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Assuming $\alpha$ is a unit speed curve, I'm trying to prove that $\alpha$ is plane. By hypothesis, I know its curvature is such that $$\kappa=1$$ I'm trying to use the torsion's formula: $$\tau=\frac{\langle \alpha'\times\alpha'',\alpha'''\rangle}{\|\alpha'\times\alpha''\|^{2}}$$ to show it's equal to zero, but I have no idea how to use it.

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  • $\begingroup$ Don't you need to define the space curve? $\endgroup$
    – Narasimham
    Oct 11 '20 at 20:50
  • $\begingroup$ @Narasimham Actually, no. I mean, the exercise just asks to prove it, just like I wrote. $\endgroup$
    – mvfs314
    Oct 11 '20 at 20:52
  • $\begingroup$ Are you asking if a unit speed curve on a sphere of radius $1$ with $\kappa = 1$ is a plane curve? $\endgroup$ Oct 11 '20 at 22:16
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    $\begingroup$ @RobertLewis the exercise I'm doing is asking to prove this: "If $\alpha$ is a curve that lays on a sphere of radius 1, show $\kappa\geq1$. Still, show if $\kappa=1$, then $\alpha$ is plane. The first part I made it, the problem is the last one. $\endgroup$
    – mvfs314
    Oct 11 '20 at 22:20
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    $\begingroup$ @mvfs314: thanks. Will see if I can come up with anything. Cheers! $\endgroup$ Oct 11 '20 at 22:22
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If $\alpha$ is a unit speed curve that lies on the sphere of radius 1 we have $\langle \alpha, \alpha\rangle = 1 = \langle \alpha^\prime, \alpha^\prime\rangle$. From this, it follows $2\langle \alpha', \alpha\rangle =0$. Then differentiating once more, we get $0=\langle \alpha'',\alpha\rangle + \langle \alpha',\alpha'\rangle = \langle \alpha'',\alpha\rangle + 1$ which implies $\alpha'' = -\alpha$, since $1=k=\lVert \alpha''\rVert$, and $\lVert\alpha\rVert = 1$. Finally we have $\alpha '''= -\alpha'\implies \alpha'''\parallel\alpha '\implies \langle \alpha'\times \alpha'', \alpha'''\rangle =0\implies \tau = 0$

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