Part II · One Graph, Four CoversChapter V

Four Reports, Six Letters

Plate V.1The kernel graph

K_4

with report 0 at the centre. Each axis is a pair of opposite edges, drawn alike:

A=\{01,23\}

in blue,

B=\{02,13\}

in gold,

C=\{03,12\}

dashed. The three nonzero displacements are the double transpositions, one per axis.

Why is the smallest complete observer a tetrahedron of reports, and what do its six edges carry?

Part I ended with a world of registers whose logs are words in six letters, each register’s log space complete only in the company of partners at the three other reports. This chapter looks at the four reports, and at the six relations between them, as one graph.

Draw the four reports as the corners of a tetrahedron. Its six edges are the pairs of reports, the letters, because every record the observer writes is a word in them. Each edge has exactly one opposite edge, its antipode, and the six edges fall into three opposite pairs, the axes. As a graph this is $K_4$ , and Part II is organised around it. A history of changes of report frame is a walk on $K_4$ , and the walk’s covering spaces keep more or less of that history: its full order, its signed tally, or only the frame it ends in. Before any cover, the graph already carries two different structures, according to whether a letter is read as an unordered pair or as an oriented edge.

The central result · One graph, two readings of its edges

The kernel graph is $K_4$ : the four reports are its vertices and the six letters its edges; a letter’s antipode is the opposite edge, and the three axes are the perfect matchings. Under the report group $S_4$ :

(i) read unsigned, the letters carry $\C^6=\mathbf 1\oplus\mathbf 3\oplus\mathbf 2$ , and the relabellings span exactly $\C\oplus M_3(\C)\oplus M_2(\C)$ ; the maps preserving the two nontrivial blocks, their Hermitian form and a common volume form are $S(\mathrm U(3)\times\mathrm U(2))$ ;

(ii) read oriented, the letters are the two-forms $\Lambda^2\R^4$ , split as gradients $\mathbf 3$ plus loops $\mathbf 3\otimes\mathrm{sign}$ ; the loops are $H_1(K_4;\Z)\cong\Z^3$ , the periods of the maximal abelian cover, and every report permutation acts on them as a proper rotation.

No nonzero equivariant linear map carries either block of (i) into the loops. The Hodge star sends each letter to $\pm$ its antipode and exchanges gradients with loops. On the observer’s report form the oriented letters are Lorentz generators: gradients generate boosts, loops generate rotations, and the star exchanges the boost along each report’s axis with the rotation about it.

Status

Everything in the central result is exact. What is not exact is any physical name: calling the first block “internal symmetry” and the loops “space” are readings, each tested in its own chapter. The covering statements of the last step are theorems as well, while the level names, branchial, spatial and rulial, are the program’s readings.

As a group, $K=S(\mathrm U(3)\times\mathrm U(2))$ is the Standard Model’s $(\SU(3)\times\SU(2)\times\mathrm U(1))/\Z_6$ , and the program’s gauge-link clause adopts it as the group of the links between neighbouring observers’ frames; the clause is an addition. Its central circle has, up to normalization, the weights $(2,-3)$ of the hypercharge generator on the five-dimensional representation of $\SU(5)$ , and the forcing theorem builds one family’s gauge representations on this group; that identification is a reading, and the program states it as one. The crystal is space only on a further clause, retention of the frame path’s circulation, which has a measurable price.

Places and displacements

Plate V.1The kernel graph

K_4

with report 0 at the centre. Each axis is a pair of opposite edges, drawn alike:

A=\{01,23\}

in blue,

B=\{02,13\}

in gold,

C=\{03,12\}

dashed. The three nonzero displacements are the double transpositions, one per axis.

The founding register’s four stable classes are canonically $\Z_2\times\Z_2$ , blind bit times consequence bit. As a set with its translations this is the affine plane $\mathrm{AG}(2,\F_2)$ : its four points are the reports, its six lines, each containing two points, are the letters, its three parallel classes are the axes, and its three nonzero translations are the double transpositions. The full symmetry group is $\mathrm{AGL}(2,\F_2)\cong S_4$ . A chosen origin is a vantage, and its stabilizer is $\GL(2,\F_2)\cong S_3$ .

