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Periodic points on T-fiber bundles over the circle.

Introduction

Let f : M [right arrow] M be a map and x [member of] M, where M a compact manifold. The point x is called a periodic point of f if there exists n [member of] N such that [f.sup.n] (x) = x, in this case x a periodic point of f of period n. The set of all {x [member of] M| x is periodic} is called the set of periodic points of f and is denoted by P(f).

If M is a compact manifold then the Nielsen theory can be generalized to periodic points. Boju Jiang introduced (Chapter 3 in [1]) a Nielsen-type homotopy invariant N[F.sub.n] (f) being a lower bound of the number of n-periodic points, for each g homotopic to f; Fix([g.sup.n]) [greater than or equal to] N[F.sub.n](f). In case dim(M) [greater than or equal to] 3, M compact PL- manifold, then any map f : M [right arrow] M is homotopic to a map g satisfying Fix([g.sup.n]) = N[F.sub.n](f), this was proved in [2].

Consider a fiber bundle F [right arrow] M [arrow.p] B where F, M, B are closed manifolds and f:M[right arrow]Ma fiber-preserving map over B. In natural way is to study periodic points of f on M, that is, given n [member of] N we want to study the set {x [member of] M | [f.sup.n](x) = x}. The our main question is; when f can be deformed by a fiberwise homotopy to a map g : M [right arrow] M such that Fix([g.sup.n]) = [empty set] ?

This paper is organized into three sections besides one. In Section 1 we describe our problem in the general context of fiber bundle with base and fiber closed manifolds.

In section 2, given a positive integer n and a fiber-preserving map f : M [right arrow] M, in a fiber bundle with base circle and fiber torus, we present necessary and sufficient conditions to deform [f.sup.n] : M [right arrow] M to a fixed point free map over [S.sup.1], see Theorem 2.3. In the Theorem 2.4 we described linear models of maps, on the universal covering of the torus, which induces fiber-preserving maps on the fiber bundle.

In section 3, in the Theorem 3.1, we used the models of maps of the section 2 to find a map g : M [right arrow] M, fiberwise homotopic to a given map f : M [right arrow] M such that [g.sup.n] : M [right arrow] M is a fixed point free map over [S.sup.1].

1 General problem

Let F [right arrow] M [arrow.p] B be a fibration and f : M [right arrow] M a fiber-preserving map over B, where F, M, B are closed manifolds. Given n [member of] N, from relation p [??] f = p, we obtain p [??] [f.sup.n] = p, thus [f.sup.n] : M [right arrow] M is also a fiber-preserving map for each n [member of] N. We want to know when f can be deformed by a fiberwise homotopy to a map g : M [right arrow] M such that Fix([g.sup.n]) = [empty set]. The the following lemma give us a necessary condition to a positive answer the question above.

Lemma 1.1. Let f : M [right arrow] M be a fiber-preserving map and n a positive integer. If the map [f.sup.k] : M [right arrow] M can not be deformed to a fixed point free map by a fiberwise homotopy, where k is a positive divisor ofn, then there is not map g : M [right arrow] M fiberwise homotopic to f such that [g.sup.n] : M [right arrow] Mis a fixed point free map.

Proof. Suppose that exists g ~B f such that Fix([g.sup.n]) = [empty set]. Since Fix([g.sup.k]) [subset] Fix([g.sup.n]) and Fix([g.sup.k]) [not equal to] [empty set] then we have a contradiction.

Therefore, a necessary condition to deform f : M [right arrow] M to a map g : M [right arrow] M by a fiberwise homotopy, such that Fix([g.sup.n]) = [empty set], is that for all positive integer k, where k divides n, the map [f.sup.k] : M [right arrow] M must be deformed by a fiberwise homotopy to a fixed point free map.

Note that for each n the square of the following diagram is commutative;

[mathematical expression not reproducible]

In our case, all topological spaces are path-connected then we will represent the generators of the groups [[pi].sub.1] (M,[f.sup.n]([x.sub.0])) for each n, with the same letters. The same thing we will do with [[pi].sub.1] (T,[f.sup.n](0)).

Let MxBM be the pullback of p : M [right arrow] B by p : M [right arrow] B and [p.sub.i] : M x BM [right arrow] M, i = 1,2, the projections to the first and the second coordinates, respectively.

The inclusion M [x.sub.B] M - [DELTA] [right arrow] MxB M, where [DELTA] is the diagonal in M x B M, is replaced by the fiber bundle q : EB (M) [right arrow] M x B M, whose fiber is denoted by F. We have [[pi].sub.m](EB(M)) [approximately equal to] [[pi].sub.m](M xBM-[DELTA]) where EB(M) = {(x,[omega]) [member of] B x [A.sup.I]|i (x) = [omega](0)}, with A = MxBM,B = M [section]B M - [DELTA] and q is given by q(x,[omega]) = [omega](1).

E. Fadell and S. Husseini in [4] studied the problem to deform the map [f.sup.n], for each n [member of] N, to a fixed point free map. They supposed that dim(F) [greater than or equal to] 3 and that F, M, B are closed manifolds. The necessary and sufficient condition to deform [f.sup.n] is given by the following theorem that the proof can be find in [4].

Theorem 1.1. Given a positive integer n, the map [f.sup.n] : M [right arrow] Mis deformable to a fixed point free map if and only if there exists a lift [sigma](n) in the following diagram;

[mathematical expression not reproducible] (1)

where [E.sub.B]([f.sup.n]) [right arrow] M is the fiber bundle induced from q by (1,[f.sup.n]).

In the Theorem 1.1 we have [[pi].sub.j-1](F) [congruent to] [[pi].sub.j](M [x.sub.B] M, M [x.sub.B] M - [DELTA]) [congruent to] [[pi].sub.j](F, F - x) where x is a point in F. In this situation, that is, dim(F) [greater than or equal to] 3 the classical obstruction was used to find a cross section.

When F is a surface with Euler characteristic [less than or equal to] 0 then by Proposition 1.6 from [5] we have necessary e sufficient conditions to deform [f.sup.n] to a fixed point free map over B. The next proposition gives a relation between a geometric diagram and our problem.

Proposition 1.1. Let f : M [right arrow] M be a fiber-preserving map over B. Then there is a map g, g ~B f, such that Fix([g.sup.n]) = [empty set] if and only if there is a map [h.sub.n] : M [right arrow] M XB M - [DELTA] of the form [h.sub.n] = (Id, [s.sup.n]), where s : M [right arrow] M, isfiberwise homotopic to f and makes the diagram below commutative up to homotopy.

[mathematical expression not reproducible] (2)

Proof ([right arrow]) Suppose that exists g : M [right arrow] M, g ~B f, with Fix([g.sup.n]) = [empty set]. Is enough to define [h.sub.n] = (Id,[g.sup.n]), that is, s = g.

