The least number of 2periodic points of a smooth selfmap of \(\varvec{S}^\mathbf{2}\) of degree 2 equals 2
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Abstract
We show that there exists a smooth selfmap of the sphere \(f:S^2\rightarrow S^2\) which has degree 2 and has only two 2periodic points.
Keywords
Fixed point periodic point Nielsen fixed point theory Dold congruences least number of periodic pointsMathematics Subject Classification
Primary 55M20 Secondary 54H25 57R991 Introduction
We fix a selfmap \(f:M\rightarrow M\) of a compact connected manifold and a natural number n. What is the least number of nperiodic points \(\#\mathrm{Fix}(h^n)\) where h runs through the homotopy class of f? If we moreover restrict to simply connected M and we allow all continuous maps homotopic to f, then the least number is always 1 or 0 [4]. However, if h runs only through the smooth homotopy class of f, then the least number may be much larger, which was noticed by Shub and Sullivan [10]. This gave rise to \(D_n^m(f)\) an algebraic lower bound of the number of nperiodic points in the smooth homotopy class of f, see [4]. In dimension \(m\ge 3\), the homotopy invariant \(D_n^m(f)\) turned out to be the best lower bound, i.e., it can be realized by a smooth map homotopic to the given f (a Weckentype theorem) [4] .
On the other hand, the sphere \(S^2\) is the unique twodimensional closed simply connected manifold. Surprisingly, the methods of reducing fixed and periodic points do not work in general on surfaces. The reason is that the Whitney trick of canceling intersection points does not hold in low dimensions, thus the Wecken theorem for periodic points works only from dimension 3 on. See [8, 9]. This makes the problem of minimizing the number of periodic points open in dimension 2, in particular for selfmaps of \(S^2\).
In [6], we started to study this case.
Theorem 1.1
(Theorem 2.6 in [6])
 (1)
\(d\le 1\)
 (2)
\(n=1\)
 (3)
\(d=2\) and \(n=2\)
In this paper and in [6] by smooth we mean \(C^{\infty }\), since the maps in Lemma 2.2, which are given explicitly in the proof of Theorem 3.7 in [1], may by represented by \(C^{\infty }\) maps.
In Sect. 3 we give the algebraic necessary condition to homotope a selfmap of \(S^2\) to a map with at most two nperiodic points. The main result of the paper is Theorem 3.2 saying that the above condition is also sufficient for \(\mathrm{deg}(f)=2\) and \(n=2\).
This must be done directly, since the techniques of reducing isolated periodic orbits with opposite indices, used in the Nielsen fixed and periodic point theory, are not available in dimension 2.
2 Indices of iterations of a smooth map
It was shown [1] that each sequence of integers \((A_k)\) satisfying Dold congruences can be realized as \(A_k=\mathrm{ind}(f^k;x_0)\), for a continuous selfmap of \({\mathbb {R}}^m\) for \(m\ge 3\). In other words, Dold congruences are the only restriction for a sequence of integers to realize the fixed point index of a continuous map.
Surprisingly, it turned out that there are much more restrictions on sequences \(A_k=\mathrm{ind}(f^k;x_0)\) when f is smooth [2, 7, 10]. In [5] it is proved that the necessary conditions given in [2] are, in dimension \(\ge 3\), also sufficient and the full description of all such sequences is given in [5]. We call them smoothly realizable in dimension m.
It is convenient to present the sequences of integers as the sum of the following elementary periodic sequences
Definition 2.1
It is easy to notice that each integer sequence \((A_n)\) can be written down uniquely in the following form of a periodic expansion: \( A_n= \sum _{l=1}^{\infty } a_l \mathrm{reg}_l(n),\) where \(a_n=\frac{1}{n} \sum _{ln} \mu (\frac{n}{l})\; A_l\) for suitable \(a_l\in {\mathbb {R}}\). Moreover, all coefficients \(a_l\) are integers if and only if the sequence \((A_n)\) satisfies Dold congruences.
