# Relation of pseudo-torsion with curvature in degenerate plane

Question: I'd like to know if there is some reference or reasonable way to develop curve theory in a plane with degenerate metric $(\Bbb R^2, {\rm d}s^2 ={\rm d}x^2)$.

Context: In Lorentz-Minkowski space $\Bbb L^3 = (\Bbb R^3, \langle \cdot,\cdot\rangle_L = {\rm d}x^2+{\rm d}y^2 - {\rm d}z^2)$ one can consider unit-speed spacelike curves $\alpha\colon I \to \Bbb L^3$ with lightlike normal direction (meaning $T(s) = \alpha'(s)$, $N(s) = \alpha''(s)$). Assume that $(T(s),N(s))$ is a positive basis of the plane they span (for example, assume that the Euclidean cross product $T(s)\times_E N(s)$ is future-directed).

Say I now complete that frame with a lightlike "binormal vector" $B(s)$ orthogonal to $T(s)$ that makes the basis $(T(s),N(s),B(s))$ positive (with normalization constraint $\langle N(s),B(s)\rangle_L = -1$). We have a Frenet-like system $$\begin{pmatrix} T'(s) \\ N'(s) \\ B'(s)\end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & \tau(s) & 0 \\ 1 & 0 & -\tau(s)\end{pmatrix}\begin{pmatrix} T(s) \\ N(s) \\ B(s) \end{pmatrix},$$where $\tau(s)$ is the so-called pseudo-torsion of $\alpha$.

Assuming $\alpha(0) = 0$ and calling $\mathcal{F} = (T(0),N(0),B(0))$, Taylor expansion gives $$\alpha(s) - R(s) = \left(s, \frac{s^2}{2} + \tau(0) \frac{s^3}{6},0 \right)_{\mathcal{F}},$$for some $R(s)$ with $R(s)/s^3 \to 0$. This in particular hints that every curve in these conditions is a plane curve (this is actually true, not that hard to prove).

Projecting that in the osculating plane of the curve, we can consider $\gamma\colon I \to \Bbb R^2$ given by $\gamma(s) = (s, \frac{s^2}{2} + \tau(0) \frac{s^3}{6})$, where the metric in this $\Bbb R^2$ is only ${\rm d}x^2$.

I'd like to know if there is any reasonable notion of curvature for $\gamma$ related to the pseudo-torsion $\tau(0)$. Naively I could compute $$\kappa_\gamma(s) = \frac{\det(\gamma'(s),\gamma''(s))}{\|\gamma'(s)\|^3} = 1+\tau(0)s,$$but this is far from satisfactory for me.

• The difficulty is that the group of symmetries of the 'metric' $\mathrm{d}x^2$ on $\mathbb{R}^2$ is the set of transformations of the form $$\phi(x,y) = \bigl(\pm x + c, f(x,y)\bigr).$$ where $c$ is a constant and $f_y(x,y)$ is nowhere vanishing. In particular, any curve of the form $\bigl(s,g(s)\bigr)$ can be transformed into $(s,0)$ by the symmetry $\phi(x,y) = \bigl(x,y-g(x)\bigr)$. Thus, all curves on which $\mathrm{d}x$ is nonvanishing are locally equivalent under the symmetry group, so there is no local invariant that can distinguish them. – Robert Bryant Apr 9 '17 at 10:33

The difficulty is that the group of symmetries of the 'metric' ${\rm d}x^2$ on $\Bbb R^2$ is the set of transformations of the form $$\phi(x,y) = (\pm x+c, f(x,y)),$$where $c$ is a constant and $f_y(x,y)$ is nowhere vanishing. In particular, any curve of the form $(s,g(s))$ can be transformed into $(s,0)$ by the symmetry $\phi(x,y) = (x,y-g(x))$. Thus, all curves on which ${\rm d}x$ is nonvanishing are locally equivalent under the symmetry group, so there is no local invariant that can distinguish them.