PERIODIC POINTS FOR SPHERE MAPS PRESERVING MONOPOLE FOLIATIONS

Let S be a two-dimensional sphere. We consider two types of its foliations with one singularity and smooth maps f : S → S preserving these foliations, more and less regular. We prove that in both cases the lower growth rate of the number of fixed points of f is at least log |deg(f)|, where deg(f) is a topological degree of f , confirming the Shub’s conjecture in these classes of maps.


Introduction
Estimating the growth rate of the number of fixed points of the n-th iterate of a smooth map of the n-dimensional sphere to itself is a challenging problem. It was conjectured by Michael Shub in 1974 that it must be (asymptotically) exponential (cf. [16,17]): where deg(f ) denotes the degree of f . On the other hand, (1.1) does not hold for every continues map f (cf. an example of a map with only two periodic points and deg(f ) = 2 in [16]), while it is known that the smoothness implies that the growth rate of number of fixed points of f n is at least linear [1] and indeed could be linear up to any fixed period, cf. [4,5]. Proving or disproving the Shub's conjecture would substantially extend our knowledge about the role of the differentiability assumption in periodic point theory.
Even in the simplest case of two-dimensional sphere S 2 the conjecture still remains unsolved, although there has been a lot of progress in some particular cases. During the last few years the exponential growth was obtained under some topological conditions for S 2 and annulus ( [2,8,9,10,11]).
Recently the problem for S 2 has been studied in [14,15] (see also [6,7] for higher dimensional spheres) under the assumption that the map preserves some "geographical" (singular) foliation. In this context a natural question arises, what happens when we change the "geography." In this paper we consider foliations with one singularity. In geographical terms this means that our planet, call it Monopole, has only one pole. After all, if physicists can admit magnetic monopoles (see, e.g. [3]), there is no reason for not admitting geographical monopoles.
The natural coordinates on the surface of Monopole are easy to describe. Suppose that our sphere S 2 is given by the equation x 2 + y 2 + (z − 1) 2 = 1. The pole P is the origin. Let us denote the xy-plane by π. A given point Q = (x, y, z) ∈ S 2 belongs to the half plane π x containing the x-axis and to the half-plane π y containing the y-axis. We denote by α (β) the angle between π y and π (π x and π, respectively). We introduce α and β as new coordinates of Q, both varying from 0 to 180 degrees. We will call them x-titude and y-titude.
Thus, the lines of constant x-titude (respectively, y-titude), which we call merallels (respectively, paridians), are circles that are the intersections of the sphere with the planes passing through the y-axis (respectively, the x-axis).
Locally, in a neighborhood of the pole, we can easily draw the foliation by the paridians. For this, we use the stereographic projection from the antipole (0, 0, 2). The picture will be on the xy-plane, and the projections of the paridians will be the x-axis and the circles tangent to it at the origin (see Figure 1). We will consider paridianal maps of the sphere, that is, C 1 maps preserving the foliation by the paridians.
We also propose a much less regular foliation, with two rabbit-like "ears." The corresponding picture in the xy-plane will be the same as for the y-paridianal case in the lower half-plane. In the upper half-pane the leaves will be graphs given by the polar equation r = c| sin(2θ)|, where θ ∈ (0, π], and then the smooth deformations of the circles, being more and more geometrically circle-like as the radius goes to the infinity (see Figure 2). We will call this foliation the rabbit foliation, and the smooth maps preserving this foliation the rabbit maps.
In our approach we consider some type of open subsets of the base of our foliation with non-zero Brouwer degree called preband and prove a "fixed point theorem" stating the existence of a fixed point of f (or f n for iterations) in every preband. Now in case of homogeneous foliation (paridianal maps) we get that exponential growth of global degree of iterations implies the same growth of the number of prebands, thus also the number of fixed points of f n , so the Shub's conjecture is valid (Theorem 2.12).
If the foliation is not homogeneous, i.e. we consider rabbit maps, the answer is the same. The Shub's conjecture holds, but for a completely different reason: all maps preserving such foliation have degree 0 or ±1 (Theorem 3.5).

Paridianal maps
We will use the general scheme from [14]. However, many elements of this scheme, in particular the discussion of the behavior of our map close to the pole, will be different. We use the classical definition of Brouwer degree, cf. [13], where the degree for a C 1 map f is defined as a sum of signs of the Jacobian of f at a finite set f −1 (y) for y being a regular value.
