type law

Floor 4, Les doubles vies · introduced in Chapter 6, Quatre groupes à double vie

How does a Galois conjugation of a lattice’s field appear on the finite geometries the lattice reduces to?

A twisted Galois symmetry of a lattice is seen on its residues by how the prime decomposes: as a seam between two residues where the prime splits, semilinearly where it is inert, linearly where it ramifies.

Plate 4.3The type law at Klein’s lattice: complex conjugation is split at 2, a polarity between two Fano planes; ramified at 7, a Möbius map of non-square determinant; inert at 3, the Frobenius of

\F_9

Definition(Lattices, residues, twisted Galois symmetry)

Let $K\subset\C$ be a finite Galois extension of $\Q$ with ring of integers $\mathcal O_K$ and Galois group $\Gamma$ . A $G$ -lattice over $\mathcal O_K$ is a finitely generated projective $\mathcal O_K$ -module $L$ with an $\mathcal O_K$ -linear action $\rho$ of $G$ such that $V=L\otimes K$ is absolutely irreducible; $\chi$ is its character. Its residue at a prime $\mathfrak P$ is the $k(\mathfrak P)G$ -module $V_{\mathfrak P}=L/\mathfrak PL$ , and $L$ has good reduction at $\mathfrak P$ if $V_{\mathfrak P}$ is absolutely irreducible.

The group of twisted Galois symmetries of $L$ is $\hat A(L)=\{(\sigma,\alpha)\in\Gamma\times\Aut(G):\sigma\circ\chi=\chi\circ\alpha\}$ , and $L$ is twisted Galois stable if the projection $\hat A(L)\to\Gamma$ is onto. Then, when the automorphisms fixing $\chi$ are the inner ones, $\sigma\mapsto\alpha_\sigma$ is a homomorphism $\Gamma\to\operatorname{Out}(G)$ with kernel $\mathrm{Gal}(K/\Q(\chi))$ .

Theorem(The type law) proved

Let $L$ have good reduction at $\mathfrak P$ and $(\sigma,\alpha)\in\hat A(L)$ . Then $L$ has good reduction at $\sigma\mathfrak P$ , and $(V_{\sigma\mathfrak P})^\alpha\cong\bar\sigma V_{\mathfrak P}$ . If $\sigma\notin D(\mathfrak P)$ , $\alpha$ is realized by a $\bar\sigma$ -semilinear seam from the residue at $\mathfrak P$ to the residue at $\sigma\mathfrak P$ (split type). If $\sigma\in D(\mathfrak P)\setminus I(\mathfrak P)$ , it is realized by a $\bar\sigma$ -semilinear bijection of the residue, linear exactly when $\bar\sigma$ fixes every trace (inert type). If $\sigma\in I(\mathfrak P)$ , it is realized by a linear map normalizing the group, not in it when $G$ is perfect and $\alpha$ outer (ramified type). If $\sigma\circ\chi=\bar\chi$ , $\alpha$ carries every residue of good reduction to its dual.

Proof

The conjugate lattice $L^\sigma$ and the lattice $L$ twisted by $\alpha$ have equal characters $\sigma\circ\chi=\chi\circ\alpha$ , so they are stable lattices in one representation; by Brauer–Nesbitt their reductions at $\sigma\mathfrak P$ , $\bar\sigma V_{\mathfrak P}$ and $(V_{\sigma\mathfrak P})^\alpha$ , have the same composition factors, and the first is absolutely irreducible. The three types are this isomorphism read according to whether $\bar\sigma$ is a map between residue fields, a field automorphism or the identity.

Corollary(The residue forms of complex conjugation) proved

Let $c$ act on $\Q(\chi)$ as complex conjugation, $(c,\alpha_c)\in\hat A(L)$ , and $L$ of good reduction at $\mathfrak P$ . If $c\notin D(\mathfrak P)$ , the residues at $\mathfrak P$ and $c\mathfrak P$ are dual and $\alpha_c$ acts as a duality. If $c\in D(\mathfrak P)\setminus I(\mathfrak P)$ , the residue carries a nondegenerate invariant hermitian form and $\alpha_c$ is realized semilinearly. If $c\in I(\mathfrak P)$ , the residue carries a nondegenerate invariant symmetric or alternating form and $\alpha_c$ is realized by a similitude of it outside the group. The trichotomy of forms is Gross’s; the reading of $\alpha_c$ in each case is what the law adds.

Example

Klein’s lattice is a rank-3 lattice over $\Q(\sqrt{-7})$ with automorphism group of order 336. At $2=\mathfrak p\bar{\mathfrak p}$ complex conjugation splits: the two residues are Fano planes, dual to each other, and $\alpha_c$ is a polarity, a seam from points to lines. At 7 it ramifies: $\alpha_c$ is an isometry of the conic form, odd on the eight points of $\Proj^1(\F_7)$ , a Möbius map in $\PGL(2,7)\setminus\PSL(2,7)$ . At 3 it is inert: $G\subset\mathrm{PSU}(3,3)$ and $\alpha_c$ is the Frobenius of $\F_9$ . So the outer automorphism that exchanges the points and the lines of the Fano plane, and is a non-square Möbius map on the line, is one Galois conjugation seen at two primes.

Built from: double lifelifeseam over an automorphismtwisting element
Objects: the seven pointsthe seven linesthe sky
In the Esquisse: 5Courte marche à travers la théorie de Galois7La trinité de Galois16Une loi de réciprocité18Exceptionnel veut dire relevableÉp.L’horizon : dessins d’enfants
In the volume: XVIIITwo Parents of the Sky