University:
Amherst College
Course:
MATH 121 | Intermediate Calculus
Academic year:

2022
Views:

453

Pages:

21
Author:

Emilee Weeks

Misiurewicz polynomials and dynamical units 2022 Notation I K is a characteristic zero field, usually a number field I K is the algebraic closure of K I f (z) ∈ K [z] is a polynomial of degree d ≥ 2 I f n = f ◦ f ◦ · · · ◦ f is the n-th iterate of f {z } | n Definition We say x ∈ P1 (K ) is I periodic if f n (x) = x for some n ≥ 1, I preperiodic if f n (x) = f m (x) for some n > m ≥ 0, I strictly preperiodic if x is preperiodic but not periodic. Postcritically Finite Maps Definition We say f ∈ K [z] is postcritically finite, or PCF, if every critical point c ∈ P1 (K ) of f is preperiodic under f . Example. f (z) = z 2 − 3 is not PCF, since the critical point x = 0 is not preperiodic. x = 0 7→ −3 7→ 6 7→ 33 7→ 1086 7→ · · · Examples of Postcritically Finite Maps Example. f (z) = z d : ∞ 7→ ∞ 0 7→ 0 Example. f (z) = z 2 − 1 : ∞ 7→ ∞ 0 7→ −1 7→ 0 Example. f (z) = z 2 − 2 : ∞ 7→ ∞ 0 7→ −2 7→ 2 7→ 2 Example. f (z) = z 2 + i : ∞ 7→ ∞ 0 7→ i 7→ i − 1 7→ −i 7→ i − 1 Example. f (z) = −2z 3 + 3z 2 : ∞ 7→ ∞ 0 7→ 0 1 7→ 1 Why should we care about PCF maps? One of many reasons: Let K be a global field and x0 ∈ K . Define I Kn := K (f −n (x0 )), S I K∞ := n≥1 Kn , I G∞ := Gal(K∞ /K ). If f is PCF, then: I The tower · · · K3 /K2 /K1 /K is ramified over only finitely many primes. (Aitken, Hajir, Maire 2005). I G∞ has infinite index in the otherwise expected automorphism group Aut(Td,∞ ) of an infinite rooted d-ary tree. Idea: Elliptic Curve torsion point CM curve ←→ ←→ Dynamical System preperiodic point PCF map The Quadratic Polynomial Family Define fc (z) = z 2 + c. Critical points are ∞ (fixed) and 0. 0 7→ c 7→ c 2 + c 7→ (c 2 + c)2 + c 7→ · · · We say c is a PCF parameter if fcn (0) = fcm (0) for some n > m ≥ 0. m = 0, n = 1 : c = 0 0 7→ 0 m = 0, n = 2 : c 2 + c = 0, so (c = 0), c = −1 0 7→ −1 7→ 0 m = 2, n = 3 : (c 2 + c)2 + c = c 2 + c, so c 4 + 2c 3 = 0, so (c = 0), c = −2 0 7→ −2 7→ 2 7→ 2 2 m = 2, n = 4 : (c 2 + c)2 + c + c = c 2 + c, so c 3 (c + 1)2 (c + 2)(c 2 + 1) = 0, so (c = 0, −1, −2), c = ±i 0 7→ i 7→ i − 1 7→ −i 7→ i − 1 PCF parameters and the Mandelbrot set Let K = C and let fc (z) = z 2 + c. The Mandelbrot set is M = {c ∈ C |{fcn (0)}n≥0 is bounded} Facts: • The PCF parameters c for which 0 is periodic lie in the interior of M. • The other (strictly preperiodic) PCF parameters c are dense in ∂M. The Unicritical Polynomial Family Fix d ≥ 2. Define fd,c (z) = z d + c. Critical points are ∞ (fixed) and 0. 