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One of the consequences of the Hopf invariant one problem is that $\mathbb{R}^n$ is a division algebra only for $n=1,2,4,8$. A division algebra structure $\odot$ on $\mathbb{R}^n$ need not play nicely with the norm: we could have $|x \odot y| \neq |x| |y|$. However, in the proofs I've read (for instance in Hatcher's notes), it seems to be taken for granted that the division algebra plays nicely with the norm.

Is there a way to lose this assumption?

Eric Auld
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    Where does Hatcher use the assumption that $|xy| = |x|,|y|$? He's not putting any restriction on the $H$-space multiplication map $\mu(x, y) = xy/|xy|$ beyond $\mu(x, 1), \mu(1, x) = x$. – anomaly Jul 04 '17 at 19:08
  • There is more than one classification theorem concerning real division algebras. As Dietrich mentions, Hurwitz's theorem is about normed division algebras. On the other hand, it is a also theorem that the only division algebras occur with dimension $1,2,4,8$ with no assumption of a norm. Sources that prove that claim do not assume a norm. You should have no issue finding more of these sources - you even link to one, Hatcher! One issue is that some sources may hastily present what is essentially Hurwitz's theorem without mentioning the norm in the theorem statement. – anon Jul 04 '17 at 19:34

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The proof presented in Hatcher's notes makes no assumption that $|x\odot y|=|x||y|$. Given any division algebra structure $\odot$ on $\mathbb{R}^n$, you get an H-space structure on $S^{n-1}$ defined by $(x,y)\mapsto \frac{x\odot y}{|x\odot y|}$. That is, you can always just rescale the multiplication so that it preserves the norm, and that will give you an H-space structure on $S^{n-1}$ (though it may no longer give a division algebra structure since multiplication may no longer be bilinear).

Eric Wofsey
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Indeed, Hurwitz Theorem only holds under the assumption that $\|x\cdot y\|=\|x\|\|y\|$ for all $x,y\in A$, i.e., for composition algebras.

Theorem(Hurwitz 1898): Suppose that $A$ is a real finite-dimensional unital algebra, and suppose that $A$ is a Hilbert space with $\|x\cdot y\|=\|x\|\|y\|$ for all $x,y\in A$. Then $A$ is one of the division algebras $\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}$.

Dietrich Burde
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