# A fixed point theorem for twist maps

Peizheng Yu, Zhihong Xia  Department of Mathematics, Southern University of
Science and Technology, Shenzhen, China  Department of Mathematics,
Northwestern University, Evanston, IL 60208 USA<EMAIL_ADDRESS><EMAIL_ADDRESS>
(Date: version: June 11, 2021)

###### Abstract.

Poincaré’s last geometric theorem (Poincaré-Birkhoff Theorem [2]) states that
any area-preserving twist map of annulus has at least two fixed points. We
replace the area-preserving condition with a weaker intersection property,
which states that any essential simple closed curve intersects its image under
$f$ at least at one point. The conclusion is that any such map has at least
one fixed point. Besides providing a new proof to Poincaré’s geometric
theorem, our result also has some applications to reversible systems.

##  1\. Introduction

In 1913, Birkhorff proved Poincaré’s Last Geometric Theorem [2] stating that
an area-preserving homeomorphism of an annulus satisfying the twist condition
has at least two fixed points. The proof has been subsequently improved by
Barrar [1], Carter [5] and many others. In this note, we replace the area-
preserving condition with a weaker intersection property and obtain a weaker,
but sharp, result. We show that there is at least one fixed point under the
intersection condition. Combining with Slaminka’s [14] result on removing
isolated fixed point with zero index, our method provides another proof of
Poincaré’s last geometric theorem for area-preserving case. It also provides a
new proof to Carter’s theorem [5]. One of the interesting applications of our
result is for the so-called reversible systems. It is interesting to note that
both the KAM theory and Aubry-Mather theory are applicable to reversible
systems (Siegel & Moser [13], Chow & Pei [6]). Here we also have a positive
answer for Poincaré’s geometric theorem, albeit with just one fixed point.

Our proof is more of a simple standard dynamical systems approach, using
techniques of Franks [9, 10], and Brouwer’s plane translation theorem (Brown
[4]). We analyse the chain recurrent set, where Conley’s fundamental theorem
for dynamical systems [7] provided a fine structure on the chain recurrent
set. The key observation is very simple: the intersection property implies
that two boundary components are in the same chain transitive component of the
chain recurrent set. The techniques used in the proof are very much standard
in dynamics.

More precisely, let $\mathbb{A}=S^{1}\times[0,1]$ be an annulus with
boundaries $A_{0}=S^{1}\times\\{0\\}$ and $A_{1}=S^{1}\times\\{1\\}$. Let
$f:\mathbb{A}\rightarrow\mathbb{A}$ be a homeomorphism satisfying the twist
condition and intersection property.

We say that $f$ satisfies the intersection property if any essential simple
closed curve intersects its image under $f$ at least at one point. A simple
closed curve is said to be essential if it is not contractible. As for the
twist condition, we need to consider the covering space
$\widetilde{\mathbb{A}}=\mathbb{R}\times[0,1]$ of $\mathbb{A}$, and the lift
of $f$ denoted by
$\widetilde{f}:\widetilde{\mathbb{A}}\rightarrow\widetilde{\mathbb{A}}$. We
say that $f$ satisfies the twist condition provided that $\widetilde{f}$ moves
the two boundary components of $\widetilde{\mathbb{A}}$ in opposite
directions. We may assume that $\widetilde{f}(x,y)=(x+r_{1}(x),y)$ for $y=1$
and $\widetilde{f}(x,y)=(x-r_{0}(x),y)$ for $y=0$, where
$r_{0}(x),r_{1}(x)>0$. Our main result is following theorem.

######  Theorem 1.1.

If $f:\mathbb{A}\rightarrow\mathbb{A}$ is a homeomorphism satisfying twist
condition and intersection property. Then $f$ has at least one fixed point.

Note that the intersection property is weaker than the area-preserving
property, and Carter gives an example (Figure 1) in [5] that such $f$ may have
exactly one fixed point. So, our weaker conclusion, $f$ has at least one fixed
point instead of two, is actually sharp.

Figure 1. Carter’s example

There are many different proofs and extensions of Poincaré’s geometric
theorem. Some proofs are less clear and convincing than others. Our statement
of theorem shares some common features with the works of Birkhoff [3] and
Carter [5]. Carter obtained two fixed points, using a stronger intersection
property.

