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This is a cross-post. While working on a variational problem, I have reached to the following question.

Let $0<\sigma_1<\sigma_2$ satisfy $\sigma_1\sigma_2=1$, and let $D \subseteq \mathbb{R}^2$ be the closed unit disk.

Question: Does there exist a smooth map $f:D \to D$ such that $df$ has everywhere the fixed singular values $\sigma_1,\sigma_2$ and $\det(df)=1$? Is there such a diffeomorphism of $D$?

The linear map $x \to \begin{pmatrix} \sigma_1 & 0 \\\ 0 & \sigma_2 \end{pmatrix}x$ does not satisfy the requirement; it gets outside of $D$, as $ \sigma_2 > 1$. Exclude a ray from $D$, there is such a map, given by $re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$.


Edit:

Here is a summary of the results from Dmitri Panov's great answer:

  1. For every choice of $\sigma_1 \in (0,1)$ he constructs an example for a diffeomorphism $D\setminus \{0\} \to D \setminus \{0\}$ with the singular values $(\sigma_1, \frac{1}{\sigma_1})$:

$f_c: (r,\theta)\to r(\cos(\theta+c\log(r)), \sin(\theta+c\log (r))),\;\; $

(for every non-zero $c ֿ\in \mathbb R$, $f_c$ is an example).

I still want to know whether there exits a diffeomorphism defined on all of $D$.

  1. He constructs an example for a (non-injective) smooth map $D \to D$ that satisfies the requirements, whenever $\sigma_1 < \frac{1}{2}$. Here are the details:

Let $D_0$ be the unit disk centered around zero, and let $D_a$ be the unit disk centered around $(a,0)$ where $a>1$. (so $D_a$ does not contain the origin). Consider the map $f: re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$. $f(D_a)$ is contained in the disk of radius $\sigma_1(1+a)$, centred at $(0,0)$. Thus, if $\sigma_1(1+a)\le 1$, the map $x \to f(x+(a,0))$ sends $D_0$ to $D_0$ and has the desired properties.

Since any $a>1$ will do, and we want to optimize $a$ in order to maximize the range for $\sigma_1$, we can take $a \searrow 1$, and construct an example for any $\sigma_1 < \frac{1}{2}$.

I still want to know whether there exits such maps defined on all of $D$, for all values of $\sigma_1 \ge \frac{1}{2}$.

  1. Panov proves that whenever $\sigma_1 \ge \frac{1}{\sqrt{2}}$, every smooth map $D \to D$ with the singular values $(\sigma_1, \frac{1}{\sigma_1})$ must be a diffeomorphism. (but we still don't know whether such diffeomorphisms exist).

Conclusion from items $(2),(3)$:

For $\sigma_1 < \frac{1}{2}$ there are non-diffeomorphic examples. For $\sigma_1 \ge \frac{1}{\sqrt{2}}$ every potential example is a diffeomorphism. We still don't know what happens when $\sigma_1 \in [\frac{1}{2},\frac{1}{\sqrt{2}})$.

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    $\begingroup$ This reminds me of the study of quasi-conformal homeomorphisms. $\endgroup$ Feb 18, 2020 at 16:32

2 Answers 2

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This answers the first (simple) half of the question, asking just about a smooth map. In fact, you've already given an answer to it, in some sense. Apply the map $f: re^{i \theta} \to \sigma_1re^{i(\sigma_2/\sigma_1) \theta}$ to a unit disk that doesn't contain $(0,0)$, say radius $1$ disk $D$, centred at $(2,0)$. Then, the image $f(D)$ is contained in the disk of radius $3\sigma_1$, centred at $(0,0)$. So if $\sigma_1<\frac{1}{3}$, the map $f+(2,0)$ sends $D$ to $D$ and has the desired properties.

PS. Concerning the second part of the question about a diffeomorphism, I can't give an example, but can give something that looks almost like an example (though one can show that this can not be perturbed to an example by a perturbation which is $C^1$-small close to the boundary of the disk). I'll construct a one parameter family of maps $f_c:D\to D$ (two maps $f_c$ for each $\sigma_1\in (0,1)$). These maps are diffeos on the complement to $(0,0)\in D$, and have all required properties on $D\setminus (0,0)$, though they are not differentiable at $(0,0)$. In the radial coordinates the formula is as follows:

$$f_c: (r,\theta)\to r(\cos(\theta+c\log(r)), \sin(\theta+c\log (r))),\;\; c\in \mathbb R.$$

In order to see that these maps satisfy the necessary conditions, it is enough to notice that they have the following two properties:

1) Each circle $x^2+y^2=r^2$ is sent isometrically to itself.

2) Each radial segment $(\cos(\theta)t, \sin(\theta)t)$ ($t\in [0,1]$) is sent to a spiral $e^s(\cos(\psi+cs),\sin(\psi+cs))$ parameterized by $s$ (where $\psi$ is a constant that depends on $\theta$).

