# Is there an area-preserving diffeomorphism of the disk which is nowhere conformal?

This question is a cross-post; it is related to this former question of mine. Let $$D \subseteq \mathbb{R}^2$$ be the closed unit disk.

Does there exist a smooth volume-preserving diffeomorphism $$f:D \to D$$ that does not have conformal points?

i.e. I want the singular values $$\sigma_i$$ of $$df$$ to be everywhere distinct, and $$\det(df)=1$$.

(An equivalent requirement to $$\sigma_1 \neq \sigma_2$$ is that the sum $$\sigma_1+\sigma_2>2$$-this follows from the AM-GM inequality together with $$\sigma_1\sigma_2=1$$).

Edit 2:

It is proven here that for any map $$f:D \to \mathbb R^2$$, the condition of being with distinct singular values is 'generic' in the following sense:

There exist $$f_n \in C^{\infty}(D, \mathbb{R}^2)$$ such that $$f_n \to f$$ in $$W^{1,2}(D, \mathbb{R}^2)$$ and $$df_n$$ has everywhere distinct singular values on $$D$$.

However, it seems to me that this approximation procedure, applied to a map $$f \in \text{Diff}(D)$$, does not guarantee that the $$f_n$$ will map $$D$$ into $$D$$, let alone be diffeomorphisms. (e.g. I think that the convergence $$f_n \to f$$ cannot be made uniform in general).

However, perhaps this genericity phenomena can still be used somehow for this question.

This answer provides the following example for a one-parameter family of such diffeomorphisms $$D\setminus \{0\} \to D \setminus \{0\}$$:

$$f_c: (r,\theta)\to (r,\theta+c\log r).$$ (this description is given in terms of polar coordinates both in the domain and the range. For each non-zero $$c ֿ\in \mathbb R$$ we get a diffeomorphism, with fixed distinct singular values whose product is $$1$$.)

Edit 1-I describe below a possible topological obstruction: (this is probably a too naive argument, but perhaps it can be upgraded somehow.)

Set $$\mathcal{NC}:=\{ A \in M_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$$, where by a non-conformal matrix, I refer to a matrix whose singular values are distinct. (I allow non-zero singular matrices in $$\mathcal{NC}$$).

Suppose that an area-preserving and nowhere conformal $$f \in \text{Diff}(D)$$ exists. Then $$df|_{\partial D}:\partial D \to \mathcal{NC}$$ is nullhomotopic.

$$df|_{\partial D}$$ maps $$T\partial D$$ to itself, and in particular, at a point $$\theta \in \mathbb{S}^1$$, $$(df|_{\partial D})_{\theta}(T_{\theta}\partial D)=T_{f(\theta)}\partial D$$. So, thinking on $$e_2$$ as an element of $$T_{(0,1)}\partial D$$, we have

$$R_{f(\theta)}^{-1} \circ df_{\theta} \circ R_{\theta}(e_2)=\lambda(\theta) e_2$$, for some positive factor $$\lambda(\theta)$$.

Setting $$A_{\theta}:=R_{f(\theta)}^{-1} \circ df_{\theta} \circ R_{\theta}$$, and $$\mathcal{F}:=\{ A \in \text{SL}_2(\mathbb{R}) \, | \, Ae_2 \in \operatorname{span}(e_2) \, \, \text{ and } \, \, A \, \text{ is not conformal} \,\},$$ we get $$df_{\theta} =R_{f(\theta)} \circ A_{\theta} \circ R_{-\theta}, \, \, \, A_{\theta}: \partial D \to \mathcal{F}. \tag{1}$$

If $$\mathcal{F}$$ were contractible in $$\mathcal{NC}$$, we could deform $$A_{\theta}$$ to a constant map $$\partial D \to \mathcal{NC}$$.

Thus, by equation $$(1)$$, $$df|_{\partial D}$$ would be homotopic to the map $$\theta \to R_{f(\theta)} \circ A \circ R_{-\theta}$$ for some constant non-conformal matrix $$A \in \mathcal{NC}$$.

