Où se rencontrent les deux parents

Le réseau de Klein chez ses voisinsKlein’s lattice at its neighbours

Read the Fano plane as $\Proj(L_\infty/\alpha L_\infty)$, the link of $v_0$ in the building at 2. For a point $p$ let $M_p$ be the neighbour of $L_\infty$ at $(\alpha)$ through $p$. It is a standard lattice, with an orthonormal basis $u_1,u_2,u_3$, and the stabilizer of $p$, a group $S_4$, turns it as the rotations of the cube with vertices $\pm u_1\pm u_2\pm u_3$.

The cube’s axes and diagonals are vectors of Klein’s lattice. Each $\bar\alpha u_k$ has norm 2, and its reflection is the half-turn about $u_k$; read modulo $(\alpha)$ that half-turn is a transvection of the Fano plane, and modulo $\sqrt{-7}$ a half-turn of the conic. The four diagonals have norm 3 and are antiflags, and the pair of conic points of an antiflag is the pair its elements of order three fix. So the flags are the pairs of vectors of norm 2, the antiflags the pairs of vectors of norm 3, and both lives of the group of order 168 are visible at one vertex.

Les demi-tours des facesFace half-turns

The faces of the congruence link complement meet this dictionary through their holonomy. A face $F$ is an ideal triangle of the tessellation; take it with a base pair $x=\{a,b\}$ of its cusps and third cusp $c$, and let $\sigma(F,x)\in\PSL(2,7)$ be the reduction modulo $\mathfrak p$ of the product of the three parabolic elements fixing the cusps of $F$, taken around $F$ from $x$. It is the half-turn that Gauss and Bonnet give a triangle of area $\pi$.

At the face $\{0,1,\infty\}$ with base pair $\{0,\infty\}$ the half-turn is $z\mapsto-1/z$. It exchanges 0 and $\infty$ and sends 1 to $-1=6$, the harmonic conjugate of 1 with respect to $\{0,\infty\}$. The pair $\{1,6\}$ is a neighbour of $\{0,\infty\}$ in the Coxeter graph, and $z\mapsto-1/z$ is the involution of that edge: in Klein’s lattice, the reflection in a vector of norm 2 orthogonal to both antiflags.

Un doublet et son carré symétriqueA doublet and its symmetric square

Both parents have a tree at 7 whose link at a vertex is $\Proj^1(\F_7)$. One step beyond the link they differ, and the difference is classical. On Mumford’s tree let $N=\Lambda_0^\vee/\Lambda_0$ with its alternating form, let $R$ be the group that $\Gamma_1$ induces on the ball of radius two about $\Lambda_0$, and let $O$ be the Sylow 7-subgroup of the kernel of $R$ on the link. On the tree of the link complement the kernel on the ball of radius two is the kernel of $\SL(2,\Z/49)\to\SL(2,7)$, the elements $I+7X$ with $X$ of trace zero over $\F_7$: it has order $7^3$, against $2\cdot7^2$ on Mumford’s.

So the second layer of the one tree is the symmetric square of the second layer of the other. Read at $\sqrt{-7}$, Klein’s lattice is the first layer at its own vertex, and it is the second layer of the other parent’s tree. Weil’s construction turns the doublet $N$ into a quartet together with Klein’s three-dimensional representation.

L’objet des points sur les deux arbresThe object of the points on the two trees

For a vertex $x$ of one of the trees, its link $\mathrm{Lk}(x)$ is an incarnation of $\Proj^1(\F_7)$. Let $K_x$ be the elements of the vertex group of $x$ that act trivially on it. For a step $x\to y$ the mixed-square group $M(x\to y)$ is the image of $K_x$ on $\mathrm{Lk}(y)$: the holonomy of a square made of a loop at $x$ that $x$ cannot see, followed by the step.