All the conditions are satisfied because the family of spirals and the family of circles form the same angle at all points in $D\setminus (0,0)$, and the map is obviously a symplectomorphism and has unit norm on all the circles. For each $\sigma_1<1$ correspond exactly two $c$ (that differ by a sign).

PPS. 22/02/2020 I would like to propose one more statement (of which I am very happy), concerning the maps with $\sigma_1\in (\frac{1}{2},1]$.

Lemma. Suppose that we have a smooth map $f: D\to \mathbb R^2$ from the unit disk to $\mathbb R^2$ with fixed $\sigma_1<1$ and $\sigma_2=\frac{1}{\sigma_1}$. Then $f$ is univalent (i.e. a diffeo on its image) in case $\sigma_1>\frac{1}{\sqrt{2}}$.

The main tool of the proof is the isomperimetric inequality that says that a simple closed curve $\eta$ on $\mathbb R^2$ bounds area nor more than $\frac{l(\eta)^2}{4\pi}$. It is also surprising, that the constant $\frac{1}{\sqrt{2}}$ is exact, i.e. for $\sigma_1<\frac{1}{\sqrt{2}}$ the map doesn't need to be univalent!

Proof. Assume the converse. Denote by $D_r\subset D$ a disk of radius $r\le 1$ centred at $(0,0)$. Clearly, for $r$ small enough the restriction map $f: D_r\to D$ is a diffeomorphism onto its image. Hence there is a minimal $t\in (0,1]$, such that this map is not a diffeo on its image. Let $S_t$ be the boundary of $D_t$ (a circle of radius $t$). Clearly, the curve $f(S_t)$ touches itself at some point. There could be more than one point where it touches itself, but the argument will not change, so we will assume that $f(S_t)$ self-touches once.

Let $x$ and $y$ be the two points in $S_t$ such that $f(x)=f(y)$. Let $(xy)$ and $(yx)$ be the two arcs into which $x$ and $y$ cut $S_t$. Without loss of generality we assume that the arc $(xy)$ is longer than $(yx)$. Denote by $\gamma_{xy}$ and $\gamma_{yx}$ the images $f((x,y))$ and $f((y,x))$. Both these images are simple closed loops. Let's prove first that the loop $\gamma_{xy}$ contains the loop $\gamma_{yx}$ in its interior.

Indeed, assume the converse. Note that by definition, the length $l(\gamma_{yx})$ satisfies $$l(\gamma_{yx})\le \frac{1}{\sigma_1}l([yx])\le \sqrt{2}\cdot \pi t.$$

One the other hand, the curve $\gamma_{yx}$ encloses the whole image $f(D_t)$ of $D_t$ and the disk bounded by $\gamma_{xy}$, that doesn't belong to $f(D_t)$. So, since $f$ is area preserving, we see that $\gamma_{yx}$ encloses area more than $\pi t^2$. This contradicts the isomperimetric inequality.

We conclude, that $\gamma_{xy}$ encloses $\gamma_{yx}$, and moreover $l((xy))>\pi t$.

Let's go on to get a contradiction. Let $z$ be the midpoint of the chord $[xy]$ in $D_t$ that joins $x$ and $y$ (don't confuse it with the curvy arc $(x,y)$ that lies in $S_t$!). Let $z$ be the midpoint of $[xy]$. Consider the circle $S_z$ centred at $z$ that passes through $x$ and $y$. Then one half of this circle lies inside $D_t$. Denote this half-circle by $\eta$. Obviously, $l(\eta)=\pi\frac{l([xy])}{2}$. Denote further by $D_t'$ the connected component of $D_t\setminus \eta$ that contains the shorter arc $(y,x)$ of $S_t$. Note finally, that $${\rm area}(D')>\frac{1}{2}\pi\left (\frac{l([xy])}{2}\right)^2,$$ since $D'$ contains a half-disk of radius $\frac{l([x,y])}{2}$ centred at $z$.

Now, to get the contradiction, we apply again the isoperimetric ineqaulity, this time to the disk bounded by the simple curve $f(\eta)$. By construction, the disk bounded by $f(\eta)$ contains in its interior the image $f(D')$, so it has area more than ${\rm area}(D')$. At the same time the length of $f(\eta)$ is less than $\frac{\pi}{\sqrt{2}} l(xy)$. END of proof.

Moral. If there are self-maps with $\sigma_1>\frac{1}{\sqrt{2}}$, they are diffeos... But I still don't know if such diffeos exist:)

PPPS. 25/02. I would like to address Asaf's question, stated in the comments. Namely, is it possible to find any symplectomorphism $f:D\to D$ ($D$ is the unit disk) that has distinct singular values at any points. It seems to me that the infinitesimal version of this question is equivalent to the following funny question:

Question. Is it possible to construct a smooth function $H$ of $D$, vanishing at the boundary of $D$, and such that the system of equations $$\frac{\partial^2 H}{\partial x^2}-\frac{\partial^2 H}{\partial y^2}=0=\frac{\partial^2 H}{\partial x\partial y}$$ has no solutions in the disk?

