Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
What do you mean by the identity element of S7? Are you assuming some Lie group structure? – Helmut Apr 20 '12 at 12:59
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.