HAUSDORFF DIMENSION OF CLOSURE OF CYCLES IN d-maps ON THE CIRCLE

HAUSDORFF DIMENSION OF CLOSURE OF CYCLES IN d-maps ON THE CIRCLE SPUR FINAL PAPER, SUMMER 014 MRUDUL THATTE MENTOR: XUWEN ZHU PROJECT SUGGESTED BY CURTIS MCMULLEN AND DAVID JERISON JULY 30, 014 Abstract. We study the dynamics of the map x dx(mod 1) on the unit circle. We characterize the invariant finite subsets of this map which are called cycles and are graded by their degrees. By looking at the combinatial properties of the base-d expansion of the elements in the cycles, we prove a conjecture of Curt McMullen that the Hausdff dimension of the closure of degree-m cycles is equal to log m/ log d. We also study the connection between this map and the degree-d maps on the hyperbolic disk. Date: July 30, 014. 1

1. Introduction Degree-d maps on the unit disk have many interesting geometric, topological and analytic properties, which are closely related to hyperbolic geometry. The dynamics of these maps has an imptant application in classifying dynamical systems generated by polynomials in single complex variable [3] and it provides useful infmation about Julia sets and Mandelbrot sets [, 5]. To get an understanding of this family of maps, people studied the behaviour of these maps restricted to the boundary unit circle [3, 4]. In the paper [6], McMullen gave a complete description of simple (i.e. degree 1) cycles of these boundary maps, showing that those simple cycles are an analogue of simple closed geodesics on hyperbolic surfaces in the sense that the closure of union of simple cycles has Hausdff Dimension 0 and the closure of union of simple closed geodesics has Hausdff Dimension 1. Here, we consider degree-d holomphic maps from the unit disk onto itself. We focus on one such map z z d. The restriction of this map to the boundary circle is equivalent to the map x dx (mod 1) on R/Z. In this paper, we study cycles of higher degrees in this map, and in particular, calculate the Hausdff dimension of their closure. Definition 1 (d-map). Let d N. We define d-map as the map on the unit circle S 1 = R/Z given by (1) x dx (mod 1), x S 1 In other wds, if the base-d expansion of a point is (0.b 1 b b 3...) d, then d-map takes it to the point whose base-d expansion is (0.b b 3 b 4...) d Definition (Cycle). Let d be a positive integer greater than 1. A finite set C S 1 is called a cycle f d-map, iff d-map restricted to C is a transitive permutation. In terms of the base-d expansion, C is given by the following collection of points. () C = {(0.b i b i+1...b n b 1 b...b i 1 ) d 1 i n} where n is a fixed positive integer and b 1, b,..., b n are fixed digits in {0, 1,..., d 1}. Here are some examples of cycles. Consider the case d =. Here, {0} = {(0.0) } is the only 1-element cycle and { 1 3, 3 } = {(0.01), (0.10) } is the only -element cycle. F an integer n >, there are multiple n-element cycles. F example, there are two 3-element cycles: { 1 7, 7, 4 7 } and { 3 7, 5 7, 6 7 }. Now we define an imptant invariant of a cycle, called degree. Definition 3 (Degree of a cycle). Let d be a positive integer greater than 1. Let C be a cycle f d-map. The degree of C is the smallest non-negative integer m f which there exists a degree-m map f : S 1 S 1 such that f and d-map agree on C. It is denoted by deg(c). Remark 1. Note that 0 deg(c) d because d-map is a degree-d map. Remark. The cycles which contain only one element are fixed points of d-map and have degree 0. All other cycles have positive degree. Now we give two examples. Consider d = and C = { 3 7, 5 7, 6 7 }. Under -map, 3 7 6 7, 5 7 3 7, 6 7 5 7 i.e. the cyclic der of the points in C is preserved. So, C has degree 1. In other wds, as we jump on the points in the cycle from 3 7 5 7 6 7 (jump over 0) and back to 3 7, we complete one

