Galois gap

Floor 3, Ce qui est su · introduced in Chapter 17, L’écart de Galois

What can no count of fixed points hear, and what can no finite set express?

The number of conjugacy classes minus the number of rational classes: the dimension of the class functions that no combination of finite G-sets reaches, measured by the same Galois orbits that decide which twists no count can hear.

Plate 3.8The character table of the group of order 168 with the row

\chi_3-\bar\chi_3

beneath it: zero on four classes and

\mp\sqrt{-7}

7A

and

7B

. It spans the gap, and no combination of finite

G

-sets reaches it.

Definition

Let $e$ be the exponent of $G$ and $\Gamma=(\Z/e\Z)^\times$ , acting on the conjugacy classes by power maps and on the irreducible characters by $\chi^{(k)}(g)=\chi(g^k)$ . A rational class is a $\Gamma$ -orbit of classes. The Galois gap of $G$ is the number of conjugacy classes minus the number of rational classes. An automorphism is Galois-like if it maps every class into its rational class, and the image of the Galois-like automorphisms in $\operatorname{Out}(G)$ is the group of inaudible twists.

Theorem(What counting cannot hear) proved

For an automorphism $\alpha$ of a finite group $G$ the following are equivalent.

(1) $\alpha$ is Galois-like.

(2) For every subgroup $H$ , the $G$ -sets $G/H$ and $G/\alpha(H)$ have the same permutation character: every $g\in G$ has as many fixed points on one as on the other, equivalently $\Q[G/H]\cong\Q[G/\alpha(H)]$ as $\Q[G]$ -modules.

(3) $\alpha$ fixes every rational-valued character of $G$ .

(4) $\alpha$ maps every irreducible character into its $\Gamma$ -orbit.

Proof

The permutation character of $G/H$ is $\pi_H(g)=|C_G(g)|\,|H\cap g^G|/|H|$ , so (2) says that $|H\cap C|=|\alpha(H)\cap C|$ for every class $C$ .

(1) $\Rightarrow$ (2). We have $|\alpha(H)\cap C|=|H\cap\alpha^{-1}(C)|$ , and $\alpha^{-1}(C)=C^{(k)}$ for some $k$ prime to the exponent $e$ . The map $x\mapsto x^k$ is a bijection from $H\cap C$ onto $H\cap C^{(k)}$ , with inverse $x\mapsto x^{k'}$ where $kk'\equiv1\pmod e$ . (2) $\Rightarrow$ (1). Take $H=\langle g\rangle$ . The group $\alpha(H)=\langle\alpha(g)\rangle$ meets the class of $\alpha(g)$ , so $\langle g\rangle$ meets it too. A conjugate of $\alpha(g)$ in $\langle g\rangle$ has the order of $g$ , so it is a generator $g^k$ .

(1) $\Leftrightarrow$ (3). Write $f^{(k)}(g)=f(g^k)$ for a class function $f$ . If $f=\sum c_\chi\chi$ is $\Gamma$ -invariant, comparing coefficients in $f^{(k)}=f$ gives $c_{\chi^{(k)}}=c_\chi$ , so the $\Gamma$ -invariant class functions are spanned by the $\Gamma$ -orbit sums of irreducible characters, which are rational-valued characters. They are the functions constant on rational classes, among them the indicator function of each rational class. Now $\alpha$ acts by $f\mapsto f\circ\alpha$ , and it fixes every function constant on rational classes exactly when it maps each rational class to itself.

(3) $\Leftrightarrow$ (4). $\alpha$ permutes the irreducible characters and commutes with $\Gamma$ . Since the irreducible characters are linearly independent, $\alpha$ fixes an orbit sum exactly when it maps the orbit to itself.

Theorem(What finite sets cannot say) proved

Over $\Q$ , the permutation characters of the finite $G$ -sets span exactly the class functions that are constant on rational classes. The dimension of this span is the number of rational classes, which equals the number of $\Gamma$ -orbits on the irreducible characters. A complement in the space of class functions is spanned by the differences $\chi-\chi^{(k)}$ of Galois-conjugate irreducible characters. Its dimension, the number of conjugacy classes minus the number of rational classes, is the Galois gap of $G$ .

Proof

Permutation characters take rational values. By Artin’s induction theorem every rational-valued character is a rational combination of the permutation characters $1_C^G$ , $C$ cyclic. By the proof of the theorem on counting, the rational-valued characters span the functions constant on rational classes, and the orbit sums form a basis of that space. Averaging over $\Gamma$ projects onto it, and the kernel of the projection is spanned by the $f-f^{(k)}$ , hence by the $\chi-\chi^{(k)}$ .

Corollary(One partition, two blindnesses) proved

The $\Gamma$ -orbits on the irreducible characters govern both theorems. Finite sets express exactly the combinations of characters that are constant along these orbits. Counting fails to hear exactly the twists that preserve each orbit. For every finite $G$ -set $X$ and every Galois-like $\alpha$ , the twisted set $X_\alpha$ has the permutation character of $X$ , because $\pi_X$ is constant on rational classes.

When all characters of $G$ are rational, the gap is zero: finite sets express every character, and a twist is inaudible only if it fixes every conjugacy class. This holds for the symmetric groups and for every finite Weyl group, so it holds for the Weyl family of Chapter 8.

Corollary(The sign of

\sqrt{-7}

cannot be counted) proved

For $G=\PSL(2,7)$ the gap is one-dimensional. It is spanned by $\chi_3-\bar\chi_3$ , whose values are $\pm\sqrt{-7}$ on $7A$ and $7B$ and 0 elsewhere. The outer automorphism acts on the irreducible characters as complex conjugation, so it is inaudible.

No count tells the points of the Fano plane from its lines, or $G/V_4^a$ from $G/V_4^b$ , or $G/A_4^a$ from $G/A_4^b$ . No combination of $G$ -sets equals $\chi_3-\bar\chi_3$ .

For the double cover $\SL(2,7)$ the gap is three-dimensional, spanned by $\chi_3-\bar\chi_3$ , $\chi_4-\bar\chi_4$ and the difference of the two faithful characters of degree 6. The characters $\chi_4$ and $\bar\chi_4$ are the Weil quartet and its conjugate. The two characters of degree 6 are exchanged by $\sqrt2\mapsto-\sqrt2$ . The outer automorphism again acts as complex conjugation, so it exchanges the quartets and fixes the last pair.

Examplecomputed

$A_6=\PSL(2,9)$ has three twists. The field twist, which $S_6$ realizes, is inaudible, by $\sqrt5\mapsto-\sqrt5$ , and the gap is 1. The diagonal twist, from $\PGL(2,9)$ , and the product, from $M_{10}$ , are audible: they move six pairs of subgroup classes, $C_3$ , $V_4$ , $S_3$ , $A_4$ , $S_4$ and $A_5$ , of which only the $V_4$ pair is Gassmann. They exchange 3-cycles with products of two 3-cycles, and the cyclic groups of order 3 witness it.

Built from: objectseam over an automorphismrefuted
Objects: the object of size 42, class athe object of size 42, class bthe object of size 14, class athe object of size 14, class bthe 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: XVSpin from the Double CoverXVIIITwo Parents of the SkyXXIIILight, Vacuum and Handedness