OF QUADRATIC POLYNOMIALS
TIMO ERKAMA
It is an open question whethern-cycles of complex quadratic polynomials can be contained in the fieldQ(i) of complex rational numbers forn≥5. We prove that if such cycles exist, they cannot contain arithmetic progressions and that they are never superattracting.
AMS 2000 Subject Classification: Primary 37F10; Secondary 11G30, 37F45.
Key words: polynomial iteration, rational cycle, Mandelbrot set.
1. INTRODUCTION
Ann-cycleof a quadratic polynomialP is a setS ofncomplex numbers cyclically permuted by P. Thus, for each x ∈ S the numbers x0 = x, x1 = P(x0), x2 = P(x1), . . . , xn−1 = P(xn−2) are distinct, and x0 = P(xn−1).
A complex rational n-cycle of P is an n-cycle contained in the field Q(i) of complex rational numbers.
It is not uncommon that three points xi, xj and xk of an n-cycle of a quadratic polynomial form an arithmetic progression, so thatxj−xi =xk−xj. This can occur for arbitrarily large values ofn. However, for complex rational cycles the situation is much more special. We namely have
Theorem 1. Let r, s and t be distinct points of a complex rational n- cycle of a quadratic polynomial P, and suppose thatt−s=s−r. Thenn= 3 and P is linearly conjugate to x2−29/16.
Theorem 1 solves a special case of the following
Conjecture. Quadratic polynomials do not have complex rational n- cycles for any n≥4.
So far, this conjecture has been proved only for real rational four- and five-cycles (see [4] and [2]), and for complex four-cycles (see [1]).
A nonrationaln-cycle of a quadratic polynomial can contain arithmetic progressions for each n > 3. For example, let v be any root of the cubic equation 2v3 −v2 + 1 = 0. Then the polynomial x2 +vx+v −1 has a
REV. ROUMAINE MATH. PURES APPL.,54(2009),5–6, 441–450
4-cycle {1−v,0, v −1,2v(v −1)} which contains an arithmetic progression {1−v,0, v−1}. However, four consecutive points of a cycle of a quadratic polynomial cannot form an arithmetic progression; this follows immediately from the Lagrange interpolation formula.
To establish our results, it suffices to consider the special case of polyno- mials of the form Pc(x) =x2+cwherecis a complex constant. The dynamics of such polynomials can be studied by using a two-dimensional model intro- duced in [1]; here, we consider this model in homogeneous coordinates. In Section 3 we show that complex rational n-cycles of quadratic polynomials are never superattracting for n≥3; it then follows that the centers of hyper- bolic components of period nof the Mandelbrot set are not contained inQ(i) forn≥3. We also prove that none of the points of a complex rationaln-cycle of Pc can be a Gaussian integer if n≥3.
2. HOMOGENEOUS COORDINATES
In [1] we showed that the dynamics of the family {Pc}c∈C is equivalent to the dynamics of a single two-dimensional quadratic polynomial map F defined by
(1) F(x, y) = (y, y2+y−x2)
in the complex 2-space C2. In particular, x is a periodic point of Pc if and only if (x, Pc(x)) is a periodic point of F with the same period. Thus, the map x7→(x, Pc(x)) maps cycles ofPc to cycles ofF.
In the study of rational cycles it is more convenient to consider this model in homogeneous coordinates. Let U = {(x, y, z) ∈ C3;z 6= 0}, and define G:U →U and ψ:U →C2 by
G(x, y, z) = (yz, yz+y2−x2, z2) and ψ(x, y, z) =x z,y
z
for each (x, y, z) ∈U. Two points of U are called equivalent if they have the same image under ψ. Such an imageψ(x, y, z) can be thought as the equiva- lence class of (x, y, z), and we shall denote it by [x, y, z]. ThenF becomes the quotient map of G, so that the diagram
U −→G U
yψ
yψ C2 −→F C2
is commutative. A periodic point of F has rational coordinates if and only if it is of the form [x, y, z] where x, y and z are integers. Similarly, points of a complex rational cycle of F are of the form [x, y, z] where x, y and z are
Gaussian integers. If xy 6= 0, we can then always choosex,y and zsuch that their greatest common divisor g.c.d.(x, y, z) in Z[i] is one.
