Frobenius theorem (real division algebras)

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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.
  • For any Template:Mvar in C define the following real quadratic polynomial:
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 Template:Mvar of all elements Template:Mvar of Template:Mvar such that a2 ≤ 0 is a vector subspace of Template:Mvar of codimension 1. Moreover D = RV as R-vector spaces, which implies that Template:Mvar generated Template:Mvar as an algebra.

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

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

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

Since p(x) is the characteristic polynomial of Template:Mvar 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(zj) = 0, in other words tr(a) = 0 if and only if a2 = −|zj|2 < 0.

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

The finish

For a, b in Template:Mvar 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 Template:Mvar is a positive definite symmetric bilinear form, in other words, an inner product on Template:Mvar.

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

If n = 0, then Template:Mvar is isomorphic to R.

If n = 1, then Template:Mvar is generated by 1 and e1 subject to the relation eTemplate:Su = −1. Hence it is isomorphic to C.

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

These are precisely the relations for H.

If n > 2, then Template:Mvar 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). If Template:Mvar were 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 Template:Mvar. This contradicts the minimality of Template:Mvar.

Remarks and related results

  • The fact that Template:Mvar is generated by e1, ..., en subject to the above relations means that Template:Mvar 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.

References