General dimension hairy ball theorem and division algebras 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 $\mathbb{R}^{n}$.
 A: Let $A$ be an $n$-dimensional division $\mathbb{R}$-algebra in the sense that you describe.  If I'm understanding the article linked to in this MSE answer, every $A$ is isotopic to a unital algebra.  In particular, if there is an $n$-dimensional division $\mathbb{R}$-algebra, then there is a unital one.  So, I will assume $A$ is unital.
So, let $1\in A$ be the unit.  We identify $\mathbb{R}\subseteq A$ as the $\mathbb{R}$-multiples of $1$.
Proposition  The elements in $\mathbb{R}$ associate and commute with everything in $A$.  That is, if $r\in \mathbb{R}$ and $x,y\in A$, then $rx = xr$ and $r(xy) = (rx)y$.
Proof:  First, $1x = x1$ because both are just $x$.  Then bilinearity of multiplication gives $r(1x) = (r1)x = x(r1)$, so $rx = xr$.
Bilinearity also gives $r(xy) = (rx)y$.  $\square$
Now, fix a basis $\{e_1, e_2, ..., e_n\}$ of $A$ with $e_1 = 1$.  Note that for any real numbers $\lambda_i$ and any $v\in A$, we have $$\left(\sum_{i=1}^n \lambda_i e_i\right) v = \sum_{i=1}^n (\lambda_i e_i) v = \sum_{i=1}^n \lambda_i(e_i v)$$ where the first equality is the distributive property and the second is from the fact that real numbers associate.
Proposition:  If $v\in A$ with $v\neq 0$, then the elements $e_i v$ are linearly independent.
Proof.  Assume $\sum_{i=1}^n \lambda_i (e_i v) = 0$.  As mentioned above, this is the same as $(\sum_{i=1}^n \lambda_i e_i)v=0$.  Now, setting $a = \sum_{i=1}^n \lambda_i e_i$, we have $L_a(v) = L_a(0) = 0$.  Thus, since we're in a division algebra, $a = 0$.  That is, $\sum_{i=1}^n \lambda_i e_i = 0$.  Since the $e_i$ are a basis, we conclude all $\lambda_i = 0$, so the $e_i v$ are linearly independent.  $\square$
Now, put an arbitrary inner product on $A$.  (What I mean by arbitrary is that I'm not making any assumptions about how the inner product interacts with the multiplication on $A$)  Having an inner product, we define $S^{n-1} = \{v\in A: \langle v,v\rangle = 1\}$.  Since all inner products on a finite dimensional space are equivalent, $S^{n-1}$ is diffeomorphic to the usual $S^{n-1}$.
Proposition  The space $S^{n-1}$ has a family of $n-1$ continuous vector fields which are linearly independent at every point.
Proof:  Let $p \in S^{n-1}$.  For any $i = 2,.., n$, define the vector field $V_i(p)$ as the projection of $e_i p$ onto the codimension one hyperplane $p^\bot\subseteq A$.
Bilinearity means the multiplication is given by some degree $2$ of polynomial in the coordinates, so multiplication is continuous.  Projection is also continuous, so the $V_i(p)$ are continuous.
Now, assume $\sum_{i=2}^{n} \lambda_i V_i(p) = 0$.  This means that $\sum_{i=2}^n \lambda_i(e_i p)$ is parallel to $p$, i.e., $sum_{i=2}^n \lambda_i(e_i p) = \lambda p$ for some real number $\lambda$.
But $\lambda p = \lambda (e_1 p)$, so this is equivalent to $\sum_{i=2}^n \lambda_i(e_i p) - \lambda (e_1 p) = 0$.  Since $p\in S^{n-1}$ implies $p\neq 0$, the previous proposition now shows that for any $i$, $\lambda_i = \lambda = 0$.  Thus, the $V_i(p)$ are linearly independent.  $\square$
Now we use a topological result (that I would have attributed to Adams, though perhaps Kevaire, Milnor, and Bott were the first to tease it out of what Adams proved?)
Theorem:  If $S^{n-1}$ has $n-1$ linearly independent vector fields, then $n-1 = 0,1,3,7$.
Thus, if there is an $n$-dimensional division $\mathbb{R}$-algebra, $n = 1,2,4,8$.
