
> 
> 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 $\sigma\_1\sigma\_2$).
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> 


**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.



Are all maps $\mathbb{R}^2 \to \mathbb{R}^2$ with fixed singular values affine?
**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.