Indeed, if we consider the Hamiltonian flow for $H$, satisfying the above conditions, I believe that for small time the flow map will have distinct singular values. What is good about this system, is that it should not be hard to program (for someone who unlike me knows how to do this), to look for examples. Namely, one can fix a degree $d>0$ and consider all polynomials $H=(x^2+y^2-1)P_d$, where $P_d$ is a poly of degree at most $d$. The space of such polynomials (for all $d$) will probably be dense in the space of all smooth functions. So if there is a counterexample to the question, it should be possible to find it among polynomials. Maybe its degree will not be very large (according to Arnold's "Topological Economy Principle in Algebraic Geometry", ). Or, on the contrary, if there is no counterexample among polynomials, there will be no among all functions, this would be quite exiting. It is easy to check that for $d=1$ indeed there is no counterexamples.

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  • $\begingroup$ Dear Asaf, indeed it is true that one of two curves $\gamma_{xy}$ or $\gamma_{yx}$ should contain the other in the interior. I am using here the following lemma: Suppose $f: D_t\to \mathbb R^2$ is an immersion and it is a one-to-to one map onto the image apart from two points $x,y\in \partial D_t$ that are mapped to the same point. Then the point $f(x)=f(y)$ cuts $f(\partial D_t)$ into two loops one of which contains the other. In particular, the bouquet of two circles can never be the boundary of an immersed disk. Are you happy with the lemma? Does this answer your question? $\endgroup$ Feb 23, 2020 at 16:41
  • $\begingroup$ Yes, thank you. I am happy with the lemma. I reviewed more thoroughly your argument again, and I find it fantastic, really. Last two remarks: (1) Are you sure that the constant $\frac{1}{\sqrt{2}}$ is sharp? As far as I can see your original construction can produce such non-diffeomorphic maps whenever $\sigma_1 < \frac{1}{2}$ (see my edit in the question), while your claim that such maps must be diffeomorphisms holds whenever $\sigma_1 >\frac{1}{\sqrt{2}}$. So, we are still left with the regime $(\frac{1}{2},\frac{1}{\sqrt{2}})$ where we don't know whether such non-diffeomorphic maps exist $\endgroup$ Feb 24, 2020 at 15:21
  • $\begingroup$ Concerning sharpness, there are two ways to state it, for your original question, indeed I don't know how to construct maps $D\to D$ with $\sigma_1>\frac{1}{2}$. But as you see, I have removed the condition from the lemma that we have a map $D\to D$. I prefer to state this lemma just for maps $D\to \mathbb R^2$. In this case, if we take a disk of radius $1$ centred at the point $(1+\varepsilon, 0)$ and apply the standard map with $\sigma_1<\frac{1}{\sqrt{2}}$, then for small $\varepsilon$ the image $f(D)$ will overlap with itself. Though it will not be contained in a disk of radius $1$. $\endgroup$ Feb 24, 2020 at 16:33
  • $\begingroup$ Concerning just constructing a diffeo with distinct singular values, it is an interesting question, I agree that one should first think about it. Probably I would first try to construct an example. If this doesn't work, I would try to connect it to $2$-dimensional symplectic geometry. Anyway, if an example comes to my mind, I'll tell you. One immediate idea that comes to mind, is to try to construct a symplectomorphism $D\to D$ that dilatetes (stretches) a certain non-zero vector field $v$ in $D$. It is not clear why such a symplectomrphism would not exist. But who knows... $\endgroup$ Feb 24, 2020 at 17:00
  • $\begingroup$ I realised, that this question about constructing an infinitesimal symplectomorphism resembles a bit an infinitesimal version of en.wikipedia.org/wiki/Carath%C3%A9odory_conjecture . This conjecture studies umbilic points of a convex surface in $\mathbb R^2$. Namely if you consider the graph of the Hamiltonian multiplied by $\varepsilon$, $z=\varepsilon H(x,y)$, then in the limit $\varepsilon\to 0$, the umbilic points on the graph should converge (it seems) to the points with two equal singular values of $H$ in the disk $D$. I don't know whether this is just a superficial connection. $\endgroup$ Mar 2, 2020 at 9:16
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Did you find out the answer to the original question ?

I came across this* work (pg. 775, conjecture 7.1) where precisely that question is formulated as a conjecture (I ignore whether or not the author managed to prove or refute it (What he calls cps-self homeomorphism is precisely your mapping with constant singular values). 2. and 3. in your summary are very interesting contributions but still, I'd be glad to know if there are examples of diffeomorphism).

*Gevirtz, Julian, Boundary values and the transformation problem for constant principal strain mappings, Int. J. Math. Math. Sci. 2003, No. 12, 739-776 (2003). ZBL1015.35062.

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  • $\begingroup$ Thank you, that is an interesting reference. No, at the moment I don't know whether there exist a diffeomorphism of the entire disk with constant, distinct singular values. It seems that there is no 'continuous family' of such diffeomorphisms though, see the end of the answer here: mathoverflow.net/a/389718/46290 $\endgroup$ May 3, 2021 at 21:02

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