Cantor bouquets, explosions, and Knaster continua : dynamics of complex exponentials

Publicacions Matem`atiques, Vol 43 (1999), 27–54. CANTOR BOUQUETS, EXPLOSIONS, AND KNASTER CONTINUA: DYNAMICS OF COMPLEX EXPONENTIALS Robert L. Devaney Abstract We describe some of the interesting dynamical and topological properties of the complex exponential family λezand its associated Julia sets. 1. Introduction Our goal in this paper is to describe some of the topology and dynamics of the complex exponential family Eλ(z)=λez. We will restrict to λ-values that are real and positive, mainly because all of the interesting phenomena present for other complex λ-values is already present in this situation. For a complex analytic function E, the interesting orbits lie in the Julia set, which we denote by J(E). For the exponential family, the Julia set of Eλhas three characterizations: 1. J(Eλ) is the set of points at which the family of iterates of Eλ, {En λ}is not a normal family in the sense of Montel. 2. J(Eλ) is the closure of the set of repelling periodic points of Eλ. 3. J(Eλ) is the closure of the set of points whose orbits tend to ∞. Note that condition 3 differs markedly from the case of polynomial iterations, where J(E) is the boundary of the set of escaping orbits. The reason for the difference is Eλhas an essential singularity at ∞, while polynomials have superattracting fixed points at ∞. The equivalence of 1 and 2 was shown by Baker, see [Ba2]. The equivalence of 1 and 3 is shown in [DT]. The complement of the Julia set is called the stable set. In J(Eλ), there are two very interesting topological structures, Cantor bouquets and Knaster-like continua. We will describe the construction of each in detail. 28 R. L. Devaney The Julia set for Eλundergoes a remarkable transformation as λpasses through 1/e. We will show in Section 1 that J(Eλ)isaCantor bouquet for 0 <λ≤1/e. Roughly speaking, a Cantor bouquet has the property that all points in the set lie on a curve homeomorphic to a closed half line. Each of these curves in J(Eλ) extend to ∞in the right half-plane. All repelling periodic points and points with bounded orbits lie at the endpoints of the curves, while points that do not lie at the endpoints have unbounded orbits. Since repelling periodic points are dense in J(Eλ), the endpoints of the Cantor bouquet must be dense in J(Eλ). Indeed, we will show that the set of endpoints is a totally disconnected set, but that the set of endpoints together with the point at infinity forms a connected set. At λ=1/e,Eλundergoes a simple saddle-node bifurcation. An attracting fixed point merges with a repelling fixed point at this λ-value, producing a neutral fixed point. When λ>1/e, this neutral fixed point gives way to a pair of repelling fixed points. This apparently simple bifurcation has profound global ramifications. When λ≤1/e, we will show that the Cantor bouquet that forms the Julia set is a nowhere dense subset of the right half plane. However, when λ>1/e,J(Eλ) suddenly becomes the whole plane. No new repelling periodic points (except the two fixed points involved in the saddle-node) are born in this bifurcation; all simply move continuously as λcrosses through 1/e. Yet somehow, as soon as λexceeds 1/e, the repelling periodic points become dense in C. At this bifurcation, the attracting fixed point and its entire basin of attraction disappear. Most of the structure of the Cantor bouquet remains in the Julia set. However, a new and interesting topological invariant set arises. We will show that this set is an indecomposable continuum on which most orbits cycle toward the orbit of 0 and ∞. 