Closed Operations in Spherical Space

Question

Euclidean vector arithmetic (addition/subtraction) is not closed under a spherical space. For example:

$\mathbb{S}^2=\{v \in{\mathbb{R}^3}|\|v\|=1\}$
We have $(1, 0, 0)\in\mathbb{S}^2$ and $(0, 1, 0)\in\mathbb{S}^2$.
However, $(1, 0, 0)+(0, 1, 0)=(1, 1, 0)\notin\mathbb{S}^2$.

Are there "similar" arithmetic operators that are closed under spherical spaces?

I'm interested in high-dimensional spaces, so an answer for any dimension $n$ would be greatly appreciated.

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

1

Yes; one can view the sphere as the complex numbers plus infinity, and then both addition and multiplication are closed operations.

Also, for points that are not opposite of each other, you can take their sum as you did above and then divide that by its length to get a new point on the sphere.

You can also view two points on the sphere as unit vectors. If they are not perpendicular, you can take their cross product and normalize the resulting vector to be a new unit vector.

Edit: You can't define a nice addition on spheres except those of dimension 1 or 3:Is there an easy way to show which spheres can be Lie groups?. So associativity or inverses or an identity must not work out for the 2-sphere.

edited Apr 13 '17 at 12:20

Community

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answered Jan 15 '14 at 13:49

Brian Rushton

13,375

How can the complex numbers (plus infinity) be viewed as the unit sphere? – Omer Jan 15 '14 at 15:15
Allow me to refine - how can this be extended to $n$ dimensions? – Omer Jan 16 '14 at 06:39
Regarding your suggestion for normalizing after addition - this operator is neither associative nor commutative. – Omer Jan 16 '14 at 06:59
I edited my answers. The complex numbers plus infinity is called the Riemann sphere. – Brian Rushton Jan 16 '14 at 12:40

Closed Operations in Spherical Space

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