The moving needle problem

November 22, 2010November 22, 2010 ConanLeave a comment

First, let’s clarify that this post has nothing to do with the Kakeya conjecture (except for the word ‘needle’ in it). Anyways, I was asked the following question via an e-mail from Charles this summer: (It turns out that the question was invented by Jonathan King and then communicated to Morris Hirsch and that’s where Charles heard about it from) In any case, I find the problem quite cute:

Problem: Given a smoothly embedded copy of $\mathbb{R}$ in $\mathbb{R}^3$ containing $\{ (x,0,0) \ | \ x \in (-\infty,-C] \cup [C, \infty) \}$ . Is it always possible to continuously slide a unit length needle lying on the ray $(-\infty, -C]$ to the ray $[C, \infty)$ , while keeping the head and tail of the needle on the curve throughout the process?

i.e. the curve is straight once it passes the point $(-C, 0)$ and $(C,0)$ , but can be bad between the two points:

We are interested in sliding the needle from the neigative $x$ -axis to the positive $x$ -axis:

Exercise: Try a few examples! It’s quite amusing to see that sometimes both ends of the needle needs to go back and forth along the curve many times, yet it always seem to get through.

One should note that this is not possible if we just require the curve to be eventually straight and goes to infinity at both ends. As we can see on a simple ‘hair clip’ curve:

The curve consists of two parallel rays of distance $<1$ apart, connected with a semicircle. A unit needle can never get from one ray to the other since the needle would have to rotate $180$ degrees and hence it has to be vertical at some point in the process, but no two points on the curve has vertical distance $1$ .

After some thought, I think I can show for each given curve, 'generic' needle length can pass through:

Claim: For any $C^2$ embedding as above, there is a full measure and dense $G_\delta$ set $\mathcal{L} \subseteq \mathbb{R}$ of lengths where the needle of any length $L \in \mathcal{L}$ can slide through the curve.

Proof: Let $\gamma: \mathbb{R} \rightarrow \mathbb{R}^3$ be a smooth parametrization of the curve s.t. $\gamma(t) = t$ for $t \in (-\infty, -C] \cup [C, \infty)$ .

Define $\varphi: \mathbb{R}^2 \rightarrow \mathbb{R}$ where $\varphi: (s, t) \mapsto d(\gamma(s), \gamma(t))$ .

Hence $\varphi$ vanishes on the diagonal and takes positive value everywhere else.
Since $\gamma(t) = t$ for $t \notin (-C, C)$ , hence $\varphi(s,t)=|s-t|$ for $|s|, |t|>C$ . $\varphi^{-1}(L)$ contains four rays $\{ |s-t| = L \ | \ s, t \notin (-C, C) \}$ :

Observation: a needle of length $L$ can slide through the curve iff there is a continuous path in the level set $\varphi(p) = L$ connecting the two rays above the diagonal. (This is merely projection onto the $x$ and $larex y$-axis.)

We also have $\varphi$ is $C^2$ other than on the diagonal. (It behaves like the absolute value function near the diagonal). By Sard’s theorem, since $\varphi$ is $C^2$ on $\{s < t\}$ , the set of critical values is both measure $0$ and first category.

Let $\mathcal{L}$ be the set of regular values of $\varphi$ . For any $L \in \mathcal{L}$ , by implicit function theorem, the level set $\varphi^{-1}(L)$ is a $C^2$ sub-manifold.

Since the arc $\gamma([-C, C])$ is compact, we can find large $R$ where $\gamma([-C, C]) \subseteq B(\bar{0}, R)$ .

Hence for $|t| > R + L$ and $s \in [-C, C]$ , we have

$\varphi (s, t) = d(\gamma(s), \gamma(t)) > d(\gamma(t), B(\bar{0}, R)) > L$

The same holds with $|s| > R + L$ and $t \in [-C, C]$

i.e. $\varphi$ takes value $>L$ in the shaded region below:

Hence for $L \in \mathcal{L}$ , $\varphi^{-1}(L) \cap \{x \leq y\}$ is a $1$ dimensional sub-manifold, outside a bounded region it contains only two rays. We also know that the level set is bounded away from the diagonal since $\varphi$ vanishes on the diagonal. By an non-ending arc argument, one connected component of $\varphi^{-1}(L)$ must be a curve connecting the end points of the two rays. Establishes the claim.

Remarks: This problem happens to come up at the very end (questio/answer part) of Charles’s talk in the midwest dynamics conference last month (where he talked about our joint work about funnel sections). A couple weeks later Michal Misiurewicz e-mailed us a counter-example when the curve is not smooth (only continuous).

