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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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    @ZhenLin Yeah, can see it's wrong now. I have no clue how to prove it now.2012-04-21