([??]) If there is [h.sub.n] : M[right arrow]MxBM - [DELTA] such that [h.sub.n] = (Id, [s.sup.n]), where s ~B f, then [s.sup.n](x) [not equal to] x for all x [member of] M. Thus, takes g = s.

2 Torus fiber-preserving maps

Let T be, the torus, defined as the quotient space R x R/Z x Z. We denote by (x,y) the elements of R x R and by [(x, y)] the elements in T.

Let [mathematical expression not reproducible] be the quotient space, where A is a homeomorphism of T induced by an operator in [R.sup.2] that preserves ZxZ. The space MA is a fiber bundle over the circle [S.sup.1] where the fiber is the torus. For more details on these bundles see [5].

Given a fiber-preserving map f : MA [right arrow] MA, i.e. p [??] f = p, we want to study the set Fix([g.sup.n]) for each map g fiberwise homotopic to f.

Consider the loops in MA given by; a(t) =< [(t, 0)],0 >,b(t) = < [(0, t)],0 > and c(t) =< [(0,0)], t > for t [member of] [0,1]. We denote by B the matrix of the homomorphism induced on the fundamental group by the restriction of f to the fiber T. From [5] we have the following theorem that provides a relationship between the matrices A and B, where

[mathematical expression not reproducible].

From [5] the induced homomorphism f# : [[pi].sub.1](MA) [right arrow] [[pi].sub.1](MA) is given by; [mathematical expression not reproducible]. Thus

[mathematical expression not reproducible].

Theorem 2.1. (1) [[pi].sub.1](MA,0) = (a,b,c|[a,b] = 1,ca[c.sup.-1] = [a.sup.a1]b[a.sup.2],cb[c.sup.-1] = [a.sup.a3][b.sup.a4])

(2) B commutes with A.

(3) Iff restricted to the fiber is deformable to a fixed point free map then the determinant of B - I is zero, where I is the identity matrix.

(4) If v is an eigenvector of B associated to 1 (for B [not equal to] Id) then A(v) is also an eigenvector of B associated to 1.

(5) Consider w = A(v) if the pair v, w generators ZxZ, otherwise let w be another vector so that v, w span ZxZ. Define the linear operator P:RxR[right arrow]RxRby [mathematical expression not reproducible] and [mathematical expression not reproducible]. Consider an isomorphism of fiber bundles, also denoted by P, P : MA [right arrow] M([A.sup.1]) where [A.sup.1] = P * A * [P.sub.-1]. Then MA is homeomorphic to M([A.sup.1]) over [S.sup.1]. Moreover we have one of the cases of the table below with [B.sup.1] = P * A * [P.sub.-1] and [B.sup.1] [not equal to] Id, except in case I:
Case I    [mathematical expression not reproducible]
Case II   [mathematical expression not reproducible]
Case III  [mathematical expression not reproducible]
Case IV   [mathematical expression not reproducible]
Case V    [mathematical expression not reproducible]


From Theorem 4.1 in [5], we have necessary and sufficient conditions to deform f to a fixed point free map over [S.sup.1]. The next theorem is equivalent to Theorem 4.1 in [5], this equivalence was made in [6].

Theorem 2.2. A fiber-preserving map f : MA [right arrow] MA can be deformed to a fixed point free map by a homotopy over [S.sup.1] if and only if one of the cases below holds:

(1) MA is as in case I and f is arbitrary

(2) MA is as in one of the cases II or III and c1 (b4 - 1) - c2b3 = 0

(3) MA is as in case IV and [b.sub.4]([b.sub.3] + 1) - 1 - [c.sub.1] ([b.sub.4] - 1) + [b.sub.3][c.sub.2] = 0 mod 2 except when:

[a.sub.3] is odd and [mathematical expression not reproducible] or

[a.sub.3] is even and [([c.sub.1],[c.sub.2])] = [(0,0)], with [mathematical expression not reproducible].

(4) MA is as in case V and either

[a.sub.3] is even and [mathematical expression not reproducible] mod 2, except when [mathematical expression not reproducible] and [mathematical expression not reproducible] are odd, or

[a.sub.3] is odd and [mathematical expression not reproducible] mod 2 except when 1 + [c.sub.2] and [mathematical expression not reproducible] are odd, where L := gcd([b.sub.4] - 1, [c.sub.2]).

Given n [member of] N we denote the induced homomorphism [mathematical expression not reproducible] by [mathematical expression not reproducible] and [mathematical expression not reproducible], where [b.sub.j1] = [b.sub.j],j = 1,...,4 and [c.sub.j1] = [c.sub.j,j] = 1,2. Thus the matrix of the homomorphism induced on the fundamental group by the restriction of [f.sup.n] to the fiber T is given by:

[mathematical expression not reproducible],

where [B.sub.1] = B is the matrix of ([f.sub.|T])# and [B.sub.n] = [B.sup.n]. From [8] we have

N([h.sup.n]) = |L([h.sup.n])| = |det([[[h.sub.#]].sup.n] - I)|,

for each map h : T [right arrow] T on torus, where [[h.sub.#]] is the matrix of induced homomorphism and I is the identity.

Since ([B.sup.n] - I) = (B - I)([B.sup.n-1] +... + B + I) then det([B.sup.n] - I) = det(B - I)det([B.sup.n-1] +... + B + I). Therefore, if [f.sub.|T] is deformable to a fixed point free map then [mathematical expression not reproducible] is deformable to a fixed point free map.

Remark 2.1. C.Y.You in [10] proved that if h : X [right arrow] X is a map, where X is a torus, then there exist g homotopic to h such that N[F.sub.n](h) = #Fix([g.sup.n]). Note that we do not have yet the Nielsen Jiang number defined for a map f : M [right arrow] M in a fiber bundle over B. This work investigates when there exist a such map g,fiberwise homotopic to f, with Fix([g.sup.n]) = [empty set], with n > 1.

In the Theorems 2.1 and 2.2, putting [f.sup.n] in the place of f we will get conditions to [f.sup.n]. The conditions in Theorem 2.1 to [f.sup.n] is the same of f but the conditions to [f.sup.n] in the Theorem 2.2 are different of f and are in the Theorem 2.3.

Given a fiber-preserving map f : MA [right arrow] MA, if f [~.sub.S1] g then [f.sup.n] [~.sub.S1] [g.sup.n]. Therefore, if Fix ([g.sup.n]) = [empty set] then the homomorphism [[f.sub.#].sup.n] : [[pi].sub.1](M) [right arrow] [pi] (M) satisfies the condition of deformability gives in [5].