Theorem2. Let[x, y, z]be a periodic point ofF such thatx,yandz6= 0 are Gaussian integers. Then there exists w ∈ Z[i] such that F([x, y, z]) = [y, w, z]. If in addition xy 6= 0 and g.c.d.(x, y, z) = 1, then g.c.d.(x, z) = g.c.d.(y, z) = 1.
Proof. Denote by (xk, yk, zk) the points of the orbit of (x, y, z) underG, so that (x, y, z) = (x0, y0, z0) andG(xk, yk, zk) = (xk+1, yk+1, zk+1) for eachk.
Note that here the sequences{xk},{yk} and{zk}are in generalnotperiodic.
Let pbe any prime in Z[i].
Lemma 1. If z≡0 (modp), theny2 ≡x2 (modp).
Proof. We first prove that
(2) xk ≡0 (mod p) and yk ≡(y2−x2)2k−1 (mod p).
for each positive integer k. The case k = 1 is obvious because x1 = yz and y1 = yz+y2 −x2. By induction, assume that (2) holds for some k. Since evidently zk=z2k, we have
xk+1=ykzk≡0 (modp) and yk+1 =ykzk+y2k−x2k≡yk2−x2k (modp), and by the induction hypothesis
yk+1 ≡yk2≡(y2−x2)2k (modp).
Hence (2) holds for each k.
If [x, y, z] has period n, we have [xn, yn, zn] = [x, y, z]. In particular, ynz=zny=z2ny, so that yn=z2n−1y. Using (2), we conclude that
(y2−x2)2n−1 ≡z2n−1y≡0 (mod p), and the assertion follows.
Corollary 1. [0, y, z] is periodic only if y= 0 or y =−z.
Proof.Suppose that y6= 0; we may then assume that g.c.d.(y, z) = 1. Ifp is a prime ofZ[i] dividingz, by Lemma 1y2 ≡0 (modp). Since g.c.d.(y, z) = 1, we conclude thatzis a unit ofZ[i]. It follows that the orbit of [0, y, z] is integral, i.e., the coordinates of each point of the orbit inC2 are Gaussian integers. In [1, Theorem 2] we proved that such an orbit must have period n≤ 2. Now, n = 2, because y 6= 0, and we conclude that F([0, y, z]) = [yz, yz+y2, z2] = [y,0, z]. This is possible only ify=−z.
Lemma 2. If z≡0 (modpν) for some ν≥1, theny2 ≡x2 (modpν).
Proof. The caseν= 1 follows from Lemma 1. By induction, assume that the assertion holds for ν, and suppose thatz ≡0 (modpν+1). Then there is µ∈Z[i] such thatz=µpν+1, and by the induction hypothesisy2−x2 =mpν for some m∈Z[i]. Therefore,
[x1, y1, z1] = [yz, yz+mpν, z2] = [yµpν+1, yµpν+1+mpν, µ2p2ν+2]
= [yµp, yµp+m, µ2pν+2].
Since [x1, y1, z1] is also periodic, by Lemma 1,
(yµp+m)2≡(yµp)2 (modp).
It follows that m2≡0 (modp), and we conclude that y2−x2 =mpν ≡0 (mod pν+1).
This completes the proof.
We can now prove Theorem 2. By Lemma 2, y2−x2 is divisible by all terms appearing in the prime factorization of z. Hence there exists m ∈Z[i]
such that y2−x2 =mz, and
F[x, y, z] = [yz, yz+y2−x2, z2] = [y, y+m, z].
Thus, we may choose w=y+m.
To prove the last assertion, assume that g.c.d.(x, y, z) = 1. Letpbe any prime in Z[i] dividing z. Then y2 −x2 = mz implies that p divides either y −x or y +x. This is possible only if p divides neither x nor y, because g.c.d.(x, y, z) = 1. Hence g.c.d.(x, z) = g.c.d.(y, z) = 1.
