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Context: Consider the Lorentz-Minkowski space $\Bbb L^3$, with the metric: $$ds^2 = dx^2 + dy^2 - dz^2$$

(which I'll denote just by $\langle \cdot, \cdot \rangle$)

If $\alpha$ is spacelike and ${\bf N}(s)$ is lightlike for all $s \in I$, the Frenet equations are:

$$\begin{pmatrix} {\bf T}' \\ {\bf N}' \\ {\bf B}' \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & \tau & 0 \\ -1 & 0 & -\tau \end{pmatrix}\begin{pmatrix} {\bf T} \\ {\bf N} \\ {\bf B} \end{pmatrix}$$ where $\alpha$ is assumed parametrized by arc-length, the tangent is ${\bf T}(s) := \alpha'(s)$, the normal is ${\bf N}(s):= {\bf T}'(s)$ and the binormal ${\bf B}(s)$ is the only lightlike vector such that $\langle {\bf B}(s), {\bf T}(s) \rangle = 0$ and $\langle {\bf B}(s), {\bf N}(s)\rangle = 1$.


The problem:

Definition: A curve $\alpha: I \to \Bbb L^3$ is a $\bf T$-helix if exists a non-zero, constant vector $\bf v$ such that $\langle {\bf T}(s), {\bf v}\rangle$ is a constant $ c \in \Bbb R$.

I'm trying to characterize all $\bf T$-helices in $\Bbb L^3$, and I'm having trouble in this last case: when $\alpha$ is spacelike and ${\bf N}(s)$ is lightlike for all $s \in I$. My strategy in all of the other cases was to simply suppose that $\alpha$ is a $\bf T$-helix, find some condition on the curvature (when it exists) and torsion, and then try to "make my way back". I'm failing here.

Suppose $\alpha$ is a $\bf T$-helix. Then exists a constant non-zero vector $\bf v$, and a constant $c$ such that $\langle {\bf T}(s), {\bf v}\rangle = c$. Taking derivatives, we have $\langle {\bf N}(s), {\bf v}\rangle = 0$. If we differentiate this again, we conclude nothing, because in the end, we're just multiplying the last expression by $\tau(s)$.

It is easy to check, by applying the Lorentzian inner product with $\bf T, N$ and $\bf B$, that: $${\bf v} = \langle{\bf T}(s), {\bf v}\rangle {\bf T}(s) + \langle{\bf B}(s), {\bf v}\rangle {\bf N}(s) + \langle{\bf N}(s), {\bf v}\rangle {\bf B}(s)$$ So, in our situation, we have that: $${\bf v} = c {\bf T}(s) + \langle{\bf B}(s), {\bf v}\rangle {\bf N}(s)$$ Differentiating this, we get $0 = 0$, that is, nothing.

Can someone help me out of this trap?

Ivo Terek
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1 Answers1

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The key is that every spacelike curve with lightlike normal vector is a T-helix; see Theorem 2.13 in R. López, Differential geometry of curves and surfaces in Lorentz-Minkowski space, IEJGEO 7 (2014), 44-107. https://doi.org/10.36890/iejg.594497

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    While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. – user37238 Jan 21 '16 at 09:55