Proposition 2.1. Let f : MA [right arrow] MA be a fiber-preserving map, where MA is a T-bundle over [S.sup.1]. Suppose that f restricted to the fiber can be deformed to a fixed point free map. This implies L([f.sub.|T]) = 0. From Theorem 2.1 we can suppose that the induced homomorphism f# : [[pi].sub.1](MA) [right arrow] [[pi].sub.1](MA) is given by; [mathematical expression not reproducible], Given n [member of] N then from relation [mathematical expression not reproducible] we obtain;

[mathematical expression not reproducible],

Proof. In fact, [mathematical expression not reproducible] and [mathematical expression not reproducible]. Suppose [mathematical expression not reproducible] and [mathematical expression not reproducible]. Then,

[mathematical expression not reproducible]

We will denote; [mathematical expression not reproducible] and [mathematical expression not reproducible]. m

Theorem 2.3. Let f : MA [right arrow] MA be a fiber-preserving map, where MA is a T-bundle over [S.sup.1]. Suppose that f restricted to the fiber can be deformed to a fixed point free map and that the induced homomorphism f# : [[pi].sub.1] (MA) [right arrow] [[pi].sub.1] (MA) is given by; f#(a) = a, [mathematical expression not reproducible] as in cases of the Theorem 2.2. If n is a positive integer, then [f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1] if and only if the following conditions are satisfies;

1) MA is as in case I and f is arbitrary.

2) MA is as in cases II, III [mathematical expression not reproducible]

3) MA is as in case IV and n([b.sub.4]([b.sub.3] + 1) - 1 - [c.sub.1]([b.sub.4] - 1) + [b.sub.3][c.sub.2]) - (n - 1) ([b.sub.4] - 1) = 0 mod 2 except when:

[a.sub.3] is odd and [mathematical expression not reproducible] or

[a.sub.3] is even and [mathematical expression not reproducible].

4) MA is as in case V and either [a.sub.3] is even and [mathematical expression not reproducible] mod 2, except when [mathematical expression not reproducible] and [mathematical expression not reproducible] are odd, or [a.sub.3] is odd and [mathematical expression not reproducible] mod 2 except when (1 + [c.sub.2]) (1 + (n - 1)[b.sub.4]) and [mathematical expression not reproducible] are odd, where L := gcd([b.sub.4] - 1, [c.sub.2]).

Proof. By Proposition 2.1 we know [mathematical expression not reproducible]

(1) From Theorem 2.2 each map f : MA [right arrow] MA is fiberwise homotopic to a fixed point free map over [S.sup.1] in particular that happens with [f.sup.n] : MA [right arrow] MA for each n [member of] N.

(2) If [b.sub.4] = 1 then the assumption of the Theorem means [c.sub.2][b.sub.3] = 0. Moreover [mathematical expression not reproducible] and [c.sub.2n] = n[c.sub.2]. In this sense, following Theorem 2.2, in cases II and III, [f.sup.n] can be deformed, by a fiberwise homotopy, to a fixed point free map if and only if [mathematical expression not reproducible]. However, [mathematical expression not reproducible], and [mathematical expression not reproducible] if and only if [c.sub.2][b.sub.3] = 0.

For [b.sub.4] [not equal to] 1 we have [mathematical expression not reproducible] and [mathematical expression not reproducible].

Note that; [mathematical expression not reproducible]

Therefore, [mathematical expression not reproducible]. In fact,

[mathematical expression not reproducible].

Therefore,

[mathematical expression not reproducible].

(3) Following Theorem 2.2, in cases IV, [f.sup.n] can be deformed, by a fiberwise homotopy, to a fixed point free map iff [mathematical expression not reproducible] 0 mod 2 except when a3 even and [mathematical expression not reproducible], or [a.sub.3] odd and [mathematical expression not reproducible].

As in (2), we hace [mathematical expression not reproducible] and [mathematical expression not reproducible]. Thus, [mathematical expression not reproducible].

The exceptions holds for [a.sub.3] even and [mathematical expression not reproducible], or a3 odd and [mathematical expression not reproducible].

In this sense, we have [mathematical expression not reproducible]. If [a.sub.3] is odd then [mathematical expression not reproducible] and [mathematical expression not reproducible]. If [a.sub.3] is even then [mathematical expression not reproducible] mod 2 and [mathematical expression not reproducible].

(4) From Theorem 2.2 the map [f.sup.n] can be deformed, over [S.sup.1], to a fixed point free map if and only if the following condition holds: [a.sub.3] is even and [mathematical expression not reproducible], except when [mathematical expression not reproducible] and [mathematical expression not reproducible] are odd, or [a.sub.3] is odd and [mathematical expression not reproducible] mod 2 except when 1 + [c.sub.2n] and [mathematical expression not reproducible] are odd, where L := gcd([b.sub.4n] - 1,[c.sub.2n]).

Note that if [b.sub.4] = 1 then from Theorem 2.1 we must have [b.sub.3] = 0 and this situation return in the case I. Therefore let us suppose [b.sub.4] [not equal to]1.

From previous calculation we have; [mathematical expression not reproducible] and [mathematical expression not reproducible]. From Theorem 2.1 we have [a.sub.3]([b.sub.4] - 1) = 2[b.sub.3].

Suppose [a.sub.3] even. Since [mathematical expression not reproducible]. Then [mathematical expression not reproducible]. In fact,

[mathematical expression not reproducible].

We have defined L := gcd([b.sub.4] - 1,[c.sub.2]). Therefore, kL = gcd(k([b.sub.4] - 1),k[c.sub.2]). We also define L := gcd([b.sub.4n] - 1,[c.sub.2n]). Now [mathematical expression not reproducible], since [mathematical expression not reproducible] and [mathematical expression not reproducible]. Furthermore, [mathematical expression not reproducible]. With these calculations we obtain the conditions statements on the theorem.

In the case [a.sub.3] odd we must have:[mathematical expression not reproducible] mod 2 except when [mathematical expression not reproducible] and [mathematical expression not reproducible] are odd, where L := gcd([b.sub.4] - 1,[c.sub.2]).

Note that [mathematical expression not reproducible] is even if and only if 1 + (n - 1)[b.sub.4] is even, and [b.sub.4] - 1 is even if and only if [b.sub.4] - 1 is even, for all n [member of] N. With this we obtain the enunciate of the theorem.

Corollary 2.1. From Theorem 2.3, if f : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1] and n is a odd positive integer, then [f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1].

Proof. If f : MA [right arrow] MA is deformed to a fixed point free map over [S.sup.1] then the conditions of the Theorem 2.2 are satisfied. Suppose n odd then the conditions of the Theorem 2.3 also are satisfied. Thus [f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1].

In the corollary above if n is even the above statement may not holds, for example in the case V of the Theorem 2.3 if n, [b.sub.4], [a.sub.3] and [mathematical expression not reproducible] are even then f : MA [right arrow] MA is deformed to a fixed point free map over [S.sup.1] but [f.sup.n] is not.

Proposition 2.2. Let f : MA [right arrow] MA be a fiber-preserving map. Suppose that for some n, odd positive integer, [f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1], as in Theorem 2.3. If k is a positive divisor of n then the map [f.sup.k] : MA [right arrow] MA can be deformed, by afiberwise homotopy, to a fixed point free map over [S.sup.1].

