b 0 + b 1 z b d z d

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1 I. Introduction Definition 1. For z C, a rational function of degree d is any with a d, b d not both equal to 0. R(z) = P (z) Q(z) = a 0 + a 1 z a d z d b 0 + b 1 z b d z d It is exactly the rational functions which are meromorphic throughout C. More precisely, if R is a rational function of positive degree d, then R is a d-fold map of C onto itself, i.e., for any w C, the equation R(z) = w has d solutions (counting multiplicities). In general, we will be concerned with sequences which consist of the image of a point under repeated application of a rational map: (1) Choose z 0 C. (2) Define a sequence {z n } n=1 = {R n (z 0 )} n=1, so we obtain z 0 z 1 = R(z 0 ) z 2 = R(z 1 ) = R(R(z 0 )) = R 2 (z 0 ). z n = R n (z 0 ) The central idea in iteration theory pertains to the sensitivity of R to the initial conditions; if w 0 B ε (z 0 ), then what about {w n }? For example, if we wish to investigate {z n } for z 0 = 2, will it be at all useful to consider {R n (1.414)}? The answer clearly depends on the choice of z 0. Accordingly, we divide C into two sets. Definition 2. For a given rational function R, the Fatou set F is the set for which, if z 0 is sufficiently close to w 0, then z n and w n will have similar behavior. The Julia set is the complement of the Fatou set, and it consists of those points z 0 whose near neighbours w 0 have iterates w n bearing no relation to the z n.

2 II. Conjugacy Definition 3. Two rational maps S, R are conjugate iff there is some Möbius map g(z) = az + b cz + d such that S = grg 1. In this case, S R. If S R then deg S = deg R. If S R then S n R n. If S R then S fixes g(z) R fixes z. Classify the polynomials within the class of rational maps: Let R be a nonconstant rational map. Then R is a polynomial iff R has a pole at and nowhere else. ( ) A polynomial is continuous on C, so clear. ( ) If R has exactly the pole, then R(z) = P (z) Q(z) = Q(z) = c. A nonconstant rat map is a poly iff R 1 { } = { }. A nonconstant rational map R is conjugate to a polynomial iff w C with R 1 {w} = {w}. Conjugation helps deal with, and frequently reduces the amount of work required when proving theorems. We will be primarily concerned with conjugation-invariant properties of maps. III. Valency Definition 4. For any f that is non-constant and holomorphic near z 0 C, f has a Taylor expansion at z 0 : f(z) = a 0 + a k (z z 0 ) k +... where a k 0 and k N + is uniquely determined by the existence of the finite and nonzero limit f(z) f(z 0 ) lim. z z 0 (z z 0 ) k

3 This k is denoted v f (z 0 ) and is the valency of f at z 0. For rational functions, we write f(z) f(z 0 ) = R(z) = P (z) Q(z) = (z p 1)(z p 2 ) (z p d ) (z q 1 )(z q 2 ) (z q d ), where the polynomials split completely because C is a algebraically complete. Then the valency of f is the number of times the linear factor (z f(z 0 )) appears in the top, i.e., the number of p j for which p j = f(z 0 ). v f (z 0 ) is the number of solutions of f(z) = f(z 0 ) at z 0. f is injective near z 0 iff v f (z 0 ) = 1. v f (z) satisfies the chain rule: v fg (z) = v f (g(z))v g (z). The fundamental fact that any non-constant rational map R of degree d is an d-fold map of C onto itself is easily expressible in terms of valency. For any z 0 R 1 {w}, v R (z 0 ) is the number of solutions of R(z) = w at z 0, and so for each w C, v R (z) = deg(r). z R 1 {w} Note: univalent = injective. Univalent implies v f (z) = 1, but not vice versa. Example: z 2 : C 0 C 0 has valency 1 everywhere but is not injective. IV. Fixed Points Of primary interest will be the fixed points of R, as well as the behavior near these fixed points. For example, suppose that for some z 0 C, we have lim n z n = w. Then w = lim z n+1 n (lim z n = lim z n+1 ) = lim R(z n ) n (defn of z n ) ( = R = R(w) lim n z n so that w is a fixed point of R. ) (R is continuous at w) Definition 5. If ζ C is a fixed point of R, then R (ζ) is defined and describes the behavior of R near ζ. ζ is:

4 (1) an attracting fixed point iff R (ζ) < 1, (2) a repelling fixed point iff R (ζ) > 1, and (3) a indifferent fixed point iff R (ζ) = 1. To justify this terminology, consider that for z very close to ζ, we can make the approximation R(z) ζ = R(z) R(ζ) R (ζ) z ζ. (1) Thus, ζ is an attracting fixed point iff the distance from R(z) to ζ is less than the distance from z to ζ, etc. The behavior of iterates of rational functions can be (and generally is) extremely complex. While for z 0 sufficiently close to an attracting fixed point ζ, we will have z n ζ; the behavior of z near a repelling fixed point ζ may be much more difficult to predict. Although z will initially be repelled away from ζ, it may return to the vicinity of ζ, and z n may in fact actually converge to ζ. Example. For R(z) = 3z 2 2z 1, consider z 0 = 1 ε. R 1 (z) = (2z 1) = 1 2 z n moves away from 1 but eventually converges to it. Proposition 6. The only way for z n to converge to a repelling fixed point ζ is to have z n = ζ, n N, where N is some number. n Proof. Suppose that z n ζ, but z n ζ for any n. Then for infinitely many n, we have z n+1 ζ < z n ζ. (2) On the other hand, since ζ is a repelling fixed point, we have R (ζ) > 1. Hence we can pick k such that R (ζ) > k > 1, and a neighbourhood B ε (ζ) such that z B ε (ζ) = R(z) ζ = R(z) R(ζ) > k z ζ, using the approximation (1). Putting z = z n, we obtain contradicting (2). R(z) ζ = z n+1 ζ > z ζ,

5 IV.1. Locating the fixed points of a rational map. Let R = P/Q be a nonconstant rational function, where P and Q are coprime polynomials. R( ) = deg P > deg Q ζ, R(ζ) = ζ = Q(ζ) 0 P (ζ) = ζq(ζ) So the fixed points of R in C are the solutions of if they exist. P (z) zq(z) = 0, Example. For example, z z + z 1 = z2 +1 z has no fixed points in C. IV.2. Counting fixed points of a rational map. If we count fixed points in the same way that we count the zeros of an analytic map, then: Theorem 7. A rational map R of degree d 1 has precisely d + 1 fixed points in C. In fact, for ζ : f has k fixed points at ζ f(z) z has k zeros at ζ. Proof. Need definitions about fixed points at, via conjugacy. Also, need that grg 1 has same number of fixed points at g(ζ) as R has at ζ. Any rat map R is conj to a rat map S which does not fix. The number of f.p. s of S is the same as for R, as is their degree. Hence, it suffices to consider R which do not fix. Let ζ be a f.p. of R = P/Q. As Q(ζ) 0, the number of zeros of R(z) z is exactly the same as the number of zeros of P (z) zq(z) at ζ. Hence the number of f.p. of R is exactly the number of solutions of P (z) = zq(z) in C. Since R does not fix, we have Thus, deg(p ) deg(q) = deg(r). deg (P (z) zq(z)) = deg(zq(z)) = deg(zr(z)) = deg(r) + 1.

6 Example. z + z 3 has three fixed points at the origin and valency 1 there, whereas z 3 has one fixed point at the origin and valency 3 there. R(z) = z + z 3 = P (z) = z + z 3, Q(z) = 1 f.p.(0) : P (z) zq(z) = z 3 = zzz = 3 f.p. at 0 v R (0) : R(0) = = 0 = R(z) R(0) = z + z 3 = z(1 + z 2 ) 1 R(z) = z 3 = P (z) = z 3, Q(z) = 1 f.p.(0) : P (z) zq(z) = z 3 z = (z 2 1)z = 1 f.p. at 0 v R (0) : R(0) = = 0 = R(z) R(0) = z 3 = zzz 3 V. Critical Points Definition 8. z is a critical point of R iff R fails to be injective in any neighbourhood of z. If R is not constant, z critical v R (z) > 1. w is a critical value of R iff R(z) = w for some z. deg(r) = d, w not a critical value = R 1 {w} = {z 1,..., z d } Significance: {z 1,..., z d } are distinct. Would have {z 1,..., z k }, k < d if w were critical. Since none of these z j are critical, there are neighbourhoods N of w and N j of each z j with bijections R j := R Nj : N j N. Definition 9. R 1 j are the branches of R 1 at w. R is injective when R has neither a zero nor a pole, so v R (z) = 1 for all but a finite set of z. Thus, (v R (z) 1) < +. z C This sum gives a measure of the number of multiple roots of R, and hence also the difficulty involved in defining R 1. In fact, we have

