I’ve been going through Thurston’s book ‘The Geometry and Topology of Three-Manifolds‘ in a reading course with Amie Wilkinson. In Chapter 3, p32, when he’s constructing a hyperbolic structure on the Whitehead link complement, there is a picture on how to glue the 2-cells to the knot, to quite Thurston, ‘the attaching map for the two-cells are indicated by the dotted lines.’ However, for me it’s impossible to see where are the dotted lines going. So I reconstruct it here with some more clear pictures. The construction itself was a cool reading that I wish to share.

First, we have the Whitehead link, looking like the first figure below:

We attach three 1-cells (line segments) as in the second figure, note that the ‘x’ in the middle represents a line segment orthogonal to the screen, connecting the top and bottom line in the figure ‘8’ loop.

Now we will start to attach four 2-cells to the 1-complex above: First, we attach a 2-cell spanning the top part of the figure ‘8’ loop, spanning one side of the middle segment and two sides of the top segment (denote this by cell A):

Do the same with the bottom half (cell B). Note that each cell is attached to three edges, hence they are triangles without vertices in the knot complement with three one-cells attached.

For the other two cells, we attach as follows (cells C and D):

Combining the four 2-cells, we get something like the figure showed below. Note that at the top, cell A is under cell C in the left, intersecting the surface spanned by cells C and D at the edge, and comes above cell D to the right of the edge.

It’s easy to see that the complement of the above 2-complex does not separate $\mathbb{R}^3$, hence it’s a 3-cell with eight faces (i.e. it has to go through both sides of each 2-cell in order to fill the 3-space) each of its face has three edges. Hence we may glue an octahedron to the 2-complex after the gluing, pairs of faces of the octahedron will be identified groups of four edges will be identified to single edges. Hence to put a hyperbolic structure on the link complement, it suffice to put an hyperbolic structure to the octahedron with vertices deleted.

Since each edge is glued up by four edges of the octahedron, it suffice to find an octahedron (without vertices) in the hyperbolic 3-space that has all adjacent faces intersect in dihedral angle $2 \pi / 4$ i.e. all adjecent faces are orthogonal in the hyperbolic space. But this is achieved if we inscribe the regular octahedron into the Klein model (also called projective model of hyperbolic 3-space.

The gluing map for the faces are merely rotations and reflections of the ball which are certainly hyperbolic isometries. Hence this gives a hyperbolic structure to the link complement.

# On Alexander horned sphere

As I was drawing pictures for some stuff that should be done a year ago, I found this part would make a cool blog post, so here it is ^^ (well I admit that I mainly just want t show off the picture)

For kids who doesn’t know, let’s first talk a bit about what this ‘sphere’ is:

This is an embedded topological sphere in $\mathbb{R}^3$ which has non-simply connected exterior. Also, Since the surface is compact, through inversion about any point bounded away from infinity by the surface, we obtain a ‘sphere’ that bounds a non-simply connected region inside. This shows that the topology of the complement of a compact surface depends on the embedding, which is not true for embeddings of  compact $1$-dimensional manifolds in $\mathbb{R}^2$. (i.e. all Jordan curves separates the plane into two simply connected open sets, via the Jordan curve theorem)

The construction, as shown in the beautiful 2 page article by Alexander, goes as follows:

Take an ordinary sphere (stage 0), stretch and bend it like a banana so that the two ‘end caps’ are supported on a pair of parallel circles such that one lies vertically on top of the other (state 1). Next, on each cap we develop a banana shape, the banana shape on the two caps link though each other and again has their caps supported on a pair of parallel circles (stage 2).  Continue the process to add successively smaller bananas on the caps produced in the immediate preceding stage.

Claim: The limit is a topological sphere.

To see this, we build homeomorphisms from $S^2$ to each sphere in the intermediate stages. i.e. let

$h_n: S^2 \rightarrow S^2_n$

be a homeomorphism where $S^2_n \subseteq \mathbb{R}^3$ is the embedded sphere at stage $n$.

