Part IV · The Exceptional InteriorChapter XV

Spin from the Double Cover

Plate XV.1The groups of the chapter: double covers down, inclusions across.

XV.1
XV.2
XV.3
XV.4
XV.5
XV.6
XV.7
XV.8
XV.9
XV.10

Where does spin come from when the relativity group is finite?

In special relativity spin is not an extra hypothesis. The proper Lorentz group acts on the celestial sphere as $\PSL(2,\C)$ , its double cover $\SL(2,\C)$ acts linearly on $\C^2$ , and a spinor is a vector of that action, on which the turn through $2\pi$ , $-I$ , acts as $-1$ . The observers of this program are related by a finite group of the same kind, $\PSL(2,\F_7)$ on eight points, and it too has a double cover, $\SL(2,\F_7)$ , and no larger one: its Schur multiplier has order two.

The question is whether the fiber $\C\otimes\Oct\cong\C^8$ feels that cover, and in which representation. The answer separates two structures on the same eight dimensions. One is the fiber’s Clifford module, the space on which the letters act as anticommuting operators with the octonion product forgotten; the cover acts on it exactly, and that action is spin. The other is the fiber as matter’s interior, whose lepton and colour labels are fixed by the product and its unit. The cover’s action moves those labels, so the program carries the labels by relabellings that keep the product, the automorphisms of the octonions that permute the units with signs, and puts spin on a factor of its own.

The central result

Lift each report permutation $g$ of each of the seven clocks to the fiber by its Clifford intertwiner $N_g$ , the orthogonal matrix, unique up to sign, with $N_gL_aN_g^{-1}=L_{g(a)}$ . These lifts generate $\SL(2,\F_7)$ , of order 336 with centre $\{\pm I\}$ : the double cover of the relativity group. Each lift is the collineation’s unsigned permutation of the units followed by right multiplication by one unit, a shift; read on the grading by $\F_2^3$ , the shift represents the nonzero class of $H^1(\GL(3,2),\F_2^3)\cong\Z/2$ , and it moves the octonion 1 for 147 of the 168 collineations.

The double cover acts on $\C^8=\mathbf{4}\oplus\overline{\mathbf{4}}$ by two inequivalent, complex-conjugate, faithful irreducible quartets, the $\pm i$ eigenspaces of $J_*=L_u$ with $u=(e_1+\dots+e_7)/\sqrt7$ , so $\SL(2,\F_7)\subset\Spin(6)_u\cong\SU(4)$ ; the quartet is the even half of the Weil representation. At each clock the fiber restricts to the binary octahedral group as $F_{3/2}\oplus F_{3/2}$ , with $F_{3/2}\cong E_{1/2}\otimes E_{\mathrm{rod}}$ , spin one-half tensored with the permutation doublet of the three rods, and leaves no complex line invariant.

In the lift the fiber is compared across clocks by one forced transport, velocity space’s parallel transport lifted through this cover. Each of its steps fixes one complex line of the quartet, one for each light direction, and these eight sky lines form the Paley frame of the Weil representation; the fiber’s scattering matrix between the light directions is the Paley matrix of order eight, whose two eigenspaces are the quartet and its conjugate. These lifts and this transport carry the Clifford module, that is, spin; the program carries matter’s labels instead by the relabellings that preserve the product, which act honestly up to colour signs, and puts spin on the report qubit.

Status

The first sentences are an exact machine computation in one fixed frame, an explicit isomorphism with the $2\times2$ matrices of determinant one over $\F_7$ checked on all $112{,}896$ products; the form of each lift as a permutation and a shift follows in one line from its definition, and the cohomology behind the shift was checked by direct computation. The splitting and the restrictions are exact character theory, and the Weil identification is a comparison of characters with a classical construction. The last sentence records a decision of the program: the octonion product with its unit is physical. That the product’s relabellings then act honestly up to colour signs is exact.

There is no qubit on which all observers agree, since $\SL(2,\F_7)$ has no two-dimensional representation; the spin factor over all observers is a bundle, the report qubit carried clock by clock and glued by the double cover, in two forms that differ by the sign of $\sqrt2$ in their characters, and nothing yet chooses between them, the lift included. The label $F_{3/2}$ does not decide between a spin-three-halves particle and spin one-half carried along the rods. And the double cover is not yet the spin of anything moving in space: the lift supplies its transport between frames and the fiber’s first couplings, not a law of motion or the strength of any coupling.