7 Theorem 10. (Riemann-Hurwitz relation) For any nonconstant rational map R : C C, (v R (z) 1) = 2 deg(r) 2. z C This may be proved using topological techniques (Euler characteristic) and may be adapted for any analytic map between compact Riemann surfaces. The general term in the sum is positive only when z is a critical point. For a nonconstant polynomial P, v P ( ) = deg(p ), hence: Corollary 11. A rational map of degree d has at most 2d 2 critical points in C. A polynomial of positive degree d 1 has at most d 1 critical points in C. VI. Rational Functions of Degree 1 A rational function of degree 1 looks like R(z) = az + b cz + d, with the usual convention that c 0 = R( ) = a c ad bc 0, a, b, c, d C and R ( d ) =, c but R( ) = when c = 0. Thus, the rational functions of degree 1 are the Möbius functions: a collection of conformal mappings whose iterates are relatively simple (though not entirely trivial) and may be computed explicitly. Example. R(z) = 3z 2 2z 1 = Rn (z) = (2n + 1)z 2n 2nz (2n 1) = 1 + z 1 2nz (2n 1). Any degree 1 rational map has a single (repeated) fixed point, or two distinct fixed points. az + b cz + d = z = az + b = cz2 + dz = cz 2 + (b a)z d = 0

8 case i) R has a single fixed point ζ. If ζ =, then R(z) = z + b, some b C 0. Thus R n (z) = z + nb = R n (z) n. If ζ, then let g(z) = 1/(z ζ) (A Möbius map taking ζ to ), and define S(z) = grg 1 (z). Now S fixes only, so S is a translation, and hence S n (z) as before. Also, S n (z) = (grg 1 )(grg 1 ) (grg 1 )(z) = gr n g 1 (z). Hence, replacing z by g(z) and applying g 1, S n = gr n g 1 = g 1 S n g = R n. Thus, R n n (z) g 1 ( ) = ζ. So if ζ is the lone fixed point of R, R n (z) n ζ, z C. n case ii) R has two distinct fixed points. First suppose that R fixes 0 and. Then R(z) = kz and R n (z) = k n z. Then R n fixes 0 and, and R n (z) 0 k < 1, R n (z) = z k = 1, R n (z) k > 1, When k = 1, either k is an n th root of unity, in which case R n = id, or else the points R n (z) are dense in S 1. Now let ζ 1 ζ 2 be the fixed points of R and take g(z) = z ζ 1 z ζ 2. This is so ζ 1 0 and ζ 2. If ζ 1, ζ 2 are not finite, then find something else. Then for S = grg 1, S fixes 0,. Since g maps lines/circles to lines/circles, R n n (z) ζ 1, ζ 2 or R n (z) moves cyclically through a finite set of points, or {R n (z)} is dense in some circle. So much for the rational functions of degree 1.

9 VII. The Fatou and Julia Sets Definition 12. Let (X, ρ) be a metric space. A family of maps F = {f : (X, ρ) (X, ρ)} is equicontinuous at x 0 iff ε > 0, δ > 0 such that ρ(x 0, x) < δ = ρ(f(x 0 ), f(x)) < ε, f F. So every function in F maps the open ball of radius δ into a ball of radius at most ε. Of course, F is equicontinuous on a subset iff it is equicontinuous at every point in that set. Note: F equicont on {D α } = F equicont on α D α. Theorem 13. Let (X, ρ) be a metric space and F = {f : X X}. Then there is a maximal open subset of X on which F is equicontinuous. Proof. Take {D α } to be the collection of open subsets of X on which F is equicontinuous. Definition 14. Let R be a nonconstant rational function. The Fatou set F = F (R) is the maximal open subset of C on which {R n } is equicontinuous. The Julia set J = J(R) is the complement of F in C. Note: F is open, and J is compact, by defn. Two important properties of F and J: Theorem 15. Let R be a nonconstant rational map and let g be a Möbius map. (1) For S = grg 1, F (S) = g(f (R)), J(S) = g(j(r)). (2) For p N, F (R p ) = F (R), J(R p ) = J(R). Proof. (1) A Möbius map g and its inverse each satisfy a Lipschitz condition w/r spherical or chordal metric on C. This allows equicontinuity to be transferred via conjugation in the obvious way. Proof. (2) Let S = R p. Since S = {S n. n 1} {R n. n 1} = R, S is equicontinuous wherever R is; thus F (R) F (S). Next, each R k sats a Lipschitz cond, so F k = {R k S n. n 0} is equicontinuous wherever S is. In partic, each F k is equicontinuous on the Fatou set F (S), and hence so too is the finite union F 0 F 1... F p 1 = R. Thus R is equicontinuous on F (S) and F (S) = F (R).

