Continuous addition and multiplication on Euclidean space (dimension > 2) making it into a field?

Question

While TAing a linear algebra class, I happened upon the following question:

Question: for $n\geq 3$, are there continuous operations $+, \cdot : \mathbb R^n \times \mathbb R^n \to \mathbb R^n$ that give $\mathbb R^n$ the structure of a field?

(Obviously if I omit the word "continuous", one can choose a set bijection with, say, $\mathbb R$, and inherit the field structure of $\mathbb R$. And if the operations restrict to the familiar ones on $\mathbb R \subseteq \mathbb R^n$, creating a real field extension, then Frobenius' theorem tells us the answer is NO.)

I asked this to my fellow graduate students, and one of them was so kind as to give me the following wonderful response:

I think it's impossible for $\mathbb R^3$, and more generally for $\mathbb R^n$ for if $n\neq 1,2,4,8$. There's something called the Hopf invariant, it's a homomorphism $H: \pi_{2n-1}(S^n) \to \mathbb Z$, defined as follows: a map $f: S^{2n-1}\to S^n$ gives you instructions to make a cell complex $X$ with a cell in dimensions $0$, $n$, and $2n$. Then $H^n(X; \mathbb Z)$ is generated by some class $\sigma$ and $H^{2n}(X; \mathbb Z)$ is generated by a class $\tau$. You define $H(f)$ as the integer so that $\sigma^2 = H(f) \tau$.

Note that $H(f)$ for a map $f: S^{2n-1} \to S^n$ is $0$ if $n$ is odd, just by the anticommutativity of the cup product. On the other hand, if there exists a continuous map $\mu:\mathbb R^n \times \mathbb R^n \to \mathbb R^n$ ["multiplication"] and distinct elements $e, z \in \mathbb R^n$ ["1", "0" resp.] so that $\mu(e, x) = \mu(x, e) = x$ and $\mu(x, y) \neq z$ when $x, y \neq z$, I think you can construct a map $f: S^{2n-1} \to S^n$ of Hopf invariant $1$. This is outlined in Mosher & Tangora chapter 4, in the exercises. The hardest exercise is maybe 4.2, but this is Proposition 15.15 in Bredon. More generally, a map of Hopf invariant $1$ is known to exist if and only if $n = 2, 4, 8$.

Perhaps someone has a proof with other technology, or a proof that works for $n=4,8$, the only remaining cases.

EDIT: I suppose the above proof is essentially that in Detail in the proof of $\mathbb{R}^n$ is a division algebra only for $n=1,2,4,8$, and also similar to Can we turn $\mathbb{R}^n$ into a field by changing the multiplication?

Answer 1

First of all, suppose $+^{\prime}\colon\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}^n$ is any map that makes $(\mathbb{R}^n,+^{\prime})$ into a topological group. Then, $\mathbb{R}^n$ is a manifold, so Hilbert's 5th implies that $(\mathbb{R}^n,+^{\prime})$ is isomorphic as a topological group to a Lie group. However, the connected abelian Lie groups are products of Euclidean spaces and tori, hence uniquely determined by their underlying manifold, so $(\mathbb{R}^n,+^{\prime})\cong(\mathbb{R}^n,+)$, where $+$ is the usual addition on $\mathbb{R}^n$.

Thus, in the given scenario, we may suppose w.l.o.g. that $+$ is the usual addition on $\mathbb{R}^n$ and $\cdot\colon\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}^n$ is some map turning $(\mathbb{R}^n,+,\cdot)$ into a topological field. Now, let $e$ denote the multiplicative identity of $\mathbb{R}^n$ and juxtaposition denote scalar multiplication. I claim $\lambda e\cdot x=\lambda x$ for all $x\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}$. Indeed, for rational $\lambda$, this is a direct consequence of the usual additive structure and field axioms. For real $\lambda$, it is then a consequence of continuity (note that this only requires continuity of $\cdot$ if one variable is held fixed, which is weaker than joint continuity).

This implies that $\mathbb{R}^n$ is a field extension of $\mathbb{R}$. Thus, $n=1$ or $2$ and the corresponding field is isomorphic to either $\mathbb{R}$ or $\mathbb{C}$ with their usual multiplication for purely algebraic reasons (essentially the FTA).

Answer 2

If $0$ is the zero element of the field structure on $\mathbb{R}^n$, then $\mathbb{R}^n\setminus\{0\}$ is a topological abelian group under multiplication. Any topological abelian group is weak homotopy equivalent to a product of Eilenberg-Mac Lane spaces. Since the only nontrivial reduced cohomology of $\mathbb{R}^n\setminus\{0\}$ is in degree $n-1$, this implies $\mathbb{R}^n\setminus\{0\}\simeq K(\mathbb{Z},n-1)$. However, this is false if $n\geq 3$, since for instance $\pi_n(\mathbb{R}^n\setminus\{0\})\cong \pi_n(S^{n-1})$ is nontrivial.

(Alternatively, you can use the closely related Dold-Thom theorem. The abelian group structure on $\mathbb{R}^n\setminus\{0\}$ gives a retraction $SP(\mathbb{R}^n\setminus\{0\},1)\to \mathbb{R}^n\setminus\{0\}$. By Dold-Thom, $\pi_i(SP(\mathbb{R}^n\setminus\{0\},1))\cong \tilde{H}_i(\mathbb{R}^n\setminus\{0\})$, so this retraction means that if $\pi_i(\mathbb{R}^n\setminus\{0\})$ is nontrivial then $\tilde{H}_i(\mathbb{R}^n\setminus\{0\})$ is nontrivial as well. But this is false if $n\geq 3$.)

Answer 3

There's a relevant result from algebraic number theory, which could be called "the topological classification/characterization of local fields":

Suppose that $K$ is any topological field that is locally compact and does not carry the trivial topology nor the discrete topology. Then $K$ is isomorphic as a topological field, to one of the following:

• $\Bbb R$
• $\Bbb C$
• A finite extension of $\Bbb Q_p$ for some $p$
• $\Bbb F_q((T))$ for some prime power $q$.

As you can see, the only connected ones on the list are $\Bbb R$ and $\Bbb C$, which proves the result that you're after, since a homeomorphism of manifolds preserves the dimension.