Two later results place the spin factor on the lift. The report qubit is the lift’s own spinor, with the lift’s hand, which every step of a history keeps: this is proved, with its remaining items checked by exact enumeration. And the records carry Wigner’s representations, massive spin one-half on records and the massless representation on reports, checked exactly over $\Q(\omega)$ . What they lack is a field: a vacuum shared by all observers, and a law of motion in space.

Double covers of Möbius groups

Plate XV.1The groups of the chapter: double covers down, inclusions across.

For a field $K$ , $\SL(2,K)$ acts on the projective line by Möbius maps with kernel $\pm I$ . As over the reals, where $\PSL(2,\R)\cong\SO^+(2,1)$ , the finite group is an orthogonal group of a three-dimensional quadratic form, $\PSL(2,\F_7)\cong\Omega_3(7)$ , so by the dimension of its form the finite sky looks like the circle. The form itself decides otherwise: it is the crystal’s definite metric read modulo seven, whose null lines are complex, as on the sphere, and Chapter XIII shows that the program’s report counts select the $3+1$ lift.

As for $\SL(2,\C)$ , $\SL(2,\F_7)$ has exactly one element of order two, namely $-I$ , so the preimage of any subgroup containing an involution has a single involution, and the preimage of a clock’s octahedral group $S_4$ is the binary octahedral group $2O$ , of order 48: the same double cover that the rotation group of a cube has inside $\SU(2)$ . The consequence is simple: a spinor that every observer agrees on has at least four components. The same facts show, in Chapter XIII, that $\PSL(2,\F_7)$ is not a subgroup of the real Lorentz group.

Proposition(Representations of the double cover)

The irreducible complex representations of $\SL(2,\F_7)$ have degrees 1,3,3,4,4,6,6,6,7,8,8. Those on which $-I$ acts as $-1$ , the faithful ones, have degrees 4,4,6,6,8, and the two quartets are complex conjugates of each other and not self-conjugate. In particular there is no nontrivial two-dimensional representation, and $\PSL(2,\F_7)$ has no nontrivial projective representation on a qubit.

Proof

The degrees are in the ATLAS. A projective action on $\C^2$ can be taken unitary, so it is a map to $\mathrm{PU}(2)\cong\SO(3)$ ; a simple group maps injectively or trivially, and the three-dimensional representations of $\PSL(2,\F_7)$ are genuinely complex, so it has no faithful real one of dimension three.

The lift at one clock

Plate XV.2Clock

e_7

and its four reports; the report transposition

124\leftrightarrow235

swaps

e_1\leftrightarrow e_5

and

e_3\leftrightarrow e_4

and fixes

e_2

and

e_6

Fix a clock $p$ . Its six letters have left multiplications with $L_aL_b+L_bL_a=-2\delta_{ab}I$ , the generators of $\mathrm{Cl}(0,6)$ , which they generate as the whole matrix algebra $M_8(\R)$ . A permutation of the generators that respects their relations is an automorphism of $M_8(\R)$ , and by the Skolem–Noether theorem every such automorphism is inner: for each report permutation $g$ there is an orthogonal $N_g$ with $N_gL_aN_g^{-1}=L_{g(a)}$ , unique up to sign because only scalars commute with all letters. It can be written as a product of an even number of Clifford reflections $(L_a-L_b)/\sqrt2$ , and it then also commutes with $L_p$ . The relations would allow a sign on each image; the lift takes none.

The 48 lifts at one clock close into the binary octahedral group, with class sizes 1,1,6,6,6,8,8,12 and centre $\{\pm I\}$ , and the fiber is two copies of its four-dimensional spinorial representation $F_{3/2}$ . One clock sees the double cover of its own rotation group; the question is what seven clocks see together.

Example(A report transposition lifts to an element of order four)

Take $p=7$ with the oriented triples 124,235,346,457,561,672,713. Swapping the reports 124 and 235 fixes the letters 2 and 6 and swaps $1\leftrightarrow5$ and $3\leftrightarrow4$ . Put $U=\tfrac12(L_1-L_5)(L_3-L_4)$ . Conjugation by $(L_a-L_b)/\sqrt2$ sends a generator $L_v$ to $-L_{s(v)}$ , with $s$ the reflection swapping $e_a$ and $e_b$ ; the two sign changes cancel, so $UL_1U^{-1}=L_5$ and $UL_3U^{-1}=L_4$ , with $L_2$ , $L_6$ , $L_7$ fixed. Since the two factors anticommute, $U^2=-\tfrac14(L_1-L_5)^2(L_3-L_4)^2=-I$ . The report transposition has order two, and its lift has order four.