We may take $h_n$ s.t. $h_n^{-1}$ restricting to the complement of the $(n-1)$th stage caps (denoted by $C_{n-1}$) agree with $h_{n-1}^{-1}$. Hence union the maps $h_n|_{C_{n-1}}$ gives a continuous map on the complement of a Cantor set on the sphere. (Since $C_n$ is increasing and the caps gets smaller) This map can be extended continuously to the whole sphere because any neighborhood of points in the Cantor set contains pre-image of some sufficiently small cap.

The extension $h$ is injective since any two points in the Cantor set will be separated by a pair of disjoint pre-image of small caps. Since the sphere is compact, we conclude $h$ is a homeomorphism. i.e. the limiting surface is a topological sphere.

The exterior of the surface is not simply connected as a loop just outside the ‘equator’ can’t be contracted to a point. In fact, it’s also easy to show that the fundamental group of the exterior is not finately generated.

For some reason, Charles and I wanted to create a diffeotopy of from the standard sphere to an Alexander horned sphere. (with differentiability failing only at time one, and this is necessary since there can be no diffeomorphism from the sphere to the horned sphere, otherwise it would extend to a neighbourhood of the surfaces and hence the whole $\mathbb{R}^3$, but the exteriors of the two are not homeomorphic.)

The above figure is in fact a particular kind of Alexander horned sphere we needed. i.e. it has the property that each cap in the $(n+1)$th stage has diameter less than $1/2$ of that in the $n$th stage, and the distance between the parallel circles is also less than $1/2$ of that in the previous stage. Spheres at each stage is differentiable.

This would allow us to construct a diffeotopy that achieves stage $n$ at time $1-1/2^n$, the diffeotopy is of bounded speed as all horns are half as large as the pervious stage, hence once we get to the first stage with bounded speed, making all points traveling at that maximum speed would get one to the next stage using $1/2$ as much time.

However, we do not know if all horned sphere can be achieved by s diffeotopy from the standard sphere. i.e. does the property of being a ‘diffeotopic sphere’ depend on the embedding in $\mathbb{R}^3$.

Many thanks to Charles Pugh for forcing me to look at this business. It is indeed very fun~

# Fundemental Theorem of Dynamical Systems (Part 2)

Now we begin to prove the theorem.

4.Attractor-repeller pairs

Definition: A compact set $A \subseteq X$ is an attractor for $f$ if there exists $U$ open, $f(\bar U) \subseteq U$ and $\displaystyle \bigcup_{i=0}^\infty f^i(U) = A$. $U$ is called a basin of attraction.

For any attractor $A \subseteq X$, $U$ be a basin for $A$, let $U^\ast = X \backslash \bar U, \ U^\ast$ is open. By definition, $f(A) = A$ and $f(A^\ast) = A^\ast$. We also have

$f^{-1}(\overline{U^\ast}) = f^{-1}(\bar{X \backslash \bar U}) \subseteq X \backslash f^{-1}(U) \subseteq X \backslash U \subseteq \overline{U^\ast}$

Definition: A repeller for $f$ is an attractor for $f^{-1}$. A basin of repelling for $f$ is a basin of attracting for $f^{-1}$.

Hence $A^\ast$ is a repeller for $f$ with basin $U^\ast$.

It’s easy to see that $A^\ast$ is defined independent of the choice of basin for $A$. (Exercise)

We call such a pair $A, \ A^\ast$ an attracting-repelling pair.

The following two properties of attracting-repelling pairs are going to be important for our proof of the theorem.

Proposition 1: There are at most countably many different attractors for $f$.

Proof: Since $X$ is compact metric, there exists countable basis $\mathcal{U} = \{U_i\}_{i \in \mathbb{N}}$ that generates the topology.

For any attractor $A$, any attracting basin $\mathcal{B}$ of $A$ is a union of sets in $\mathcal{U}$, i.e. $\displaystyle \mathcal{B} = \bigcup_{i=1}^\infty U_{n_i}$latex for some subsequence $(n_i)$ of $\mathbb{N}$.

