Question: Can someone please give a clear explanation, or point to a clear visual, that explains how the existence (or non-existence) of a non-vanishing continuous $n$-vector field on an $n$-sphere relates to division algebras over the Reals in $n+1$ dimensions?
As different sources use slightly different definitions of the term "division algebra" (for example some assuming an identity element, or some assuming associativity unless explicitly stating as a "non-associative" algebra, etc.), let me define a division algebra according to this review article (which does not assume an identity element or associativity):
Let $k$ be a field. A $k$-algebra is understood to be a vector space $A$ over $k$, endowed with a bilinear multiplication mapping $A \times A \to A$, $(x,y) \mapsto xy$. The algebra $A$ is said to be a division algebra if $A \ne \{0\}$ and the linear endomorphisms $L_a : A \to A$, $x \mapsto ax$ and $R_a : A \to A$, $x \mapsto xa$ are bijective for all $a \in A \setminus \{0\}$. In case $A$ is finite-dimensional, this is equivalent to saying that $A$ has no zero divisors, i.e. $xy=0$ only if $x=0$ or $y=0$.
The usual "hairy ball theorem" proves there is no non-vanishing continuous tangent vector field on the $2$-sphere. I have heard there is a more general version which concludes that the only dimensions which allow a non-vanishing continuous $n$-vector field on the $n$-sphere are: $n=1, 3, 7$ (and maybe $n=0$ as a trivial case depending on definitions). The review paper gives two references: Bott and Milnor, and Kervaire in 1958. I do not currently understand these proofs, but am willing to take it as a given.
What I am interested in is the connection between the existence (or non-existence) of such a $n$-vector field on an $n$-sphere, and the existence of an $(n+1)$-dimesional division algebra over the reals. This connection is even mentioned briefly in the wikipedia article on division algebras. But currently I do not see the connection.
First, is the ultra basics: is this just a necessary requirement, i.e. it shows an $n+1$ dimensional real division algebra is possible, but alone does not mean that one does exist. Or is the relationship strong enough that given such a vector field on an $n$-sphere, I could "extract" the division algebra that corresponds to this.
Second, I am having trouble seeing the connection because the dimension of the vector field is one less than the division algebra.
It is easy to visualize "combing the n-hair" on a circle, and seeing that it doesn't work on a sphere. But I do not understand how to relate this to a division algebra. Such a tangent field would only give a map on some patch from $\mathbb{R}^n \to \mathbb{R}^n$, where as some $L_a$ for the division algebra would give me $\mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$. Where did the other dimension go? I can see that if I knew $L_a$ on just the sphere, I could use bilinearity to get the rest, but that would still require input information that looks more like $\mathbb{R}^n \to \mathbb{R}^{n+1}$. And I don't see why non-vanishing on $\mathbb{R}^{n+1}$ leads to non-vanishing when truncating (projecting?) to $\mathbb{R}^{n}$.