Let $D\subset\mathbb{R}^2$ be a planar domain (maybe simply connected) and consider all the mappings $f:D\to\mathbb{R}^2$ with constant, fixed, positive singular values. Let $E=(E\_1,E\_2)$ be the orthonormal frame on $D$, such that at each point $p$ the image vectors $\mathrm{d}f\bigl(E\_i(p)\bigr)$ are orthogonal and have constant norm.


Calling $\kappa\_1$ and $\kappa\_2$ the curvatures of the $E\_1$ and $E\_2$ integral curves, then



> 
> Can $\kappa\_1$ and $\kappa\_2$ change sign on $D$ ?
> 
> 
> 


(When $D$ is simply connected) the nets of such integrals lines (called HP-nets and/or slip-line nets) have important applications in elasticity and plasticity theory, and there's a great deal of physical evidence towards a negative answer ($\kappa\_1$ and $\kappa\_2$ cannot change sign), however I have no knowledge of a proof.



Signs of curvatures of integrals lines of frames with constant principal values
Here's how one can construct a specific example to illustrate what can happen:


First, recall from my answer to [this question](https://mathoverflow.net/questions/351546/are-all-maps-mathbbr2-to-mathbbr2-with-fixed-singular-values-affine) that, if you have a smooth map $f:D\to\mathbb{R}^2$ with constant positive singular values, then, letting $(e\_1,e\_2)$ be the orthonormal frame field on $D$ such that $\bigl(f'(e\_1),f'(e\_2)\bigr)$ are orthogonal with $|f'(e\_i)|=\sigma\_i$ with $0<\sigma\_1<\sigma\_2$, then the dual $1$-forms $(\omega\_1,\omega\_2)$ satisfy
$$
\mathrm{d}\omega\_1 = \kappa\_1\,\omega\_1\wedge\omega\_2\qquad
\mathrm{d}\omega\_2 = \kappa\_2\,\omega\_2\wedge\omega\_1\,\tag1
$$
where $\kappa\_i:D\to\mathbb{R}$ is the curvature function of the flow lines of $e\_i$. Moreover, we know that
$$
\mathrm{d}(\kappa\_1\,\omega\_1) = \mathrm{d}(\kappa\_2\,\omega\_2) = 0.\tag2
$$


Conversely, given a surface $S$ endowed with a coframing $\omega = (\omega\_1,\omega\_2)$ and functions $\kappa\_1$ and $\kappa\_2$ that satisfy equations (1) and (2), one can construct a corresponding $f:D\to\mathbb{R}^2$ with constant singular values $\sigma\_i$.


Now, it's not hard to construct such data $(\omega\_1,\omega\_2,\kappa\_1,\kappa\_2)$ for which $\kappa\_1$ and $\kappa\_2$ vanish along some curves. For example, on a domain $S$ in the $uv$-plane that contains the origin, assume that $\lambda\_1$ and $\lambda\_2$ are nonzero functions that satisfy the linear hyperbolic system
$$
\frac{\partial\lambda\_1}{\partial v} + u\lambda\_2 =
\frac{\partial\lambda\_2}{\partial u} + v\lambda\_1 = 0\,.\tag3
$$
For example, $\lambda\_1 = \lambda\_2 = \mathrm{e}^{-uv}$ satisfies (3) on the entire $uv$-plane. Then the data
$$
\omega\_1 = \lambda\_1\,\mathrm{d}u,\quad
\omega\_2 = \lambda\_2\,\mathrm{d}v,\quad
\kappa\_1 = \frac{u}{\lambda\_1}\,,\quad
\kappa\_2 = \frac{v}{\lambda\_2}
$$
satisfy (1) and (2) while $\kappa\_1$ vanishes along $u=0$ and $\kappa\_2$ vanishes along $v=0$. Applying the integration method from the above-mentioned answer, one can now use this data to construct the desired $f$.


