window

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

Where can the excluded thing still happen?

For a graded absence, the set of parameter values that actually occur; it is usually small, and its edges carry the structure.

Plate 3.9Two windows on the primes: Galois’s

\{5,7,11\}

, where

\PSL(2,p)

acts on

p

points, and the spinor window

\{3,5\}

, where

\SL(2,p)

has a two-dimensional representation.

Definition(Absence, window, witness)

An absence is graded when it comes with a function $\nu\colon\mathcal K\to\R$ , such as a dimension, a characteristic, a prime or a ratio of orders, and determines the set of values $W=\nu(\mathcal K)$ that actually occur. $W$ is the window of the absence.

The window is where the excluded thing can still happen; it is usually small, and its edges carry the structure. For Hurwitz’s theorem on composition algebras, $\nu$ is the dimension and $W=\{1,2,4,8\}$ .

Theorem(Galois’s window)

Let $p\ge5$ be prime. Then $\PSL(2,p)$ has a subgroup of index $p$ if and only if $p\in\{5,7,11\}$ . The subgroups of index $p$ are isomorphic to $A_4$ , $S_4$ and $A_5$ respectively; they form one conjugacy class for $p=5$ and two for $p=7$ and $p=11$ .

Galois stated this without proof in his letter to Auguste Chevalier of 29 May 1832. The last member of the window, $p=11$ , has no second life among the linear groups: no group $\PSL(n,q)$ other than $\PSL(2,11)$ , and no alternating group, has order 660.

Proof

A subgroup $H$ of index $p$ has order $(p^2-1)/2$ , prime to $p$ . In Dickson’s list, a subgroup of the stabilizer of a point with order prime to $p$ has order dividing $(p-1)/2$ , and the cyclic and dihedral subgroups have order at most $p+1<(p^2-1)/2$ . So $H$ is $A_4$ , $S_4$ or $A_5$ , and $(p^2-1)/2\in\{12,24,60\}$ , that is $p\in\{5,7,11\}$ . Existence and the numbers of classes were checked by computation.

Theorem(The spinor window)

Let $p$ be an odd prime. Then $\SL(2,p)$ has a nontrivial two-dimensional complex representation if and only if $p\in\{3,5\}$ ; in these cases $\SL(2,3)\cong2T$ and $\SL(2,5)\cong2I$ . In particular $\SL(2,7)$ has none.

Proof

For $p\ge5$ the group is perfect, so a nontrivial two-dimensional representation has determinant 1, and its image is a non-solvable finite subgroup of $\SL(2,\C)$ of order $p(p^2-1)$ or $p(p^2-1)/2$ , at least 60. By Klein’s list it is conjugate to $2I$ , of order 120; $p(p^2-1)=120$ forces $p=5$ , and $p(p^2-1)/2=120$ has no prime solution.

Theorem(p = 7: the Fano plane)

Let $G=\PSL(2,7)$ , let $H_1$ , $H_2$ be subgroups isomorphic to $S_4$ from the two conjugacy classes, and let $\Omega=G/H_1$ . Then $G$ acts 2-transitively on the seven points of $\Omega$ ; $H_2$ fixes no point and has orbits of sizes 3 and 4; the $G$ -orbit $\mathcal L$ of its 3-orbit has seven members, and $(\Omega,\mathcal L)$ is a Fano plane; and the action is an isomorphism $G\to\GL(3,2)$ . So the middle member of Galois’s window carries the plane, and its two classes of subgroups of index 7 are the objects of size 7.

Proof

The permutation character takes the values 7,3,1,1,0,0 on the classes of elements of orders 1,2,3,4,7,7, so its norm is $\tfrac1{168}(49+21\cdot9+56+42)=2$ , and the action is 2-transitive. An orbit of $H_2$ of size 2 would give an $A_4$ normal in $H_2$ and in a conjugate of $H_1$ , hence normal in the group they generate, which is $G$ , since $H_2$ has prime index and is maximal; this contradicts simplicity, so the orbits are 3 and 4. By 2-transitivity every pair of points lies in the same number $\lambda$ of members of $\mathcal L$ , and $7\cdot3=21\lambda$ gives $\lambda=1$ .

Remark

Read through descriptions, the window is the image of the absence’s invariant $\nu$ : the values whose fibre is not empty.

Built from: absence
Builds: imprint
Objects: the seven pointsthe seven lines
In the Esquisse: 5Courte marche à travers la théorie de Galois7La trinité de GaloisÉp.L’horizon : dessins d’enfants
In the volume: VFour Reports, Six LettersXIVWhy OctonionsXVSpin from the Double CoverXVIThe Quartet