Proof. It is enough to verify that if the conditions of the Theorem 2.3 are satisfied for some n > 1 odd then those conditions are also satisfied for n = 1. The validity of the conditions for any k which divides n follows of the Corollary 2.1. We will analyze each case of the Theorem 2.3.

Case I. In this case for each n [member of] N the fiber-preserving map can be deformed over [S.sup.1] to a fixed point free map.

Cases II and III. In these cases if for some n odd the fiber-preserving map [f.sup.n] : MA [right arrow] MA is deformed to a fixed point free map over [S.sup.1] then we must have; [c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0. Thus, for all k [less than or equal to] n, [f.sup.k] can be deformed to a fixed point free map over [S.sup.1], in particular when k divides n.

Case IV. Suppose that for some odd positive integer n the fiber-preserving map [f.sup.n] : MA [right arrow] MA is deformed to a fixed point free map over [S.sup.1], then n([b.sub.4]([b.sub.3] + 1) - 1 - [c.sub.1]([b.sub.4] - 1) + [b.sub.3][c.sub.2]) - (n - 1)([b.sub.4] - 1) [equivalent to] 0 mod 2 and if [a.sub.3] is odd then [mathematical expression not reproducible] or if [a.sub.3] is even then [mathematical expression not reproducible].

Suppose [a.sub.3] is odd. If f : MA [right arrow] MA can not be deformed to a fixed point free map over [S.sup.1], then we must have [mathematical expression not reproducible] or [mathematical expression not reproducible] odd, that is, [c.sub.2] - 2[c.sub.1] [equivalent to] 0 mod 4 or ([b.sub.4]([b.sub.3] + 1) - 1 - [c.sub.1] ([b.sub.4] - 1) + [b.sub.3][c.sub.2]) odd. Note that ([b.sub.4]([b.sub.3] + 1) - 1 - [c.sub.1]([b.sub.4] - 1) + [b.sub.3][c.sub.2]) odd iff n([b.sub.4]([b.sub.3] + 1) - 1 - [c.sub.1]([b.sub.4] - 1) + [b.sub.3][c.sub.2]) - (n - 1)([b.sub.4] - 1) odd for any n odd. Now, if [c.sub.2] - 2[c.sub.1] [equivalent to] 0 mod 4 then we have [c.sub.2] even and therefore [c.sub.2] - 2[c.sub.1] - (n - 1)[b.sub.3][c.sub.2] = 0 mod 4. Thus, we have [c.sub.2] - 2[c.sub.1] - (n - 1)[b.sub.3][c.sub.2] [equivalent to] 0 mod 4 or n([b.sub.4]([b.sub.3] + 1) - 1 -[c.sub.1] ([b.sub.4] - 1) + [b.sub.3][c.sub.2]) - (n - 1)([b.sub.4] - 1) odd. These two conditions guarantee that[f.sup.n] can not be deformed to a fixed point free map over [S.sup.1], which is a contradiction by hypothesis.

If [a.sub.3] is even then

[mathematical expression not reproducible].

Then, f : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1].

Case V. Suppose that for some n odd, n [member of] N the fiber-preserving map [f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1].

If [a.sub.3] is even then [f.sup.n] can be deformed if [mathematical expression not reproducible] mod 2, except when [mathematical expression not reproducible] and [mathematical expression not reproducible] are odd, where L := gcd([b.sub.4] - 1, [c.sub.2]). But [mathematical expression not reproducible] even implies [mathematical expression not reproducible] even, and [mathematical expression not reproducible] odd implies [mathematical expression not reproducible] odd. Therefore, f : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1]. The case [a.sub.3] odd is analogous.

Proposition 2.3. Letf : MA [right arrow] MA be a fiber-preserving map. If m, n are odd positive integers, then [f.sup.m] is deformable to a fixed point free map over [S.sup.1] if and only if [f.sup.n] is deformable to a fixed point free map over [S.sup.1].

Proof. If m,n are odd and [f.sup.m] is deformable to a fixed point free map over [S.sup.1] then by Proposition 2.2 f is deformable to a fixed point free map over [S.sup.1]. From Corollary 2.1 [f.sup.n] is deformable to a fixed point free map over [S.sup.1].

We have a analogous result for even numbers;

Proposition 2.4. Let f : MA [right arrow] MA be a fiber-preserving map, where MA is a T-bundle over [S.sup.1]. Suppose that the induced homomorphism f# : [[pi].sub.1](MA) [right arrow] [[pi].sub.1](MA) is given by; [mathematical expression not reproducible] as in cases of the Theorem 2.2. Given an even positive integer n such that [f.sup.n] is deformable to a fixed point free map over [S.sup.1], as in Theorem 2.3, then [f.sup.k] is deformable to a fixed point free map over [S.sup.1], for all even positive integer k divisor of n.

Proof. Is enough to verify that if the conditions of the Theorem 2.3 are satisfied for some n even then those conditions are also satisfied by every even k. We will analyze each case of the Theorem 2.3.

Case I. In this case for each n [member of] N the fiber-preserving map can be deformed over [S.sup.1] to a fixed point free map.

Cases II and III. In these cases if for some n even the fiber-preserving map [f.sup.n] : MA [right arrow] MA is deformed to a fixed point free map over [S.sup.1] then we must have; [c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0 or [b.sub.4] = - 1. Thus, for all even k, [f.sup.k] can be deformed to a fixed point free map over [S.sup.1].

Case IV. If n is an even positive integer and [f.sup.n] : MA [right arrow] MA is deformed to a fixed point free map over [S.sup.1], then n([b.sub.4]([b.sub.3] + 1) - 1 - [c.sub.1]([b.sub.4] - 1) + [b.sub.3][c.sub.2]) - (n - 1) ([b.sub.4] - 1) = 0 mod 2 and

if [a.sub.3] is odd then [mathematical expression not reproducible] or

if [a.sub.3] is even then [mathematical expression not reproducible].

Note that [b.sub.4] is odd when n is even. If [a.sub.3] is odd then [b.sub.4] = 1 and

[mathematical expression not reproducible];

If [a.sub.3] is even we have

[mathematical expression not reproducible]

[mathematical expression not reproducible] and [b.sub.3][c.sub.2] [equivalent to] 1 mod 2.

Note that, if [f.sup.n] can be deformed to a fixed point free map over [S.sup.1] then n [equivalent to] 2 mod 4. Let k be an even positive integer, then

k([b.sub.4]([b.sub.3] + 1) - 1 - [c.sub.1]([b.sub.4] - 1) + [b.sub.3][c.sub.2]) - (k - 1)([b.sub.4] - 1) = 0 mod 2.

Hence, [f.sup.k] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1] except when k [equivalent to] 0 mod 4 since;

if [a.sub.3] is odd then

[mathematical expression not reproducible]

because [c.sub.2] - 2[c.sub.1] - (k - 1)[b.sub.3][c.sub.2] [equivalent to] 1 mod 2 if [a.sub.3] is even then

[mathematical expression not reproducible].

