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I've been trying to prove the non existence of a semi-conjugacy (intuitively assuming this is true) between the circle maps $E_k$ and $E_{-k}$ for $k = 2,3,\dots$ defined as $$ E_k: S^1 \to S^1,\qquad E_k(x \mod 1) = kx \mod 1. $$ where $S^1 \simeq [0,1)$. Conjugacy is clearly not an option as the number of fixed points varies, i.e. $P_n(E_{k}) = |k^n -1|$, where $P_n(f)$ is the number of fixed points of $f^n$. What about semi-conjugacy however?

Any help is more than welcome!

BB3C
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1 Answers1

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Assume $$ h \circ E_{-2}= E_2 \circ h,$$ where $h$ is the semi cojnugacy. Applying the degree both sides you get $deg(h)=0$. Let $H$ denote the lifting of h such that $$H(0) \in [0,1), $$ then, $$H(x+1)=H(x) $$ i.e H is periodic. Moreover, lifting both sides of the above equation, we easily obtain that $2H(x) $ and $H(2-2x)$ are both lifing of the same map. A well known result implies that $$2H(x)= H(2-2x) + c$$ where c is an integer. Periodicity implies $$2H(0)=H(2)+c= H(0)+c$$ and the only possibile solution to this is $c=0$ and $H(0)=0$. Applying weiestrass theorem and periodicity we get that H attains a global maximum and minimum and the first is nonnegative, the second is nonpositive. Since the global extrema of $2H(x) , H(2-2x) $ must coincide, we get their extrema are zero. In details: let $M,m$ be resp. the maximum and minimum of H. Assume $M>0$. Then, since $c=0$, for proper $y$: $$2M=2H(y)=H(2-2y)>M, $$ contradicting the fact that M is a maximum. This implies $M=0.$ Now assume $m<0$,say $m=-q$, with $q$ strictly positive. Again, for proper $y$: $$-2q=2H(y)=H(2-2y)<-q.$$ This implies $c=0$. Then, H is constant, contradiction. This is generalizable for different $k$, but not so trivially , though.

Giuseppe Tenaglia
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  • "Since the global extrema of $2H(x), H(2-2x)$ must coincide, we get their extrema are zero...". Why exactly do they need to be the same? – BB3C Nov 23 '20 at 11:25
  • I edited again the post, in an attempt to clarify things! – Giuseppe Tenaglia Nov 24 '20 at 02:05
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    I see, thanks! Last thing, where exactly did you use the map $E_2$ in your proof and how come isn’t this method generalizable to any integer $k$? – BB3C Nov 25 '20 at 07:47
  • $E_k$ admits $kx$ as a lift, $E_{-k}$ admits $k-kx$. Forcing $H(0) \in [0,1)$, and using the fact that composition of lifting is a lifting, I can prove that $(k-1)H(0)=c$, which gives me $H(0)=c=0$ only when $k=2$. In the most general case, I guess that funtional equation still admits no solution but you have to modify the argument. The key point is that such H must be unbounded, contradicting periodicity. However, once you prove that for $k=2$ you do not have semi-conjugacy, It is very intuitive you do not have it for any k. So you (at least I do not) do not really wanna generalize it – Giuseppe Tenaglia Nov 25 '20 at 12:01