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A problem was given to me to prove that if the torsion of a curve is 0, then the curve lies on a plane. I proved instead that the curve must lie on a line (which obviously means that the curve then lies on a plane) but I'm a bit cautious as to whether this proof works because I've not been able to see anyone online prove the same thing. So I'm guessing I went wrong somewhere

Let $\gamma(t) : I\to\mathbb{R}^3$ be a regular space curve with $\tau(t)=0$ where $\tau$ represents the torsion of the curve. Now

$$\tau(t) = -\frac{b'(t)\cdot n(t)}{|v(t)|^2}=0$$ where $b(t), n(t)$ and $v(t)$ are the binormal, normal and velocity vectors of the curve respectively. By the fernet equations we know that $b'(t) = -|v(t)|\tau(t)n(t)$ so then $$-\frac{-|v(t)|\tau(t)n(t)\cdot n(t)}{|v(t)|^2}=0$$

so $n(t)$ must be the 0 vector which means that the velocity vector and acceleration span each other and lie on the same line as the perpendicular component of acceleration must be 0

This means that $$(v(t)\cdot v(t))' = (|v(t)|^2)'=2v(t)\cdot a(t) = 2|v(t)||a(t)| = 2|v(t)||v(t)|'$$

and thus

$$|a(t)| = |v(t)|'$$

Now consider the unit speed reparameterization of $\gamma(t)$ which must exist by the reparameterization theorem

in this case $|v(t)|$ = 1 and thus $|a(t)|$ = 0 so $$a(t)=0$$ as only the 0 vector has 0 magnitude and therefore $$v(t) = \vec{c}$$ for some vector $\vec{c}\in\mathbb{R}^3$ and therefore $$\gamma(t) = \vec{c}t+\vec{d}$$ for some vectors $\vec{c}, \vec{d} \in \mathbb{R}^3$

is there something wrong with this proof? and if not why do authors not specify zero torsion implies a curve lies on a line?

Dan Lupu
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    Are you asking specifically to review your proof or for simply a proof that 0-torsion curves are planar? If the latter, does this answer your question: https://math.stackexchange.com/questions/127894/prove-that-curve-with-zero-torsion-is-planar?noredirect=1&lq=1 ? – Derek Allums Oct 07 '24 at 09:18
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    "I proved instead that the curve must lie on a line". This is not true, that is true for curves with zero curvature, not zero torsion. – RicardoMM Oct 07 '24 at 11:46
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    Also, you write $\tau = 0$ and then say that a product that has $\tau$ in it implies $n \cdot n = 0$. – RicardoMM Oct 07 '24 at 11:49

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The proof has a mistake. You assumed that $\tau = 0$, so when you write:

$$-\frac{-|v(t)|\tau(t)n(t)\cdot n(t)}{|v(t)|^2}=0, $$ you can't conclude that $n(t) = 0$.

A way to do it would be to show that $b(t)$ is a constant, and from there conclude that the movement is in a plane (the plane defined by $r \cdot b = 0$).

RicardoMM
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