Case V. If n is an even positive integer and [f.sup.n] : MA [right arrow] MA is deformed to a fixed point free map over [S.sup.1], then

if [a.sub.3] is odd then [mathematical expression not reproducible] mod 2 and at least one of (1 + [c.sub.2])(1 + (n - 1)[b.sub.4]) and [mathematical expression not reproducible] is even, where L := gcd([b.sub.4] - 1, [c.sub.2]), or if [a.sub.3] is even then [mathematical expression not reproducible] mod 2 and at least one of [mathematical expression not reproducible] and [mathematical expression not reproducible] is even, where L := gcd([b.sub.4] - 1, [c.sub.2]).

Let [a.sub.3] odd and k an even positive integer then

[mathematical expression not reproducible].

Then, [f.sup.k] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1] for [a.sub.3] odd. Let [a.sub.3] even and k an even positive integer then

[mathematical expression not reproducible].

Then, [f.sup.k] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1] for [a.sub.3] even.

Given n [equivalent to] N and f : MA [right arrow] MA a fiber-preserving. If [f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1], then from Propositions 2.3 and 2.4 the conditions to deform f and [f.sup.n] to a fixed point free map over [S.sup.1] are enough to deform [f.sup.k] to a fixed point free map over [S.sup.1] for all k divisor of n.

Theorem 2.4. Let f : T x I [right arrow] T x I be the map defined by;

f(x,y,t) = (x + b3y + c1t + [epsilon],[b.sub.4] y + c2t + [delta],t).

Denoting [f.sup.n] : T x I [right arrow] T x I by [f.sup.n] (x, y, t) = ([x.sub.n], [y.sub.n], t), then [x.sub.n] and [y.sub.n] are given by

[mathematical expression not reproducible]

If for each positive integer n and [epsilon], [delta] satisfying the following conditions, in each case of the Theorem 2.1,

Case I) [a.sub.1][member of] + [a.sub.3][delta] = [member of] + k and [a.sub.2][member of] + [a.sub.4] [delta] = [delta] + l where k,l [member of] Z

Case II) [a.sub.3][delta] [member of] Z

Case III) [a.sub.3][delta] [member of] Z and [delta] =[2.sup.k], k [member of] Z

Case IV) [mathematical expression not reproducible] and [mathematical expression not reproducible] where m,k [member of] Z

Case V) [mathematical expression not reproducible] where k [member of] Z

then the map f : T x I [right arrow] TxI induces a fiber-preserving map in the fiber bundle MA, as in Theorem 2.1, such that the induce homomorphism f# is given by; [mathematical expression not reproducible]. Moreover, the map [f.sup.n] :T x I [right arrow] T x I induces a fiber-preserving map in the fiber bundle MA, which we will represent by [f.sup.n](< x,y,t >) =< [x.sub.n],[y.sub.n],t >, such that the induces homomorphism ([f.sup.n])# is as in the Proposition 2.1.

Proof. Denote [f.sup.n](x,y,t) = ([x.sub.n],[y.sub.n],t) for each positive integer n. We have [x.sub.2] = [x.sub.1] + [b.sub.3][y.sub.1] + [c.sub.1]t + [epsilon] = (x + [b.sub.3]y + [c.sub.1]t + [epsilon]) + [b.sub.3]([b.sub.4]y + [c.sub.2]t + [delta]) + [c.sub.1]t + [epsilon] = x + [b.sub.3]y([b.sub.4] + 1) + (2[c.sub.1] + b3[c.sub.2])t + [b.sub.3][delta] + 2[epsilon]. Also, [y.sub.2] = [b.sub.4][y.sub.1] + [c.sub.2]t + [delta] = [b.sub.4]([b.sub.4] y + [c.sub.2]t + [delta]) + [c.sub.2]t + [delta] = [b.sup.2.sub.4]y + [c.sub.2]([b.sub.4] + 1)t + ([b.sub.4] + 1)[delta].

Suppose that [f.sup.n] (x, y, t) = ([x.sub.n], [y.sub.n], t) as in hypothesis, then

[f.sup.n+1] (x,y,t) = ([x.sub.n] + [b.sub.3][y.sub.n] + [c.sub.1]t + [epsilon], [b.sub.4][y.sub.n] + [c.sub.2]t + [delta], t) = ([x.sub.n+1],[y.sub.n+1],t),

where; [x.sub.n+1] = [x.sub.n] + [b.sub.3][y.sub.n] + [c.sub.1]t + [epsilon]

[mathematical expression not reproducible];

as we wish. Now, we will verify that [mathematical expression not reproducible].

We have; [mathematical expression not reproducible], and [mathematical expression not reproducible].

Now, we will analyze each case of the Theorem 2.1.

Case I. In this case we need consider [b.sub.3] = 0 and [b.sub.4] = 1. Thus, in MA we have f(< x,y, 0 >) =< x + [member of],y + [delta],0 > =< [a.sub.1]x + [a.sub.3]y + [a.sub.1][member of] + [a.sub.3][delta],[a.sub.2]x + [a.sub.4]y + [a.sub.2][member of] + [a.sub.4][delta],1] > and [mathematical expression not reproducible]. Therefore,[mathematical expression not reproducible] if [a.sub.1][member of] + [a.sub.3][delta] = [member of] + k and [a.sub.2][member of] + [a.sub.4][delta] = [delta] + l where k,l [member of] Z.

Case II. In this case we have [a.sub.1] = [a.sub.4] = 1, [a.sub.2] = 0 and [a.sub.3]([b.sub.4] - 1) =0. Therefore, f(< x,y,0 >) =< x + [b.sub.3]y + [member of],[b.sub.4]y + [delta],0 >=< x + ([a.sub.3] + [b.sub.3])y + [member of] + [a.sub.3][delta],[b.sub.4]y + [delta], 1 >= < x+ ([a.sub.3] + [b.sub.3])y + [member of] + [a.sub.3][delta],[b.sub.4]y + [delta],1 >,and [mathematical expression not reproducible]. Thus, [mathematical expression not reproducible] if [a.sub.3][delta] [member of] Z.

Case III. In this case we have [a.sub.1] = 1, [a.sub.4] = - 1, [a.sub.2] = 0 and [a.sub.3]([b.sub.4] - 1) = -2[b.sub.3]. Therefore, [mathematical expression not reproducible], and [mathematical expression not reproducible]. Then, [mathematical expression not reproducible] if [a.sub.3][delta] [member of] Z and [mathematical expression not reproducible].

Case IV. In this case we have [a.sub.1] = - 1, [a.sub.4] = - 1, [a.sub.2] = 0 and [a.sub.3]([b.sub.4] - 1) = 0.Thus, [mathematical expression not reproducible], and [mathematical expression not reproducible]. Therefore, [mathematical expression not reproducible] and [mathematical expression not reproducible] where m,k [member of] Z.