10 Definition 16. A family of maps F = {f : X X} is said to be normal or a normal family iff every infinite sequence of functions from F contains a subsequence which converges locally uniformly on X. f n : X Y converges locally uniformly on X iff each point of X has a neighbourhood on which f n f uniformly. In this case, the convergence is uniform on each compact subset of X. Theorem 17. (Arzela-Ascoli) Let D be a subdomain of the complex sphere, and let F be a family of continuous functions of D into the sphere. Then F is equicontinuous F is normal. Theorem 18. (The Mighty) D C is a domain. (1) Let Ω = C \{0, 1, }. Then the family F of all analytic maps {f : D Ω} is normal in D. (2) Let F be a family of maps, all analytic in D. Suppose that for each f there are a f, b f, c f C such that (a) f does not take the values a f, b f, c f, and (b) min{ρ(a f, b f ), ρ(a f, c f ), ρ(b f, c f )} m, where m > 0 is some fixed constant. Then F is normal in D. (3) Let ϕ 1, ϕ 2, ϕ 3 : D C be analytic and such that the closures of the images ϕ j (D) are disjoint. Let F = {f : D C } be analytic and such that f(z) ϕ j (z), z D, f F. Then F is normal in D. Proof. The sketch alone takes a full page. The actual proof takes 5 pages and a bunch of hyperbolic geometry. VIII. Invariant Sets Definition 19. If g : X X, then E X is (i) forward invariant if g(e) = E, (ii) backward invariant if g 1 (E) = E, (iii) completely invariant if g(e) = E = g 1 (E).

11 If g is surjective, then backward invariance and complete invariance coincide. Surjectivity gives g(g 1 (E)) = E. Then g 1 (E) = E gives g(e) = E. Since rational functions map C onto itself, backward invariance coincides with complete invariance for us. Since g 1 commutes with, the intersection of a family of completely invariant set is itself completely invariant. Thus, we can take any subset E 0 and form E, the smallest compl inv subset that contains E 0. We say E 0 generates E. Theorem 20. Let R be a rational map of degree d 2. If E is a finite set, E compl.inv. under R = E has at most 2 elements. Proof. Suppose E has k elements. Because E is finite, and R : E E, R must act as a permutation of E. For a suitable integer q, R q = id E. Suppose deg(r q ) = d. Then R q (z) = w has d solutions, for each w E. Then by the Riemann-Hurwitz relation applied to R q, (v R q(w) 1) = k(d 1) 2d 2. w E Dividing both sides by (d 1) yields k 2. Definition 21. We define an equivalence relation x y n, m N such that R n (x) = R m (y). Denote the equivalence class containing x by [x], call it the orbit of x. Theorem 22. [x] is the completely invariant set generated by the {x}. Proof. Messing about with set inclusions. Corollary 23. E completely invariant E = [x]. Corollary 24. E completely invariant X E, E 0, E, Ē are, too. Theorem 25. Let R be any rational map. Then F (R), J(R) are completely invariant.

12 Proof. Suffices to show F is backward invariant. Much messing around with ε and δ and the metric σ to show equicontinuity around points in R 1 (F ), etc. Corollary 26. Let P be a polynomial with deg(p ) 2. F (P ) and F, the component of F containing, is completely invariant under P. Proof. More messing around with ε and δ and the metric σ. Definition 27. z C is exceptional for R when [z] is finite. The set of the exceptional points of R is E(R). Theorem 28. deg(r) 2 = R has at most two exceptional points. Proof. Flawed. I mailed the author about this, still haven t heard back. Corollary 29. deg(r) 2 = E(R) F. Definition 30. The backward orbit of z is O (z) = {w. R n (w) = z, some n} = n 0 R n {z}. The points of O (z) are the predecessors of z. Theorem 31. O (z) is finite iff z is exceptional. Proof. ( ) Assume O (z) is finite. Define the nonempty sets B n by so R 1 (B n ) = B n+1, and B n = m n R m {z}, [z] O (z) = B 0 B 1 B 2. Each B n is finite, so there is some m with B m = B m+1. This means that R 1 (B m ) = B m and so B m is completely invariant. Then B m must contain some equivalence class [w], and as it is a subset of [z], it must be [z]. Thus, [z] is finite. ( ) As O (z) [z], any exceptional point has a finite backward orbit.