full circle (we jump over 0 exactly once). As we jump on the images of these points, 6 7 (jump over 0 to) 3 7 5 7 and back to 6 7, again we complete one circle (we jump over zero exactly once). This means that the degree of C is 1. Now consider d = 3 and C = { 1 5, 5, 3 5, 4 5 }. Under 3-map, 1 5 3 5, 5 1 5, 3 5 4 5, 4 5 5 So, as we jump on the points in the cycle from 1 5 5 3 5 4 5 (jump over 0) and back to 1 5, we complete one full circle (we jump over 0 exactly once). As we jump on the images of these points, 3 5 (jump over 0 to) 1 5 4 5 (jump over 0 to) 5 and back to 3 5, we complete two circles (we jump over zero twice). So, the degree of C is. Definition 4 (Closure of cycles). Let m, d N with 1 < d and 1 m d. We define E m,d as the closure of the union of degree-m cycles f d-map. (3) E m,d = {c S 1 c is in a cycle of degree-m f d-map} In the paper [6] Curtis McMullen discussed the simple (i.e. degree 1) cycles f the d-map and and computed the Hausdff Dimension of their closure E 1,d. Recall that Hausdff Dimension is defined as follows: Definition 5 (Hausdff Dimension). The Hausdff Dimension of a set E is defined as: (4) dim H (E) = inf {δ 0 inf { r δ i E B(x i, r i )} = 0} Theem 1 (McMullen). Let d be a positive integer greater than 1. Then dim H (E 1,d ) = 0. In this paper, we characterize the degree-m cycles f d-map and generalize the above result. Theem. Let m, d N with 1 < d and 1 m d. Then dim H (E m,d ) = log m This paper is ganized in the following way: In Section, we get a lower bound on the Hausdff Dimension of E m,d by calculating the Hausdff Dimension of a subset of E m,d. In Section 3, we prove the upper bound by imitating the proof of Theem 1 in paper [6], using combinatial arguments. In Section 4, we futher investigate the relation with hyperbolic geometry by looking into a conjecture proposed in [6] and give partial results.. Lower Bound In this section, we define and study two useful invariants of a cycle called crossing number and Digit Ptrait, which are directly related to the degree of the cycle. Then we use these properties to construct a subset of E m,d which has Hausdff Dimension log m Definition 6 (Crossing). Let d be a positive integer greater than 1. Let C = {c 1, c,..., c n } be a cycle f d-map such that 0 < c 1 < c <... < c n < 1 Let c n+1 = c 1. F any 1 i n, the pair (c i, c i+1 ) is called a crossing generated by C ( simply a crossing) iff (5) 0 < dc i+1 (mod 1) < dc i (mod 1) < 1 The total number of such crossings is called the crossing number of C. (6) η(c) = # (crossings generated by C) 3

Remark 3. As we jump on S 1 in the counterclockwise direction and trace the points of the cycle from c 1 c... c n and (jump over 0) back to c 1, we complete one full circle. When we trace the images of these points, dc 1 (mod 1) dc (mod 1)... dc n (mod 1) and back to dc 1 (mod 1), we may jump over 0 multiple times. Each time we jump over 0, we have a crossing. The crossing number of the cycle is equal to its degree. Remark 4. If n = 1, then there are no crossings. The crossing number is 0 which is the degree of 1-element cycles. Lemma 1. Let d be a positive integer greater than 1. Let C = {c 1, c,..., c n } be a cycle f d-map with 0 < c 1 < c <... < c n < 1. Then, the crossing number of C is equal to its degree η(c) = deg(c) Proof. The case n = 1 is obvious. Assume n > 1. Let c n+k = c k, k N. Let 1 i 1 deg(c). We can divide S 1 [0, 1) in intervals I 1, I,..., I deg(c) such that on each I r, there exists a continuous non-decreasing map g r to [0, 1) which agrees with d-map on I r C. Note that there are exactly η(c) sets of the type {c it, c it+1,...c it+1 }. Since η(c) > deg(c), we can find t such that {c it, c it+1,...c it+1 } I r f some r. So, 0 < g r (c it+1) = dc it+1(mod 1) < dc it (mod 1) = g r (c it ) < 1 which contradicts the non-decresing nature of g r. So, deg(c) η(c). Now we prove that deg(c) η(c) by constructing an appropriate map of degree η(c). Define f : S 1 S 1 as: f(x) = dc it +1(mod 1) c it +1 c i t (x c i t +c it +1 ) if x [ c i t +c it +1, c it+1] dc i (mod 1) + dc i+1(mod 1) dc i (mod 1) c i+1 c i (x c i ) if x [c i, c i+1 ] [c it+1, c it+1 ] dc it+1 (mod 1) + 1 dc i t+1 (mod 1) c it+1 +1 c i t+1 (x c it+1 ) ifx [c it+1, c i t+1 +c it+1 +1 ] where intervals are taken in counter-clockwise direction. Here, we have divided S 1 into η(c) intervals of the type [ c i t +c it +1, c i t+1 +c it+1 +1 ], each of which is mapped onto S 1.f maps the endpoints of the interval to 0, points of C to their images under d-map. In between, f is linear. So, f is a continuous function and t, and restriction of f gives a bijection between [ c i t +c it +1 and [0, 1) S 1. Thus, f has degree η(c). Also, f C agrees with d-map. This completes the proof., c i t+1 +c it+1 +1 ) Now that we have established the relation between the degree and the crossing number of a cycle, we need a tool to estimate the crossing number. We observe that the crossing number of a cycle is related to the der of points in the cycle, and hence the digits in the base-d expansion of points of the cycle. Now we define an invariant of the cycle called Digit Ptrait which characterizes these digits. Definition 7 (Digit Ptrait). Let d be a positive integer greater than 1. Let C be a cycle f d-map. The Digit Ptrait of C is the non-decreasing map F : {0, 1,,..., (d 1)} {0, 1,,..., C } 4