The affine reading separates two kinds of object. The reports are places, points of the plane where an observer can stand, none of them an origin. The translations are displacements, steps from place to place, each moving every report by the same step. A place has no preferred member; a displacement does, the zero displacement, which is no step at all. Choosing an origin matches the two sets, each place with the step that reaches it from the origin, and that matching is what a vantage adds. The four displacements return as the four classes of an observer’s octonion units modulo its clock, and as the four lines of the Pati–Salam quartet, where the zero displacement is the lepton’s line.

Proposition(Places and displacements)

(i) The report group $S_4\cong\mathrm{AGL}(2,\F_2)$ permutes the four places transitively.

(ii) The displacements form its normal Klein subgroup $V_4$ : the identity and the three double transpositions, one per axis, whose two cycles are the two letters of that axis. The report group acts on $V_4$ by conjugation with two orbits, the identity alone and the three double transpositions, and it induces all of $S_3\cong\GL(2,\F_2)$ on the three.

(iii) The dressing $T$ and the swap $S$ of Chapter I generate the stabilizer of the place $(+,+)$ . Every relabelling is uniquely an element of this stabilizer followed by a displacement, and it moves $(+,+)$ exactly when that displacement is not the identity.

Proof

Write the reports additively as $\F_2^2$ , with $(+,+)$ as 0, so that the displacements are the translations $t_v(x)=x+v$ , acting simply transitively. A nonzero $t_v$ fixes no point and squares to the identity, so it is a double transposition; its cycles $\{x,x+v\}$ are the two lines of direction $v$ , the two letters of one axis. For an affine map $g(x)=Ax+b$ one has $g\,t_v\,g^{-1}=t_{Av}$ , so conjugation fixes $t_0$ and moves the nonzero translations as $A$ moves the nonzero vectors, which $\GL(2,\F_2)$ does faithfully. Finally $g=t_b\circ A$ with $b=g(0)$ , and the factors are unique because $t_0$ is the only translation fixing 0.

Antipodes exist only for four

Plate V.2The double transposition

(01)(23)

fixes the two letters of axis

A

and swaps the letters of the other two:

02\leftrightarrow13

and

03\leftrightarrow12

. It fixes every axis.

Among $n\ge2$ reports permuted by $S_n$ , call two letters antipodal when they are disjoint and together use every report. The complement of a two-element subset of an $n$ -set has $n-2$ elements, and it is a letter exactly when $n=4$ . In words, “which axis” is a coarser distinction than “which letter”, and no other $S_n$ has a proper quotient of this kind.

There is also a uniqueness statement in the other direction. $S_4$ has exactly three subgroups of index three, all conjugate Sylow 2-subgroups, and the stabilizer of a matching is one of them; so every three-element set on which the report group acts transitively is, as an $S_4$ -set, the set of axes. Each nontrivial element $(ij)(kl)$ of $V_4$ fixes the two letters of the matching $\{ij,kl\}$ and swaps the letters of the other two, so the three axes correspond to the three double transpositions. The third road to four, exact return of pair memory, splits the pair space into sectors of dimensions 1, $n-1$ and $n(n-3)/2$ , and only $n=4$ survives.

Proposition(Why four)

Let $n\ge2$ reports be permuted by $S_n$ . A letter has an antipode if and only if $n=4$ . The number $(n-1)!!$ of perfect matchings equals the number $n-1$ of independent report differences only for $n\in\{2,4\}$ . For $n=4$ the antipodal pairs are the three perfect matchings, the normal Klein subgroup $V_4\triangleleft S_4$ fixes every axis, and $S_4$ acts on the axes through $S_4/V_4\cong S_3$ .

