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I'm studying the book titled, The Topology of Chaos: Alice in Stretch and Squeezeland by Robert Gilmore and Marc Lefranc (Wiley, 2011).

CONFUSION : I am unable to understand the use of the notation $f^{-1}(\cdot)$ because $f(\cdot)$ is a chaotic map and from the symbol it seems that $f^{-1}$is the inverse of the chaotic map. But I think a map that has an inverse cannot be chaotic. Also, $f^{-1}(\cdot)$ is not chaotic map. (please correct me wherever I am wrong).

Based on this confusion, I am unable to understand what $f^{-1}(\cdot)$ denotes in Eqs (2.49)-- (2.51) given in Chapter 2, Discrete Dynamical Systems. On page 44 Eq(2.43) presents the commutative diagram.

The crux of the matter is that with the use of backward iteration of an inverse function $\Theta^{-1}(\cdot)$ (in the book, this is the function $f^{-1}(\cdot)$) guided by a symbolic sequence, can converge towards an initial condition $x_0$.

This is my interpretation where I am not using the symbol $f^{-1}$. Can somebody please let me know if my understanding is correct or not?


A nonlinear map, $f(\cdot)$ which has chaotic dynamics is described by the following equation \begin{align} x_n &= f(x_{n-1},\wp), x_n \in \mathcal{R} \label{Eq2} \end{align} where $f : \mathcal{R} \rightarrow \mathcal{R}$, $\wp$ is the bifurcating parameter of the chaotic map. A symbolic sequence can be generated from a chaotic sequence by partitioning of the state space, $I$ of the chaotic map into a finite number of $m$ disjoint regions (partitions), $I = \{I_1,I_2,\ldots,I_m\}$. For any sequence generated by iterating the above map, if we an assign $m$ alphabets, $a_1,a_2,\ldots,a_m$ to each of the disjoint regions, the dynamics of the system can be represented by a sequence of the finite alphabet. In other words, each partition is denoted by a symbol $s_n = \Theta(x_n) = a_i, x_n \in I_{a_i}, (i = 1,2,\ldots, m)$. As an example, for $m=2$ symbols, we would get \begin{align} s_n = \Theta(x_n) &= 0, \text{if } x_n \in I_0, \nonumber \\ &= 1, \text{if } x_n \in I_{1} \end{align} where the symbols $a_1 = 0$ and $a_2 = 1$.

A symbolic sequence can be governed by a left shift operation $\sigma(\cdot)$ where the leftmost symbol is discarded at each iteration, shown as follows: \begin{align} \mathbf{s}_n = \sigma(s_n s_{n+1} \ldots) = s_{n+1}s_{n+2}\ldots \end{align} where $s_n s_{n+1} \ldots$ denotes a character string, and the symbol $s_n$ depends on which interval $I_p$, $x_n$ falls within.

The map $\sigma : \Sigma \rightarrow \Sigma$

By knowing $x_n$ and $s_n$ it is sufficient to determine $x_{n-1}$ by the following operation expressed mathematically as: \begin{align} x_{n-1} &= \Theta^{-1}(s_n,x_n) \end{align} where $\Theta^{-1}(\cdot)$ is the inverse of $\Theta(\cdot)$ given a symbol $s_n$. This means that

\begin{align} x_0 = \Theta^{-1}(s_1,x_1) \circ \Theta^{-1}(s_2,x_2) \circ \dotsb\circ \Theta^{-1}(s_{N-1},x_{N-1}) \end{align} The above Equation is my understanding of Eq(2.51) in the book.