2. Exponential Dynamics As in the often-studied quadratic family Qc(z)=z2+c, it is the orbit of 0 that plays a crucial role in determining the dynamics of Eλ. For the exponential family, 0 is an asymptotic value rather than a critical point. Nevertheless, any stable domain in the complement of the Julia set of Eλmust be associated with the orbit of 0 in the following sense: Theorem 2.1. Suppose Eλhas an attracting or parabolic periodic point. Then En λ(0) must tend to the attracting or parabolic cycle. If, on the other hand, En λ(0) →∞, then J(Eλ)=C. Dynamics of complex exponentials 29 The proof of the first statement in this theorem is a classical fact that goes back to Fatou. The second follows from the Sullivan No Wandering Domains Theorem [Su], as extended to the case of the exponential by Goldberg and Keen [GK] and Eremenko and Lyubich [EL]. Rather than rely on this big machinery, we will give a bare-hands approach due to Misiurewicz [Mi] to show that J(Eλ)=Cwhen λ>1/e in Section 5. The exponential family undergoes a saddle node bifurcation at λ=1/e since, when λ=1/e, the graph of E1/e is tangent to the diagonal at 1. See Figure 1. We have E1/e(1) = 1 and E 1/e(1) = 1. When λ>1/e, the graph of Eλlies above the diagonal and all orbits (including 0) tend to ∞. When λ<1/e, the graph of Eλcrosses the diagonal twice, at an attracting fixed point aλand a repeling fixed point rλ. For later use note that 0 <a λ<1<r λ. Note also that the orbit of 0 tends to aλ,as it must by Fatou’s theorem. See Figure 1. Eλ(x) 2 1 λ=1/e λ<1/e −1aλνλrλ x 1 Figure 1. The graphs of Eλfor λ=1/e and λ<1/e. 3. Cantor Bouquets In this section, we begin the study of the dynamics of Eλby considering the case where λ≤1/e. We show here that J(Eλ)isaCantor bouquet. In Figure 2, we display the Julia set for E1/e. The complement of the Julia set is displayed in black. It appears that this Julia set contains large open sets, but this in fact is not the case. The Julia set actually consists of uncountably many curves or “hairs” extending to ∞in the right half plane. Each of the “fingers” in this figure seems to have many smaller fingers protruding from them. As we zoom in to this image, we see more and more of the self-similar structure, as each finger generates 30 R. L. Devaney more and more fingers. In fact, each of these fingers consists of a cluster of hairs that are packed together so tightly that the resulting set has Hausdorff dimension 2. Figure 2. The Julia set for λ=1/e. Figure 3. Magnification of the Julia set for λ=1/e. Dynamics of complex exponentials 31 3.1. The Idea of the Construction. Here is a rough idea of the construction of a Cantor bouquet. We will “tighten up” these ideas in following sections. Let E(z)=(1/e)ez. We have E(1) = 1 and E(1) = 1. If x0∈R and x<1, then En(x0) tends to the fixed point at 1. If x0>1, then En(x0)→∞as n→∞. This can be shown using the web diagram as shown in Figure 4. E(x) 2 1y=x −1 x 12 Figure 4. The graph of E(x)=(1/e)ex. The vertical line Re z= 1 is mapped to the circle of radius 1 centered at the origin. In fact, Eis a contraction in the half plane Hto the left of this line, since |E(z)|=1 eexp(Re z)<1 if z∈H. Consequently, all points in Hhave orbits that tend to 1. Hence this half plane lies in the stable set, i.e., in the complement of the Julia set. We will try to paint the picture of the Julia set of Eby painting instead its complement. Since the half plane His forward invariant under E, we can obtain the entire stable set by considering all preimages of this half plane. Now the first preimage of Hcertainly contains the horizontal lines Im z= (2k+1)π,Rez≥1, for each integer k, since Emaps these lines to the negative real axis which lies in H. Hence there are open neighborhoods of each of these lines that lie in the stable set. The first preimage of H is shown in Figure 5. The complement of E−1(H) consists of infinitely 32 R. L. Devaney many “fingers”. The fingers are 2kπi translates of each other, and each is mapped onto the complementary half plane Re z≥1. HC1 C0 C−1 Figure 5. The preimage of Hconsists of Hand the shaded region. We denote the fingers in the complement of E−1(H)byCjwith j∈Z, where Cjcontains the half line Im z=2jπ,Rez≥1, which is mapped into the positive real axis. That is, the Cjare indexed by the integers in order of increasing imaginary part. Note that Cjis contained within the strip −π 2+2jπ ≤Im z≤π 2+2jπ. Now each Cjis mapped in one-to-one fashion onto the entire half plane Re z≥1. Consequently each Cjcontains a preimage of each other Ck. Each of these preimages forms a subfinger which extends to the right. See