Galois theory of quadratic rationaljhs/JMMSF2010/Manes.pdf · Strands of work Maps with...

Strands of work Maps with automorphisms Arboreal Galois representations Primitive divisors

Galois theory of quadratic rationalfunctions with a non-trivial automorphism1

Michelle Manes

University of Hawai‘i at Manoa

January 15, 2010

1Joint work with Rafe Jones

Strands of work

Maps with automorphisms Work started in my thesis, thatI’ve been trying to continue.

Arboreal Galois representations Jones, Boston & Jones.Primitive divisors Classical problems, up through recent work

of Silverman & Ingram; Faber & Granville; Rice;Jones; Flatters; Everest, Mclaren, & Ward, . . .

Strands of work

Arboreal Galois representations Jones, Boston & Jones.

Primitive divisors Classical problems, up through recent workof Silverman & Ingram; Faber & Granville; Rice;Jones; Flatters; Everest, Mclaren, & Ward, . . .

Strands of work

Arboreal Galois representations Jones, Boston & Jones.Primitive divisors Classical problems, up through recent work

of Silverman & Ingram; Faber & Granville; Rice;Jones; Flatters; Everest, Mclaren, & Ward, . . .

Rational Functions with Automorphisms

Let φ : P1 → P1 be a rational function, so φ(x) = P(x)/Q(x),where P,Q ∈ Z[x ]. Let f (x) ∈ PGL2(Q).

φn(x) = φ ◦ φ ◦ · · · ◦ φ︸︷︷︸n times

(x), and

φf (x) = f−1 ◦ φ ◦ f (x).

DefinitionGiven a rational map φ, we define the automorphism group of φ:

Aut(φ) = {f ∈ PGL2 : φf = φ}.

φn(x) = φ ◦ φ ◦ · · · ◦ φ︸︷︷︸n times

(x), and

φf (x) = f−1 ◦ φ ◦ f (x).

Aut(φ) = {f ∈ PGL2 : φf = φ}.

φn(x) = φ ◦ φ ◦ · · · ◦ φ︸︷︷︸n times

(x), and

φf (x) = f−1 ◦ φ ◦ f (x).

Aut(φ) = {f ∈ PGL2 : φf = φ}.

Most rational functions of degree d ≥ 2 have no nontrivialautomorphisms, but some do.

Exampleφ(z) = 2z + 5/z has a nontrivial PGL2 automorphismf (z) = −z. We see that f−1(z) = −z as well. So

φf = −φ(−z) = φ(z).

Most rational functions of degree d ≥ 2 have no nontrivialautomorphisms, but some do.

Exampleφ(z) = 2z + 5/z has a nontrivial PGL2 automorphismf (z) = −z. We see that f−1(z) = −z as well. So

φf = −φ(−z) = φ(z).

More specifically, degree-d functions with a nontrivialautomorphism form a Zariski-closed subset of Ratd .

If d = 2

Most maps no nontrivial autmorphisms except on a cuspidalcubic inM2

∼= A2.

Some maps For all maps on that cubic except the map at thecusp, Aut(φ) ∼= C2.

One map At the cusp, Aut(φ) = S3 (conjugate toφ(z) = 1/z2).

If d = 2

∼= A2.

Some maps For all maps on that cubic except the map at thecusp, Aut(φ) ∼= C2.

If d = 2

∼= A2.Some maps For all maps on that cubic except the map at the

cusp, Aut(φ) ∼= C2.

If d = 2

∼= A2.Some maps For all maps on that cubic except the map at the

cusp, Aut(φ) ∼= C2.One map At the cusp, Aut(φ) = S3 (conjugate to

φ(z) = 1/z2).

A nice normal form: If φ(x) ∈ Q(x) has degree 2 andAut(φ) 6= id, then φ(x) is conjugate (over Q) to a unique map ofthe form

ψ(x) = k(x + 1/x) with k ∈ Q∗.

If Aut(φ) ∼= C2, then φ(x) is conjugate (over Q) to a unique mapof the form

ψ(x) = k(x + b/x) with k ∈ Q∗ and b ∈ Q∗/(Q∗)2.