Proof

The complement of a letter has $n-2$ elements, and is a letter exactly when $n=4$ . Over even $n$ the sequence $(n-1)!!$ runs $1,3,15,105,\dots$ and meets $n-1$ only at $n=2$ and $n=4$ . A double transposition $(ij)(kl)$ either fixes a matching letter-wise or swaps its two letters, so it fixes every matching; $V_4$ is the kernel of the action on the three matchings, and the quotient is $S_3$ .

Frames, clocks and what only a partner writes

Plate V.3An observer anchored at report 0. Its star, the rods 01, 02, 03, meets every axis once; the odd letters 23, 13, 12 lie opposite them.

An observer that stands on one report, its anchor or vantage, breaks the report group from $S_4$ to the stabilizer $S_3$ of that report. The letters then split three and three: the vantage’s rods, one per axis, which form the star of the report, and the three odd letters opposite them. Standing on report $v$ and re-anchoring to $w$ moves the vantage along the letter $vw$ , so a history of re-anchorings is a walk on $K_4$ ; this is why Part II calls $K_4$ the rulial frame graph. The same graph is one clock of the Fano plane, and the plane’s seven points give seven clocks, each with its own $K_4$ .

The octahedron of letters has diameter two, so two consecutive letters of a record are equal, adjacent or antipodal, and nothing else: persistence, the register’s own distinction, or a received exchange. This is the precise content of the program’s phrase “letters as links”. The stronger reading, that a letter is a relation held jointly by the two registers anchored at its reports, fails as dynamics: no ready exchange configuration can be represented by literal joint ownership. The letters are links of the kernel graph, not jointly held tokens.

Unsigned letters: the lock

Plate V.4The letter 01 in three sectors: the constant part

\tfrac16

on every edge; the report part

\tfrac12

on 01 and

-\tfrac12

on 23; the matching part

\tfrac13

on 01 and 23 and

-\tfrac16

on the other four. They add to 1 on 01 and 0 elsewhere.

Read a letter as an unordered pair. On the letter space $\C^6$ let $\mathbf 1$ be the all-ones vector and $(\delta x)_{ij}=x_i+x_j$ , and put $L=\C\mathbf 1$ , $W_3=\delta(\mathbf 1_4^\perp)$ and $W_2=\ker\delta^*$ . Since $\delta^*\delta=2I+J$ , the three are orthogonal of dimensions 1,3,2. The report block $W_3$ is spanned by the matching differences $e_{01}-e_{23}$ , $e_{02}-e_{13}$ , $e_{03}-e_{12}$ , and the constant together with the matching block is spanned by the matching sums $m_A=e_{01}+e_{23}$ , $m_B$ and $m_C$ . The letter 01 decomposes as $e_{01}=\tfrac16\mathbf 1+\tfrac12(e_{01}-e_{23})+\tfrac16(2m_A-m_B-m_C)$ , with squared lengths $\tfrac16,\tfrac12,\tfrac13$ : every letter spreads over the sectors in the proportion $1:3:2$ of their dimensions.

The forcing theorem then fixes a group. The complex-linear maps of $W_3\oplus W_2$ that preserve the two blocks, their Hermitian form and a volume form are exactly $K=S(\mathrm U(3)\times\mathrm U(2))$ , and $(z,A,B)\mapsto(z^2A,z^{-3}B)$ maps $\mathrm U(1)\times\SU(3)\times\SU(2)$ onto it with kernel $\{(\zeta,\zeta^{-2}I_3,\zeta^3I_2):\zeta^6=1\}$ . As a group $K$ is the Standard Model’s, and the gauge-link clause adopts it, as an addition. On the program’s own carriers its physical status is limited, and the lock’s $\mathbf 3$ is not the colour of the fiber: as modules the program has two colours, and identifying them is an added step, while their labels, indexed by the axes, are identified.

Theorem(The lock)

The span of the report permutations acting on the six unsigned letters is $\C\,I_L\oplus\operatorname{End}(W_3)\oplus\operatorname{End}(W_2)\cong\C\oplus M_3(\C)\oplus M_2(\C)$ , an algebra of dimension 14. For $n$ reports the same argument gives $\C\oplus M_{n-1}(\C)\oplus M_{n(n-3)/2}(\C)$ .