Figure 6. The complement of these subfingers necessarily lies in the stable set. Now we continue inductively. Each subfinger is mapped onto one of the original fingers by E. Consequently, there are infinitely many sub-subfingers which are mapped to the Cj’s by E2. So at each stage we remove the complement of infinitely many subfingers from each remaining finger. This process is reminiscent of the construction of the Cantor set in the dynamics of polynomials when all critical points tend to ∞. In that construction, the complements of disks are removed at each stage; here we remove the complement of infinitely many fingers. As a result, after performing this operation infinitely many times, we do not end up with points. Rather, as we will see, the intersection of all of these fingers is a simple curve extending to ∞. Dynamics of complex exponentials 33 Figure 6. The second preimage of Hin one of the fingers Cj. This collection of curves forms the Julia set. Epermutes these curves and each curve consists of a well-defined endpoint together with a “hair” which extends to ∞. It is tempting to think of this structure as a “Cantor set of curves”, i.e., a product of the set of endpoints and the half-line. However, this is not the case as the set of endpoints is not closed. Note that we can assign symbolic sequences to each point on these curves. We simply watch which of the Cj’s these orbit of the point lies in after each iteration and assign the corresponding index j. That is, to each hair in the Julia set we attach an infinite sequence s0s1s2... where sj∈Zand sj=kif the jth iterate of the hair lies in Ck. The sequence s0s1s2... is called the itinerary of the curve. For example, the portion of the real line {x|x≥1}is an invariant curve in the Julia set since all points (except 1) tend to ∞under iteration, not to the fixed point. These points all have itinerary 000 ... . One temptation is to say that there is a hair corresponding to every possible sequence s0s1s2... . This, unfortunately, is not true, as certain sequences simply grow too quickly to correspond to orbits of E. So this is J(E): a “hairy” object extending toward ∞in the righthalf plane. We call this object a Cantor bouquet. We will see that this bouquet has some rather interesting topological properties as we investigate further. 3.2. Straight Brushes. To describe the structure of a Cantor bouquet, we need to introduce the notion of a straight brush. 34 R. L. Devaney To each irrational number ζ, we assign an infinite string of integers n0n1n2... as follows. We will break up the real line into open intervals In0n1...nkwhich have the following properties 1. In0...nkstrictly contains In0...nk+1 . 2. The endpoints of In0...nkare rational. 3. ζ=∞ k=1 In0...nk. Now there are many ways to do this. We choose the following inductive method based on the Farey tree. For any k∈Z, we first define Ik= (k,k+1). Given In0...nkand j∈Z, we define In0...nkjas follows. Suppose In0...nk=α β,γ δ and let p0/q0=(α+γ)/(β+δ), the Farey child of α/β and γ/δ. Let pn/qn be the Farey child of pn−1/qn−1and γ/δ for n>0, and let pn−1/qn−1 be the Farey child for α/β and pn/qnfor n≤0. We then set In0...nkjto be the open interval (pj/qj,p j+1/qj+1). Example. I0=(0,1). The Farey child of 0/1 and 1/1is1/2, so p0/q0=1/2. Then p1/q1=1 2⊕1 1=2/3, p2/q2=2 3⊕1 1=3 4, and pn/qn=n+1/n + 2 for n>0. For the remaining nwe have p−1/q−1=0 1⊕1 2=1 3 p−2/q−2=0 1⊕1 3=1 4 p−n/q−n=1 n+2. Therefore, if n≥0, I0n=n+1 n+2,n+2 n+3 and if n<0, I0n=1 −n+2,1 −n+1. Dynamics of complex exponentials 35 See Figure 7. Note that we exhaust all of the rationals via this procedure, so each irrational is contained in a unique In0n1.... Note that the endpoints of the successive intervals In0,I n0n1,... correspond to the usual sequence of convergents in the continued fraction expansion of ξ. 1/5 1/40/1 1/3 1/2 2/3 3/4 4/5 1/1 I0−3I0−2I0−1I00 I01 I02 Figure 7. Construction of I0n. We now define a straight brush, a notion due to Aarts and Oversteegen [AO]. Definition 3.1. A straight brush Bis a subset of [0,∞)×N, where Nis a dense subset of irrationals and Bhas the following 3 properties. 