which satisfies F (j) = C [0, (j + 1)/d) 0 j (d 1) F (j) = # (elements of C whose base-d expansion starts with a digit less than j + 1) Let dig(c) be the number of distinct positive values taken by F. Note that if a digit j is absent in the base-d expansion of a point in C, then F (j) = 0 F (j) = F (j 1). So, dig(c) is also the number of distinct digits which appear in the base-d expansion of any point in C. To estimate the crossing number of C, we need the second interpretation of dig(c). Here is an example. Consider d = 4 and the cycle C = {(0.001) 4, (0.010) 4, (0.100) 4, (0.001) 4 }. Note that 0 = (0.0) 4 < (0.001) 4 < (0.010) 4 < 1 4 = (0.1) 4 < (0.100) 4 < 4 = (0.) 4 < (0.001) 3 < 3 4 = (0.3) 4 The Digit Ptrait of C is the map F : {0, 1,, 3} {0, 1,, 3, 4} given by: F (0) =, F (1) = 3, F () = 4, F (3) = 4 F takes the values, 3 and 4. So, dig(c) is 3. There are exactly 3 digits (0, 1 and ) which appear in the base-4 expansion of the points in C. Now we establish the relation between dig(c) and the crossing number of C: Lemma. Let d be a positive integer greater than 1. Let C = {c 1, c,..., c n } be a cycle f d-map with 0 < c 1 < c <... < c n < 1 and n > 1. Then, the crossing number of C is at most the number of distinct digits which appear in the base-d expansion of a point in C η(c) dig(c) Proof. Let 1 i < n. Let c i [j 1 /d, (j 1 +1)/d), c i+1 [j /d, (j +1)/d). Note that j 1 j because c i < c i+1. If j 1 = j = j, then 0 < dc i (mod 1) = dc i j < dc i+1 j = dc i+1 (mod 1) < 1 In this case, (c i, c i+1 ) cannot be a crossing. So, (c i, c i+1 ) is a crossing only if j 1 < j i.e. c i is the largest element of C [j 1 /d, (j 1 + 1)/d) and c i+1 is the smallest element of C [j /d, (j + 1)/d). Thus, there are at most (dig(c) 1) i s f which 1 i < n and (c i, c i+1 ) is a crossing. F some cycles, (c n, c 1 ) is an additional crossing. So, there are at most dig(c) i s f which 1 i n and (c i, c i+1 ) is a crossing. Lemma 1 and Lemma together give a way of estimating the degree of a cycle by looking at the digits in the base-d expansion of a point in the cycle. Now we use this to get a sufficient condition f a point to be in the closure of the cycles of fixed degree. Lemma 3. Let m, d N with 1 m d and 1 < d. Then any point in S 1 whose base-d expansion contains atmost m distinct digits lies in E m,d. Proof. Let α S 1. Let α = (0.α 1 α α 3...) d such that r, α r {b 1, b,..., b m } {0, 1,..., (d 1)}. Here, b 1, b,..., b m are fixed digits in base-d such that b 1 < b <... < b m. To prove that α lies in E m,d, we will show that q N, there exists a degree-m cycle C f d-map which intersects d q neighbhood of α. Any periodic point whose base-d expansion contains exactly m digits is in a cycle of degree at most m. To get the maximum possible degree, we need maximum possible crossings. This can be achieved with the following construction: 5