Since $A$ is compact, $U_{n_i}$ is an open cover of $A$, we have some $\{ m_1, \ m_2, \ \cdots \ m_k \} \subseteq \{n_i\}_{i \in \mathbb{N}}$ s.t. $\{ U_{m_1}, \ U_{m_2}, \ \cdots, \ U_{m_k} \}$ covers $A$.
Let $B' = U_{m_1} \cup U_{m_2} \cup \cdots \cup U_{m_k}$ hence $A \subseteq B' \subseteq B$. We have

$\displaystyle A \subseteq \bigcap_{n \in \mathbb{N}} f^n(B') \subseteq \bigcap_{n \in \mathbb{N}} f^n(B)$

Since $B$ is an attracting basin for $A$, all three sets are equal. Hence $A = \bigcap_{n \in \mathbb{N}} f^n(B')$. i.e. any attractor is intersection of foreward interates of come finite union of sets in $\mathcal{U}$. Since $\mathcal{U}$ is countable, the set of all finite subset of it is coubtable.

Hence there are at most countably many different attractors. This establishes the proposition.

By proposition 1, we let $(A_n)_{n \in \mathbb{N}}$ be a list of all attractors for $f$. Now we are going to relate the arrtactor-repeller pairs to the chain recurrent set and chain transitive components.

Proposition 2:

$\mathcal{R}(f) = \displaystyle \bigcap_{n \in \mathbb{N}}(A_n \cup A^\ast_n)$

Proof: i) $\mathcal{R}(f) \subseteq \bigcap_{n \in \mathbb{N}}(A_n \cup A^\ast_n)$

This is same as saying for any attractor $A$, $\mathcal{R}(f) \subseteq (A \cup A^\ast)$.

For all $x \notin (A \cup A^\ast)$, let $B$ be a basin of $A$, then there is $N \in \mathbb{N}$ for which $x \notin (f^N(B) \cup f^{-N}(B^\ast))$ (recall that $B^\ast$ is the dual basin of $B$ for $A^\ast$). Since $B^\ast = X \backslash \overline{B}$ and $f(\overline{B}) \subseteq B$ we conclude

$X \backslash f^{-N}(B^\ast) = f^{-N}(\overline{B}) \subseteq f^{-N-1}(f(\overline{B})) \subseteq f^{-N-1}(B)$

Hence $x \in f^{-N-1}(B)$. Let $M$ be the smallest integer for which $x \in f^{-M}(B)$. Hence $x \in f^{-M}(B) \backslash f^{-M+1}(B)$. Let $U = f^{-M}(B)$ is also a basin for $A$.

Now we show such $x$ cannot be chain recurrent: Since $X \backslash f(U)$ and $\overline{f^2(U)}$ are compact and disjoint, we may let

$\varepsilon_1 = \frac{1}{2} \min\{ d(a, b) \ | \ a \in X \backslash f(U), \ b \in \overline{f^2(U)} \}$

Since $f(x) \in f(U)$, there exists some $\varepsilon_2$ s.t.

$\overline{B(f(x), \varepsilon_2)} \subseteq f(U)$

$f(\overline{B(f(x), \varepsilon_2)}) \subseteq f^2(U)$ so there exists $\varepsilon_3$ s.t.

$N(f(\overline{B(f(x), \varepsilon_2)}), \varepsilon_3) \subseteq f^2(U)$

(Here $B(p, r)$ denotes the ball of radius $r$ around $p$ and $N(C, r)$ denotes the $r$-neignbourhood of compact set $C$)

