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I would like to show that $S^7$, the 7-sphere, is a parallelizable manifold. Let $\mathcal{O}$ be the octonions, the normed division algebra (noncommutative, nonassociative) over $H\times H$, where $H$ is the quaternion algebra. Using $\mathcal{O}$ we can define a parallelization of $S^7$. Would someone explain this construction?

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The sphere $S^7$ can be identified with the unit octonions (the octonions of norm 1). If $v$ is any nonzero tangent vector at the identity element $1 \in S^7$, then we can define a continuous nonzero vector field on $S^7$ by assigning to each point $x \in S^7$, the vector $v$ "multiplied" by $x$ using the octonion multiplication. I'll leave it to you to make this precise.

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What do you mean by the identity element of S7? Are you assuming some Lie group structure? –  Helmut Apr 20 '12 at 12:59
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A Ted himself said, the sphere can be viewed as the set of unit octonions, which has a (non-associtive) multiplication with a unit element. –  Mariano Suárez-Alvarez Apr 20 '12 at 13:35
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Octonions have the form $x_0 + x_1 e_1 + x_2 e_2 + \ldots + x_7 e_7$, which can be identified with the vector $(x_0, x_1, \ldots, x_7) \in \mathbb{R}^8$. The unit octonions are those for which $\sum x_i^2 = 1$, which make up an $S^7$ in $\mathbb{R}^8$. The identity octonion is 1, which corresponds to the point $(1,0,\ldots,0)$. –  Ted Apr 20 '12 at 16:09

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