Can a division algebra over $\mathbb{R}^3$ be used to construct a counterexample to the hairy ball theorem? Suppose (for contradiction) that there is a (if necessary associative and/or normed) division algebra over $\mathbb{R}^3$. Is there a simple way to use this to construct a nonvanishing continuous tangent vector field on $\mathbb{S}^2$, and thus contradict the hairy ball theorem?
 A: If I'm not mistaken, you can fairly explicitly construct nowhere vanishing continuous tangent vector fields on $S^{n-1}$ from sufficiently nice multiplications on $\mathbb{R}^n$.

Theorem.  Let $n \geq 2$ be a positive integer, and suppose that $*$ is a bilinear map on $\mathbb{R}^n$ with the property that there is a two-dimensional subspace $W$ of $\mathbb{R}^n$ with the property that for all nonzero $y \in W$, the map $\mathbb{R}^n \to \mathbb{R}^n$ given by $x \mapsto y*x$ is invertible.  Then there is a nowhere vanishing continuous tangent vector field on $S^{n-1}$.

(Note that the hypothesis on $*$ is far weaker than the assumption that $*$ turns $\mathbb{R}^n$ into a division algebra.)
Some preliminaries before the proof.  Regard $\mathbb{R}^n$ as a vector space in the usual way, and let $\langle \cdot,\cdot\rangle$ denote the usual inner product on $\mathbb{R}^n$.  Identify $S^{n-1}$ with the subset $\{x \in \mathbb{R}^n: \langle x, x\rangle = 1\}$ of $\mathbb{R}^n$.  With this identification, for any $y \in S^{n-1}$ we can identify the tangent space to $S^{n-1}$ at $y$ with a subspace of $\mathbb{R}^n$; under this identification, the tangent space to $S^{n-1}$ at $y$ is precisely the subset $\{w \in \mathbb{R}^n: \langle w,y\rangle = 0\}$ of $\mathbb{R}^n$.  (This might be clearest to see when $n=3$: the set of all vectors tangent to the $2$-sphere at $y \in S^2$ is precisely the plane consisting of all vectors orthogonal to $y$.)

Proof of theorem. Choose any basis $\{e_1, e_2\}$ for $W$ and for $j = 1,2$ let $L_j$ denote the map $\mathbb{R}^n \to \mathbb{R}^n$ given by $x \mapsto e_j * x$.  Note that by our hypotheses on $*$ the maps $L_j$ are linear bijections.
Fix $y \in S^{n-1}$, define $X(y)$ in $\mathbb{R}^n$ as follows:
  $$
X(y) = L_2(L_1^{-1}(y)) - \frac{\langle L_2(L_1^{-1}(y)), y\rangle}{\langle y,y\rangle} y.
$$
  (This does define an element of $\mathbb{R}^n$, as we identify $S^{n-1}$ with a subset of $\mathbb{R}^n$, and $y \in S^{n-1}$ is nonzero.) 
I claim that $X(y)$ is tangent to $S^{n-1}$ at $y$.  As observed before the proof, it suffices to show that $X(y)$ is orthogonal to $y$ (in the usual sense of the inner product on $\mathbb{R}^n$).  But it clearly is; just do a calculation with the definition of $X(y)$ and use the bilinearity of $\langle \cdot,\cdot\rangle$.  (Note: $X(y)$ is the second vector in the two-element list that results from applying the usual Gram-Schmidt process, without normalization, to the two-element list $y, L_2(L_1^{-1}(y))$, so of course it is going to be orthogonal to $y$.)
I claim that $X(y)$ is nonzero.  In fact, any vector of the form $L_2(L_1^{-1}(y)) - \lambda y$ (for some scalar $\lambda$ and some nonzero $y \in \mathbb{R}^n$) will be nonzero.  To see this, note that since $y$ is nonzero and $L_1$ is a bijective linear map, there is a nonzero $z \in \mathbb{R}^n$ with $y = L_1(z)$.  Since $\{e_1, e_2\}$ is linearly independent, $e_1 - \lambda e_2$ is nonzero, and so by our hypothesis on $*$, the linear map $L: \mathbb{R}^n \to \mathbb{R}^n$ given by $x \mapsto (e_2 - \lambda e_1) * x$ is invertible.  And from the bilinearity of $*$ we have that
  $$
X(y) = L_2(L_1^{-1}(y)) - \lambda y = L_2(z) - \lambda L_1(z) = (e_2 * z) - (\lambda e_1) * z = (e_2 - \lambda e_1) * z = L(z)
$$
  is the result of applying the invertible linear map $L$ to the nonzero vector $z$.  Thus $X(y)$ is indeed nonzero.
The formula for $X$ therefore defines a nowhere-vanishing tangent vector field on $S^{n-1}$.  It is clear from the definition (since $L_1^{-1}$ and $L_2$ are linear maps, and the vector space operations on $\mathbb{R}^n$ and the inner product on $\mathbb{R}^n$ are continuous) that this vector field is continuous.  End of proof.

