Does this equation admit a solution $f:\mathbb{R}^2 \to \mathbb{R}^2$ with non-constant Jacobian?

Question

Does there exist a smooth map $f:\mathbb{R}^2 \to \mathbb{R}^2$ such that:

1. $\det df$ is not constant.
2. $d(\det df)(e_1)df(e_2)=d(\det df)(e_2)df(e_1)$.

Such a map cannot be an immersion, since if $df$ is invertible, then $df(e_1),df(e_2)$ are linearly independent, hence $d(\det df)=0$.

Of course, that does not rule out $\det df \neq 0$ at some points. It only means that $\det df \neq 0 \Rightarrow d(\det df) = 0$.

Answer 1

You're basically there - your last deduction by itself implies that $\det df$ is constant:

Suppose you have a $C^1$ function $g : \mathbb R^n \to \mathbb R$ satisfying $g(x) \ne 0 \implies dg(x) = 0$. If $g$ is not constant, then there is some point $y$ and direction $e_i$ such that $\partial_i g(y) \ne 0$, and thus by continuity of $\partial_i g$ there is some $\epsilon > 0$ such that the restriction of $g$ to the line segment $[y -\epsilon e_i, y+\epsilon e_i]$ has derivative with constant sign. Thus $g$ takes on uncountably many values on an interval where $dg \ne 0$, a contradiction.

Answer 2

This is just an elaboration of Anthony Carapetis's answer.

Lemma:

Let $g : \mathbb R^n \to \mathbb R$ be a $C^1$ function satisfying $g(x) \ne 0 \implies dg(x) = 0$. Then $g$ is constant.

Proof:

Suppose that $g$ is not constant, then there exist $x=(x_1,\dots,x_n) \in \mathbb{R}^n$ such that $dg(x) \neq 0$. Thus, there exist a direction $e_i$ such that $\frac{\partial g}{\partial x_i}(x) \ne 0$.

Denote by $f:\mathbb{R} \to \mathbb{R}$ the function $f(s)=g(x_1,\dots,x_{i-1},s,x_{i+1},\dots,x_n)$.

$f \in C^1$, and $f'(x_i)=\frac{\partial g}{\partial x_i}(x) \neq 0$. Thus, by continuity $f'$ has a constant sign on some interval $[x_i -\epsilon , x_i+\epsilon]$.

So, $f$ is strictly monotone on $[x_i -\epsilon , x_i+\epsilon]$, so the image $g([x -\epsilon e_i, x+\epsilon e_i])$ contains an open interval, which we can assume W.L.O.G that does not contain zero. But we also know that $dg \neq 0$ on $[x -\epsilon e_i, x+\epsilon e_i]$, which is a contradiction to the assumption.