Seven clocks generate the double cover

Plate XV.3The lift

U

with its signs forgotten: the Möbius involution

m(x)=(3x+4)/(x-3)

, swapping the sky points in four pairs.

The correspondence can be checked on the example. As a matrix, $U$ sends $1\mapsto e_2$ , $e_2\mapsto-1$ , $e_1\mapsto e_3$ , $e_3\mapsto-e_1$ , $e_4\mapsto-e_5$ , $e_5\mapsto e_4$ , $e_6\mapsto-e_7$ , $e_7\mapsto e_6$ . Without signs it swaps the sky points $6\leftrightarrow5$ , $1\leftrightarrow0$ , $4\leftrightarrow2$ and $\infty\leftrightarrow3$ , which is the Möbius involution $m(x)=(3x+4)/(x-3)$ , of determinant $-13\equiv1$ . Its matrix has trace zero, so it squares to $-I$ in $\SL(2,\F_7)$ , just as $U^2=-I$ . Both conjugacy classes of octahedral subgroups were checked: the seven clocks and the seven vantage lines each give preimages $2O$ .

Unsigned, the lifts are the relativity group moving eight sky points, and the fiber is the space $\mathbf{1}\oplus\mathbf{7}$ of functions on them, on which $-I$ is invisible; signed, the same matrices form the double cover, and the central element reverses every vector. The two are different representations, not two bases of one. The spin is in the signs. The movement of the octonion 1 is not: unsigned, the lifts already carry 1 to other units, because each of them ends with a right multiplication.

Theorem(The double cover in the fiber) computed

The lifts of the seven clocks’ report permutations generate a group of 336 orthogonal $8\times8$ matrices, isomorphic to $\SL(2,\F_7)$ , with centre $\{\pm I\}$ . Every element is a signed permutation matrix in the basis $(1,e_1,\dots,e_7)$ , and forgetting the signs gives the Möbius action of $\PSL(2,\F_7)$ on $\Proj^1(\F_7)$ under the correspondence $1\mapsto6$ , $e_1\mapsto1$ , $e_2\mapsto5$ , $e_3\mapsto0$ , $e_4\mapsto4$ , $e_5\mapsto2$ , $e_6\mapsto\infty$ , $e_7\mapsto3$ .

A permutation and a shift

Plate XV.4The shift of

U

, right multiplication by

e_2

, carries the gold lepton line

\{1,e_7\}

onto the colour line

\{e_2,e_6\}

Grade the units by $\F_2^3$ , with 1 at 0, and write $P_g$ for the unsigned permutation of the units by the collineation $g$ , which fixes 1, and $R_w$ for right multiplication by a unit $w$ . Since $e_xw=\pm e_{x+w}$ , right multiplication by a unit moves every label by the same vector, a shift. The lift $U$ of the example sends $1\mapsto e_2$ , so its shift is $w=e_2$ : its relabelling part fixes 1, $e_2$ , $e_6$ and $e_7$ , and right multiplication by $e_2$ gives $e_7\mapsto e_7e_2=e_6$ . At the clock 7 it therefore carries the lepton line $\operatorname{span}\{1,e_7\}$ to $\operatorname{span}\{e_2,e_6\}$ , the line of the letter pair $\{2,6\}$ on the Fano line 672 through the clock: a colour line.

As long as a clock’s rotations act spinorially on the dimensions that carry the lepton and colour labels, they move the labels. That is why the program, keeping the octonion product and its unit, carries matter’s labels by relabellings and puts spin on a factor of its own.

Theorem(The lifts as a permutation and a shift)

(1) Each lift is $N_g=R_w\circ P_g$ , the unsigned permutation $P_g$ of the units followed by right multiplication by $w=N_g(1)$ , $\pm$ a unit; on the labels it is the affine map $x\mapsto g(x)+w$ . Exactly 21 of the 168 collineations have $w=\pm1$ , a Borel subgroup, the stabilizer of one light direction; the other 147 move the octonion 1. (2) At a clock $p$ the lifts of its report group commute with $L_p$ , which makes the fiber a $\C^4$ , and permute its four complex lines, the classes $\{x,x+p\}$ , as the affine map does; its linear part fixes the unit’s class $\operatorname{span}\{1,e_p\}$ , the lepton line, and permutes the three colour lines, and the lepton line is kept exactly when $w$ lies in the unit’s class. (3) The map $g\mapsto w$ is a cocycle representing the nonzero class of $H^1(\GL(3,2),\F_2^3)\cong\Z/2$ . Of the sixteen complements to the translations in $\mathrm{AGL}(3,2)$ , eight fix a point and eight act transitively on the eight units, and the lifts’ affine maps form a transitive one; no choice of which unit is called 1 removes the shift. (4) Spin forces the shift: in the binary octahedral group $-I$ is a commutator, and it acts irreducibly on the clock’s $\C^4$ , so no complex line is invariant.