Case V. In this case we have [a.sub.1] = -1, [a.sub.4] = 1, [a.sub.2] = 0 and [a.sub.3]([b.sub.4] - 1) = 2[b.sub.3]. Therefore, [mathematical expression not reproducible] and [mathematical expression not reproducible]. Thus, [mathematical expression not reproducible] where k [member of] Z.

In an analogous way we obtain the following conditions for [f.sup.n], in each case of the Theorem 2.1.

Case I) [mathematical expression not reproducible]

Case II) [mathematical expression not reproducible]

Case III) [mathematical expression not reproducible]

Case IV) [mathematical expression not reproducible],

Case V) [mathematical expression not reproducible]

where k, l [member of] Z. Thus for each n [member of] N and [member of], [delta] satisfying the conditions above the map [f.sup.n] : T x I [right arrow] T x I induces a fiber-preserving map on MA which will be represent by the same symbol.

Proposition 2.5. Let n, [b.sub.3], [b.sub.4], [c.sub.1], [c.sub.2] [member of] Z, n [greater than or equal to] 1. If [c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0 then for all [epsilon], [delta] [member of] R there are [k.sub.n], [l.sub.n] [member of] Z such that [x.sub.n] = x + [k.sub.n] and [y.sub.n] = y + [l.sub.n] has solution (x,y,t) [member of] [R.sup.2] x I, where:

[mathematical expression not reproducible];

Proof Suppose [b.sub.4] = 1 and [b.sub.4] = -1 with n even ([b.sub.4] = - 1 with n odd is allowed) and [c.sub.1] (b4 - 1) - [b.sub.3][c.sub.2] = 0 then given [epsilon], [delta] [member of] R we have the solutions x[member of]R and:

[mathematical expression not reproducible];

Thus, we need to find [k.sub.n], [l.sub.n] [member of] Z such that 0 [less than or equal to] t [less than or equal to] 1. Let [mathematical expression not reproducible] and [mathematical expression not reproducible]. If 0 [less than or equal to] [[DELTA].sub.1] [less than or equal to] [[DELTA].sub.0] or [[DELTA].sub.0] [less than or equal to] [[DELTA].sub.1] < 0 let [k.sub.n] = [l.sub.n] = 0, then [mathematical expression not reproducible]. If 0 < [[DELTA].sub.0] [less than or equal to] [[DELTA].sub.1] or [[DELTA].sub.1] [less than or equal to] 0 < [[DELTA].sub.0] then there are d, q [member of] Z such that [[DELTA].sub.1] = d[[DELTA].sub.0] + q with 0 [less than or equal to] q < [[DELTA].sub.0]. Let [k.sub.n] = n[c.sub.1]d and [l.sub.n] = n[c.sub.2]d, then

[mathematical expression not reproducible].

If [[DELTA].sub.1] [less than or equal to] [[DELTA].sub.0] < 0 or [[DELTA].sub.0] < 0 [less than or equal to] [[DELTA].sub.1] then there are d, q [member of] Z such that [[DELTA].sub.1] = d[[DELTA].sub.0] + q with 0 [less than or equal to] q < |[[DELTA].sub.0]|. Let k [member of] Z the least integer greater than [mathematical expression not reproducible] and [l.sub.n] = n[c.sub.2](d - k), then

[mathematical expression not reproducible].

Then, 0 [less than or equal to] t [less than or equal to] 1. If [b.sub.4] = 1 and [c.sub.1]([b.sub.4] - 1) - [b.sub.3][c.sub.2] [not equal to] 0 then [b.sub.3][c.sub.2] [not equal to] 0. Thus, given [epsilon], [delta] [member of] R we have the solutions x [member of] R and:

[mathematical expression not reproducible].

We need to find [l.sub.n] [member of] Z such that 0 [less than or equal to] t [less than or equal to] 1. If [c.sub.2] > 0 take n[delta] [less than or equal to] [l.sub.n] [less than or equal to] n(c2 + [delta]) and if [c.sub.2] < 0 take n[delta] [greater than or equal to] [l.sub.n] [greater than or equal to] n(c2 + [delta]).

Remark 2.2. Note that the hypothesis f,[f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1], is equivalent to require that the induced homomorphisms f# and f# satisfy the conditions of the Theorem 2.3 in each case of the fiber bundle MA. But if f# and [[f.sub.#].sup.n] satisfy the conditions of the Theorem 2.3 then, by Propositions 2.2, 2.3 and 2.4, the induced homomorphism [[f.sub.#].sup.k] satisfies the conditions of the Theorem 2.3 for each k positive divisor ofn. Thus, the hypothesis f, [f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1] implies that [f.sup.k] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1], for each k positive divisor of n.

3 Fixed points of [f.sup.n]

In this section we will give the proof of the main result.

Theorem 3.1 (Main Theorem). Let f : MA [right arrow] MA be a fiber-preserving map, where MA is a T-bundle over [S.sup.1] as in the Theorem 2.1, and n > 1 a positive integer. Suppose [mathematical expression not reproducible] and f#(c) = [[a.sup.c].sup.1][b.sup.c2]c, and f,[f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1]. If the following conditions are satisfied in each case bellow then f isfiberwise homotopic to a g so that [g.sup.n] is fixed point free.

Case I

i) ([a.sub.1] - 1)([a.sub.4] - 1) - [a.sub.2][a.sub.3] [not equal to] 0 andgcd((a4 + [a.sub.2] - 1), ([a.sub.3] + [a.sub.1] - 1)) > 1.

ii) ([a.sub.1] - 1)([a.sub.4] - 1) - [a.sub.2][a.sub.3] = 0, c2 [not equal to] 0 and a1 = 1.

Case II

i) [c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0, |[b.sub.3]| + |[b.sub.4] - 1| [not equal to] 0 and [b.sub.4] [not equal to] 1

ii) [c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0, |[b.sub.3]| + |[b.sub.4] - 1| [not equal to] 0,[b.sub.4] = 1 and [a.sub.3] not divides n.

iii) [c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0, |[b.sub.3]| + |[b.sub.4] - 1| [not equal to] 0,[b.sub.4] = 1 and [a.sub.3] = 0.

Case III

[c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0, |[b.sub.3]| + |[b.sub.4] - 1| [not equal to] 0.

Case IV

[c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0, |[b.sub.3]| + |[b.sub.4] - 1| [not equal to] 0.

Case V

[c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0, |[b.sub.3]| + |[b.sub.4] - 1| [not equal to] 0.

Remark 3.1. Note that in th e Case III, the condition [c.sub.1] ([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0 is necessary and sufficient to deform f and [f.sup.n] to a fixed point free map. Thus, if [c.sub.1] ([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0 can not exist g fiberwise homotopic to f such [g.sup.n] is fixed point free. The condition |[b.sub.3]| + |[b.sub.4] - 1| = 0 in the cases II, III, IV and V is only to guarantee that the matrix B = [(f|T)#] is not the identity matrix is these cases.