Now set $\varepsilon = \min\{ \varepsilon_1, \ \varepsilon_2, \ \varepsilon_3\}$, for any $\varepsilon$-chain $x, x_1, x_2, \cdots$, we have: Since $\varepsilon < \varepsilon_2$ and $\varepsilon_3$, $x_1 \in B(f(x), \varepsilon_2)\subseteq f(U)$ and $x_2 \in B(f(x_1), \varepsilon_3) \subseteq N(f(\overline{B(f(x), \varepsilon_2)}), \varepsilon_3) \subseteq f^2(U)$. Hence the third term of any such chain must be in $f^2(U)$. Since $\varepsilon < \varepsilon_1$, no $\varepsilon$-chain starting at $x_2$ can reach $X \backslash f(U)$, in particular, the chain $x, x_1, x_2, \cdots$ does not come back to $x$. Hence we conclude that $x$ is not chain recurrent. ii) $\bigcap_{n \in \mathbb{N}}(A_n \cup A^\ast_n) \subseteq \mathcal{R}(f)$ Suppose not, there is $x \in \bigcap_{n \in \mathbb{N}}(A_n \cup A^\ast_n)$ and $x \notin \mathcal{R}(f)$. i.e. for some $\varepsilon > 0$ there is no $\varepsilon$-chain from $x$ to itself. Let $U$ be the open set consisting all points that can be connected from $x$ by an $\varepsilon$-chain.

We wish to generate an attractor by $V$, to do this all we need to check is $f(\overline{V}) \subseteq V$:
For any $y \in \overline{V}$ there exists $y' \in V$ with $d(f(y), f(y')) < \varepsilon$. Since $y' \in V$, there is $\varepsilon$-chain $x, x_1, \cdots, x_n, y'$ which gives rise to $\varepsilon$-chain $x, x_1, \cdots, x_n, y', f(y)$. Therefore $f(\overline{V}) \subseteq V$.

Hence $\displaystyle A = \bigcap_{n \in \mathbb{N}} f^n(V)$ is an attractor with $V$ as a basin.

By assumption, $x \in A \cup A^\ast$, since $A \in V$ and there is no $\varepsilon$-chain from $x$ to itself, $x$ cannot be in $A$. i.e. $x \in A^\ast$ Take any limit point $y$ of $(f^n(x))_{n\in \mathbb{N}}$, since $A^\ast$ is compact $f$-invariant we have $y \in A^\ast$. But since we can find $N$ where $d(f^N(x), y)<\varepsilon$, $x, f(x), \cdots, f^{N-1}(x), y$ gives an $\varepsilon$-chain from $x$ to $y$, hence $y \in V$.

Recall that $V$ is a basin of $A$ hence $A^\ast \cap V$ is empty. Contradiction. Establishes proposition 2.

This proposition says that to study the chain recurrent set is the same as studying each attractor-repeller pair of the system. But the dynamics is very simple for each such pair as all points not in the pair will move towards the attractor under foreward iterate. We can see that such property is goint to be of importantce for our purpose since the dynamical for each attractor-repeller pair is like the sourse-sink map.

5. Main ingredient

Here we are going to prove a lemma that’s going to produce for us the ‘building blocks’ of our final construction. Namely a function for each attracting-repelling pair that strickly decreases along the orbits not in the pair. In light of proposition $2$, we should expect to put those functions together to get our complete Lyapunov function.

Lemma1: For each attractor-repeller pair $A, A^\ast$ there exists continuous function $g: X \rightarrow [0,1]$ s.t. $g^{-1}(0)=A, \ g^{-1}(1)=A^\ast$ and $g(f(p)) < g(p)$ for all $p \notin (A\cup A^\ast)$.

Proof: First we define $\varphi: X \rightarrow [0,1]$ s.t.

$\varphi(x) = \frac{d(x,A)}{d(x, A)+d(x, A^\ast)}$

Note that $\phi$ takes value $0$ only on $A$ and $1$ only on $A^\ast$. However, $\phi$ can’t care less about orbits of $f$.

Define $\bar\varphi(x) = \sup\{\varphi(f^n(x)) \ | \ n\in\mathbb{N} \}$. Hence automatically for all $x$, $\bar \varphi(f(x)) \leq \bar \varphi(x)$. Since no points accumulates to $A^\ast$ under positive iterations, we still have the $\bar \varphi^{-1}(0)=A$ and $\bar \varphi^{-1}(1) = A^\ast$.

