Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Arboreal Galois Representations
Rafe Jones
Carleton college

November 9, 2014
UNC-Greensboro

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

OutlineI. (Dynamical) Arboreal Galois Representations: introduction
II. Image: examples and conjectures
III. Ramification
IV. Image: number of orbits.

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjecturesRamification
Image: number of orbits

Arboreal Galois representations

Let K be a number field with absolute Galois group GK .
An arboreal Galois representation is a continuous homomorphismρ : GK → Aut(T ),where T is a locally finite rooted tree, and Aut(T ) is the group oftree automorphisms of T .

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Arboreal Galois representations from dynamics
Let φ ∈ K (z) have degree d ≥ 2, let b ∈ P1 (K ), and denote by φnthe n-fold composition of φ with itself. Put
φ−i (b) = {β ∈ Q : φi (β) = b}.
The preimage tree of φ with root b isG
T∞ :=φ−i (b),i≥0with two elements connected iff φ maps one to the other.
Denote the truncation of T∞ to the nth level by Tn .
Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits!
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First two levels of preimage tree of f (x) = x x+1 , b = 0.

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Then GK acts naturally on T∞ as tree automorphisms (since φ isdefined over K ), giving
ρφ,b : GK → Aut(T∞ ).
Denote the image of ρφ,b by G∞ , and note thatG∞ = lim Gn ,←−where Gn = Gal(K (φ−n (b))/K ).
Put Kn = K (φ−n (b)) and K∞ =Rafe JonesSn≥1 Kn .

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Example 1: Lattès maps

E[`]

/Exx



P1

φE ,`


/ P1nLet φ = φE ,` and b = ∞. Then φ−nE ,` (∞) = x(E [` ]).

Hence Gn is a subgroup of index at most two in Gal(K (E [`n ])/K ).
Thus the arboreal and `-adic representations have nearly identicalimage.
Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Example 2: A map with no special structure at all
Let K = Q, φ(x) = x 2 + 1, and b = 0.
In the 80s, Odoni conjectured that G∞ = Aut(T∞ ), or
equivalently Gn ∼= Aut(Tn ) for all n ≥ 1.

He showed that Gn ∼= Aut(Tn ) provided that in the sequence
φ2 (0), φ3 (0), . . . , φn (0), each term has a primitive prime divisor
appearing to odd multiplicity.
In 1990, Cremona used this to show Gn ∼= Aut(Tn ) for n = 5 · 107 .
log2 |Aut(T5·107 )| = 3210000000 − 1.
In 1992, Stoll used an ingenious trick to prove Odoni’s conjecture.
Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Example 3: A map with a small critical orbit

Let K = Q, φ(x) = (x + 1)2 − 2, and b = 0.
Then K∞ contains Q(ζ2∞ ), and Gal(K∞ /Q(ζ2∞ )) is generated bytwo elements.
Reason: the orbit under f of the critical point is−1 → −2 → −1 → · · · . Thus φ is post-critically finite (PCF)
One can show [Aut(T∞ ) : G∞ ] = ∞.
2 (#Gn )Conjecture (Boston-RJ): loglog(#Aut(Tn )) → 2/3.
2

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Example 4: A map with a non-trivial automorphism
2

Let K = Q, φ(x) = x x+1 , and b = 0.
Then µ : x → −x commutes with φ, and µ(b) = b.
Thus G∞ commutes with the action of µ on T∞ , and we haveG∞ ≤ C ,where C is the centralizer in Aut(T∞ ) of the action of µ.
[Aut(T∞ ) : C ] = ∞, but C contains an index-two subgroupisomorphic to Aut(T∞ ).
Theorem (RJ-Manes 2014]): G∞ = C .
Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Open image conjecture for quadratics

Conjecture (RJ 2013)
Let φ ∈ K (x) have degree two and critical points γ1 , γ2 ∈ P1 (K ).
Then [Aut(T∞ ) : G∞ ] = ∞ iff one of the following holds:
1. There is a non-trivial Möbius transformation that commuteswith φ and fixes b.
2. φ is post-critically finite (this includes Lattés maps).
3. γ1 and γ2 of φ have a relation of the form
φr +1 (γ1 ) = φr +1 (γ2 ) for some r ≥ 1.