A nice normal form: If φ(x) ∈ Q(x) has degree 2 andAut(φ) 6= id, then φ(x) is conjugate (over Q) to a unique map ofthe form

ψ(x) = k(x + 1/x) with k ∈ Q∗.

If Aut(φ) ∼= C2, then φ(x) is conjugate (over Q) to a unique mapof the form

ψ(x) = k(x + b/x) with k ∈ Q∗ and b ∈ Q∗/(Q∗)2.

Motivaton

Two sources of totally disconnected, locally compact groups:matrix groups over local fieldsautomorphism groups of locally finite trees

Galois representations of the former have been studiedextensively (Serre’s Theorems, e.g.), but the latter have onlyrecently been considered.

Question: How can we find instances of Galois groups actingon locally finite trees?

Motivaton

Trees from dynamical systems

For n ≥ 1, put Kn = Q(φ−n(0)). We have

Q ⊆ K1 ⊆ K2 ⊆ · · ·

If we write φn(x) = Pn(x)/Qn(x) with Pn,Qn ∈ Z[x ], then Kn isthe splitting field of the polynomial Pn(x).

Let K∞ =⋃n≥1

Kn and G∞ = Gal(K∞/Q).

For n ≥ 1, put Kn = Q(φ−n(0)). We have

Q ⊆ K1 ⊆ K2 ⊆ · · ·

If we write φn(x) = Pn(x)/Qn(x) with Pn,Qn ∈ Z[x ], then Kn isthe splitting field of the polynomial Pn(x).

Let K∞ =⋃n≥1

Kn and G∞ = Gal(K∞/Q).

Let T0 =⋃n≥1

φ−n(0).

Example

The first two levels of T0 for φ(x) = x2+1x .

Galois action on the tree

G∞ acts on T0 as automorphisms, giving an injection

G∞ ↪→ Aut(T0).

This is the arboreal Galois representation associated to (φ,0).

Natural questions:

For which φ(x) ∈ Q(x) can we determine G∞?

When does G∞ have finite index in Aut(T0)?

Galois action on the tree

G∞ acts on T0 as automorphisms, giving an injection

G∞ ↪→ Aut(T0).

This is the arboreal Galois representation associated to (φ,0).

Natural questions:

For which φ(x) ∈ Q(x) can we determine G∞?When does G∞ have finite index in Aut(T0)?

Summary of known results

[Aut(T0) : G∞] <∞ forφ(x) = x2 + a, for a ≡ 1,2 (mod 4), and a < 0, a ≡ 0(mod 4) (Stoll, 1992).φ(x) = x2 − ax + a for a ∈ Z andφ(x) = x2 + ax − 1 for a ∈ Z r {0,2} (Jones, 2008).

These are the only families of quadratic rational maps wheresuch a result has been established.

Summary of known results

[Aut(T0) : G∞] <∞ forφ(x) = x2 + a, for a ≡ 1,2 (mod 4), and a < 0, a ≡ 0(mod 4) (Stoll, 1992).φ(x) = x2 − ax + a for a ∈ Z andφ(x) = x2 + ax − 1 for a ∈ Z r {0,2} (Jones, 2008).

These are the only families of quadratic rational maps wheresuch a result has been established.

Dynamical “complex multiplication”

If φ(x) commutes with some map f (x) ∈ Q(x) which fixes 0,then the action of G∞ on T0 must commute with the action of fon T0.

Let φ(x) = k(x + 1/x) with f (x) = −x .

Then G∞ ↪→ C(f ), where C(f ) is the centralizer in Aut(T0) ofthe involution induced by f .

Example

Let T be the complete binary rooted tree of height 2, and labelthe vertices at the top level of T by 1,2,3,4.

Example

Aut(T ) = {(12), (34), (12)(34), (1324), (1423), (13)(24), (14)(23)}= D4.

Example

G2 = {e, (12)(34), (13)(24), (14)(23)}.

Maximality Criterion

Theorem (Maximality Criterion)Suppose that degφ = 2, φ(∞) =∞ and φ(x) 6∈ Q[x ]. Let γ1, γ2be the critical points of φ. If Pn−1(x) is irreducible and there is aprime p with vp (Pn(γ1)Pn(γ2)) odd and p - Disc Pn−1, thenGal(Kn/Kn−1) is maximal.

