Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a smooth map whose differential has fixed distinct singular values $0<\sigma_1<\sigma_2$ and an everywhere positive determinant (which is the product $\sigma_1\sigma_2$).

Must $f$ be affine?

My assumption is equivalent to $df_x \in \text{SO}(2) \cdot \text{diag}(\sigma_1,\sigma_2) \cdot \text{SO}(2)$ for every $x \in \mathbb{R}^2$.

If we were only allowing a copy of $\text{SO}(2)$ from one of the sides of $ \text{diag}(\sigma_1,\sigma_2)$, then the answer would be positive. (This reduces to the case of isometries).

Similarly, if we had $\sigma_1=\sigma_2$, the answer would also be positive.


Answer modified on 1 February 2020:

It's not true 'locally' in the sense that non-affine $f$'s satisfying this system of PDE can be constructed on some open sets in $\mathbb{R}^2$. This first order, determined PDE system is hyperbolic, so there are many local solutions. However, it turns out (see below) that all $C^3$ solutions with domain equal to $\mathbb{R}^2$ are affine. (The proof I give below does not work for solutions of lower regularity.)

Let $D\subset\mathbb{R}^2$ be a $1$-connected open domain on which there exists a $C^3$ mapping $f:D\to\mathbb{R}^2$ whose differential $\mathrm{d}f$ has constant, distinct singular values $0<\sigma_1<\sigma_2$. Because $D$ is simply connected, one can choose an orthonormal frame field $E_1,E_2$ on $D$ such that, at each point $p\in D$, the image vectors $F_i(p) = \mathrm{d}f\bigl(E_i(p)\bigr)$ are orthogonal and satisfy $|F_i(p)|=\sigma_i$.

Let $\omega = (\omega_1,\omega_2)$ be the dual coframing on $D$, which is of regularity type $C^2$. The $1$-forms $\eta_i = \sigma_i\,\omega_i$ for $i=1,2$ have the property that $(\eta_1)^2+(\eta_2)^2$, being the $f$-pullback of the flat metric on $\mathbb{R}^2$, must also be a flat metric.

Let $\omega_{12}$ be the connection $1$-form associated to the coframing $\omega$, i.e., it satisfies the structure equations $$ \mathrm{d}\omega_1 = -\omega_{12}\wedge\omega_2 \qquad\text{and}\qquad \mathrm{d}\omega_2 = \omega_{12}\wedge\omega_1\,.\tag1 $$ Write $\omega_{12} = -\kappa_1\,\omega_1 + \kappa_2\,\omega_2$. The function $\kappa_i$ is the curvature of the $E_i$-integral curve. Since $\omega_{12}$ is $C^1$, so are the functions $\kappa_i$. A straightforward computation shows that the $1$-form $\eta_{12}$ that satisfies the corresponding structure equations $$ \mathrm{d}\eta_1 = -\eta_{12}\wedge\eta_2 \qquad\text{and}\qquad \mathrm{d}\eta_2 = \eta_{12}\wedge\eta_1\,.\tag2 $$ is given by $$ \eta_{12} = -(\sigma_1/\sigma_2)\,\kappa_1\omega_1 + (\sigma_2/\sigma_1)\,\kappa_2\omega_2\,. $$ Since $\sigma_1\not=\sigma_2$, the conditions $\mathrm{d}\omega_{12} = \mathrm{d}\eta_{12}=0$ (which hold because the domain metric and the $f$-pullback of the range metric are both flat) are equivalent to $$ 0 = \mathrm{d}(\kappa_i\,\omega_i) = \bigl(\mathrm{d}\kappa_i - {\kappa_i}^2\,\omega_{3-i}\bigr)\wedge\omega_i\,\qquad i = 1,2.\tag3 $$

Proposition: If $D = \mathbb{R}^2$, then $\kappa_1 \equiv \kappa_2 \equiv 0$, and $f$ is an affine map.

Proof: Suppose that, say, $\kappa_1$ be nonzero at some point $p\in\mathbb{R}^2$ and consider the value of $\kappa_1$ along the $E_2$ integral curve through $p$, which, since $E_2$ has unit length, is necessarily complete. Let $p(s)$ be the flow of $E_2$ by time $s$ starting at $p = p(0)$. Then (3) implies that the function $\lambda(s) = \kappa_1\bigl(p(s)\bigr)$ satisfies $\lambda'(s) = \lambda(s)^2$. Consequently, $$ \kappa_1\bigl(p(s)\bigr) = \frac{\kappa_1\bigl(p(0)\bigr)}{1-\kappa_1\bigl(p(0)\bigr)s}. $$ Hence $\kappa_1$ cannot be continuous along this integral curve, which is a contradiction. Thus, $\kappa_1$ and, similarly, $\kappa_2$ must vanish identically when $D = \mathbb{R}^2$. In particular, $\mathrm{d}\omega_i = 0$, from which one easily concludes that $f$ is affine. QED

More interesting, locally, is what happens near a point where $\kappa_1\kappa_2\not=0$. (There is a similar analysis when one of $\kappa_i$ vanishes identically that can safely be left to the reader, but see the note at the end.) One might as well assume that $\kappa_1\kappa_2$ is nowhere vanishing on $D$. Then one can write $$ \kappa_1\,\omega_1 = \mathrm{d}u \qquad\text{and}\qquad \kappa_2\,\omega_2 = \mathrm{d}v $$ for some $C^2$ functions $u$ and $v$ on $D$, uniquely defined up to additive constants.

