MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Manneville-Pomeau map

$f:[0,1]\rightarrow [0,1]: x+x^{1+s} mod 1$ is topologically conjugate to a full one-sides shift on two symbols. Can you give the corresponding homeomorphism? Many thanks.

share|cite|improve this question

It should be 'almost 1 to 1' conjugate to one-side shift. First take $s=0$, then $f_0(x)=2x\mod 1$ has a Markov partition $\alpha=\lbrace I_0,I_1\rbrace$, where $I_0=[0,0.5],I_1=[0.5,1]$.

The partition $\alpha_n=\alpha\vee\cdots \vee f^{1-n}\alpha$ can be indexed by $w\in\lbrace 0,1\rbrace^n$, with $f^k(A_w)\subset I_{w_k}$ for all $k=0,\cdots,n-1$.

Then the related factor map is $\pi:\lbrace 0,1\rbrace^{\mathbb{N}}\to\mathbb{T}$, $\mathbf{w}=(w_n)\to\bigcap_{n\ge1 }A_{w_0\cdots w_n}$.

It is easy to see $\pi\circ \sigma=f_0\circ \pi$, and $\pi$ is 1 to 1 expect countable many points.

For other $s$, we solve $1=f_s(x)=x+x^{1+s}$, say $x_s$. Then $\alpha_s=\lbrace I_0^s,I_1^s\rbrace$ is a Markov partition, where $I^s_0=[0,x_s],I^s_1=[x_s,1]$.

Accordingly we assign indices to elements in $\alpha_{s,n}$ for all $n$ and then let $\pi_s:\lbrace 0,1\rbrace^{\mathbb{N}}\to\mathbb{T}$, $\mathbf{w}=(w_n)\to\bigcap_{n\ge1 }A^s_{w_0\cdots w_n}$.

In the case $s=0$, we know $\pi(\mathbf{w})=\sum_{n\ge1}\frac{w_n}{2^n}$. I do not know the explicit formula of $\pi_s$ for general $s$.

share|cite|improve this answer

Pengfei's answer gives you the general form of the semi-conjugacy $\pi_s$ (basically, the answer is the proof that there is a semi-conjugacy that is 1-1 except at preimages of the endpoints). If you're looking for more detailed information about the structure of $\pi_s$, then it depends on exactly what your goal is, but the most useful thing to have is usually an estimate on the points $\beta_n := \pi_s(0^n 1^\infty)$ -- that is, the points which lie in the first branch of $f$ for $n$ iterates and then are mapped into the endpoint $1$. (This is usually the hard part of the estimates because the map is uniformly expanding everywhere else, which makes life easier.)

Many useful estimates on these points can be found in Section 6.2 of Lai-Sang Young, Recurrence times and rates of mixing, Isr. J. Math. 110 (1999), 153-188. In particular, one has $\beta_n \in [(n+1)^{-1/s}, n^{-1/s}]$.

You can also derive useful estimates by approximating the first branch of the Manneville-Pomeau map with the time-1 map of the ODE $x' = x^{1+s}$. (If you just want to study a map with an indifferent fixed point, then you can use this as your definition, so there's no approximation required.) This ODE can be explicitly solved by observing that $(x^{-s})' = -s$, and so $x(t) = (x_0^{-s} - st)^{-1/s}$.

I'm not sure how feasible it is to write down an explicit expression for the semi-conjugacy, even armed with these estimates, but these give you most of the quantitative tools you need when you're working with the Manneville-Pomeau map.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.