1. Bis “hairy” in the following sense. If (y,α)∈B, then there exists ayα≤ysuch that (t, α)∈Biff t≥yα. That is, the “hair” (t, α) is contained in Bwhere t≥yα.yαis called the endpoint of the hair corresponding to α. 2. Given an endpoint (yα,α)∈Bthere are sequences βnαand γnαin Nsuch that (yβn,β n)→(yα,α) and (yγn,γ n)→ (yα,α). That is, any endpoint of a hair in Bis the limit of endpoints of other hairs from both above and below. 3. Bis a closed subset of R2. The following facts are easily verified: 1. For any rational number vand any sequence of irrationals αn∈N with αn→v, show that the hairs [yαn,α n] must tend to [∞,v]in {∞} × R. 2. Condition 2 above may be changed to: if (y,α)isanypointinB (yneed not be the endpoint of the α-hair), then there are sequences βnα,γnαso that (yβn,β n)→(y,α) and (yγn,γ n)→(y,α)inB. 3. Let (y,α)∈Band suppose yis not the endpoint yα. Then (y,α) is inaccessible from R2\Bin the sense that there is no continuous curve γ:[0,1] →R2such that γ(t)/∈Bfor 0 ≤t<1 and γ(1) = (y,α). 4. On the other hand, the endpoint (yα,α) is accessible from R2\B. 42 R. L. Devaney Theorem 4.1. Γis an indecomposable continuum. Moreover, we will see that Λ is constructed in similar fashion to a family of indecomposable continua known as Knaster continua. As we will show in Section 4.2, the topology of Λ is quite intricate. Despite this, we will show that the dynamics of Eλon Λ is quite tame. Specifically, we will prove: Theorem 4.2. The restriction of Eλto Λ−{orbit of 0}is a homeomorphism. This map has a unique repelling fixed point wλ∈Λ, and the α-limit set of all points in Λis wλ. On the other hand, if z∈Λ,z=wλ, then the ω-limit set of zis either 1. The point at ∞,or 2. The orbit of 0under Eλtogether with the point at ∞. Thus we see that Eλpossesses an interesting mixture of topology and dynamics in the case where the Julia set is the whole plane. In the plane the dynamics of Eλare quite chaotic, but the overall topology is tame. On our invariant set Λ, however, it is the topology that is rich, but the dynamics are tame. 4.1. Topological Preliminaries. In this section we review some of the basic topological ideas associated with indecomposable continua. See [Ku] for a more extensive introduction to these concepts. Recall that a continuum is a compact, connected space. A continuum is decomposable if it is the union of two proper subcontinua. Otherwise, it is indecomposable. One famous example of an indecomposable continuum is the Knaster continuum, K. One way to construct this set is to begin with the Cantor middle-thirds set. Then draw the semi-circles lying in the upper half plane with center at (1/2,0) that connect each pair of points in the Cantor set that are equidistant from 1/2. Next draw all semicircles in the lower half plane which have for each n≥1 centers at (5/(2 ·3n),0) and pass through each point in the Cantor set lying in the interval 2/3n≤x≤1/3n−1. The resulting set is partially depicted in Figure 12. Dynamics of complex exponentials 43 Figure 12. The Knaster Continuum. For a proof that this set is indecomposable, we refer to [Ku]. Dynamically, this set appears as the closure of the unstable manifold of Smale’s horseshoe map (see [Ba], [Sm]). Note that the curve passing through the origin in this set is dense, since it passes through each of the endpoints of the Cantor set. It also accumulates everywhere upon itself. Such a phenomenon gives a criterion for a continuum to be indecomposable, as was shown by S. Curry. Theorem 4.3. Suppose Xis a one-dimensional nonseparating plane continuum which is the closure of a ray that limits upon itself. Then X is indecomposable. We refer to [Cu] for a proof. Another view of the Knaster continuum which is intimately related to our own construction is as follows. Begin with the unit square S0in the plane. Next remove a “canal” C1from S0whose boundary lies within a distance 1/3 of each boundary point of S0as depicted in Figure 2. Call this set S1. Next remove a new canal C2from S1. This time the boundary of C2should be within 1/9 of the boundary of S1as depicted in Figure 13. It is possible to continue this construction inductively in such a way that the resulting set is homeomorphic to the Knaster continuum. 