Initially I tried to use the above argument to get the length $1$ needle. Everything works fine until a point where one has a continuum in the level set connecting the end-points of the two rays. We want the continuum to be path connected. I got stuck on that. In the continuous curve case, Michal’s counter-example corresponds to the continuum containing a $\sin(1/x)$ curve, hence is not path connected.

I believe such thing cannot happen for smooth. The hope would be that the length $1$ needle can slide through any $C^2$ (or $C^1$ ) curve. (Note that once the length $1$ needle can pass through, then all length can pass through just by rescaling the curve.) In any case, still trying…

On C^1 closing lemma

May 25, 2010May 27, 2010 Conan1 Comment

Let $f: M \rightarrow M$ be a diffeomorphism. A point $p$ is non-wandering if for all neighborhood $U$ of $p$ , there is increasing sequence $(n_k) \subseteq \mathbb{N}$ where $U \cap f^{n_k}(U) \neq \phi$ . We write $p \in \mathcal{NW}(f)$ .

Closing lemma: For any diffeomorphism $f: M \rightarrow M$ , for any $p \in \mathcal{NW}(f)$ . For all $\varepsilon>0$ there exists diffeomorphism $g$ s.t. $||f-g||_{C^1} < \varepsilon$ and $g^N(p) = p$ for some $N \in \mathbb{N}$ .

Suppose $p \in \mathcal{NW}(f)$ , $\overline{\mathcal{O}(p)}$ is compact, then for any $\varepsilon>0$ , there exists $x_0 \in B(p, \varepsilon)$ , $k \in \mathbb{N}$ s.t. $f^k(x) \in B(p, \varepsilon)$ .

First we apply a selection process to pick an appropriate almost-orbit for the closing. Set $x_i = f^i(x_0), \ 0 \leq i \leq k$ .

If there exists $0 < j < k$ where

$\min \{ d(x_0, x_j), d(x_j, x_k) \} < \sqrt{\frac{2}{3}}d(x_0, x_k)$

then we replace the origional finite sequence by $(x_0, x_1, \cdots, x_j)$ or $(x_j, \cdots, x_k)$ . Iterate the above process. since the sequence is at least one term shorter after each shortening, the process stops in finite time. We obtain final sequence $(p_0, \cdots, p_n)$ s.t. for all $0 < i < n$ ,

$\min \{ d(p_0, p_i), d(p_i, p_n) \} \geq \sqrt{\frac{2}{3}}d(p_0, p_n)$ .

Since the process is applied at most $k$ times, $x_0, x_k \in B(p, \varepsilon)$ , after the first shortening, $d(p, x_{i_1}) \leq \max \{d(p, x_0), d(p, x_k) \} + \sqrt{\frac{2}{3}}d(x_0, x_k)$ $\leq \varepsilon + 2 \sqrt{\frac{2}{3}} \varepsilon$ .

i.e. both initial and final term of the sequence is at most $(\frac{1}{2}+ \sqrt{\frac{2}{3}}) 2 \varepsilon$ . Along the same line, we have, at the $i$ -th shortening, the distance between the initial and final sequence and $p$ is at most $(\frac{1}{2} + \sqrt{\frac{2}{3}} + (\sqrt{\frac{2}{3}})^2 + \cdots (\sqrt{\frac{2}{3}})^i) 2 \varepsilon$ . Hence for the final sequence $p_0, p_n \in B(p, 1+2 \sqrt{\frac{2}{3}}/(1-\sqrt{\frac{2}{3}}) \varepsilon) \subseteq B(p, 10 \varepsilon)$ .

There is a rectangle $R \subseteq M$ where $p_0, p_n \in \sqrt{\frac{3}{4}}R$
(i.e. shrunk $R$ by a factor of $\sqrt{\frac{3}{4}}$ w.r.t. the center) and for all $0 < i < n, \ p_i \notin R$ .

Next, we perturb $f$ in $R$ i.e. find $h: M \rightarrow M$ with $||h||_{C^1} < \delta$ and $h|_{M \backslash R} =$ id. Hence $||h \circ f - f ||_{C^1} < \delta$ .

