# Moser’s theorem with boundary

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 $B$.

We apply the lemma to $\mu$ and $\lambda$, obtain $\psi$.

Hence $\lambda(B) = \mu(\psi(B)) = \lambda(\hat{g}^{-1} \circ \psi(B))$ $\hat{g}^{-1} \circ \psi$ is a $C^1$ diffeo that preserved Lebesgue volume, so is its inverse $\psi^{-1} \circ \hat{g}$.

Let $\hat{f} = \psi^{-1} \circ \hat{g}, \ \hat{f}|_{S^1} = \hat{g}|_{S^1} = f$.

Hence $\hat{f}$ is a volume preserving extension of $f$.

## 2 thoughts on “Moser’s theorem with boundary”

1. […] A simple case of this type of question was discussed in my earlier Moser’s theorem post: […]

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