Proof

A matrix commuting with every pair permutation has entries that depend only on whether the two pairs share zero, one or two reports, so the commutant has dimension three. The three invariant projections already span it, so the three sectors are inequivalent irreducibles, each of multiplicity one. By the double commutant theorem the span of the group is the commutant of its commutant, $\operatorname{End}(L)\oplus\operatorname{End}(W_3)\oplus\operatorname{End}(W_2)$ .

Oriented letters and the Hodge star

Plate V.5Oriented letters. The star of report 0,

e_{01}+e_{02}+e_{03}

(gold), whose negative is the gradient of report 0, goes under the Hodge star to the triangle

1\to2\to3\to1

opposite it (blue):

\star(-e_{01}-e_{02}-e_{03})=-(e_{12}+e_{23}-e_{13})

Now read a letter as an oriented edge, with $e_{ji}=-e_{ij}$ . These are the integral one-chains of $K_4$ , and over $\R$ the two-forms $\Lambda^2\R^4$ on the four reports. The boundary map $\partial e_{ij}=e_j-e_i$ gives the graph’s Hodge decomposition into gradients of report functions and closed loops. The loops are $H_1(K_4;\Z)\cong\Z^3$ , with integral basis the three triangles through report 0: $c_1=e_{01}+e_{12}-e_{02}$ , $c_2=e_{01}+e_{13}-e_{03}$ and $c_3=e_{02}+e_{23}-e_{03}$ .

The lock’s $\mathbf 3$ is the space of report functions, which transforms like the tetrahedron with its mirrors; the loops are its orientation twin, which transforms like the proper rotations of a cube. The same six oriented letters carry both, as complementary Hodge parts, and the program calls this the solder theorem. With the reports as an orthonormal frame of $\R^4$ , the Hodge star $\star e_{ij}=\varepsilon_{ijkl}e_{kl}$ sends each letter to $\pm$ its antipode, satisfies $\star^2=1$ , and exchanges gradients with loops, $\star P\star=I-P$ . It sends the star of a report to the triangle opposite it.

Proposition(The two faces of a letter)

As representations of $S_4$ , the unsigned letters are $\mathbf 1\oplus V\oplus W$ and the oriented letters are $V\oplus V\otimes\mathrm{sign}$ , with the gradients carrying $V$ and the loops carrying $V\otimes\mathrm{sign}$ . On $H_1(K_4;\R)$ every report permutation acts by an orthogonal matrix of determinant $+1$ . Consequently $\operatorname{Hom}_{S_4}(\mathbf 1\oplus V\oplus W,\,H_1)=0$ , and the only equivariant linear maps from oriented letters to loops are the multiples of the projection onto the loops.

Proof

The oriented character counts fixed letters with the sign of their orientation; on the classes of the identity, a transposition, a double transposition, a three-cycle and a four-cycle it is $(6,0,-2,0,0)$ , which is $V+V\otimes\mathrm{sign}=(3,1,-1,0,-1)+(3,-1,-1,0,1)$ . The gradients are the report functions modulo constants, which is $V$ , so the loops carry the rest. $V\otimes\mathrm{sign}$ has trace $-1$ on involutions, 0 on three-cycles and 1 on four-cycles, and in each case the eigenvalues have product $+1$ . It shares no irreducible with $\mathbf 1\oplus V\oplus W$ , and occurs once in the oriented letters, so Schur’s lemma gives both statements about maps.

Letters as Lorentz generators

Plate V.6On the report qubit’s Minkowski space the star of report 0 generates the boost along report 0‘s axis, and the triangle opposite it generates the rotation about the same axis. The Hodge star exchanges the two.