That is the whole difference between the trees, and the chapter’s theorem follows from it. On the one tree every mixed square stays inside $\PSL(2,7)$, so the rigid object $X_7$ is carried in exactly one way; on the other an improper involution carries $X_7$ to $X_7'$, and no seam joins them. Parity is properness: two Klein vertices $v_0$ and $xv_0$ have the same parity exactly when $x$ acts properly, and when they differ an element of order seven acting on the link by a given Möbius map has trace $\alpha$ on one Klein lattice and $\bar\alpha$ on the other. The two vertices carry Klein’s representation and its conjugate.

Le parent jointThe joint parent

Write $\mathcal O=\Z[\omega]$, $\mathfrak p=(3+\omega)$ and $\bar{\mathfrak p}$ for the two primes above 7, and $\Gamma_S=\SL(2,\mathcal O[1/\mathfrak p])$ for the group of the link complement with its prime inverted, which acts on $\mathbb H^3$ and on the tree $T_{\mathfrak p}$ of $\SL(2,\Q_7)$. For a vertex $t$ let $\rho_t$ be the action of its stabilizer on the plane whose eight lines are $\mathrm{Lk}(t)$, and $\tilde\varpi\colon\Gamma_1\to\SL(2,7)$ Mumford’s reduction at $\sqrt{-7}$.

By Goursat’s lemma a subgroup of $\Gamma_1\times\Gamma_S$ that projects onto both factors is the fiber product over a common quotient, so the parents are joined only that way. The fiber products are non-cocompact lattices of index 336, modulo scalars, in $\PGL_3(\Q_2)\times\SL(2,\C)$ and in $\PGL_3(\Q_2)\times\SL(2,\C)\times\SL(2,\Q_7)$, and they are reducible: by Margulis’s arithmeticity theorem an irreducible lattice there would come from one absolutely almost simple group, of one Dynkin type, while $\PGL_3$ has type $A_2$ and $\SL_2$ type $A_1$. An amalgam over $F_{21}$ is not available, since $F_{21}$ is not a finite subgroup of $\PSL(2,\C)$. By Serre’s solution of the congruence subgroup problem every homomorphism of $\Gamma_S$ onto $\PSL(2,7)$ is reduction modulo $\bar{\mathfrak p}$ followed by an automorphism.

At one scale the attachment is forced. Fix $t$ and an isomorphism $\iota$ of the two copies of $\SL(2,7)$ keeping the class of $S_4^a$, and let $\Gamma^{(t)}$ be the pairs $(\gamma,\delta)$ with $\iota\tilde\varpi(\gamma)=\rho_t(\delta)$. It has index 336; the two incarnations of $X_7$ are joined by exactly one equivariant seam, a seam system over $\Delta_2\times\mathbb H^3$ with trivial gauge group; the other class of $\iota$ exchanges the classes of $S_4$; and $\Gamma^{(t)}$ is the stabilizer of $t$ in no subgroup of finite index. The one lattice that joins the parents across scales, $\Gamma_1\times_{\SL(2,7)}\Gamma_S$, attaches Mumford’s finite line to the line at $\bar{\mathfrak p}$, which does not move with the scale.

Flèches et boucles sur l’arbre des échellesArrows and loops on the scale tree

A choice of one parent at each vertex is the remaining datum of a seam system along a tree, and the lemma says what such choices look like. No choice is invariant under a group that fixes no vertex, edge or end; in particular none is invariant under either parent.

Now carry the octonion table around the loops of $\Gamma_S$ at the base vertex $t_0$ of the scale tree $T_{\mathfrak p}$. Fix the edge $e_0=(t_0,t_\infty)$, whose stabilizer is the Iwahori subgroup $\Gamma_I$, root the parents at $e_0$, and choose frames in which every parent sits at $\infty$. The holonomy $h(\gamma,t_0)\in\PSL(2,7)$ of a loop acts on the seven points of the Fano plane through the double life read in Klein’s lattice, with the point $a^*=\Lambda_0$ at $\infty$ in every frame, and the signed relabellings of the table $e_xe_{x+1}=e_{x+3}$, a group $2^3{\cdot}L_3(2)$ of order 1344, realize each collineation in eight ways.