Let N N such that t, N is greater than the number of times b t appears in the first q digits of α. Let denote b t b t b t...b t (N times). Consider the following point: (7) c = (0.α 1 α α 3...α q ... b m ) d Clearly, c is in d q neighbhood of α. It is a point of cycle C = {c 1, c,...c q+n(m 1)+1 } f d-map with 0 < c 1 < c <... < c q+n(m 1)+1. Now we need to prove that deg(c) = m. F each 1 t m, let i t be such that the largest element of C whose base-d expansion starts with the digit b t is c it. Note that c i1 is at least (0.b 1 b m ) d. F t > 1, c it is at least (0. ) d. F each 1 t < m, c it+1 is the smallest element of C whose base-d expansion starts with the digit b t+1. Note that the base-d expansion of c it+1 starts with 0.b t . So, 1 t < m, 0 < dc it+1(mod 1) < dc it (mod 1) < 1 i.e. (c it, c it+1) is a crossing. Note that c q+n(m 1)+1 = c im and c im+1 = c 1 = c i1. So, i.e. (c im, c im+1) is a crossing. Thus, the crossing number of C is at least m. 0 < dc im+1(mod 1) < (0.b ) d < dc im (mod 1) < 1 η(c) m From Lemma, we know that the crossing number of C is at most the number of distinct digits in the base-d expansion of a point C, which in this case, is m. So, deg(c) = η(c) = m. This immediately gives the following result: η(c) dig(c) = m Proposition 1. E d,d = S 1. Thus, dim H (E d,d ) = 1 = log d Proof. Note that since base-d has only d digits, the set of all points of S 1 whose base-d expansion contains at most d digits is S 1 itself. So, we have S 1 E d,d S 1 E d,d = S 1. Let m and d be positive integers satisfying d > 1 and 1 m d. Now we use Lemma 3 to construct a subset of E m,d. Let A m,d be the set of points in S 1 whose base-d expansion contains the digits from 0 to (m 1) only. Clearly, A m,d E m,d. The structure of A m,d is similar to the structure of Cant s set. Here, we start with X 0 = [0, 1]. Write X 0 in d closed intervals of length 1/d. Let X 1 be union of first m of these intervals. X 1 = m 1 i=0 [i/d, (i + 1)/d] X 1 is the union of m intervals of length 1/d. Divide each such interval in d equal parts and take the first m in X. X is the union of m intervals of length 1/d. Repeat the process. If X i is the union of m i intervals of length d i, then divide each such interval in d equal parts and take the first m in X i+1. A m,d = i=0 6 X i

Now we calculate the Hausdff Dimension of A m,d. Lemma 4. dim H (A m,d ) = log m Proof. Let β > 0. Note that i, A m,d X i. So, f each i, we have m i intervals of length d i that fm a covering of A m,d. If β = log m log d + ε f some ε > 0, lim i mi (d i ) β = lim m i (d iε )((d log m log d ) i ) = lim d iε = 0 i i This means that, f any β > log m log d, we can cover A m,d such that the summation of β powers of the lengths of the intervals in the cover is as small as we like. So, dim H (A m,d ) log m Now we need to prove that the Hausdff dimension of A m,d is at least log m log d. Note that we can consider A m,d as a subset of [0, m/d] which is a compact set. So, f any countable cover (U r ) of A m,d, we can find finitely many open sets V 1, V,..., V p such that p U r and r=0 r=0 p V r β r=0 V r U r β Now we will get a lower bound on p r=0 V r β. Let b N such that r, d b V r. 1 i b,let N i be the number of V r s which satisfy d i V r < d i+1. Note that if V r < d i+1, then V r can intersect at most two intervals in X i 1. Hence, V r can contain at most m b i+1 intervals in X b. X b has m b intervals. So, b m b i+1 N i m b F β = log m log d, p V r β r=0 i=0 b i=0 r=0 m i N i 1 m b (d i ) β N i = i=0 b i=0 m i N i 1 m i.e. the summation of log m log d powers of the lengths of the intervals which fm a cover of A m,d has a positive lower bound. Thus, dim H (A m,d ) log m Now we have a lower bound on the Hausdff Dimension of E m,d. Theem 3. Let m, d N with 1 < d and 1 m d. Then, dim H (E m,d ) log m log d Proof. A m,d E m,d. So, dim H (E m,d ) dim H (A m,d ) = log m 7

