Part II · One Graph, Four CoversChapter VIII

Space as a Tally

Plate VIII.1The kernel graph with the tree letters at report 0 and the chords 12, 13, 23 counted by

z_1

z_2

z_3

. The triangle

0\to1\to2\to0

(gold) closes in

K_4

and crosses one chord.

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

What remains of a history when its order is forgotten but its tally is kept?

An observer stands at one of four reports and can change where it stands. A change of report, a re-anchoring, carries the observer along a letter, so a history of such changes is a walk on $K_4$ . Chapter VII kept all of such a walk. Keep much less: for each letter record only how many times it was crossed in each direction, and forget when. Call this record the tally. Walking the triangle $0\to1\to2\to0$ and then the triangle $0\to2\to3\to0$ leaves the same tally as walking them in the other order, and the same as walking the square $0\to1\to2\to3\to0$ once.

A tally looks like a poor memory. This chapter shows that the tallies form a three-dimensional crystal, which the program now reads as its space, and that this space differs from the one the program used to draw in a way that matters: relabelling two reports acts on it as a rotation, not a mirror. It then places the crystal in the observer’s own Lorentz geometry, where its periods turn out to be the generators of rotations, and reads the crystal’s metric modulo the primes 2, 3 and 7: the Fano plane with one point marked, the four axes of the reports under the rotations of a cube, and the eight points of the finite sky.

The central result

Call the tally of a walk on $K_4$ the integral chain that counts, with sign, how often each oriented letter was crossed. Walks from a fixed report with the same endpoint and the same tally are exactly the vertices of the maximal abelian cover of $K_4$ , with periods $H_1(K_4;\Z)\cong\Z^3$ . Realized by the harmonic projection, the only report-covariant linear map from oriented letters to periods, this cover is Sunada’s $K_4$ crystal: trivalent, three-dimensional, of girth ten, strongly isotropic and chiral. Its periods carry the report group as $\Lambda^2V\cong V\otimes\mathrm{sgn}$ , so all twenty-four report permutations act as proper rotations, and the crystal’s edges along complementary letters are perpendicular, not opposite as the old mirror theorem assumed. Report-related crossings of a covariant walk therefore carry equal charges, and at the symmetric momenta an isolated triplet crossing is forced to be exactly isotropic while it may carry a charge.

On the report qubit’s Minkowski space each letter generates a boost between its two reports, and a tally’s boost part is half its boundary. The periods are therefore exactly the tallies that generate rotations, and the crystal’s metric is the norm of the rotation algebra. Read modulo 2, this one metric is the Fano plane with its marked point, a clock. Read modulo 3, it picks out the four report axes, permuted by the rotation group of a cube. Read modulo 7, its null directions are the eight points of the finite celestial sphere, the sky of the twenty-eight observers. Read as positions rather than circulations, the same lattice is the arena of report counts of Chapter IV.

Status

The geometry is exact, and so is the crystal’s place in the observer’s Lorentz geometry. That this crystal is physical space is a reading, and it costs one added clause: the circulation of the observer’s live path of report changes must be retained as live content. That clause is not the unchanged register written in new coordinates; it changes a finite statistic already at fourth order in time. The walks on which isotropic charged crossings have been exhibited use further supplied ingredients.

The arena and the crystal are one lattice, which settles which of them is space on the condition that the lift’s symmetries are the symmetries of space; it does not say which reading a law of motion uses. The crystal’s chirality is not yet a hand of matter: the law does not prefer one enantiomer, and choosing one is an added background. Whether the integral tallies commute with the dictionary between clocks has not been checked.

The tally of a walk

Plate VIII.1The kernel graph with the tree letters at report 0 and the chords 12, 13, 23 counted by

z_1

z_2

z_3

. The triangle

0\to1\to2\to0

(gold) closes in

K_4

and crosses one chord.