If I haven't lost my mind, with a similar idea you can explicitly show that if $*$ is bilinear on $\mathbb{R}^n$ with the property that the map $\mathbb{R}^n \to \mathbb{R}^n$ given by $x \to y*x$ is invertible for all nonzero $y$ in a $k$-dimensional subspace of $\mathbb{R}^n$, then there is a linearly independent set of $k-1$ nowhere vanishing vector fields on $S^{n-1}$. (Define $L_1, \dots, L_k$ appropriately, and consider the last $k-1$ vectors resulting from doing Gram-Schmidt on the list $y, L_2(L_1^{-1}(y)), \dots, L_k(L_1^{-1}(y))$.)  This would of course establish the well-known fact that if $\mathbb{R}^n$ is a division algebra, then the tangent bundle $TS^{n-1}$ not only has a nowhere vanishing section, but is in fact trivial.
A: Assume $n>1$ and that there exists a bilinear map $\mathbb{R}^n \times \mathbb{R}^n\to \mathbb{R}^n$ such that for all $a\ne 0$ the map $L_a\colon b\mapsto a\circ b$ is a linear isomorphism. Since $\mathbb{R}^n \backslash\{0\}$ is connected, all the linear maps $L_a$ have the same sign of the determinant. But $L_{-a} = - L_{a}$. This implies that $n$ is even.
This argument can be made for $S^{n-1}= \mathbb{R}^n \backslash\{0\}/ (0,\infty)$. Assume that we have a continuous multiplication $\mu\colon S^{n-1} \times S^{n-1}\to S^{n-1}$ such that $L_{a} \colon b\mapsto a\cdot b$ is a homeomorphism for all $a \in S^{n-1}$, and moreover, $L_{-a} = - L_a$. Since $L_a$, $L_{-a}$ are homotopic, we conclude that $L_{a}$, $-L_{a}$ are homotopic, so multiplying on the right by $L_{a}^{-1}$, $Id$ and $-Id$ are homotopic. One might produce from here in the smooth case, with some care, a vector field ( direction the opposite point!).
Let us just mention that if we are in the classical normed algebras, we can produce $n-1$ vector fields tangent to $S^{n-1}$, as follows: for every $i\ne 1$ in the standard basis of the algebra,  consider the vector field $X_i$ defined by $X_i(v) = i\cdot v$ ( very similar to the case of $\mathbb{C}$ with basis $1$, $i$, or $\mathbb{H}$ with basis $1$, $i$, $j$, $k$.
It is worth studying ( never did :=|) the Hurwitz algebras that produce the maximal number of independent tangent vector fields to $S^{n-1}.
$\bf{Added:}$ If there exist a structure of a division algebra on $\mathbb{R}^n$ then there exist a structure of a smooth division magma on $S^{n-1}$ : for every $a$ $b$ there exist unique $x$, $x'$ such that $a x = b$, $x'a=b$. Moreover, $x$, $x'$ depend smoothly on $a$, $b$.  Now  any manifold $M$ with such a multiplication is parallelizable. Indeed, take $a_0$ a fixed point, and $v_1$, $\ldots$, $v_n$ tangent vectors at $a_0$. For every $a\in M$ consider the unique translation $t$ taking $a_0$ to $a$. Use  $d t$ to map $v_i$ to tangent vectors at $a$, for all $a$. We get in this way $n$ smooth vector fields on $M$ that are linearly independent at each point.