0 7→ c 7→ c d + c 7→ (c d + c)d + c 7→ · · · m+n m (0) for some (0) = fd,c We say c ∈ K is a PCF parameter if fd,c m ≥ 0 and n ≥ 1. In that case, I c is a Gleason parameter if 0 is periodic under fd,c , or I c is a Misiurewicz parameter if 0 is strictly preperiodic under fd,c . Y m+n m+n−1 m−1 m Note: fd,c (0) − fd,c (0) = fd,c (0) − ζfd,c (0) ζ d =1 0 7→ a1 7→ a2 · · · 7→ am−1 7→ am 7→ · · · 7→ am+n−1 7→ am+n am+n = am =⇒ am+n−1 = ζam−1 Gleason and Misiurewicz Polynomials k (0). For the family fd,c (z) = z d + c, with ak = fd,c (Note: deg ak = d k−1 ) The n-th Gleason polynomial is Gd,0,n (c) := Y µ(n/k) . ak (c) k|n Fix a d-th root of unity ζ 6= 1. The (m, n)-Misiurewicz polynomial is ( Y µ(n/k) (Gd,0,n )−1 if n|m − 1, ζ · Gd,m,n := am+k−1 − ζam−1 1 if n - m − 1. k|n Simple Roots fd,c (z) = z d + c, Y µ(n/k) Gd,0,n (c) := ak (c) k ak = fd,c (0) k|n ζ Gd,m,n := Y k|n am+k−1 − ζam−1 µ(n/k) ( (Gd,0,n )−1 · 1 if n|m − 1, if n - m − 1. Theorem Let d ≥ 2, m ≥ 2, and n ≥ 1. Let ζ 6= 1 be a d-th root of unity. Then I Gd,0,n is a monic polynomial in Z[c] with only simple roots. ζ I Gd,m,n is a monic polynomial in Z[ζ][c] with only simple roots. Irreducibility? fd,c (z) = z d + c, Y µ(n/k) Gd,0,n (c) := ak (c) k ak = fd,c (0) k|n ζ Gd,m,n := Y k|n am+k−1 − ζam−1 µ(n/k) ( (Gd,0,n )−1 · 1 if n|m − 1, if n - m − 1. Question/Conjecture: For every d ≥ 2, m ≥ 2, and n ≥ 1, I Is Gd,0,n irreducible over Q? ζ I Is Gd,m,n irreducible over Q(ζ)? Fact: (Buff 2017): If d ≡ 1 (mod 6), then Gd,0,3 is divisible by c 2(d−1) + c d−1 + 1 But in all other cases, computations suggest not only irreducibility, but that the Galois group of the polynomial is SN . Iterates of Misiurewicz parameters fd,c (z) = z d + c, k ak = fd,c (0) Fix d ≥ 2, m ≥ 2, n ≥ 1, and ζ 6= 1 with ζ d = 1. ζ Roots of Gd,m,n satisfy an+m−1 = ζam−1 Theorem Let c0 be a root of Gd,m,n , let K = K (c0 ), and let v ∈ MK . I If v (d) = 0, then v (ai (c0 )) = 0 for all i ≥ 1 I If d = p e and v (d) > 0, then v (ai (c0 )) = 0 for n - i but v (ai (c0 )) > 0 for n|i. (And there’s an exact formula, v (ai (c0 )) ≈ d −m v (p).) Proof considers v (ai − aj ) and v (ζ − 1) in: · · · 7→ am−1 7→ am 7→ · · · 7→ am+n−1 = ζam−1 7→ am Evaluating a special polynomial at special parameter Example: Let Φm be the m-th cyclotomic polynomial, and let ζn be a primitive n-th root of unity, with m > n ≥ 1. Then Φm (ζn ) is not a unit ⇔ m = p k n (some prime power p k ) Essentially: when is ζm − ζn an algebraic unit? Example (Morton, Silverman, 1995): Let f ∈ Z[z] be a monic polynomial of degree d ≥ 2. Let a, b ∈ Q be periodic points of exact periods m, n ≥ 1 (respectively), with m - n and n - m. Then a − b is an algebraic unit. Y µ(n/d) defining I.e., plugging a into the polynomial f d (z) − z d|n b gives an algebraic unit. ζ So what about Gd,j,` (c0 ) if c0 is a root of a different Misiurewicz polynomial? ζ When is Gd,j,` (c0 ) a