BOUNDS ON THE RADIUS OF THE P-ADIC MANDELBROT SET JACQUELINE ANDERSON 2 1 Abstract. Let f(z) = zd +a zd−1 +···+a z ∈ C [z] be a 0 d−1 1 p degree d polynomial. We say f is post-critically bounded, or PCB, 2 if all of its critical points have bounded orbit under iteration of f. l u It is known that if p ≥ d and f is PCB, then all critical points J of f have p-adic absolute value less than or equal to 1. We give 1 a similar result for 1d ≤ p < d. We also explore a one-parameter 2 1 family of cubic polynomials over Q to illustrate that the p-adic 2 Mandelbrot set can be quite complicated when p < d, in contrast ] T with the simple and well-understood p≥d case. N . h t a 1. Introduction m [ In complex dynamics, the Mandelbrot set is a source of inspiration 1 for much current research. This set, v 8 M = {c ∈ C : the critical orbit of f (z) = z2 +c is bounded}, c 2 6 is a complicated and interesting subset of the moduli space of degree 2 twopolynomials. Inthepasttwodecades, muchresearchhasbeendone . 7 on dynamical systems in a nonarchimedean setting. See, for example, 0 2 [2, 5, 7, 9]. For a survey of the subject, see [3] or [10]. If one examines 1 the Mandelbrot set over a p-adic field, one finds the object to be much : v less inspiring. For any prime p, the p-adic Mandelbrot set for quadratic i X polynomials as defined above, replacing C with C , is simply the unit p r disk. But when we consider an analogous set for polynomials of higher a degree, the p-adic Mandelbrot set for p < d can have a complicated and interesting structure. LetP denotetheparameterspaceofmonicpolynomialsf ofdegree d,p d defined over C with f(0) = 0. Note that every degree d polynomial p can be put in this form via conjugation by an affine linear transforma- tion. We call a map f post-critically bounded (PCB) if all of its critical points have bounded orbit under iteration of f. Let M denote the d,p subset of P that is PCB. We define the following quantity, which d,p Date: July 12, 2012. 1 2 JACQUELINE ANDERSON measures the radius of the p-adic Mandelbrot set in P : d,p r(d,p) = sup max {−v (c)}. (1) p f∈Md,pfc(cid:48)(∈cC)=p0 Knowing r(d,p) can be useful in searching for all post-critically finite polynomials over a given number field, as is done for cubic polynomials over Q in [6]. For small primes, in particular p < d, the set M may d,p be complicated and have a fractal-like boundary. We use r(d,p) as a way to measure its complexity. Just as the critical points for quadratic polynomials in the classical Mandelbrot set over C are contained in a disk of radius 2 [1, Theorem 9.10.1], the critical points for polynomials in M are contained in a disk of radius pr(d,p). For p > d or d = pk, d,p it is known that r(d,p) = 0, but for lack of a suitable reference we will provide an elementary proof. The following is the main result of this paper, which gives the exact value of r(d,p) for certain values of p < d. Theorem 1. For 1d < p < d we have 2 p r(d,p) = . d−1 Further, for p = 1d we have r(d,p) = 0. 2 It may also be interesting to pursue such questions in Berkovich space. For some work related to critical behavior for polynomials in Berkovich space, see [11]. In Section 2, we describe the notation and tools used throughout this paper. Section 3 consists of some lemmas that are frequently employed in the proofs that follow. In Section 4, we discuss the known results in this realm and provide elementary proofs for when p > d or d = pk. We prove our main result in Section 5. Finally, we conclude the paper in Section 6 with a study of a one-parameter family of cubic polynomials overC toillustratethefactthatM canindeedbequitecomplicated. 2 d,p Acknowledgements. Theauthorwouldliketothankheradvisor, Joseph Silverman, for his guidance and advice, Robert Benedetto for an email exchangethatpromptedthisproject,andeveryoneinvolvedinICERM’s semester program in complex and arithmetic dynamics for providing the stimulating research environment in which much of this work was completed. 