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I want to find all the periodic points of the following diffeomorphism of the circle:

$f(x) = x + \frac{1}{4} + \frac{1}{10} \sin(8 \pi x) \mod 1$

Where a periodic point is $p$ such that $f^n(p) = p$ for some $n \in \mathbb{N}$. I have shown the rotation number $\rho (f) = \frac{1}{4}$ (I think) so I know that periodic points exist (as the rotation number is rational) but that none are fixed points, i.e. no such $p$ for $n = 1$ (as rotation number is not zero).

I tried working from the definition but I get a horrible nest of sine functions. In general finding periodic points of functions like this doesn't appear to be well documented in my course notes, the recommended book or indeed the internet (from what I can find) so any help at all would be appreciated. I'm looking for a general method, not just the solution for this specific diffeomorphism, but really any help is great.

This isn't homework if anyone was wondering, it's revision and I'm more interested in the mathematics behind the general method for a solution than getting the specific solution of this question. Hints definitely welcome.

EDIT: Forgot to add, as for a lift $F$ of $f$ we have $deg(f) = F(x + 1) - F(x) = 1$, we know the diffeomorphism is orientation preserving.

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  • $\begingroup$ Apologies. $f^n(p) = f(f( \dots (f(p)) \dots )$, i.e. $f$ applied $n$ times. A lift $F$ of $f$ is a map such that $\pi \circ F = f \circ \pi$ where $\pi(x) = x \mod 1$, in the case of circle homeomorphisms like this $F$ will be a function from $\mathbb{R}$ to itself and $f$ is a function from $\frac{\mathbb{R}}{\mathbb{Z}}$ to itself. $\endgroup$ – D. P Mar 27 '16 at 19:58
  • $\begingroup$ and you have shown that $f(x+k/4) = f(x)+k/4$. and the "lifts" are of the form $F(x) = x + \frac14 + \frac1{10} \sin(8 \pi x) + k$. but I don't see what it tells us on the fixed points of $p \to f^n(p)$ $\endgroup$ – reuns Mar 27 '16 at 20:07
  • $\begingroup$ We can use the lift $F$ of $f$ to calculate the degree and rotation number of $f$ which tells us that there are periodic points ($f$ has periodic points if and only if $\rho (f)$ is rational) and that there are no periodic points for $n = 1$/fixed points (as $\rho(f) = \frac{1}{4} \neq 0$). I can't see what the first part will tell us either, sadly. Working through would be fairly simple without the nested sine function, which can be bound but that doesn't seem to help much. $\endgroup$ – D. P Mar 27 '16 at 20:32
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Since you know what is the rotation number, there is simple way. Recall that the rotation number is $p/q$ with $p$ and $q$ coprime if and only if $f$ has a periodic point of period $q$. So you should concentrate on finding a point of period $4$, and a simple inspection shows $0$ has period $4$.

Other than that, the only question remaining is whether there are other orbits with period $4$ (for sure there are no orbits with other periods!). Indeed there are three other such orbits, that of $1/8$ and two others.

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  • $\begingroup$ Thank you, this helps a lot. Do you know where $\frac{p}{q}$ with $p$ and $q$ coprime if and only if $f$ has a periodic point of period $q$ comes from? I'm sure I'm missing something stupid. $\endgroup$ – D. P Mar 27 '16 at 21:20
  • $\begingroup$ Not immediate, depends on what you have shown and in what order. But it goes like this: $f$ has a $q$-periodic point if and only if the rotation number is $p/q$ for some $q$. In order that the period is $q$ it is thus necessary and sufficient that $p$ and $q$ are coprime, since otherwise $p/q$ could be written as $p'/q'$ with $q'<q$ (and you would have a $q'$-periodic point). $\endgroup$ – John B Mar 27 '16 at 21:36
  • $\begingroup$ We have not directly stated the $p / q$ coprime argument, however I think I have created an equivalent argument (probably the same): I have proven $\rho(f^n) = n \rho(f)$, that $\rho(f) = 0 \iff f$ has a fixed point and if $f$ has a periodic point of period $n$ then $f^n$ has a fixed point; so from this I can conclude that all periodic points of $f$ are of period $4$ (as any other value that gives $\rho(f^n) = 0$ is a multiple of 4). Does this argument hold (assuming $deg(f) = 1$)? I'm not sure how you concluded that there are precisely 4 periodic points, I'll have a look. $\endgroup$ – D. P Mar 28 '16 at 17:50
  • $\begingroup$ So I found that $f^4$ has countably infinite fixed points, $\frac{m}{8} \forall m \in \mathbb{N}$, however you said there are only 4 periodic points of $f$. Am I missing something? $\endgroup$ – D. P Mar 28 '16 at 18:11
  • $\begingroup$ @D. P On your first comment: indeed, you can also proceed as you describe. On your second comment: I said that there are $4$ orbits of period $4$, which gives sixteen $4$-periodic points; the points of the form $m/8$ are not infinite, but $8$ (since they are computed$\bmod1$); so, besides these $8$ points there are $8$ more, which I obtained looking at the monotonicity between the "initial" ones, but the computations are very much complicated, and I cannot suggest any straightforward approach (other than drawing the graph of $f^4$). $\endgroup$ – John B Mar 28 '16 at 18:45

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