Proof

Since $e_a=L_a1$ , $N_g(e_a)=N_gL_aN_g^{-1}N_g(1)=e_{g(a)}w$ , so $N_g=R_w\circ P_g$ , and $e_xw=\pm e_{x+w}$ gives the affine map. Unsigned, the lifts permute the eight directions transitively, so those fixing the direction of 1 are the stabilizer of the sky point 6, of order $168/8=21$ . A lift commuting with $L_p$ carries the line of each class $\{x,x+p\}$ to the line of the image class; the linear part fixes $p$ and the class of 0, and the shift carries that class to the class of $w$ . A coboundary $w=g(b)+b$ would make every affine map fix the label $b$ , and no direction is fixed by all the lifts; calling another unit 1 conjugates by a translation, which preserves the class, and the sixteen complements were checked by direct computation. In $2O$ the lifts of two half-turns about perpendicular axes anticommute, as $i$ and $j$ do, so $-I$ is a commutator. A complex line invariant under $2O$ would carry a one-dimensional character, trivial on commutators and so on $-I$ , while $-I$ acts on every line as $-1$ .

The democratic unit splits the fiber

Plate XV.5Seven anticommuting chiralities, one per clock, permuted by every change of clock; only their diagonal

D_*

, at the same angle

\arccos(1/\sqrt7)

to each, is fixed.

Which representation of the double cover is the fiber’s Clifford module? Not the irreducible faithful $\mathbf8$ , which is of quaternionic type and would restrict at a clock as $E_{1/2}\oplus E_{5/2}\oplus F_{3/2}$ . Every lift permutes the seven imaginary units under conjugation without signs, so it fixes their sum. Put $u=(e_1+\dots+e_7)/\sqrt7$ and $J_*=L_u$ ; since the $L_a$ anticommute and square to $-I$ , $J_*^2=-I$ , a complex structure commuting with the whole double cover, whose $\pm i$ eigenspaces are invariant subspaces of complex dimension four.

The splitting belongs to no single observer. Each clock’s chirality $D_p=-iL_p$ splits $\C^8$ too, but a change of clock moves it, $N_gD_pN_g^{-1}=D_{g(p)}$ , and the seven $D_p$ are orthonormal directions of a Clifford frame whose democratic diagonal is $D_*=-iJ_*$ . No observer can measure the common splitting: an observer’s own law keeps only its own clock’s splitting, read with the parity of its count of records, and the common splitting meets each clock’s at one fixed angle, with $\cos^2=\tfrac12(1\pm1/\sqrt7)$ (Chapter XVI). Each lift is an even product of reflections in vectors $e_a-e_b$ orthogonal to $u$ , so $\SL(2,\F_7)\subset\Spin(6)_u\cong\SU(4)$ , and $\Lambda^2\mathbf{4}=\mathbf{6}$ : the letters are bilinears of the spinor quartet. $\Spin(6)\cong\SU(4)$ is also the group in which Pati and Salam placed the lepton as a fourth colour, and since the relativity group acts irreducibly on the quartet, under the lifts a change of clock mixes the quark and lepton components of any one observer’s decomposition; that mixing is the shift, spin acting on the dimensions that carry the labels, not a symmetry of the labels, and the program does not adopt Pati and Salam’s group. The quartet is the even half of the Weil representation, the finite oscillator, whose odd half is the triplet on whose projective plane Klein found his quartic.

Proposition(The two quartets)

The eigenspaces of $J_*$ are inequivalent, complex-conjugate, faithful irreducible representations of $\SL(2,\F_7)$ , and the commutant of the double cover on $\C^8$ is $\C I\oplus\C J_*$ . The quartet’s character is 4, $-4$ , 1, $-1$ , 0, 0 on the elements of orders 1,2,3,6,4,8, and $(1\pm i\sqrt7)/2$ and $(-1\pm i\sqrt7)/2$ on the two pairs of classes of orders 7 and 14.