Suppose $R = I_1 \times I_2; L_1, L_2$ are the lengths of $I_1, I_2$ , $L_1 < L_2$ .
By main value theorem, for all $x \in M, \ d(x, h(x)) < \delta L_1$ .
On the other hand, since $p_0 \in \sqrt{\frac{3}{4}}R$ , it's at least $\frac{1}{2}(1-\sqrt{\frac{3}{4}})L_1$ away from the boundary of $R$ . i.e. there exists bump function $h$ satisfying the above condition and $d(p_0, h(p_0)) > \frac{\delta}{8}(1-\sqrt{\frac{3}{4}})L_1$ .

Hence in order to move a point by a distance $L_1$ , we need about $1/ \delta$ such bump functions, to move a distance $L_2$ , we need about $\frac{L_2}{\delta L_1}$ bumps.

For simplicity, we now suppose $M$ is a surface. By starting with an $\varepsilon$ (and hence $R$ ) very small, we have for all $0 \leq i \leq N+M, \ f^i(R)$ is contained in a small neighbourhood of $p_i$ . Hence on $f^i(B), f^i$ is $C^1$ close to the linear map $p_i + Df^i(p_0)(x-p_0)$ . Hence mod some details we may reduce to the case where $f$ is linear in a neighborhood of $\mathcal{O}(p_0)$ .

By choosing appropiate coordinate system in $R$ , we can have $f$ preserving the horizontal and vertical foliations and the horizontal vectors eventually grow more rapidly than the vertical vectors.

It turns out to be possible to choose $R$ to be long and thin such that for all $i \leq 40 / \delta$ , $f^i(R)$ has height greater than width. (note that $M = \lfloor 40/ \delta \rfloor$ bumps will be able to move the point by a distance equal to the width of the original rectangle $R$ . Since horizontal vectors eventually grow more rapidly than the vertical vectors, there exists $N$ s.t. for all $N \leq i \leq N+M$ , $f^i(R)$ has width greater than its height.
For small enough $\epsilon$ , the boxes $f^i(R)$ are disjoint for $0 \leq i \leq N+40/ \delta$ . Construct $h$ to be identity outside of

$\displaystyle \bigsqcup_{i=0}^M f^i(R) \sqcup \bigsqcup_{i=N}^{N + M} f^i(R)$

For the first $M$ boxes, we let $h$ preserve the horizontal foliation and move along the width so that $g = h \circ f$ has the property that $g^M(p_n)$ lies on the same vertical fiber as $f^M(p_0)$ .

On the boxes $f^{N+i}(R), \ 0 \leq i \leq M$ , we let $h$ pushes along the vertical direction so that

$g^{N+M}(p_n) = f^{N+M}(p_0)$

Since iterates of the rectangle are disjoint, for $N+M \leq i \leq n, \ h(p_i) = p_i$ , $g(p_i) = f(p_i)$ .

Hence $g^n(p_n) = g^{n-(N+M)} \circ g^{N+M}(p_n)$ $= g^{n-(N+M)} f^{N+M}(p_0) = g^{n-(N+M)} (p_{N+M}) = p_n$ .

Therefore we have obtained a periodic point $p_n$ of $g$ .

Since $p_n \in B(p, 10 \varepsilon)$ , we may further perturb $g$ to move $p_n$ to $p$ . This takes care of the linear case on surfaces.

On plaque expansiveness

May 5, 2010November 1, 2011 ConanLeave a comment

This note is mostly based on parts of (RH)^2U (2006) and conversations with R. Ures while he was visiting Northwestern.

Let $\mathcal{F}$ be a foliation of the manifold $M$ , for $p \in M$ , a plaque in of $\mathcal{F}$ through $p$ is a small open neighborhood of $p$ in the leaf $\mathcal{F}_p$ that’s pre-image of a disc via a local foliation chart. (i.e. plaques stuck nicely to make open neighborhoods where the foliation chart is defined.) For $\varepsilon$ small enough, whenever the leaves of $\mathcal{F}$ are $C^1$ , the path component of $B(p, \varepsilon)$ containing $p$ is automatically a plaque, we denote this by $\mathcal{F}_\varepsilon(p)$ .

Given a partially hyperbolic diffeomorphism $f: M \rightarrow M$ , suppose the center integrates to foliation $\mathcal{F}^c$ .

Definition: An $\varepsilon$ -pseudo orbit w.r.t. $\mathcal{F}^c$ is a sequence $(p_n)$ where for any $n \in \mathbb{Z}$ , $f(x_n) \in \mathcal{F}^c_\varepsilon(x_{n+1})$ .

i.e. $p_{n+1}$ is the $f$ -image of $p_n$ except we are allowed to move along the center plaque for a distance less than $\varepsilon$ .

