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I would like to know of any parametrization of the standard 3-sphere:

{$(x_1,x_2,x_3,x_4): x_1^2+x_2^2+x_3^2+x_4^2=1$} embedded in $\mathbb R^4$.

I know of parametrizations for $S^1$, for $S^2$ , but I cannot think of how to parametrize $S^3$ as above. The closest I found in a search was a formula using quaternions; is it possible (I would prefer, if possible) to avoid using quaternions.

Thanks for any ideas

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2 Answers 2

up vote 3 down vote accepted

If

$$\sin^2 u + \cos^2 u =1$$

then

$$(\sin v \sin u)^2 + (\sin v \cos u)^2 = \sin^2 v$$

so

$$(\sin v \sin u)^2 + (\sin v \cos u)^2 +\cos^2 v = 1.$$

Can you repeat the same procedure once more? After that, you will have to delimit the values of the parameters if you want to parametrize the sphere exactly once.


Alternatively, you can use the formula

$$(a^2-b^2-c^2-d^2)^2 + (2ab)^2 + (2ac)^2 + (2ad)^2 = (a^2+b^2+c^2+d^2)^2,$$ which gives you a parametrization of the sphere by rational functions.

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How is it possible to parametrize the 3-sphere with just two parameters $u,v$ , like you did in the top part? It seems that if you could use just 2 parameters, you would get the 2-sphere $S^2$. –  user99680 Nov 8 '13 at 1:54
    
@user99680 Exactly, what I wrote is the parametrization of the $2$-sphere. Which I got from the parametrization of the circle by a simple manipulation. Now I'm saying that you should apply the same manipulation once more: multiply both sides of the equation by $\sin^2 w$ for some third parameter $w$, and then... –  Bruno Joyal Nov 8 '13 at 1:56
    
:Oh, I see, thanks; let me try it. –  user99680 Nov 8 '13 at 1:59
    
@user99680 My pleasure. Let me know if you need more help. –  Bruno Joyal Nov 8 '13 at 2:03
    
:So I get: $(sinwsinvsinu)^2+(sinwsinvcosu)^2+(sinwcosv)^2+cos^2w=1$ . Is this correct? –  user99680 Nov 8 '13 at 2:08

You can repeat the "sines and cosines" process one more time if you want. Like our friend @Bruno Joyal said:

For the 1-sphere:

$X(\phi)=(\cos(\phi),\sin(\phi))$, so that its squared coordinates sums up to 1.

$\cos^2(\phi) + \sin^2(\phi) = 1$

For the 2-sphere:

$X(\phi,\psi)=(\cos(\phi)\cdot\cos(\psi),\sin(\phi)\cdot \cos(\psi),\sin(\psi))$, so that its squared coordinates sums up to 1.

$\cos^2(\phi)\cdot\cos^2(\psi) + \sin^2(\phi) \cdot \cos^2(\psi) + \cos^2(\psi) = $

$=(\cos^2(\phi) + \sin^2(\phi))\cdot \cos^2(\psi)+\sin^2(\psi)=1$

Repeat the process by multiplying the coordinate functions by $\cos(\theta)$ and adding another coordinate function $\sin(\theta)$, obtaining:

$X(\phi,\psi,\theta)=(\cos(\phi)\cdot\cos(\psi)\cdot \cos(\theta),\sin(\phi)\cdot \cos(\psi)\cos(\theta),\sin(\psi)\cdot\cos(\theta),\sin(\theta))$

You may quickly verify that its squared coordinates sums up to 1. This last $X$ is a parametrization of a 3-sphere in $\mathbb{R}^4$.

You may repeat this process as much as you like to obtain the parametrization of a n-sphere in $\mathbb{R}^{n+1}$.

As pointed out by @user99680, you may not parametrize a 3-sphere with just 2 parameters as a 3-sphere is what we call a 3-manifold. Hence it has 3 dimensions and every little piece of it is homeomorphic to an open set in $\mathbb{R}^3$. If there was a way to parametrize a 3-sphere with just 2 parameters, every little neighbourhood of it would be homeomorphic to an open set of $\mathbb{R}^2$. This would imply that we could establish an homeomorphism from an open set of $\mathbb{R}^3$ to an open set of $\mathbb{R}^2$, which is absurd.

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