Proof

A representation is irreducible when its character has norm one. With class sizes 1,1,56,56,42,84 and 48,48 for the last two columns, and $\lvert(\pm1\pm i\sqrt7)/2\rvert^2=2$ , the norm is $(16+16+56+56+0+0+48\cdot2+48\cdot2)/336=1$ . The value $-4$ at $-I$ says the quartet is faithful: it is a spinor.

What each clock sees, and the spin bundle

Plate XV.6The report qubit induced from one clock to all seven: a spin bundle of dimension 14, in two forms.

At one clock the faithful irreducibles restrict as $\mathbf{4}\mapsto F_{3/2}$ , $\mathbf{6}\mapsto E\oplus F_{3/2}$ and $\mathbf{8}\mapsto E_{1/2}\oplus E_{5/2}\oplus F_{3/2}$ , with $E_{1/2}\otimes E_{\mathrm{rod}}\cong E_{5/2}\otimes E_{\mathrm{rod}}\cong F_{3/2}$ . At every clock the fiber is spin one-half tensored with the rod doublet, twice, and no bare spin-one-half doublet sits inside it. The same $2O$ is the double cover of the rotations by which the report group acts on the crystal. Pair each fiber lift of a rotation with one of that rotation’s two spin lifts so that products correspond: then all 576 products agree, and there are exactly two such pairings, the second the first twisted by the sign of $S_4$ , which exchanges $E_{1/2}$ with $E_{5/2}$ .

The observer’s report qubit, whose quarter turns generate $2O$ , is an exact spinor of one clock with no native attachment to the fiber’s coloured states; the program attaches it as a tensor factor. The oldest-sign qubit of Chapter XVI is a genuine internal multiplicity, but the report group acts on it with centre $+I$ : it is not a spin one-half at all. Which bundle carries spin is the question of which lift of a quarter-turn is the quarter-turn itself and which is the quarter-turn followed by a full turn. In the rotation group a path from the identity decides it; a finite group has no such path, and nothing in the program decides it yet. The lift does not decide it either: the quarter-turns are not among its motions.

Proposition(The spin bundle)

If each clock carries its own spin-one-half doublet $E$ , the double cover acts on the sections of the induced bundle $S_E=\SL(2,\F_7)\times_{2O}E$ over the seven clocks, of dimension 14. (1) The doublet is $E_{1/2}$ or $E_{5/2}$ , exchanged by $\sqrt2\mapsto-\sqrt2$ in their characters and by no automorphism of $\SL(2,\F_7)$ , and the bundles are $S_{1/2}\cong\mathbf{6}\oplus\mathbf{8}$ and $S_{5/2}\cong\mathbf{6}'\oplus\mathbf{8}$ , with $\mathbf6$ and $\mathbf6'$ the two faithful sextets. (2) Each is an honest representation with $-I$ acting as $-1$ ; at every clock its fiber is the report qubit, and it restricts to that clock’s $2O$ as $2E\oplus E'\oplus2F_{3/2}$ , with $E'$ the other doublet. (3) Along the spinor transport the holonomy of every closed loop maps the fiber over its base clock to itself: the identity on a corner; on a face it squares to $-1$ and acts on the base clock’s qubit as a lift of a half-turn; on a decagon it has order three, with eigenvalues $\omega$ and $\omega^2$ on that qubit, a lift of a third of a turn, and the quartet’s decagon eigenvalues $\{1,1,\omega,\omega^2\}$ are these two multiplied by the rods’ three-cycle. (4) Neither bundle shares an irreducible constituent with the Clifford module $\mathbf{4}\oplus\overline{\mathbf{4}}$ , which lies once in the bundle induced from $E\otimes E_{\mathrm{rod}}=F_{3/2}$ , namely $\mathbf4\oplus\overline{\mathbf4}\oplus\mathbf6\oplus\mathbf6'\oplus\mathbf8$ for either $E$ .

Proof

Items (1), (2) and (4) are character theory. By Frobenius reciprocity the multiplicity of an irreducible $V$ in $S_E$ is that of $E$ in $V$ restricted to the clock’s $2O$ : one for $\mathbf8$ and for the sextet whose restriction contains $E$ , zero for the quartets and the other sextet; the same count with $F_{3/2}$ gives (4), and restricting back gives (2). The two classes of elements of order eight in $\SL(2,\F_7)$ have traces 3 and 4 in $\F_7$ ; conjugation preserves traces, so no automorphism exchanges the two classes, and the two doublets differ exactly there. The rest, the bundles as matrices on all 336 elements, a change of basis at each clock onto the report qubit, and the holonomies over every corner, face and decagon, was checked by direct computation.

