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$\newcommand{\GL}{\operatorname{GL}^+_2}$

Let $U \subseteq \mathbb{R}^2$ be an open connected subset, and let $f:U \to \mathbb{R}^2$ be a smooth orientation-preserving map, i.e. $df \in \GL$ everywhere on $U$. Suppose also that the singular values of $df$ are constant $\sigma_1 \neq \sigma_2$.*

How to prove that there exist continuous functions $v,w:U \to \mathbb{S}^1 \subseteq \mathbb{R}^2$ such that $v \perp w$, $df(v) \perp df(w)$, and $\|df(v)\|=\sigma_1,\|df(w)\|=\sigma_2$.

That is, I ask if we can choose continuously the singular vectors of $df$.

The problem is that if $df=U\Sigma V^T$ is the SVD of $df$, where $U,V \in \text{SO}(2)$, then $U$ and $V$ are unique up to a simultaneous change of sign.

So, I know that they can be chosen smoothly locally; the question is whether they can be chosen continuously globally on $U$.

*I think that it should suffice to assume that $\sigma_1(df) \neq \sigma_2(df)$ everywhere on $U$, not that they are constant and distinct.

Asaf Shachar
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  • The singular values correspond to orthogonal 1-dimensional distributions on $TU$ and $TF(U)$. There certainly exist such distributions which do not have nonvanishing sections (provided $U$ is not contractible), but constructing a function which serves as a counterexample is still not so easy. – Kajelad Nov 10 '20 at 23:22
  • Thanks, I thought there should be some topology involved. Just to be sure that I understand: If $U$ is contractible, then every vector bundle over it is trivial, hence in particular every $1$-dimensional distribution on $TU$ has a nonvanishing section, which can then be normalized (this is how we get our global singular vector, right?). – Asaf Shachar Nov 11 '20 at 07:45
  • Correct. Technically, one only needs that the connected components of $U$ are contractible. – Kajelad Nov 11 '20 at 15:08

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