How to construct a topological conjugacy? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:52:52Z http://mathoverflow.net/feeds/question/20246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20246/how-to-construct-a-topological-conjugacy How to construct a topological conjugacy? Akhil Mathew 2010-04-03T16:29:47Z 2010-04-03T20:32:07Z <p><strong>Background:</strong> One says that continuous maps $f: X \to X, g: Y \to Y$ are <a href="http://en.wikipedia.org/wiki/Topological_conjugacy" rel="nofollow">topologically conjugate</a> if there exists a homeomorphism $h: X \to Y$ such that $h \circ f = g \circ h$. There are many ways one can see that two maps are <em>not</em> topologically conjugate. For instance, if $f$ has fixed points and $g$ does not (more generally if the same power $f$ and $g$ have different numbers of fixed points), they cannot be topologically conjugate. <a href="http://en.wikipedia.org/wiki/Topological_entropy" rel="nofollow">Topological entropy</a> provides a fancier invariant in terms of coverings (assuming $X$ and $Y$ are compact spaces). </p> <p>I see also that there are many general theorems of that allow one to conclude that two maps are topologically conjugate (e.g. the <a href="http://en.wikipedia.org/wiki/Hartman-Grobman_theorem" rel="nofollow">Hartman-Grobman theorem</a>). </p> <p>However, I am curious:</p> <blockquote> <p>Given two discrete dynamical systems, is there a trick one can use to <em>construct</em> a topological conjugacy between them?</p> </blockquote> <p>I suppose an analogy would be a comparison between the Brouwer and Banach fixed point theorems. I'm curious if there is an iterative process as in the proof of the latter.</p> <p>(Full disclosure: I was motivated to ask this question because I got stuck on what should be a simple exercise in a book on dynamical systems. The exercise was to prove that any two $C^1$ maps $f,g: \mathbb{R} \to \mathbb{R}$ such that <code>$0&lt;f'(0)&lt;1, 0 &lt; g'(0)&lt;1$</code> are locally topologically conjugate.)</p> http://mathoverflow.net/questions/20246/how-to-construct-a-topological-conjugacy/20260#20260 Answer by Martin M. W. for How to construct a topological conjugacy? Martin M. W. 2010-04-03T20:14:27Z 2010-04-03T20:14:27Z <p>Your intuition is on target, since you mention the Banach fixed point theorem together with the Hartman-Grobman theorem. The proofs I've seen of Hartman-Grobman find the topological conjugacy as the fixed point of a certain contraction. Planet Math has a <a href="http://planetmath.org/encyclopedia/ProofOfHartmanGrobmanTheorem.html" rel="nofollow">nice write-up</a>.</p> <p>For dynamical systems with nice hyperbolicity properties, there are other types of constructions. For instance I believe you can prove structural stability of Anosov diffeomorphisms with a construction based on intersections of stable and unstable manifolds. And, while this might not be constructive enough for you, there is also the <a href="http://books.google.com/books?id=OcD9oXqmmW0C&amp;pg=PA77&amp;dq=the+shadowing+theorem&amp;cd=2#v=onepage&amp;q=the%20shadowing%20theorem&amp;f=false" rel="nofollow">shadowing theorem</a>: under the right conditions, "pseudo" orbits can be uniquely approximated by real orbits, where the approximation directly gives a homeomorphism. This is good for proving structural stability results, since a orbits of a perturbed map will be pseudo-orbits for the original map.</p> <p>Finally, you can sometimes construct conjugacies "by hand." A low-tech method of that probably works in your motivating example is to create a conjugacy by defining it on fundamental domains of $f$ and $g$ and then extending to a full neighborhood of the origin via the dynamics.</p> http://mathoverflow.net/questions/20246/how-to-construct-a-topological-conjugacy/20262#20262 Answer by Sergei Ivanov for How to construct a topological conjugacy? Sergei Ivanov 2010-04-03T20:32:07Z 2010-04-03T20:32:07Z <p>Let me throw in some speculations based on my limited involvement in dynamical systems.</p> <p>The conjugation formula $f=h^{-1}gh$ is in general not a type of a functional equation that can be solved by iterative approximations or a clever fixed-point trick. The problem is that you cannot determine how badly a particular $h$ fails by looking at the difference between LHS and RHS of this equation. The obstructions are not local and you don't see them until you consider all iterations of $f$ and $g$. Sometimes you can do approximations (e.g. Anosov system more or less survive under perturbations) but this works only in special types of systems (some kind of "hyperbolicity" is needed).</p> <p>It seems that the only "general" way for constructing a topological conjugation is to study the orbits of $f$ and $g$ and send each orbit of $f$ to a similar orbit of $g$ so that the whole map is continuous. (A dense set of orbits is sufficient, e.g. the set of periodic points of an Anosov system.) The problem is, of course, that the structure of orbits can be <em>really</em> complicated. But if it is simple, one can hope to construct a conjugation directly.</p> <p>For example, consider two homeomorphisms $f,g:\mathbb R\to\mathbb R$ satisfying $f(x)>x$ and $g(x)>x$ for all $x$. They are conjugate. To see this, consider an orbit $\dots,x_{-1},x_0,x_1,x_2,\dots$ of a point $x_0$ under the iterations of $f$. This is an increasing sequence and the intervals $[x_i,x_{i+1}]$ cover $\mathbb R$. Every other orbit "interleaves" with this one: for example, if $y_0\in(x_0,x_1)$, then $y_i:=f^i(y_0)$ lies between $x_i$ and $x_{i+1}$. So every orbit has a unique member in the interval $[x_0,x_1)$. In a sense, this interval (or rather the closed one with the endpoints glued together) naturally represents the set of all orbits.</p> <p>So take any orbit $(x_i')$ of $g$ and let $h_0$ be <em>any</em> order-preserving bijection from $[x_0,x_1]$ to $[x_0',x_1']$. This defines a unique conjugacy map $h:\mathbb R\to\mathbb R$ such that $h|_{[x_0,x_1]}=h_0$: the orbit <code>$\{f^i(y)\}$</code> of a point $y\in [x_0,x_1]$ is mapped to the $g$-orbit <code>$\{g^i(h_0(y))\}$</code> of the point $h_0(y)$. And all conjugations can be obtained this way.</p> <p>Already in this simple example, you can see how fragile things can be. Even if $f$ and $g$ are smooth and have bounded derivatives, you have no control over how big the derivatives of $h$ can grow. (And you actually lose smoothness if you do the same on a closed interval rather than $\mathbb R$.)</p> <p>If you vary the map $g$, it remains conjugate to $f$ while the condition $g(x)>x$ holds true. But it suddenly stops being conjugate once a fixed point appears. However trivial this fact is, is shows that limit of conjugacy maps does not make sense in general.</p> <p>The exercise you mention can be solved in a similar fashion as my toy example; the orbits are not much more complicated. In fact, given any two homeomorphisms $\mathbb R\to\mathbb R$, it is easy to understand whether they are conjugate or not (just study the intervals between fixed points). But the next step - homeomorphisms of the circle - is much more difficult: there are beautiful theorems, unexpected conterexamples, connections to number theory and other signs of a rich theory around such a seemingly trivial object. See Denjoy theorem and especially its smoothness requirements to get an idea how interesting these things can be.</p>