The above observations, applied in dimension 2, resulted in the full description of all possible sequences smoothly realizable in dimension 2. In the next Lemma, we reformulate Theorem 3.7 in [1] using our notations
Lemma 2.2
[1] see also Lemma 1.1 in [6].
 (1)
\(\mathrm{ind}(f^k;(0,0))=c\cdot \mathrm{reg}_1(k)\) for \(c\in {\mathbb {Z}}\)
 (2)
\(\mathrm{ind}(f^k;(0,0))=\mathrm{reg}_1(k) +c\cdot \mathrm{reg}_m(k)\) for \(c\in {\mathbb {Z}}\) , \(m\in {\mathbb {N}}\) , \(c\ne 0\)
 (3)
\(\mathrm{ind}(f^k;(0,0))=c\cdot \mathrm{reg}_2(k)\) for \(c\in {\mathbb {Z}}\) , \(c\ne 0\)
 (4)
\(\mathrm{ind}(f^k;(0,0))= \mathrm{reg}_1(k) +c\cdot \mathrm{reg}_2(k)\) for \(c\in {\mathbb {Z}}\) , \(c\ne 0\) \(\square \)
Remark 2.3
 (1)
constant
 (2)
\({\left\{ \begin{array}{ll} 1+cm\; \text {if}\;\; m \mid k\\ 1 \;\text {otherwise} \end{array}\right. }\)
 (3)
\(0,2c,0,2c,\ldots \)
 (4)
\(1,1+2c,1,1+2c,\ldots \). \(\square \)
3 Algebraic necessary condition
Lemma 3.1
Let \(f:S^2\rightarrow S^2\) be a selfmap of degree d satisfying \(d\ge 2\) and let \(n\in {\mathbb {N}}\), \(n\ne 1\). Then there exist expressions \(C_1\) , \(C_2\) of types \((1)(4)\) such that the equality \(d^k+1=C_1(k)+C_2(k)\) is satisfied for all kn if and only if n is a prime number.
Proof
\(\Rightarrow \). We assume that \(d^k+1=C_1(k)+C_2(k)\) for kn where \(C_1,C_2\) are of type \((1)(4)\). We will show that n is a prime. First we assume that one of \(C_1\) , \(C_2\) is of type (1). Then, the sum \(C_1(k)+C_2(k)\) takes at most two values. Since \(d^k+1\) takes distinct values (for a fixed \(d\ge 2\)) and equality holds for all divisors of n, n must be a prime.
On the other hand if no of \(C_1,C_2\) is of type (1) then \(2\le C_1(1)+C_2(1)\le 2\) hence \(3\le d\le 1\). This proves \(\Rightarrow \) for \(d\notin \{3,2,1,0,+1\}\). It remains to consider the cases \(d=2\) , \(d=3\) where no \(C_i\) is of type (1).
Let \(d=2\). Then \(C_1(1)+C_2(1)=1\), hence one of \(C_i\) must be of type (4) and the other of type (3). Then \(C_1+C_2\) takes at most two values.
Let \(d=3\). Then \(C_1(1)+C_2(1)=2\), hence both \(C_1,C_2\) must be of type (4 ), so their sum takes only two values.
\(\square \)
In the rest of the paper, we will assume that \(n=d=2\). We will show that then the above condition is also sufficient. We will show that
Theorem 3.2
There exists a smooth map \(f:S^2\rightarrow S^2\) of degree 2 which has only two 2periodic points.
This will be done as follows. First, we give a convenient formula of a map of degree 2. Then, we deform smoothly this map near the poles to realize their appropriate values of the fixed point index. Some extra 2periodic points appear. The last step is to remove these points.
Remark 3.3
To get a smooth map we start with a map which is smooth near the poles and we use continuous deformations which are constant near the poles. Finally we deform the obtained map, with only two 2periodic points, to a smooth map by a homotopy constant near the poles. If the last deformation is sufficiently small, the poles remain the unique 2periodic points.
