Question 1. (J. Mather [Reference Eliasson, Kuksin, Marmi, Yoccoz, Marmi and YoccozEKMY98, Problem 3.1.1])

Does there exist an example of a symplectic $C^r$ twist map with an essential invariant curve that is not $C^1$ and that contains no periodic point?

Question 2. (M.-C. Arnaud)

Does there exist a regular symplectic twist map of the annulus that has an essential invariant curve that is non-differentiable on which the restricted dynamics is minimal?

Proof of Theorem 1

To prove that Theorem 3 implies Theorem 1, we use Herman’s beautiful trick that associates to some circle homeomorphisms f a twist map of the annulus which preserves a circle that is determined by f and on which the restricted dynamics is given by f. For completeness, we state and prove Herman’s observation.

First of all, note that if $\tilde {f}$ is a lift of a circle homeomorphism f and if there exists a $C^1$ function $\phi :{\mathbb T} \to {\mathbb R}$ , which we also view as a function from ${\mathbb R}$ to ${\mathbb R}$ of period $1$ , satisfying

(1) $$ \begin{align} \mathrm{Id}_{{\mathbb R}}+ \frac{{\phi}(\cdot)}{2}= \frac{1}{2} (\tilde{f}(\cdot)+\tilde{f}^{-1}(\cdot)),\end{align} $$

then the map

(2) $$ \begin{align} g: {\mathbb A} \to {\mathbb A} : (\theta,r) \mapsto (\theta+r,r+ \phi(\theta+r))\end{align} $$

is a $C^1$ Hamiltonian twist map of the annulus (this follows from the fact that $\int _{{\mathbb T}}\phi (\theta )d\theta =0$ , as proved in [Reference HermanH83, §II.2.3]). Moreover, with $\psi :=\tilde {f}-id_{{\mathbb R}}$ , which we view as a function from ${\mathbb T}$ to ${\mathbb R}$ , or from ${\mathbb R}$ to ${\mathbb R}$ of period $1$ , the following holds.

Proof. From equation (1), we have that

$$ \begin{align*} 2 \tilde{f}+{\phi \circ \tilde{f}}= \tilde{f}\circ \tilde{f}+\mathrm{Id}_{{\mathbb R}}=\tilde{f}+\psi \circ \tilde{f}+\mathrm{Id}_{{\mathbb R}},\end{align*} $$

so

$$ \begin{align*}\psi \circ f - \psi = \phi \circ f,\end{align*} $$

and finally equation (2) implies

$$ \begin{align*} g(\theta,\psi(\theta))&=(f(\theta),\psi(\theta)+ \phi(f(\theta)))\\&=(f(\theta),\psi(f(\theta))).\\[-37pt]\end{align*} $$

The proof that Theorem 3 implies Theorem 1 is now straightforward from the claim.

To prove Theorem 3, we will work with a special class of circle diffeomorphisms that we call C-great.

Proof Proof that Lemmas 5 and 6 imply Theorem 3

Let $\alpha \in {\mathbb R} \setminus {\mathbb Q}$ . We want to construct a sequence of circle diffeomorphisms $(f_n)$ that converge to a circle homeomorphism $f_\infty $ that is topologically conjugated to $R_\alpha $ and such that $f_\infty $ is not a diffeomorphism while $f_\infty +f_\infty ^{-1}$ is $C^1$ . We will use the criterion of Lemma 4 to guarantee the non-differentiability of $f_\infty $ .

In light of Lemma 5, we can start with a $C^1$ circle diffeomorphism $f_0$ and $x_0$ such that $(f_0,x_0)$ is C-great and $f_0$ is topologically conjugate to the circle rotation of angle $\alpha $ . Using Lemma 6, we construct inductively, for $n\geq 1$ , sequences $(h_n)$ of circle diffeomorphisms, $(x_n)$ of points in ${\mathbb T}$ , and $(u_n)$ and $(v_n)$ of positive numbers, such that for

$$ \begin{align*}f_n:=h_n \circ f_{n-1} \circ h_n^{-1}, \quad F_n:=f_n+f_n^{-1}, \quad H_n:=h_n \circ \cdots \circ h_1,\end{align*} $$

the following hold:

• • $(f_n,x_n)$ is C-great;

• • $u_n+v_n\leq 2^{-n}$ ;

• • $\|x_{n+1}-x_n\|\leq 2^{-n}$ ;

• • $\Delta (f_n,x_n,u_n,v_n)>1+{\epsilon }_0$ ;

• • $\|H_n-H_{n-1}\|_{C^0}\leq 2^{-n}$ ;

• • $\|F_{n+1}-F_n\|_{C^1}\leq 2^{-n}$ .

