Galois theory of quadratic rational functions with a non-trivial automorphism 1

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1 Galois theory of quadratic rational functions with a non-trivial automorphism 1 Michelle Manes University of Hawai i at Mānoa January 15, Joint work with Rafe Jones

2 Strands of work Maps with automorphisms Work started in my thesis, that I ve been trying to continue.

3 Strands of work Maps with automorphisms Work started in my thesis, that I ve been trying to continue. Arboreal Galois representations Jones, Boston & Jones.

4 Strands of work Maps with automorphisms Work started in my thesis, that I 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,...

5 Rational Functions with Automorphisms Let φ: P 1 P 1 be a rational function, so φ(x) = P(x)/Q(x), where P, Q Z[x]. Let f (x) PGL 2 (Q).

6 Rational Functions with Automorphisms Let φ: P 1 P 1 be a rational function, so φ(x) = P(x)/Q(x), where P, Q Z[x]. Let f (x) PGL 2 (Q). φ n (x) = φ φ φ(x), and }{{} n times φ f (x) = f 1 φ f (x).

7 Rational Functions with Automorphisms Let φ: P 1 P 1 be a rational function, so φ(x) = P(x)/Q(x), where P, Q Z[x]. Let f (x) PGL 2 (Q). φ n (x) = φ φ φ(x), and }{{} n times φ f (x) = f 1 φ f (x). Definition Given a rational map φ, we define the automorphism group of φ: Aut(φ) = {f PGL 2 : φ f = φ}.

8 Rational Functions with Automorphisms Most rational functions of degree d 2 have no nontrivial automorphisms, but some do.

9 Rational Functions with Automorphisms Most rational functions of degree d 2 have no nontrivial automorphisms, but some do. Example φ(z) = 2z + 5/z has a nontrivial PGL 2 automorphism f (z) = z. We see that f 1 (z) = z as well. So φ f = φ( z) = φ(z).

10 Rational Functions with Automorphisms More specifically, degree-d functions with a nontrivial automorphism form a Zariski-closed subset of Rat d.

11 Rational Functions with Automorphisms More specifically, degree-d functions with a nontrivial automorphism form a Zariski-closed subset of Rat d. If d = 2 Most maps no nontrivial autmorphisms except on a cuspidal cubic in M 2 = A 2.

12 Rational Functions with Automorphisms More specifically, degree-d functions with a nontrivial automorphism form a Zariski-closed subset of Rat d. If d = 2 Most maps no nontrivial autmorphisms except on a cuspidal cubic in M 2 = A 2. Some maps For all maps on that cubic except the map at the cusp, Aut(φ) = C 2.

13 Rational Functions with Automorphisms More specifically, degree-d functions with a nontrivial automorphism form a Zariski-closed subset of Rat d. If d = 2 Most maps no nontrivial autmorphisms except on a cuspidal cubic in M 2 = A 2. Some maps For all maps on that cubic except the map at the cusp, Aut(φ) = C 2. One map At the cusp, Aut(φ) = S 3 (conjugate to φ(z) = 1/z 2 ).

14 Rational Functions with Automorphisms A nice normal form: If φ(x) Q(x) has degree 2 and Aut(φ) id, then φ(x) is conjugate (over Q) to a unique map of the form ψ(x) = k(x + 1/x) with k Q.

15 Rational Functions with Automorphisms A nice normal form: If φ(x) Q(x) has degree 2 and Aut(φ) id, then φ(x) is conjugate (over Q) to a unique map of the form ψ(x) = k(x + 1/x) with k Q. If Aut(φ) = C 2, then φ(x) is conjugate (over Q) to a unique map of the form ψ(x) = k(x + b/x) with k Q and b Q /(Q ) 2.

16 Motivaton Two sources of totally disconnected, locally compact groups: matrix groups over local fields automorphism groups of locally finite trees

17 Motivaton Two sources of totally disconnected, locally compact groups: matrix groups over local fields automorphism groups of locally finite trees Galois representations of the former have been studied extensively (Serre s Theorems, e.g.), but the latter have only recently been considered.

18 Motivaton Two sources of totally disconnected, locally compact groups: matrix groups over local fields automorphism groups of locally finite trees Galois representations of the former have been studied extensively (Serre s Theorems, e.g.), but the latter have only recently been considered. Question: How can we find instances of Galois groups acting on locally finite trees?

19 Trees from dynamical systems For n 1, put K n = Q(φ n (0)). We have Q K 1 K 2 If we write φ n (x) = P n (x)/q n (x) with P n, Q n Z[x], then K n is the splitting field of the polynomial P n (x).

20 Trees from dynamical systems For n 1, put K n = Q(φ n (0)). We have Q K 1 K 2 If we write φ n (x) = P n (x)/q n (x) with P n, Q n Z[x], then K n is the splitting field of the polynomial P n (x). Let K = n 1 K n and G = Gal(K /Q).

21 Trees from dynamical systems Let T 0 = n 1 φ n (0). Example The first two levels of T 0 for φ(x) = x 2 +1 x.

22 Galois action on the tree G acts on T 0 as automorphisms, giving an injection G Aut(T 0 ). This is the arboreal Galois representation associated to (φ, 0).