Writing $\omega_1 = p\,\mathrm{d}u$ and $\omega_2 = q\,\mathrm{d}v$ for some non-vanishing functions $p$ and $q$, one finds that the structure equations (1), with $\omega_{12} = -\mathrm{d}u + \mathrm{d}v$, yield the equations $$ p_v = - q \qquad\text{and}\qquad q_u = -p. $$ In particular, note that $p_v$ is $C^1$ and $p_{uv}-p = 0$.

Conversely, if $p$ be any nonvanishing $C^2$ function on a domain $D'$ in the $uv$-plane that satisfies the hyperbolic equation $p_{uv}-p=0$ and is such that $p_v$ is also nonvanishing on $D'$, then the $1$-forms $$ \omega_1 = p\,\mathrm{d}u,\quad \omega_2 = -p_v\,\mathrm{d}v,\qquad \omega_{12} = -\mathrm{d}u+\mathrm{d}v\tag4 $$ satisfy the structure equations of a flat metric, and so do $$ \eta_1 = \sigma_1\,p\,\mathrm{d}u,\quad \eta_2 = -\sigma_2\,p_v\,\mathrm{d}v,\qquad \eta_{12} = -(\sigma_1/\sigma_2)\,\mathrm{d}u+(\sigma_2/\sigma_1)\,\mathrm{d}v.\tag5 $$

Indeed, one now sees that the $1$-forms $$ \begin{aligned} \alpha_1 &= \cos(u{-}v)\,p\,\mathrm{d}u +\sin(u{-}v)\,p_v\,\mathrm{d}v\\ \alpha_2 &= \sin(u{-}v)\,p\,\mathrm{d}u -\cos(u{-}v)\,p_v\,\mathrm{d}v \end{aligned} $$ are closed, and therefore can be written in the form $\alpha_i = \mathrm{d}x_i$ for some $C^3$ functions $x_i$ on $D'$. $$ (\mathrm{d}x_1)^2 + (\mathrm{d}x_2)^2 = (\alpha_1)^2 + (\alpha_2)^2 = (\omega_1)^2 + (\omega_2)^2 $$ and, hence, they define a $C^3$ submersion $x = (x_1,x_2):D'\to\mathbb{R}^2$ that pulls back the standard flat metric on $\mathbb{R}^2$ to the metric $(\omega_1)^2 + (\omega_2)^2$ on $D'$.

Similarly, setting $\rho = \sigma_2/\sigma_1$ and $$ \begin{aligned} \beta_1 &= \cos(u/\rho{-}\rho v)\,\sigma_1\,p\,\mathrm{d}u +\sin(u/\rho{-}\rho v)\,\sigma_2\,p_v\,\mathrm{d}v\\ \beta_2 &= \sin(u/\rho{-}\rho v)\,\sigma_1\,p\,\mathrm{d}u -\cos(u/\rho{-}\rho v)\,\sigma_2\,p_v\,\mathrm{d}v, \end{aligned} $$ one finds that $\mathrm{d}\beta_i = 0$ and hence there exist $C^3$ functions $y_i$ on $D'$ such that $\beta_i = \mathrm{d}y_i$. Set $y = (y_1,y_2)$.

Restricting to a subdomain $D''\subset D'$ on which $x$ is $1$-to-$1$ onto its image $D = x(D'')$ yields a domain on which $x^{-1}:D\to D''$ is a $C^3$ diffeomorphism. Now set $f = y\circ x^{-1}:D\to\mathbb{R}^2$, and one has a $C^3$ solution of the original PDE system.

This completely determines the structure of the 'generic' local $C^3$ solutions.

The case when one of the $\kappa_i$, say, $\kappa_1$, vanishes identically (so that the corresponding integral curves are straight lines) and the other is nonvanishing can easily be reduced to the normal form $$ \omega_1 = \mathrm{d}u,\qquad \omega_2 = \bigl(p(v)-u\bigr)\,\mathrm{d}v,\qquad \omega_{12} = \mathrm{d}v\tag6 $$ where, now, $p$ is a $C^2$ function of $v$, and the rest of the analysis goes through essentially unchanged.