4 Notation
Remark 4.1
The degree of \(f_0\) equals 2, since the restrictions of \(f_0\) to \(S^2_\) and \(S^2_+\) are orientationpreserving diffeomorphisms .
Lemma 4.2
\(\mathrm{Fix}(f_0^2)\) consists of three fixed points \(N\mathrm {Pole}\), \(S\mathrm {Pole}\) and [0, 0] and two 2orbits \(\{[\frac{3\pi }{10}; 0]; [\frac{\pi }{10}; \pi ]\}\) , \(\{[\frac{3\pi }{10}; 0]; [\frac{\pi }{10}; 0]\}\).
Proof
We notice that \(f_0\) has exactly one fixed point in each sector: \(N\mathrm {Pole}\in S^2_+\) , \([0,0]\in S^2_0\) , \(S\mathrm {Pole}\in S^2_\), since the coordinate \(\theta \) is being expanded on each sector.
Now we look for 2orbits. The above argument (\(\theta \) is expanding) implies that there is no 2orbit contained in a sector. Moreover the component \(\phi \) excludes a 2orbit with a point in \(S^2_+\) and the other one in \(S^2_\). Now each 2orbit must have one element in \(S^2_0\) and the second in \(S^2_+\) or in \(S^2_\). Let \([\theta ,\phi ]\in S^2_+\) be a 2periodic point. Then \(f_0[\theta ,\phi ]=[3\theta \pi ,\phi +\pi ]\in S^2_0\) which implies \(f_0^2[\theta ,\phi ]=f[3\theta \pi ,\phi +\pi ]=[9\theta +3\pi ,0]\). Now \( [9\theta +3\pi ,0] =[\theta ,\phi ]\) implies \(9\theta +3\pi =\theta \) , \(\phi =0\), hence \(\theta =\frac{3\pi }{10}\) , \(\phi =0\). Since \(f_0[\frac{3\pi }{10}; 0]= [\frac{\pi }{10}; \pi ]\), we get the orbit \(\{[\frac{3\pi }{10}; 0]; [\frac{\pi }{10}; \pi ]\}\).
Let \([\theta ,\phi ]\in S^2_\) be a 2periodic point. Then \(f_0[\theta ,\phi ]=[3\theta +\pi ,\phi ]\in S^2_0\) which implies \(f_0^2[\theta ,\phi ]=f[3\theta +\pi ,\phi ]=[9\theta 3\pi ,0]\). Now \( [9\theta 3\pi ,0] =[\theta ,\phi ]\) implies \(9\theta 3\pi =\theta \) , \(\phi =0\), hence \(\theta =\frac{3\pi }{10}\) , \(\phi =0\). Since \(f_0[\frac{3\pi }{10}; 0]= [\frac{9\pi }{10}+\pi ; 0]= [\frac{\pi }{10};0]\), we get the orbit \(\{[\frac{3\pi }{10}; 0]; [\frac{\pi }{10}; 0]\}\). \(\square \)
In Sect. 5 we deform the map \(f_0\) near \(S\mathrm {Pole}\) to a smooth map \(f_1\) satisfying \(\mathrm{ind}(f_1;S\mathrm {Pole})=2\). Then a new fixed point \(b'\) appears and we remove simultaneously the points \(b', b, f(b),[0,0]\) by a homotopy with the carrier in a neighborhood near the meridian \(\phi =0\). We get a map \(f_2\) with \(\mathrm{Fix}(f_2^2)=\{N\mathrm {Pole}, S\mathrm {Pole}; a,f(a)\} \).