An outline of this paper is as follows: In section 2, we recall some
properties of chain recurrent set and introduce Conley’s fundamental theorem
of dynamical systems and the Brouwer planar translation theorem. In section 3,
we will first assume $f$ is rigid rotation on boundaries i.e. $r_{0}(x)$ and
$r_{1}(x)$ are constant and prove our main theorem, then we will show for the
general case, the theorem also holds true. In section 4, we show how our
result applies to reversible systems.

##  2\. Chain recurrent set and useful theorems

Chain recurrence is an important concept in dynamical systems. Given a compact
metric space $X$ and homeomorphism $f:X\rightarrow X$. We have the following
definitions.

######  Definition 2.1.

Given two points $p,q\in X$, an $\epsilon$-chain from $p$ to $q$ is a sequence
$x_{1},x_{2},...,x_{n}$, where $x_{1}=p$, $x_{n}=q$ and for all $1\leq i\leq
n-1$, $d(f(x_{i}),x_{i+1})<\epsilon$.

######  Definition 2.2.

A point $p\in X$ is chain recurrent if for all $\epsilon>0$, there is an
$\epsilon$-chain from $p$ to itself. The set of all chain recurrent points in
X is called the $chain$ recurrent set, denoted by $\mathcal{R}(f)$.

Now, we can define an equivalence relation on $\mathcal{R}(f)$ as follows: For
$p,q\in\mathcal{R}(f)$, $p\sim q$ if and only if for all $\epsilon>0$, there
are $\epsilon$-chains from $p$ to $q$ and from $q$ to $p$. And it’s easy to
see that $\sim$ is reflexive, symmetric and transitive.

######  Definition 2.3.

The equivalence classes in $\mathcal{R}(f)$ for $\sim$ is called $chain$
$transitive$ $components$. A set is called chain transitive if any two points
$p$, $q$ in this set, we have $p\sim q$.

The next lemma allows us to relate chain transitive components with connected
components.

######  Lemma 2.1.

The connected components of $\mathcal{R}(f)$ are chain transitive.

###### Proof.

Let $K$ be a connected component of $\mathcal{R}(f)$. Given any $\epsilon>0$,
we can cover $K$ by disks with radius $\frac{\epsilon}{4}$. Since $K$ is
closed and hence compact, there exists a finite subcover
$\\{B_{i}\\}_{i=1}^{m}$. Then for any $x,y\in K$, we can find a sequence
$\\{x=x_{0},x_{1},...,x_{n-1},x_{n}=y\\}$ in $\mathcal{R}(f)$ such that
$x_{i}$ and $x_{i+1}$ lies in a same disk for $i=0,1,...,n-1$. Note each
$x_{i}\in\mathcal{R}(f)$, there is a $\frac{\epsilon}{2}$-chain from $x_{i}$
to $x_{i}$. We replace the last point by $x_{i+1}$, then we get an
$\epsilon$-chain form $x_{i}$ to $x_{i+1}$. And finally we get an
$\epsilon$-chain form $x$ to $y$. Since $x$ and $y$ are arbitrary, $K$ is
chain transitive. ∎

In order to state the fundamental theorem of dynamical systems, we need to
give this definition.

######  Definition 2.4.

$g:X\rightarrow\mathbb{R}$ is a complete Lyapunov function for $f$ if:

  1. (1)

$\forall p\notin\mathcal{R}(f)$, $g(f(p))<g(p)$.

  2. (2)

$\forall p,q\in\mathcal{R}(f)$, $g(p)=g(q)$ if and only if $p\sim q$.

  3. (3)

$g(\mathcal{R}(f))$ is compact and nowhere dense in $\mathbb{R}$.

######  Theorem 2.2 (Conley’s [7] fundamental theorem of dynamical systems).

Complete Lyapunov function exists for any homeomorphism on compact metric
spaces.

This theorem states that we can get a complete Lyapunov function that stays
constant only on the chain transitive components and strictly decreases along
any orbit not in $\mathcal{R}(f)$. More details of this theorem can be found
in Franks [11].

