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First I will quote Adam's theorem (also known as the Hopf invariant theorem) from here:

"The Hopf invariant one theorem, sometimes also called Adams' theorem, is a deep theorem in homotopy theory which states that the only n-spheres which are H-spaces are $\mathbb S^0, \mathbb S^1, \mathbb S^3,$ and $\mathbb S^7$. The theorem was proved by Adams (1958, 1960)."

Now in the definition of H spaces it says (here):

An H-space, named after Heinz Hopf, and sometimes also called a Hopf space, is a topological space together with a continuous binary operation $\mu:X×X \rightarrow X,$ such that there exists a point $e$ in $X$ with the property that the two maps $x\mapsto \mu(x,e)$ and $x\mapsto \mu(e,x)$ are both homotopic to the identity map $id_X$ on $X$, through homotopies preserving the point $e$. The element e is called a homotopy identity.

Now my question is: What is the binary operation $\mu$ on those spheres? I assume it has something to to with Hurwitz Theorem/the induced cross product from the Real Numbers, the Complex Numbers, the quaternions and the octonions but I can't figure out $\mu $ itself. I tried searching for the answer but I couldn't find it.

I hope you can help me

Sincerely

Slinshady

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    $\begingroup$ For $S^0$ / $S^1$ / $S^2$ / $S^3$ / $S^7$ binary operation is the product of unit norm real numbers / complex numbers / quaternions / octonions respectively $\endgroup$ – Grigory M Dec 26 '14 at 14:41
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The operation on $S^1$ is addition of angles (or complex multiplication of numbers of modulus 1). The one on $S^3$ is unit-quaternion multiplication. The one on $S^7$ is unit-"octonion" multiplication. (On $S^0 = \{-1, 1\}$, it's mutliplication of real numbers of absolute value 1, of course.)

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  • $\begingroup$ Thank you. For some reason, I thought of $\mathbb S^n$ as $\{x \in \mathbb R^n | |x|=1\}$ and not $\{x \in \mathbb R^{n+1} | |x|=1\}$ $\endgroup$ – slinshady Dec 26 '14 at 14:46
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    $\begingroup$ That would make it much, much harder. :) $\endgroup$ – John Hughes Dec 26 '14 at 14:49
  • $\begingroup$ Indeed. now it is all crystal clear. It is a pretty obvious map if one uses the corrct definitions.... $\endgroup$ – slinshady Dec 26 '14 at 14:51

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