Proof (of the main theorem). The technique used to proof the main theorem consists to show that for appropriated [epsilon] and [delta] the map g : T x I [right arrow] T x I defined by; g((x,y,t)) = (x + [b.sub.3]y + [c.sub.1]t + [epsilon],[b.sub.4]y + [c.sub.2]t + [delta], t) induces a fiber-preserving map on MA, which we will represent by the same symbol, such that f [~.sub.S]1 g and [g.sup.n] is a fixed point free map. Note that if [c.sub.1] ([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0, then by Proposition 2.5 that map g does not works, that is, [g.sup.n] will have fixed points. Thus, will use g in the situation [c.sub.1]([b.sub.4] - 1) - [c.sub.2][b.sub.3] = 0. From Theorem 2.4, the map [g.sup.n] induces a fiber-preserving map if [epsilon], [delta] satisfy the following conditions, in each case of the Theorem 2.1,

Case I) n[a.sub.1][epsilon] + n[a.sub.3][delta] = n[epsilon] + k, and n[a.sub.2][epsilon] + n[a.sub.4][delta] = n[delta] + l

Case II) [mathematical expression not reproducible]

Case III) [mathematical expression not reproducible] and [mathematical expression not reproducible]

Case IV) [mathematical expression not reproducible] and [mathematical expression not reproducible]

Case V) [mathematical expression not reproducible]

where k, l [member of] Z. Is important to observe that our interest is in the case n > 1. Let us suppose that each map f,[f.sup.n] : MA [right arrow] MA can be deformed to a fixed point free map over [S.sup.1].

(Case I) For each map f such that ([f.sub.|T])# = Id consider the map g fiberwise homotopic tofgiven by: g (< x,y,t >) =< x + [c.sub.1]t + [member of],y + [c.sub.2]t + [delta],t >, with[member of], [delta] satisfying the conditions;

(I) [mathematical expression not reproducible]

for some [k.sub.n], [l.sub.n] [member of] Z. If det = ([a.sub.1] - 1)([a.sub.4] - 1) - [a.sub.2][a.sub.3] [not equal to] 0, we obtain

(II) [mathematical expression not reproducible]

Note that g\ is fiberwise homotopic to the map g defined by:

[mathematical expression not reproducible]

In fact, H : MA x I [right arrow] MA defined by:

[mathematical expression not reproducible]

is a homotopy between g' and g. Note that,

[mathematical expression not reproducible]

i) Suppose det [not equal to] 0 and d = gcd(([a.sub.4] + [a.sub.2] - 1),([a.sub.3] + [a.sub.1] - 1)) > 1. Choose [mathematical expression not reproducible]. This values satisfy the system (I) and [mathematical expression not reproducible]. If [g.sup.n] has a fixed point for [mathematical expression not reproducible] then we must have n[delta] [member of] Z. Also, if [g.sup.n] has a fixed point for [mathematical expression not reproducible] then we must have n[??] [member of] Z, which is a contradiction, that is, Fix([g.sup.n]) = [empty set].

ii) Suppose 0 = det = ([a.sub.1] - 1)([a.sub.4] - 1) - [a.sub.2][a.sub.3], [c.sub.2] = 0 and [a.sub.1] = 1. Thus, we must have [a.sub.2] = 0. From system (I) we obtain the equations; n[a.sub.3][delta] = [k.sub.n] and n([a.sub.4] - 1)[delta] = [l.sub.n], for some [k.sub.n], [l.sub.n] [member of] Z. This equations do not depend of [member of], therefore we can choose [??] an irrational number. Thus, we choose [member of] an irrational number and [mathematical expression not reproducible]. We observe that both g and g are fiberwise homotopic to the given map f, and (g')n(< x,y,t >) =< x + n[c.sub.1]t + n[??],y + n[c.sub.2]t + n[delta],t >, with [??],[delta] satisfying the conditions of the system (I). If [(g).sup.n] has a fixed point then we must have n[c.sub.1]t + n[member of] = [p.sub.n] and n[c.sub.2]t + n[delta] = [q.sub.n] for some [p.sub.n],[q.sub.n] [member of] Z. If [c.sub.1] =0 we have a contradiction because [member of] is an irrational number. If [c.sub.1] [not equal to] 0 and [c.sub.2] [not equal to] 0 then we have n[c.sub.2][member of] - n[c.sub.1][delta] = [c.sub.2][p.sub.n] - [c.sub.1][q.sub.n], which is a contradiction because [member of] is an irrational number and [mathematical expression not reproducible]. Therefore, [(g').sup.n] can not have a fixed point.

(Case II) Let g : MA [right arrow] MA be the map fiberwise homotopy to f given by [mathematical expression not reproducible], where [a.sub.3][delta] [member of] Z. If [b.sub.4] = 1 then [c.sub.2][b.sub.3] = 0, but if [b.sub.3] = 0 then the matrix B of (f|T)# is the identity, contradicting a hypothesis. Suppose [b.sub.4] = 1, [b.sub.3] = 0 and [c.sub.2] = 0, by Theorem 2.4, [g.sup.n](< x,y,t >) =< [x.sub.n],[y.sub.n],t > for each n [member of] N where,

[mathematical expression not reproducible]

If [g.sup.n] : MA [right arrow] MA has a fixed point < x,y,t > then [x.sub.n] = x + [k.sub.n] and [y.sub.n] = y + [l.sub.n] for some [k.sub.n], [l.sub.n] [member of] Z. By the second equation of the system above we must have n[delta] = [l.sub.n] for some [l.sub.n] [member of] Z. Therefore, [g.sup.n] : MA [right arrow] MA is fixed point free if [a.sub.3] = 0 and [delta] [member of] R - Q or if [a.sub.3] not divides n and [mathematical expression not reproducible].

Now we suppose [b.sub.4] [not equal to] 1 and we choose [delta] = 0 then [g.sup.n](< x, y, t >) = < [x.sub.n],[y.sub.n],t >, where

[mathematical expression not reproducible].

If [b.sub.4] = -1 and n is even then [g.sup.n] : MA [right arrow] MA is fixed point free for [epsilon] [member of] R - Q, otherwise we had n[epsilon] = [k.sub.n] [member of] Z. Suppose [b.sub.4] [not equal to] 1 or [b.sub.4] = -1 with n odd. If [c.sub.2] [not equal to] 0 we have:

[mathematical expression not reproducible].

Hence, then [g.sup.n] : MA [right arrow] MA is fixed point free for [epsilon] [member of] R - Q. On the other hand, if [c.sub.2] = 0 then [c.sub.1] = 0 because [b.sub.4] [not equal to] 1. Therefore,

[mathematical expression not reproducible].

So, [g.sup.n] : MA [right arrow] MA is fixed point free for [epsilon] [member of] R - Q, otherwise [mathematical expression not reproducible], [mathematical expression not reproducible] and

[mathematical expression not reproducible].