Write $e_{ab}$ for the letter $ab$ crossed from $a$ to $b$ , with $e_{ba}=-e_{ab}$ , and let the boundary map send a crossing to its change of report, $\partial e_{ab}=e_b-e_a$ . The tally $c(\gamma)\in\Z^6$ of a walk is the sum of its crossings, each counted with its sign. It forgets order completely, and $\partial c(\gamma)$ is the endpoint minus the starting point; a closed walk has a tally in $\ker\partial$ , the integral circulations, which is $H_1(K_4;\Z)$ .

So the tally of a walk is a pair $(v;z)$ : the report it ends at and three integers. Tree letters leave $z$ unchanged; crossing $1\to2$ , $1\to3$ or $2\to3$ adds $(1,0,0)$ , $(0,1,0)$ or $(0,0,1)$ , and the reverse crossing subtracts it. The triangle $0\to1\to2\to0$ ends at $(0;1,0,0)$ : back at report 0, but not back at the start. The triangle $0\to2\to3\to0$ ends at $(0;0,0,1)$ ; walking both, in either order, ends at $(0;1,0,1)$ , and so does the square $0\to1\to2\to3\to0$ . The count three belongs to four reports: the complete graph on $n$ points has $(n-1)(n-2)/2$ independent loops, which is 3 exactly when $n=4$ .

Proposition(Circulation coordinates)

$H_1(K_4;\Z)$ is free of rank $6-4+1=3$ , with basis the three triangles through report 0, $c_1=e_{01}+e_{12}-e_{02}$ , $c_2=e_{01}+e_{13}-e_{03}$ and $c_3=e_{02}+e_{23}-e_{03}$ . Every walk from 0 ending at $v$ has a tally $t_v+c_1z_1+c_2z_2+c_3z_3$ with $z\in\Z^3$ , where $t_0=0$ and $t_v=e_{0v}$ otherwise. The integers $z$ are the net signed numbers of crossings of the three letters 12, 13 and 23.

Proof

Take the three letters at 0 as a spanning tree. Each remaining letter, a chord, occurs in exactly one $c_i$ , with coefficient one. Subtracting those chord coefficients from any integral circulation leaves a circulation supported on the tree, which vanishes: a leaf’s boundary equation forces its edge coefficient to zero, and induction removes the tree. For an open walk, subtract $t_v$ first.

The crystal

Plate VIII.2The triangle

0\to1\to2\to0

lifted to the crystal. Each step rises a third of a period and turns by

120^\circ

, and after three steps the walker is back at a site of type 0, one period

(1,-1,-1)

higher. Short blue stubs are each site’s third edge.

To place a tally in space one needs a metric on periods, and the letters already carry one. Give the six oriented letters unit length and project each onto the circulation space, $P=I-\tfrac14\partial^{\mathsf T}\partial$ ; the projected letter $b_{ab}=Pe_{ab}$ is one step along the letter $ab$ , an edge of the crystal. In coordinates in which the triangle periods are $(1,-1,-1)$ , $(1,-1,1)$ and $(-1,-1,1)$ , the edges are $b_{01}=\tfrac12(1,-1,0)$ , $b_{02}=\tfrac12(-1,0,1)$ , $b_{03}=\tfrac12(0,1,-1)$ , $b_{12}=\tfrac12(0,-1,-1)$ , $b_{13}=\tfrac12(1,0,1)$ and $b_{23}=\tfrac12(-1,-1,0)$ . Every edge has squared length $\tfrac12$ , the three edges at a site meet pairwise at $120^\circ$ , edges along complementary letters are perpendicular, and the periods are the integer triples whose coordinates have equal parity.

This realized cover is a classical crystal: crystallographers know it as the Laves graph, the srs net or $(10,3)$ -a, and physicists of Kitaev models as the hyperoctagon lattice. It is trivalent, its shortest cycles have length ten, so the first four shells around a site are those of a tree, thirty oriented decagons start and end at each site, and the number of sites within distance $r$ grows as $r^3$ . Strong isotropy has a direct meaning in report language: every permutation of the three edges at a site of type $v$ is realized by a symmetry fixing that site, namely the lift of the stabilizer $S_3$ of the anchored report $v$ .

