Is there a notion of an "Lie (algebraic) skew field" and can it be shown that $\mathbb R^3$ isn't one?

Question

I was wondering what makes the skew-field structures on $\mathbb C$ (complex numbers) and $\mathbb C^2$ (cuaternions) special. A natural answer seems to be that they are manifolds such that with respect to addition $\mathbb C$ and $\mathbb C^2$ are Lie groups and also $\mathbb C \setminus \{ 0 \}$ and $\mathbb C^2 \setminus \{ 0 \}$ are Lie groups with respect to multiplication. My first question is, is this generalised to some well known-notion of "(skew-)Lie field" or something like that?

What I'm more interested in is, does there exist a (skew-)field structure on $\mathbb R^3$ such that with respect to the standard manifold structure $\mathbb R^3$ is a Lie group with respect to addition and $\mathbb R^3 \setminus \{ 0 \}$ is a Lie group with respect to multiplication? Does the answer change if instead of Lie groups we only ask them to be topological groups? I wasn't able to find anything on this anywhere.

Answer 1

This is not an answer, I just needed space for the remark I was talking about in the comments

First of all, only $\mathbb{R}^{2^m}$ can hope to be endowed with a structure of (non associative) unitary division ring. So for $\mathbb{R}^3$ it can't work.

Let me first introduce a proposition:

Proposition: Let $G$ be a Hausdorff topological group and $H$ a closed subgroup of $H$. Let $G/H$ be the space of orbits for the left action of $H$ on $G$.
Suppose the quotient map $q: G \to G/H$ has a local section at $[e](=H)$, then if $K \leqslant H$ is closed, the quotient map $p: G/K \to G/H$ is a fibre bundle with fibre $H/K$.

With this theorem and the definition of $l$-frames given by: $$V_{k,l}^\mathbb{F} = \left\{ (v_1,\dots,v_l) \in (\mathbb{F}^n)^l \mid \langle v_i,v_j\rangle = \delta_i^j\right\}$$ So the $l$-orthonormal families, and its sort of dual; the grassmanian: $$Gr_{n,l}^\mathbb{F} = \left\{ V \subset \mathbb{F}^n \mid V \text{ is a sub-vector-space of dimension } l\right\}$$ You have one first fibre bundle that links Stiefel manifolds: $$V_{n-k,l-k}^\mathbb{F} \to V_{n,l}^\mathbb{F} \to V_{n,k}^\mathbb{F}$$ And a second that defines the Grassmanian manifold: $$V_{l,l}^\mathbb{F} \to V_{n,l}^\mathbb{F} \to Gr_{n,l}^\mathbb{F}$$ Unfortunately I can't give to much detail.

This is used to solve the problem: can $\mathbb{R}^{2^m}$ be endowed with a structure of unitary division ring? Of course all the topological properties follow here.

For that we define the Hopf fiber bundles given by: $$V_{1,1}^\mathbb{F} \to V_{n+1,1}^\mathbb{F} \to Gr_{n+1,1}^\mathbb{F} \simeq \mathbb{F}P^n$$ So for $\mathbb{F}= \mathbb{R}, \mathbb{C}, \mathbb{H}$, we get $$\mathbb{S}^0 \to \mathbb{S}^1 \to \mathbb{R}P^n$$ $$\mathbb{S}^1 \to \mathbb{S}^{2n+1} \to \mathbb{C}P^n$$ $$\mathbb{S}^3 \to \mathbb{S}^{4n+3} \to \mathbb{H}P^n$$ Unfortunately for $m \geq 3$ the projective space is not defined since we lose associativity (but it can exist for $n=1$ and $(n=2, m=3)$... This leads us to one more definition: Hopf fibrations (just set $n=1$ in the previous fibre bundles): $$\mathbb{S}^0 \to \mathbb{S}^1 \to^{\cdot 2} \mathbb{S}^1 $$ $$\mathbb{S}^1 \to \mathbb{S}^{2n+1} \to^{\eta} \mathbb{S}^2$$ $$\mathbb{S}^3 \to \mathbb{S}^{4n+3} \to^{\nu} \mathbb{S}^4$$ Check on YouTube, this is cool!

And from what I know Hopf fibrations can be defined for $m = 3$ (octonions) and we get a Bundle: $$\mathbb{S}^7 \to \mathbb{S}^{15} \to^{\sigma} \mathbb{S}^8$$ I don't know exactly why, but this is one of the key to study this (I believe) open problem: can $\mathbb{R}^{2^m}$ be endowed with a structure of unitary division ring?.
A construction already exists for $m=4$ it is called sedenions.

Answer 2

I may be misinterpreting your question, but it sounds to me like you're asking the following:

Is there a continuous operation $\cdot$ on $\mathbb{R}^3$ which a) distributes over the usual addition and b) has a unit, and c) all non-zero elements have inverses?