3. Upper Bound In this section, we first find an upper bound on the number of degree-m cycles f d-map which have n elements. We extend this result to precycles. Then, we find an appropriate covering of E m,d to prove that its Hausdff Dimension is at most log m Definition 8 (Partition generated by a cycle). Let C = {c 1, c,..., c n } be a degree-m cycle f d-map such that 0 < c 1 < c <... < c n < 1. Let σ be the map on {1,,..., n} which satisfies: dc r (mod 1) = c σ(r), 1 r n Note that σ is a permutation of {1,,..., n} because C is a cycle. Let 1 i 1 σ(i t + 1) and σ(i t + 1) < σ(i t + ) <... < σ(i t+1 ), 1 t < m and σ(i m ) > σ(i m + 1) and σ(i m + 1) <... < σ(n) < σ(1) <... < σ(i 1 ) Now we construct a partition of {1,,.., n} using the above property of σ. { {σ(r) it < r i P t = t+1 } if 1 t < m {σ(r) i m < r n} {σ(r) 1 r i 1 } if t = m {P t 1 t m} is a partition of {1,,.., n}, called as the partition generated by C and it is denoted by P (C). Both P (C) and the map σ are useful counting n-element degree-m cycles. Let us consider d = 3 and C = {(0.0010) 3, (0.0100) 3, (0.0001) 3, (0.1000) 3, (0.0010) 3 } as an example. Here, n = 5. Under 3-map, c 1 = (0.0010) 3 (0.0100) 3 = c c = (0.0100) 3 (0.1000) 3 = c 4 c 3 = (0.0001) 3 (0.0010) 3 = c 5 c 4 = (0.1000) 3 (0.0001) 3 = c 3 c 5 = (0.0010) 3 (0.0010) 3 = c 1 So, σ is the permutation of {1,, 3, 4, 5} which takes 1,, 3, 4, 5 to, 4, 5, 3, 1 respectively. (c 3, c 4 ) and (c 4, c 5 ) are the crossings generated by the cycle. So, i 1 = 3, i = 4 and the partition generated by C is given by: P (C) = {P 1, P } where P 1 = {σ(r) 3 < r 4}, P = {σ(r) 4 < r 5} {σ(r) 1 r 3} P (C) = {{3}, {1,, 4, 5}} Now we will show that if some properties of a cycle such as degree, the partition it generates and its digit ptrait are given, then we can construct the cycle (find its points). Later, we will use this result to count the number of cycles of fixed degree and cardinality. Lemma 5. Given positive integers d, m, n, P = {P 1, P,..., P m } which is a partition of {1,,..., n}, positive integer i 1 < P m and a non-decreasing map F : {0, 1,,..., (d 1)} {0, 1,,..., n} with F (d 1) = n, there exists at most one cycle C f d-map such that: 8

(1) C = {c 1, c,..., c n } with 0 < c 1 < c <... < c n < 1 () deg(c) = m (3) P (C) = P P is the partition generated by C (4) If i t = i t 1 + P t 1 < t m, then (c i, c i+1 ) is a crossing i {i 1, i,..., i m }. (5) F is the Digit Ptrait of C. Proof. Suppose C is a cycle f d-map which satisfies (1), (), (3), (4), (5). Let σ be the map on {1,,..., n} given by: (r i t ) th element of P t if i t < r i t+1 and 1 t < m σ(r) = (r i m ) th element of P m if i m < r (r + n i m ) th element of P m if r < i 1 where 1 t m, the elements of P t are in increasing der. Note that σ is uniquely determined by m, n, P and i 1. Comparing this with the definition of partition generated by a cycle, we conclude that dc r (mod 1) = c σ(r), 1 r n Let b : {1,,..., n} {0, 1,,..., (d 1)} given by: { 0 if r < F (0) b(r) = j if F (j 1) < r F (j) and 1 j (d 1) Note that b is uniquely determined by n, d and k. Comparing this with the definition of Digit Ptrait, we conclude that c r [ b(r) d, b(r) + 1 ), 1 r n d b(r) = the first digit in the base-d expansion of c r, 1 r n Also, b(σ(r)) = the first digit in the base-d expansion of c σ(r) = the second digit in the base-d expansion of c r Thus, we conculde that c r = (0.b(r)b(σ(r))b(σ (r))...b(σ d 1 (r)) ) d i.e. the cycle C is uniquely determined by σ and b. Remark 5. (c i, c i+1 ) can be a crossing only if the first digits of the base-d expansions of c i and c i+1 differ. So, cycle C satisfying all conditions in the above lemma can exist only if {F (0), F (1),..., F (d 1)} {i 1, i,...i m } {n} We will use this in the proof of the following lemma. Lemma 6. Let d, m, n N with 1 m d and n, d > 1. Then the number of cycles C f d-map satisfying C = n and deg(c) = m is at most O(n d m+1 m n 1 ). Proof. d, m, n are given. Now we need P = {P 1, P,..., P m } which is a partition of {1,,..., n}, positive integer i 1 < P m and a non-decreasing map F : {0, 1,,..., (d 1)} {0, 1,,..., n} with F (d 1) = n to fix a degree-m, n-element cycle. 9

