Frobenius theorem on real division algebras

by Dan Shved

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.