We also ask that the convergence of $(H_n)$ and $(x_n)$ be sufficiently fast so that $\Delta (f_m,x_m,u_n,v_n)>1+{\epsilon }_0$ for all $m\geq n$ .

From the convergence of $H_n$ , we get that $f_n$ converges to a circle homeomorphism $f_\infty $ that is topologically conjugated to $R_\alpha $ . In addition, the $C^1$ limit of $F_n$ must be $f_\infty +f_\infty ^{-1}$ . Because $\Delta (f_m,x_m,u_n,v_n)>1+{\epsilon }_0$ for all $m\geq n$ , we guarantee that the limit point $x_\infty $ of $(x_n)$ satisfies $\Delta (f_\infty ,x_\infty ,u_n,v_n)>1+{\epsilon }_0$ for every n. Hence, $f_\infty $ is not differentiable by Lemma 4.

In conclusion, $f_\infty $ satisfies all the properties of Theorem 3.

Proof of Lemma 6

To motivate what follows, note that if $g=h \circ f \circ h^{-1}$ with $Dh(x)=1+e(x)$ , then, to have $g+g^{-1}=f+f^{-1}$ , one must have

$$ \begin{align*}h \circ f+h \circ f^{-1}=f \circ h+f^{-1} \circ h.\end{align*} $$

The latter implies that

$$ \begin{align*}e(f(x)) Df(x)+\frac {e(f^{-1}(x))} {Df(f^{-1}(x))}=e(x) \bigg(Df(h(x))+\frac {1} {Df(f^{-1}(h(x))}\bigg)+\Delta,\end{align*} $$

with

$$ \begin{align*}\Delta=Df\circ h-Df+Df^{-1}\circ h-Df^{-1}.\end{align*} $$

Assuming h is $C^0$ close to $\mathrm {id}$ , we see that for $g+g^{-1}$ to be $C^1$ -close to $f+f^{-1}$ , we should choose e such that $(e(f(x))-e(x)) Df(x) Df(f^{-1}(x))$ is close to $e(x)-e(f^{-1}(x))$ .

To construct the conjugacy h of Lemma 6, we start by defining a sequence $(e_j)_{j\in {\mathbb Z}}$ that will serve to define h along the orbit of a point that is C-good for f.

Let x be a C-good point for f. Let $b_j=Df(f^j(x))$ , $j \in {\mathbb Z}$ . Define $c_j$ , $j \in {\mathbb Z}$ by $c_0=1$ and $c_j b_j b_{j-1}=c_{j-1}$ .

Proof. We will need the following properties on the orbit of x that will be a consequence from the fact that x is C-good for f.

Claim. We have that $\sum _{j \geq 0} c_j\leq C^3$ and $\sum _{j \leq 0} c_j=\infty $ , $\liminf _{j \leq 0} c_j=0$ .

Proof. We have that for $j\geq 0$ , $c_j=({Df(x)Df(f^j(x))})/{(Df^{j+1}(x))^2}$ . Hence, it follows from Definition 1 that $\sum _{j\geq 0} c_j \leq C^3$ .

For $j\leq -1$ , we have that $c_j=({Df(x)}/{Df(f^j(x))}) {1}/{(Df^{j}(x))^2}$ . Hence, it follows from Definition 1 that $\sum _{j< 0} c_j=\infty $ as well as $\liminf _{j \leq 0} c_j=0$ .

Let us now define $e_j$ , $j \in {\mathbb Z}$ as follows. Fix N large such that $c_{-N}$ is small. Note that by taking N sufficiently large, we will have that $c_N$ is also small. We let $e_j=0$ for $j>N$ . For $|j| \leq N$ , we define $e_j$ so that $e_{j+1}-e_j=-c_j$ . We now let $N'$ be much larger than N such that $c_{-N'+1}$ is small and such that $\alpha ={\sum _{-N' \leq j <-N} c_j}/{e_{-N}}$ is large. We set then $e_{j+1}-e_j= {c_j}/{\alpha }$ for $-N' \leq j \leq -N-1$ . We now define $e_j=0$ for $j<-N'$ . By construction, $e_1-e_0=-c_0=-1$ .

Note the following.

• • For $j\in [-N,N]$ , we have that $e_j=\sum _{k=j}^N c_k$ .

• • For $j\in [-N',-N)$ , we have that $e_j=e_{-N}-({1}/{\alpha }) \sum _{k=j}^{-N-1}c_k$ .