As in [1], we say that a Gaussian integer is even if the sum of its real and imaginary parts is even; otherwise we call it odd. Obviously a nonzero Gaussian integer is even if and only if it is divisible by κ= 1 + i.
Theorem3. Suppose that [x, y, z]has periodn≥3andg.c.d.(x, y, z) = 1. Then z is even and there is an n-periodic sequence {xk} of odd Gaussian integers such that x0 =x,x1=y and
(3) F([xk, xk+1, z]) = [xk+1, xk+2, z]
for each k≥0. The sequence {xk} has the algebraic properties (xl+1−xk+1)z=x2l −x2k for eachl, k≥0, (4)
n
Y
k=1
(xk+xk+ν) =zn for eachν= 1, . . . , n−1.
(5)
Moreover, the Gaussian integers x0, . . . , xn−1 and z are pairwise relatively prime.
Proof. The existence of a periodic sequence {xk} satisfying (3) follows from Theorem 2. To prove (4), we may assume that l=k+ν for some ν >0.
Then by (3) for each µ≥0 we have
[xµ+1, xµ+2, z] =F([xµ, xµ+1, z]) = [xµ+1z, xµ+1z+x2µ+1−x2µ, z2].
Comparison of the second components shows that xµ+2z=xµ+1z+x2µ+1−x2µ and, therefore, (xµ+2−xµ+1)z =x2µ+1−x2µ. Addition of these equations for µ=k, k+ 1, . . . , k+ν−1 proves (4).
For l = k+ν it follows from (4) that (xk+ν+1 −xk+1)z = x2k+ν −x2k. Multiplication of these equations yields
n−1
Y
k=0
(xk+ν+1−xk+1)z=
n−1
Y
k=0
(x2k+ν −x2k).
Since by periodicityxn+ν−xn=xν−x0, cancellation of the differencesxk+ν− xk proves (5).
The rest of the proof depends on
Lemma 3. Let λ be a Gaussian integer dividing xj − xk such that g.c.d.(λ, z) = 1. Then λdividesxj+ν −xk+ν for each ν = 1, . . . , n−1.
Proof. The assertion follows immediately from (4) by induction on ν, for if λ divides xj+ν −xk+ν, then it also divides x2j+ν −x2k+ν = (xj+ν+1− xk+ν+1)z.
Next, we show that z is even and that each xk is odd. Since n ≥ 3, at least two of the numbers xk are congruent mod κ. Then we can choose ν ∈ {1, . . . , n−1} such that the left-hand side of (5) is even. Hence z is even. Since by hypothesis g.c.d.(x, y, z) = 1, eitherx ory is odd while by (4) y2−x2 = (x2 −x1)z is even. This is possible only if both x and y are odd.
Now using (4) for l= 0 we see that eachxk is odd.
It remains to prove that the Gaussian integers x0, . . . , xn−1 and z are pairwise relatively prime. Application of Theorem 2 to [x1, x2, z] shows that g.c.d.(x2, z) = 1, and continuation of the argument by induction proves that g.c.d.(xk, z) = 1 for each k. (Note that by Corollary 1 we need not worry about the possibility that some xk might be zero.) Finally, assume that xk and xk+ν have a common odd prime factorpfor someν >0. From (4) we see thatxk+ν+1−xk+1 is divisible byp2. Lemma 3 then shows thatxk+2ν−xk+ν
is divisible by p2, so that xk+2ν is divisible by p. Now (4) implies that (xk+2ν+1−xk+ν+1)z= (xk+2ν−xk+ν)(xk+2ν+xk+ν)
is divisible byp3. Similarly,xk+3ν+1−xk+2ν+1 will be divisible by p4 etc. By periodicity, this leads to a contradiction because a fixed difference xk+ν −xk
cannot be divisible by arbitrarily high powers of p. Thus xk+ν and xk are relatively prime, and the proof of Theorem 3 is complete.
Corollary2. Complex rationaln-cycles ofPc do not contain Gaussian integers for n≥3.