So, we have the following relations (Diagram on pg. 44 Eq(2.43))

commutating diag

where $f(\cdot)$ is conjugate to $\sigma$ via the function $\Theta(\cdot)$. The diagram is read as $f = \Theta^{-1} \circ \sigma \circ \Theta$, which can also be written as $f\Theta = \Theta \sigma$. This means that the dynamics of $\sigma$ on $\Sigma$ and $f(\cdot)$ on $\mathcal{R}$ are same. If $f(\cdot)$ is chaotic on $\mathcal{R}$, then $\sigma(\cdot)$ is chaotic on $\Sigma$ via $\Theta(\cdot)$. Given a RS $\mathbf{s} = \{s_1s_2\ldots s_N \}$, we can construct $\mathbf{x} = \{x_1 x_2 \ldots x_N\}$ such that $\Theta(\mathbf{x}) = \mathbf{s}$ using $\Theta^{-1}(\cdot)$.

Is there an alternate better way to explain or should I use $f^{-1}(\cdot)$?

UPDATE : An application of inverse interval mapping is found in section 2.3.1 of the paper

Chaos-Based Simultaneous Compression and Encryption for Hadoop. M Usama and N Zakaria. PLoS ONE 12 no. 1, e0168207 (2017).

The authors apply inverse interval mapping using the inverse function to encode a given a symbolic sequence into an initial condition. The authors are using the map $f^{-1}$ to do the operation which is given a symbolic sequence they are using the $f^{-1}$ to map the string to a single point. Then, starting from the initial condition, if the map $f(.)$ is iterated and the symbolic dynamics obtained by state space partition using the intervals computed from the encoding stage, we can obtain the same symbolic sequence.

The maps used by the authors are: For two symbols $f(\cdot)$ the Skew Tent Map and its inverse given below

\begin{equation} \begin{aligned} f^{-1}(I) = x[n] = \begin{cases} p \times I_{d[n]}, \text{ symbol } d[n] =0 \\ 1-p \times I_{d[n]}, \text {symbol } d[n] =1 \label{InverseSkewTentMap} \end{cases} \end{aligned} \end{equation} where $I_{1}$ implies interval when the symbol at $d[i]$ is = 0 and $I_{1}$ implies the interval when the symbol at $d[i]$ is = 1$.

The Skew Tent map is expressed as \begin{equation} \begin{aligned} f(x) = \begin{cases} x/p, 0\le x <p \\ (1-x)/(1-p), p \le x \le 1 \label{SkewTentMap} \end{cases} \end{aligned} \end{equation}

The Skew Tent map is related to the symbols by the choice of the partition point $p$.

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  • $\begingroup$ It may just mean preimage. I don't have the book to look at it. $\endgroup$ – user73352 May 5 '17 at 21:59
  • $\begingroup$ $f^{-1}(.)$ is the pre-image of $f(.)$ which takes in a symbol and a number from an interval. According to my understanding, this is what $\Theta$ and $\Theta^{-1}(.)$ on the left and right vertical branches are doing. Hence, I have replaced $ f^{-1}(.)$ with $\Theta^{-1}(.)$. Please correct me where wrong. $\endgroup$ – SKM May 6 '17 at 0:04
  • $\begingroup$ It makes sense to me that the property of $f^{-1}$ being chaotic or not is not immediately provable without knowing $f$. Instinctively, a chaotic mapping which has some energy-like component that is always increasing would have a non-chaotic inverse (it always stabilizes back to 0 energy states). However, it also seems natural that the inverse pendulum problem, when inverted, is the same chaotic system, just with a time reversal. $\endgroup$ – Cort Ammon May 6 '17 at 0:51
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In general, there are many maps that are diffeomorphisms that are chaotic, so both $f$ and their inverse $f^{-1}$ are chaotic. Especially in the case of phase space dimensions two and higher. So I am assuming you are referring to the case of maps on the real line or the circle. Chaos for such one dimensional maps occurs when the maps are not invertable globally, i.e. they do not have globally defined inverse. Like the map $x \mapsto \lambda \, x(1 - x)$ in the interval $[0,1]$. However, they do have a locally defined inverse maps in a small open interval around almost every point (almost!). On top of that the notation $f^{-1}$ most likely refers to taking the preimage of a set, rather than an element, like $f^{-1}\big( [a,b] \big) = \{x \in \mathbb{R} \, | \, a \leq f(x) \leq b \}$. This is just a notation for pre-image. Not an actual map. The map doesn't have a global inverse.

