest. Remember a history in full, step by step and in order, and two different histories never become the same. Every history is a path from the beginning, two histories agree up to some step and differ afterwards, and nothing closes up again. The picture is a tree.

For the kernel graph this tree has a precise name. A history of frame changes is a walk on $K_4$ , and the walks that never retrace a letter at once form the universal cover of $K_4$ , the 3-regular tree. The program calls this level branchial: at each depth the tree holds the alternatives of that age, and the branch points hold their shared past. This chapter asks what the tree keeps that the other covers forget, where the program’s own dynamics lives on it, and where the tree stops being the right picture.

The central result · The branchial tree

The universal cover of the kernel graph is the 3-regular tree $T$ of reduced histories of frame changes, with deck group $\pi_1(K_4)\cong F_3$ . Two histories from one report have the same tally, the same point of the crystal, exactly when they differ by an element of $[F_3,F_3]$ . The tally separates all histories of length at most four; the first it identifies have length five, and each such pair closes a decagon, a loop of ten steps in the crystal, such as the commutator of two triangle loops.

Within one observer the depth of the last common ancestor is an ultrametric, and the memory walk moves on such a tree on a sparse support that the tree alone does not select. The memory has two parts at two primes: each record is one digit read at the lift’s prime above three, while a register’s own logs are the reduced words of the free group on the three axes, and on two axes its rewrites form a binary counter. Between observers the ultrametric fails, because causal pasts merge.

Status

The covering statements and the ultrametric are exact. The claim that the program’s memory walk is branchial is a partial correspondence, proved in the form stated and false as an identity. The two primes of memory are proved by hand where they are identifications, and checked exactly where they are counts. The failure between observers is an exact counterexample.

The physical content assigned to the branchial level, that the internal clock operators, masses and the class constant live here and not in space, is the program’s typing, and it rests on the partial correspondence with its gaps. That the class constant is a physical mass needs a bridge the program does not have, and a derived transport of memory, since the value moves with it.

The tree of ordered histories

Plate VII.1The tree of reduced histories from report 0 to depth three; each vertex shows the report reached. Two histories that differ once stay different.

A history from report 0 is a walk $0=v_0,v_1,\dots,v_n$ on $K_4$ , reduced if $v_{i+1}\neq v_{i-1}$ for every $i$ . The reduced histories are the vertices of the tree $T$ ; the root has three children and every other vertex two, so there are $3\cdot2^{n-1}$ reduced histories of length $n\ge1$ . The fundamental group $F_3$ , freely generated by the triangle loops $\lambda_1=0\to1\to2\to0$ , $\lambda_2=0\to1\to3\to0$ and $\lambda_3=0\to2\to3\to0$ , acts on $T$ by prefixing loops and reducing. For two histories of equal length, with longest common prefix $h\wedge h'$ , set $d(h,h')=\beta^{-|h\wedge h'|}$ for a fixed $\beta>1$ .

Wolfram’s branchial graph is the first layer of this metric: at a fixed depth it joins alternatives with an immediate common predecessor. Nothing in the tree says how different two siblings are. In the forcing theorem’s two-region comparison experiment the six pair outcomes are all siblings, and their normalized quantum overlaps are $2/5$ for pairs sharing a report and $1/5$ for disjoint pairs. Branchial adjacency is a statement about ancestry, not about distinguishability: the tree carries the ancestry, and the metric of distinguishability is the kernel’s.

Proposition(Prefixes nest)

The branchial distance is an ultrametric: $d(h,h'')\le\max\{d(h,h'),\,d(h',h'')\}$ .

Proof

If $h$ and $h'$ agree for $a$ steps and $h'$ and $h''$ agree for $b$ steps, then $h$ and $h''$ agree for at least $\min\{a,b\}$ steps, because both agreements are prefixes of $h'$ and prefixes of one word are nested.

What the tally forgets

Plate VII.2The histories

A

(blue) and

B

(gold) leave by different first letters and end five steps down at different vertices of the tree, with the same tally. In the crystal they are the two halves of one decagon, the reduced commutator of

\lambda_1

and

\lambda_2^{-1}

The tally of a history is its net signed count of letters, $c(h)=\sum_ie_{v_iv_{i+1}}$ with $e_{ji}=-e_{ij}$ ; it determines the report reached, and the tallies are the vertices of the $K_4$ crystal. The histories $A=0\to1\to2\to0\to3\to1$ and $B=0\to3\to1\to2\to0\to1$ both end at report 1 with tally $e_{01}-e_{02}+e_{03}+e_{12}-e_{13}$ . $A$ runs the triangle $\lambda_1$ and then moves along $0\to3\to1$ ; $B$ makes the same move first and then runs the same triangle based at 1. The closed walk $AB^{-1}$ is $0\,1\,2\,0\,3\,1\,0\,2\,1\,3\,0$ , the commutator $[\lambda_1,\lambda_2^{-1}]$ after its one cancellation: the tally cannot tell whether a loop was run before or after a move.