From the definition of $\alpha $ , we deduce that $e_{-N'}=0$ and that $e_j \geq 0$ for all $j\in {\mathbb Z}$ . Also, $-e_0=e_{N+1}-e_0=-\sum _{0 \leq j \leq N} c_j$ , so $e_0=\sum _{0 \leq j \leq N} c_j\leq C^3$ .

Note that by construction, because $c_j b_j b_{j-1}=c_{j-1}$ , we have that

$$ \begin{align*}\eta_j=(e_{j+1}-e_j) b_j b_{j-1}+(e_{j-1}-e_j)=0,\end{align*} $$

for all $j \in {\mathbb Z}$ , except for $j=-N'$ , $j=-N$ , and $j=N+1$ . We also have, because $e_{-N'-1}=e_{-N'}=0$ , that $\eta _{-N'}=c_{-N'}/\alpha $ is small as $c_{-N'}$ is small and $\alpha $ is large. Next, we compute $\eta _{-N}=-c_{-N-1}-({c_{-N-1}})/{\alpha }$ , which is also small because $c_{N}$ is small and $\alpha $ is large. Finally, $\eta _{N+1}=c_N$ is also small by our choice of N.

We want now to modify f by conjugation along a neighborhood of the orbit of x between the times $-N'$ and N to obtain the diffeomorphism g with the required properties of Lemma 6. We will look for the conjugacy under the form $Dh(x)=1+e(x)$ . The choice of the sequence $(e_j)$ in Sublemma 7, would essentially allow to get a good control on g and $g+g^{-1}$ along the orbit of the point x if the function e takes the values $e_j$ along the orbit of x. We need however to define h in the neighborhood of the orbit of x without losing the required properties on g and $g+g^{-1}$ and this will will require some additional technicalities that we now address.

To guarantee a good control of $g+g^{-1}$ everywhere, we start by slightly modifying f to make it affine along the $-N'$ to N orbit of a small interval around x. For this, choose a small interval $I_0$ centered around x, and let $I_j=f^j(I_0)$ , $j \in {\mathbb Z}$ . We may assume that $I_j \cap I_{j'}=\emptyset $ if $0<|j-j'| \leq 2 N'$ . Then for every $\delta>0$ , we can define a diffeomorphism $h':{\mathbb R}/{\mathbb Z} \to {\mathbb R}/{\mathbb Z}$ , which is the identity outside $\cup _{-N'+1 \leq j \leq N+1} I_j$ such that $\sup _y |Dh'(y)-1|<\delta $ , and such that letting $f'=h' \circ f \circ h^{\prime \,-1}$ , we have $f'(y)=f^{j+1}(x)+b_j(y-f^j(x))$ whenever y is near $f^j(x)$ and $-N' \leq j \leq N$ .

Let us now select an interval $I^{\prime }_0 \subset I_0$ centered around x such that letting $I^{\prime }_j=f^{\prime \,j}(I^{\prime }_0)$ , $j \in {\mathbb Z}$ , we have that $f'|I^{\prime }_j$ is affine for $-N' \leq j \leq N$ .

Note that by choosing $\delta>0$ small, we get that $F'=f'+f^{\prime \,-1}$ is $C^1$ close to $F=f+f^{-1}$ .

We now define another diffeomorphism $h:{\mathbb R}/{\mathbb Z} \to {\mathbb R}/{\mathbb Z}$ which is the identity outside $\cup _{-N'+1 \leq j \leq N} I^{\prime }_j$ as follows. To specify h, it is enough to define $Dh$ on $I^{\prime }_j$ for $-N'+1 \leq j \leq N$ . We let $\phi :{\mathbb R} \to [-\delta ,1]$ be a smooth function supported on $(-1/2,1/2)$ , symmetric around $0$ , and such that $\phi (0)=1$ and $\int \phi (x) dx=0$ . We then let $Dh=1+e_j \phi \circ A_j$ , where $A_j:I^{\prime }_j \to [-1/2,1/2]$ is an affine homeomorphism. If $\delta>0$ is chosen sufficiently small, h is indeed a diffeomorphism because all the $e_j$ are positive.