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    $\begingroup$ @AsafShachar: I'll try to find some time in the next day or so to explain more fully. Interestingly, the PDE you need to solve to construct solutions can be locally transformed into the equation $p_{uv}+p=0$ for a function $p$ on the $uv$-plane. Since this linear equation is hyperbolic, the original system must be of hyperbolic character. $\endgroup$ – Robert Bryant Jan 30 '20 at 20:01
  • $\begingroup$ Thank you very much. This is a superb analysis. I have followed carefully most of the technical details; However, I am afraid that I am missing something fundamental here: How do you know that $d\eta_{12}=0$? If I understood correctly your argument, this is so because $\eta_{12}$ is the connection one-form of some flat symmetric connection (which should be the Levi-Civita connection of the pullback metric?), w.r.t a suitable (orthonormal?) frame. How can we see that this is the case? (what is the frame? Is it the one dual to $\eta_1,\eta_2$ which is orthogonal but not orthonormal?).... $\endgroup$ – Asaf Shachar Feb 2 '20 at 16:04
  • $\begingroup$ Do Cartan's first structure equations uniquely determine the connection form?... (I am trying to understand how the verification in equation $(2)$ in your answer implies that $\eta_{12}$ is the connection one-form). I am apologise if my question seems a bit unclear; I am not entirely comfortable with the application of Cartan's second equation here to $\eta_{12}$, and I feel that I don't have a clear understanding of what is the chain of reasoning here. Thank you again for this detailed and thorough analysis. $\endgroup$ – Asaf Shachar Feb 2 '20 at 16:06
  • $\begingroup$ @AsafShachar: You ask whether the structure equations uniquely determine the connection form. Yes, it does. The $f$-pullback metric, ${\eta_1}^2+{\eta_2}^2 = \sigma_1^2\,{\omega_1}^2+\sigma_2^2\,{\omega_2}^2$ is flat, so $\mathrm{d}(\eta_{12})=0$. Since the formula I claim for $\eta_{12}$ satisfies the Cartan structure equations $\mathrm{d}\eta_1 = -\eta_{12}\wedge\eta_2$ and $\mathrm{d}\eta_2 = \eta_{12}\wedge\eta_1$, it follows that $\eta_{12}$ must, indeed, be the Levi-Civita connection form for the pullback metric relative to the (orthonormal) $\eta$-coframing. $\endgroup$ – Robert Bryant Feb 2 '20 at 16:52
  • $\begingroup$ OK, thank you. I think I got it. Just to be sure-this "unique determination" of the connection form from the first structure equation is special to dimension $2$, right? The equations determine the "wedge of $\eta_{12}$ against the basis elements $\eta_1,\eta_2$", hence determine it completely. Now, you can just use the unique $\eta_{12}$ satisfying the equations (which, in this case, is given by the formula you have prescribed) to define a connection in terms of the $\eta$-orthonormal frame dual to $\eta_1,\eta_2$. The resulting connection would be metric w.r.t $\eta$, and torsion-free... $\endgroup$ – Asaf Shachar Feb 3 '20 at 7:50

I would like to propose a simple local example:

Consider the map in polar coordinates, $\mathbb C\to \mathbb C$ that takes a complex number $z=e^{2\pi i \theta}r$ to $e^{(\sigma_1/\sigma_2)\cdot 2\pi i \theta}r\sigma_2$.

(apologies for the previous wrong example, I confused in it singular values with eigenvalues...)

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    $\begingroup$ The $df$ of your map does not have the form of a product of rotation matrix, diagonal 1,2, and rotation matrix, as required in the problem. $\endgroup$ – Ben McKay Jan 30 '20 at 16:27
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    $\begingroup$ To harmonize the the notation in the original question, maybe in the exponent you want $\sigma_1 / \sigma_2$ instead of just $\sigma_1$? $\endgroup$ – Willie Wong Jan 30 '20 at 21:05
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    $\begingroup$ For appropriate $\sigma_1,\sigma_2$ I guess this gives a smooth example with domain $\mathbb R^2$ minus a point. $\endgroup$ – Will Sawin Feb 1 '20 at 13:31
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    $\begingroup$ Asaf, the map is well defined on the complement to $0$ if the ratio $\sigma_1/\sigma_2$ is integer. So one needs to rename $\sigma_1$ as $\sigma_2 $ and $\sigma_2$ as $\sigma_1$ in order to have such maps defined on $\mathbb C\setminus 0$ for which $\sigma_1<\sigma_2$ $\endgroup$ – Dmitri Panov Feb 1 '20 at 16:41
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    $\begingroup$ It's interesting to note that Dimitri's solution corresponds to the special case $p(v) = 0$ in the special case I list at the end of my analysis that covers when one of the two eigencurve foliations is a foliation by straight lines. $\endgroup$ – Robert Bryant Feb 8 '20 at 14:35

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