A complete count lists every coincidence at length five. The 48 reduced histories of that length from report 0 have 35 tallies: 24 histories are alone in theirs, nine pairs share one, all ending away from the base like $A$ and $B$ , and two triples return to 0 with the tally of the far triangle, differing only in which rod carried the observer out and back. At length six the 96 histories have 66 tallies. What the tally forgets is graded by the lower central series of the free group: the first quotient is $\Z^3$ , the periods; the second is $\Lambda^2\Z^3$ , the signed areas swept by history loops, where the decagon above maps to $-[\lambda_1]\wedge[\lambda_2]$ ; the third has rank 8.

Proposition(What the tally forgets)

Two reduced histories from report 0 have the same tally if and only if they end at the same report and the loop $h\,h'^{-1}$ lies in the commutator subgroup $[F_3,F_3]$ . The tally is injective on histories of length at most four. It is not injective on histories of length five, and every pair of length-five histories with equal tally closes a cycle of length ten in the crystal.

Proof

The first statement is $H_1(K_4;\Z)=F_3/[F_3,F_3]$ : the class of a closed walk in $H_1$ is its integral edge count. For the second, if $h\neq h'$ of length at most $n$ have equal tally, cancelling their common prefix leaves a nontrivial reduced closed walk of length at most $2n$ with zero tally, which lifts to a closed walk in the crystal without immediate reversals and so contains a cycle. The crystal is the $(10,3)$ -a net, of girth ten, so $n\ge5$ . Conversely a decagon through a lift of 0, split at its opposite vertex, gives two reduced histories of length five with equal tally.

The observer’s memory is a tree

Plate VII.3A register’s possible logs from the letter 01: five continuations at each step, the antipode 23 excluded. The depth-two rewrite joins

(01,01,02)

and

(01,02,01)

, whose common ancestor is two steps back.

An observer’s record is another tree of ordered histories, and the program’s dynamics lives on it. A register’s log is a word of letters, oldest first, and each new letter repeats the last one, is adjacent to it, or is its antipode, received. A private writer therefore has five continuations at every step; its possible logs of length $k$ form six trees, one for each first letter, with $6\cdot5^{k-1}$ leaves in all, and siblings, the five children of one parent, form a complete graph $K_5$ .

The program’s model of one register’s memory, the tower, works at a fixed age, on all $6^k$ words of length $k$ . Besides re-anchoring it has two kinds of move: a free move replaces the newest letter by an adjacent one, and a memory rewrite of depth $j$ , anchored at a vantage $v$ , joins $P\,A\,r(B)^{j-1}$ and $P\,B\,r(A)^{j-1}$ for adjacent letters $A,B$ , where $r(x)$ is the letter on the axis of $x$ that contains $v$ . The typing in the program’s axioms is that this walk is branchial: free moves are sibling moves, and memory rewrites move between possible logs of one age whose common prefix lies $j$ steps back. An exact recognition test decides how much of that holds.

Proposition(How branchial the memory walk is)

On the append histories of one register, at every age $k$ and depth $2\le j<k$ : (i) every free move between two append histories joins siblings, and of the ten sibling pairs of a parent the two antipodal pairs are not free moves; (ii) every memory rewrite of depth $j$ between append histories joins two histories whose last common ancestor is $j$ steps back; (iii) there are $60\cdot5^{k+j-3}$ such pairs and only $48\cdot5^{k-j-1}$ anchored memory rewrites among them, a fraction $4/5^{2j-1}$ ; (iv) the walk also acts on the $6^k-6\cdot5^{k-1}$ words that no append history reaches, and at $j=k$ its rewrites join histories with different first letters and no common ancestor.

Memory as a six-adic tree

Plate VII.4At age three the stationary walk’s moves first differ at the newest letter 36 times in 43, one letter back 6 times and two letters back once. As the age grows, five moves in six touch only the newest letter.

So the memory walk is branchial in type: a sparse graph organised by depth of shared past. It is not the branchial graph. The free part needs an antipodal exclusion, the memory part needs the representative rule and the reset pattern, and the weights come from elsewhere; at age three and depth two, the depth-two layer holds 1500 ancestor-related pairs, and 48 of them are memory rewrites.