2. Notation and Tools Throughout this paper, we fix a prime number p and we let f(z) = zd +a zd−1 +···+a z ∈ P d−1 1 d,p BOUNDS ON THE RADIUS OF THE P-ADIC MANDELBROT SET 3 be a degree d polynomial in C [z]. We suppress the p from our notation p for absolute values and valuations. We denote the critical points of f by c ,...,c , not necessarily distinct, labeled so that 1 d−1 |c | ≥ |c | ≥ ··· ≥ |c |. 1 2 d−1 We denote the closed disk centered at a of radius s in C by p D¯(a,s) = (cid:8)z ∈ C : |z −a| ≤ s(cid:9). p The filled Julia set of f is the set K = {z ∈ C : the f-orbit of z is bounded}. f p We let R ≥ 0 be the smallest number such that K ⊆ D¯(0,pR). (2) f Equivalently, as shown in [3], we can define R as follows: (cid:26) (cid:27) −v(a ) i R = max . 1≤i≤d d−i We also set r = −v(c ). (3) 1 We will often use the fact that d a = (−1)d−i σ , i d−i i where σ denotes the jth symmetric function of the critical points. j Whenever we count critical points, roots, or periodic points for f, we do so with multiplicity. The Newton polygon is a useful object in p-adic analysis that we will use frequently. Consider a polynomial n (cid:88) g(z) = b zi. i i=0 The Newton polygon for g is the lower convex hull of the set of points {(i,v(b ))}. If any b = 0, that point is omitted. (One can think of that i i point as being at infinity.) This object encodes information about the roots of g. In particular, it tells us that g has x roots of absolute value pm if the Newton polygon for g has a segment of horizontal length x and slope m. For proofs of these facts, see [8]. One consequence of these facts is that for polynomials, or more gen- erally, for power series over C , a disk in C is mapped everywhere p p n-to-1 onto its image, which is also a disk. The following proposition, whose proof can be found in [3, Corollary 3.11], will prove useful. 4 JACQUELINE ANDERSON Proposition 2. Let f(z) = (cid:80)d b (z−a)i ∈ C [z] be a degree d poly- i=0 i p nomial and let D = D¯(a,ps) be a disk in C . Then f(D) = D¯(f(a),pr), p where r = max{si−v(b )}. i 1≤i≤d Moreover, f : D → f(D) is everywhere m-to-1 for some positive integer m. 3. Preliminary Lemmas Lemma 3. Let f ∈ M , and let r and R be as defined by (2) and d,p (3). If r > 0 and p > 1d, then R = r. 2 Proof. First note that if f is post-critically bounded, then R ≥ r is necessary. Recall that (cid:26) (cid:27) −v(a ) i R = max . (4) 1≤i≤d d−i Since |a | = |σ | for i (cid:54)= p, and |a | = p−1|σ |, i d−i p d−p the only way that R could be strictly greater than r is if −v(a )/(d−p) p is maximal in the formula (4) for R, with (d−p)r−1 < −v(σ ) ≤ (d−p)r. d−p Inthiscase, weseethatR = −v(a )/(d−p)couldbeaslargeasr+ 1 . p d−p But if this is true, then f(c ) is dominated by a single term, namely 1 a cp with p 1 −v(f(c )) = −v(a cp) = pr+(d−p)R > R, 1 p 1 contradicting the fact that f is PCB. Thus R = r. (cid:3) Lemma 4. Let f ∈ C [x] be a degree d polynomial, let D¯(a,s) be a disk p in C , and let m be an integer with p (cid:45) m. If f maps D¯(a,s) m-to-1 p ¯ onto its image, then D(a,s) contains exactly m−1 critical points of f, counted with multiplicity. ¯ Proof. Withoutlossofgenerality,replacef withaconjugatesoD(a,s) = ¯ D(0,1). Let d (cid:88) f = b zi. i i=0 Then, counting with multiplicity, f(z)−f(0) has m roots in the unit disk, which implies that m is the largest positive integer such that v(b ) = min v(b ). m i 1≤i≤d BOUNDS ON THE RADIUS OF THE P-ADIC MANDELBROT SET 5 Now consider the Newton polygon for f(cid:48). Since m is the largest integer such that v(b ) is minimal and p (cid:45) m, we see that m is also the largest m integer such that v(mb ) is minimal among all v(ib ). Therefore, the m i Newton polygon for f(cid:48) has exactly m − 1 non-positive slopes, which implies that there are m−1 critical points, counted with multiplicity, in D¯(0,1). (cid:3) 4. Motivation: p > d Theorem 5. Let p > d. Then f ∈ P is PCB