Puiseux series dynamics and leading monomials of escape regions

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1 Puiseux series dynamics and leading monomials of escape regions Jan Kiwi PUC, Chile Workshop on Moduli Spaces Associated to Dynamical Systems ICERM Providence, April 17, 2012

2 Parameter space Monic centered cubic polynomials with marked critical points:

3 Parameter space Monic centered cubic polynomials with marked critical points: where (a, v) C 2. f a,v : z (z a) 2 (z + 2a) + v

4 Parameter space Monic centered cubic polynomials with marked critical points: f a,v : z (z a) 2 (z + 2a) + v where (a, v) C 2. After identification of (a, v) with ( a, v) one obtains the moduli space of cubic polynomials with marked critical points.

5 Parameter space Monic centered cubic polynomials with marked critical points: f a,v : z (z a) 2 (z + 2a) + v where (a, v) C 2. After identification of (a, v) with ( a, v) one obtains the moduli space of cubic polynomials with marked critical points. Critical points of f a,v are ±a. Critical value: v = f a,v (+a). Co-critical value 2a, since v = f a,v (+a) = f a,v ( 2a).

6 Periodic Curves For p 1, the periodic curve S p consists of all (a, v) such that +a has period p under f a,v.

7 Periodic Curves For p 1, the periodic curve S p consists of all (a, v) such that +a has period p under f a,v. S p C 2 is a smooth affine algebraic curve of degree d p where n p d n = 3 p 1.

8 Periodic Curves For p 1, the periodic curve S p consists of all (a, v) such that +a has period p under f a,v. S p C 2 is a smooth affine algebraic curve of degree d p where n p d n = 3 p 1. What is its topology?

9 Periodic Curves For p 1, the periodic curve S p consists of all (a, v) such that +a has period p under f a,v. S p C 2 is a smooth affine algebraic curve of degree d p where n p d n = 3 p 1. What is its topology? Is S p irreducible?

10 Periodic Curves For p 1, the periodic curve S p consists of all (a, v) such that +a has period p under f a,v. S p C 2 is a smooth affine algebraic curve of degree d p where n p d n = 3 p 1. What is its topology? Is S p irreducible? Bonifant,-,Milnor: the Euler characteristic is S p is (2 p)d p.

11 Periodic Curves For p 1, the periodic curve S p consists of all (a, v) such that +a has period p under f a,v. S p C 2 is a smooth affine algebraic curve of degree d p where n p d n = 3 p 1. What is its topology? Is S p irreducible? Bonifant,-,Milnor: the Euler characteristic is S p is (2 p)d p. What is the Euler characteristic of the smooth compactification of S p?

12 Periodic Curves For p 1, the periodic curve S p consists of all (a, v) such that +a has period p under f a,v. S p C 2 is a smooth affine algebraic curve of degree d p where n p d n = 3 p 1. What is its topology? Is S p irreducible? Bonifant,-,Milnor: the Euler characteristic is S p is (2 p)d p. What is the Euler characteristic of the smooth compactification of S p? Requires to compute the number N p of escape regions. (De Marco and Schiff, De Marco-Pilgrim).

13 Escape Regions The connectedness locus is compact. C(S p ) = { f a,v S p f n a,v( a) }

14 Escape Regions The connectedness locus C(S p ) = { f a,v S p fa,v( a) n } is compact. The escape locus E(S p ) = { f a,v S p fa,v( a) n } is open and every connected component is unbounded.

15 Escape Regions The connectedness locus is compact. The escape locus C(S p ) = { f a,v S p f n a,v( a) } E(S p ) = { f a,v S p f n a,v( a) } is open and every connected component is unbounded. A escape region U is a connected component of E(S p ): U is conformally isomorphic to punctured disk. The puncture is at U.

16

17 Asymptotics of critical periodic orbit For f a,v U, let a 0 = +a a 1 = v a p 1 a 0.

18 Asymptotics of critical periodic orbit For f a,v U, let a 0 = +a a 1 = v a p 1 a 0. Dynamical space picture. After conjugacy, f a,v (az)/a, we have:

19 Asymptotics of critical periodic orbit For f a,v U, let a 0 = +a a 1 = v a p 1 a 0. Dynamical space picture. After conjugacy, f a,v (az)/a, we have: a + o(a) or a j = 2a + o(a).

20 Asymptotics of critical periodic orbit For f a,v U, let a 0 = +a a 1 = v a p 1 a 0. Dynamical space picture. After conjugacy, f a,v (az)/a, we have: a + o(a) or a j = 2a + o(a). Bonifant and Milnor: Do the leading terms of a j a, for j = 1,..., p 1 determine U uniquely?