Definition: $f$ is plaque expansive at $\mathcal{F}^c$ if there exists $\varepsilon>0$ s.t. for all $\varepsilon$ -pseudo orbits $(p_n), (q_n)$ w.r.t. $\mathcal{F}^c$ , $d(p_i, q_i)<\varepsilon$ for all $i \in \mathbb{Z}$ then $p_0 \in \mathcal{F}^c_\varepsilon(q_0)$ .

i.e. any two pseudo-orbits in different plagues will eventually (under forward or backward iterates) be separated by a distance $\varepsilon$ .

In the book Invariant Manifolds (Hirsch-Pugh-Shub), it’s proven that

Theorem: If a partially hyperbolic system has plaque expansive center foliation, then the center being integrable and plaque expansiveness are stable under perturbation (in the space of diffeos). Furthermore, the center foliation of the perturbed system $g$ is conjugate to the center foliation of the origional system $f$ in the sense that there exists homeomorphism $h: M \rightarrow M$ where

1) $h$ sends leaves of $\mathcal{F}^c_f$ to leaves of $\mathcal{F}^c_g$ i.e. for all $p \in M$ ,

$h(\mathcal{F}^c_f(p)) = \mathcal{F}^c_g(p)$

2) $h$ conjugates the action of $f$ and $g$ on the set of center leaves i.e. for all $p \in M$ ,

$h \circ f \ (\mathcal{F}^c_f(p)) = g \circ h \ ( \mathcal{F}^c_f(p))$

(both sides produce a $\mathcal{F}^c_g$ leaf)

Morally this means plaque expansiveness implies structurally stable in terms of permuting the center leaves.

It’s open whether or not any partially hyperbolic diffeomorphism with integrable center is plaque expansive w.r.t. its center foliation.

Another problem, stated in HPS about plaque expansiveness is:

Question: If $f$ is partially hyperbolic and plaque expansive w.r.t. center foliation $\mathcal{F}_c$ , then is $\mathcal{F}_c$ the
unique $f$ −invariant foliation tangent to $E^c$ ?

(RH)^2U has recently gave a series of super cool examples where the 1-dimensional center bundles of a $C^1$ partially hyperbolic diffeomorphism 1) does not integrate OR 2) integrates to a foliation but leaves through a given point is not unique (there is other curves through the point that’s everywhere tangent to the bundle). I will say a few words about the examples without spoil the paper (which is still under construction).

Start with the cat map on the $2$ -torus (matrix with entries $( 2, 1, 1, 1)$ , take the direct product with the source-sink map on the circle, we obtain a diffeo on the $3$ torus. For the purpose of our map, we make the expansion in the source-sink map weaker than that of the cat map and the contraction stronger.

Then we perturb the map by adding appropriate small rotations to the system, the perturbation vanish on the $\mathbb{t}^2$ fibers corresponding to the two fixed points in the source-sink map. This will make our system partially hyperbolic, with center bundles as shown below:

To construct a non-integrable center, we make a perturbation that gives center boundle (inside the unstable direction of the cat map times the circle):

For intergrable but have non-unique center leaves, we simply rotate the upper and bottom half in opposite directions and obtain:

Note that in this case, all center leaves are merely copies of $S^1$ . The example is plaque expansive due to to fact that all centers leaves are compact (and of uniformly bounded length). However, although the curve through any given point tangent to the bundle is non-unique, there is only one possible foliation of the center. Hence this does not give a counter example to the above mentioned question in HPS.

I think there are hopes to modify the example and make one that has similar compact leafs but non-unique center foliation, perhaps by making the unique integrability fail not only on a single line.

On Alexander horned sphere

April 18, 2010April 18, 2010 Conan4 Comments

As I was drawing pictures for some stuff that should be done a year ago, I found this part would make a cool blog post, so here it is ^^ (well I admit that I mainly just want t show off the picture)

For kids who doesn’t know, let’s first talk a bit about what this ‘sphere’ is:

This is an embedded topological sphere in $\mathbb{R}^3$ which has non-simply connected exterior. Also, Since the surface is compact, through inversion about any point bounded away from infinity by the surface, we obtain a ‘sphere’ that bounds a non-simply connected region inside. This shows that the topology of the complement of a compact surface depends on the embedding, which is not true for embeddings of compact $1$ -dimensional manifolds in $\mathbb{R}^2$ . (i.e. all Jordan curves separates the plane into two simply connected open sets, via the Jordan curve theorem)

The construction, as shown in the beautiful 2 page article by Alexander, goes as follows:

Take an ordinary sphere (stage 0), stretch and bend it like a banana so that the two ‘end caps’ are supported on a pair of parallel circles such that one lies vertically on top of the other (state 1). Next, on each cap we develop a banana shape, the banana shape on the two caps link though each other and again has their caps supported on a pair of parallel circles (stage 2). Continue the process to add successively smaller bananas on the caps produced in the immediate preceding stage.