Taking the solution $\lambda\_1=\lambda\_2 = \mathrm{e}^{-uv}$ of (3), one finds that, letting $F:\mathbb{C}\to\mathbb{C}$ be the entire holomorphic function that satisfies $F(0)=0$ and $F'(z) = \mathrm{e}^{iz^2/2}$, and, for each real $\sigma>1$, letting $S\_\sigma:\mathbb{C}\to\mathbb{C}$ satisfy $S\_\sigma(x+iy)= x + i\sigma y$, and letting $D\subset\mathbb{C}$ be a domain containing $0\in\mathbb{C}$ on which $F$ has an inverse $F^{-1}:D\to\mathbb{C}$ satisfying $F^{-1}(0)=0$, then the mapping $f\_\sigma:D\to\mathbb{C}$ defined by $f\_\sigma(z) = F\bigl(S\_\sigma\bigl(F^{-1}(z)\bigr)\bigr)$ has constant singular values $1$ and $\sigma$ while the corresponding functions $\kappa\_1$ and $\kappa\_2$ change sign along curves that meet transversely at the origin in $D$.


More generally, it is useful to switch points of view and let $F$ be the set of triples $(x;e\_1,e\_2)$ where $x\in\mathbb{R}^2$ and $(e\_1,e\_2)$ are an oriented orthonormal basis of $\mathbb{R}^2$. Then $F$ is a smooth $3$-manifold embedded naturally in $(\mathbb{R}^2)^3 = \mathbb{R}^6$.


Define the canonical $1$-forms $\omega\_i = e\_i\cdot \mathrm{d}x$ and $\omega\_{12} = e\_1\cdot\mathrm{d}e\_2$, and note that they satisfy
$$
\mathrm{d}\omega\_1 = -\omega\_{12}\wedge\omega\_2\,,\quad
\mathrm{d}\omega\_2 = \omega\_{12}\wedge\omega\_1\,,\quad
\mathrm{d}\omega\_{12} = 0.\quad
$$
Now, on $F\times\mathbb{R}^2$ with projection $(\kappa\_1,\kappa\_2):F\times\mathbb{R}^2\to\mathbb{R}^2$ onto the second factor, consider the exterior differential ideal $\mathcal{I}$ generated by the $1$-form $\theta = \omega\_{12}+\kappa\_1\,\omega\_1 -\kappa\_2\,\omega\_2$ (which is a contact form on the $5$-manifold $F\times\mathbb{R}^2$) and the pair of $2$-forms
$$
\Upsilon\_1 = (\mathrm{d}\kappa\_1-{\kappa\_1}^2\,\omega\_2)\wedge\omega\_1
\qquad
\Upsilon\_2 = (\mathrm{d}\kappa\_2-{\kappa\_2}^2\,\omega\_1)\wedge\omega\_2
$$
Then $\mathrm{d}\theta \equiv \Upsilon\_1 - \Upsilon\_2 \mod \theta$. Then $\mathcal{I}$ is involutive for the independence condition $\omega\_1\wedge\omega\_2\not=0$.


In particular, each of the integral curves of the rank 4 system
$$
\theta = \mathrm{d}\kappa\_1-\omega\_1-{\kappa\_1}^2\,\omega\_2
 = \mathrm{d}\kappa\_2-\omega\_2-{\kappa\_2}^2\,\omega\_1
 = \omega\_1-\omega\_2 = 0
$$
on $F\times\mathbb{R}^2$ can be thickened to an integral surface of $\mathcal{I}$ on which $\omega\_1\wedge\omega\_2$ is nonvanishing. For example, we have
the integral curve
$$
(x;e\_1,e\_2,\kappa\_1,\kappa\_2) = \bigl((t,t);\,\partial/\partial x,\,\partial/\partial y, \,\arctan(t), \,\arctan(t)\bigr),
$$
which extends to an integral manifold of $\mathcal{I}$ on an open domain $D$ containing the origin $(0,0)$. The corresponding solution $f$ will have each of $\kappa\_1$ and $\kappa\_2$ changing sign along a curve that meets the 'diagonal' transversely. (One could probably solve this initial value problem explicitly, but I don't have time to attempt that right now.)