In Sect. 6 we consider again the original map \(f_0\) (not \(f_2\)) and we deform it near \(N\mathrm {Pole}\) to get a map \(f_3\) satisfying \(\mathrm{ind}(f_3;N\mathrm {Pole})=1\) , \(\mathrm{ind}(f_3;N\mathrm {Pole})=3\). This will give a new 2orbit \(\{w_0,w_2\}\). In the next deformation the orbit a, f(a) reduces with the new one. We get a map \(f_4\) with \(\mathrm{Fix}(f_4^2)=\{N\mathrm {Pole}, S\mathrm {Pole};b', b, f(b),[0,0]\} \). Finally the maps \(f^2\) and \(f^4\) define a map \({{\tilde{f}}}\), also of degree 2 satisfying \(\mathrm{Fix} ({\tilde{f}}^2)=\{N\mathrm {Pole},S\mathrm {Pole} \}\).
5 Removing the orbit \(\{b,f(b)\}\)
In this section, we will remove the orbit \(\{b,f(b)\}\) and the fixed point [0, 0] by a deformation with the carrier in a neighborhood of the arc \(<b,f(b)>\). Here is the sketch of the deformation. We start by a smooth deformation of \(f_0\) near \(S\mathrm {Pole}\) adding an additional fixed point \(b'\). The orbit \(\{b,f_0(b)\}\), the points [0, 0] and \(b'\) belong to the arc \(<S\mathrm {Pole},[\pi /6,0]>\). The map \(f_0\) sends the ends of the last arc to \(S\mathrm {Pole}\), the middle point of the arc goes to \(N\mathrm {Pole}\) and the \(f_0\) is linear on each half of the arc. The restrictions of \(f_0\) to neighbor arcs \(<S\mathrm {Pole}, [\pi /6,\phi _0]>\) look similar. We consider the region \(V_1=\{[\theta ,\phi ]; \pi /2\le \theta \le \pi /6 , \phi \le \epsilon \}\) for a small \(\epsilon >0\). We notice that \(f_0\) sends \(V_1\) to the region \(V_2=\{[\theta ,\phi ]; \pi /2\le \theta \le \pi /2 , \phi \le \epsilon \}\). We consider the restriction \(f_{0}:V_1\rightarrow V_2\). Now if we denote \(f_0[\theta ,\phi ]= [\theta ', \phi ']\) then \(\phi '=\phi \) for \(\pi /2\le \theta \le \pi /6\) and \(\phi '=0\) for \(\pi /6\le \theta \le \pi /6\). We deform the restriction of \(f_0\) to the arc \(<S\mathrm {Pole},[\pi /6,0]>\), keeping the end points fixed, by squeezing the arc to a neighborhood of \(S\mathrm {Pole}\) so that there is no periodic point inside the arc. Then, we extend this deformation to \(V_1\) by a homotopy which keeps the boundary \(\mathrm{bd}V_1\) fixed, the meridians are sent into themselves, or to the meridian 0. Finally, we compose the obtained deformation with a homeomorphism of \(V_1\) which is constant on the boundary and for \(\phi =0\) and which makes \(\phi \) smaller elsewhere. The final map is a selfmap of degree 2 with no periodic points inside \(V_1\). This gives the maps \(f_2\) with \(\mathrm{Fix}(f_2^2)=\mathrm{Fix}(f_2^2){\setminus } \{b,f(b)\} =\{N\mathrm {Pole}, S\mathrm {Pole}, a , f_2(a)\} \).
 (1)
\(f_2=f_1\) in \([\pi /2;b''] \times [\epsilon ,\epsilon ]\), for some \(b''\in (\pi /2,b')\), and on the boundary of \([\pi /2,\pi /6] \times [\epsilon ,\epsilon ]\).
 (2)
\(f_2([\pi /2,\pi /6] \times [\epsilon ,\epsilon ])\subset [\pi /2,\pi /2]\times [\epsilon ,\epsilon ]\).
 (3)
\(f^2_2 (x)\ne x\) for \(x\in (\pi /2,\pi /6] \times [\epsilon ,\epsilon ]\).
It remains to construct a map \(f_2\) satisfying (1)(3).