Finally, let

$$ \begin{align*}g:=h \circ f^{\prime} \circ h^{-1}=h\circ h^{\prime} \circ f \circ h^{\prime\, -1} \circ h^{-1}.\end{align*} $$

Proof. Let us show that $G=g+g^{-1}$ is $C^1$ close to $F'=f'+f^{\prime \,-1}$ . Note that $G=F'$ in the complement of $\bigcup _{-N' \leq j \leq N} I^{\prime }_j$ , so it is enough to show that $DG-DF'$ is small in each $I^{\prime }_j$ , $-N' \leq j \leq N$ . Indeed for $y \in I^{\prime }_j$ , letting $\kappa =\phi \circ A_j(y)$ , we have

$$ \begin{align*}DG(h(y))-DF'(h(y))=\frac {\kappa} {(1+e_j \kappa) b_{j-1}} ((e_{j+1}-e_j) b_j b_{j-1}+e_{j-1}-e_j),\end{align*} $$

which is small because the term ${\kappa }/{(1+e_j \kappa ) b_{j-1}}$ is bounded. Indeed, $-\delta \leq \kappa \leq 1$ , and $1+e_j \kappa \geq 1-e_j \delta \geq 1/2$ provided $\delta $ is chosen sufficiently small, and finally $b_{j-1}>C^{-1}$ because $\sup _{x\in {\mathbb T}} |Df(x)+Df^{-1}(x)|<C$ .

Moreover, we have $Dg(x)-Df(x)= ({(1+e_1) b_0})/({1+e_0})-b_0= {-b_0}/({1+e_0}) <-({1}/{C (1+C^3)})$ . Because $(({g(x+u)-g(x)})/{u}) \sim Dg(x)$ for $u \ll |I_0|$ , while $(({g(x+v)-g(x)})/{v}) \sim Df(x)$ for $1\gg u \gg |I_0|$ , this allows to exhibit u and v such that $\Delta (g,x,u,v) \geq 1+{\epsilon }_0$ with ${\epsilon }_0= {1}/({2C^2 (1+C^3)})$ .

To conclude, we must show that there exists arbitrary close to x a point y such that $(g,y)$ is C-great. It suffices for this to show that in any interval J around x, there is an uncountable set of C-good points for g. By definition of x, we know that the latter is true for f. Note that if y is C-good for f, then $y'=h(h'(y))$ is C-good for g if $\sum _{n \geq 0} |Dg^n(y')|^{-2}<C$ .

Fix some $\Lambda>0$ much larger than $\sup e_j$ . Notice that for small $\lambda>0$ , we can choose $m>0$ such that there is an uncountable set $K' \subset J$ of y such that

$$ \begin{align*}\sum_{n \geq 0} |Df^n(y)|^{-2}<C-\lambda, \quad \text{and} \quad \sum_{n \geq m} |Df^n(y)|^{-2}<\lambda/\Lambda.\end{align*} $$

Now, notice that $h \circ h'=\mathrm {id}$ except in $\cup _{-N'+1 \leq j \leq N+1} I_j$ , and the derivative of $h \circ h'$ is bounded by $(1+\delta ) (1+\sup e_j)$ . In particular, if $y \in K'$ , then $h(h'(y))$ will be C-good for g provided $g^n(y) \notin \cup _{-N'+1 \leq j \leq N+1} I_j$ for $0 \leq n \leq m$ . If we choose the size of $I_0$ sufficiently small, we will have the latter property for uncountably many $y \in K' \subset J$ .

The proof of Lemma 6 is thus accomplished.

Proof of Lemma 5

Given an interval $I=[a,b]$ , let $l(I)=a+\tfrac {3} {8}(b-a)$ and $r(I)=a+\tfrac {5} {8} (b-a)$ .

Our construction will depend on a sequence of integers $m_n \in {\mathbb {N}}$ , $n \geq 0$ , such that $m_0$ is large and $m_{n+1}$ is much larger than $m_n$ .

Let $\Omega $ be the set of all finite sequences $\omega =(\omega _1,\ldots ,\omega _n)$ of l and r and length $|\omega | \geq 0$ . For $\omega \in \Omega $ , we define intervals $I_\omega $ inductively as follows. Let $n=|\omega |$ . If $n=0$ , we let $I_\omega $ be the interval of length $2^{-m_0}$ centered on $0$ . If $n \geq 1$ , let $\omega '$ consist of $\omega $ stripped of its last digit, and let $I_\omega $ be the interval of length $2^{-m_n}$ centered on $t(I_{\omega '})$ , where $t \in \{l,r\}$ is the last digit of $\omega $ .

Let $\phi :{\mathbb R} \to [0,1]$ be a smooth function symmetric around $0$ , supported on $[-1/2,1/2]$ and such that $\phi |[-1/4,1/4]=1$ .