if and only if |c | ≤ 1 d,p i for all critical points c of f. i This is a known result, but since it does not appear in the literature, we present an elementary proof here. Proof. First, suppose |c | ≤ 1 for all i. Then |a | = |dσ | ≤ 1 for all i i i d−i ¯ ¯ i, and so f(D(0,1)) ⊆ D(0,1). Therefore, f is PCB. Now let f be PCB and suppose, for contradiction, that −v(c ) = r > 1 0. Let m be maximal such that −v(c ) = r. (In other words, there are m exactly m critical points with absolute value pr.) First, we show that there are exactly m roots z ,z ,...,z of f such that −v(z ) = r. 1 2 m i Since p > d, the Newton polygons for f and f(cid:48) are the same, up to horizontal translation. Thus, the rightmost segment of the Newton polygonforf hasthesameslopeandhorizontallengthastherightmost segment of the Newton polygon for f(cid:48), and therefore, f has exactly m roots z such that −v(z ) = r. i i Next,weuseLemma4toreachacontradiction. Considerf−1(D¯(0,pr)). This is a union of up to d smaller disks D¯(z ,psi), where the z are the i i roots of f. Note that, since f is PCB, each critical point must lie in one of these disks. By Proposition 2, we know that each s ≤ r/d. i So, each of the m large critical points c ,...,c must lie in the 1 m following set: m (cid:91) V = D¯(z ,psi). i i=1 V is a disjoint union of n ≤ m disks. Relabel the subscripts so that we can express V as follows: n (cid:97) V = D¯(z ,psi). i i=1 6 JACQUELINE ANDERSON Let D¯(z ,psi) map d -to-1 onto D¯(0,r). Then, since V contains exactly i i m preimages of 0, counted with multiplicity, we have n (cid:88) d = m. i i=1 Let b be the number of critical points in D¯(z ,psi). Then, Lemma 4 i i tells us that b = d −1, and so the number of critical points, counted i i with multiplicity, in V is n (cid:88) b = m−n < m. i i=1 This is a contradiction. Thus, if f is PCB, all the critical points lie in the unit disk. (cid:3) In particular, Theorem 5 implies that r(d,p) = 0 for p > d. The same is also true if d = pk for some positive integer k. This result follows immediately upon comparing the Newton polygons for f and f(cid:48). A proof of this result can be found in [4], but we present a proof below that is simple and tailored to the normal form used in this paper. Proposition 6. Let d = pk for some positive integer k, and let f ∈ P . Then f is PCB if and only if |c | ≤ 1 for all critical points c of d,p i i f. Proof. First, suppose all the critical points for f lie in the unit disk. ¯ ¯ Thenallthecoefficientsoff arep-integral,andsof(D(0,1)) ⊆ D(0,1). Therefore, f is PCB. Now, suppose f is PCB. By comparing the Newton polygons for f and f(cid:48), one sees that the slope of the rightmost segment of the Newton polygon for f(cid:48) is greater than the slope of the rightmost segment of the Newton polygon for f. In other words, the largest critical point of f is strictly larger than the largest root. If this critical point c were outside the unit disk, then |f(c)| = |c|d > R, and f would not be PCB. Therefore, f is PCB if and only if all critical points lie in the unit disk. (cid:3) 5. The Mandelbrot Radius for Primes 1d ≤ p < d 2 So far, we have seen a few situations in which the p-adic Mandelbrot set can be very easily described: it is simply a product of unit disks. For primes smaller than d, this is often not the case. In Section 6, we investigate a one-parameter family of cubic polynomials over Q to 2 illustrate that M can be quite intricate. While we cannot hope to d,p describe M exactly for p < d, we can get a sense of the size of M d,p d,p BOUNDS ON THE RADIUS OF THE P-ADIC MANDELBROT SET 7 by calculating r(d,p). We now give a lower bound for r(d,p) when p < d. Proposition 7. Suppose that p < d and that d is not a power of p. Let k be the largest integer such that pk < d and let (cid:96) be the largest integer such that p(cid:96)|d. Write d = apk +b, where 1 ≤ a < p and 1 ≤ b < pk. Then, a(k −(cid:96))pk r(d,p) ≥ . d−1 Proof. Let α ∈ C