### Frobenius theorem on real division algebras

This post is a proof of the well known Frobenius theorem. It’s nothing groundbreaking, I’m just writing it down to organize my own thoughts.

Definition. A real division algebra is a set $A$ together with operations $+: A \times A \to A$, $\cdot : \mathbb {R}\times A \to A$ and $\ast : A \times A \to A$ such that:

1. $(A, +, \cdot )$ is a vector space over $\mathbb {R}$.
2. $(A, +, \ast )$ is a division ring.
3. $r \cdot (a_1 \ast a_2) = (r \cdot a_1) \ast a_2 = a_1 \ast (r \cdot a_1)$ for every $r \in \mathbb {R}$ and $a_1, a_2 \in A$.

Examples of real division algebras: $\mathbb {R}$ itself, the algebra of complex numbers $\mathbb {C}$, the algebra of quaternions $\mathbb {H}$.

Theorem (Frobenius). Every finite-dimensional real division algebra is isomorphic to one of $\mathbb {R}$, $\mathbb {C}$ or $\mathbb {H}$.

Before we begin, some notation and terminology conventions:

• We will omit $\cdot$ and $\ast$ everywhere and write simply $ra$ and $a_1 a_2$ insdead of $r \cdot a$ and $a_1 \ast a_2$.
• Brackets $\langle X \rangle$ mean the linear span of $X$ over $\mathbb {R}$. $\dim Y$ means the dimension of $Y$ over $\mathbb {R}$.
• “Homomorphism” always means “homomorphism of $\mathbb {R}$-algebras sending unity to unity”.
• In the same vein, a “subalgebra” of a real division algebra always means “a subset closed with respect to addition and both multiplications, and also containing the unity”.

Proof of Frobenius theorem. Let $A$ be a finite-dimensional real division algebra, and $1_ A \in A$ be its unity. Clearly, $\psi : \mathbb {R}\to A,\ r \to r 1_ A$ is a homomorphism, so $\mathbb {R}_ A = \langle 1_ A \rangle$ is a subalgebra of $A$, and it is isomorphic to $\mathbb {R}$. If $\dim A = 1$, then it follows that $A = \mathbb {R}_ A \simeq \mathbb {R}$. Therefore we can assume that $\dim A > 1$ and $\mathbb {R}_ A \subsetneq A$.

Exercise 1. Any finite field extension of $\mathbb {R}$ is isomorphic (as a real division algebra) to either $\mathbb {R}$ or $\mathbb {C}$. (Hint: any irreducible polynomial over $\mathbb {R}$ has degree at most $2$).

Exercise 2. $\mathbb {R}_ A$ is central in $A$. In other words, for every $r \in \mathbb {R}_ A$ and $a \in A$ we have $ar = ra$.

Exercise 3. Let $a \in A \setminus \mathbb {R}_ A$. Then $\langle 1_ A, a \rangle$ is a subalgebra of $A$. Moreover, $\langle 1_ A, a \rangle \simeq \mathbb {C}$.

By the last exercise we have an element $i_ A \in A$ such that $i_ A^2 = -1_ A$, and $\mathbb {C}_ A = \langle 1_ A, i_ A \rangle$ is isomorphic to $\mathbb {C}$. If $\dim A = 2$ then $A = \mathbb {C}_ A \simeq \mathbb {C}$. Let us assume then that $\dim A > 2$ and $\mathbb {C}_ A \subsetneq A$.

Exercise 4. Assume that $a \in A \setminus \mathbb {C}_ A$ and $a i_ A = i_ A a$.

1. Show that $\mathbb {C}_ A[a] = \langle c a^ n \ |\ c \in \mathbb {C}_ A, n \in \mathbb {Z}, n \geqslant 0 \rangle$ is a commutative ring.
2. Show that $\mathbb {C}_ A[a]$ is in fact a field, and therefore a finite field extension of $\mathbb {C}_ A$.
3. Conclude that such an $a$ cannot exist. (Hint: $\mathbb {C}_ A$ is algebraically closed and cannot have any nontrivial finite extensions).

If we take two elements $c \in \mathbb {C}_ A$ and $a \in A$, we can form their product $ca$. Thus $A$ is a vector space over the field $\mathbb {C}_ A$. Note that $A$ is not a $\mathbb {C}_ A$-algebra: that would mean that $c(a_1a_2) = (c a_1) a_2 = a_1(ca_2)$, and there’s no reason why the second equality should hold.