44 R. L. Devaney Figure 13. A different construction of the Knaster Continuum. 4.2. Construction of Λ. Recall that the strip Sis given by {z|0≤Im(z)≤π}. Note that Eλ maps Sin one-to-one fashion onto {z|Im z≥0}−{0}. Hence a branch of E−1 λis well defined taking S−{0}to S. In fact, E−n λis defined for all non S−{orbit of 0}. We will always assume that E−n λmeans E−n λ restricted to this subset of S. Define Λ={z|En λ(z)∈Sfor all n≥0}. If z∈Λ it follows immediately that En λ(z)∈Sfor all n∈Zprovided zdoes not lie on the forward orbit of 0. Our goal is to understand the structure of Λ. Toward that end we define Lnto be the set of points in Sthat leave Sat precisely the nth iteration of Eλ. That is, Ln={z∈S|Ei λ(z)∈Sfor 0 ≤i<nbut En λ(z)/∈S}. Let Bnbe the boundary of Ln. Recall that Eλmaps a vertical segment in Sto a semi-circle in the upper half plane centered at 0 with endpoints in R. Either this semicircle is completely contained in Sor else an open arc lies outside S. As a consequence, L1is an open simply connected region which extends to ∞toward the right in Sas shown in Figure 14. There is a natural parametrization γ1:R→B1defined by Eλ(γ1(t)) = t+iπ. As a consequence, lim t→±∞ Re γ1(t)=∞. Dynamics of complex exponentials 45 L3 L2L1 Figure 14. Construction of the Ln. If c>0 is large, the segment Re z=cin Smeets S−L1in two vertical segments v+and v−with Im v−>Im v+.Eλmaps v−to an arc of a circle in S∩{z|Re z<0}while Eλmaps v+to an arc of a circle in S∩{z|Re z>0}. As a consequence, if cis large, v+meets L2in an open interval. Since L2=E−1 λ(L1), it follows that L2is an open simply connected subset of Sthat extends to ∞in the right half plane below L1. Continuing inductively, we see that Lnis an open, simply connected subset of Sthat extends to ∞toward the right in S. We may also parametrize the boundary Bnof Lnby γn:R→Bnwhere En λ(γn(t)) = t+iπ as before. Again lim t→±∞ Re γn(t)=∞. Since each Lnis open, it follows that Λ is a closed subset of S. Proposition 4.4. The set ∞ i=nBiis dense in Λfor each n>0. Proof: Let z∈Λ and suppose z/∈Bifor any i. Let Ube an open connected neighborhood of z.Fixn>0. Since Ei λ(z)∈Sfor all i,we may choose a connected neighborhood V⊂Uof zsuch that Ei λ(V)⊂S for i=0,... ,n. Now the family of functions {Ei λ}is not normal on V, since zbelongs to the Julia set of Eλ. Consequently, ∞ i=0 Ei λ(V) covers C−{0}.In particular, there is m>nsuch that Em λ(V) meets the exterior of S. Since Em λ(z)∈S, it follows that Em λ(V) meets the boundary of S. Applying E−m λ, we see that Bmmeets V. 46 R. L. Devaney In fact, it follows that for any z∈Λ and any neighborhood Uof z, all but finitely many of the Bmmeet V. This follows from the fact that Eλ has fixed points outside of S(in fact one such point in each horizontal strip of width 2π—see [DK]), so we may assume that Em λ(V) contains this fixed point for all sufficiently large m. In particular, we have shown: Proposition 4.5. Let z∈Λand suppose that Vis any connected neighborhood of z. Then Em λ(V)meets the boundary of Sfor all sufficiently large m. Proposition 4.6. Λis a connected subset of S. Proof: Let Gbe the union of the boundaries of the Lifor all i. Since Λ is the closure of G, it suffices to show that Gis connected. Suppose that this is not true. Then we can write Gas the union of two disjoint sets A and B. One of Aor Bmust contain infinitely many of the boundaries of the Li.SayAdoes. But then, if b∈B, the previous proposition guarantees that infinitely many of these boundaries meet any neighborhood of b. Hence bbelongs to the closure of A. This contradiction establishes the result. We can now prove: Theorem 4.7. There is a natural compactification Γof Λthat makes Γinto an indecomposable continuum. Proof: We first compactify Λ by adjoining the backward orbit of 0. To do this we identify the “points” (−∞,0) and (−∞,π)inS: this gives E−1 λ(0). We then identify the points (∞,π) and limt→−∞ γ1(t). This gives E−2 λ(0). For each n>1 we identify lim t→∞ γn(t) and lim t→−∞ γn+1(t) to yield E−n−1 λ(0). This augmented space Γ may easily be embedded in the plane. See Figure 15. Moreover, if we extend the Biand the lines y= 0 and y=πin the natural way to include these new points, then this yields a curve which accumulates everywhere on itself but does not separate the plane. See the proposition above. By a theorem of S. Curry [Cu], it follows that Γ is indecomposable. Dynamics of complex exponentials 47 L3 L2L1 Figure 15. Embedding Γ in the plane.  