I'm studying chaos and I got stuck on following:

'the Bernoulli shift map is topologically conjugate to the tent map' e.g. Wikipedia.

I tried finding the conjugacy function C(x) such that:

C(f(x))=g(C(x)) with f(x) the tent map and g(x) the Bernoulli shift map

to proof that this indeed was valid, but I don't seem to find a valid C(x).

I searched online in the hope that I could find the answer there but without luck.

Does someone know the answer to my question? Or is it really untrivial to find such a C(x)?



The "Bernoulli shift" on 1-sided sequences of $0$'s and $1$'s cannot be topologically conjugate to the tent map on $[0,1]$, since the first acts on a totally disconnected space while the second acts on a connected space.

However, the Bernoulli shift maps onto the map $z\to z^2$ on the unit circle (a "semicomjugacy"), which we can represent by $Bt = 2t \mod 1$ on $[0,1]$, with $0$ and $1$ identified. I claim this conjugates to the tent map $$Ty = 2 \min(y,1-y) = 1 -|2y - 1|$$ on $[0,1]$.

Use the conjugating maps $y =(1/\pi) \arccos(1-2\sin^2 (\pi t/2)), t=(2/\pi) \arcsin \sqrt{(1/2)(1 - \cos (\pi y)}$.

I got these by conjugating each of $B$ and $T$ to the logistic map $Lx = 4x(1-x)$, using $x = \sin^2 (\pi t /2), y = (1/\pi) \arccos (1 - 2x), x= (1/2)(1 - \cos (\pi y)), t= (2/\pi) \arcsin \sqrt {x}$.

Here $\arcsin$ is in $[-\pi/2, \pi/2]$ and $\arccos$ is in $[0,\pi]$.

Does it work?

  • $\begingroup$ I partially edited your answer to use MathJax. Do you think you can do the rest? It's a useful skill. Also, I upvoted your post but I think it might be worth mentioning the idea of semi-conjugacy. I think that might really be what the OP meant, since he wrote $C(f(x))=g(C(x))$. Thanks for your contribution! $\endgroup$ – Mark McClure Dec 13 '16 at 16:01
  • 1
    $\begingroup$ Thanks--I did the MathJax and mentioned semiconjugacy. $\endgroup$ – Al Boldt Dec 14 '16 at 13:29

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