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I was wondering about a proof for the result that the only spheres which permit a group structure are $S^0, S^1$ and $S^3$. The idea of the proof being that a group structure on $S^{n-1}$ induces a normed division algebra on $\mathbb{R}^{n}$.

More specifically, $\mathbb{R}^{n}$ is a vector space over $\mathbb{R}$, and so we just need to equip it with an appropriate binary operation known as multiplication. Defining $x \cdot 0 = 0 \cdot x = 0$ for all $x \in \mathbb{R}^{n}$, and then for any non-zero $x, y \in \mathbb{R}^{n}$ we define multiplication by: $$ x \cdot y = N(x)N(y) \left(\frac{x}{N(x)}\cdot \frac{y}{N(y)}\right) $$ where $N(x)$ denotes the norm of $x$, and the elements inside the brackets can be multiplied using the group multiplication. Then I just need to prove that this multiplication is left and right distributive, as well as compatible with scalars. If I can prove this, then we have that: \begin{align*} N(xy) &= N\left(N(x)N(y)\left(\frac{x}{N(x)}\cdot \frac{y}{N(y)}\right)\right) \\ &= N(x)N(y) N\left(\frac{x}{N(x)}\cdot \frac{y}{N(y)}\right) \text{ by the compatibility of scalars with norms} \\ &= N(x)N(y) \text{ as } \frac{x}{N(x)}\cdot \frac{y}{N(y)} \in S^{n-1} \end{align*} So we have the normed property, and also: \begin{align*} x\cdot y = 0 \implies N(x)=0 \text{ or } N(y)=0 \implies x=0 \text{ or } y=0 \end{align*} So we also have the division property, and hence $\mathbb{R}^{n}$ must be a normed division algebra. But by Hurwitz Theorem, this means it must be that n=1, 2, 4 or 8. But the octonions are non-associative so the group structure could not be associative i.e. it could not have been a group, and hence we must have that the only group structures on spheres are $S^0, S^1$ and $S^3$.

I think the proof is okay and works, but I was hoping someone could help with proving distributivity and compatibility with scalars for the algebra, as I'm not having any luck. Also of course if there are any errors or general feedback, I would very much appreciate it.

Thanks!

YCor
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    It follows form this post. – Dietrich Burde Feb 27 '19 at 19:33
  • Mmm okay, I was hoping to prove it directly in this way though and present it as a corollary of Hurwitz theorem. Any ideas? –  Feb 27 '19 at 19:44
  • You mean topological group structures, or Lie group structure? (the answers are the same, but not of the same difficulty). This is topological/differential, so in any case it's quite absurd to tag this "abstract algebra". – YCor Feb 27 '19 at 20:39
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    I think your approach is hopeless. Indeed, take the standard group law on $S^1$. Conjugate it by a non-isometric self-homeomorphism of $S^1$. Then most likely the kind of "scalar product" constructed resulting group law will fail to be bilinear. – YCor Feb 27 '19 at 20:43
  • @DietrichBurde the question is not the classification of skew-fields over $\mathbf{R}$ or any non-associative variant, but to reduce the more difficult question of which spheres can be made a topological group, to this algebraic fact. – YCor Feb 27 '19 at 20:52
  • @YCor Yes I meant topological group structures - my apologies, it was late when I posted this. Thank you very much for fixing the tags. As for the proof itself, that's really unfortunate that it doesn't seem to work, but thank you for pointing that out to me. –  Feb 28 '19 at 03:35
  • Would there be no way to recover bilinearity, perhaps by modifying the defined multiplication? –  Feb 28 '19 at 03:46

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