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I want to prove that any $\alpha(s)$ with arc parameter $s$ lying on a sphere with radius $R$ has curvature $k\geq1/R^2$.

I tried so many times to prove it and at this point it's pissing me off really hard. The conditions of arc length parameter $$ |\alpha'(s)|=1 $$ and lying in the sphere $$|\alpha(s)|=R$$ should be enough to prove this by deriving those equations etc. The only thing I have found with this is that $$ \{\alpha,\alpha',\alpha\wedge \alpha' \} $$ is a basis of the curve at each point. This works for nothing to obtain the curvature $k$, as curvature is defined as $$k(s) = |\alpha''(s)| $$ or as the function such that $$ \alpha''(s) = k(s)N(s)$$ where $N$ is in a vector in the direction of $\alpha''$.

I also tried methods such as starting first with plane curves of maximum radius, which gives me a curvature of $1/R^2$. Then saying that plane curves of less radius have curvature $1/\tilde{R}^2\geq 1/R^2$ as $\tilde{R}$ is smaller. But I don't have any clue for the other type of curves in the sphere.

Note: I have seen many similar posts but none address this problem but other properties about the curvature of a curve in a sphere.

puradrogasincortar
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  • An idea too vague for a hint. The great circle geodesics have curvature $1/R^2$, Perhaps you can see how your curve deviates from the geodesic joining two close points. – Ethan Bolker Jun 02 '21 at 16:07
  • Also related: here, here – Arctic Char Jun 02 '21 at 16:14
  • @ArcticChar I checked before those posts. The "curve on a sphere" post states an inequality about the curvature that is different from the one I have($k>1/R^2$ and $k>1/R$), so I guess one of both is not correct. Anyways, that post tells me to check $\frac{d}{ds}\langle \alpha,\alpha'\rangle$ but that simply tells me $\langle \alpha,\alpha''\rangle=-1$ and I can not tell how is that helpful. The other thing that uses is Cauchy-Schwarts inequality and even if I can use it, my professor didn't teach me that so I guess there's another way... – puradrogasincortar Jun 02 '21 at 16:27
  • @EthanBolker I'm supposed not to know what a geodesic is, so I don't think that's the way to go :( – puradrogasincortar Jun 02 '21 at 16:28
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    The curve $\sigma (\theta) = (r\cos(\theta/r), r\sin(\theta/r), 0)$ has curvature $1/r$. So $1/r$ is the correct scaling and your problem has a typo. – Arctic Char Jun 02 '21 at 16:36
  • IIRC, the Cauchy-Schwarz inequality is elementary linear algebra coming from the quadratic formula. If it really isn't available, however, another approach is to use the Darboux frame of $\alpha$, whose elements are the unit tangent $T = \alpha'$, the surface normal at $n = \alpha(s)$, and their cross product $V = n \times T$. The components of $T'$ in this frame turn out to be the normal curvature of the surface and the geodesic curvature. – Andrew D. Hwang Jun 02 '21 at 19:26
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    Do you know the Meusnier formula relating curvature and normal curvature of a curve on any surface? – Ted Shifrin Jun 02 '21 at 23:38

2 Answers2

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Let us denote the dot product of two vectors $\vec v$ and $\vec u$ by $\langle \vec v, \vec u \rangle$.
Parameterize the curve $\alpha = \alpha(s)$ with arc length $s$. Then, $|\alpha'(s)| = 1$. Because $\alpha$ is a part of the sphere, $|\alpha(s)| = R$. Differentiating $R^2 = \langle \alpha(s), \alpha(s) \rangle$, we obtain $2 \cdot \langle \alpha(s), \alpha'(s) \rangle = 0$, $\implies \alpha(s) \perp \alpha'(s)$. Differentiating $\langle \alpha(s), \alpha'(s) \rangle$ further, we get $\langle \alpha(s), \alpha''(s) \rangle$ $+ \langle \alpha'(s), \alpha'(s) \rangle$ $=0$. But, $|\alpha'(s)| = 1$ $\implies$ $\langle \alpha'(s), \alpha'(s) \rangle = 1$. So, $\langle \alpha(s), \alpha''(s) \rangle$ $=-1$.
Now, notice that $|\langle \vec v, \vec u \rangle | \leq |\vec v| \cdot |\vec u|$. Therefore, $1 = |\langle \alpha(s), \alpha''(s) \rangle | \leq |\alpha(s)| \cdot |\alpha''(s)|$ $= R \cdot |\alpha''(s)|$.
So, $$|\alpha''(s)| \geq \dfrac{1}{R}$$ $\blacksquare$.

KVS02
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I think I found a solution without using unknown concepts such as geodesics and geodesic curvature and will post the answer to my own question.

As $\alpha$ is lying on the sphere of radius R, $|\alpha(s)|=R$ and deriving the expression we obtain $\langle \alpha, \alpha' \rangle=0$.

Using the Frenet frame, we know that $\alpha = aT + bN + cB$. As $\langle \alpha, \alpha' \rangle=0 \implies a = 0$ so $\alpha = bN + cB$.

Deriving this and using the Frenet formulas we see that $$ \alpha' = -b(kT-\tau B) + c\tau N $$ As $\alpha$ is parametrised by arc length, $-bk=1$ and using that $|b|\leq R$ and taking absolute value, $$k=\frac{1}{|b|}\geq\frac{1}{R} $$ qed

puradrogasincortar
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