Bottom line, chaotic invertable maps typically start from dimension two (not counting the restriction of maps onto some weird fractals of fractional dimensions). Plus, I am not even sure if there is a universal agreement on what the definition for chaos is.

When you do dynamics, your phase space, i.e. space of states, has some notion of how close any two points from that space are. It could be just topology (notion of open and closed sets) but in your case there is usually a metric, allowing you to measure distance. In this case a map $F : X \to Y$ of two such spaces is called a homeomorphism if it is bijective (there is well defined $F^{-1} : Y \to X$ because $F$ maps different points to different points and the whole space $X$ is mapped onto the whole space $Y$, no element left behind) and both $F$ and $F^{-1}$ are continuous, meaning they map points that are close together to points that are close together. Then $X$ and $Y$ are equivalent and whatever happens on one, the same thing can happen on the other. In addition to that, if $X$ and $Y$ are nice spaces like the line, the plane, the sphere, the three space, the three sphere, etc (these are called manifolds, generalizations of the usual 2D surfaces) you know how to take derivatives. Then the map $F$ is a diffeomorphism if on top of being bijective, both $F$ and $F^{-1}$ are continuously differentiable maps as many times as you want them to be (usually infinitely many times). In particular, you are allowed to have $X = Y$. Then $F : X \to X$ is a dynamical system.

The map $\Theta : x \to s$ is exactly a homeomorphism. Makes the two phase spaces equivalent but on top of that the property $\Theta \circ f = \sigma \circ \Theta$ makes it so that the dynamics of $f$ on one of them is absolutely equivalent to the dynamics of the other $\sigma$. You know the dynamics of $\sigma$ you know the dynamics of $f$ (and vice versa). Usually in this case the state space of your one dimensional map $f$ is in fact a Cantor set. It is complicated, but equivalent to the simpler one $\Sigma$. The commuting diagram you have looks correct, however, I do not understand the meaning of $\Theta(s_n,x_n)$. In general $\Theta(x) = s$ and in particular $\Theta(x_n) = s_n$.

The map $\Theta$ is not equal to $f$ or $f^{-1}$. \begin{align*} f(x) &= \frac{1}{p} \, x \, , \,\,\, \text{ for } x \in I_0 = [0,p)\\ f(x) &= \frac{1}{1 - p} \, (1-x) \, , \,\,\, \text{ for } x \in I_1 = [p,1] \end{align*} As you see $I = [0,1] = [0,p) \cup [p,1] = I_0 \cup I_1$. So, the space of states, also called phase space, or dynamic space, is $I$ and the dynamical system is $$f \, : \, I \, \to \, I$$ because clearly $f(I) = I$. However, the map is $2:1$ meaning for each point $y \in I$ there are two different points $x_1$ and $x_2 \in I$ such that $f(x_1) = f(x_2) = y$ (except for $x=p$). Therefore, $f$ is not invertable and it doesn't have a well defined global inverse. Locally however, it does. On $I_0$ we have $f^{-1}(y) = p\, y$ while on $I_1$ we have $f^{-1}(y) = 1-(1-p) \, y$.