The next concept and theorem are introduced in [9]. It is very useful for
showing the existence of a fixed point on plane. And it is easy to see from
the definition that $\epsilon$-chain and disk chain have a closed
relationship.

######  Definition 2.5.

Let $f:M\rightarrow M$ be a homeomorphism of a surface. A disk chain for $f$
is a finite set $U_{1},U_{2},...,U_{n}$ of embedded open disks in $M$
satisfying

  1. (1)

$f\left(U_{i}\right)\cap U_{i}=\varnothing$ for $1\leq i\leq n$.

  2. (2)

If $i\neq j$, then either $U_{i}=U_{j}$ or $U_{i}\cap U_{j}=\varnothing$.

  3. (3)

For $1\leq i\leq n$, there exists $m_{i}>0$ with
$f^{m_{i}}\left(U_{i}\right)\cap U_{i+1}\neq\varnothing$.

If $U_{1}=U_{n}$, we will say that $U_{1},U_{2},...,U_{n}$ is a $periodic$
$disk$ $chain$.

######  Theorem 2.3 (Brouwer planar translation theorem).

Let $f:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ be an orientation preserving
homeomorphism which possesses a periodic disk chain. Then there is a simple
closed curve $\gamma$ in $\mathbb{R}^{2}$ such that $I(\gamma,f)=1$. If $f$
has only isolated fixed points, then $f$ has a fixed point of positive index.

The original version of Brouwer’s planar translation theorem states that a
fixed point free homeomorphism of $\mathbb{R}^{2}$ can be viewed locally as
plane translations.

##  3\. Proof of the Main Theorem

With the above preparation, we can prove our main theorem.

The rough idea of the proof is as follows: we first consider the case that $f$
is rigid rotation on the boundaries of the annulus i.e. $r_{0}(x)$ and
$r_{1}(x)$ are both constant. The general case can be embedded in this case
and we will explain this in more details in the end. In the simple case, note
that $A_{0}$ and $A_{1}$ are both in $\mathcal{R}(f)$, we will first show, in
lemma 3.1, that we can find an $\epsilon$-chain from $A_{0}$ to $A_{1}$ and an
$\epsilon$-chain from $A_{1}$ to $A_{0}$. Suppose, on th econtrary, that
$A_{0}$ and $A_{1}$ are not in the same chain transitive component, by
Conley’s fundamental theorem of dynamical systems, there is a complete
Lyapunov function $g$ defined on $\mathbb{A}$, such that $g(A_{0})\neq
g(A_{1})$. And this allows us to find an essential closed curve, close to some
level curves of $g$, that doesn’t satisfies the intersection property.

With Lemma 3.1 and the twist condition on the boundary, there is a periodic
$\epsilon$-chain in the covering space. Then, similar to Franks [9], we can
use this $\epsilon$-chain to construct a periodic disk chain. And the
existence of a fixed point follows from the Brouwer plane translation theorem.

We now proceed with details.

######  Lemma 3.1.

If $f$ is rigid rotation on the boundaries of the annulus, there is a chain
transitive component $D$ of $\mathcal{R}(f)$ such that $D\cap
A_{0}\neq\varnothing$ and $D\cap A_{1}\neq\varnothing$

###### Proof.

We prove by contradiction. First, by Conley’s fundamental theorem of dynamical
system, there is a complet Lyaponuv function $g$ on $\mathbb{A}$ with the
following three properties:

  1. (1)

$\forall p\notin\mathcal{R}(f)$, $g(f(p))<g(p)$.

  2. (2)

$\forall p,q\in\mathcal{R}(f)$, $g(p)=g(q)$ if and only if $p\sim q$.

  3. (3)

$g(\mathcal{R}(f))$ is compact and nowhere dense in $\mathbb{R}$.

Suppose that, contrary to the conclusion of the lemma, $A_{0}$ and $A_{1}$ are
not in the same chain transitive component, then, by property (2), the
function $g$ takes different values on $A_{0}$ and $A_{1}$. Assume
$g|_{A_{0}}=a<g|_{A_{1}}=b$, where $a$, $b$ are real numbers.