\(\eta : [\pi /2,\pi /6]\times [\epsilon ,+\epsilon ] \rightarrow [0,1]\) is a Urysohn function satisfying \(\eta ^{1}(0)= \mathrm{bd}([\pi /2,\pi /6]\times [\epsilon ,+\epsilon ])\cup ([\pi /2,b'']\times [\epsilon ,\epsilon ]\)) , \(\eta ^{1}(1)= [b'_,b'_+]\times 0\) and

\(\mathrm{sign}(\phi ') =\mathrm{sign}(\phi )\),

\(\phi '=\phi \) on the boundary and for \( \theta \le b''\)

\(\phi '<\phi \) for (\(0<\phi  < \epsilon \) and \(\theta \ge b''\)).
First we assume that \(\phi \ne 0\) and we show \((\theta ,\phi )\in (\pi /2,\pi /6]\times [\epsilon ,+\epsilon ]\) is not a periodic point. First we assume that \(\theta \le b''\). Then \(f_2(\theta ,\phi )=f_1(\theta ,\phi )\). If we denote \(f_2(\theta ,\phi )=(\theta ',\phi ')\) then \(\phi '<\phi \) because \({{\hat{f}}}\) has also this property near 0 (Remark 7.3). If \(b''<\theta \) then \(R(\theta ,\phi )\) makes \(\phi '<\phi \). This proves that there is no periodic point for \(\phi \ne 0\).
Now we consider a point \((\theta ,0)\).
If \(\pi /2\le \theta \le b''\) then \(p_1f_(\theta ,0)<\theta \), since the similar inequality holds for the map \({{\hat{f}}}\).
If \(b''\le \theta \le b_+\) then \(p_1f_(\theta ,0)=\bar{f}_(\theta )=\min (C,p_1f_1(\theta ))\le C<b''\le \theta \).
If \(b_+\le \theta \le \pi /6\) then \(p_1f_(\theta ,0)<b_+\le \theta \). \(\square \)
6 Removing the orbit a, f(a)
In this section, we deform the map \(f_0\) (not \(f_2\)) and we remove the other 2orbit \(\{a,f(a)\}\). The carrier of the deformation will be disjointed from the carrier of the previous deformation. At the end of the section, we will show that the two deformations give a map of degree 2 whose 2periodic points are only \(N\mathrm {Pole}\) and \(S\mathrm {Pole}\). This will end the proof of Theorem 3.2
We will say that a subset \(A\subset S^2\) is \(S^2\)convex if A does not contain antipodal points and for each \(a,a'\in A\) the geodesic joining the points is contained in A.
Let \(f_3\) be the induced map of \(S^2\). We will cancel simultaneously the orbits; \(\{a=[\frac{3\pi }{10}; 0]; f_0(a)=[\frac{\pi }{10}; \pi ]\}\) and \(\{w_0, w_2\}\) by a homotopy with the carrier in an arbitrarily prescribed neighborhood of the arc \(< f_3(a) ; w_2>\subset S^2\).
We consider the arc \(<w_0,a>\), its images \(f_3<w_0,a>=<w_2,f_3(a)>\) and \(f_3^2<w_0,a>=f_3<w_2,f_3(a)>=<w_0,S\mathrm {Pole}>.\) See Fig. 2. Since the above arcs contain no antipodal points, we can choose an \(S^2\)convex neighborhood \(V_3\subset <w_0,S\mathrm {Pole}>\) then an \(S^2\)convex neighborhood \(V_2\supset <w_2,f_3(a)>\) satisfying \(f_3(\mathrm {cl}(V_2))\subset V_3\) and an \(S^2\)convex neighborhood \(V_1\supset <w_0,a>\) satisfying \(f_3(\mathrm {cl}(V_1))\subset V_2\).
 1.
\(\mathrm {cl}(V_1)\cap f_3(\mathrm {cl}(V_1))=\emptyset \), since the elements of both sets have different longitudes
 2.
\(f_{3\mathrm {cl}(V_1)}\) is a homeomorphism, since \(\mathrm {cl}(V_1) \subset int( S^2_+)\) and \(f_3\) is a homeomorphism on \(int( S^2_+)\).