This is a refinement of Theorem 2 of [1], where we proved that a complex rationaln-cycle of Pc cannot consistof Gaussian integers for n≥3.
Proof. If [x, y, z] is a point of a complex rationaln-cycle ofPc withn≥3, by [1, Theorem 2] the orbit of [x, y, z] is not integral, i.e., |z|>1. Thenx/z cannot be a Gaussian integer, because g.c.d.(x, z) = 1.
Our next result shows that the numbers xk cannot have small prime factors if nis large.
Theorem 4. Let I be a prime ideal of Z[i] containing one of the num- bers xk of Theorem 3. Then the quotient ring Z[i]/I contains at least 2n−1 elements.
Proof. Let π : Z[i] → Z[i]/I denote the canonical projection, and let x0, . . . , xn−1 and z be as in Theorem 3. We first prove that the restriction of π to the set S={x0, . . . , xn−1} is one-to-one.
Suppose that π(xi) = π(xj) for some i, j, and let p be a prime of Z[i]
generating I. Then p divides xk, because xk ∈ I, so that g.c.d.(p, z) = 1 by Theorem 3. In particular, p is odd. Since π(xi) = π(xj), either xi = xj
or p divides xi−xj. Choose ν ∈ {0, . . . , n−1} such that xj+ν = xk. If p divides xi−xj, by Lemma 3 it also divides xi+ν −xj+ν =xi+ν −xk, and we conclude that p divides xi+ν. This is possible only if xi+ν =xk, because the elements of S are relatively prime, and we conclude that xi+ν = xj+ν and, therefore, xi =xj.
Since π is a homomorphism, the restriction of π to the set −S = {−x0, . . . ,−xn−1}is also one-to-one. Hence it remains to show that the inter- section of the sets π(S) and π(−S) consists of the point π(xk) only. Indeed, if π(xi) = π(−xj) for some i, j, we have π(xi+xj) = 0, so that xi +xj is divisible by p. Sincep-z, by (5) this is possible only if xi=xj =xk.
3. SUPERATTRACTING CYCLES
Let P(x) = αx2+βx+γ be an arbitrary quadratic polynomial where α, β and γ are complex and α 6= 0. An n-cycle of P is superattracting if it contains a point x such thatP0(x) = 0.
Theorem 5. Superattracting n-cycles of quadratic polynomials are not contained in Q(i) for n≥3.
Proof. Letξ0, . . . , ξn−1 be points of ann-cycle of a quadratic polynomial P(x) =αx2+βx+γ withn≥3, and suppose thatξk∈Q(i) for 0≤k≤n−1;
we wish to prove that P0(ξk)6= 0 for eachk.
We may of course assume that ξk+1 = P(ξk) for 0 ≤ k ≤ n−1, by defining ξn = ξ0. For 0 ≤ k ≤ 2 these equations then form a nonsingular linear system with a unique solution {α, β, γ}. Since the coefficients of this system are contained in Q(i), the same must be true of the coefficients α, β and γ of P.
The linear polynomial ζ(x) = αx+β/2 conjugates P to a normalized polynomial Pc of the quadratic family, so that,
(6) ζ◦P =Pc◦ζ.
In addition, the pointsxν =ζ(ξν) are distinct points ofQ(i) for 0≤ν≤n−1, and (6) shows that Pc(xν) = xν+1 for each ν. Hence the points xν form an n-cycle of Pc. Sincen≥3, by Corollary 1,xν 6= 0 for each ν. Differentiation of (6) then shows thatP0(ξν)6= 0 for each ν. We conclude that the cycle{ξν} is not superattracting, and the proof is complete.
The postcritical limit set of Pc is the set of limit points of the sequence 0, Pc(0), Pc(Pc(0)), . . . of iterated images of the critical point 0 of Pc. The Mandelbrot set M consists of all c ∈ C such that the points of the above sequence form a bounded subset ofC. An equivalent definition ofM in terms of F was given in [1].
A component of the interior ofM is calledhyperbolic if for each pointc of this component the postcritical limit set of Pc is ann-cycle of Pc for some n≥1. It is conjectured that each interior component of M is hyperbolic but a complete proof has not yet been given [3].