We now show that $\bar\varphi$ is continuous:

For $x \in A^\ast$ and $(x_i)\rightarrow x$, $\varphi(x_i) \leq \bar\varphi(x_i) \leq 1$ and $(\varphi(x_i)) \rightarrow 1$ hence $\bar\varphi(x_i)\rightarrow 1$ i.e. $\bar\varphi$ is continuous on $A^\ast$.

For $x \in A$ we use the fact that $A$ is attracting. Let $B$ be a basin of $A$. For all $(x_i) \rightarrow x$, for any $\varepsilon>0$, there is $N \in \mathbb{N}$ s.t. $f^N(B) \subseteq N(A, \varepsilon)$. Therefore for some $x_i \in f^N(B)$, all $f^n(x_i)$ are in $N(A, \varepsilon)$ i.e. $\varphi(f^n(x_i))\leq \frac{\varepsilon}{\varepsilon+C}$ hence $\bar\varphi(x_i)\leq \frac{\varepsilon}{\varepsilon+C}$. But for some $M \in \mathbb{N}$ and all $m>M$, $x_m \in B$. Therefore $\bar\varphi(x_i) \rightarrow 0$. $\bar\varphi$ is continuous on $A$.

Let $T = \overline{B} \backslash f(B)$, for any $\displaystyle x \in T, \ r=\inf_{x \in T} \varphi(x)$, since $f^n(T) \subseteq f^n(\overline{B})$, there exists $N>0$ s.t. for all $n>N$ $\varphi(f^n(T))\subseteq [0,r/2]$. i.e. $\displaystyle \bar\varphi9x) = \max_{0\leq n\leq N} \varphi(f^n(x))$ which is countinous. Since those ‘bands’ $T$ partitions the whole $X$ (by taking $B_n$ to be $f^n(B) \backslash f^{n+1}(B)$), hence we have proven $\bar\varphi$ is continuous on the whole $X$.

Finally, we define

$\displaystyle g(x) = \sum_{n=0}^\infty \frac{\bar\varphi(f^n(x))}{2^{n+1}}$

We check that $g$ is continuous since $\bar\varphi$ is. $g$ takes values $0$ and $1$ only on $A$ and $A^\ast$, respectively. For any $x \notin (A \cup A^\ast)$,

$\displaystyle g(f(x))-g(x) = \sum_{n=0}^\infty \frac{\bar\varphi(f^{n+1}(x))-g(f^n(x))}{2^{n+1}}$

therefore $g(f(x))-g(x) = 0$ iff $\bar\varphi(f^{n+1}(x)) = \bar\varphi(f^n(x))$ for all $n$ i.e. $\bar\varphi$ is constant on the orbit of $x$. But this cannot be since there is a subsequence of $(f^n(x))$ converging to some point in $A$, continuity of $\bar\varphi$ tells us this constant has to be $0$ hence $x \in A$.

Therefore $g$ is strictly decreasing along orbits of $f$ not in $(A \cup A^\ast)$.

Establishes lemma 1.

6.Proof of the main theorem

The proof of the main theorem now follows easily from what we have established so far.

First we restate the fundamental theorem of dynamical systems:

Theorem: Complete Lyapunov function exists for any homeomorphisms on compact metric spaces.

Proof: First we enumerate the countably many attractors as $(A_n)_{n \in \mathbb{N}}$. For each $A_n$, we have function $g_n: X \rightarrow R$ where $g_n$ is $0$ on $A_n$, $1$ on $A_n^\ast$ and is strictly decreasing on $X \backslash ( A_n \cup A_n^\ast)$.
Define $g: X \rightarrow \mathbb{R}$ by

$\displaystyle g(x) = 2 \cdot \sum_{n=1}^\infty \frac{g_n(x)}{3^n}$

Since each $g_n$ is bounded between $0$ and $1$, the sequence of partial sums converge uniformly. Hence the limit function $g$ is continuous.