Theorem(Sunada)

Among three-dimensional crystal nets with injective standard realization, the strongly isotropic ones are the diamond crystal and the $K_4$ crystal, the latter together with its mirror image. The $K_4$ crystal is chiral: it is not carried to its mirror image by any orientation-preserving isometry.

Why a change of report is a rotation

Plate VIII.3What the transposition

(1\,2)

does in each dictionary. Old: the six letters at the vertices of an octahedron, complementary letters opposite, and

(1\,2)

is the reflection in the plane

x=y

. Crystal: the four face periods

h_v

form a tetrahedron in the cube, and

(1\,2)

sends

h_v

-h_{\pi v}

, a mirror followed by an inversion: the half-turn about

(0,1,-1)

The report group relabels the four reports and so permutes the oriented letters, with a sign whenever a relabelling reverses a letter’s orientation; because $\partial$ commutes with relabelling, it acts on the periods by integral matrices and on space by orthogonal ones. The old dictionary placed the six unoriented letters on three perpendicular axes, the two letters of each perfect matching at opposite ends. There every odd relabelling acts on space as an improper transformation: the transposition $(1\,2)$ is the reflection in the plane $x=y$ .

The crystal carries a different representation. An oriented letter $e_a\wedge e_b$ is a bivector in $\R^4=\R\oplus V$ , so the letters span $\Lambda^2(\R\oplus V)=V\oplus\Lambda^2V$ : the first summand is what $\partial$ sees, the change of report, and the second is its kernel, the circulations. In three dimensions a bivector is an axial vector, $\Lambda^2V\cong V\otimes\mathrm{sgn}$ , so an odd relabelling reverses the orientation of $V$ and of every loop at once, and the two reversals cancel. No report-covariant linear map sends unoriented letters to spatial directions at all; the oriented letters admit exactly one up to scale, the harmonic projection that built the crystal. The old dictionary sends complementary letters to opposite vectors, the harmonic map sends them to perpendicular ones, and no change of coordinates turns one into the other.

Proposition(Every relabelling is a rotation)

For every $\pi\in S_4$ , $\det R_\pi=+1$ . The twenty-four matrices $R_\pi$ are the rotation group $O$ of the cube, a faithful representation of $S_4$ . For $\pi=(1\,2)$ , $R_{(1\,2)}:(x,y,z)\mapsto(-x,-z,-y)$ , a half-turn about the axis $(0,1,-1)$ . Of the 192 affine maps that could carry the crystal to itself with one of the 48 signed permutation matrices as linear part, exactly 24 do, and all are proper.

Proof

On the circulations, $\pi$ acts by $\Lambda^2$ of its action on $V$ , so $\det R_\pi=(\det R_\pi|_V)^2=\mathrm{sgn}(\pi)^2=1$ . The displayed matrices and the count of affine symmetries are exact finite computations.

Tallies are Lorentz generators

Plate VIII.4On the report qubit the triangle opposite a report generates the rotation about that report’s axis: for

0\to1\to2\to0

\omega=\tfrac13(1,1,-1)=-n_3/\sqrt3

. Relabellings move the report directions with mirrors and the rotation vectors without.

The observer who walks the crystal has a geometry of its own, on its report qubit: the Hermitian $2\times2$ matrices with $\langle A,B\rangle=\tfrac12(\operatorname{Tr}A\operatorname{Tr}B-\operatorname{Tr}AB)$ , of signature $(1,3)$ , with the identity as time axis. The four reports are the null projectors $\Pi_v=\tfrac12(I+n_v\cdot\sigma)$ , with $n_v$ the vertices of a regular tetrahedron on the Bloch sphere, and the letter crossed from $a$ to $b$ acts by $M_{ab}X=\langle\Pi_b,X\rangle\,\Pi_a-\langle\Pi_a,X\rangle\,\Pi_b$ , a generator of Lorentz transformations; a tally acts by the corresponding sum, $M_c$ .

