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 $\pi$.

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.