23 Galois action on the tree G acts on T 0 as automorphisms, giving an injection G Aut(T 0 ). 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(T 0 )?

24 Summary of known results [Aut(T 0 ) : G ] < for φ(x) = x 2 + a, for a 1, 2 (mod 4), and a < 0, a 0 (mod 4) (Stoll, 1992). φ(x) = x 2 ax + a for a Z and φ(x) = x 2 + ax 1 for a Z {0, 2} (Jones, 2008).

25 Summary of known results [Aut(T 0 ) : G ] < for φ(x) = x 2 + a, for a 1, 2 (mod 4), and a < 0, a 0 (mod 4) (Stoll, 1992). φ(x) = x 2 ax + a for a Z and φ(x) = x 2 + ax 1 for a Z {0, 2} (Jones, 2008). These are the only families of quadratic rational maps where such a result has been established.

26 Dynamical complex multiplication If φ(x) commutes with some map f (x) Q(x) which fixes 0, then the action of G on T 0 must commute with the action of f on T 0. Let φ(x) = k(x + 1/x) with f (x) = x. Then G C(f ), where C(f ) is the centralizer in Aut(T 0 ) of the involution induced by f.

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

28 Example Aut(T ) = {(12), (34), (12)(34), (1324), (1423), (13)(24), (14)(23)} = D 4.

29 Example G 2 = {e, (12)(34), (13)(24), (14)(23)}.

30 Maximality Criterion Theorem (Maximality Criterion) Suppose that deg φ = 2, φ( ) = and φ(x) Q[x]. Let γ 1, γ 2 be the critical points of φ. If P n 1(x) is irreducible and there is a prime p with v p (P n (γ 1 )P n (γ 2 )) odd and p Disc P n 1, then Gal(K n /K n 1 ) is maximal.

31 Maximality Criterion Theorem (Maximality Criterion) Suppose that deg φ = 2, φ( ) = and φ(x) Q[x]. Let γ 1, γ 2 be the critical points of φ. If P n 1(x) is irreducible and there is a prime p with v p (P n (γ 1 )P n (γ 2 )) odd and p Disc P n 1, then Gal(K n /K n 1 ) is maximal. Problem: Can t apply this to φ(x) = k(x + /1x) since the critical points are ±1 and P n (x) is always even.

32 Maximality Criterion Theorem (Maximality Criterion Redux) If φ(x) = k(x + 1/x). If n 3 and P n 1 (x) is irreducible, then Gal(K n /K n 1 ) is maximal provided that there exists a prime p with v p (kp n (1)) odd and v p (kp j (1)) = 0 for 1 j n 1.

33 Maximality Criterion Theorem (Maximality Criterion Redux) If φ(x) = k(x + 1/x). If n 3 and P n 1 (x) is irreducible, then Gal(K n /K n 1 ) is maximal provided that there exists a prime p with v p (kp n (1)) odd and v p (kp j (1)) = 0 for 1 j n 1. In other words, look at the sequence kp 1 (1), kp 2 (1),..., kp n (1),... We want Gal(K n /K n 1 ) to be maximal for each n. This happens if the sequence above has a primitive divisor appearing to an odd power for each term after the third.

34 Example Let φ(x) = (x 2 +1) x (so take k = 1).

35 Example Let φ(x) = (x 2 +1) x (so take k = 1). Then P n (1) is the first coordinate in the recurrence give by (r 0, s 0 ) = (1, 1) ) (r n, s n ) = (r n s2 n 1, r n 1s n 1 The first few terms of the recurrence are (1, 1), (2, 1), (5, 2), (29, 10), (941, 290),...

36 Example Let φ(x) = (x 2 +1) x (so take k = 1). Then P n (1) is the first coordinate in the recurrence give by (r 0, s 0 ) = (1, 1) ) (r n, s n ) = (r n s2 n 1, r n 1s n 1 The first few terms of the recurrence are (1, 1), (2, 1), (5, 2), (29, 10), (941, 290),... Since P n (1) is relatively prime to P j (1) for all j < n and P n (1) 2 (mod 3) for all n 2, the Theorem applies.

37 Consequences of the Maximality Criterion Theorem Let φ(x) = k(x + 1/x) with k Z. Suppose that for all n 2 kp n (1) is not a square in Z. Then [C(f ) : G ] <.

38 Consequences of the Maximality Criterion Theorem Let φ(x) = k(x + 1/x) with k Z. Suppose that for all n 2 kp n (1) is not a square in Z. Then [C(f ) : G ] <. Theorem Suppose that φ(x) = k(x 2 +1) x for k Z. Let P n (1) be the numerator of the n th term in the orbit of x = 1 as before, and let R n (1) be the numerator in the n th term of the orbit of 1 for the map ψ(x) = (x 2 +1) x. Assume that for all primes l dividing some R j (1) we have l k and that kp n (1) is not a square for all n 2. Then G = C(f ).

Arboreal Galois Representations

Carleton college November 9, 2014 UNC-Greensboro Outline I. (Dynamical) : introduction II. III. IV.. Arboreal Galois representations Let K be a number field with absolute Galois group G K. An arboreal