imprint

Floor 3, Ce qui est su · introduced in Chapter 2, L’espace en creux

What structure does an absence force to exist?

A structure that exists, with a theorem characterizing it by an absence: terminal (the survivors at the edge of a window), carrier (what carries local data that do not globalize) or separating (a finer invariant).

Plate 3.10The complement of the line 246 is the quadrangle 1,3,5,7. Its three pairs of opposite sides meet at 2, 4 and 6: the diagonal points are the line itself.

Definition(Imprint)

Let $A$ be an absence. An imprint of $A$ is a structure $I$ , which exists, together with a theorem that characterizes $I$ by means of $A$ in one of three ways. Then $A$ forces $I$ .

Terminal: $A$ is graded, and $I$ is the list, up to isomorphism, of the members of $\mathcal K$ whose value of $\nu$ is an extreme point of the window. When the list has a single member it is the last survivor.

Carrier: $A$ asserts that certain local data are not the restriction of a global structure of a prescribed type, and $I$ is a structure of a larger type that carries the local data and is characterized by a universal property.

Separating: $A$ asserts that two structures sharing an invariant are not isomorphic, and $I$ is a finer invariant taking different values on them.

The word is meant literally. A seal is cut in negative and leaves a positive figure in the wax; the absence is the cut, the imprint is the figure. The definition classifies pairs of an absence and a theorem, not structures, and which structure is singled out is a stated choice. Forces means only that the characterizing theorem uses the absence essentially, not that the imprint is constructed out of it.

Proposition(The diagonal line)

In $\mathrm{PG}(2,2)$ the complement of every line $\ell$ is a complete quadrangle, and its diagonal points are exactly the three points of $\ell$ . The map $\ell\mapsto\mathrm{PG}(2,2)\setminus\ell$ is a $\GL(3,2)$ -equivariant bijection from the seven lines to the seven complete quadrangles; the six sides of the quadrangle are the six lines other than $\ell$ , and each point of $\ell$ is the common point of a pair of opposite sides.

Read on the complete graph $K_4$ on the four points of the quadrangle, the sides are the six edges and the diagonal points are the three perfect matchings. In characteristic 2 the three matchings are forced onto one line, and $K_4$ together with that line is the whole Fano plane: the terminal imprint of the failure of Fano’s axiom.

Proof

A line other than $\ell$ meets $\ell$ in one point and its complement in two, so no three points of the complement are collinear. In characteristic 2 the diagonal points of a complete quadrangle are collinear, and they lie off the quadrangle; seven points minus four leaves the three points of $\ell$ .

Theorem(The spin bundle)

For $p\in\{5,7,11\}$ the point stabilizer of the $p$ -point action of $\SL(2,p)$ is $2T$ , $2O$ , $2I$ , a group of spinors, and a spin representation $\rho$ of it gives the spin bundle $S_\rho$ over the $p$ points. $S_\rho$ is equivariantly trivial if and only if $p=5$ . For $p=7$ its sections form a 14-dimensional representation,

\operatorname{Ind}_{2O}^{\SL(2,7)}\rho_\pm\cong6_\pm\oplus8,

all three quaternionic; the two forms of the bundle share the 8 and differ in the 6. Every representation of $\SL(2,7)$ whose restriction to $2O$ contains $\rho_+$ contains $6_+$ or 8, so has dimension at least 6: the carrier imprint of the absence of a two-dimensional representation.

Proof

An equivariant bundle $G\times_HW$ is trivial exactly when $W$ is a restriction from $G$ , and for $p=7,11$ there is no nontrivial two-dimensional representation of $\SL(2,p)$ . The decomposition comes from the induced characters, computed, and the last statement is Frobenius reciprocity.

Proposition

The triangle group $\Delta(2,3,7)$ has exactly one normal subgroup with quotient $\PSL(2,7)$ , so there is exactly one compact Riemann surface of genus 3 with 168 automorphisms, the Klein quartic $x^3y+y^3z+z^3x=0$ . It is the first survivor at the Hurwitz bound $84(g-1)$ , a terminal imprint.

Proof

Epimorphisms with torsion-free kernel correspond to generating pairs of elements of orders 2 and 3 with product of order 7. There are 336 such pairs, found by computation, and $\Aut\PSL(2,7)$ , of order at least 336, acts freely on them, so there is one orbit and the kernel is unique.

Example

The three Gassmann pairs of $\PSL(2,7)$ are separating absences: each pair shares its permutation character, and its members are not isomorphic. One imprint separates all three at once, the column $V_4^a$ of the table of marks:

\begin{gathered}|(G/V_4^a)^{V_4^a}|=6,\ |(G/V_4^b)^{V_4^a}|=0;\quad|(G/A_4^a)^{V_4^a}|=2,\ |(G/A_4^b)^{V_4^a}|=0;\\|(G/S_4^a)^{V_4^a}|=1,\ |(G/S_4^b)^{V_4^a}|=3.\end{gathered}

Likewise no point of the complex projective plane has stabilizer $V_4$ under Klein’s representation, and the imprint of that absence is the self-polar triangle: the object $G/V_4$ , which no point can carry, is carried by ordered pairs of vertices of a self-polar triangle.

Proposition(Carriers are induced from kernels)

Let $H$ be the kernel of the orbit description of a point $y$ of a transitive $G$ -set $Y$ , and $W$ a representation of $H$ . Then $\operatorname{Ind}_H^GW$ , the space of sections of the bundle $G\times_HW$ over $Y$ , is characterized by $\mathrm{Hom}_G(\operatorname{Ind}_H^GW,V)\cong\mathrm{Hom}_H(W,\mathrm{Res}_HV)$ for all representations $V$ of $G$ , and the bundles over $Y$ correspond to the representations of $H$ . So the induced object is the universal structure over $Y$ carrying the data $W$ given on the kernel. The permutation representation $\C[Y]$ is induced from the trivial representation, the spin bundle from the binary octahedral kernel, and a family of algebras over an object from the algebra over one point and its stabilizer.

Proof

Frobenius reciprocity. Classical; no novelty is claimed.

Examplecomputed

A carrier imprint inside the Weil representation of $\SL(2,7)$ . Read through the octonions, the stabilizer of the pair $\{\infty,0\}$ preserves the algebra $\mathfrak{su}(3)_0$ of derivations killing $e_0$ , and transporting it defines an algebra over each of the 28 pairs of points; the family is equivariant over the object of size 28, and so built. Exactly seven of the 28 algebras consist of derivations of the octonions, those over the pairs $\{\infty,x\}$ , because only the odd lift of the stabilizer of $\infty$ acts by automorphisms. So the octonionic data at seven pairs are not the restriction of an octonionic structure over all 28, and the equivariant family carries them.