The lift’s transport and the sky in the quartet

Plate XV.7The Paley matrix on the sky: an arrow from

a

b

wherever the entry is

+1

The lift’s parallel transport reduces at each step to an element of $\PSL(2,\F_7)$ of order seven, which has exactly one lift of order seven in the double cover; acting through the Clifford lifts this is the spinor transport, velocity space’s Thomas precession carried on the fiber’s Clifford module. It is covariant under all 168 collineations, its face holonomies square to $-I$ , and its decagon holonomies have order three. No transport by octonion automorphisms is exactly covariant, because the group of signed automorphisms does not split over $\PSL(2,7)$ ; those that act linearly in each clock’s frame are covariant up to colour signs.

Each step of the spinor transport fixes exactly one complex line of the quartet, depending only on the light direction the step keeps, and the eight lines are permuted as the sky is: an equiangular tight frame at the Welch bound, $\lvert\langle s|t\rangle\rvert^2=\tfrac17$ , the Paley frame. In this frame every observer’s quartet splits as $1+1+2$ , two lines in the plane of its ends’ sky states and a doublet. On the little group of a letter’s rest frame, the binary dihedral group of order twelve, the quartet has the character of spin three-halves, and this split is the spin split along the letter’s axis; the same character belongs to spin one-half tensored with the rod doublet, so no finite group of the program tells the two apart.

The lift’s first cohomology with spinor coefficients is $\mathbf4\oplus\overline{\mathbf4}$ , the Clifford module; how it relates to matter’s labelled fiber, $\mathbf1\oplus\mathbf7$ under the product’s relabellings, is open. As a field it has a scattering matrix between the light directions, and the quartet and its conjugate scatter with opposite amplitudes, $-\sqrt{-7}\,y$ and $+\sqrt{-7}\,y$ , told apart by $\sqrt{-7}$ , the number that separates their characters. The constant is not a convention: the lift’s integral structure, unique up to scale, together with the program’s labels fixes $y=(22-4\omega)/49$ , of modulus $\sqrt{12}/7$ (Chapter XVIII). The opposite signs are a charge, not a chirality: a rotation of the lift that exchanges the two quartets carries the whole scattering to itself, as the exchange of particle and antiparticle carries electromagnetism to itself. Light’s table is a different one, built from a cubic character.

Proposition(The fiber scatters by the Paley matrix) computed

The fiber’s scattering matrix between the eight light directions is $L_V=y\,DPD$ , with $D$ a diagonal matrix of signs, $y$ a complex constant that no structure of the lift normalizes, and $P$ the Paley matrix of order eight: zero diagonal, entry $\bigl(\frac{b-a}{7}\bigr)$ at finite $a\ne b$ , and border $+1$ along the row of $\infty$ and $-1$ down its column. $P$ is antisymmetric with $P^2=-7I$ , its eigenspaces are $\mathbf{4}$ and $\overline{\mathbf{4}}$ , and the eight light-direction states projected onto one eigenspace form the equiangular tight frame of sky lines, with overlaps $\tfrac17$ and triple products $\pm i\,7^{-3/2}$ signed by the Legendre orientation.

Colour carried as frame data

Plate XV.8The seven observers through the light direction of the unit 1, the point 6: each one’s clock is the unit at its far end, and their colour groups are the seven that are symmetries of the octonion table.

The sky lines are the units’ own lines. The sky line of each point is the $J_*$ -line $\operatorname{span}\{e,J_*e\}$ of the basis direction the correspondence assigns to it, with no phase between the two frames. So the octonion unit 1 is the sky state of one light direction, the point 6, whose stabilizer lifts with $w=\pm1$ , and the seven imaginary units are the sky states of the other seven points. Each of these is the far end of an observer that shares the light direction of 1, and that observer’s clock is the unit at its far end: the one unit that all six of the observer’s symmetries fix.

The spinor transport is not a symmetry of the product: it moves the unit, and its commutant on the fiber is spanned by $I$ and $J_*$ , so it commutes with no one colour group. What it does is carry colour as frame data, taking each observer’s colour group to the colour group of the observer it takes that observer to, while an observer’s own symmetries keep its colour. Only the octonion table is tied to one light direction: in the table’s own frame the colour groups of the seven observers through that direction are groups of symmetries of the product, and the others are the same groups carried along. Chapter XVII takes the gauge group’s colour from this family.