For points $p \in \mathcal{R}(f)$, we have $p \in (A_n \cup A_n^\ast)$ for all $n \in \mathbb{N}$. i.e. $latex \ \forall n \in \N, \ g_n(p) = 0$ or $g_n(p) = 1$. Hence we have

$g(p) = \sum_{n=1}^\infty \frac{2 \cdot g_n(p)}{3^n} = \sum_{n=1}^\infty \frac{a_n}{3^n}$

where each $a_n$ is in $\{0, 2 \}$. This is same as saying the base-$3$ expansion of $g(p)$ only contains digits $0$ and $2$. We conclude $g(\mathcal{R}(f)) \subseteq \mathcal{C}$ where $\mathcal{C}$ is the standard middle-third Cantor set in $[0, 1]$. i.e. $g(\mathcal{R}(f))$ is compact and nowhere dense in $\mathbb{R}$.

For $p \notin \mathcal{R}(f)$, there exists $n \in \mathbb{N}$ such that $p \notin (A_n \cup A_n^\ast)$, hence $g_n(f(p)) < g_n(p)$. This implies $g(f(p)) < g(p)$ since $g_i(f(p)) \leq g_i(p)$ for all $i \in \mathbb{N}$. i.e. $g$ is strictly decreasing along orbits that are not chain recurrent.

To show $g$ is constant only on the chain-transitive components, we need the following lemma:

Claim: $p, \ q \in \mathcal{R}(f)$ are in the same chain-transitive component iff there is no attracting-repelling pair $A, \ A^\ast$ where one of $p, \ q$ is in $A$ while the other in $A^\ast$.

Proof (of claim):$\Rightarrow$” Suppose $x, y \in \mathcal{R}(f)$ and $x \sim y$, for any attractor $A$, if $x \notin A$ and $y \notin A$, then $x, y$ are both in $A^\ast$ and we are done. Hence suppose at least one of $x, y$ is in $A$. W.L.O.G. suppose $x \in A$. Let $B$ be a basin of $A$. Since $\overline{f(B)}, \ X\backslash B$ are closed and disjoint, we may choose $\varepsilon<\min\{ d(a, b) \ | \ a \in (X\backslash B), \ b \in \overline{f(B)} \}$. By the same arguement as in proposition $2$, there are no $\varepsilon/2$-chain (with length $>1$) from any point in $f{B}$ to any point in $X \backslash B$. Hence there is also no $\varepsilon/2$-chain from any point in $A$ to any point in $A^\ast$. Hence $y \notin A^\ast$ i.e. $y \in A$.

$\Leftarrow$” Suppose for any attractor $A$, $x \in A$ iff $y \in A$. For any $\varepsilon>0$, let $U$ be the set of all points $y$ for which there is an $\varepsilon$-chain from $x$ to $y$, as defined in proposition $2$. We have showed in proposition $2$ that $U$ is a basin of some attractor $A'$. Since $x \in \mathcal{R}(f) \subseteq (A' \cup {A'}^\ast)$ and $x \in U$, hence $x \in A'$. Hence by our assumption, $y$ must be also in $A'$. Hence $y \in U$ i.e. there is an $\varepsilon$-chain from $x$ to $y$. Since the construction is symetric, we may also show there is an $\varepsilon$-chain from $y$ to $x$. i.e. $x\sim y$.

Establishes the claim.

Finally, for $p, q \in \mathcal{R}(f)$, $g(p) = g(q)$ means $g(p)$ and $g(q)$ has the same base-$3$ expansion in the Cantor set. This is same as saying $\forall i \in \mathbb{N}, \ g_i(p) = g_i(q) \in \{0, 2\}$, which is to say there is no $i \in \mathbb{N}$ for which one of $p, \ q$ is in $A_i$ while the other in $A_i^\ast$. Hence by Lemma, we conclude that $g(p) = g(q)$ iff $p, \ q$ are in the same chain transitive component.

