Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Here I was shown how to prove that $TS^1$ is a trivial bundle.

Similarly, I can show that $TS^3$ is a trivial bundle.

Identify $\mathbb{H}$ with $\mathbb{R}^4$ and that that for $x \in S^3$ we have $x \mapsto ix$, $x \mapsto jx$ and $x \mapsto kx$ as orthogonal vectors, which are also orthogonal to $x$. Then we can see that there is an ismorphism $S^3 \times \mathbb{R}^4 \to S^3 \times \mathbb{R}^3$ given by $(x,z) \mapsto (x,iz_1/x \times iz_2/x \times iz_3/x)$ where $z_1 = \{ t_1 i x| t_1 \in \mathbb{R} \},z_2 = \{ t_2 j x| t_2 \in \mathbb{R} \},z_3 = \{ t_3 k x| t_3 \in \mathbb{R} \}$ (Possibly I haven't written that last bit so nicely).

It is also possibly to do similar construction for $S^7$ using Octonions.

This then leads onto the general question:

Prove that $TS^{n-1}$ is a trivial bundle, if $\mathbb{R}^n$ may be provided with an $\mathbb{R}$-algebra structure without zero divisors.

Of course, now my construction kind of fails - I can't very well construct the map $x \mapsto ix$ (for example), because I don't have a multiplication table! So what is the more general way to approach this problem? (From a simple algebra approach - let's not invoke Bott periodicity or anything just yet!)

Edit: Maybe the question should be - What property of $\mathbb{R}$-algebra structures without zero divisors, helps to solve this problem?

Edit 2: (The following is probably wrong). So assume that $\mathbb{R}^n$ has an $\mathbb{R}$ algebraic structure without zero divisors. We seek an isomorphism $$\phi:S^{n-1} \times \mathbb{R}^n \to S^{n-1} \times \mathbb{R}^{n-1}.$$ Denote our algebraic structure by $\mathbb{T}$, and let it has orthonormal basis $\{ e_0, \cdots, e_{n-1} \}$. Then $S^{n-1}$ is the subset of $\mathbb{T}$ with (Euclidean norm) = 1 (I think I can still define the Euclidean norm?). Then given $z \in S^{n-1}$, $\{ ze_0, \cdots, ze_{n-1} \}$ is still an orthonormal basis (maybe). Then define the function

$$(x,z) \mapsto (x,iz_1/x \times \cdots\times iz_{n-1}/x)$$ where $z_i = \{ t_i e_i x| t_i \in \mathbb{R} \}$

(the division exists because $x$ is orthogonal to $z$ and we have no zero divisors)

share|cite|improve this question
up vote 7 down vote accepted

Suppose that $\mathbb R^n$, with basis $(e_1,...,e_n)$, has the structure of a division ring and denote by $a.b$ the product of the vectors $a$ and $b$. Consider $x\in S^{n-1}$; then the vectors $e_1.x, ...,e_n.x$ are linearly independant (cf. right multiplication by $x^{-1}$) and you can apply the Gram-Schmidt process to them, obtaining $v_1(x),...,v_n(x)$ with $v_1(x)=e_1.x/||e_1.x||$. The vectors $v_2(x),...,v_n(x)$ are an orthonormal basis of the tangent space to $S^{n-1}$ at the point $v_1(x)$ and so the $(n-1)$ vectors $ v_i(v_1^{-1}(x) \quad (i=2,...,n)$ are an orthonormal basis of the tangent space $T_x(S^{n-1})$ of the sphere at $x$. This is the desired trivialization of the tangent bundle $TS^{n-1} $.

1) To be perfectly explicit , the formula for $v_1^{-1}$ is $ v_1^{-1}(y)=e_1^{-1}.y/||e_1^{-1}.y||$.
2) The euclidean scalar product and norm are the usual ones on $\mathbb R^n$ and have absolutely no interference with the division algebra structure: let us be sure not to confuse the vector $a.b \in \mathbb R^n$ and the scalar product $< a, b>\in \mathbb R$ !
3) However the quaternions have a supplementary structure: the conjugation which is an $\mathbb R$-linear anti-involution. This allows the quaternions to see the the euclidean structure of $\mathbb R^4$ thanks to the pleasant formula linking the division ring structure and the metric structure $$v.\bar v=||v||^2 1_{\mathbb H} \quad ( v\in \mathbb H)$$

share|cite|improve this answer
thank you for the nice answer! This is the picture I had in my head, just written down properly. If we started with $(e_1,\ldots,e_n)$ as an orthonormal basis, then multiply by $x$, does this give us the orthonormal basis? (Of course as you point out, we can just use G-S anyway, but I am just interested) – Juan S May 4 '11 at 23:27
No, because multiplication by $x$ in the division ring is not an orthogonal linear map. You see, $e_1.x$ is what the division ring decides it is and if you decree that $||e_1||=1$ the division ring doesn't see that, doesn't care and won't oblige you by arranging that $||e_1.x||=1$ also. Similarly the division ring doesn't see orthogonality. In functional analysis however, for example in Banach algebra theory, axioms are introduced relating the algebraic structure to the metric structure: for example $||x.y|| \leq ||x||||y||$. – Georges Elencwajg May 5 '11 at 0:27
OK, thanks for clearing that up. There is probably some background to this question I am not familar with! – Juan S May 5 '11 at 0:39
I upvoted this ages ago, but coming back, I'm missing two things (which are probably obvious). First, in general, having no zero divisors is weaker than being a division algebra. Are you simply assuming this stronger condition in order to get the desired result or does the stronger condition follow from the weaker in this particular case? Second: Why is the given trivialization smooth (or continuous)? – Jason DeVito Feb 27 '13 at 20:00

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.