By property (3), there is a number $c\in[a,b]$ such that the set
$C=g^{-1}(c)\subset\mathbb{A}$ does not intersect the chain recurrent set of
$f$, therefore, by property (1), $g(f(x))<g(x)=c$, therefore, $f(C)\cap
C=\varnothing$, and $C$ seperates $A_{0}$ and $A_{1}$

Let

$\epsilon=\inf_{x\in C}(g(x)-g(f(x)))=\inf_{x\in C}(c-g(f(x))),$

then $\epsilon>0$, by the compactness of $C$.

The open set $B=g^{-1}((c-\epsilon,c))$ seperates $A_{0}$ and $A_{1}$.
Moreover, $f(B)\cap B=\varnothing$. One can easily choose an essential simple
closed curve $\gamma$ inside $B$. We have $f(\gamma)\cap\gamma=\varnothing$.
This contradicts to the intersection property.

∎

One can prove the above lemma directly by using the so-called attractor-
repeller pair. However, using Lyapunov function is more conceptual.

To prove our main theorem, all we need to do is to construct a periodic disk
chain in the covering space.

###### Proof of theorem 1.1.

First, we consider a simple case that $f$ is rigid rotation on the boundaries
of the annulus. Let’s extend
$\widetilde{f}:\widetilde{\mathbb{A}}\rightarrow\widetilde{\mathbb{A}}$ to
$h:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ as follows:

$h(x,y)=\begin{cases}(x+r_{1},y),&\text{ if }y\geq 1\\\
\widetilde{f}(x,y)&\text{ if }0<y<1\\\ (x-r_{0},y),&\text{ if }y\leq
0\end{cases}$

By lemma 3.1, there exists $\widetilde{p_{0}}\in\widetilde{A_{0}}$,
$\widetilde{p_{1}}\in\widetilde{A_{1}}$ s.t. for any $\epsilon>0$, we have
$\epsilon$-chains with respect to $h$ from $\widetilde{p_{0}}$ to
$\widetilde{p_{1}}$ and from $\widetilde{p_{1}}+(l_{1},0)$ to
$\widetilde{p_{0}}+(l_{0},0)$ for some $l_{0},l_{1}\in\mathbb{Z}$. Because of
our boundary assumption, we can choose a suitable $l\in\mathbb{Z}$ such that
there exist $\epsilon$-chains from $\widetilde{p_{1}}$ to
$\widetilde{p_{1}}+(l_{1}+l,0)$ and from $\widetilde{p_{0}}+(l_{0}+l,0)$ to
$\widetilde{p_{0}}$. Note that $h(p+(1,0))=h(p)+(1,0)$. We obtain an
$\epsilon$-chain from $\widetilde{{p_{0}}}$ to $\widetilde{{p_{0}}}$ passing
through $\widetilde{p_{1}}$, $\widetilde{p_{1}}+(l_{1}+l,0)$ and
$\widetilde{p_{0}}+(l_{0}+l,0)$.

Prove by contradiction. Suppose
$\widetilde{f}:\widetilde{\mathbb{A}}\rightarrow\widetilde{\mathbb{A}}$
doesn’t have fixed points, since $\mathbb{A}$ is compact, it follows that
$\exists\delta>0$ s.t. $\|\widetilde{f}(p)-p\|\geq\delta$ for all
$p\in\widetilde{\mathbb{A}}$. Hence, $\|h(p)-p\|\geq\delta$ for all
$p\in\mathbb{R}^{2}$.

Take $\epsilon<\frac{\delta}{4}$, and assume
$\\{z_{0}=\widetilde{{p_{0}}},z_{1},...,z_{n-1},z_{n}=z_{0}\\}$ is the above
$\epsilon$-chain from $\widetilde{{p_{0}}}$ to $\widetilde{{p_{0}}}$. We will
construct a periodic disk chain and apply the Brouwer planar translation
theorem to get a contradiction.

Let
$U_{0}=B_{\epsilon}(z_{0}),U_{1}=B_{\epsilon}(z_{1}),...,U_{n-1}=B_{\epsilon}(z_{n-1}),U_{n}=B_{\epsilon}(z_{n})=U_{0}$,
where $B_{\epsilon}(z)=\\{q:\|q-z\|<\epsilon\\}$. Then we have the following
properties.