Now, consider a map $\varphi : A \to A$ defined by $\varphi (a) = i_ A a i_ A^{-1}$.

Exercise 5. $\varphi$ is an involution, i.e. $\varphi ^2 = \mathrm{id}_ A$.

Exercise 6. $\varphi$ preserves multiplication, i.e. $\varphi (a_1 a_2) = \varphi (a_1) \varphi (a_2)$ for any $a_1,a_2 \in A$.

Exercise 7. $\varphi$ is $\mathbb {C}_ A$-linear, i.e. $\varphi (c_1 a_1 + c_2 a_2) = c_1 \varphi (a_1) + c_2 \varphi (a_2)$ when $c_1,c_2 \in \mathbb {C}_ A$ and $a_1,a_2 \in A$.

Here comes the main idea. From exercises 5 and 7 it follows that eigenvalues of $\varphi$ can only be equal to $1_ A$ or $-1_ A$, and $A$ (as a vector space over $\mathbb {C}_ A$) is the direct sum of the corresponding eigenspaces: $A = U_1 \oplus U_{-1}$ where

$\displaystyle \begin{array}{rcl} U_1 & =& \left\{ a \in A \ |\ \varphi (a) = a \right\} ,\\ U_{-1} & =& \left\{ a \in A \ |\ \varphi (a) = -a \right\} . \end{array}$

Clearly, $U_1 \supset \mathbb {C}_ A$. Exercise 4 shows that in fact $U_1 = \mathbb {C}_ A$.

Since $\mathbb {C}_ A \subsetneq A$, $U_{-1} \neq 0$. Pick an arbitrary $b \in U_{-1} \setminus \{ 0\}$. From exercise 3 we have $b^2 = r_1 1_ A + r_2 b$, where $r_1, r_2 \in \mathbb {R}$. Acting by $\varphi$ on both sides and using exercises 6 and 7, we get:

$\displaystyle \begin{array}{rcl} \left(\varphi (b)\right)^2 & = & r_1 1_ A + r_2 \varphi (b). \end{array}$

By choice of $b$ we have $\varphi (b) = -b$, therefore

$\displaystyle b^2 = r_1 1_ A - r_2 b.$

It follows that $r_2 = 0$. Since $b \not\in \mathbb {R}_ A$, $r_1$ must be negative. Introduce new elements

$\displaystyle \begin{array}{rcl} j_ A & =& \frac{1}{\sqrt {-r_1}} b, \\ k_ A & =& i_ A j_ A. \end{array}$

Exercise 8. Elements $1_ A, i_ A, j_ A$ and $k_ A$ are linearly independent over $\mathbb {R}$.

Exercise 9. Establish all the equalities below that haven’t been established yet:

$\displaystyle \begin{array}{c} i_ A^2 = j_ A^2 = k_ A^2 = -1_ A, \\ i_ A j_ A = -j_ A i_ A = k_ A, \\ j_ A k_ A = - k_ A j_ A = i_ A,\\ k_ A i_ A = - i_ A k_ A = j_ A. \end{array}$

(Hint: remember that $j_ A \in U_{-1}$, which means that $i_ A j_ A i_ A^{-1}= -j_ A$).

It follows that $\mathbb {H}_ A = \langle 1_ A, i_ A, j_ A, k_ A \rangle \subset A$ is a subalgebra of $A$, and $\mathbb {H}_ A \simeq \mathbb {H}$. All that’s left to prove is that $A = \mathbb {H}_ A$.

Take an arbitrary $a \in U_{-1}$. Using exercise 6 we get

$\displaystyle \varphi (j_ A a) = \varphi (j_ A) \varphi (a) = (-j_ A)(-a) = j_ A a,$

therefore $j_ A a \subset U_1 = \mathbb {C}_ A$. But then $a \in j_ A^{-1} \mathbb {C}_ A \subset \mathbb {H}_ A$. So, $U_{-1} \subset \mathbb {H}_ A$ and $U_1 \subset \mathbb {H}_ A$. It means that $A = U_1 \oplus U_{-1} \subset \mathbb {H}_ A$, which implies $A = \mathbb {H}_ A$, as required.