4. The root b of T∞ is periodic under φ.

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

What is known

I If any of conditions 1-4 hold, then [Aut(T∞ ) : G∞ (φ)] = ∞(Pink, RJ-Manes).

I Conjecture holds for φ(x) = x 2 − kx + k, k ∈ Z \ {0, 2} andφ(x) = x 2 − kx + 1, k ∈ Z \ {0, 2} (RJ 2008).
I If φ(x) = x 2 + c with c 6= 2, −c not a square, then ABC

implies ρφ,b is surjective (Gratton-Nguyen-Tucker 2013).

If φ satisfies one of conditions 1-4, then there is an over-group inwhich G∞ ought to have finite index. This group is
well-understood (RJ-Manes, Pink, Swaminathan).

Rafe Jones

Arboreal Galois RepresentationsDynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

More of what is known

Theorem (Hindes 2014)
Replace K by K (t). Let φ(x) = x 2 + c ∈ K (t)[x] with c a
non-constant polynomial in t. Then
[Aut(T∞ ) : G∞ ] ≤ 265519 .

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Application: prime divisors of recurrence sequences

Let φ(x) ∈ K [x] and a0 ∈ K . Let
P(φ, a0 ) = {p : p | φn (a0 ) for at least one n ≥ 1}.
Suppose that b = 0 and#{g ∈ Gn : g fixes at least one element of φ−n (0)}= 0.
n→∞#GnlimThen the natural density of P(φ, a0 ) is zero (independent of a0 ).

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Odoni showed that if G∞ = Aut(T∞ ), the limit on the previous
slide is indeed zero.
Problem: give conditions on a subgroup G ≤ Aut(T∞ ) that ensure
this limit is zero.
Some well-known orbits:
I φ(x) = (x − 1)2 + 1, a0 = 3:

3, 5, 17, 257, 65537, . . .

I φ(x) = x 2 − x + 1, a0 = 2:
2, 3, 7, 43, 1807, . . .
I φ(x) = x 2 − 2, a0 = 4:4, 14, 194, 37634, . . .
Lucas-Lehmer test: Mp is prime iff Mp | φp−2 (a0 ).

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Frobenius conjugacy classes
Problem: given a prime p of K , associate a meaningful invariant to
the image of the Frobenius conjugacy class Frob p in G∞ .
One idea (Boston-RJ): study the cycle type of the action of Frob pon φ−n (b), and let n → ∞.
Consider the set of “ends” E := lim (φ−n (b)) of T∞ , which has a←−natural probability measure that is the inverse limit of the uniform
probability measure on φ−n (b) for each n ≥ 1.
We conjecture that every orbit of the action of Frob p on E has
positive measure, giving a (possibly infinite) partition of 1.

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Ramification
Let φ(x) = p(x)/q(x) with p, q relatively prime.
Assume for simplicity that ∞ is not a critical point of φ and b = 0.
Then K∞ can ramify only over primes of K dividing one of thefollowing:Q
1. γ φi (γ), where the product is over the critical points γ of φ,and over i with 1 ≤ i ≤ n.
2. The leading coefficient of pq 0 − qp 0 .
3. The leading coefficients of p and q.
4. The resultant of p and q.
When φ is a monic polynomial, (3) and (4) in the above list areboth 1, and (2) is just the degree of φ.
Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Finitely ramified representations

Recall that φ is post-critically finite (PCF) if for every critical point
γ of φ, the orbit {γ, φ(γ), φ2 (γ), . . .} is finite.

Theorem (Aitken-Hajir-Maire 2005, Hajir-Cullinan 2012)
Let K be a number field and φ ∈ K (x). If φ is PCF, then theextension K∞ /K is ramified over only finitely many primes of K .
Already known for Lattès maps φE ,` with b = ∞: K∞ can ramify
only at ` and the primes of bad reduction for E .

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Example Let K = Q and φ(x) = x 2 − 2. 0 7→ −2 7→ 2 7→ 2
K∞ is ramified over Q only at the prime 2.
n∼∼
Kn = Q(ζ2n+2 + ζ2−1n+2 ). So Gn = Z/2 Z, G∞ = Z2 .

