gistfile1.txt

Let D be a finite-dimensional real division algebra.

For any element x of D, we have an associated endomorphism on D that acts by left multiplication.
Since the endomorphism algebra over a finite-dimensional vector space is itself finite dimensional,
the elements 1, x, ..., x^m must eventually develop a linear dependency if we choose m sufficiently
large. Thus there exists real numbers a_i such that

    a_0 + a_1 x + ... + a_m x^m = 0

Considering this as a real polynomial, we get a factorization into linear and irreducible quadratic
parts. As D is a division algebra, a product being zero means one of the factors must be zero. If
we normalize by dividing by the leading coefficient, we have x - a = 0 or x^2 - 2bx + c = 0 for
real a, b, and c. In the first case, x = a is just a real number sitting inside D. In the second
case, we may set y = x + b to get y^2 = d for d = -(c + b^2). Thus y has a real square, and as the
quadratic is irreducible we must have d < 0 so that y is a square root of a negative real number.

This means that any x can be written in a unique way as a sum of a real element Re(x) and an imaginary
element Im(x) where an element is defined as imaginary if it has a non-positive real square.

Suppose D contains a non-real element x. Then x has a non-zero imaginary component Im(x) that
generates an isomorphic copy <Im(x)> of the complex numbers inside D. Letting <Im(x)> act by left
multiplication gives D the structure of a complex vector space. Since <Im(x)> is itself a
two-dimensional real vector space, this implies that D is an even-dimensional real vector space.

An immediate corollary is that there are no odd-dimensional real division algebras other than R itself.

Consider the set of imaginary elements Im(D). Because of the unique decomposition into real and
imaginary components, if x and y are imaginary then x + y must be imaginary, and similarly s x
is imaginary for real s since (s x)^2 = s^2 x^2. This shows that the imaginary elements form a
vector subspace; they are not a subalgebra since x^2 is non-imaginary.

We have (x + y)^2 = x^2 + y^2 + xy + yx, so xy + yx = (x + y)^2 - x^2 - y^2 is a real number if
x and y are imaginary. The bilinear pairing <x, y> = -(xy + yx)/2 is therefore an inner product on
Im(D). Take any minimal generating subspace of Im(D). This is an inner product space which must
therefore have some orthonormal basis e_1, ..., e_n. Observe that <e_i, e_i> = 1 is equivalent to
e_i^2 = -1 and <e_i, e_j> = 0 is equivalent to e_i e_j = -e_j e_i.

These equations should look very familiar. They are the defining relations for quaternions when
n = 2 if we identify i with e_1, j with e_2, and k with e_1 e_2. We already saw that n = 1 gives
us the complex numbers and n = 0 are obviously the real numbers. But what happens for n > 2?

If there are three or more orthonormal imaginaries then set x = e_1 e_2 e_3 and note that x^2 = 1.
Indeed, the square of a product of any odd number of orthonormal imaginaries must be positive due
to the anti-commutation relations.  But x^2 = 1 factors into (x - 1)(x + 1) = 0 which in a division
algebra implies x = +/- 1. But this means that e_3 = -/+ e_1 e_2 in contradiction of the minimality
of the generating imaginaries.

Therefore the only finite-dimensional real division algebras are R, C and H.