# 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?

• Sure: assume there exists a division algebra over $\mathbb R^3$. It is also true tat a division algebra does not exist. Therefore, "False" is a true statement, and since false$\implies A$ is true for any statement $A$, just take "the hairy ball theorem is false" as your statement $A$. – 5xum Apr 8 '14 at 12:42
• And there is no such thing as a division algebra over $\Bbb R^3$. Any such algebra would contain a copy of $\Bbb R^3$, which has zero divisors. – rschwieb Apr 8 '14 at 12:43
• Why are you guys playing dumb? I assume the OP would like a proof of the equivalence "there is a division algebra structure on $\mathbb{R}^3$ iff there is a nowhere vanishing vector field on $S^2$" without assuming either theorem. It's a common question on Hopf invariant type problems, even though we now know exactly for which $n$ there is a Hopf invariant one map... – Najib Idrissi Apr 8 '14 at 12:45
• Since there's no division algebra of dimension $3$, one might wonder whether a counterexample can be used to construct a nowhere vanishing vector field on $S^3$, perhaps in a manner less silly than @5xum's answer. – Dustan Levenstein Apr 8 '14 at 12:50
• If $\mathbb{R}^n$ had the structure of a division algebra over $\mathbb{R}$ then $\mathbb{R}^n- \{0\}$ would be a Lie group under multiplication. Moreover, we have a copy of $\mathbb{R}^*$ inside the center of this group acting by scalar multiplication. If we quotient by this subgroup we get a Lie group structure on $S^{n-1}$, so in particular $S^{n-1}$ must be parallelizable and therefore have lots of nonvanishing vector fields (that's where the hairy ball theorem could come in). – Nate Apr 8 '14 at 12:59

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.