21 Leading monomials There exists µ 1 and a local coordinate ζ for U near such that a = 1 ζ µ.

22 Leading monomials There exists µ 1 and a local coordinate ζ for U near such that Then, a = 1 ζ µ. a j a a = holomorphic(ζ) = c k ζ k = k k 0 k k 0 c k ( ) k/µ 1. a

23 Leading monomials There exists µ 1 and a local coordinate ζ for U near such that Then, a = 1 ζ µ. a j a a = holomorphic(ζ) = c k ζ k = k k 0 k k 0 c k ( ) k/µ 1. a Leading monomial m j = c k0 ( 1 a ) k0 /µ.

24 Theorem. The leading monomial vector m = (m 1,..., m p 1, 0) determines the escape region uniquely.

25 Theorem. The leading monomial vector m = (m 1,..., m p 1, 0) determines the escape region uniquely. Corollary. (Bonifant,-,Milnor) Assume that the leading monomial vector of U is Then, for all j, (c 1 a k 1/µ,..., c p 1 a k p 1/µ ). a j Q(c 1,..., c p 1 )((a 1/µ )).

26 Equations For an escape region U, is a solution of v = a 1 = (+a or 2a) + a f p a,v(+a) = +a (*) k k 0 c k ( ) k/µ 1 a in some extension of Q((1/a)).

27 Field Put t instead of 1/a to get Q((t)) and study solutions of (*) in the algebraic closure of Q((t)): Q a t = Q a ((t 1/m )).

28 Field Put t instead of 1/a to get Q((t)) and study solutions of (*) in the algebraic closure of Q((t)): Q a t = Q a ((t 1/m )). Q a t is endowed with z = c k t k/m = e ord0(z) = e k0/m. k k 0

29 Field Put t instead of 1/a to get Q((t)) and study solutions of (*) in the algebraic closure of Q((t)): Q a t = Q a ((t 1/m )). Q a t is endowed with z = c k t k/m = e ord0(z) = e k0/m. k k 0 Complete it to get L.

30 Non-Archimedean problem For ν L, consider ψ ν (z) = t 2 (z 1)(z + 2) + ν L[z].

31 Non-Archimedean problem For ν L, consider ψ ν (z) = t 2 (z 1)(z + 2) + ν L[z]. ν is periodic parameter if ω + = +1 is periodic under ψ ν.

32 Non-Archimedean problem For ν L, consider ψ ν (z) = t 2 (z 1)(z + 2) + ν L[z]. ν is periodic parameter if ω + = +1 is periodic under ψ ν. In this case: ω + = +1 ω + 1 = ν ω+ p 1 ω+.

33 Non-Archimedean problem For ν L, consider ψ ν (z) = t 2 (z 1)(z + 2) + ν L[z]. ν is periodic parameter if ω + = +1 is periodic under ψ ν. In this case: ω + = +1 ω + 1 = ν ω+ p 1 ω+. It follows, ω c t k/m + h.o.t. = j 2 + d t l/n + h.o.t.

34 Non-Archimedean problem For ν L, consider ψ ν (z) = t 2 (z 1)(z + 2) + ν L[z]. ν is periodic parameter if ω + = +1 is periodic under ψ ν. In this case: ω + = +1 ω + 1 = ν ω+ p 1 ω+. It follows, ω c t k/m + h.o.t. = j 2 + d t l/n + h.o.t. The corresponding leading monomial are m(ω + ω + c t k/m ) = j 3 respectively.

35 Theorem. The leading monomial vector m = (m 1,..., m p 1, 0) determines the periodic parameter ν uniquely.

36 Dynamical Space The filled Julia set of ψ ν is K(ψ ν ) = {z L ψ n ν(z) }.

37 Dynamical Space The filled Julia set of ψ ν is K(ψ ν ) = {z L ψ n ν(z) }. ν 1 = ψ n ν(z) when z > 1.

38 Dynamical Space The filled Julia set of ψ ν is K(ψ ν ) = {z L ψ n ν(z) }. ν 1 = ψ n ν(z) when z > 1. Level 0 dynamical ball D 0 = { z 1}.

39 Dynamical Space The filled Julia set of ψ ν is K(ψ ν ) = {z L ψ n ν(z) }. ν 1 = ψ n ν(z) when z > 1. Level 0 dynamical ball D 0 = { z 1}. Level n set {ψ n ν(z) D 0 } is a disjoint union of finitely many dynamical balls of level n.