Proposition(The transport carries each observer’s colour) computed

Let an observer share the light direction of 1, let $e_p$ be its clock, and give it the colour group of $e_p$ , the stabilizer of $e_p$ among the automorphisms of the octonions. (1) The lifts of the observer’s six symmetries commute with $L_{e_p}$ and carry its colour group to itself; the three that exchange the observer’s two ends move the unit, with $w=\pm e_p$ , so they keep the lepton line $\operatorname{span}\{1,e_p\}$ . (2) Carrying this colour group by the lifts gives each of the twenty-eight observers a colour group, independent of the lift used. Exactly seven of the twenty-eight consist of automorphisms of the octonions, those of the observers that share the light direction of 1.

Proof

Both items are exact computations on the lifts. Item (2) follows from item (1), since a family carried by a group action is well defined exactly when each member is kept by its own stabilizer.

The report qubit is the lift’s spinor

Plate XV.9The base tetrahedron of the lift: at each cusp the report state is the line the lift’s null rotations there leave fixed, the twelve even permutations of the cusps are motions of the lift, and the odd ones and the quarter-turns are not.

The spin factor was placed beside the fiber as the report qubit, a spinor of each clock’s rotations. The lift has spinors of its own, the two-component spinors of velocity space, on which the Bianchi group acts through $\SL(2,\Z[\omega])$ ; call that system $V$ , and its mirror image $\bar V$ , on which each element acts through its complex conjugate, the spinor system of the mirror lift. The two spinor systems, the observer’s and the lift’s, are one, and the report states of the base tetrahedron’s four cusps are where the identification starts. The theorem settles the spacetime side of the spin factor and leaves the global question where it was: the even part of the clock’s symmetry is honest geometry of the lift, and the quarter-turns move the qubit only by transport, so the lift does not decide which of the bundle’s two forms is spin.

A register’s history keeps this hand. Each record is a rest frame of the lift and each new record a boost of the last: an own move is the null rotation about the shared report’s light direction that carries the departing report to the arriving one, and a received exchange is a half-turn of the tetrahedron’s binary group. Every one of these is a motion of the lift, a proper Lorentz transformation, so the spinor’s hand never changes along a history. A reflection of the tetrahedron would carry frames to frames and reverse the hand, but it is not a motion of the chiral lift. Matter, the report qubit tensored with the fiber, therefore carries the lift’s hand on every record of every history.

Theorem(The report qubit is the lift’s spinor)

(1) At each cusp $c$ of the base tetrahedron the null rotations $u_c(t)=I+t\,\xi_c\xi_c^{\mathsf T}J$ , with $t\in\Z[\omega]$ and $J=\bigl(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\bigr)$ , lie in the Bianchi group, and for $t\neq0$ each fixes exactly one line of spinors, the report state $\xi_c$ : the report state of a light direction is the line its motions leave fixed. (2) Of the twenty-four permutations of the tetrahedron’s cusps exactly the twelve even ones are motions of the lift; their lifts to $\SL(2,\Z[\omega])$ form the binary tetrahedral group, which acts on the report qubit as the even part of the clock’s $2O$ does, for either doublet. The twelve odd ones are reflections, which reverse the lift’s orientation. (3) The clock’s quarter-turns are not motions of the lift: their lifts in $\SL(2,\F_7)$ have traces $\pm\sqrt2$ modulo seven, $\sqrt2$ is not in $\Q(\omega)$ , and $\SL(2,\Z[\omega])$ has no element of order eight. On the lift they act without fixed points, carrying spinors from one fiber to another. (4) The report states follow the cusps under $\psi\mapsto g\psi$ and not under the conjugate action, and the two actions are inequivalent: the report qubit is the fiber of $V$ at the cusps and carries the lift’s hand. In the mirror lift it would carry the other.

Proof

For (1), $\xi^{\mathsf T}J\xi=0$ gives $u_c(t)\xi_c=\xi_c$ , $\det u_c(t)=1$ , and $u_c(t)-I$ has rank one with image the line of $\xi_c$ . For (3), $(a+b\omega)^2=2$ with $a,b$ rational would need $a^2-b^2=2$ and $b(2a-b)=0$ , which has no solution. For (4), $g=\bigl(\begin{smallmatrix}1&\omega\\0&1\end{smallmatrix}\bigr)$ moves the cusp 0 to $\omega$ while $\bar g$ moves it to $\bar\omega$ , and $h=\bigl(\begin{smallmatrix}1&\omega\\1&1+\omega\end{smallmatrix}\bigr)$ has trace $2+\omega$ , which differs from its conjugate, so $V$ and $\bar V$ are inequivalent. The remaining items were checked by exact enumeration.