Let $A^\omega :I_\omega \to [-1/2,1/2]$ be an affine homeomorphism. Let us now define functions smooth functions $\phi ^\omega :{\mathbb R}/{\mathbb Z} \to {\mathbb R}$ supported inside the intervals $I_\omega +k \alpha $ , $1 \leq k \leq 2 n-1$ , such that

$$ \begin{align*}\phi^\omega(x+k \alpha)=(n-|n-k|) \phi(A^\omega(x)).\end{align*} $$

Note that by selecting $m_n$ sufficiently large, we may assume the following property.

Low recurrence. For any $n\in {\mathbb {N}}$ , for all $|\omega |=|\omega '|=n$ for any $0 \leq |k| \leq 2^{2^n}$ , $I_\omega +k \alpha $ does not intersect $I_{\omega '}$ . In particular, the supports of $\phi ^\omega $ and $\phi ^{\omega '}$ do not intersect.

We let $\phi _n=\sum _{|\omega |=n} \phi _\omega $ . Note that low recurrence implies that, for any $x\in {\mathbb R} / {\mathbb Z}$ , the following hold:

1. (L1) $|\phi _n(x+\alpha )-\phi _n(x)| \leq 1$ ;

2. (L2) $\phi _n(x-m \alpha )=0$ if $m \geq 0$ and $2n-1+m \leq 2^{2^n}$ ;

3. (L3) $\phi _n(x)\leq n$ .

Finally, define a non-decreasing sequence of smooth function $\Phi _N:{\mathbb R}/{\mathbb Z} \to {\mathbb R}$ by

$$ \begin{align*}\Phi_N=\sum_{0 \leq n \leq N} \frac {1} {(n+1)^{4/3}} \phi_n.\end{align*} $$

Define the uncountable set $K:=\bigcap _n \bigcup _{|\omega |=n} I_\omega $ . Note that for $x\in K$ , we have the following for $N\geq ~1$ :

1. (K1) $\Phi _N(x)=0$ ;

2. (K2) $\Phi _N(x+N \alpha ) \geq ({1}/ {10}) N^{2/3}$ ;

3. (K3) for $m\in [0, 2^{2^n}-2n+1]$ , $\Phi _N(x-m \alpha ) \leq {n (n+1)}/ {2}$ .

To see the last item, just observe that if $\Phi _N(x-m \alpha ) \geq {n (n+1)}/ {2}$ , then $\phi _{n'}(x-m\alpha )>0$ for some $n \leq n' \leq N$ [because $\phi _n \leq n$ by item (L3)], and by item (L2), we must have $m>2^{2^n}-2n+1$ .

Next, observe that item (L1) implies that $\Phi _N(\cdot +\alpha )-\Phi _N(\cdot )$ converges in the $C^0$ topology to some continuous function $\Theta : {\mathbb R} / {\mathbb Z} \to {\mathbb R}$ .

Introduce $\xi _N:=\int e^{\Phi _N(x)} dx$ and $\Psi _N:=e^{\Phi _N}/\xi _N$ . Again, by letting the integers $m_n$ grow fast, we obtain that $\xi _N$ is close to $1$ for all $N \geq 1$ , and that if we consider the circle homeomorphism $h_n:{\mathbb R}/{\mathbb Z} \to {\mathbb R}/{\mathbb Z}$ given by $Dh_n=\Psi _n$ and $h_n(0)=0$ , we get that $h_n$ converges in the $C^0$ topology to some homeomorphism $h:{\mathbb R}/{\mathbb Z} \to {\mathbb R}/{\mathbb Z}$ .

Because $\Phi _N(x+\alpha )-\Phi _N(x)$ converges in $C^0$ to a continuous function $\Theta $ , we get that $f_n(x)=h_n(h_n^{-1}(x)+\alpha )$ is converging in the $C^1$ topology to some f satisfying $f(x)=h(h^{-1}(x)+\alpha )$ and $\ln Df=\Theta \circ h^{-1}$ . In particular, f is minimal.

Note that all $x \in h(K)$ are C-good for some absolute C, because $Df^n(x)\geq e^{{n^{2/3}}/{10}}$ for $x\in h(K)$ by item (K2).

However, item (K2) also implies that for each x, from time to time, $h^{-1}(x)-n \alpha $ , for $n\geq 0$ , will visit regions where $\sup _N \Phi _N$ is large, so for $x \in h(K)$ , $Df^{-n}(x)$ will be large, because $\Phi _N(h^{-1}(x))=0$ by item (K1). Moreover, if $x \in h(K)$ , item (K3) implies that $\Phi _N(h^{-1}(x)-n \alpha )$ is at most of order $(\ln \ln n)^2$ , so $\sum _{n \geq 1} |Df^{-n}(x)|^{-2}=\infty $ .