Let T be the set of dered (m + 1)-tuples (P 1, P,..., P m, i 1 ) such that P = {P 1, P,..., P m } is a partition of {1,,..., n}, i 1 N and i 1 P m. We want an upper bound on #(T ). #(T ) = = = a 1 +a +...+a m=n a i N a 1 +a +...+a m=n a i N a 1 +a +...+a m=n a i N ( )( ) n n a1 a m... a 1 a a m a 1 +a +...+a m 1 +(a m 1)=n 1 a i N {0} = n m n 1 n! a 1! a!...a m! n! a 1! a!...a m 1! (a m 1)! n ( ) n a1 a... a m 1 a m (n 1)! a 1! a!...a m 1! (a m 1)! After fixing an element of T, we need to fix a non-decreasing map F from {0, 1,,..., (d 1)} to {0, 1,,..., n} with F (d 1) = n and {F (0), F (1),..., F (d 1)} {i 1, i,...i m } {n}. We have (n + 1) (d m 1) (n + 1) (d m) choices f F, depending on if n is in {i 1, i,...i m } not. Now, using Lemma 5, we conclude that the number of degree-m n-element cycles f d-map is at most O(n d m+1 m n 1 ). Definition 9 (Precycle). Let d be a positive integer greater than 1. A finite set C P S 1 is called a precycle f d-map, iff C P = {c d i (mod 1) i N {0}} f some c S 1. So, a precycle is the fward bit of a rational point in S 1. It can be written in terms of base-d expansion of its points as follows: C P = {(0.b r b r+1...b n1 b 1 b...b n ) d 1 r n 1 } {(0.b k b k+1...b n b 1...b k 1 ) d 1 k n } where n 1 is a fixed non-negative integer, n is a fixed positive integer and all b r, b k are fixed digits in {0, 1,,..., d 1}. Remark 6. Every precycle C P includes a cycle. Note that if n 1 = 0, then C P itself is a cycle. Here is an example of a precycle. Consider d = and C P = { 1 3, 3, 5 6 }. C P can be written as C P = { 5 6 di (mod 1) i N {0}} C P = { 5 6 } {1 3, 3 } = {(0.110) } {(0.01), (0.10) } Remark 7. We can define degree, crossing, crossing number, digit ptrait and partition f a precycle simply by replacing C by C P in the respective definitions. Note that if C P is a precycle which is not a cycle, then the map σ in the definition of partition (Definition 8) is not a permutation. Under σ, one element of {1,,..., n} has no preimages, one has two preimages and all other elements have exactly one preimage. Now by using the logic which proves Lemma 6, we can prove the following result: Lemma 7. Let d, m, n N with 1 m d. Then the number of precycles C P f d-map satisfying C P = n and deg(c P ) = m is at most O(n d m+3 m n 1 ). 10

Remark 8. The increase in the exponent of n is due to the small change in the nature of the map σ in the definition of partition (Definition 8). Now we find a suitable covering of E m,d and get an upper bound on its Hausdff Dimension. Theem 4. Let m, d N with 1 < d and 1 m d. Then dim H (E m,d ) log m log d Proof. Fix a positive integer n > 1. C P (n) = {C P C P is a precycle f d-map, deg(c P ) m, C P n} Lemma 7 implies that C P is at most O(n d m+4 m n ). Let C be a degree-m cycle f d-map. Let c be a point in C. There exists a degree-m map f : S 1 S 1 which agrees with d-map on C. Note that (f n ) (c) = d n. This means that there exists a point x in O(d n ) neighbhood of c f which two of x, f(x), f (x),..., f n (x) coincide. So, c is in O(d n ) neighbhood of an element in C P. Each point of any degree-m cycle f d-map lies in O(d n ) neighbhood of an element in C P. So, E m,d lies in O(d n ) neighbhood of C P which has at most O(n d m+4 m n ) elements. Let β = log m log d + ε f some ε > 0. lim n (d n ) β n d m+4 m n = lim (d log m log d ) n d nε n d m+4 m n n = lim n d nε n d m+4 = 0 F any β > log m log d, we can cover E m,d such that the summation of β powers of the lengths of the intervals in the cover is as small as we like. So, dim H (E m,d ) log m 4. Maps on the disk Now that we have studied the boundary behaviour, we would like to study some properties of the maps on hyperbolic space. As in paper [6], we study the following functions: Let d be a positive integer greater than 1. Let B d be the space of maps f on the unit disk which are of the fm d 1 f(z) = r=1 z( z a r 1 a r z ) where a r < 1, 1 r d 1. Note that if a r = 0, 1 r d 1, we have a map on the disk which takes z to z d. We denote this map by p d. As mentioned in [6], f each map f B d, there is a unique homeomphism φ f : S 1 S 1 such that - (1) f(z) = φ f 1 p d φ f (z), z S 1 () φ f varies continuously with f (3) φ pd is the identity map on S 1. Remark 9. Here S 1 is the set of complex numbers z satisfying z = 1. 11