Writing $$A=R_{\alpha} \Sigma R_{\beta}$$ where $$\Sigma$$ is non-negative and diagonal, we would get that $$df|_{\partial D}$$ is homotopic to $$\theta \to R_{f(\theta)+\alpha} \circ \Sigma \circ R_{-\theta+\beta}.$$

On the space of non-conformal matrices $$\mathcal{NC}$$, there is a continuous map* $$H:\mathcal{NC} \to \mathbb{S}^1$$, given by $$H(R_{\phi} \Sigma R_{\theta})= R_{2\theta}$$. This leads to a contradiction to $$df|_{\partial D}$$ being nullhomotopic:

Indeed, if it were nullhomotopic, then so would the map $$\theta \to R_{f(\theta)+\alpha} \circ \Sigma \circ R_{-\theta+\beta}.$$ Composing it with $$H$$, we obtain the map $$\theta \to R_{-2\theta+2\beta}$$, or $$\theta \to -2\theta$$, which is not nullhomotopic.

Since $$\mathcal{F}$$ is not contractible in $$\mathcal{NC}$$, this argument fails. However, perhaps a more refined topological argument could obtain more, I don't know.

*The map $$H:\mathcal{NC} \to \mathbb{S}^1$$ is well-defined, since $$U\Sigma V^T=(-U)\Sigma (-V)^T$$, and this is the only ambiguity in the SVD of a matrix in $$\mathcal{NC}$$. Thus $$\theta$$ is well defined up to an addition of $$\pi$$.

• Maybe a fixed point theorem could be used. – Sylvain JULIEN Mar 29 at 12:18
• I don't know the concepts you refer to, but is a nonconformal map the same as a map with no conformal points? – Sylvain JULIEN Mar 30 at 16:53

Consider $$V_{k}$$ defined as the set of compact subsets of $$D$$ whose any open subset is a domain and that share the same volume $$k$$. The restriction of $$f$$ on any element thereof is a permutation of $$V_{k}$$. Taking the intersection of all $$V_{k}$$ for all possible values of $$k$$ gives you a fixed point of $$f$$ (which exists by Brouwer theorem). The sequence of restrictions $$f_{n_k}$$ on a family of compact subsets of decreasing volume $$k$$, with the considered volume tending to $$0$$ as $$n_k$$ tends to $$\infty$$, converges to the differential of a similitude, hence a conformal map. So the considered fixed point should be conformal.
Edit: here's an explanation of why I think the limit is a conformal map. For two elements $$A$$ and $$B$$ of $$V_k$$, define a distance $$\delta(A,B)$$ between the shapes of $$A$$ and $$B$$ as $$\displaystyle{\inf_{s}\dfrac{\mu(A\Delta s(B))}{\mu(A\cap s(B))}}$$ where $$s$$ runs over the isometries of the complex plane, where $$\Delta$$ denotes the symetric difference and $$\mu$$ the $$2$$-dimensional Lebesgue measure. As $$k$$ tends to $$0$$, so does $$\delta(A_{k},B_{k})$$ with $$A_{k}$$ and $$B_{k}$$ elements of $$V_{k}$$. Hence their shapes converge to the same limit shape, making $$f$$ a similitude locally.
• Thanks, but I am not sure that I follow your argument. What do you mean by taking the intersection of all the $V_k$? Each $V_k$ is a collection of subsets... Which of those subsets exactly do you choose to intersect for different values of $k$? and how does that produce a fixed point of $f$? Finally, I am also not sure why $\lim_{k \to 0}\delta(A_{k},B_{k}) =0$. – Asaf Shachar Mar 29 at 15:47
• I conceive the elements of $V_{k}$ only up to isometries, as neighborhoods of the fixed point, call it $u_{f}$. That way for a given $k$ the elements of $V_{k}$ have a non empty intersection – Sylvain JULIEN Mar 29 at 15:56
• I agree that for given $k$ the intersection $I_{k}$ of the elements $U_{k}$ of $V_{k}$ is not uniquely defined, but the limit of $\bigcap_{l\leq k}I_{l}$ as $k$ tends to $0$ is $u_{f}$. – Sylvain JULIEN Mar 29 at 16:03
• And as you require $det(df)=1$, $df(u_{f})$ must be a rotation, hence $\lim_{k\to 0}\delta(A_{k},B_{k})=0$. – Sylvain JULIEN Mar 29 at 16:11