At the same time, you have another dynamical system, defined as follows. Let $$\Sigma = \{ \, s \, : \{0,1,2,...\} \, \to \{0,1\} \, | \, \text{ i.e. $s$ is a sequence of the symbols $0$ and $1$ }\}$$ be the set of all sequences of zeroes and ones infinite in one direction. In other words, $s \in \Sigma$ exactly when $$s = \big(s_0, s_1, s_2, ..., s_n, ...\big)$$ where $s_k \in \{0,1\}$ for $k = 0, 1, 2, ...$. The map on $\Sigma$ is $$\sigma \, : \, \Sigma \, \to \, \Sigma$$ $$\sigma \big(s_0, s_1, s_2, ..., s_n, ...\big) = \big(s_1, s_2, ..., s_{n+1}, ...\big)$$ The claim is that the dynamics of $f$ on $I$ is equivalent (indistinguishable, whatever happens for one system, the same happens for the other) to the dynamics of $\sigma$ on $\Sigma$. The map that establishes this equivalence is $$\Theta \, : \, I \, \to \, \Sigma$$ and it has a very well defined inverse map $$\Theta^{-1} \, : \, \Sigma \, \to \, I $$ For this map, we can show that for any $x \in I$ $$\Theta \circ f(x) = \sigma \circ \Theta(x) $$ (this is equivalent to the commuting diagram in your post) and in fact, we can define a metric (a way to measure distance on $\Sigma$) so that $\Theta$ is in fact a homeomorphism. The map $\Theta$ is a translating map, that translates the dynamics written in the language of $I, f$ to the same dynamics written in the language of $\Sigma, \sigma$. Thus, as you can see, $\Theta$ is a completely different map from $f$. It also has a completely different purpose: $f$ describes the evolution of the system, $\Theta$ translates from one language to another.

Edit. The map $\Theta : I \to \Sigma$ is not the map $f^{-1}$ from your post and I don't think it is featured in the article you posted. The map $\Theta$ is defined as follows: Consider the partition $I = I_0 \cup I_1$. Pick a point $x \in I$. Set $s_0 = 0$ whenever $x \in I_0$ and $s_0 = 1$ whenever $x \in I_1$. Then take $x_1 = f(x)$. Set $s_1 = 0$ whenever $x_1 \in I_0$ and $s_1 = 1$ whenever $x_1 \in I_1$. Continue like that. Assume you have the orbit $x_0, x_1, x_2, ..., x_n$ and you have set the sequence $s_0, s_1, s_2, ..., s_n$. Then take $x_{n+1} = f(x_n)$ and again Set $s_{n+1} = 0$ whenever $x_{n+1} \in I_0$ and $s_{n+1} = 1$ whenever $x_{n+1} \in I_1$. Thus, given a point $x \in I$ you can generate a one sided infinite sequence $s = \big(s_0, s_1, s_2, ..., s_n, ...\big) \in \Sigma$. The map $\Theta : I \to \Sigma$ is defined as $\Theta(x) = s$. It is immediate to check that $$\Theta\big(f(x)\big) = \Theta\big(x_1\big) = (s_1, s_2,...) = \sigma(s_0,s_1,s_2...) = \sigma\big(\Theta(x)\big)$$ One needs to check whether $\Theta$ is injective, i.e. if $x\neq y \in I$ then $\Theta(x) \neq \Theta(y) \in \Sigma$, and whether it's surjective, i.e. for any $s \in \Sigma$ there exists $x \in I$ such that $\Theta(x) = s$. I think verifying the injectivity of $f$ is easier. It relies on the observation that the distance between any two points on $I$ is increased uniformly by the map $f$, so eventually the orbits of any two points starting from the same subinterval $I_i$ for $i=1,2$ will end up in two different subintervals, one in $I_0$ an the other in $I_1$ so the corresponding sequences of zeros and ones will be different. The surjectivity requires a bit more careful analysis, leading to a unique binary expansion of every element $x \in I$. Finally, to turn $\Sigma$ from a set to a metric space, i.e. to be able to measure distance on it, you simply push forward the usual distance on $I$, defined as $d_I(x,y) = |x-y|$, to the distance $d_{\Sigma}(s, t) = |\Theta^{-1}(s) - \Theta^{-1}(s)|$. Thus, the map $\Theta : I \to \Sigma$ is a homeomorphism (in fact an isometry, i.e. it preserves distances between the two spaces) that conjugates the two dynamical systems $\Theta \circ f = \sigma \circ \Theta$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind May 9 '17 at 10:40

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