Problem: Can’t apply this to φ(x) = k(x + /1x) since thecritical points are ±1 and Pn(x) is always even.

Theorem (Maximality Criterion)Suppose that degφ = 2, φ(∞) =∞ and φ(x) 6∈ Q[x ]. Let γ1, γ2be the critical points of φ. If Pn−1(x) is irreducible and there is aprime p with vp (Pn(γ1)Pn(γ2)) odd and p - Disc Pn−1, thenGal(Kn/Kn−1) is maximal.

Problem: Can’t apply this to φ(x) = k(x + /1x) since thecritical points are ±1 and Pn(x) is always even.

Theorem (Maximality Criterion Redux)If φ(x) = k(x + 1/x). If n ≥ 3 and Pn−1(x) is irreducible, thenGal(Kn/Kn−1) is maximal provided that there exists a prime pwith vp(kPn(1)) odd and vp(kPj(1)) = 0 for 1 ≤ j ≤ n − 1.

In other words, look at the sequence

kP1(1), kP2(1), . . . , kPn(1), . . .

We want Gal(Kn/Kn−1) to be maximal for each n. This happensif the sequence above has a primitive divisor appearing to anodd power for each term after the third.

Theorem (Maximality Criterion Redux)If φ(x) = k(x + 1/x). If n ≥ 3 and Pn−1(x) is irreducible, thenGal(Kn/Kn−1) is maximal provided that there exists a prime pwith vp(kPn(1)) odd and vp(kPj(1)) = 0 for 1 ≤ j ≤ n − 1.

In other words, look at the sequence

kP1(1), kP2(1), . . . , kPn(1), . . .

We want Gal(Kn/Kn−1) to be maximal for each n. This happensif the sequence above has a primitive divisor appearing to anodd power for each term after the third.

Example

Let φ(x) = (x2+1)x (so take k = 1).

Then Pn(1) is the first coordinate in the recurrence give by

(r0, s0) = (1,1)

(rn, sn) =(

r2n−1 + s2

n−1, rn−1sn−1

)The first few terms of the recurrence are

(1,1), (2,1), (5,2), (29,10), (941,290), . . .

Since Pn(1) is relatively prime to Pj(1) for all j < n andPn(1) ≡ 2 (mod 3) for all n ≥ 2, the Theorem applies.

Example

(r0, s0) = (1,1)

(rn, sn) =(

r2n−1 + s2

n−1, rn−1sn−1

(1,1), (2,1), (5,2), (29,10), (941,290), . . .

Example

(r0, s0) = (1,1)

(rn, sn) =(

r2n−1 + s2

n−1, rn−1sn−1

(1,1), (2,1), (5,2), (29,10), (941,290), . . .

Consequences of the Maximality Criterion

TheoremLet φ(x) = k(x + 1/x) with k ∈ Z. Suppose that for all n ≥ 2kPn(1) is not a square in Z. Then [C(f ) : G∞] <∞.

Theorem

Suppose that φ(x) = k(x2+1)x for k ∈ Z. Let Pn(1) be the

numerator of the nth term in the orbit of x = 1 as before, and letRn(1) be the numerator in the nth term of the orbit of 1 for themap ψ(x) = (x2+1)

x . Assume that for all primes ` dividing someRj(1) we have ` - k and that kPn(1) is not a square for all n ≥ 2.Then G∞ ∼= C(f ).

Consequences of the Maximality Criterion

TheoremLet φ(x) = k(x + 1/x) with k ∈ Z. Suppose that for all n ≥ 2kPn(1) is not a square in Z. Then [C(f ) : G∞] <∞.

Theorem

Suppose that φ(x) = k(x2+1)x for k ∈ Z. Let Pn(1) be the

numerator of the nth term in the orbit of x = 1 as before, and letRn(1) be the numerator in the nth term of the orbit of 1 for themap ψ(x) = (x2+1)

x . Assume that for all primes ` dividing someRj(1) we have ` - k and that kPn(1) is not a square for all n ≥ 2.Then G∞ ∼= C(f ).

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