satisfy the following equation: p dd αd−1 = . (−apk)apkbb Then, the lower bound given in this proposition is realized by the following map: f(z) = zb(z −α)apk. This map has either two or three critical points: α, bα, and possibly d 0. (Zero is a critical point if b (cid:54)= 1.) We choose α above so that f(bα) = α,f(α) = 0, and f(0) = 0. Thus, f is post-critically finite, d and therefore PCB, with a(k −l)pk −v(α) = . d−1 (cid:3) Note that this gives a positive lower bound for r(d,p) in most situations in which p < d. It does not give a positive lower bound when d = pkq, where q < p. The next proposition is the beginning of an exploration of that situation. Proposition 8. Let f ∈ P and suppose d = 2p. Then f is PCB if d,p and only if |c | ≤ 1 for all critical points c of f. i i Proof. Again, one direction is straightforward. If all the critical points are in the unit disk, then all the coefficients of f are p-integral, and f is PCB. Now let f be PCB and suppose for contradiction that f has a critical point outside the unit disk, with −v(c ) = r > 0. By comparing the 1 rightmost segments of the Newton polygons for f and f(cid:48), since the rightmost vertex for f(cid:48) is one unit above the rightmost vertex for f(cid:48), we get that in most cases the largest critical point for f is larger than its largest root, and thus f cannot be PCB. The only situation in which this does not happen is if the rightmost segment of the Newton polygon for f has horizontal length equal to p, in which case it is 8 JACQUELINE ANDERSON possible for the largest root of f to have the same absolute value as the largest critical point. In this situation, there are exactly p roots z with −v(z ) = r, and there are at least p critical points c with i i i −v(c ) = r. Suppose there are exactly k such critical points, counted i with multiplicity, where p ≤ k ≤ 2p−1. Then by Lemma 4, this is only possible if they are all contained in a disk centered at a root z that 1 maps p-to-1 onto D¯(0,pr). Let this disk have radius ps, where s ≤ r 2p by Proposition 2. Writing c = c +(cid:15) for 2 ≤ i ≤ k, we calculate f(c ): i 1 i 1 (cid:18) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19) 2p k 2p k k f(c ) = c2p 1− + −···+2p 1 1 2p−1 1 2p−2 2 2p−1 +(cid:15),where −v((cid:15)) < 2pr. We will reach a contradiction if the coefficient of c2p is a p-adic unit, 1 because this would imply that −v(f(c )) = 2pr > r = R, and thus 1 that f is not PCB. Modulo p, the coefficient of c2p is congruent to 1 (cid:18) (cid:19) 2p k 1− ≡ 1−2 ≡ −1 (mod p). p p (cid:106) (cid:107) This is because (cid:0)k(cid:1) ≡ k (mod p) and p ≤ k < 2p. Thus, we reach p p the desired conclusion, that f is PCB if and only if all the critical points lie in the unit disk. (cid:3) In particular, this implies that r(2p,p) = 0. A similar but more elaborate argument shows that r(3p,p) = 0 as well, but the techniques used do not generalize to arbitrary r(kp,p). Now we turn our attention to the case where 1d < p < d to prove 2 the remainder of Theorem 1. Proof. Suppose that 1d < p < d. Note that Proposition 7 shows that 2 r(d,p) ≥ p . It remains to show that p is also an upper bound d−1 d−1 for r(d,p). Suppose there is a polynomial f ∈ M with a critical d,p point c such that −v(c ) = r > 0. Let d = p+k, where 1 ≤ k ≤ p−1. 1 1 Lemma 3 implies that the critical orbits for f are all contained in D¯(0,pr). Let m denote the number of critical points with absolute value pr (with multiplicity), i.e., m = max{i : −v(c ) = r}. i We break the proof into two cases. The first case we consider is m < p. We will refer to {c ,c ,...c } as the large critical points. Each large 1 2 m critical point must lie in one of the disks in the following set, where BOUNDS ON THE RADIUS OF THE P-ADIC MANDELBROT SET 9 f(z ) = 0 and s ≤ r/d: i i d (cid:91) f−1(D¯(0,pr)) = D¯(z ,psi). i i=1 By Lemma 4, we must have more than m roots z such that −v(z ) = r. i i Since the Newton polygons for f and