As a consequence of this theorem, Λ must contain uncountably many composants (see [Ku, p. 213]). In fact, in [DK] it is shown that Λ contains uncountably many curves. 4.3. Dynamics on Λ. In this section we describe completely the dynamics of Eλon Λ. Proposition 4.8. There exists a unique fixed point wλin Sif λ>1/e. Moreover, wλis repelling and, if z∈S−orbit of 0,E−n λ(z)→ wλas n→∞. Proof: First consider the equation λeycot ysin y=y. Since ycot y→1asy→0 and λe > 1, we have λeycot ysin y>yfor ysmall and positive. Since the left-hand side of this equation vanishes when y=π, it follows that this equation has at least one solution yλin the interval 0 <y<π. Let xλ=yλcot yλ. Then one may easily check that wλ=xλ+iyλ is a fixed point for Eλin the interior of S. Since the interior of S is conformally equivalent to a disk and E−1 λis holomorphic, it follows from the Schwarz Lemma that wλis an attracting fixed point for the restriction of E−1 λto Sand that E−n λ(z)→wλfor all z∈S. 48 R. L. Devaney Remarks. 1. Thus the α-limit set of any point in Λ is wλ. 2. The bound λ>1/e is necessary for this result, since we know that Eλhas two fixed points on the real axis for any positive λ<1/e. These fixed points coalesce at 1 as λ→1/e and then separate into a pair of conjugate fixed points, one of which lies in S. We now describe the ω-limit set of any point in Λ. Clearly, if z∈Bn then En+1 λ(z)∈Rand so the ω-limit set of zis infinity. Thus we need only consider points in Λ that do not lie in Bn. We will show: Theorem 4.9. Suppose z∈Λand z=wλ,z/∈Bnfor any n. Then the ω-limit set of zis the orbit of 0under Eλtogether with the point at infinity. To prove this we first need a lemma. Lemma 4.10. Suppose z∈Λ,z=wλ. Then En λ(z)approaches the boundary of Sas n→∞. Proof: Let hbe the uniformization of the interior of Staking Sto the open unit disk and wλto 0. Recall that E−1 λis well defined on S and takes Sinside itself. Then g=h◦E−1 λ◦h−1is an analytic map of the open disk into itself with a fixed point at 0. This fixed point is therefore attracting by the Schwarz Lemma. Moreover, if |z|>0we have |g(z)|<|z|. As a consequence, if {zn}is an orbit in Λ, we have |h(zn+1)|>|h(zn)|, and so |h(zn)|→1asn→∞. The remainder of the proof is essentially contained in [DK] (see pp. 45– 49). In that paper it is shown that there is a “quadrilateral” Qcontaining a neighborhood of 0 in Ras depicted in Figure 5. The set Qhas the following properties: 1. If z∈Λ−nBnand z=wλ, then the forward orbit of zmeets Qinfinitely often. 2. Qcontains infinitely many closed “rectangles” Rk,R k+1,R k+2,... for some k>1 having the property that if z∈Rj, then Ej λ(z)∈Q but Ei λ(z)/∈Qfor 0 <i<j. 3. If z/∈∞ j=kRj, then z∈Lnfor some n. 4. Ej λ(Rj) is a “horseshoe” shaped region lying below Rjin Qas depicted in Figure 5. 5. limj→∞ Ej λ(Rj)={0}. Dynamics of complex exponentials 49 Q Rj Rj+k Ej λ(Rj) Figure 16. The return map on Q. As a consequence of these facts, any point in Λ has orbit that meets the ∪Rjinfinitely often. We may thus define a return map Φ: Λ ∩(∪jRj)→Λ∩(∪jRj) by Φ(z)=Ej λ(z) if z∈Rj. By item 4, Φ(z) lies in some Rkwith k>j. By item 5, it follows that Φn(z)→0 for any z∈Λ∩Q. Consequently, the ω-limit set of zcontains the orbit of 0 and infinity. For the opposite containment, suppose that the forward orbit of z accumulates on a point q. By the Lemma, qlies in the boundary of S. Now the orbit of qmust also accumulate on the preimages of q.Ifq does not lie on the orbit of 0, then these preimages form an infinite set, and some points in this set lie on the boundaries of the Ln. But these points lie in the interior of S, and this contradicts the Lemma. Thus the orbit of zcan only