Claim: The limit is a topological sphere.

To see this, we build homeomorphisms from $S^2$ to each sphere in the intermediate stages. i.e. let

$h_n: S^2 \rightarrow S^2_n$

be a homeomorphism where $S^2_n \subseteq \mathbb{R}^3$ is the embedded sphere at stage $n$ .

We may take $h_n$ s.t. $h_n^{-1}$ restricting to the complement of the $(n-1)$ th stage caps (denoted by $C_{n-1}$ ) agree with $h_{n-1}^{-1}$ . Hence union the maps $h_n|_{C_{n-1}}$ gives a continuous map on the complement of a Cantor set on the sphere. (Since $C_n$ is increasing and the caps gets smaller) This map can be extended continuously to the whole sphere because any neighborhood of points in the Cantor set contains pre-image of some sufficiently small cap.

The extension $h$ is injective since any two points in the Cantor set will be separated by a pair of disjoint pre-image of small caps. Since the sphere is compact, we conclude $h$ is a homeomorphism. i.e. the limiting surface is a topological sphere.

The exterior of the surface is not simply connected as a loop just outside the ‘equator’ can’t be contracted to a point. In fact, it’s also easy to show that the fundamental group of the exterior is not finately generated.

For some reason, Charles and I wanted to create a diffeotopy of from the standard sphere to an Alexander horned sphere. (with differentiability failing only at time one, and this is necessary since there can be no diffeomorphism from the sphere to the horned sphere, otherwise it would extend to a neighbourhood of the surfaces and hence the whole $\mathbb{R}^3$ , but the exteriors of the two are not homeomorphic.)

The above figure is in fact a particular kind of Alexander horned sphere we needed. i.e. it has the property that each cap in the $(n+1)$ th stage has diameter less than $1/2$ of that in the $n$ th stage, and the distance between the parallel circles is also less than $1/2$ of that in the previous stage. Spheres at each stage is differentiable.

This would allow us to construct a diffeotopy that achieves stage $n$ at time $1-1/2^n$ , the diffeotopy is of bounded speed as all horns are half as large as the pervious stage, hence once we get to the first stage with bounded speed, making all points traveling at that maximum speed would get one to the next stage using $1/2$ as much time.

However, we do not know if all horned sphere can be achieved by s diffeotopy from the standard sphere. i.e. does the property of being a ‘diffeotopic sphere’ depend on the embedding in $\mathbb{R}^3$ .

Many thanks to Charles Pugh for forcing me to look at this business. It is indeed very fun~

Moser’s theorem with boundary

January 29, 2010March 3, 2010 Conan2 Comments

In the process of constructing a diffeo with a uniformly hyperbolic set with intermediate measure, I came across the following problem which I find interesting in its own right:

Given a $C^1$ diffeo $f:S^1 \rightarrow S^1$ , when does it extend to a volume preserving diffeo of the unit disc to itself?

Some people suggested me to look into Moser’s theorem for volume forms on compact manifolds, I found it pretty cool, so here it is (taken from Moser’s paper):

Theorem: Given two smooth volume forms $\tau, \sigma$ with same total volume on a compact connected $C^\infty$ manifold $M$ , there exists diffeomorphism $\phi: M \rightarrow M$ s.t. $\sigma = \phi^* \tau$ .

However this does not directly apply to the boundary problem as we are dealing with manifolds (disc) with boundary rather than entire compact manifolds…Hence to make it applicable to the case, I would have to get some kind of ‘relative’ version of the theorem.