 3.
Follows from the definitions of both functions \(h_t\).
 4.
\(f_3(\mathrm {cl}(V_1))\cap h_1(\mathrm {cl}(V_1))=\emptyset \), since \(h_1(\mathrm {cl}(V_1))\subset V_3\) and the last is disjointed from \(f_3(\mathrm {cl}(V_1)\).
Lemma 6.1
 (1)
\(\mathrm {cl}(A)\cap f(\mathrm {cl}(A))=\emptyset \)
 (2)
the restriction \(f_:\mathrm{{cl}(A)} \rightarrow f(\mathrm {cl}(A))\) is a homeomorphism
 (3)
\(h_t:\mathrm {cl}(A)\rightarrow X\) is a homotopy satisfying: \(h_0=(f^2)_{\mathrm {cl}(A)}\) , \(h_t\) is constant on the boundary, \(h_1(a)\ne a\) for all \(a\in \mathrm {cl}(A)\).
 (4)
\(f(\mathrm {cl}(A))\cap h_1(\mathrm {cl}(A))=\emptyset \)
 (1)
\(f_1\) is homotopic to f by a homotopy constant outside \(f(\mathrm {cl}(A))\)
 (2)
\(f_1^2(a)\ne a\) for all \(a\in \mathrm {cl}(A)\cup f(\mathrm {cl}(A))\)
 (3)
\(\mathrm{Fix}(f_1^2)\subset \mathrm{Fix}(f^2){\setminus } (\mathrm {cl}(A)\cup f(\mathrm {cl}(A)))\).
Proof
To get (2) we fix \(a\in \mathrm {cl}(A)\). Then \( f_1^2(a)= f_1 f(a)= h_1 f^{1}_{\mathrm {cl}(A)} f(a)= h_1(a)\ne a\).
To show (3) we first prove that \(\mathrm{Fix}(f_1^2)\subset \mathrm{Fix}(f^2){\setminus } \mathrm {cl}(A)\).
By (2) it is enough to show that \(\mathrm{Fix}(f_1^2)\subset \mathrm{Fix}(f^2)\). Let \(f_1^2(x)=x\). If moreover no of the points x, f(x) belongs to \(\mathrm{cl}(A)\) then \(f^2(x)=f_1^2(x)=x\). Otherwise we may assume that \(x\in \mathrm{cl}(A)\). But now (2) implies \(f_1^2(x)\ne x\) which is a contradiction. \(\square \)
Proof of Theorem (3.2)
It remains to show that \(\mathrm{Fix}({\tilde{f}}^2)=\{N\mathrm {Pole}, S\mathrm {Pole}\}\). \(\supset \) is evident. To prove \(\subset \) we consider an orbit in \(\mathrm{Fix}({\tilde{f}}^2)\) which contains no pole. If the orbit is disjointed from \((\pi /2,\pi /6)\times (\epsilon ,\epsilon )\) then \(x={\tilde{f}}^2(x)=f^2_4(x)\). But \(\mathrm{Fix}(f_4^2)=\{N\mathrm {Pole},S\mathrm {Pole},b, f_0(b)\}\) implies that the orbit coincides with \(\{b,f_0(b)\}\). But \({\tilde{f}}\) and \(f_2\) coincide in \((\pi /2,\pi /6]\times [\epsilon ,\epsilon ]\) and \(f_2\) has no periodic points there.
\(\square \)
7 Lemmas
We consider the complex plane as the union of sectors \(\displaystyle {\mathbb {C}}=\bigcup \nolimits _{k=0}^{2n1}S_k\) where \(S_k=\{z; \frac{k\pi }{n}\le \arg (z)\le \frac{(k+1)\pi }{n}\}\). See Fig. 3 for \(n=4\).