We consider the natural system of coordinates on the sphere S 2 , defined in the introduction. We include the pole P , so the y-titude (after the affine rescaling) is an element of the circle T = R/Z. To denote specific points on this circle we will usually use the numbers from [0, 1). We will denote the y-titude of a point x ∈ S 2 by (x). The function is well defined and continuous on S 2 {P }.  Observe that all paridians, except for the pole, are circles, so the pole belongs to all paridians. However, the singleton of P is also a paridian. We will call paridians other than {P } proper paridians. Note that if we used the convention that the pole belongs only to the paridian {P }, our class of maps preserving paridians would be much smaller.
Proof. If f (P ) = P then by the continuity of f , since every paridian contains P , the image of the whole sphere would be contained in one paridian. Therefore f would not be injective and thus deg(f ) = 0.
From now on, we will assume that deg(f ) = 0. Fix a paridianal map f . Since f maps paridians to paridians, there exists a map ϕ : T → T such that for x, f (x) = P As we will consider only maps with non-zero degree, by Lemma 2.2 we may assume that f (P ) = P . Since P belongs to all paridians, ϕ is not defined uniquely. To make it unique, whenever the whole paridian −1 (y) is mapped by f to P , we set ϕ(y) = 0 and define (P ) = 0. Then we get the following commutative diagram: Proof. Let us take a parametrization η y : S 1 → S y , which depends continuously on y. Then, d Sy (f ) is the degree of the self-map of the circle As a consequence by changing y we obtain a homotopy between f y and f y for all y , y ∈ B and the result follows from homotopical invariance of the degree.
We will call the common value of d Sy (f ) for y ∈ B the paridian degree of B, and denote it d(B). Proof. Assume that d(B) = 0 and B = (y − , y + ). Continuity of ϕ on B follows from continuity of f . We will show that for y > y − which tends to y − , ϕ(y) tends to 0, i.e., we have right-continuity at y − (the argument for left-continuity at y + is the same). Notice that f (S y ) = S ϕ(y) and that f (S y ) covers the whole S ϕ(y) because d(B) = 0. By continuity of f , f (S y )(= S ϕ(y) ) tends to f (S y − ) = {P }, which implies that ϕ(y) tends to 0.
Thus, in case of d(B) = 0, we can define the sign of B, which we will denote ∆(B).  Proof. Suppose that the number of prebands is infinite. Then, there exists a sequence of prebands (B i ) i which converges to some y 0 ∈ A. We choose a point Q ∈ S 2 such that Q = P . As a consequence of the fact that deg(B i ) = 0 we may find a sequence of points (Q i ) i such that On the other hand, (Q i ) i (or its subsequence) is convergent and we get: However, (Q 0 ) = y 0 , which leads to the following contradiction: where in the third equality we use the formula (2.1).
Note that outside the pole P the map f can be locally written down in the following form where α and β are the x-and y-titude, respectively.
Lemma 2.6. If f : S 2 → S 2 is a paridianal map, then its degree is Proof. We choose Q = P , a regular value of f . Then by the definition In the neighborhood of any point x i ∈ f −1 (Q) the map f has the form (2.4) for On the other hand, f is a paridianal map and thus ϕ is a function of only one variable y. Thus where a i ∈ R is the derivative of ϕ at α i and b i = ∂f | Sy ∂β (α i , β i ). For each fixed α i take the finite set of β ij such (α i , β ij ) ∈ f −1 (Q) and all β ij belong to the same paridian S α i = −1 (α i ). Then, by the formula (2.6) and taking into account that by Lemma 2.3 d Sα i (f ) is the same for all α i ∈ B and is equal to d(B) we get: where the summation is taken over a finite number of bands B, (for some of them perhaps deg(B) = 0). This gives us the formula (2.5).
Studying the derivative Df (P ) we consider the local planar coordinate system near P given by the stereographic projection from the antipole (0, 0, 2). In this system 0 is a fixed point representing P and circles represent paridians (see Figure 1). The x-axis (with the point at infinity, which we will not mention later) is also a paridian. To simplify the notation we use the same letter f for our map in this coordinate system. Lemma 2.7. Assume that one of the following conditions hold: (1) there is a preband B = (0, 1), (2) B = (0, 1) is a preband and |d(B)| > 1. Then Df (P ) = 0.
Proof. We use the planar coordinates, so P becomes 0, and proper paridians become the x-axis and circles tangent to it at 0 (see Figure 1).
