# 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

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)

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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 $TS^{n-1}$.

Remarks
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 $$v.\bar v=||v||^2 1_{\mathbb H} \quad ( v\in \mathbb H)$$

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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