Property of parameterized by arc length curves

Question

Problem:

Suppose $\gamma:[0,L] \to \mathbb{R}^3$ is a closed and regular curve parameterized by arc length. Thus we have that $\|\gamma'(s)\|=1$ for all $s \in [0,L]$. Define $t(s)=\gamma'(s)$ for $s \in [0,L]$ where $t:[0,L] \to \mathbb{S}^2$. Prove that $t([0,L])$ is not contained in an open hemisphere of $\mathbb{S}^2$.

Attempt:

I tried proving that there must exist $x,s \in [0,L]$ such that $t(s)=-t(x)$. In this case obviously $t([0,L])$ is not contained in an open hemisphere of $\mathbb{S}^2$. But I am not sure if do exist $x,s \in [0,L]$ such that $t(s)=-t(x)$. Any help will be appreciated.

Is your curve closed? Other wise just taking a line segment might not work , as there is a unit speed parametrization such that the tangent is constant . Rectify me if I am wrong. — Noob mathematician, Aug 12 '21 at 21:23
What you're trying to prove certainly need not be true. But the statement is not true, as it stands. What if $\gamma$ is a line? — Ted Shifrin, Aug 12 '21 at 21:23
@Noobmathematician I edited my question. The curve is closed as you said. — Filippo Giovagnini, Aug 12 '21 at 21:56

score 3 · Accepted Answer · answered Aug 12 '21 at 22:11

3

HINT: Suppose the $z$-coordinate of $\gamma’$ is always positive. What goes wrong?

answered Aug 12 '21 at 22:11

Ted Shifrin

115,160

Then the $z$-coordinate of $\gamma$ is strictly increasing? – Filippo Giovagnini Aug 13 '21 at 07:21
This is more a genuinely question @Ted that not an hint: with a similar mindset could you consider a function "distance of the curve from $\gamma(0)=\gamma(L)$" and get to a similar consideration? – Augusto Matteini Aug 13 '21 at 09:37
@SMC Could you please add more details? – Filippo Giovagnini Aug 13 '21 at 10:03
Honestly I don't even know if this could lead to a clean proof while I'm almost sure that Ted hint is the "neat way to prove it" but here is my reasoning: what I want to use is as the curve is closed "sometimes somewhere" my tangent vector has to "change orientation" (here note that orientation is not used in the mathematical sense), to me this "feels" related to distance from the $\gamma(0)$ point and I was wondering if we could make this more rigorous, using that the distance "i walk in a way" i must make in the other as the curve is closed. I repeat that is by far from a mathematical 1/2 – Augusto Matteini Aug 13 '21 at 10:07
argument and that Ted's hint is much more direct and easier to compute. I suspect they are "morally" the same thing. On a note I introduced the norm cause I wanted an elementary proof via Lagrange Theorem but i'm still trying. – Augusto Matteini Aug 13 '21 at 10:10
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@SMC The Mean Value Theorem does not hold in the form of an equality for vector-valued functions! Yes, Filippo, that’s all there is to it. – Ted Shifrin Aug 13 '21 at 13:56
Yes that was clear to me and why I was trying to use the norm in some way looking to my composition as a map from $\mathbb{R}\to \mathbb{R}$. But "I can believe" that this does not lead to anything useful. – Augusto Matteini Aug 13 '21 at 14:30
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You both might want to take a look at the discussion on pp. 24-25 of my differential geometry text (linked in my profile). The gist is that the tangent indicatrix of a closed curve must be "balanced" in every direction. – Ted Shifrin Aug 13 '21 at 15:21

Property of parameterized by arc length curves

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