In [6], McMullen defined the following Length function: Definition 10 (Length of a cycle). The length on f of a cycle C f p d is given by: (8) L(C, f) = log (f q ) (z) where q = C and z = φ f 1 (c) f some c C. Note that f(φ f 1 (c)) = φ f 1 p d φ f (φ 1 (c)) = φ f 1 (c d ). In general, f i (φ f 1 (c)) = φ f 1 (c di ), c S 1 Now by applying chain rule to calculate (f q ), we get that (9) L(C, f) = log f (φ 1 f (c)) c C The Length function is an interesting invariant of f, which suggests some properties of the cycles of p d. F example: Theem 5 (McMullen). Let C be a cycle f p d. If L(C, f) < log, then C is a simple cycle, i.e. deg(c) = 1. There is a conjecture suggested by McMullen: Conjecture 1. Let C be a degree-1 cycle f p d. Then, L(C, f) has no critical points. Now we focus on the case d =. Since there is only one parameter a which determines f, we write f a instead of f and φ a instead of φ f. We have (10) f a (z) = z( z a 1 az ), z and (11) f a(z) = 1 + 1 a z a, z S1 When we choose C with less than 3 elements, we can calculate φ a 1 (c) f each c in C and compute L(C, f a ). We get the following result: Proposition. Let C be a cycle f p. If C < 3, then L(C, f a ) has no critical points. Proof. F detailed calculations, see Appendix A. Case 1: C = {1} We know that f a (φ 1 a (1)) = φ 1 a (1 ) = φ 1 a (1) and φ 1 a (1) S 1 This leads to φ 1 a (1) = 1 + a 1 + a and L(C, f a ) = log f (φ 1 f (1)) = log( + a + a 1 a ) Write a as x + iy. + x (1) L(x, y) = L(C, f x+iy ) = log( 1 x y ) Now we compute partial derivatives of L w.r.t. x and y and observe that L x is 0 iff (1 + x) = ±y is 0 iff y = 0. Thus, both partial derivatives of L are zero iff a = 1 which is impossible and L y since a < 1. Thus, L(C, f a ) does not have any critical points. 1

Case : C = {ω, ω } where ω = e π 3 i We know that and f a (φ a 1 (ω)) = φ a 1 (ω ) f a (φ a 1 (ω )) = φ a 1 (ω 4 ) = φ a 1 (ω) This means that φ a 1 (ω), φ a 1 (ω ), φ a 1 (1) and 0 satisfy the degree-4 equation: f a (f a (z)) = z By simplifying this equation and facting out [z(z φ 1 a (1))], we obtain a quadratic equation which has φ 1 a (ω) and φ 1 a (ω ) as its roots. From the coefficients of the quadratic, we get: We use this to show that φ a 1 (ω) + φ a 1 (ω ) = a 1 and φ a 1 (ω)φ a 1 (ω ) = 1 a 1 a L(C, f a ) = log f a(φ a 1 (ω)) f a(φ a 1 (ω )) = log(4 + a (a + a) Write a as x + iy. Now we have: (13) L(C, f a ) = L(x, y) = log(4 4x + x + y + Now we compute partial derivatives of L w.r.t. x and y. L y y 1 x y ) (a a) 1 a ) is 0 iff y = 0. Thus, L x = L y = 0 iff x = and y = 0 which is impossible since a < 1. Thus, L(C, f a ) does not have any critical points. We are still studying cycles with me elements. Here are some difficulties that we need to overcome. Let q N. Note that if C is a cycle and C divides q, then f any point c C, φ 1 f (c) satisfies the degree-d q equation f q (z) = z F example, f d = and q = 3, f q a(z) = z has 8 roots two of which are 0 and φ a 1(1). We can reduce it to a degree-6 equation whose roots are preimages of two different cycles under φ a. In this case, we cannot apply the method used above to calculate L and check its critical points because we first need to separate the roots in two groups - one f each cycle. F q > 3, the problem is even harder since the roots of the cresponding equation may include preimages (under φ a ) of cycles of higher degree. Due to our calculations in the proof of Proposition, we belive that in the case d =, L(C, f a ) is an algebraic function of a and a, and Conjecture 1 is true. Acknowledgment First, I would like to thank Prof. Curtis McMullen f suggesting the problems and giving valuable comments. I would also like to thank Prof. David Jerison and Prof. Pavel Etingof f their guidance and insightful discussions. I thank my ment Xuwen Zhu f her help and suppt. 13

