# Proving that $\mathbb R^3$ cannot be made into a real division algebra (and that extending complex multiplication would not work)

I am trying to solve the following exercise:

Prove that complex multiplication does not extend to a multiplication on $\mathbb R^3$ so as to make $\mathbb R^3$ into a real division algebra.

I am aware of Frobenius' theorem that there are only three finite dimensional real associative division algebras. But the exercise would be pointless if the theorem was assumed.

Let $x,y \in \mathbb R^3$. Since $x y$ has to extend the complex multiplication:

$$xy = (x_1y_1-x_2y_2, x_1y_2 + x_2 y_1, ?)$$

I have no idea how I can proceed from here. Could someone tell me how to do this? It should be easy as the other exercises I did so far were also easy.

Assume that $D$ is a 3-dimensional division algebra over $\Bbb{R}$. Let $a\in D\setminus\Bbb{R}$. Consider the linear mapping $\rho_a:z\mapsto az, z\in D,$ from $D$ to itself. Let $M$ be the matrix representing $\rho_a$ (with respect to some basis). The eigenvalue polynomial $$\chi_a(x)=\det(xI_3-M)\in\Bbb{R}[x]$$ is monic of degree three. Thus $\lim_{x\to\pm\infty}\chi_a(x)=\pm\infty$. Therefore, by continuity of $\chi_a(x)$, there exists a real number $r$ such that $\chi_a(r)=0$. This means that the mapping $$L:D\to D, z\mapsto az-rz=(a-r)z$$ has a non-trivial kernel. Therefore the element $a-r$ cannot be invertible. Because $a\notin\Bbb{R}$ we have $a-r\neq0_D$. This contradicts the assumption that $D$ is a division algebra.

Edit: The above was a bit of overkill for the task at hand (=proving that complex multiplication cannot be extended to a 3D-space). A simpler argument follows.

If the multiplication of $D$ is an extension of the multiplication of $\Bbb{C}$, then $D$ has a subalgebra isomorphic to $\Bbb{C}$. Therefore $D$ has a structure of a (left) vector space over $\Bbb{C}$. Thus $D$ is a finite dimensional vector space over $\Bbb{C}$. But this implies that the dimension of $D$ as a vector space over $\Bbb{R}$ is an even number.

• Thank you! You use $D \setminus \mathbb R$ to mean that $a = (a,0,0)$ or $(0,a,0)$ or $(0,0,a)$, right? – learner Jul 14 '14 at 16:42
• Uhm wait, do I understand correctly that you proved "there cannot exist a real division algebra of dimension $3$"? Which is slightly stronger than "there does not exist a real division algebra of dimension $3$ that extends complex multiplication"? – learner Jul 14 '14 at 16:45
• 1) I meant that we identify the 1-dimensional subspace spanned by $1_D$ with $\Bbb{R}$. If $1_D=(1,0,0)$, then $\Bbb{R}$ is identified with the real axis $(r,0,0),r\in\Bbb{R}$. So $a$ can be any 3-tuple $a=(a_1,a_2,a_3)$ such that either $a_2\neq0$ or $a_3\neq0$ (or both). 2) Correct. The argument shows that there is no 3-dimensional division algebra over $\Bbb{R}$. – Jyrki Lahtonen Jul 14 '14 at 18:20
• Thank you. Do you see any way for a much easier argument to just prove the weaker statement that the exercise asks me to prove? I'm of course very happy with your answer, it's great. – learner Jul 14 '14 at 19:14
• @user161650: Sorry about wanting to show off a bit first :-) I added a shorter argument to settle your precise question. – Jyrki Lahtonen Jul 14 '14 at 19:42

I will also add a more "topological" proof but requires some familiarity with manifolds:

For a multiplication to "expand" the complex multiplication we would require that $|u_1\times u_2|=|u_1||u_2|$ which is saying that $S^2$ would take the structure of a group (as is the case in $\mathbb{C}$ and $S^1$). But since $S^n$ are manifolds , this would $S^2$ a Lie group and we know that for every Lie Group $TM=M\times M$ , namely is is parallelizable.

Using the (very deep) fact that only $S^1,S^3,S^7$ are parallelizable we get that the only multiplications like $\mathbb{C}$ exist in $\mathbb{R^2},\mathbb{R^4},\mathbb{R^8}$ which ofcourse correspond to the complex numbers, quartenions and octonions.

Notice that for tha case of $S^2$ and in general $S^{2k}$ we can just use the hairy-ball theorem to show that they aren't parallelizable!