accumulate in the finite plane on points on the orbit of 0. Since the “preimage” of 0 is infinity, the orbit also accumulates at infinity. It is known that there are uncountably many curves in the λ-plane having the property that, if λlies on one of these curves, then En λ(0) →∞. Consequently, for such a λ-value, the Julia set of Eλis again the complex plane. For these λ-values, a variant of the above construction also yields invariant indecomposable continua in the Julia set. Whether these continua are homeomorphic to any of those constructed above is an open question. We plan to discuss these constructions in a later paper. 50 R. L. Devaney Douady and Goldberg [DoG] have shown that if λ, µ > 1/e, then Eλand Eµare not topologically conjugate. Each such map possesses invariant indecomposable continua Λλand Λµin S, and the dynamics on each are similar, as shown above. In fact, one can show that each pair of these invariant sets is non-homeomorphic. As a final remark, M. Lyubich has shown that each Λλis a set of measure 0 in S. Indeed, it follows from his work [Ly] that the set of points in Cwhose orbits have arguments that are equidistributed on the unit circle have full measure. In Λλ, the arguments of all orbits tend to 0 and/or π, and so Λλhas measure 0 in S. 5. After the Explosion As we have mentioned, when λ>1/e, the Julia set of Eλis the entire plane. In 1981, Misiurewicz showed that J(E1)=C, answering a sixty-year-old question of Fatou. We present his proof of this fact below, generalizing it to the case λ>1/e. The following proposition highlights one of the differences between Eλ(z) and polynomials: points which tend to ∞under iteration of Eλ need not be in the stable set. Proposition 5.1. The real line is contained in J(Eλ)and hence all preimages of the real line lie in J(Eλ). Proof: Let Sdenote the strip |Im(z)|≤π. Suppose Ej λ(z)∈R. Hence En λ(z)→∞. Let Ube a neighborhood of z. Then Ei λ(U) meets the real line for all sufficiently large i. Drawing on the results of the previous section, there are points arbitrarily close to Ei λ(z) whose images eventually lie in the far left half plane, and so their next images lie in the unit disk about 0. Thus the family of iterates {En λ}is not a normal family on Uand so z∈J(Eλ). Thus to show that J(Eλ)=C, it suffices to show that inverse images of the real line are dense in C. For this, we need several lemmas. Lemma 5.2. |Im(En λ(z))|≤|(En λ)(z)|. Proof: If z∈Rthe inequality is trivial. Hence we assume z/∈R.If z=x+iy, we have |Im(Eλ(z))|=λex|sin y| ≤λex|y| =|E λ(z)|| Im(z)| Dynamics of complex exponentials 51 so that |Im(Eλ(z))| |Im(z)|≤|E λ(z)| since z/∈R. More generally, if En λ(z)/∈R, we may apply this inequality repeatedly to find |Im(En λ(z))| |Im(Eλ(z))|= n−1  i=1 |Im Eλ(Ei λ(z))| |Im(Ei λ(z))| ≤ n−1  i=1 |E λ(Ei λ(z))|. Since |Im(Eλ(z))|≤|Eλ(z)|=|E λ(z)|we may write |Im(En λ(z))|≤ n−1  i=0 |E λ(Ei λ(z))| =|(En λ)(z)|. Now let W={z||Im z|≤π/3}. We have seen that the orbits of an open set of points in W⊂Sleave Sand hence Wunder iteration. The next lemma shows, however, that the orbits of most of these points must eventually return. Lemma 5.3. Let Ube an open connected set. Then only finitely many of the En λ(U)can be disjoint from W. Proof: Let us assume that infinitely many of the images of Uare disjoint from W. If there is an nfor which En λis not a homeomorphism taking Uonto its image, then there exist z1,z 2∈U,z1=z2, for which En λ(z1)=En λ(z2). Consequently, there is a jfor which Ej λ(z1)=Ej λ(z2)+2kπi for some k∈Z−{0}. But then Ej λ(U) must meet a horizontal line of the form y=2mπ for m∈Zand so Ej+1 λ(U) meets R. Hence Ej+α λ(U) meets Rfor all α>0 and only finitely many of the images of Ucan be disjoint from W. We thus conclude that each En λmust be a homeomorphism on U. Now suppose there is a sequence njsuch that for each j,Enj λ(U)∩W= φ. By the previous lemma, |(Enj λ)(z)|≥(π/3)njfor each jand all z∈U. It follows that, if Ucontains a disk of radius δ>0, then Enj λ(U) contains a disk of radius δ(π/3)nj. Hence for jlarge enough, Enj λ(U) must meet a line of the form y=2πand again we are done. We can now prove