We will start by proving that Df (0) maps paridians that are circles to paridians. Let L r be a paridian of diameter r. Fix ε > 0. By the definition of the derivative, there exists δ 0 such that for every vector For δ ∈ (0, δ 0 ), define the map g δ by g δ (v) = f (δv)/δ. Since f maps paridians to paridians, the set g δ (L r ) is contained in a paridian. For v ∈ L r we have δv ≤ δr, and Df (0) is linear, so Since ε > 0 was arbitrary, this means that Df (0) is the uniform limit of maps g δ as δ → 0. Therefore, Df (0)(L r ) is contained in a paridian.
In case (1) there is always a proper paridian that is mapped by f to P , but then the derivative Df (0) in x-direction is 0. This direction is an eigendirection of Df (0), so the only possibility for paridians to be mapped by Df (0) to paridians is that Df (0) = 0.
Let us now consider case (2). Then sufficiently small proper paridians cannot be mapped to the x-axis, unless they are mapped to 0. Therefore, for sufficiently small δ, the set g δ (L r ) is either a proper paridian other than the x-axis, or {0}. Hence, the same is true for Df (0) instead of g δ . This means that either det(Df (0)) = 0 or Df (0) = 0.
If det(Df (0)) = 0, then f is one-to-one in a small neighborhood of 0. However, since |d(B)| > 1, in such neighborhood there are paridians, which are mapped by f not in the one-to-one manner. This is a contradiction, so we must have Df (0) = 0. Now we want to compare the number of fixed points of a paridianal map f with its degree if | deg(f )| > 1.
We have to show that in each preband the map ϕ has a fixed point (which must be different from 0 and 1 which are in the set A). Proof. We have to show that the graph of ϕ on B crosses the diagonal. This is clear if the closure of B does not contain 0 and 1. Suppose now that one endpoint of B is 0 (with 1 the situation is analogous).
If ∆(B) = −1 then clearly ϕ has a fixed point in B. Suppose that ∆(B) = 1. By Lemma 2.7, Df (P ) = 0. Therefore, ϕ (0) = 0, so the point (ε, ϕ(ε)) lies below the diagonal for small ε > 0. thus, again the graph of ϕ on B crosses the diagonal. A fixed point of ϕ in a preband gives us a proper paridian that is mapped by f to itself. The next step is to estimate the number of fixed points in this paridian, other than P , compared to the degree of f restricted to this paridian. In general, any continuous map of a circle to itself of degree d has at least |d − 1| fixed points (cf. [12]). Thus, if d ≤ 0, this number is at least |d| + 1. Taking into account that in our case one of those points is P , we get for any preband B at least |d(B)| other fixed points. This leaves us with the case when d(B) > 0, when we have to take into account the local behavior of the map at P . Proof. By Lemma 2.7, Df (P ) = 0. In our planar coordinate system this fact implies that the derivative of f at 0 in the x-direction is also 0. This means that f y : S 1 → S 1 (conjugate to f restricted to the paridian −1 (y)) has the derivative 0 at 0 (here 0 ∈ T corresponds to P ). Now considerf y : [0, 1] → R, the lift of f y . The fact that the derivative is 0 implies that the graphf y intersects d(B) straight lines of the form y = x + k, k = 0, . . . , d(B) − 1, and thus there must be at least d(B) fixed points of f y other than 0. This completes the proof.
By Lemma 2.2, P is always a fixed point of f in −1 (y), where y is a fixed point of ϕ. Thus, as a straightforward consequence of Lemma 2.10, we get the following corollary.  Proof. Using Lemma 2.6 and Corollary 2.11 we get Taking into account Lemma 2.2 in case | deg(f )| ≤ 1, we get for arbitrary degree the following corollary, which is the main results of this section. Corollary 2.13. If f is a paridianal map, then it has at least | deg(f )| fixed points.
If f is a paridianal map then f n is also a paridianal map. Moreover, deg(f n ) = deg(f ) n . As a consequence, we obtain the conclusion, in which we obtain (1.1) in a stronger version with the upper limit replaced by the lower limit.
Corollary 2.14. If f is a paridianal map, then the lower growth rate of the number of fixed points of f n is at least log | deg(f )|.

Rabbit maps
Here we use the rabbit foliation, defined in the introduction, and consider smooth (of class C 1 ) sphere maps preserving this foliation, the rabbit maps. We denote the family of all rabbit maps by R.