f(cid:48) can only differ at one place (namely, at the pth place), this is only possible if there are exactly k roots of f (and at most k − 1 critical points) with absolute value pr. This implies that −v(a ) = kr. Let c be the largest critical point p m+1 such that −v(c ) < r and let t = −v(c ). Since a = dσ , we m+1 m+1 p p k must have −v(σ ) = kr −1, which implies that t ≥ r −1. Looking at k f(c ), the sole largest term is a cp , which implies that m+1 p m+1 −v(f(c )) = kr+pt ≥ dr−p. m+1 If f is PCB, then −v(f(c )) ≤ r, which gives the inequality dr−p ≤ m+1 r, and the desired bound follows. Now suppose the number of large critical points is m ≥ p. Then, by analysis of the Newton polygons for f and f(cid:48), either f has a root z 1 with −v(z ) > r, or f has exactly m roots of absolute value pr. The 1 first possibility does not occur, because if −v(z ) > r, then z must 1 1 be in the basin of infinity, by Lemma 3. This is a contradiction, since z is preperiodic, as 0 is a fixed point for f. So, the largest root z of 1 1 f satisfies −v(z ) = r and the number of large critical points is equal 1 to the number of roots of absolute value pr. By Lemma 4, the only way for f to be PCB is if there is a disk D¯(c ,ps) mapping p-to-1 onto 1 D¯(0,pr) containing at least p of the large critical points, where s ≤ r/d by Proposition 2. We will again divide into two cases. First, suppose −v(c − c ) ≤ max{0,s} for all critical points c ,c . i j i j Let c = c +(cid:15) , where −v((cid:15) ) ≤ max{0,s}. Then we have i 1 i i d d f(c ) = cd − σ cd−1 +···+(−1)d−1 σ c . 1 1 d−1 1 1 1 d−1 1 We will use the fact that (cid:18) (cid:19) p+k −1 σ = ci +δ ,where −v(δ ) < ir i i 1 i i to simplify our expression for f(c ) to the following: 1 10 JACQUELINE ANDERSON (cid:18) (cid:18) (cid:19) (cid:18) (cid:19) d d−1 d d−1 f(c ) = cd 1− + 1 1 d−1 1 d−2 2 (cid:18) (cid:19)(cid:19) d d−1 −···+ (−1)d−1 +(cid:15) (5) 1 d−1 It remains to check that the coefficient of cd is a p-adic unit and to 1 determine the largest possible absolute value for (cid:15). First, we look at the coefficient of cd in (5). This coefficient can be rewritten as follows: 1 (cid:18) (cid:19) (cid:18) (cid:19) d−1 (cid:18) (cid:19) d d−1 d d−1 (cid:88) d 1− +···+(−1)d−1 = (−1)i . d−1 1 1 d−1 i i=0 Since the full alternating sum from 0 to d of binomial coefficients is alwayszero, weseethatthecoefficientofcd iseither1or−1, depending 1 on whether d is even or odd. Either way, it is a p-adic unit, and so the first term in f(c ) has absolute value pdr. Since we must have 1 −v(f(c )) ≤ r in order for f to be PCB, it is necessary that −v((cid:15)) = dr 1 as well. The only term that can possibly be that large is the one corresponding to a cp. Let σ ((cid:15) ) denote the jth symmetric function on p 1 j i the (cid:15) . Then, the portion of a cp contributing to (cid:15) is: i p 1 (cid:18)(cid:18) (cid:19) (cid:18) (cid:19) d d−2 d−3 (−1)k σ ((cid:15) )ck−1 + σ ((cid:15) )ck−2 p k −1 1 i 1 k −2 2 i 1 (cid:18) (cid:19) (cid:19) p +···+ σ ((cid:15) )c +σ ((cid:15) ) cp. 1 k−1 i 1 k i 1 Note that since (cid:0)d−i−1(cid:1) is a multiple of p for all i < k, the last term is k−i the only one that can possibly realize the absolute value pdr. Looking at x = (−1)kdσ ((cid:15) )cp, we see that p k i 1 −v(x) ≤ pr+1+ks ≤ 1+r(p+k/d). Since −v(x) = dr, we have dr ≤ 1 + pr + kr/d, which implies that r ≤ d . This is strictly smaller than p for k > 1, and we obtain k(d−1) d−1 the desired result. We will now treat the k = 1 case separately. Suppose d = p+1 and all p critical points are in a disk centered at c of radius ps. The above 1 argumentshowsthat,iff isPCB,thenwemusthave−v(σ ((cid:15) )) = r−1. 1 i We will improve our upper bound for s to prove the result in this case. We know that D¯(c ,ps) maps p-to-1 onto D¯(0,pr). Writing f(z) so 1 that it is centered at c , we have 1

Bounds on the radius of the p-adic Mandelbrot set

Similar Bounds on the radius of the p-adic Mandelbrot set