1

$\mathbf{Definition:}$ Let $ X,Y $ be two metric spaces and $ f:X\to X$ and $g:Y \to Y$ be two mappings, $f $ and $g$ are said to be topologically conjugated (denoted by $f\sim g$) if there exist $h:X\to Y$ homeomorphism s.t $h\circ f = g\circ h$

$\mathbf{Question:}$ I am given with two mappings, $f$ and $g$, both from $\Bbb T^2 \to \Bbb T^2$, s.t $$f\begin{pmatrix}x\\y\\\end{pmatrix} = \begin{pmatrix}2&1\\1&1\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix} + \begin{pmatrix}a\\b\\\end{pmatrix}$$ Where $a,b\in \Bbb R$. And the second is "Arnold's cat map" given as $$g\begin{pmatrix}x\\y\\\end{pmatrix} = \begin{pmatrix}2&1\\1&1\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}$$ And I have asked to prove $f\sim g$. I tried a lot, but I didn't succeeded. Please if any one help me regarding finding that homeomorphism '$h$'. Thanks

Alp Uzman
  • 12,209
Badshah Khan
  • 376
  • 1
    You should not expect something so immediate. Note however that there are dense orbits. So it suffices to take one and extend the map by continuity, which gives easily a semiconjugacy. Then you prove that it is invertible. – John B Jan 16 '20 at 16:00
  • Yes Sir... You are right. I do the same thing for many other proofs of Conjugacy. But it didn't work here... – Badshah Khan Jan 16 '20 at 16:15
  • Like in many other questions, i find 'h' for just positive end, and than take the negation with possible way to make it homeo.... But here, i even not able to find atleast one way map... – Badshah Khan Jan 16 '20 at 16:17
  • 1
    Actually there is some properties of this homeo map 'h', that is, fixed point of h is fixed point of g, and periodic point of h is also the periodic point of g... – Badshah Khan Jan 16 '20 at 16:18

1 Answers1

1

When stuck, an important method in dynamics is to try to prove something stronger. In this case,

$$\Phi(x,y) = (x-b,y+b-a) \mod 1$$

is an affine automorphism of the torus such that $\Phi\circ g = f\circ \Phi$.

Jokes aside, below are two ways to see why an affine automorphism might suffice; once one realizes an affine conjugacy works then one can use your observation that any bijective conjugacy must preserve sets of fixed points (in this case a simple calculation gives that both $f$ and $g$ have unique fixed points). This gives the translation part of the automorphism. Next, differentiating the conjugacy equation one realizes that one can take the linear part to be identity.

First approach is through general ergodic theory. A classical theorem of Adler-Palais states that for $A,B$ two affine automorphisms of the $n$-torus (see e.g. Example of a continuous affine group action), if both $A$ and $B$ are ergodic (relative to Haar), then any homeomorphism conjugating them must be affine. It's well-known that $g$ is ergodic, and one can check that $f$ too is ergodic (for instance using the Hoare-Parry criterion). See Walters' paper "Topological Conjugacy of Affine Transformations of Tori" for details.

Second approach is through hyperbolic dynamics. A computation shows that the stable and unstable manifolds of $f$ and $g$ (as discussed in e.g. Is the square root of a hyperbolic map hyperbolic?) are translations of images of eigenspaces of the two-one-one-one matrix, and any topological conjugacy must send stable manifolds to stable manifolds and unstable manifolds to unstable manifolds. Products of pieces of stable and unstable manifolds give neighborhoods ("local product structure"), and after some work one can show that identity up to a translation is a conjugacy.

Alp Uzman
  • 12,209