Let us be more precise. The pole is still P and there are two ears, which are the sets given (in the plane, after the stereographic projection, see Figure 2) in the polar coordinates by {(r, θ) : r = c sin(2θ), c ∈ [0, 1]} in the first quadrant and {(r, θ) : r = −c sin(2θ), c ∈ [0, 1]} in the second quadrant. We denote them by E 1 and E 2 , and their union by E. Each ear is foliated by the appropriate curves r = ±c sin(2θ); P is a singular point and belongs to all leaves. We divide the rest of the sphere, G = (S 2 E) ∪ {P } into two areas G 1 which is the part of G in the first and the second quadrant and G 2 which is the part in the third and fourth quadrant and define the foliation in the following way. In G 1 its leaves are given by r = 2 sin θ √ cos 2 θ + c, where c ≥ 0. In G 2 the leaves are circles r = c sin θ. Additionally, the x-axis (with the point at infinity) is also a leaf.
Two leaves are very special. They are boundaries of the ears, ∂E 1 and ∂E 2 . The leaves other than ∂E 1 , ∂E 2 and {P } will be called regular.
Fix a rabbit map f . Proof. If f (P ) = P then, since every leaf contains P , the image of the whole sphere is contained in one leaf. Therefore deg(f ) = 0.
In the rest of the paper we assume that deg(f ) = 0. In particular, f (P ) = P . We will call the leaves of the rabbit foliation simply leaves.
As in the preceding section, we want to define the map . However, now we need it only in a one-sided (from the side of G 1 ) neighborhood U of ∂E. There we parametrize the set of leaves by an interval [0, α), for a small α > 0, where 0 corresponds to the union of two leaves ∂E 1 and ∂E 2 . Thus, we have a projection : U → [0, α). We want to show that there exists ε ∈ (0, α) and ϕ : We have the orientation of proper paridians consistent throughout the sphere. Therefore if L, K are leaves other than {P }, and f (L) ⊂ K, the degree d L (f ) of the map f | L : L → K is well defined.
We start with three lemmas.
Lemma 3.2. Let (L n ) and (K n ) be sequences of leaves other than {P }, such that f (L n ) ⊂ K n , d Ln (f ) = 0, and the sequence (K n ) converges to ∂E from the side of G 1 . Then the sequence (L n ) also converges to ∂E from the side of G 1 .
Proof. Observe that since d Ln (f ) = 0, we have f (L n ) = K n . If there is a subsequence of (L n ) convergent to a leaf L, then f (L) = ∂E, but ∂E is not contained in any leaf, a contradiction. However, the only possibility that there is no subsequence of (L n ) convergent to a leaf is that (L n ) converges to ∂E from the side of G 1 . Proof. Choose a point x ∈ ∂E and a sequence (x n ) convergent to x from the side of G 1 , which are regular values of f . Since deg(f ) = 0, for each n there is a point y n ∈ f −1 (x n ) such that if L n is the leaf containing y n , then d Ln (f ) = 0 (if d Ln (f ) = 0 then the sum of the signs of the Jacobian of f over all elements of L n ∩ f −1 (x n ) is zero). Then, by Lemma 3.2, the sequence (L n ) converges to ∂E from the side of G 1 . The leaf K n = f (L n ) contains x n , so the sequence (K n ) also converges to ∂E from the side of G 1 . Therefore, f (∂E) = ∂E.
The second statement of the lemma follows from the fact that f maps leaves to leaves. The third statement follows from the second one.  Proof. We may assume that deg(f ) = 0. Choose a point x ∈ G 1 , which is a regular value of f , and lies sufficiently close to ∂E (but far from P ). By Lemma 3.2, all elements of f −1 (x) which belong to leaves L such that d L (f ) = 0, lie in −1 ((0, ε)). By Lemma 3.4 (f) and the same arguments as in the proof of Lemma 2.6, deg(f ) is equal to the common value of d −1 (t) (f ) (which has modulus 1) multiplied by the sum of the signs of ϕ at the points of ϕ −1 ( (x)). This sum has modulus not larger than 1, so | deg(f )| ≤ 1.
If deg(f ) = 0, then by Lemma 3.1 f has a fixed point. Therefore, f has at least | deg(f )| fixed points.
From this theorem we get an obvious corollary. Corollary 3.6. If f is a rabbit map, then the lower growth rate of the number of fixed points of f n is at least log | deg(f )|.