One loop in eight is carried by a relabelling that reverses no unit: exactly those with $\gamma(t_\infty)$ the parent of $\gamma t_0$. The condition depends only on the coset $\gamma\Gamma_I$; in the decomposition of $\Gamma_S$ by the infinite dihedral group of $s_0$ and $s_1$ it holds exactly when the reduced word ends in $s_1$, and at distance $2k$ the oriented edges reached number $8\cdot7^{2k-1}$ toward $e_0$ and $8\cdot7^{2k}$ not, with generating function $56u^2/(1-49u^2)$ for the first. The best relabellings for the other loops reverse exactly two units. The residual is not a class function, since $\mathrm{diag}(\pi^{-1},\pi)$ and $\mathrm{diag}(\pi,\pi^{-1})$ are conjugate with residuals 2 and 0, so the signed table defines no homomorphism on the loop group and no twist of Ihara’s zeta function.

Le résidu et les deux transportsThe residual and the two transports

Call a holonomy misaligned, or broken, when it moves $a^*$, and let $g$ be its collineation. Two ways of carrying the table across a broken loop present themselves. There is an orthogonal map $A_g$ of $\Oct$, unique up to sign, with $A_gL_{e_x}A_g^{-1}=L_{e_{g(x)}}$ for every $x$; it commutes with the complex structure $J_0=L_u$, $u=(e_0+\dots+e_6)/\sqrt7$, and it is an automorphism of $\Oct$, up to sign, exactly when the holonomy fixes $a^*$. Every relabelling over $g$ is $\pm A_gL_{e_{x_1}}\cdots L_{e_{x_k}}$ over the units it reverses, and of the 128 realizations of $g$ in $\mathrm{Spin}(7)$ exactly two keep $J_0$, namely $\pm A_g$, exactly eight keep the product, the relabellings, and none keeps both.

The table $A_g$ carries across the loop has the same metric, the same $J_0$ and the same Cayley form of left type, and a different associative 3-form. Its unit is $A_g(1)=\pm e_{c_1}$ with $c_1=g(c_0)$, and the two tables share exactly the $\mathrm{SU}(3)$ that fixes $e_{c_1}$. Everything else at one scale agrees for the two: the Coxeter meetings, the Fano orientation up to the signs of the units, the face half-turns, the products of the units of each of the 70 four-sets of $\Proj^1(\F_7)$, $J_0$ and the Cayley form of left type. Only the unit tells the two transports apart.

La charge, observateur par observateurCharge, observer by observer

Each of the twenty-eight antiflags of the Coxeter graph carries its own copy of the table, and copies are compared only along the graph’s edges, the meetings. Write $c(o)=p$ for the clock of $o=(p,L)$. Over a collineation $g$ a per-observer transport is a choice of relabellings $R_o$ over $g$, one at each observer, with clock signs given by $R_oe_{c(o)}=s_oe_{g\,c(o)}$. It keeps the clocks when every $s_o=+1$, and it is a single map when every $R_o$ is the same.

A single map cannot satisfy the meetings. Every pair of distinct clocks is joined by exactly two meetings and no meeting joins equal clocks, so a single map that reverses a set $S$ of clocks fails exactly the $2|S|(7-|S|)$ meetings that join $S$ to its complement: 20 for the three single maps of residual two, 24 for four, 12 for one.

Nor can a loop reverse a pattern. With its 24 heptagons as faces the Coxeter graph has $H_1=0$ and is simply connected, so a comparison system trivial around every heptagon is trivial around every closed path; for one broken loop the flat $\Z/2$ data form $\Z/2$, a reversal of every observer together. And $\operatorname{Hom}(\SL(2,\mathcal O),\Z/2)=\operatorname{Hom}(\Gamma_S,\Z/2)=0$, since $\mathcal O$ is Euclidean and conjugation by $\mathrm{diag}(\omega^2,\omega)$ multiplies the entry of an elementary matrix by $\omega$. Over the stabilizer $S_3$ of an observer the extension $2^3{\cdot}L_3(2)$ splits by clock-keeping lifts, which form an honest action of the 168 collineations on the 28 copies.