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I realize this isn’t a well-put mathematical question, but it has been bugging me and so I ask it in hope of getting some feedback.

The one point compactification of 3-dimensional Real space $\mathbb{R}^3$ is homeomorphic to the 3-sphere $ S^3$. I emphasize that here $\mathbb{R}^3$ has no metric defined on it - it is not Euclidean space with the standard metric on it, but only the topological space $\mathbb{R}^3$.

Let’s say we one-point compactify $\mathbb{R}^3$ to $S^3$.

Now $S^3$ is isomorphic to the unit quaternions $H(u)$ with an isomorphism $F: H(u) \rightarrow S^3$. So any element “a” of $H(u)$ is represented by some point on $S^3.$

$H(u)$ contains the division algebraic binary operation * from which $a*b=c $ is also an element of $H(u).$ Thus $S^3$ should also contain an isomorphic operation *' on it for relation of the elements $ F(a) *'F(b) = F(c) $ in the same way that $H(u) $ relates its elements.

But assuming all this, then where does the operation *' come from on $S^3$ in its origination from the one-point compactification of $\mathbb{R}^3$? The operation doesn’t exist on $\mathbb{R}^3$ as a purely topological space. Furthermore, the operation isn't simply a topological property.

Where in my reasoning am I wrong? What branch of mathematics would deal with this kind of issue – differential topology? algebraic topology?

How much structure (the least possible) on $\mathbb{R}^3$ do I have to have in order that its one-point compactification to $S^3$ would produce an $S^3$ isomorphic to H(u)?

As an aside, this question can be stated identically using the one-point compactification of the real line $\mathbb{R}$ to the unit circle $S^1$ which is isomorphic to C(u) - the unit complex numbers.

Thanks.

Wolf
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    Why do you think that binary operation wouldn't exist on $\mathbb{R}^3$? There are plenty of binary operations on $\mathbb{R}^3$. – Steven Stadnicki Apr 06 '19 at 06:40
  • To Steven Stadnicki: Because I am taking $\mathbb{R}^3$ as existing only in it being a topological space with no such algebraic operations defined on it. I could define various operations on $\mathbb{R}^3$, but I want it to have the least amount of structure as possible. Even without this additional algebraic structure, it can be one-point compactified to the 3-sphere. – Wolf Apr 06 '19 at 07:01
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    There does not seem to be a well-formulated question here, but you do have to be careful when you restrict the algebra operations from $S^3$ to $S^3\setminus{p}$. You can multiply two unit quaternions in $S^3\setminu{p}$ and end up at $p$, so the operations don't restrict to binary operations on the subset. – Cheerful Parsnip Apr 06 '19 at 07:06
  • To Cheerful Parsnip: I'm not considering $S^3$/{p}. I'm considering $S^3$ which is generated from the one-point compactification of $\mathbb{R}^3$, in which $\mathbb{R}^3$ has not been given algebraic structures but is a purely topological space. – Wolf Apr 06 '19 at 07:27
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    You say that addition and multiplication make $H(u)$ a ring. Multiplication makes it a group, but addition does not work in $H(u)$. – Paul Frost Apr 06 '19 at 12:01
  • Paul Frost: I edited it to address this. The question is the same. – Wolf Apr 06 '19 at 12:25

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The answer is that $S^3$ one-point compactified from $\mathbb{R}^3$ is not in itself isomorphic to the unit quaternions H(u).

People often assume that it is because they see the term $S^3$. But the one-point compactified $S^3$ only has topological structure, and has no binary operation on it.

It is however the natural topological carrier space of H(u), and through the canonically isomorphic structure transport of both differential smoothness and the smooth division algebraic binary operation from the $S^3`$ which in fact is the differentiable manifold isomorphic to H(u) the topological $S^3$ inherits this structure.

See, for instance, Jack Lee's comments here: Why is there no natural metric on manifolds?

Wolf
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