Each hyperbolic component of M has a unique center c such that the postcritical limit set of Pc is a superattracting n-cycle of Pc for some n≥1.
Thus, the result below is a special case of Theorem 5.
Corollary 3. Let c be the center of a hyperbolic component of the Mandelbrot set. Then c ∈ Q(i) only if the superattracting cycle of Pc has period ≤2.
We conclude that M has only two hyperbolic components with centers in Q(i); the corresponding values ofcare 0 and −1.
4. PROOF OF THEOREM 1
LetP be as in Theorem 1. The linear conjugation in (6) maps complex rational cycles of P to complex rational cycles of Pc and arithmetic progres- sions to arithmetic progressions. Thus we may assume thatP =Pc for somec.
Assume thatr,sand tare distinct points of a complex rational n-cycle ofPc forming an arithmetic progression as in Theorem 1, so thatt−s=s−r.
We have to prove that n= 3 and that Pc(x) =x2−29/16.
Let [x, y, z] be an expression of (s, Pc(s)) in homogeneous coordinates, so that (s, Pc(s)) =ψ(x, y, z) and g.c.d.(x, y, z) = 1. Theorem 3 implies that there exists a periodic sequence of Gaussian integers xk such that x0 = x, x1 =y and F([xk, xk+1, z]) = [xk+1, xk+2, z] for each k. Moreover, xk and z are relatively prime in Z[i] for eachk.
Let i and j be subscripts such that r = xi/z and t = xj/z. Then the Gaussian integers xi,x0 andxj form an arithmetic progression, so that
(7) xj−x0 =x0−xi =m
for some Gaussian integer m. Now, (5) shows that 2x0 =xi+xj divides zn. However, sincex0 andz are relatively prime inZ[i], we conclude thatx0 must be a unit of Z[i]. Multiplyingz and eachxk by a unit, we may then obviously assume thatx0= 1.
Let Γ be the multiplicative semigroup of Gaussian integers generated by i andκ= 1 + i. Then Γ consists of nonzero Gaussian integers of the formuκn, where u is a unit of Z[i] and n is a nonnegative integer. Note that the real and imaginary parts of elements of Γ are contained in Γ∪ {0}.
Lemma 4. Let p be an odd prime factor of z. Then p dividesm and the numbers x0+xi = 2−m andx0+xj = 2 +m are in Γ.
Proof. From (4) we see thatp divides each number of the formx2l −x2k. Then p must divide either xl−xk or xl+xk but not both, because xl and xk are relatively prime. If p - m then, by (7), we conclude that p divides xj+x0 = 2 +m as well asx0+xi = 2−m; so, p also divides the difference 2m, a contradiction. Thus p dividesm. The last assertion of the lemma now follows, because the numbers x0 −xi and x0 −xj are divisible by m and, consequently, x0+xi and x0 +xj cannot have common odd prime factors with z.
Since m is even, it is divisible by κ. Denote by a and b the real and imaginary parts of m/κ, respectively. Then (x0+xi)/κ= 1−a−i(1 +b) and (xj+x0)/κ= 1 +a−i(1−b) are in Γ.
It follows that the real and imaginary parts of 1−a−i(1 +b) and 1 + a−i(1−b) are contained in Γ∪ {0}. The same is then true of their products 1−a2 and 1−b2. This is possible only ifaas well asb is one of the numbers 0, ±1 and ±3. We conclude that either m ∈Γ or m is of the form m = γd, where γ divides 2 anddis one of the three numbers 2±i and 3.
DefineX = 12κ(xj−1−xi−1) andY = 12κ(xj−1+xi−1). Since each xk is odd, X andY are inZ[i], and
(8) X2+Y2= i(x2j−1+x2i−1).
Since xj−x0=x0−xi=m, by (4) we havex2j−1−x2n−1=x2n−1−x2i−1. So, x2j−1+x2i−1 = 2x2n−1 and, consequently,
(9) X2+Y2 = (X+ iY)(X−iY) = 2ix2n−1.
