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 $\mathbb{R}^{n}$.

  • $\begingroup$ I don't know about the connection with the hairy ball theorem, but it seems like you should first understand (i) frobenius' classification of division algebras over R, and (ii) how quaternions and octonions relate to special orthogonal groups of one less dimension. $\endgroup$
    – Kimball
    Commented May 9, 2020 at 13:12
  • $\begingroup$ @Kimball You appear to be using a more restrictive definition of the phrase "division algebra". I wrote out the definition of a division algebra to try to avoid these issues. The classification of division algebras over R is still an open problem. math.stackexchange.com/questions/2892818/… In short, R,C,H,O are not the only division algebras over the reals. Frobenius restricts enough to only get R,C,H. Hurwitz relaxed associativity to also get O, but assumed an identity element and looked at normed algebras. $\endgroup$
    – PPenguin
    Commented May 12, 2020 at 0:17
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    $\begingroup$ First, all odd dimensional spheres have non-vanishing vector fields on them. An odd dimensional sphere naturally lives in $\mathbb{C}^n$, and multiplication by $i$ gives rise a non-vanishing vector field. What is special about dimensions $1,3,$ and $7$ is that the have $1$, $3$, and $7$ vector fields which are linearly independent at every point. There are no other $n$s for which $S^n$ has $n$ everywhere-linearly-independent vector fields (except $n=0$). For example, $S^5$ has a non-vanishing vector field, but any two non-vanishing vector fields must be linearly dependent at some point. $\endgroup$ Commented May 12, 2020 at 0:48
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    $\begingroup$ @JohnSmall: The only reference I know is jstor.org/stable/… , which is Frank Adams annals paper on it. The short version is as follows. Given a positive integer $n$, write it as $A\cdot 2^B$ with $A$ odd. Writing $B = c + 4d$ with $0\leq c\leq 3$, define $\rho(n)=2^c + 8d$. Using Clifford algebras, one can show that a sphere $S^{n-1}$ admits $\rho(n)-1$ linearly independent vector fields. (continued) $\endgroup$ Commented Mar 15, 2021 at 18:58
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    $\begingroup$ Frank Adams shows that it cannot admit more than $\rho(n)-1$ linearly independent vector fields. When $n=6$ (so we're talking about $S^5$), $6 = 3\cdot 2^1$, and $1= 1 + 4(0)$, so $\rho(6) = 2^1 +8(0) = 2$. Thus, $S^5$ admits $2-1 = 1$ linearly independent vector field. According to Adams, the non-existance result was already known for $B\leq 3$ from work of Steenrod and Whitehead. I don't know if the result for $S^5$ goes back farther than that. $\endgroup$ Commented Mar 15, 2021 at 19:02

1 Answer 1


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, $n = 1,2,4,8$.


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