The answer to this is yes. In fact, there is commutative example on any $\mathbb{R}^n$.

Consider the elements $\{e, x_1, x_2,..., x_n\}$ as a basis of $\mathbb{R}^{n+1}$. We define $e$ to be the multiplicative identity, $x_i^2 = -1$ and $x_i x_j = x_j x_i = 0$ for $i\neq j$. We extend this definition to all of $\mathbb{R}^{n+1}$ bilinearly.

This resulting operation obviously distributes over addition and has a unit. What about inverses?

Well, given a non-zero $\alpha e + \sum \beta_i x_i$, it is easy to verify that the inverse is $\displaystyle\frac{\alpha e - \sum \beta_i x_i}{\alpha^2 + \sum \beta_i^2}.$

Of course, in the above example, for any $n\geq 3$ there are zero-divisors. We will show that this is necessary, at least if $n$ is odd.

Let's prove this.

Proposition 1: If $\cdot$ is continuous and distributes over addition, then for any $\alpha\in \mathbb{R}$, $x,y\in \mathbb{R}^n$, we have $$(\alpha x)\cdot y = x\cdot(\alpha y) = \alpha(x\cdot y).$$

Proof: We will only prove that $(\alpha x)\cdot y = \alpha (x\cdot y)$, the other case being similar.

We first prove it for $\alpha = 0$. Here, note that $0\cdot x = (0+0)\cdot x = 0\cdot x + 0\cdot x$. Subtracting $0\cdot x$ from both sides, we find $0\cdot x = 0$.

Next, we prove it for $\alpha \in \mathbb{N}$. We have $$(\alpha x)\cdot y = (x + x + ... + x) \cdot y = x\cdot y + x\cdot y + ... + x\cdot y = \alpha (x\dot y).$$

Next, we prove it for $\alpha \in \mathbb{Z}$. We have $$0 = 0\cdot y = (\alpha x - \alpha x)\cdot y = (\alpha x)\cdot y + (-\alpha x)\cdot y,$$ so $(-\alpha x)\cdot y = -(\alpha x\cdot y).$ Since we know the result for $\mathbb{N}$, it now follows for $\mathbb{Z}$.

Next, we prove it for $\alpha \in \mathbb{Q}$. Write such an $\alpha$ as $\alpha = \frac{\beta}{\gamma}$. Then $$\gamma \alpha (x\cdot y) = (\gamma \alpha x)\cdot y = ( \underbrace{\alpha \cdot x + ... + \alpha \cdot x}_{\gamma \text{ copies}}\cdot y = (\alpha x)\cdot y + (\alpha x)\cdot y + ... + (\alpha x)\cdot y = \gamma((\alpha x)\cdot y).$$ So $\gamma\alpha(x\cdot y) = \gamma((\alpha x)\cdot y).$ Now divide both sides by $\gamma$.

Lastly, we prove it for $\alpha\in \mathbb{R}$. Let $\alpha_n$ be a sequence of rationals which converge to $\alpha$. Then $\alpha_n(x\cdot y) = (\alpha_n x)\cdot y$ for all $n$. Taking $\lim_{n\rightarrow\infty}$ of both sides and using the fact that $\cdot$ is continuous, we find $\alpha (x\dot y) = (\alpha x)\cdot y.$ $\square$

Let $L_x:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be given by $L_x(y) = x\cdot y$.

Proposition 2: The map $L_x$ is a linear transformation.

Proof: We have $$L_x(\alpha y + \beta z) = L_x(\alpha y) + L_x(\beta z) = \alpha L_x(y) + \beta L_x(z),$$ where the first equality is because $\cdot$ distributes over $+$ and the second is from Proposition 1. $\square$

Let $e\in \mathbb{R}^n$ denote the unit.

Proposition 3: Suppose $n\geq 3$ is odd and suppose that we have a multiplication $\cdot$ satisfying a) multiplication is continuous, b) multiplication distributes over $+$, and c) all non-zero elements have multiplicative inverses. Then there are zero divisors.

Proof: From Proposition 2, for each $x\in \mathbb{R}^n$, $L_x$ is a linear transformation.

Now, select $x \in \mathbb{R}^n$ with $\{x,e\}$ linearly independent, which is possible because $n>1$.

Because $n$ is odd, the linear transformation $L_x$ has an eigenvector $0\neq y\in \mathbb{R}^n$, so $xy = \alpha y$ for some $\alpha \in \mathbb{R}$.

Now, consider the element $z:=x-\alpha e$. Because $\{x,e\}$ is linearly independent, $z\neq 0$. But then $z\cdot y = x\cdot y - \alpha e\cdot y = \alpha y - \alpha y= 0$. Thus, there are zero-divisors. $\square$