Let the age grow. In the tower’s stationary walk the fraction of moves whose endpoints first differ $d$ letters below the newest one is $5\cdot6^{k-1-d}/(6^k-1)$ , which is $(36,6,1)/43$ at $k=3$ and tends to $5\cdot6^{-(d+1)}$ : five moves in six touch only the newest letter. The oldest records freeze, and the infinite memory is an element of the six-adic integers $\Z_6$ , with rate and jump size inverse: the level law of a hierarchical, Vladimirov-type process with exponent one. The exponent does not select a measure. The commit’s own record stream is the Parry measure of the five-way subshift, of entropy $\log5$ ; the depth expansion has entropy $\log6$ ; and a measure must be chosen separately.

Proposition(The level law)

Put any positive symmetric conductances $c_e$ on the tower’s edges. There are exactly $12\cdot6^s$ edges whose endpoints first differ at position $s$ ; let $C_s$ be their total conductance and $a_s=C_s/6^s$ . The stationary frequency of level $s$ is $r_s=C_s/\sum_tC_t$ , and $\alpha_s=\log_6(r_{s+1}/r_s)=1+\log_6(a_{s+1}/a_s)$ . So $\alpha_s=1$ at every level exactly when the average conductance per free prefix is the same at every level.

Proof

Detailed balance for the walk $P(x,y)=c_{xy}/\sum_zc_{xz}$ makes each directed edge carry stationary flow $c_e/(2\sum_ec_e)$ ; summing over a level gives $r_s$ . The rest is the definition of $a_s$ .

What a record is: the prime above three

Plate VII.5The four points of

\Proj^1(\F_3)

as the vertices of

K_4

. The thirteen lines of

\mathfrak{sl}_2(\F_3)

are its four vertices (nilpotent lines: reports), its six edges (split lines: letters) and its three perfect matchings (non-split lines: axes).

The six-adic tree says how far back two memories agree, not what the digits are. The lift’s Lorentz transformations are $2\times2$ matrices whose entries are Eisenstein integers $a+b\omega$ , $\omega=e^{2\pi i/3}$ , and among these numbers $\sqrt{-3}$ is a prime; reading modulo it leaves $\F_3$ . Read there, the lift’s four ends are the four reports and its symmetry is the report group. The transformations that become the identity at $\sqrt{-3}$ can be written $I+\sqrt{-3}\,X$ , and $X$ read modulo $\sqrt{-3}$ is a traceless $2\times2$ matrix over $\F_3$ : these form the Lie algebra $\mathfrak{sl}_2(\F_3)$ , which has 26 nonzero elements, so 13 lines.

So what a record is, which report, letter or axis, is one digit read at the lift’s prime above three. Chapter XI meets the same layer from the other side: it holds the next digit of a history loop’s holonomy.

Proposition(The alphabet at the prime above three)

The thirteen lines of $\mathfrak{sl}_2(\F_3)$ are of three kinds, and the kinds are the observer’s alphabet on the four points of $\Proj^1(\F_3)$ . Four lines are nilpotent, each picking out one point: a report. Six lines are split, with two eigenlines, each picking out two points: a letter. Three lines are non-split, with no eigenline, each pairing the four points off: an axis. The correspondence commutes with $\PGL(2,3)\cong S_4$ , the report group.

Proof

Write $X=\left(\begin{smallmatrix}a&b\\c&-a\end{smallmatrix}\right)$ . Then $X^2=(a^2+bc)I$ , so $X$ is nilpotent, has the two eigenvalues $\pm1$ , or has no eigenvalue in $\F_3$ , according as $a^2+bc$ is 0, 1 or 2. Counting nonzero solutions gives 8, 12 and 6 matrices, hence 4, 6 and 3 lines. A nilpotent $X$ has one eigenline and a split $X$ has two. A non-split $X$ is invertible with $X^2$ scalar, so it acts on the four points as an involution with no fixed point, which pairs them off. Conjugation by $\GL(2,3)$ , which acts on $\Proj^1(\F_3)$ as $S_4$ , preserves each kind and carries the picked-out points along.

How records pile up: a binary counter

Plate VII.6Words of length three on two axes, read as binary numbers. Free moves join

2m

and

2m+1

; depth-two rewrites join 1,2 and 5,6; the depth-three rewrite joins 3 and 4. Together they join every

u

u+1

: the dyadic adding machine, cut off at the word’s length.

