Table Of ContentBOUNDS 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 fol-
lowing 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 situ-
ations 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
