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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 $\pi$.

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  • $\begingroup$ Maybe a fixed point theorem could be used. $\endgroup$ – Sylvain JULIEN Mar 29 at 12:18
  • $\begingroup$ I don't know the concepts you refer to, but is a nonconformal map the same as a map with no conformal points? $\endgroup$ – Sylvain JULIEN Mar 30 at 16:53
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Too long for a comment even though I'm not sure the following argument is valid, but let's give it a try.

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.

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  • $\begingroup$ 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 $\lim_{k \to 0}\delta(A_{k},B_{k}) =0$. $\endgroup$ – Asaf Shachar Mar 29 at 15:47
  • $\begingroup$ 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 $\endgroup$ – Sylvain JULIEN Mar 29 at 15:56
  • $\begingroup$ 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 $u_{f}$. $\endgroup$ – Sylvain JULIEN Mar 29 at 16:03
  • $\begingroup$ And as you require $det(df)=1$, $df(u_{f})$ must be a rotation, hence $\lim_{k\to 0}\delta(A_{k},B_{k})=0$. $\endgroup$ – Sylvain JULIEN Mar 29 at 16:11
  • $\begingroup$ The general idea is that similitudes are exactly the shape-preserving bijections, or to say it differently, the shape is what remains invariant under the actions of the group of all the similitudes. See also page 170 of books.google.fr/… map differential similitude&f=false. $\endgroup$ – Sylvain JULIEN Mar 29 at 16:19

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