This establishes our theorem.

# Fundemental Theorem of Dynamical Systems (Part 1)

This article was written as a homework of professor Wilkinson’s dynamical systems course. Since the content is expository and detailed presentation of the theorem is missing from many books, I decided to post it here. I have mostly followed a set of notes by John Franks, with additional discussions on the intuition and ideas behind the statement and the proof.

1.Introduction

So far we have discussed various different kinds of dynamical systems ranging from topological, smooth to hyperbolic and partially hyperbolic. One might wonder if there is a united theme to the subject as a whole. Indeed, as in many other subjects, there is a so-called fundamental theorem of dynamical systems. This theorem is first stated and proved by Conley in where he studies attractor and repellers. The theorem, loosely speaking, gives a universal decomposition of any systems on compact metric spaces into invariant compact sets wandering orbits that travels between such sets.

To state this more precisely we make an analogy with Morse theory: When looking at the gradient flow on a compact embedded manifold, we ‘decompose’ the manifold into critical points and orbits that originates and ends at critical points. In a similar spirit, given any homeomorphisms on a compact metric space, we may look at it’s ‘indecomposible’ compact invariant sets and how they are ‘connected’ by wandering orbits, we then ‘place’ those compact sets on different ‘heights’ and have all other point going between the minimal sets they originates and ends at. The theorem guarantees that we can ‘place’ the space in a way that all wandering orbits are going ‘down’ at all times.

In light of the theorem, we have descried the global structure of the system except for what happens on the ‘indecomposible’ sets. i.e. The problem of understanding general topological systems on compact manifolds is reduced to understanding ‘transitive’ homeomorphisms on compact sets. The latter, unfortunately, could still be quite complicated as we have seen in the Horseshoe example.

The theorem is proposed to be the Fundemental theorem of dynamical systems because of its nature in giving concise description of all possible behaviors of a system in the given setting. In some sense, dynamics is the study of limiting behrviors of all points under interation, the theorem breaks the system down into a recurrent part and a wandering part where the behavior of the wandering part is gradient-like. Since we have developped sets of different tools for studying systems that exhibits a lot of recurrence as well as for studying gradient-like systems, this allows us to connect combine the tool sets and treat any systems in the setting.

2.Background

In this section, we define $\varepsilon$-chains, chain recurrent sets and chain transitive components for a homeomorphism on a compact metric space. Those concepts will come up in the statement of the fundamental theorem. In fact, those are going to be the ‘minimal compact sets’ we decompose our metric space into.

Given compact metric space $X$ and homeomorphism $f: X \rightarrow X$,
Definition: Given two points $p, q \in X$, an $\varepsilon$-chain from $p$ to $q$ is a sequence $x_1, x_2, \cdots x_n$, $n>1$ where $x_1 = p, \ x_n = q$ and for all $1 \leq i \leq n-1$, $d(f(x_i), x_{i+1}) < \varepsilon$.

i.e. we take a point and start applying $f$ to it, but at each iterate, we are allowed to perturb the resulting point by $\varepsilon$. Such ‘pseudo-orbits’ are in general much easier to obtain than true orbits.

More generally, $\varepsilon$-chains can be taken infinite. i.e. if we have a (possibly infinite) subinterval $I \subseteq \mathbb{Z}$, an $\varepsilon$-chain indexed by $I$ is a set of points $(p_i)_{i \in I}$ s.t. $d(f(p_i), p_{i+1}) < \varepsilon$ whenever $i, i+1$ are both in $I$.

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

Note that non-wandering points are necessarily chain recurrent: If $p \in X$ is non-wandering, we may take the neighborhood to be the $\varepsilon$-ball around $p$, since $p$ is non-wandering, we have some $n>1$ where $f^n(B_\varepsilon(p)) \cap B_\varepsilon(p) \neq \phi$, we pick $q$ in the intersection and define $\varepsilon$-chain $p, f^{-n}(q), f^{-n+1}(q), \cdots, q, p$.

