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http://www.math.uwaterloo.ca/~karigian/talks/CUMC-2010.pdf

In this pdf it says that if you have a $S^{n-1}$ that admits a group structure, then you can get normed divison algebra.

How do you prove this? As I need to have a result like this as I'm trying to show there are only four Hopf maps. I'm trying to find an easy proof that there are only four Hopf maps, the easiest way I see is to prove that there are only four H-spaces that are spheres.

Anyone got any ideas how you prove the above.

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Here's an attempt that needs a bit of filling in.

Suppose $S^{n-1}$ admits a group structure. We will (try to) define a normed division algebra structure on $\mathbb{R}^n$. To start with, the addition is the standard vector addition and the norm is the usual length.

What about multiplication?

First, $0\cdot x = x\cdot 0 = 0$.

For $x$ and $y$ both nonzero, both $\frac{x}{|x|}$ and $\frac{y}{|y|}$ are points on the sphere so we can use the group structure on $S^{n-1}$ to define $\frac{x}{|x|}\cdot\frac{y}{|y|}$. Then define $x\cdot y = |x||y|(\frac{x}{|x|}\cdot \frac{y}{|y|})$.

To divide by $x\neq 0$ on the left, multiply by $\frac{1}{|x|}(\frac{x}{|x|})^{-1}$ on the left, ($\frac{x}{|x|}$ lives on $S^{n-1}$, so has an inverse).

This verifies ALL of the desired properties, except I can't seem to show it's distributive.

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    Thanks you are a big help. Especially since you helped me yesterday aswell.2012-04-20
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    Don't think you need distributive as I think that's too strong.2012-04-20
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    @simplicity: If you don't have distributivity then you don't have an algebra of any reasonable kind.2012-04-20
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    @Simplicity - I'm still planning on doing that other question, but it's going to take longer.2012-04-20
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    @ZhenLin I think associativity is trivial. Can't you just start with associative on the sphere and do the same argument. So you have $x \ast y \ast z = |x||y||z|(\frac{x}{|x|} \cdot \frac{y}{|y|} \cdot \frac{z}{|z|})$, but I thought it wouldn't matter but can't get inverses to work without associativity.2012-04-20
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    @JasonDeVito I think I understand the other question now, but have trouble checking it's a fibration. Thanks for the help through.2012-04-20
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    @simplicity: Associativity and distributivity are two different things.2012-04-20
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    @ZhenLin Yeah, can see it's wrong now. I have no clue how to prove it now.2012-04-21