The crystal’s metric came from its edges, with no reference to the report qubit, and it is the norm of the observer’s rotation algebra. A period is an angular velocity of the report frame, not a direction on the report sphere: relabellings move the report sphere by the tetrahedral group with its mirrors, and angular velocities, being axial, pick up the determinant, which cancels every mirror. What the tally forgets has the same kind of reading. Compose the flows of the letters along a decagon with a small parameter per step: the first-order term vanishes because the tally is zero, and for each of the thirty decagons the leading term is a pure rotation about one of the six two-fold axes of the cube, the Thomas–Wigner rotation that a closed sequence of boosts leaves behind.

Proposition(Tallies as generators)

(i) Each letter generates a pure boost in the plane of its two reports: $M_{ab}\Pi_a=\tfrac13\Pi_a$ and $M_{ab}\Pi_b=-\tfrac13\Pi_b$ , and $M_{ab}$ vanishes on the spacelike plane orthogonal to both.

(ii) For every tally, $M_cI=-\tfrac12\,\iota(\partial c)$ , where $\iota$ sends report $v$ to $\Pi_v$ : the boost part is half the boundary. So $M_c$ fixes the time axis exactly when $c$ is closed, and the tallies that generate rotations are exactly the periods.

(iii) For a period $c$ , $|\omega_c|^2=\tfrac19|c|^2$ , where $|c|^2$ is the crystal’s metric. So $c\mapsto\omega_c$ is a similarity from the period lattice onto a lattice in the rotation algebra $\mathfrak{so}(3)\cong\mathfrak{su}(2)$ of the report qubit.

(iv) A relabelling of reports moves the directions $n_v$ by the full tetrahedral group, whose odd elements are reflections, and moves the rotation vectors $\omega_c$ by $\det(R)\,R$ . These twenty-four matrices are the rotation group $O$ .

Proof

For (i), $\langle\Pi_a,\Pi_a\rangle=0$ and $\langle\Pi_a,\Pi_b\rangle=\tfrac13$ give the two eigenvalues. For (ii), $M_{ab}I=\tfrac12\Pi_a-\tfrac12\Pi_b$ , and both sides are linear in $c$ . For (iii), the rotation part of one letter is $\omega_{ab}=\tfrac14\,n_a\times n_b$ ; the triangle $0\to1\to2\to0$ has $\omega=\tfrac13(1,1,-1)$ , of squared length $\tfrac13$ , against the triangle’s $|c|^2=3$ . The ratio on all periods, and (iv), are exact finite computations.

The mirror obstruction falls

Plate VIII.5An isolated triplet crossing at a point fixed by the whole rotation group: three bands

-|\alpha||q|

, 0 and

|\alpha||q|

, the same in every direction. On the old dictionary the comparable crossings were anisotropic by factors of ten to six thousand.

The old dictionary had a sharp dynamical consequence. If an odd report acts as an unbroken unitary symmetry of a translation-invariant walk, it sends an isolated charged crossing $(k,\omega,q)$ to $(R_\pi k,\omega,-q)$ , because the momentum map has degree $\det R_\pi=-1$ : every complete finite set of crossings at one quasienergy has total charge zero, and a crossing fixed by a mirror carries none. In that law symmetry never forced a charged crossing to be isotropic, and none was: the twelve representative crossings established with rigorous bounds on the old cubic dictionary were anisotropic by factors between 10.3 and $6{,}257$ .