(Case III) The proof in this case is similar to the case (2), but here we consider [mathematical expression not reproducible] with [a.sub.3][delta] [member of] Z and k [member of] Z. If [b.sub.4] = 1 then [b.sub.3] = 0, and this situation we will have |[b.sub.3]| + |[b.sub.4] - 1| = 0 contradicting a hypothesis. If [b.sub.4] = 1 then [g.sup.n] : MA [right arrow] MA is fixed point free for [epsilon] [member of] R-Q and the proof is the same of the case II.

(Case IV) Suppose [mathematical expression not reproducible] such that [mathematical expression not reproducible] and [mathematical expression not reproducible]. Thus, given n [greater than or equal to] 1 and [g.sup.n](< x, y, t >) =< [x.sub.n], [y.sub.n], t > we want to know when [g.sup.n] has a fixed point, i.e., there are [k.sub.n], [l.sub.n] [member of] Z such that [x.sub.n] = x + [k.sub.n] and [y.sub.n] = y + [l.sub.n].

Note that the expression [b.sub.4](n[b.sub.3] +1) [equivalent to] 1 mod 2 follows from item 3 of Theorem 2.3 as below

[mathematical expression not reproducible].

If [b.sub.4] = 1 and n is odd then we must have [c.sub.2] = 0 because if [b.sub.3] = 0 then we would have |[b.sub.3]| + |[b.sub.4] - 1| = 0. So, [g.sup.n] : MA [right arrow] MA has not a fixed point < x, y, t > for [mathematical expression not reproducible], otherwise we had [mathematical expression not reproducible] and [mathematical expression not reproducible]. Note that we have a exception if [b.sub.4] = 1 and n even, because [c.sub.2] = 0. Hence, [g.sup.n] is fixed point free if [b.sub.4] = 1 and [mathematical expression not reproducible].

Suppose [b.sub.4] = 1. From expression [b.sub.4](n[b.sub.3] + 1) [equivalent to] 1 mod 2, proved above, we must have [b.sub.4] odd. Thus, we have [a.sub.3] = 0 and [mathematical expression not reproducible]. If [g.sup.n] : MA [right arrow] MA has a fixed point < x, y, t > then

[mathematical expression not reproducible].

So, [k.sub.n] [??] Z for appropriates [delta] and [epsilon], n [member of] N. Therefore, gn : MA [right arrow] MA is fixed point free.

(Case V) Suppose g(< x,y,t >) =< x + [b.sub.3]y + [c.sub.1]t + [epsilon],[b.sub.4]y + [c.sub.2]t + [delta],t > such that [mathematical expression not reproducible]. We must consider [b.sub.4] [not equal to] 1, otherwise we will obtain [b.sub.3] = 0 since [a.sub.3]([b.sub.4] - 1) = 2[b.sub.3], therefore |[b.sub.3]| + |[b.sub.4] - 1| = 0 contradicting our hypothesis. Suppose [b.sub.4] = 1. From Theorem 2.4 we have two equations;

(I) [mathematical expression not reproducible]

(II) [mathematical expression not reproducible].

If [b.sub.4] = - 1 and n is even then [g.sup.n] : MA [right arrow] MA has not a fixed point < x, y, t > for [delta] [member of] R - Q and [mathematical expression not reproducible], otherwise

[mathematical expression not reproducible].

Now suppose n > 1 any natural number with [b.sub.4] [not equal to] 1, (except [b.sub.4] = - 1 and n even, which was already made). In this situation [g.sup.n] : MA [right arrow] MA has not a fixed point < x,y,t > for [delta] [member of] R - Q and [mathematical expression not reproducible], otherwise we will obtain [x.sub.n] = x + [k.sub.n] and [y.sub.n] = y + [l.sub.n] with [k.sub.n], [l.sub.n] [member of] Z. From equation (II) we obtain

[mathematical expression not reproducible].

Replacing the value of y of the last equation into equation (I), and using [mathematical expression not reproducible], we will obtain;

[mathematical expression not reproducible]

Replacing this value into the equation [x.sub.n] = x + [k.sub.n] we obtain;

[mathematical expression not reproducible]

When [b.sub.3] [not equal to] 0 we have a contradiction because [delta] [member of] R - Q. When [b.sub.3] = 0 we have a contradiction because [k.sub.n] [member of] Z. Therefore, gn : MA [right arrow] MA is a fixed point free map.

Acknowledgments. We would like to thank the referee by your comments and suggestions, which helped to improve the manuscript.

References

[1] B.J. Jiang; Lectures on the Nielsen Fixed Point Theory, Contemp. Math., vol. 14, Amer. Math. Soc., Providence, 1983.

[2] J. Jezierski; Weckens theorem for periodic points in dimension at least 3, Topology and its Applications 153 (2006) 18251837.

[3] J. Jezierski and W. Marzantowicz; Homotopy Methods in Topological Fixed and Periodic Point Theory, vol. 3, Topological Fixed Point Theory and Its Applications, Springer, 2006.

[4] E. Fadell and S. Hussein; A fixed point theory for fibre-preserving maps Lectures Notes in Mathematics, vol.886, Springer Verlag, 1981, 49-72.

[5] D. L. Goncalves, D.Penteado and J.P Vieira; Fixed Points on Torus Fiber Bundles over the Circle, Fundamenta Mathematicae, vol.183 (1), 2004, 1-38.

[6] G. L. Prado; Deformabilidade sobre [S.sup.1] a livre de ponto fixo para auto-aplicac oes de T-fibrados e Reidemeister sobre [S.sup.1], Master dissertation, USP, 2010.

[7] B. Halpern, Periodic points on tori, Pacific J. Math. 83 (1979), no. 1, 117133.

[8] Edward Keppelmann, Periodics points on nilmanifolds and solvmanifolds, Pacific Journal of Mathematics, vol.164 (1) (1994), 105-128.

[9] G. W. Whitehead; Elements of Homotopy Theory, Springer-Verlag, 1918.

[10] C.Y. You; The least number of periodic points on tori, Adv. Math. (China) 24 (2) (1995) 155-160.

Weslem Liberato Silva

Rafael Moreira de Souza

Received by the editors in October 2016 - In revised form in September 2017.

Communicated by Dekimpe, Goncalves, Wong.

2010 Mathematics Subject Classification : Primary 55M20; Secondary 37C25.

Key words and phrases : Periodic points, fiber bundle, fiberwise homotopy.

Dept. de Ciecias Exatas e Tecnologicas Universidade Estadual de Santa Cruz, Campus Soane Nazare de Andrade Rodovia Jorge Amado, Km 16, Bairro Salobrinho CEP 45662-900 Ilheus-Bahia, Brazil

email: wlsilva@uesc.br

Universidade Estadual de Mato Grosso do Sul, Cidade Universitaria de Dourados Caixa postal 351 CEP: 79804-970, Dourados-MS, Brazil

email: moreira@uems.br
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Author:Silva, Weslem Liberato; de Souza, Rafael Moreira
Publication:Bulletin of the Belgian Mathematical Society - Simon Stevin
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Date:Oct 1, 2017
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