Case 1: Appendix A. Calculating L(C, f a ) L(C, f a ) = log f (φ f 1 (1)) = log(1 + 1 a ( 1+a 1+a ) ) a = log(1 + 1 + a 1 a ) = log( + a + a 1 a ) + x L(x, y) = L(C, f a ) = log( 1 x y ) L x = (1 x y )( (1 x y ) + ( + x)(x) + x (1 x y ) ) 1 + x + x y = ( ( + x)(1 x y ) ) L y = (1 x y ( + x)(y) )( + x (1 x y ) ) y = 1 x y Case : φ a 1 (ω), φ a 1 (ω ), φ a 1 (1) and 0 satisfy the degree-4 equation: f a (f a (z)) = z z( z a z a z( 1 az )( ) a 1 az 1 az( z a 1 az )) = z z( z a 1 az )( z az a + aaz 1 az az + aaz ) = z So, φ 1 a (ω), φ 1 a (ω ) and φ 1 a (1) satisfy the following cubic: (z a)(z + (aa a)z a) = (1 az)(1 + (aa a)z az ) (1 a )z 3 + (aa a + a a + aa )z + (a a a a + a aa)z + (a 1) = 0 [(1 + a)z (1 + a)] [(1 a)z + (1 a a + aa)z + (1 a)] = 0 We divide both sides by [(1 + a)z (1 + a)] = (1 + a)(z φ 1 a (1)) to get the desired quadratic equation. Let φ a 1(ω) = z 1 and φ a 1(ω ) = z. We have z 1 + z = a 1 and z 1 z = 1 a 1 a 14

Now we can compute the modulus of the product of the derivatives of f at φ a 1(ω) and φ a 1(ω ). f a(φ a 1ω) f a(φ a 1(ω )) = f a(z 1 ) f a(z ) = (1 + 1 a 1 a )(1 + z 1 a z a ) = 1 + 1 a (z 1 a)(z a) ( z 1 a + z a + 1 a ) 1 a = 1 + 1 a 1 a + a(1 a) + a (1 + a az 1 az 1 + 1 + a az az + 1 a ) 1 a = 1 + 1 a (3 + a + a(1 a) + a(1 a)) = 1 + 1 a 1 a (3 + a + a a ) = 1 + 1 a + 1 a ( + a + a) 1 a = + a a a + (1 + a a a) 1 a ( + a + a) = 4 + a + ( a a a)( + a + a) 1 a = 4 + a + 4 a (a + a)(1 a ) (a + a) 1 a = 4 + a (a + a) (a a) 1 a Now we take log of both sides and replce a by x + iy. L(x, y) = L(C, f a ) = log(4 4x + x + y + y 1 x y ) L x = 1 8xy ( 4 + x + 4 4x + x + y + y (1 x y ) ) 1 x y L y = 1 4y (y)(1 + 4 4x + x + y + y (1 x y ) ) 1 x y 15

References [1] Joan S Birman and Caroline Series, Geodesics with bounded intersection number on surfaces are sparsely distributed, Topology 4 (1985), no., 17 5. [] Adrien Douady, John H Hubbard, and P Lavaurs, Etude dynamique des polynômes complexes (1984). [3] Lisa R Goldberg, Fixed points of polynomial maps. i. rotation subsets of the circles, Annales scientifiques de l ecole nmale supérieure, 199, pp. 679 685. [4] Lisa R Goldberg and John Miln, Fixed points of polynomial maps. part ii. fixed point ptraits, Annales scientifiques de l ecole nmale supérieure, 1993, pp. 51 98. [5] Karsten Keller, Invariant facts, julia equivalences and the (abstract) mandelbrot set, Springer, 000. [6] C McMullen, Dynamics on the unit disk: sht geodesics and simple cycles, Comment. Math. Helv 85 (010), 73 749. [7] Curtis T McMullen, Hausdff dimension and confmal dynamics, iii: Computation of dimension, American journal of mathematics (1998), 691 71. [8], Hausdff dimension and confmal dynamics i: Strong convergence of kleinian groups (1999). [9], Hausdff dimension and confmal dynamics ii: Geometrically finite rational maps, Commentarii Mathematici Helvetici 75 (000), no. 4, 535 593. 16