13 IX. Properties of the Julia Set Throughout the following, we take deg(r) 2. J(R) is infinite. Proof. If J is empty, then {R n } is normal on all of C. Then there is some subsequence of {R n } in which each map has the same degree (deg is continuous). Then deg(r n ) = (deg(r)) n = deg(r) = 1. So there must be some ζ J. J is completely invariant, so J finite = ζ is an exceptional point. But E(R) F. < (Minimality) If E is closed and completely invariant, then case i) E has at most two elements and E E(R) F (R), or case ii) E is infinite and J(R) E. Proof. Based on (18.2). Thus J is the smallest closed, completely invariant set with at least 3 points. Either J = C or J has empty interior. Proof. C = J o J F, each of which are completely invariant. If F, then F J is an infinite, closed, completely invariant set and so, by the minimality of J, it contains J. Thus J J. Example. For R(z) = (z2 +1) 2 4z(z 2 1), J(R) = C. 4pg proof + 3pg appendix. J is a perfect set, and hence is uncountable. Proof. Claim: J 0 (the accumulation pts of J) is infinite, closed, and completely invariant. Then by the minimality of J, J J 0, and hence J = J 0. So J has no isolated points. Pf of Claim: As J is infinite, J 0, and as a derived set, J 0 is automatically closed. Since R is continuous and of finite degree, R(J 0 ) J 0, and hence J 0 R 1 (J 0 ). Then since R is an open map, R 1 (J 0 ) J 0 and J 0 is completely invariant. <

14 Uncountability of a perfect set is a standard consequence of Baire s Theorem. Let W be any open set which meets J. Then (i) n=0r n (W ) C \E(R), and (ii) J R n (W ) for all sufficiently large n. Hence, R n (W ) eventually covers almost all of C. Proof. Takes a full page; uses (18.2). If R and S commute, then J(R) = J(S). Proof. Lengthy, with ε, δ, σ. J is contained in the closure of the set of periodic points of R. ζ is a periodic point of R iff it is fixed by some iterate R n. Note: this shows that R has infinitely many periodic points. Proof. Takes a full page; uses (18.3). R is a rat map, deg(r) 2. Then (i) z not exceptional = J O (z). (ii) z J = J = O (z). Note: This shows that for any non-exceptional z, the backward orbit O (z) is a countable set which accumulates at every point of J. Proof. Take any non-exceptional z and any nonempty open set W that meets J. As W meets J, z lies in some R n (W ) and so O (z) meets W. This shows (i). If z J, then the closed completely invariant set J contains the closure of the backward orbit O (z). Together with (i), this shows (ii). Let E be compact in C with the property that for any point z F (R), the sequence {R n (z)} does not accumulate at any point of E. Then given any open set U which contains J(R), R n (E) U for all sufficiently large n. In this case, we say R n (E) converges to J: R n (E) J.

15 E.g., suppose F 0 is a component of the Fatou set that contains an attracting fixed point ζ. Then let E be any compact subset of F 0 \{ζ}. X. Properties of the Fatou Set J(R) is connected iff each component of F (R) is simply connected. If F 0 is a completely invariant component of F, then (i) F 0 = J, (ii) F 0 is simply connected or infinitely connected, (iii) all other components of F are simply connected, and (iv) F 0 is simply connected iff J is connected. If F is connected, then either (i) F is simply connected and J is connected, or (ii) F is infinitely connected and J has an infinite number of components. Let F be the Fatou set of a polynomial of degree at least 2. Then (i) F is either simply connected or infinitely connected, and (ii) all other components of F are simply connected. F has 0,1,2, or infinitely many components. At most two are completely invariant. If there are two, they are simply connected. XI. The Mandelbrot Set Consider the quadratic iterators R c (z) = z 2 + c, where c is some constant complex number. Definition 32. The Mandelbrot set is M = {c. J(R c ) is connected}.