On the crystal an odd relabelling is a proper rotation and preserves Berry charge, so report-related crossings have the same charge, and at the momenta $\Gamma$ and $R$ fixed by the whole point group the little group is all of $O$ , acting irreducibly on space. On one walk, with a supplied rule that leaves a letter idle when its edge does not meet the current site, 664 isolated charged triplets at $\Gamma$ and $R$ have been established with rigorous error bounds, every one exactly isotropic and untilted. Three limits come with this: they are three-band, charge-two crossings, not two-band Weyl particles; their speeds are not common; and a chiral involution still pairs a crossing at $\omega$ with one of opposite charge at $-\omega$ , with winding number zero. What the crystal removes is pairing at the same energy.

Proposition(Symmetry forces isotropy and permits charge)

Let a report-covariant walk on the crystal have an isolated crossing at $\Gamma$ or $R$ that carries an ordinary three-dimensional representation of $O$ . Its linear germ is, up to a constant change of basis, $H_1(q)=\alpha\,q\cdot J$ with $(J_i)_{jk}=-i\,\epsilon_{ijk}$ , with no tilt. Its eigenvalues are $-|\alpha||q|$ , 0 and $|\alpha||q|$ , the same in every direction. If $\alpha\neq0$ the three ordered bands carry charges $(2\,\mathrm{sgn}\,\alpha,0,-2\,\mathrm{sgn}\,\alpha)$ .

Proof

The germ is a Hermitian map from the vector representation into the endomorphisms of the triplet, and $\operatorname{End}(3)$ and $\operatorname{End}(3')$ each contain the vector representation of $O$ exactly once, in their antisymmetric parts, where the intertwiner is $q\cdot J$ . One- and two-dimensional representations admit no linear germ at all.

What keeping the tally costs

Plate VIII.6On

K_4

the six oriented squares

0\to a\to b\to c\to0

return to report 0 after four re-anchorings. A square has the tally of two triangles, so in the crystal it does not come back.

For the crystal to be space, the register’s own report changes must move it there. Attach to each state of one register an address $R\in\Z^3$ , and let a re-anchoring $v\to u$ add the chord increment of the letter $vu$ while hops and memory rewrites leave $R$ alone. The one added clause is this: the circulation of the live vantage path is retained as an unbounded coherent coordinate. Nothing in the adopted frame forces it. Retention is not a record, since the coordinate is live content that later moves can bring back together, and it is not free.

With the register’s own hops and rewrites restored, the order-four difference below is unchanged, and resolved by the letter the observer reports in its moving frame, the probabilities of two outcomes move by $\mp t^4/7203+O(t^6)$ , with $7203=3\cdot7^4$ . So keeping the tally is live which-path information, and it changes the experiment. Records do not choose the quotient either: filling the four triangles kills every period, keeping the ordered paths gives the tree, and the permanent records do not contain the tally.

Example(What the register forgets and space remembers)

Isolate the re-anchoring part: let $A$ be the adjacency operator of $K_4$ , start at report 0, and write $a$ for the elapsed time in units in which one re-anchoring has unit amplitude. On $K_4$ , with eigenvalues $3,-1,-1,-1$ , $|\langle0|e^{-iaA}|0\rangle|^2=\tfrac{10+6\cos4a}{16}=1-3a^2+4a^4+O(a^6)$ . On the crystal, closed walks of length 2 and 4 from a site number 3 and 15, the counts on the three-regular tree, because the girth is ten, and the probability of being found again at report 0, in any cell, is $1-3a^2+\tfrac72a^4+O(a^6)$ . The difference is $-\tfrac12a^4$ . On $K_4$ there are 21 closed walks of length four from a report; the six extra ones are the oriented squares $0\to a\to b\to c\to0$ . A square has the tally of two triangles, so in space it does not come back.

Space read at its primes

Plate VIII.7Read modulo seven, the crystal’s null directions are the eight points of the sky

\Proj^1(\F_7)

, and an observer is a pair of them. The base observer

\{0,\infty\}

is in gold; the three observers whose axes are orthogonal to its axis,

\{1,6\}

\{2,5\}

and

\{3,4\}

, the pairs harmonic to it, are in blue.

