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 $\pi$.