40 Dynamical Space The filled Julia set of ψ ν is K(ψ ν ) = {z L ψ n ν(z) }. ν 1 = ψ n ν(z) when z > 1. Level 0 dynamical ball D 0 = { z 1}. Level n set {ψ n ν(z) D 0 } is a disjoint union of finitely many dynamical balls of level n. Each level n + 1 ball is contained in, and maps onto, a level n ball.

41 Branches Maximal open balls in a closed ball D are parametrized by Q a. If ψ ν maps D onto D by degree d, then it induces a polynomial map of degree d in Q a.

42 Computing ω + j ω +. Fact. If B is a maximal open ball of a dynamical ball D l of level l, then B contains at most one dynamical ball of level l + 1.

43 Computing ω + j ω +. Fact. If B is a maximal open ball of a dynamical ball D l of level l, then B contains at most one dynamical ball of level l + 1. Consequence. If D l (ω + ) = D l (ω + j ) but D l+1 (ω + ) D l+1 (ω + j ), then ω + j ω + is the diameter of D l (ω + ) = D l (ω + j ).

44 Parameter space The level 0 parameter ball is D 0 = { ψ ν (ω + ) = ν 1}.

45 Parameter space The level 0 parameter ball is D 0 = { ψ ν (ω + ) = ν 1}. The parameters of level n { ψ n 1 ν (ω + ) 1} are a disjoint union of closed balls D n called level n parameter balls.

46 Parameter ball, dynamical balls and branch dynamics Proposition. Assume D l is a level l parameter ball. Then: D l = D l (ν) for all ν D l.

47 Parameter ball, dynamical balls and branch dynamics Proposition. Assume D l is a level l parameter ball. Then: D l = D l (ν) for all ν D l. The level l + 1 j dynamical ball ψ j ν(d l (ν)) is independent of ν D l.

48 Parameter ball, dynamical balls and branch dynamics Proposition. Assume D l is a level l parameter ball. Then: D l = D l (ν) for all ν D l. The level l + 1 j dynamical ball ψ j ν(d l (ν)) is independent of ν D l. The ψ ν action on maximal open balls is also independent of ν D l.

49 Parameter ball, dynamical balls and branch dynamics Proposition. Assume D l is a level l parameter ball. Then: D l = D l (ν) for all ν D l. The level l + 1 j dynamical ball ψ j ν(d l (ν)) is independent of ν D l. The ψ ν action on maximal open balls is also independent of ν D l. If p l is the smallest integer q such that ω + ψν q 1 (D l (ν)), then every periodic parameter in D l has period at least p l.

50 Centers Proposition. There exists a unique ν in D l which is periodic with period p l.

51 Centers Proposition. There exists a unique ν in D l which is periodic with period p l. Proof. The map T : D l D l defined by T(ν) = ( ψ p l 1 ν Dl (ν)) 1 (ω + ) is a strict contraction. Such ν is called the center of D l.

52 Level l + 1 correspondence Proposition. Let B be a maximal open ball of a parameter ball D l and consider any ν D l. B contains a parameter ball of level l + 1 if and only if B contains a dynamical ball of level l + 1. In this case, the level l + 1 balls are unique.

53 Leading monomials determine parameter balls Assume ν, ν have the same leading monomial vector.

54 Leading monomials determine parameter balls Assume ν, ν have the same leading monomial vector. Let us prove by induction on l. D l (ν) = D l (ν )

55 Leading monomials determine parameter balls Assume ν, ν have the same leading monomial vector. Let us prove by induction on l. D l (ν) = D l (ν ) Take ν l the center of level l.

56 Leading monomials determine parameter balls Assume ν, ν have the same leading monomial vector. Let us prove by induction on l. Take ν l the center of level l. D l (ν) = D l (ν ) There is a unique maximal open ball B in D l (ν l ) whose orbit is compatible with the leading monomial vector.

57 Leading monomials determine parameter balls Assume ν, ν have the same leading monomial vector. Let us prove by induction on l. Take ν l the center of level l. D l (ν) = D l (ν ) There is a unique maximal open ball B in D l (ν l ) whose orbit is compatible with the leading monomial vector. That is, ν, ν belong to B Thus ν, ν belong to the unique level l + 1 parameter ball contained in B. Hence, D l+1 (ν) = D l+1 (ν ).

58 Homework Given a sequence D 0 D 1. With centers ν 0, ν 1,. Which lie in L 0 ((t 1/m 0 )) L 1 ((t 1/m 1 )) Branner and Hubbard tell us how to compute m k.

59 Homework Given a sequence D 0 D 1. With centers ν 0, ν 1,. Which lie in L 0 ((t 1/m 0 )) L 1 ((t 1/m 1 )) Branner and Hubbard tell us how to compute m k. Compute L k.

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