  1. (1)

$\forall 0\leq i\leq n$, $h(U_{i})\cap U_{i+1}\neq\varnothing$.

  2. (2)

$\forall 0\leq i\leq n$, $h(U_{i})\cap U_{i}=\varnothing$.

Now, let’s consider two cases.

One case is that any $i\neq j\in\\{0,1,...,n-1\\}$ implies $U_{i}\cap
U_{j}=\varnothing$, then $\\{U_{0},U_{1},...,U_{n-1},U_{n}=U_{0}\\}$ is a
periodic disk chain and it follows from the Brouwer planar translation theorem
that $h$ has a fixed point, which contradicts to our assumption. This
completes the proof, in this case, when $f$ is rigid rotation on the
boundaries.

The other case is that $\exists i_{0}\neq j_{0}\in\\{0,1,...,n-1\\}$ s.t.
$U_{i_{0}}\cap U_{j_{0}}\neq\varnothing$. Here, we can take $i_{0}<j_{0}$ and
consider the sequence $\\{U_{i_{0}},U_{i_{0}+1},...,U_{j_{0}-1},U_{j_{0}}\\}$.
We may assume that this sequence has the property: for any $i\neq
j\in\\{i_{0},i_{0}+1,...,j_{0}-1,j_{0}\\}$ except for $i=i_{0}$ and $j=j_{0}$
simultaneously, we have $U_{i}\cap U_{j}=\varnothing$. Because if not,
$\exists i_{1}\neq j_{1}\in\\{i_{0},i_{0}+1,...,j_{0}-1,j_{0}\\}$ s.t.
$U_{i_{1}}\cap U_{j_{1}}\neq\varnothing$ and
$\\{i_{1},i_{1}+1,...,j_{1}-1,j_{1}\\}$ is a shorter sequence than
$\\{i_{0},i_{0}+1,...,j_{0}-1,j_{0}\\}$. Then we can let the new $i_{0}=i_{1}$
and new $j_{0}=j_{1}$. Moreover, since $U_{i}\cap U_{i+1}=\varnothing$,
$\forall i=0,1,...,n-1$, we can always find the shortest sequence satisfying
the above property and this guarantees the existence of such $i_{0}$ and
$j_{0}$.

Let $V_{0}:=U_{i_{0}}\cup U_{j_{0}}$, $V_{1}:=U_{i_{0}+1}$, …,
$V_{k-1}:=U_{j_{0}-1}$, and $V_{k}:=V_{0}$. We will show that
$\\{V_{0},V_{1},...,V_{k}\\}$ is a periodic disk chain.

Firstly, since $U_{i_{0}}\cup U_{j_{0}}\neq\varnothing$, we have

$\begin{split}d(h(U_{i_{0}}),U_{j_{0}})&:=\inf_{p\in U_{i_{0}},q\in
U_{j_{0}}}\|h(p)-q\|\geq\inf_{p\in U_{i_{0}},q\in
U_{j_{0}}}|\|h(p)-p\|-\|q-p\||\\\ &\geq\delta-\sup_{p\in U_{i_{0}},q\in
U_{j_{0}}}\|p-q\|\geq\delta-4\epsilon>0.\end{split}$

and similarly, $d(h(U_{j_{0}}),U_{i_{0}})>0.$ Thus,

$h(V_{0})\cap V_{0}=h(U_{i_{0}}\cup U_{j_{0}})\cap(U_{i_{0}}\cup
U_{j_{0}})=(h(U_{j_{0}})\cap U_{i_{0}})\cup(h(U_{i_{0}})\cap
U_{j_{0}})=\varnothing$

Along with the property that $\forall 0\leq i\leq n$, $h(U_{i})\cap
U_{i}=\varnothing$, we have $h\left(V_{i}\right)\cap V_{i}=\varnothing$ for
$0\leq i\leq k$.

Secondly, according to the property that for any $i\neq
j\in\\{i_{0},i_{0}+1,...,j_{0}-1,j_{0}\\}$ except for $i=i_{0}$ and $j=j_{0}$
simultaneously we have $U_{i}\cap U_{j}=\varnothing$, it is easy to see that
if $i\neq j\in\\{0,1,...,k-1\\}$, then $V_{i}\cap V_{j}=\varnothing$.