At this point, it’s perhaps worthwhile to mention our completed ordering of different notions of recurrence:

$\overline{\mbox{Per}(f)} \subseteq \mbox{Rec}(f) \subseteq \mbox{NW}(f) \subseteq \mathcal{R}(f)$

Each of the above inclusion can be made strict (see Exercises). Chain recurrence is perhaps the weakest notion I’ve seen for a point to be, in any sense, recurrent. A Friendly challenge to the reader: think of a case where you feel confortable calling a point ‘recurrent’ while it’s not in the chain recurrent set of the system.

We now define equivlence relation on $\mathcal{R}(f)$ as follows:

For $p, q$ in $\mathcal{R}(f)$, $p \sim q$ iff for all $\varepsilon > 0$, there are $\varepsilon$-chains from $p$ to $q$ and from $q$ to $p$. $\sim$ is reflexive since all points in $\mathcal{R}(f)$ are chain recurrent; symmetric by definition and transitive by the obvious composition of $\varepsilon$-chains.

Definition: The equivalence classes in $\mathcal{R}(f)$ for $\sim$ are called chain transitive components.

It’s easy to check that chain transitive components are compact and $f$-invariant. Those are components that’s transitive in a very weak sense. i.e. any two points are connected by a ‘pseudo-orbit’, or equivalently, there is a dense (infinite) pseudo-orbit. (see exercises)

As mentioned above, throughout the rest of the chapter, we will consider chain transitive components as ‘indecomposible parts’ of our system. Those are the parts for which all points are ‘recurrently’ and each component is ‘transitive’, both in a very weak sense. We further specify how does the points that are not in the chain-recurrent set iterates between those components.

3.Statement of the theorem

Given compact metric space $X$ and homeomorphism $f: X \rightarrow X$,

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

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

$\forall \ p, q \in \mathcal{R}(f), \ g(p) = g(q) \ \mbox{iff} \ p \sim q$

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

Hence this is a function that stays constant only on the chain transitive components and strictly decreases along any orbit not in $\mathcal{R}(f)$. We also require the image of $\mathcal{R}(f)$ to be compact and nowhere dense which cooresponds to the ‘critical values’ of a gradient function being compact nowhere dense as a result of Sard’s theorem.

Fundemental theorem of dynamical systems:

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

As a historical remark, the theorem first appeared in Charles Conley’s CBMS monograph Isolated Invariant Sets and the Morse Index in 1978 [C]. In the book he developed the theory of attractor-repeller pairs in relation to Morse decomposition and index theory. The above theorem was one of the major results. Although Conley was originally more focused on the setting where instead of a homeomorphism, we have a continuous flow on the manifold (which makes it even more similar to the gradient flow), but this discrete formulation became more popular as the theory develops. The theorem is later proposed by D. Norton as the fundamental theorem of dynamical in 1995.

The proof is going to be a specific construction: First, we define a family of partitions of the chain recurrent set, each divides the set into two pieces (i.e. a attractor-repeller pair intersected with $\mathcal{R}$). Then we prove the family is countable and points in the same chain transitive component are not separated by any partition in the family. Furthermore, each chain-transitive component is uniquely defined by specifying which set does it belong to in each partition. i.e. the smallest common refinement for the family exactly partitions $\mathcal{R}$ into chain-transitive components. (section 4)

Next, for each attractor-repeller pair, we prove the existence of a function that takes value $0$ on the attractor and $1$ on the repeller and strictly decreases along orbits of points that’s not in $\mathcal{R}(f)$. We should also mention the fact that all points that are contained in one of the sets in each pair must be in chain recurrent. (section 5)

The complete Lyapunov function is then constructed by taking an appropriate infinite sum of such functions. This way we get a function that separates all chain transitive components, stays constant on each component and strictly decreases along all orbits which are not in $\mathcal{R}$. The image of the chain recurrent set will be contained in the middle-third Cantor set. (section 6)

(see part 2 for sections 4-6)