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Let's say a normed division algebra is a real vector space $A$ equipped with a bilinear product, an element $1$ such that $1a = a = a1$, and a norm obeying $|ab| = |a| |b|$.

There are only four finite-dimensional normed division algebras: the real numbers, the complex numbers, the quaternions and the octonions. This was proved by Hurwitz in 1898:

Adolf Hurwitz, Über die Composition der quadratischen Formen von beliebig vielen Variabeln, Nachr. Ges. Wiss. Göttingen (1898), 309-316.

Are there any infinite-dimensional normed division algebras? If so, how many are there?

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The infallible wikipedia states that there are only four normed division algebras over the reals, period. – David Roberts Nov 11 '10 at 6:21
Note that a normed division algebra is a composition algebra if and only if its norm arises from an inner product via $|a| = \sqrt{\langle a,a \rangle}$. This is in fact always true in the finite-dimensional case, and I bet it's true in the infinite-dimensional case as well, but I haven't checked it, so there's a slight gap here, at least in my brain. – John Baez Nov 12 '10 at 7:31
up vote 11 down vote accepted

A MathSciNet search reveals a paper by Urbanik and Wright (Absolute-valued algebras. Proc. Amer. Math. Soc. 11 (1960), 861–866) where it is proved that an arbitrary real normed algebra (with unit) is in fact a finite-dimensional division algebra, hence is one of the four mentioned in the OP. A key piece of the argument (Theorem 1) is to show that such an algebra $A$ is algebraic, in the sense that if $x \in A$, then the subalgebra of $A$ generated by $x$ is finite-dimensional. The authors then invoke a theorem of A. A. Albert stating that a unital algebraic algebra is a finite-dimensional division algebra.

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Thanks very much! By "real normed algebra" (with unit)" you mean what I'd called "normed division algebra (with unit)". (The authors actually use the term "absolute-valued algebra", which is nonstandard.) Also, when you write "unital algebraic algebra", I'd write "unital algebraic normed division algebra" - we need $|ab| = |a||b|$ in Albert's theorem. But anyway: you gave me what I wanted! And it's a very cute skinny PDF file, too – John Baez Nov 12 '10 at 7:18

Hurwitz's theorem is stated here (section 2.6 of 'A taste of Jordan algebras' by Kevin McCrimmon) as:

Any composition algebra $C$ over a field $\Phi$ of characteristic not 2 has finite dimension $2^n$ for $n=0,1,2,3$ and is one of the following ...

it goes on then to describe generalisations of the usual normed division algebras.

The wikipedia page composition algebra tells us that one only has a 1-dimensional composition algebra when the characteristic of the base field is not 2, but otherwise you can start from a 2-dimensional composition algebra over a characteristic 2 field and perform the usual Cayley-Dickson construction.

Edit: The following theorem was proved by Kaplansky (Proc AMS 1953), which finishes off the classification. A quadratic form $g$ on an algebra $A$ over a field $F$ in this context is a function $g:A \to F$ such that $g(kx) = k^2g(x)$ for $k\in F$ and $x\in A$.

Theorem. Let $A$ be an algebra with unit element over a field $F$. Suppose that $A$ carries a nonsingular quadratic form $g$ satisfying $g(xy) = g(x)g(y)$ for all $x, y \in A$. Then:
(a) A is alternative,
(b) except for the case where $A$ has characteristic two and is a purely inseparable field over $F$, $A$ is finite dimensional and of dimension 1, 2, 4, or 8,
(c) $A$ is either simple or the direct sum of two copies of $F$,
(d) $g(x) = x^\ast x$ where $x\mapsto x\ast$ is an involution of $A$.

So unless your base field has characteristic 2, and your division algebra is a purely inseparable extension of the base field, your division algebra has to be finite dimensional.

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Note that normed division algebras are special cases of composition algebras... – David Roberts Nov 11 '10 at 6:30
Thanks, David. In the finite-dimensional case I know how to show that normed division algebras must have a norm arising from an inner product, and are thus composition algebras. I would need to check this in the infinite-dimensional case. But the result you state is a very nice purely algebraic result along the general lines I sought. – John Baez Nov 12 '10 at 7:37

The associative case follows from Mazur's Theorem (see here). He proved that there are up to isomorphism precisely three Banach division algebras, namely $\mathbb R,\mathbb C$ and $\mathbb H$. This applies to the completion of any normed division algebra, which still verifies the identity $|ab|=|a||b|$, and hence is a division algebra.

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The question isn't assuming associativity, so I'm not sure if Mazur's theorem still applies. – Faisal Nov 11 '10 at 7:08
Sorry, I did not notice. I changed the text accordingly. – Andreas Thom Nov 11 '10 at 7:14

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