Thirdly, since $\forall 0\leq i\leq n$, $h(U_{i})\cap U_{i+1}\neq\varnothing$,

$h(V_{0})\cap V_{1}=(h(U_{i_{0}})\cup h(U_{j_{0}}))\cap U_{i_{0}+1}\supset
h(U_{i_{0}})\cap(U_{i_{0}+1})\neq\varnothing$

and similarly, $h(V_{k-1})\cap V_{k}\neq\varnothing$. Thus, $\forall 0\leq
i\leq k$, $h\left(V_{i}\right)\cap V_{i+1}\neq\varnothing$.

Therefore, $h:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ possesses a periodic
disk chain, and hence $h$ has at least one fixed point. By the definition of
$h$, the fixed point must lie in the interior of $\widetilde{\mathbb{A}}$, so
this point is also fixed by $\widetilde{f}$, a contradiction to our
assumption. We have thus proved the theorem for the case that $f$ is rigid
rotation on the boundaries.

We now consider the general case where $f$ on the boundaries are not rigid
rotations. Recall that $\widetilde{f}(x,1)=(x+r_{1}(x),1)$ and
$\widetilde{f}(x,0)=(x-r_{0}(x),0)$ with $r_{0}(x),r_{1}(x)>0$. Suppose the
rotation numbers for $\widetilde{f}|_{\mathbb{R}\times\\{0\\}}$ and
$\widetilde{f}|_{\mathbb{R}\times\\{1\\}}$ are $\alpha$ and $\beta$
respectively. Take a small $\delta>0$, we will extend $\widetilde{f}$ to
$\widetilde{f}_{\delta}:\mathbb{R}\times[-\delta,1+\delta]\rightarrow\mathbb{R}\times[-\delta,1+\delta]$
in the following way:

$\widetilde{f}_{\delta}(x,y)=\begin{cases}\frac{1+\delta-y}{\delta}\widetilde{f}(x,1)+\frac{y-1}{\delta}(x+\beta,1+\delta),&\text{
if }1\leq y\leq 1+\delta\\\ \widetilde{f}(x,y)&\text{ if }0<y<1\\\
\frac{y+\delta}{\delta}\widetilde{f}(x,0)-\frac{y}{\delta}(x+\alpha,-\delta),&\text{
if }-\delta\leq y\leq 0\end{cases}$

We can project $\widetilde{f}_{\delta}$ through the covering map to the
extended annulus, we have a map $f_{\delta}$ that is rigid rotation on both
boundaries of the extended annulus. Let
$\mathbb{A}_{\delta}:=S^{1}\times[-\delta,1+\delta]$ with two boundaries
$A_{0,\delta}$ and $A_{1,\delta}$; the two extended regions are
$D_{0}:=S^{1}\times[-\delta,0]$ and $D_{1}:=S^{1}\times[1,1+\delta]$; the
covering space of them are $\widetilde{\mathbb{A}_{\delta}}$,
$\widetilde{A_{0,\delta}}$, $\widetilde{A_{1,\delta}}$,$\widetilde{D_{0}}$,
and $\widetilde{D_{1}}$ respectively.

Now, We should check that $f_{\delta}$ still satisfies the intersection
property on $\mathbb{A}_{\delta}$. For any essential simple closed curve
$\gamma$, there are three cases.

The first one is the curve $\gamma$ is fully contained in $\mathbb{A}$. Then
it must satisfy the intersection property.