The record is the form

Plate XV.10Each record measures the spin it carries: a step of a history carries the record

T

gTg^\dagger

and its form

\xi^\dagger T^{-1}\eta

onto the next record’s, and at the light directions the report states turn by phases.

A register’s spinor carries the lift’s group by matrices that keep no inner product, so a single spin state has no invariant length, and Wigner’s theory of particles needs one. The records supply it. Each record is a rest frame, a positive Hermitian matrix $T$ of unit determinant, and $T$ itself measures spin: $\langle\xi,\eta\rangle_T=\xi^\dagger T^{-1}\eta$ . A Lorentz transformation $g$ carries the record $T$ to $gTg^\dagger$ , and since $g^\dagger(gTg^\dagger)^{-1}g=T^{-1}$ it carries the form at $T$ exactly onto the form at $gTg^\dagger$ .

In these unitary spaces the conjugate representation exists. A state missing from a filled set of light-like states transforms by the conjugate phase, the opposite helicity, while massive states have real characters and carry both helicities, as a massive particle of spin one-half does. So the program’s matter has native relativistic kinematics: its records are particle states, and its reports carry the hand as helicity. What it lacks is a field, a vacuum that all observers share and a locality that makes each particle come with its partner of the opposite hand (Chapter XVI).

Theorem(Records and reports as particle states)

(1) Every step of a register’s history, each own move and each received exchange of a clock’s tetrahedron, is an isometry from the form at the old record to the form at the new. (2) On spinor-valued functions of the records, each measured by its own record’s form, the lift’s group acts unitarily: Wigner’s representation of a particle of spin one-half and unit mass, in the discrete form the lift’s group allows. (3) On the light directions, with each report state taken as a primitive integral spinor, the lift’s group acts unitarily too, by phases at each direction: Wigner’s representation of a massless particle. The motions about a light direction multiply its report state by a non-real root of unity, and the mirror lift’s spinors carry the conjugate phase; on light-like states the hand is the helicity. (4) Records and reports all point to the future. There is no past half to fill, so neither space carries a sea that every frame agrees on.

Proof

Item (1) is the identity above, applied to the own moves and received exchanges of a history. Item (2) is induction from the stabilizer of one record, which acts unitarily for that record’s form. Item (3) holds because the lift’s group carries primitive spinors to unit multiples of primitive spinors, and units have modulus one; the phase about $\infty$ is $\omega$ , which is not real. Item (4) holds because $gTg^\dagger$ is positive with $T$ . All four were checked exactly over $\Q(\omega)$ .

In relativity, spinors are forced by the double cover of the Möbius group of the celestial sphere. Here the same holds for the finite sphere, the fiber’s Clifford module is such a spinor, and the smallest one the finite group admits has four components where the continuous group admits two. The lift fixes how it is carried between frames, by velocity space’s own spinor transport, and finds the sky inside the quartet as a frame of eight lines, which is also how the fiber scatters between the light directions; the fiber’s first coupling, its absorption of light, leads to the chiral octet one degree up. Seams counts the missing qubit among its absences, with a window and, at $p=7$ , the spin bundle $\mathbf{6}_\pm\oplus\mathbf{8}$ as its carrier imprint; there too the seven octonionic colour algebras among the twenty-eight carried ones are an absence whose carrier imprint is the carried family. Its chapter on orientation and charge states the spinor system at the cusps, the fixed scattering constant and the records’ forms with neutral names.

Whether this finite spinor becomes the spin of particles in space is a question of dynamics, open together with which degree of the lift carries matter; on the crystal an added, colour-blind rotor gives isotropic charged spin-one-half doublets, but at two speeds, and the rotor is not derived. The same eight dimensions carry matter’s labels, which the octonion product fixes and the spinor transport would move. The program keeps the product and carries the labels by its $1{,}344$ signed automorphisms, which, with the record carrying the product’s signs, act honestly up to colour signs when chosen clock by clock, and exactly when chosen observer by observer: an honest action of all 168 collineations on the twenty-eight observers’ fibers. Spin goes on the report qubit, which is the lift’s own spinor with the lift’s hand, and whose global form is the bundle with its open choice. The next chapter reads the eight dimensions as matter: a quark–lepton quartet and its conjugate, with the lepton at the unit’s class.