Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\gamma_0$ and $\gamma_1$ be two paths in $\mathbb{S}^2$ (the 2-d unit sphere in $\mathbb{R}^3$). Let both $\gamma_0$ and $\gamma_1$ start at $p \in \mathbb{S}^2$ and end in $q \in \mathbb{S}^2$. What is an explicit formula that is a homotopy from $\gamma_0$ to $\gamma_1$, using intermediate curves that must all lie on $\mathbb{S}^2$ and connect $p$ to $q$.

share|improve this question
1  
stereographic projection to the plane, $t\gamma_1+(1-t)\gamma_0$ go from the plane back to the sphere. –  butt Aug 9 '12 at 21:45
    
@butt But what if one of the curves passes through the north pole? I don't think there is a nice explicit formula unless you assume the curves aren't surjective. –  MartianInvader Aug 10 '12 at 0:03
    
This seems a good case where the Seifert-van Kampen Theorem tells you that a homotopy exists, but does not give an explicit formula, as the proof of the theorem involves subdivisions. Why should an explicit formula be expected? –  Ronnie Brown Aug 10 '12 at 8:36
add comment

2 Answers

If you're sure that, for every $t$,

$$ \gamma_0(t) \neq - \gamma_1(t) \ , $$

that is, for every $t$, $\gamma_0(t)$ and $\gamma_1(t)$ are not antipodal points, then you can use the straight line homotopy in disguise:

$$ H(s,t) = \dfrac{(1-s) \gamma_0(t) + s \gamma_1(t)}{\vert\vert (1-s) \gamma_0(t) + s \gamma_1(t) \vert\vert} \ . $$

share|improve this answer
add comment

The following works when $\gamma_0$ and $\gamma_1$ both miss a point $x$ of $S^2$. Which is nearly always - if you really want to use a space-filling curve or something, then you'll have to first construct a homotopy that moves the curve away from a point you choose on the sphere. This is easy to describe but hard to construct explicitly.

First of all, let's solve the problem in $\mathbb R^2$. There, it's simple: just move along the straight line between $\gamma_0(t)$ and $\gamma_1(t)$, i.e. \[H(s,t)=s\gamma_0(t)+(1-s)\gamma_1(t)\]. Then for your paths on the sphere, use stereographic projection, with the point $x$ that both paths missed as your projection point, solve the problem in the plane, and project back. There are explicit formulae for that projection, but finding them and applying them I leave an exercise to the reader.

It's mostly unenlightening in any case to have the explicit formulae. Algebraic topologists are more often than not satisfied with descriptions or even pictures - usually you're far more interested in if a homotopy exists than what it precisely is.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.