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.

  • $\begingroup$ Thank you! You use $D \setminus \mathbb R$ to mean that $a = (a,0,0)$ or $(0,a,0)$ or $(0,0,a)$, right? $\endgroup$ – learner Jul 14 '14 at 16:42
  • 1
    $\begingroup$ 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"? $\endgroup$ – learner Jul 14 '14 at 16:45
  • $\begingroup$ 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 $\Bbb{R}$. $\endgroup$ – Jyrki Lahtonen Jul 14 '14 at 18:20
  • $\begingroup$ 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. $\endgroup$ – learner Jul 14 '14 at 19:14
  • 2
    $\begingroup$ @user161650: Sorry about wanting to show off a bit first :-) I added a shorter argument to settle your precise question. $\endgroup$ – 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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.