Because the Gaussian integersxk are relatively prime,X and Y cannot have common odd prime factors. The same must then be true of X+iY and X−iY and, because their product is even, both must be even but not divisible by 2. It follows that there exists a unit u of Z[i], and relatively prime odd Gaussian integers α and β such that
X+ iY =uκα2 (10)
X−iY = 1 uκβ2. (11)
Then, by (9), x2n−1 =α2β2. Obviously, we may chooseα and β such that
(12) xn−1 =αβ.
Sinceu4= 1, by squaring (10) and (11) and subtracting we find that (13) 4iXY = iκ2(x2j−1−x2i−1) =u2κ2(α4−β4).
Here, by (4), x2j−1−x2i−1 = (xj−xi)z= 2mz, and we conclude that
(14) α4−β4 =κ2u2mz.
It remains to solve this equation for possible values of m and z. For this purpose we can use the result below which was proved in [1].
Lemma 5. Let α, β, γ andδ be Gaussian integers such that γ divides2 and
(15) α4−β4=γδ2.
Then αβδ= 0.
This lemma implies easily that z ∈ Γ. In fact, we have seen earlier that either m ∈ Γ or m is of the form m = γd, where γ divides 2 and d is one of the three numbers 2±i and 3. On the other hand, because z divides x2j −x2i = 2m(xj+xi) = 4m, either z ∈Γ ormz is of the form ρd2 for some ρ∈Γ. In the latter case the right-hand side of (14) has the same form as the right-hand side of (15), in contradiction with Lemma 5. Hence z∈Γ.
If m ∈ Γ as well, the right-hand sides of (14) and (15) have again the same form, contradicting the fact that αβmz6= 0. Thus, it remains to study the case when z∈Γ and m=γdfor someγ dividing 2.
In this caseddivides one, and only one of the four factorsα±βandα±iβ of the left-hand side of (14), while the remaining three factors are in Γ. At least two of these three factors must have modulus ≤2, because if e.g. α+β and α−iβ were both divisible byκ3, thenα= κ1[i(α+β) + (α−iβ)] would be even. We conclude that α as well as β has modulus ≤2√
2, so that the only possible prime factors ofαandβare 2 + i and 2−i. Sinceα4−β4 6= 0, at least one of the numbers α4 and β4 cannot be a unit and is therefore contained in a prime ideal I generated by either 2 + i or 2−i. But thenxn−1=αβ also is contained in I.
By Theorem 4, the quotient ringZ[i]/I contains at least 2n−1 elements.
Hence 2n−1≤5, and we conclude that n≤3.
It remains to show that c=−29/16. By interchangingi and j if neces- sary, we may assume that P(r) =s, P(s) =t and P(t) =r. It then follows from the Newton interpolation formula that
P(x) =s+ (x−r)− 3z
2m(x−r)(x−s)
= [−3z2x2+ (6−m)zx+ 2m2+ 3m−3]/2mz.
The normalization of P = Pc now implies that −3z2 = 2mz and 6−m = 0.
Then m= 6 and z=−4, so that c = (2m2+ 3m−3)/2mz =−29/16. This completes the proof of Theorem 1.
Ifr,sandtform an arithmetic progression as in Theorem 1, they satisfy a linear relationr−2s+t= 0. There are many similar relations which cannot be satisfied by points of complex rational n-cycles of the quadratic family for n≥4. For example, the relation 3r+ 4s−2t= 0 is impossible by Theorem 3, because each xk is odd. A less obvious example is given by the relation r+ 2s−t= 0, whose study is left for the interested reader.
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[2] E.V. Flynn, Bjorn Poonen and Edward F. Schaefer,Cycles of quadratic polynomials and rational points on a genus-2curve. Duke Math. J.90(1997),3, 435–463.
[3] Curtis T. McMullen,Frontiers in complex dynamics. Bull. Amer. Math. Soc. (N.S.)31 (1994),2, 155–172.
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Received 17 December 2008 University of Joensuu
FI-80101 Joensuu Finland [email protected]