The second case is the curve $\gamma$ is fully contained in $D_{0}$ or
$D_{1}$. We may assume $\gamma$ is in $D_{0}$. Consider the covering space
$\widetilde{\mathbb{A}_{\delta}}=\mathbb{R}\times[-\delta,1+\delta]$, let
$\pi_{y}:\mathbb{R}\times[-\delta,1+\delta]\rightarrow[-\delta,1+\delta]$ and
$\pi_{x}:\mathbb{R}\times[-\delta,1+\delta]\rightarrow\mathbb{R}$ be the
projections. And we can find points $p,q\in\widetilde{\gamma}$ such that
$\pi_{y}(p)=\min{\pi_{y}(\widetilde{\gamma})}$ and
$\pi_{y}(q)=\max{\pi_{y}(\widetilde{\gamma})}$. Let $\widetilde{\gamma}_{p,q}$
be the part of $\widetilde{\gamma}$ connecting $p$ and $q$, then
$\widetilde{f}(\widetilde{\gamma}_{p,q})\cap\widetilde{\gamma}\neq\varnothing$,
since
$\pi_{y}(\widetilde{f}(p))\leq\pi_{y}(\widetilde{\gamma}|_{\pi_{x}(\widetilde{f}(p))})$
and
$\pi_{y}(\widetilde{f}(q))\geq\pi_{y}(\widetilde{\gamma}|_{\pi_{x}(\widetilde{f}(q))})$.
Therefore, the curve $\gamma$ satisfies the intersection property.

The third case is that part of the curve $\gamma$ is contained in $\mathbb{A}$
denoted by $\gamma_{0}$ and part of $\gamma$ is contained in $D_{0}$ or
$D_{1}$ denoted by $\gamma_{1}$. We first construct $\gamma^{\prime}$ in the
way that we fix $\gamma_{0}$ in $\mathbb{A}$ and replace $\gamma_{1}$ with
parts of $A_{0}$ or $A_{1}$ fixing the endpoints s.t. $\gamma^{\prime}$ is an
essential simple closed curve. Then
$f(\gamma^{\prime})\cap\gamma^{\prime}\neq\varnothing$. If there is an
intersection point is in $int(\mathbb{A})$, then it is also the intersection
point of $f(\gamma)\cap\gamma$. Otherwise, parts of $\gamma^{\prime}$ in
$A_{0}$ or $A_{1}$ must intersect its image under $f$. Similar arguement as
the second case implies $f(\gamma_{1})\cap\gamma_{1}\neq\varnothing$. So, the
curve $\gamma$ still satisfies the intersection property.

In conclusion, we show that $f_{\delta}$ still satisfies the intersection
property on $\mathbb{A}_{\delta}$. Follow the proof of rigid rotation case, we
can show
$\widetilde{f_{\delta}}:\widetilde{\mathbb{A}_{\delta}}\rightarrow\widetilde{\mathbb{A}_{\delta}}$
has a fixed point. And it’s easy to see $\widetilde{f_{\delta}}$ has no fixed
point on $D_{0}$ and $D_{1}$ since $\widetilde{f_{\delta}}$ in these regions
just moves all points in $D_{0}$ or $D_{1}$ in one direction. Thus, the fixed
point we get must lie in $\mathbb{A}$. Hence, it is fixed by $f$.

This completes the proof of our main theorem. ∎

##  4\. An application to reversible systems

We first introduce the definition and some properties of reversible systems.

######  Definition 4.1.

A map $R:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is called an involution if
it satisfies $R^{2}=\text{Id}$.

An easy example of involution is $R:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$
s.t. $R(x,y)=(-x,y)$.

Next, we define general reversible system for both continuous and discrete
dynamical systems on $\mathbb{R}^{n}$.

######  Definition 4.2.

A vector field in $\mathbb{R}^{n}$

$\dot{x}=F(t,x)$

is said to be reversible if there exists an $C^{1}$ involution
$R:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ such that

$\text{D}R\circ F(-t,R(x))=-F(t,x)$

where $\text{D}R$ is the derivative of $R$.

A homeomorphism $f$ is said to be reversible if there is a continuous
involution $R:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ such that

$f^{-1}=R\circ f\circ R$

Intuitively, a system is reversible if under some involution it is transformed
to a system which is the same as the original one except that the time
direction is reversed. Reversible system naturally arises in mechanics
(Devaney [8]). A simple non-Hamiltonian example is the system derived
following second order equation

$\ddot{x}+f(t,x)\dot{x}+g(t,x)=0$

where

$f(-t,-x)=-f(t,x),\;\;g(-t,-x)=-g(t,x).$

On the annulus, we take the standard involution for reversible systems. Let
$R:\mathbb{A}\rightarrow\mathbb{A}$ be the involution that takes $(x,y)$,
where $x\in\mathbb{R}/\mathbb{Z},\;y\in[0,1]$ to $(-x,y)$. A map on the
annulus $f:\mathbb{A}\rightarrow\mathbb{A}$ is said to be reversible if
$f^{-1}=R\circ f\circ R$.