In this section we will show
Lemma 7.1
 (1)
\(K_1\) maps each sector \( S_k=\{[r,\phi ]; \frac{k\pi }{2n}\le \phi \le \frac{(k+1)\pi }{2n}\}\) into itself,
 (2)
\(K_1\) maps each halfline \( L_k =\{[r,\phi ]\in S^2; \frac{k\pi }{n}=\phi \}\) into itself,
 (3)
\(K_1\) has exactly \(n+1\) fixed points \(w_0,w_2,\ldots ,w_{2n2}\) (lying on lines \(L_0\) , \(L_2\) ,..., \(L_{2n2}\), respectively) and (0, 0) . See Fig. 4.
 (4)
\(K_1\) has no periodic points different than \(w_0,w_2,\ldots ,w_{2n2}\) and (0, 0)
 (5)
\(\mathrm{ind}(K_1^m;\omega _i)=1\) for all \(m\in {\mathbb {N}}\) , \(i=0,2,\ldots ,2n2\).
 (6)
\(\mathrm{ind}(K_1^m;0)=1+n\)
We are going to define map \(K_1\). We consider vector field \(\Phi \) given by Fig. 3.
Lemma 7.2
 (1)
0 is the unique fixed point
 (2)
\(\phi (S_k)\subset S_k\) for each sector \(S_k=\{z; \frac{k\pi }{n}\le \arg (z)\le \frac{(k+1)\pi }{n}\}\) for \(k=0,\ldots ,2n1\).
 (3)
In particular \(\phi \) maps each halfline \(L_k=\{z; \arg (z) = \frac{k}{n}\}\) into itself.
 (4)
Points \(0,z, \phi (z)\) belong to a line \(\iff \) \(z\in L_k=\{z; \arg (z) = \frac{k}{n}\}\) for a \(k=0,\ldots ,2n1\). Moreover
\(z\in (0,\phi (z))\) \(\iff \) \(z\in L_k\) for k odd ;
\(\phi (z)\in (0,z)\) \(\iff \) \(z\in L_k\) for k even .
 (5)
If \(z\in intS_k\) then \(\mathrm{arg}(\phi (z))< \mathrm{arg}(z)\) (\(\mathrm{arg}(\phi (z))> \mathrm{arg}(z))\) for k even (k odd). \(\square \)
Let us fix two numbers \(0<\epsilon _1<\epsilon _2\) and a smooth Urysohn function \(\eta :[0,\infty )\rightarrow {\mathbb {R}}\) satisfying \(\eta (t)= 1\) for \(0\le t<\epsilon _1\) , \(\eta (t)= 0\) for \(t\ge \epsilon _2\) and moreover \(t\le t'\) implies \(\eta (t)\ge \eta (t')\).
We define a new vector field on \({\mathbb {C}}\) as the convex combination \(K(z)=\eta (z)\cdot \Phi _1(z) +(1\eta (z))\cdot az\), where \(a>1\).
Proof of Lemma 7.1
The first two properties follow from Lemma 7.2, since the map \(\Phi _1\) preserves sectors \(S_k\) and halflines \(L_k\). To prove the third property we notice that the vectors \(\phi (z)\) , az are collinear \(\iff \) \(z\in L_k\) for a \(k=0,\ldots ,2n1\). Moreover they have the same direction for k odd and are opposite for k even. Now, for k odd, their convex combination never vanishes \(K(z)\ne 0\). Similarly, for k even, K has exactly one zero for \(\epsilon _1\le z\le \epsilon _2\) , \(z\in L_k\), since \(\eta \) is nonincreasing. Now we prove (5). We notice that in each fixed point \(z_{2k}\) the map \(K_1\) is expanding the line \(L_{2k}\) and is squeezing at the orthogonal direction (since so does \(\Phi _1\)). Now fixed point index equals \((1)\cdot (+1)=1\). The same argument works for all iterations of \(K_1\). To prove (6) we notice that the total index must be \(+1\), hence \(\mathrm{ind}(K_1^m;(0,0))=1n(1)=1+n\) for any \(m\in {\mathbb {N}}\). \(\square \)
Notes
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