# Showing a curve which lays on a sphere of radius 1 is plane

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.

• Don't you need to define the space curve? Oct 11 '20 at 20:50
• @Narasimham Actually, no. I mean, the exercise just asks to prove it, just like I wrote. Oct 11 '20 at 20:52
• Are you asking if a unit speed curve on a sphere of radius $1$ with $\kappa = 1$ is a plane curve? Oct 11 '20 at 22:16
• @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. Oct 11 '20 at 22:20
• @mvfs314: thanks. Will see if I can come up with anything. Cheers! Oct 11 '20 at 22:22

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 $$\alpha '''= -\alpha'\implies \alpha'''\parallel\alpha '\implies \langle \alpha'\times \alpha'', \alpha'''\rangle =0\implies \tau = 0$$