Let $D \subseteq \mathbb{R}^2$ be the closed unit disk.
Let $f:D \to \mathbb{R}^2$ be a smooth map, such that the singular values of $df$, $\sigma_1(df),\sigma_2(df)$ satisfy
$$ \sigma_1(df)+\sigma_2(df)=1 \tag{1} $$ everywhere on $D$.
Question: Must the Jacobian $\det(df)$ be constant?
Edit:
If we remove the origin from the disk, then I can build such an $f$ with non-constant Jacobian on $\{ x \in \mathbb R^2 \, | \, 0<\|x\|< \frac{1}{3\lambda} \}$ into the disk with radius $\lambda$. Can we build such an example on the entire disk- with the origin included?
(I need to check if my solution extends smoothly to the origin).
Here are the details:
Let $f$ be the map described in polar coordinates by
$$ \big(r,\theta\big )\mapsto \big(\psi(r),\theta+\phi(r)\big). $$
We have
$$ df\big(\frac{\partial}{\partial \theta}(r)\big)=\frac{\partial}{\partial \theta}\big(\psi(r)\big), df(\frac{\partial}{\partial r})=\psi'(r)\frac{\partial}{\partial r}+\phi'(r)\frac{\partial}{\partial \theta}\big(\psi(r)\big),$$
so, w.r.t the orthonormal frame $\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}$,
$$ df\big(\frac{1}{r}\frac{\partial}{\partial \theta}(r)\big)=\frac{\psi(r)}{r}\cdot \bigg(\frac{1}{\psi(r)}\frac{\partial}{\partial \theta}\big(\psi(r)\big)\bigg)$$ and
$$ df(\frac{\partial}{\partial r})=\psi'(r)\frac{\partial}{\partial r}+\big(\phi'(r)\psi(r) \big) \bigg(\frac{1}{\psi(r)} \cdot \frac{\partial}{\partial \theta}\big(\psi(r)\big)\bigg).$$
In other words, $$ [df]_{ \frac{\partial}{\partial r},\frac{\partial}{\partial \theta}}=\begin{pmatrix} \psi' & 0 \\\ \phi'\psi & \frac{\psi}{r}\end{pmatrix}. $$
Now, for a matrix $A=\begin{pmatrix} a & 0 \\\ b & c\end{pmatrix}$, we have that $\sigma_1(A)+\sigma_2(A)=1 \iff 1= (\sigma_1(A)+\sigma_2(A))^2 \iff$
$$1=\sigma_1(A)^2+\sigma_2(A)^2+2\sigma_1(A)\sigma_2(A)=|A|^2+2\det A.$$
So, $\sigma_1(A)+\sigma_2(A)=1 \iff (a+c)^2+b^2=1.$ Thus, $ \sigma_1(df)+\sigma_2(df)=1 $ if and only if
$$ (\psi'+\frac{\psi}{r})^2+(\phi'\psi)^2=1. \tag{2}$$
Thus, taking $\psi(r)=\lambda r^n$ (this makes $ \big(r,\theta\big )\mapsto \big(\psi(r),\theta+\phi(r)\big)$ smooth since we removed the origin), equation $(2)$ becomes
$$ \big((n+1)r^{n-1}\big)^2+(\phi' r^n)^2=\frac{1}{\lambda^2 }.$$
Thus, we can find an explicit expression for $\phi'(r)$ and integrate it, thus obtaining a solution $\phi$. Note that a solution exists in the domain where $\big((n+1)r^{n-1}\big)^2 \le \frac{1}{\lambda^2 }$ or
$$ r \le \sqrt[n-1]{\frac{1}{(n+1)\lambda}}.$$ In particular, taking $n=2$ results in $r \le \frac{1}{3\lambda}$.