Accumulation is a different matter. Pair the two letters of each axis as a generator and its inverse. A private writer may follow a letter by any letter except its antipode, just as a reduced word never follows a generator by its inverse, so a private writer’s logs are exactly the reduced words of the free group on three generators, one for each axis. This free group should not be confused with the fundamental group $F_3$ of the first step, whose generators are the triangle loops.

The obvious guess is that the lift’s structure at $\sqrt{-3}$ also organizes the accumulation. A prime carries a natural tree, its Bruhat–Tits tree, which at $\sqrt{-3}$ branches four ways, one branch for each report. The guess fails on counts: the tree has $4\cdot3^{k-1}$ non-backtracking paths of length $k$ , against $6\cdot5^{k-1}$ private logs and $6^k$ words of the tower, and equals neither for any $k$ . Nor is the six-way tree of logs the tree of any Eisenstein prime: it would need a residue field with five elements, and 5 stays prime there. The counter below is 2-adic, and the prime 2 is not a place of the lift. The transport of memory that the mass number needs cannot come from the tree at $\sqrt{-3}$ ; the program’s two 2-adic candidates for it have not been tested.

Proposition(Memory counts in binary)

Take words whose letters lie on two of the three axes, write 0 for a letter on the first and 1 for a letter on the second, and read the newest letter as the lowest digit. Then every free move and every memory rewrite of the tower changes the number by exactly one: a free move of the newest letter adds or subtracts one with no carry, and a rewrite of depth $j$ adds one with a carry through $j-1$ digits. For words of length up to seven, every pair of numbers $u$ and $u+1$ is joined, except where adding one would overflow the word.

Proof

Letters on different axes are always adjacent, so a free move between the two axes flips the lowest digit. With $A$ on the first axis and $B$ on the second, the depth- $j$ rewrite joins $P\,A\,r(B)^{j-1}$ and $P\,B\,r(A)^{j-1}$ , which read $\dots0\,1^{j-1}$ and $\dots1\,0^{j-1}$ in binary, oldest digit first: a number ending in $j-1$ ones, and the number one larger. The last statement was checked exactly for all three pairs of axes.

Between observers the tree fails

Plate VII.7Four registers on a path

A

–

C

–

B

–

D

A

knows the first meeting,

D

the second, and

B

and

C

both, through their bridge; so the closeness of shared pasts is not an ultrametric.

Within one observer prefixes nest, and the tree is exact. Between two observers, measure closeness by their latest shared past: let $\ell$ be the causal height of the latest occurrence in the common causal past of both, and put $d=6^{-\ell}$ . A register’s past is a prefix of one word, and prefixes of one word are nested; two registers’ common past is the intersection of two down-sets of a partial order, and down-sets are not nested. The world’s order is a partial order with commuting independent occurrences, not a tree.

The program’s working answer for what relates two observers is that the relation is rulial: it is the relative frame of two anchored observers, and its metric is the Coxeter graph on the twenty-eight of them, in which the distance between two observers is fixed by the cross-ratio of their celestial pairs. The ultrametric of the tower’s memory is branchial, the geometry of one observer’s alternatives. Shared history is not a tree, and no lateral geometry is selected by it.

Example(Merging pasts)

Place four registers on a path $A$ – $C$ – $B$ – $D$ of the crystal. Let $A$ meet $C$ and, independently, $B$ meet $D$ , both at causal height 1. Then $C$ and $B$ each make a private move at height 2 and meet each other at height 3. At the end $A$ knows the first meeting, $D$ knows the second, and $B$ and $C$ know both through their bridge. So $\ell_{AB}=1$ , $\ell_{BD}=1$ and $\ell_{AD}=0$ , and $d_{AD}=1>\max\{d_{AB},d_{BD}\}=\tfrac16$ : the ultrametric inequality fails. The history is legal, and a genuine prefix tree passes the same test: the common-prefix metric on the 16 binary words of length four satisfies all 4096 triangle inequalities.

Composition joins this level to the rulial one: the representative choice that is a rule for a register is a history for a composite, which cannot tell which member realized a move, so composition turns rulial blindness into branchial blindness. In the tower’s composites the mass gap comes from the blind sums of Chapter II, sums over siblings that the records cannot tell apart; it leaves the class constant $g=0.021386958933918\ldots$ , with $y=-10g$ a root of $3y^4+5y^3-16y^2-27y-5$ , a branchial quantity in this exact sense and a mass only as a reading. The tree keeps too much to be space: the next chapter forgets the order and keeps the tally.

Words defined here: branchialtower
Concepts: objectincarnationseambridgestatusrefutedcompletionabsence
Objects: the seven points