The expectation is that this can be done by plugging in the case into Moser’s proof and see if it can be modified.. I’ll update this pose when I got around to do that (hopefully in the next few days)

Okay…I think I have finally figured out how to do this. (with a huge amount of hints and directions from Keith Burns). But temporarily I have to lose one degree of regularity when making the extension (i.e. starting with a $C^2 f$ , extend it to a $C^1$ volume preserving. But at this point I strongly believe that we can in fact do it with $f$ being $C^1$ . (the regularity is lost when I extended the diffeo locally)

Theorem: Given $C^2$ diffeomorphism $f:S^1 \rightarrow S^1$ , we can find (Lebesgue) volume preserving diffeomorphism $\hat{f}: \mathbb{D} \rightarrow \mathbb{D}$ that extends $f$ . (where $\mathbb{D}$ is the closed unit disc and $S^1$ is its boundary)

Proof:

First we extend the diffeo $f:S^1 \rightarrow S^1$ to a neignbourhood of $S^1$ in $\mathbb{D}$ .

Let $A = \{ \ r e^{i \theta} \ | \ 1/2 < r \leq 1 \}$ be the half-open annuals with radius $1/2$ and $1$ .

Let $C = \mathbb{R} / 2 \pi \mathbb{Z} \times [0, \infty)$ be the half-cylinder.

Define $\phi: A \rightarrow C$ s.t. $\phi (r e^{i \theta}) = (\theta, 1/2 - r^2/2)$

$\phi$ is a volume preserving diffeo from $A$ to $\mathbb{R} / 2 \pi \mathbb{Z} \times [0, 3/8)$

Consider $h = \phi \circ f \circ \phi^{-1}: \mathbb{R} / 2 \pi \mathbb{Z} \times \{0\} \rightarrow \mathbb{R} / 2 \pi \mathbb{Z} \times \{0\}$

$(h^{-1})'$ is continuous hence bounded.

Choose $\epsilon < 1/(3 \max_\theta | (h^{-1})' (\theta) | )$

Define $\hat{h}: \mathbb{R} / 2 \pi \mathbb{Z} \times [0, \epsilon) \rightarrow C$ s.t.

$\hat{h}(\theta, y) = ( \pi_1(h(\theta, 0) ), y |(h^{-1})'(\theta) | )$

$y |(h^{-1})'(\theta) | < \epsilon \max_\theta | (h^{-1})'(\theta) | 0 \ \mathbb{R} / 2 \pi \mathbb{Z} \times [0, \delta) \subseteq \hat{h}(\mathbb{R} / 2 \pi \mathbb{Z} \times [0, \epsilon/4))$ .

Let $\epsilon'=1-\sqrt{1-2\epsilon}, \ \delta'=1-\sqrt{1-2\delta}$ . Let $g:N_{\epsilon'}(S^1) \rightarrow A, \ g := \phi^{-1} \circ \hat{h} \circ \phi$ .

Hence $g$ is volume preserving, $g|_{S^1} = f$ and $N_{\delta'}(S^1) \subseteq g(N_{\epsilon'/2}(S^1))$ .

Hence we have successfully extended $f$ to a neighborhood of $S^1$ in a volume preserving way.

Now we further extend $g|_{N_{\epsilon'/2}(S^1)}$ to a (non-volume-preserving) diffeo of $\mathbb{D}$ to itself.

This can be done by first take $f \times I$ on $C$ , average it with $g$ by a $C^\infty$ bump function that is $1$ on $\mathbb{D} \backslash N_{\epsilon'}(S^1)$ and vanishes on $N_{\epsilon'/2}(S^1)$ .

Since both functions preserve vertical segments, taking the resulting map back to $A$ will produce a diffeo except for at the point $0$ . We may smooth out the map at $0$ . Call the resulting diffeo $\hat{g}: \mathbb{D} \rightarrow \mathbb{D}$

Define measure $\mu$ on $\mathbb{D}$ by $\mu(B) = \lambda(\hat{g}^{-1}(B))$ where $B$ is any Lebesgue measurable set and $\lambda$ is the Lebesgue measure.

Since $\hat{g}^{-1}$ is volume preserving on $N_{\delta'}(S^1)$ , hence $\mu$ is equal to $\lambda$ on $N_{\delta'}(S^1)$ .

Also we have $\mu(\mathbb{D}) = \lambda(\hat{g}^{-1}(\mathbb{D}) = \lambda(\mathbb{D})$ .

Hence we can apply lemma 2 in Moser’s paper:

Lemma: If two $C^r$ volume forms $\mu_1, \ \mu_2$ agree on an $\epsilon$ neighbourhood of the boundary of the cube (disc in our case) with the same total volume, then there exists $C^r$ diffeo $\psi$ from the cube (disc) to itself s.t. $\psi$ is identity on the $\epsilon$ neighbourhood of the boundary and $\mu_1(\psi(B)) = \mu_2(B)$ for all measurable set $B$ .