Example Let K = Q and φ(x) = (x + 1)2 − 2. −1 7→ −2 7→ −1
K∞ is ramified over Q only at 2 and ∞.
#G2 = 23
#G3 = 26
#G4 = 211
#G5 = 222
#G6 = 243 (J. Klüners)#G7 = 286 (J. Klüners, M. Watkins)#G8 = 2171 ?

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Observation: If φ(x) = x 2 + c(c ∈ Q) is PCF, then taking
K = Q(c), we have that K∞ /K is unramified away from primes
over 2.
Question: Does there exist a number field K and a PCF mapφ ∈ K (x) of degree 2 such that K∞ is unramified at 2?

Theorem (Benedetto-Ingram-RJ-Levy, 2014)
Let d, B ∈ Z with d ≥ 2 and B ≥ 1. Up to conjugacy, there are
only finitely many PCF rational functions of degree d defined overa number field of degree at most B, except for flexible Lattès maps.

Rafe JonesArboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Corollary (Manes-Lukas-Yap, 2014)
Suppose that φ ∈ Q(x) is quadratic and PCF. Then φ isMöbius-conjugate to one of the following:x2x2 − 2x2 − 1

1/x 2

1(x−1)212(x−1)22(x−1)2−1
4x 2 −4x−49x 2 −12x2x+1
4x−2x 2−2x2x 2 −4x+1

3x 2 −4x+11−4xMoreover, none of these twelve is conjugate to any of the others.

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Definitions
Definition
Let φ ∈ K (x), b ∈ K and write φn (x) = pn (x)/qn (x) with(pn , qn ) = 1. The pair (φ, b) is stable if pn (x) − bqn (x) is
irreducible over K for all n ≥ 1, and eventually stable if thenumber of irreducible factors of pn (x) − bqn (x) remains bounded
as n grows.
Equivalently, (φ, b) is stable (resp. eventually stable) if GK acts on
E with a single orbit (resp finitely many orbits).
If φ is a polynomial, we say φ is stable if (φ, 0) is stable, i.e. alliterates of φ are irreducible.

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Some results on stability
Fun exercise: if φ ∈ Z[x] is Eisenstein, then so is φn (x) for alln ≥ 1, and hence φ is stable.

Theorem (Fein-Danielson 2001)
Let φ(x) = x d − b ∈ Z[x]. If φ is irreducible, then φ is stable.

Theorem (RJ 2012)
Let φ(x) ∈ Z[x] be quadratic with critical point γ ∈ Z. If φ(x) isirreducible and φ(γ) is odd, then φ is stable.
Main tool: fact that if none of φ2 (γ), φ3 (γ), . . . φn (γ) is a square,then φn (x) is irreducible.

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Cautionary examples
1. φ(x) = x 2 + 10x + 17 is irreducible over Q. But
φ2 (x) = (x 2 + 8x + 14)(x 2 + 12x + 34).
2. φ(x) = x 2 − x − 1 is irreducible, and so is φ2 (x). But
φ3 (x) = (x 4 − 3x 3 + 4x − 1)(x 4 − x 3 − 3x 2 + x + 1)
3. φ(x) = x 2 − 169 has




22221032222φ (x) = x − 2x +x + 2x +x −x −.
9999Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

Eventual stability conjecture

Conjecture (RJ-Levy 2014)
Let K be a number field. If φ ∈ K (x) and b ∈ K is not periodic
under φ, then (φ, b) is eventually stable.

Rafe Jones

Arboreal Galois Representations

Dynamical arboreal Galois representations
Image: examples and conjectures
Ramification
Image: number of orbits

A few results
Theorem (Ingram 2013)
Let φ ∈ K [x] be monic polynomial of degree d ≥ 2. If there is aprime p of K with p - d and |φn (b)|p → ∞ as n → ∞. Then (φ, b)
is eventually stable.

Theorem (Hamblen-RJ-Madhu 2014)
Let φ(x) = x d + c ∈ K [x] with c 6= 0. If there is a prime p of K
with |c|p < 1, then (φ, 0) is eventually stable over K .
Using similar ideas, can show that if φ(x) ∈ Z[x] andφ(x) ≡ x d mod p for some p, then (φ, 0) is eventually stable.
Question: let φ(x) = x 2 + 1/k for k ∈ Z \ {0}. Is (φ, 0) eventuallystable?
Rafe Jones

Arboreal Galois Representations