The periods are an integral lattice with an integral metric, in cube coordinates $Q=x^2+y^2+z^2$ , the only report-covariant quadratic form on the periods up to scale and the norm of the report qubit’s rotation algebra. Over the complex numbers the null vectors of a rotation algebra form the celestial sphere. The rotations about a report’s axis have exactly two null eigenvectors, along $n_v$ and $-n_v$ , the report and its anti-report, so the eight celestial points of a clock are its four reports and their four antipodes; these eight null lines are defined over the Eisenstein integers.

The primes 2, 3 and 7 are those of 168, the order of the group relating the clocks, and at each the one lattice gives an object that the program had found by another road: modulo 2 the clock, modulo 3 one clock’s four reports with the rotations of a cube, modulo 7 the finite celestial sphere with its seven clocks and twenty-eight observers, and the Coxeter graph, which here is simply orthogonality of the observers’ axes. The sky of the whole network is therefore not added to space: it is space’s own null cone, read at seven. Why three reductions of one lattice should carry the program’s three finite structures is a question about the object they are reductions of, which Chapter XIII identifies as a hyperbolic manifold.

Proposition(The crystal at its primes)

Read the period lattice $H_1(K_4;\Z)$ with its metric modulo a prime.

(i) Modulo 2 the metric is the parity of a loop’s length, a linear function. The seven nonzero classes are the four triangles, each labelled by the report it misses, and the three squares, each labelled by the axis whose letters it misses; they are the seven lines of the clock’s Fano plane, three classes summing to zero exactly when their lines pass through one point. The parity is odd on the four reports and even on the three axes: it is the clock point itself, read as a linear function on lines that vanishes exactly on the lines through it.

(ii) Modulo 3 the null lines of $Q$ are exactly the four body diagonals, the axes of the four reports, and $\SO(3,\F_3)$ is exactly the cube group.

(iii) Modulo 7 the lattice with its metric is report-covariantly isometric to $\mathfrak{sl}_2(\F_7)$ with its determinant, uniquely up to scale, and $\SO(3,\F_7)\cong\PGL(2,7)$ . Its eight null lines are the eight points of the sky $\Proj^1(\F_7)$ , and each report’s two axis ends are its observer’s pair of points. With the seven clocks taken as the seven conjugates of the cube group under $\PSL(2,7)$ , their 28 three-fold axes are exactly the 28 points off the null conic that lie on two of its tangents, the observers, and their 21 four-fold axes are the 21 points on no tangent. Two observers’ axes are orthogonal exactly when their pairs of points are harmonic, and this relation is the Coxeter graph.

Proof

For (i), $Q(c)=\sum_ec_e^2$ , which modulo 2 is the number of letters in $c$ counted with multiplicity: three for a triangle, four for a square. The labels are the clock’s own, and the triangle missing report $v$ uses exactly the letters that are not on the line $v$ . The rest of (i), and (ii) and (iii), are exact finite computations. That the eight axis ends fill the whole null cone only at 7 is a count: over $\F_p$ a nondegenerate conic has $p+1$ points, and eight distinct ends exist only when $\F_p$ contains the cube roots of unity, that is, when $p\equiv1\pmod3$ .

The crystal is the middle of the covering tower. The tree above it keeps the order of a history of report changes, and $K_4$ below it keeps only the current report, the frame. Position is reversible content that can go back down, as the backward helix shows, while elapsed time only grows. The crystal belongs to one clock; each of the seven clocks has its own, and read modulo seven each is the same quadratic space $\mathfrak{sl}_2(\F_7)$ , with the seven clocks as seven cube groups inside its rotation group. What the tally forgets, the loops with zero tally, is the subject of the next chapter.

Words defined here: crystaltally
Concepts: objectincarnationstabilizer classbridgerefuted
Objects: the seven pointsthe seven linesthe twenty-eightthe skythe object of size 21