The oriented letters have a third face, on the observer’s own geometry. On the report qubit’s Minkowski space the four reports are null vectors $\Pi_0,\dots,\Pi_3$ , and the letter crossed from $a$ to $b$ acts as the bivector of its two reports, $M_{ab}X=\langle\Pi_b,X\rangle\,\Pi_a-\langle\Pi_a,X\rangle\,\Pi_b$ , a generator of Lorentz transformations. It is a pure boost in the plane of its two reports, stretching one and shrinking the other by the same factor: a Doppler shift between them.

Relative to the time axis of the report form every generator splits into a boost part and a rotation part, and the Hodge decomposition is that split. A gradient generates a pure boost, and the star of a report the boost along that report’s axis; a loop generates a pure rotation, and the triangle opposite a report the rotation about that report’s axis; the star exchanges the two parts letter by letter. The map from oriented letters to generators is an isomorphism onto the Lorentz algebra. The star squares to one, so it is not the electric–magnetic duality of the Lorentz algebra, which squares to minus one; it is the plain exchange of the two parts. Across the seven clocks, the letters of one clock are the restriction of the irreducible six-dimensional representation of $\PSL(2,7)$ , the exterior square of the faithful quartet of the double cover $\SL(2,7)$ .

One graph, four covers

Plate V.7The covering tower of the kernel graph. The tree keeps the full order of a history of frame changes. Dividing by the commutator subgroup keeps only the signed tally: the crystal, whose shortest closed loops are decagons. Dividing further by the periods

\Z^3

keeps only the current frame. Time sits apart.

A history of frame changes is a walk on $K_4$ , and coverings of $K_4$ are the natural ways to remember more or less of it. Part II is built on four objects: $K_4$ itself, its universal cover, its maximal abelian cover, and the kernel of the map between the last two. Two histories from report 0 to report 2, the direct move and the detour $0\to1\to2$ , are the same to $K_4$ . The crystal tells them apart, since their tallies differ by the loop $c_1$ , and so does the tree. Two histories that the crystal cannot separate but the tree can must differ by a commutator loop; the shortest such pair has length five.

The covering statements are theorems; the names on the levels are readings. The program’s working picture is that the kernel of the tally map, the loops that space cannot see, is where interference and gauge holonomy live. The crystal is space only on a further clause. And time is not a cover: the covers are quotients of the frame path, which is live content and can be undone, while elapsed time is the count of committed occurrences, which only grows. For $K_n$ the loops have rank $(n-1)(n-2)/2$ , three at $n=4$ , so “why four” fixes the lock’s $(3,2)$ and the three periods of space together.

Proposition(The covers of the kernel graph)

Fix the base report 0. (i) The fundamental group $\pi_1(K_4,0)$ is free of rank three, freely generated by the triangles $0\to1\to2\to0$ , $0\to1\to3\to0$ and $0\to2\to3\to0$ . (ii) The universal cover is the 3-regular tree $T$ , whose vertices are the reduced walks from 0, and $F_3=\pi_1(K_4,0)$ acts on it freely with quotient $K_4$ . (iii) The maximal abelian cover is $X=T/[F_3,F_3]$ , with deck group $H_1(K_4;\Z)\cong\Z^3$ and vertices the pairs $(v,c)$ with $c$ an integral one-chain and $\partial c=v-0$ : this is Sunada’s $K_4$ crystal. (iv) The map $T\to X$ is again a covering, with deck group $[F_3,F_3]=\pi_1(X)$ , a free group of infinite rank, and $1\to[F_3,F_3]\to F_3\to\Z^3\to1$ is exact.

Proof

A connected graph with $V$ vertices and $E$ edges has free fundamental group of rank $E-V+1$ , here $6-4+1=3$ ; contracting the star of 0, each chord 12,13,23 gives one free generator, and these are the three triangles. The universal cover of a graph is a tree, here regular of degree three. Covers of a connected graph correspond to subgroups of its fundamental group, and the maximal abelian cover to the commutator subgroup: two walks from 0 end at the same vertex of $X$ exactly when they have the same integral edge count.

One row of the volume’s table of levels is not a cover at all, and the next chapter takes it up: time, the count of the occurrences that write a history on the graph.

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