power

Floor 2, Les sutures · introduced in Chapter 4, La monodromie des sutures

How can automorphisms of incarnations in different theories be compared?

For a self-centralizing cyclic stabilizer, the residue k such that every seam to a conjugacy class turns the automorphism into the k-th power map; it depends on no seam, class or marking.

Plate 2.9A cyclic labelling of the Fano plane and its Singer cycle (ink). Hall’s multiplier 2 replaces the cycle by its fourth power (blue); on the flexes,

\tau

does the same.

Definition

Let $x\in G$ generate $H=\langle x\rangle$ with $C_G(x)=H$ , of order $n$ , and let $K_H=\{k\in(\Z/n)^\times: x^k\text{ is conjugate to }x\}$ . For an automorphism $a$ of an incarnation $Y$ of $G/H$ , the power of $a$ is the unique $k\in K_H$ such that $s\circ a\circ s^{-1}$ is the $k$ -th power map for every seam $s$ from $Y$ to a conjugacy class $(x^j)^G$ .

Lemma(Power)

(a) The automorphisms of the conjugacy class $x^G$ , as a $G$ -set, are the power maps $y\mapsto y^k$ , $k\in K_H$ , and $K_H\cong N_G(H)/H$ . (b) Every automorphism of an incarnation of $G/H$ has a power. The power is a homomorphism $\Aut_G(Y)\to K_H$ , and it does not change when the marking of $Y$ is changed by an automorphism of $G$ .

Proof

(a) The class $x^G$ is an incarnation of $G/C_G(x)=G/H$ . A power map commutes with conjugation and maps $x^G$ to itself exactly when $x^k$ is conjugate to $x$ . $N_G(H)$ acts on $H$ with kernel $C_G(H)=H$ , and $nxn^{-1}=x^k$ defines an injective homomorphism $N_G(H)/H\to(\Z/n)^\times$ with image $K_H$ ; so the power maps are $|N_G(H):H|$ distinct automorphisms, hence all of them.

(b) Another seam to $(x^j)^G$ differs from $s$ by a power map, and power maps commute; $y\mapsto y^j$ is a seam $x^G\to(x^j)^G$ commuting with power maps; and changing the marking by $\beta$ conjugates $y\mapsto y^k$ to itself.

Remark

For $\PSL(2,7)$ the lemma applies to $C_3$ and $C_4$ , with $K=\{\pm1\}$ , and to $C_7$ , with $K_{C_7}=\{1,2,4\}$ , the squares modulo 7. The class $C_2$ does not satisfy $C_G(t)=\langle t\rangle$ . On the object of size 24, two automorphisms of two incarnations correspond under some, equivalently every, seam if and only if they have the same power.

Theorem(Monodromy of the object of size 24) computed

(c) The three role seams from the cyclic labellings to the Heawood matchings differ pairwise by automorphisms; relative to the role 0, the roles 1 and 3 have powers 2 and 4.

(d) Squaring on $7A$ and on $7B$ has power 2; Hall’s multiplier $\ell\mapsto2\ell$ on the labellings has power 4; scaling vectors by $\lambda$ has power $\lambda^2$ ; and $\tau$ on the flexes has power 4. Consequently, under every seam between the flexes and the cyclic labellings, $\tau$ corresponds to Hall’s multiplier 2, and both correspond to the fourth-power map, the inverse of squaring, on $7A$ and $7B$ .

Proof

By machine, through the natural seams and the lemma. For scaling, the transvection of $\lambda v$ is the $\lambda^2$ -th power of that of $v$ ; for Hall’s multiplier, the Singer map of $2\ell$ is $\ell^{-1}\circ(x\mapsto x+4)\circ\ell$ , the fourth power of that of $\ell$ .

Example

Hall’s multiplier theorem explains why 2 is a multiplier of the difference set $\{0,1,3\}$ modulo 7: 2 is the order of the plane. So the flex-tangent map of the Klein quartic and Hall’s multiplier on the cyclic labellings of the Fano plane, facts of two different geometries, are the same automorphism of the object of size 24, under every seam.

Built from: seamincarnation
Builds: twisting element
Objects: the object of size 24the object of size 56the object of size 42, cyclic
In the Esquisse: 10La table en deux, en sept et à l’infini16Une loi de réciprocité