Reversible homeomorphisms may not necessarily have the general intersection
property (cf. Sevryuk [12]). Given an essential closed curve $\gamma$,
$f(\gamma)$ may not intersect with $\gamma$. So our theorem does not apply
directly. However, the proof of our theorem is based on analysis of the chain
recurrent set $\mathcal{R}(f)$. It is easy to see that the chain recurrent set
is symmetric with respect $R$ for reversible systems, i.e.,
$R(\mathcal{R}(f))=\mathcal{R}(f)$, furthermore, the intersection property
holds for any symmetric simple closed curve $\gamma$. More precisely, if
$R(\gamma)=\gamma$, then by reversibility,

$R(f^{-1}(\gamma))=f(R(\gamma))=f(\gamma).$

Suppose, without loss of generality, $f(\gamma)$ is inside the annulus bounded
by $A_{0}$ and $\gamma$, then $f^{-1}(\gamma)$ must be outside of the annulus
bounded by $A_{0}$ and $\gamma$ which contains $f(\gamma)$, a contradiction.

Our proof works under the intersection property for symmetric simple closed
curves.

We conclude that, by the proof of our theorem, any reversible twist map on the
annulus has at least one fixed point. It turns out that there must be two
fixed points in this case. One can remove index zero fixed point (Slaminka
[14]) by some local modifications without breaking the intersection property.
Carter’s [5] result does not apply in this case.

## References

  * [1] Barrar, R. B. Proof of the fixed point theorems of Poincaré and Birkhoff.  Canadian J. Math. 19 (1967), 333–343.
  * [2] Birkhoff, G. D. Proof of Poincaré’s geometric theorem.  Trans. Amer. Math. Soc. 14, 1 (1913), 14–22.
  * [3] Birkhoff, G. D. An extension of Poincaré’s last geometric theorem.  Acta Math. 47, 4 (1926), 297–311.
  * [4] Brown, M. A new proof of Brouwer’s lemma on translation arcs.  Houston J. Math. 10, 1 (1984), 35–41.
  * [5] Carter, P. H. An improvement of the Poincaré-Birkhoff fixed point theorem.  Trans. Amer. Math. Soc. 269, 1 (1982), 285–299.
  * [6] Chow, S.-N., and Pei, M. L. Aubry-Mather theorem and quasiperiodic orbits for time dependent reversible systems.  Nonlinear Anal. 25, 9-10 (1995), 905–931.
  * [7] Conley, C. Isolated invariant sets and the Morse index, vol. 38 of CBMS Regional Conference Series in Mathematics.  American Mathematical Society, Providence, R.I., 1978.
  * [8] Devaney, R. L. Reversible diffeomorphisms and flows.  Trans. Amer. Math. Soc. 218 (1976), 89–113.
  * [9] Franks, J. Generalizations of the Poincaré-Birkhoff theorem.  Ann. of Math. (2) 128, 1 (1988), 139–151.
  * [10] Franks, J. Geodesics on $S^{2}$ and periodic points of annulus homeomorphisms.  Invent. Math. 108, 2 (1992), 403–418.
  * [11] Franks, J. Notes on Chain Recurrence and Lyapunonv Functions.  http://arxiv.org/abs/1704.07264 (2017), 1–8.
  * [12] Sevryuk, M. B. Reversible systems, vol. 1211 of Lecture Notes in Mathematics.  Springer-Verlag, Berlin, 1986.
  * [13] Siegel, C. L., and Moser, J. K. Lectures on celestial mechanics.  Springer-Verlag, New York-Heidelberg, 1971.  Translation by Charles I. Kalme, Die Grundlehren der mathematischen Wissenschaften, Band 187.
  * [14] Slaminka, E. E. Removing index $0$ fixed points for area preserving maps of two-manifolds.  Trans. Amer. Math. Soc. 340, 1 (1993), 429–445.