unit? fd,c (z) = z d + c Let d, m ≥ 2, n ≥ 1, and ζ 6= 1 a d-th root of unity. ζ Let c0 be a root of Gd,m,n . m+n m So fd,c (0) = fd,c (0) 0 0 ζ Question: For which integers j ≥ 2 and ` ≥ 1 is Gd,j,` (c0 ) an algebraic unit? Theorem (B., Goksel) With notation as above, if j 6= m, then Gd,j,` (c0 ) is a unit ⇐⇒ ` 6= n. And when ` = n, we have a precise description of hGd,j,` (c0 )i. ζ Proof ideas for Gd,j,` (c0 ) with j 6= m ··· Actual: .. .. . am+n−1 0 a1 a2 ··· am−1 ζ Modulo v , if v Gd,j,` (c0 ) > 0: am+1 am ··· .. .. . aj+`−1 0 a1 a2 ··· . aj−1 . aj+1 aj ζ When is Gd,j,` (c0 ) a unit if j = m? fd,c (z) = z d + c Let d, m ≥ 2, n ≥ 1, and ζ 6= 1 a d-th root of unity. ζ Let c0 be a root of Gd,m,n . m+n m So fd,c (0) = fd,c (0) 0 0 Theorem (B., Goksel) ζ With notation as above, if 1 ≤ ` ≤ n and ` - n, then Gd,m,` (c0 ) is an algebraic unit. Conjecture ζ (c0 ) is With notation as above, if 1 ≤ ` ≤ n and `|n, then Gd,m,` not an algebraic unit. ζ Proof ideas for Gd,m,` (c0 ), i.e., j = m ··· Actual: .. .. . am+n−1 0 a1 a2 ··· am−1 Modulo v : . am+1 am ··· .. .. . aj+1 am+`−1 0 a1 a2 ··· am−1 . am The hard case: j = m and `|n ··· ··· .. .. . am+n−1 0 a1 a2 ··· am−1 am . .. . am+1 am+`−1 0 a1 a2 ··· am−1 .. . aj+1 am ζ ζ Let c0 be a root of Gd,m,n , and let α0 be a root of Gd,m,` ζ Goal: ∃ finite place v s.t. v (Gd,m,` (c0 )) > 0 ζ Assume: Gd,m,n irreducible over Q(ζ). Then: ζ (Goal) ⇐= ∃ finite place v s.t. v (Gd,m,n (α0 )) > 0 ⇐= the multiplier λ(α0 ) of the periodic cycle in the critical orbit of fd,α0 is v -adically close to ζn/` ⇐= v (Φn/` λ(α0 ) > 0 The multiplier polynomial Definition Let d, m ≥ 2 and n ≥ 1 be integers, and let ζ 6= 1 be a d-th root ζ of unity. The multiplier polynomial associated with Gd,m,n is ζ Pd,m,n (x) = Y x − λ(cj ) ∈ Z[ζ][x], j where λ(cj ) = m+n−1 Y d(ai (cj ))d−1 is the multiplier of the periodic i=m cycle of fd,cj . Conjecture For d, m, n, ζ as above, and for every i ≥ 1, the resultant ζ Res(Pd,m,n , Φi ) is not a unit in Z[ζ]. The case d = 2: the quadratic family fc (z) = z 2 + c ζ ζ 2 = 1 with ζ 6= 1 means ζ = −1, so write Gm,n instead of Gd,m,n Prior results (Goksel et al.): Gm,n is irreducible for all m ≥ 1 and 1 ≤ n ≤ 3. Theorem (B., Goksel) If f = x k + Ak−1 x k−1 + · · · + A0 ∈ Z[x] is 2-special, i.e., I v2 (Ak−1 ) > v2 (2), and I v2 (Aj ) > v2 (Ak−1 ) for j = 0, 1, . . . , k − 2, then for all ` ≥ 1, we have |Res(f , Φ` )| > 1. Theorem (B., Goksel) For every m ≥ 2, both Pm,1 and Pm,2 are 2-special. Combining various implications Corollary Let d = 2, and let m ≥ 2. For n = 1, 2, 3: For any root c0 of Gm,n , and any integer 1 ≤ ` ≤ n, Gm,` (c0 ) is a unit ⇐⇒ `-n

Misiurewicz Polynomials and Dynamical Units

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