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I've recently been thinking about various problems involving two points on the surface of a unit sphere. Let's specify them with a pair of unit 3-vectors ${\bf \hat a}$ an ${\bf \hat b}$. Is there some aid to thinking about this 4-dimensional space? In particular:

  1. What is its topology? (In terms suitable for dumb engineers like me.)
  2. How do I integrate over parts of this space? The measure would derive from surface area on the original sphere but the integrands of interest are probably all simple functions the scalar ${\bf \hat a\cdot \hat b}$
  3. Is there a coordinate space or other model that makes it easy to deal with questions like (1) and (2).
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  • $\begingroup$ I don't see how you define a 4-D space with two vectors in the 3-sphere. Would you explain? $\endgroup$
    – user99680
    Jun 7 '14 at 5:51
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    $\begingroup$ The 3-sphere is a two dimensional space (just a surface). Two independent points on a two dimensional space gives four dimensions in total. $\endgroup$ Jun 7 '14 at 6:31
  • $\begingroup$ That's not how dimensions work. $\endgroup$ Jun 7 '14 at 7:21
  • $\begingroup$ @user99680: He is talking about two points in the 2-sphere, not in the 3-sphere. $\endgroup$
    – Lee Mosher
    Jun 7 '14 at 12:48
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This is $S^2 \times S^2$; you can give it the product topology, so that a basis for the topology is given by $\{U \times V \mid U, V \text{ open } \subset S^2\}$. In order to think about 1 and 2, it might help to think of it as a manifold; you can use polar coordinates to homeomorphically biject open sets to open sets of $\mathbb{R}^4$, and you can integrate by using the change of coordinates formula:

https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables

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  • $\begingroup$ That's a very nice first answer. Keep it up! $\endgroup$
    – M. Vinay
    Jun 7 '14 at 8:45

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