report counts

Part III, Observers of Observers · defined in Chapter XIII, The Level-Seven Shadow

Integer combinations of the four reports.

Plate W.25The plane at infinity with its Eisenstein lattice and the base tetrahedron’s finite cusps 0, 1 and

1+\omega

, with

\infty

above: their null matrices, with that of

\infty

, form a basis of the report counts.

With the reports as null vectors, the report counts of one observer form the Eisenstein lattice $\mathrm{Herm}_2(\Z[\omega])$ in Minkowski space. The null matrices of the base tetrahedron’s four cusps form a $\Z$ -basis of it, and the determinant is the null report form of Chapter I, there derived from the reports alone and here from the unimodularity of four Eisenstein spinors. So the Bianchi group $\SL(2,\Z[\omega])$ , acting by $X\mapsto gXg^\dagger$ , is the integral Lorentz group of report counts, and the lift is built from it. The letters sit in the same lattice as frames.

The lattice carries a fact about reflection. In the continuum a proper Lorentz transformation can carry any spacelike separation to its negative, and field theory uses that to show that a local field needs antiparticles. The lattice’s own Lorentz group cannot always do it: at the separations marked by the prime above three no transformation of the lattice reverses the separation, so a field on the report counts is not local there. Read as positions, the counts that a register’s records add up to are the arena.

Proposition(report counts are Eisenstein) proved

Take the base tetrahedron’s four cusps $\infty$ , 0, 1 and $1+\omega$ , with primitive spinors $\xi_\infty=(1,0)$ , $\xi_0=(0,1)$ , $\xi_1=(1,1)$ , $\xi_{1+\omega}=(1+\omega,1)$ , and the null matrices $N_a=\xi_a\xi_a^\dagger$ . Then (a) $N_\infty,N_0,N_1,N_{1+\omega}$ form a $\Z$ -basis of $\mathrm{Herm}_2(\Z[\omega])$ , and (b) $\det\bigl(\sum_ax_aN_a\bigr)=\tfrac12\bigl[(\sum_ax_a)^2-\sum_ax_a^2\bigr]$ . So the Bianchi group $\SL(2,\Z[\omega])$ , acting by $X\mapsto gXg^\dagger$ , is the integral Lorentz group of report counts.

Proof

For two spinors, $\det(\xi\xi^\dagger+\eta\eta^\dagger)=|\det(\xi,\eta)|^2$ , so $\langle N_a,N_b\rangle=\tfrac12|\det(\xi_a,\xi_b)|^2$ and $\det N_a=0$ . The six determinants are 1 for each pair containing $\infty$ , then $-1$ , $-(1+\omega)$ and $-\omega$ , all units of $\Z[\omega]$ since $1+\omega=-\omega^2$ . So every $\langle N_a,N_b\rangle$ is $\tfrac12$ , which gives (b). For (a), the diagonal entries of $N_\infty$ and $N_0$ generate the two integer diagonals, and the off-diagonal entries 1 and $1+\omega$ of $N_1$ and $N_{1+\omega}$ generate $\Z[\omega]$ .

Theorem(The prime above three blocks a reflection) proved

Let $X$ be a spacelike vector of the report lattice with $3\mid\det X$ , not divisible by the prime $\pi=1-\omega$ above three. Reduced modulo $\pi$ , $X$ is $\lambda\,\ell\ell^{\mathsf T}$ over $\F_3$ , and whether $\lambda$ is a square modulo three is unchanged by every Lorentz transformation of the lattice, $X\mapsto gXg^\dagger$ with $g\in\GL(2,\Z[\omega])$ , and by the mirror. Since $-1$ is not a square modulo three, no such transformation carries $X$ to $-X$ .

Proof

Since $\omega\equiv\bar\omega\equiv1$ modulo $\pi$ , reduction turns $X\mapsto gXg^\dagger$ into a congruence over $\F_3$ , which multiplies $\lambda$ only by a square.

As mathematics

$\Lambda=\mathrm{Herm}_2(\Z[\omega])$ with the form $\det$ , on which $\SL(2,\Z[\omega])$ acts by $X\mapsto gXg^\dagger$ . For the cusps $\infty,0,1,1+\omega$ of the base tetrahedron, with primitive spinors $\xi_\infty=(1,0)$ , $\xi_0=(0,1)$ , $\xi_1=(1,1)$ , $\xi_{1+\omega}=(1+\omega,1)$ , all six determinants $\det(\xi_a,\xi_b)$ are units, so $\langle N_a,N_b\rangle=\tfrac12$ for $a\ne b$ and the $N_a$ form a $\Z$ -basis. The records $T=N_a+N_b$ span the even sublattice, of index two.

The kernel $\Gamma(\mathfrak p)$ of reduction modulo $\mathfrak p$ keeps exactly one quadratic form on the Hermitian matrices up to scale, the Minkowski form. Every nonzero vector of $\Lambda$ has an infinite orbit under $\SL(2,\Z[\omega])$ , so no locally finite set of bonds on $\Lambda$ is invariant under both translations and the Lorentz group.

Its name in another field	Bridge
$\mathrm{Herm}_2(\Z[\omega])$ , the Eisenstein lattice in Minkowski space	built
the lattice of which the Bianchi group $\SL(2,\Z[\omega])$ is the integral Lorentz group	classical
a body-centred cubic lattice, in their spatial parts	built
integral Lorentz vectors	a reading

Built from: report
Builds: arenalift
In the volume: IThe Founding SentenceIVWorld, Kernel, ObserverVITime as CountVIIISpace as a TallyXIIThe Coxeter GraphXIIIThe Level-Seven ShadowXXIVA Number Nature Could RefuteEp.Forcing, Not Sacred Geometry