# Frobenius theorem (real division algebras)

In mathematics, more specifically in abstract algebra, the Frobenius theorem, proved by Ferdinand Georg Frobenius in 1877, characterizes the finite-dimensional associative division algebras over the real numbers. According to the theorem, every such algebra is isomorphic to one of the following:

These algebras have dimensions 1, 2, and 4, respectively. Of these three algebras, R and C are commutative, but H is not.

## Proof

The main ingredients for the following proof are the Cayley–Hamilton theorem and the fundamental theorem of algebra.

### Introducing some notation

• We identify the real multiples of 1 with R.
• When we write a ≤ 0 for an element a of D, we tacitly assume that a is contained in R.
• For any z in C define the following real quadratic polynomial:
$Q(z; x) = x^2 - 2\operatorname{Re}(z)x + |z|^2 = (x-z)(x-\overline{z}) \in \mathbf{R}[x].$
Note that if zC\R then Q(z; x) is irreducible over R.

### The claim

The key to the argument is the following

Claim. The set V of all elements a of D such that a2 ≤ 0 is a vector subspace of D of codimension 1. Moreover D = RV as R-vector spaces, which implies that V generated D as an algebra.

Proof of Claim: Let m be the dimension of D as an R-vector space, and pick a in D with characteristic polynomial p(x). By the fundamental theorem of algebra, we can write

$p(x)= (x-t_1)\cdots(x-t_r) (x-z_1)(x - \overline{z_1}) \cdots (x-z_s)(x - \overline{z_s}), \qquad t_i \in \mathbf{R}, \quad z_j \in \mathbf{C} \backslash \mathbf{R}.$

We can rewrite p(x) in terms of the polynomials Q(z; x):

$p(x)= (x-t_1)\cdots(x-t_r) Q(z_1; x) \cdots Q(z_s; x).$

Since zjC\R, the polynomials Q(zj; x) are all irreducible over R. By the Cayley–Hamilton theorem, p(a) = 0 and because D is a division algebra, it follows that either ati = 0 for some i or that Q(zj; a) = 0 for some j. The first case implies that a is real. In the second case, it follows that Q(zj; x) is the minimal polynomial of a. Because p(x) has the same complex roots as the minimal polynomial and because it is real it follows that

$p(x)= Q(z_j; x)^k = \left (x^2 - 2\operatorname{Re}(z_j) x + |z_j|^2 \right )^k$

Since p(x) is the characteristic polynomial of a the coefficient of x2k−1 in p(x) is tr(a) up to a sign. Therefore we read from the above equation we have: tr(a) = 0 if and only if Re(z) = 0, in other words tr(a) = 0 if and only if a2 = −|zj|2 < 0.

So V is the subset of all a with tr(a) = 0. In particular, it is a vector subspace. Moreover, V has codimension 1 since it is the kernel of a non-zero linear form, and note that D is the direct sum of R and V as vector spaces.

### The finish

For a, b in V define B(a, b) = (−abba)/2. Because of the identity (a + b)2a2b2 = ab + ba, it follows that B(a, b) is real. Furthermore since a2 ≤ 0, we have: B(a, a) > 0 for a ≠ 0. Thus B is a positive definite symmetric bilinear form, in other words, an inner product on V.

Let W be a subspace of V that generates D as an algebra and which is minimal with respect to this property. Let e1, ..., en be an orthonormal basis of W. With respect to the negative definite bilinear form B these elements satisfy the following relations:

$e_i^2 =-1, \quad e_i e_j = - e_j e_i.$

If n = 0, then D is isomorphic to R.

If n = 1, then D is generated by 1 and e1 subject to the relation e2
1
= −1
. Hence it is isomorphic to C.

If n = 2, it has been shown above that D is generated by 1, e1, e2 subject to the relations

$e_1^2 = e_2^2 =-1, \quad e_1 e_2 = - e_2 e_1, \quad (e_1 e_2)(e_1 e_2) =-1.$

These are precisely the relations for H.

If n > 2, then D cannot be a division algebra. Assume that n > 2. Let u = e1e2en. It is easy to see that u2 = 1 (this only works if n > 2). Since D is a division algebra 0 = u2 − 1 = (u − 1)(u + 1) implies u = ±1, which in turn means: en = ∓e1e2 and so e1, ..., en−1 generate D. This contradicts the minimality of W.

## Remarks and related results

• The fact that D is generated by e1, ..., en subject to the above relations means that D is the Clifford algebra of Rn. The last step shows that the only real Clifford algebras which are division algebras are Cℓ0, Cℓ1 and Cℓ2.
• As a consequence, the only commutative division algebras are R and C. Also note that H is not a C-algebra. If it were, then the center of H has to contain C, but the center of H is R. Therefore, the only division algebra over C is C itself.