Frobenius Theorem (real Division Algebras) - Proof

Proof

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

We can consider D as a finite-dimensional R-vector space. Any element d of D defines an endomorphism of D by left-multiplication and we will identify d with that endomorphism. Therefore we can speak about the trace of d, the characteristic and minimal polynomials. Also, 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. 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.

To see that, we pick an a ∈ D. Let m be the dimension of D as an R-vector space. Let p(x) be the characteristic polynomial of a. By the fundamental theorem of algebra, we can write

for some real t_i and (non-real) complex numbers z_j. We have 2s + r = m. The polynomials are irreducible over R. By the Cayley–Hamilton theorem, p(a) = 0 and because D is a division algebra, it follows that either a − t_i = 0 for some i or that, z = z_j for some j. The first case implies that a ∈ R. In the second case, it follows that 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

and m=2k. The coefficient of in p(x) is the trace of a (up to sign). Therefore we read from that equation: the trace of a is zero if and only if Re(z) = 0, that is .

Therefore 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 (nonzero) linear form. Also note that D is the direct sum of R and V (as vector spaces). Therefore, V generates D as an algebra.

Define now for Because of the identity, it follows that is real and since if 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 which generated D as an algebra and which is minimal with respect to that property. Let be an orthonormal basis of W. These elements satisfy the following relations:

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

If n = 1, then D is generated by 1 and e₁ subject to the relation . 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 and . These are precisely the relations for H.

If n > 2, the D cannot be a division algebra. Assume that n > 2. Put . It is easy to see that u2 = 1 (this only works if n > 2). Therefore 0 = u2 − 1 = (u−1)(u+1) implies that u = ±1 (because D is still assumed to be a division algebra). But if u= ±1, then and so generates D. This contradicts the minimality of W.

Remark: The fact that D is generated by subject to above relation can be interpreted as the statement 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.

Remark: 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.