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Complex numbers are essentially 2-dimensional vectors with product defined such that the set is a field and extends real product. So, is there an analogous definition for 3, or higher, dimensional vectors such that the Euclidean space becomes a field also?

Daniel Li
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  • https://en.wikipedia.org/wiki/Hypercomplex_number – JMoravitz Jul 22 '19 at 19:35
  • quaternion multiplication is non-commutative, so quaternions are not a field – J. W. Tanner Jul 22 '19 at 19:40
  • An Irishman beat you to this question (in dimension 3) over 150 years ago, and it took him about a decade to resolve it. The result was finding an answer in dimension 4 that isn't commutative. It was one of the first steps toward ring theory. Then your question is pretty much completely answered by the Frobenius theorem. – rschwieb Jul 22 '19 at 20:30

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Yes; probably the most important are the quaternions, but there are also octonions, sedenions, and others - although as we go further, we get worse and worse, algebraically speaking (the quaternions aren't commutative, the octonions aren't associative, and the sedenions are "even less associative"). There are also interesting impossibility results, most famously Hurwitz' theorem.

A general process for whipping up structures like this is the Cayley-Dickson construction.

J. W. Tanner
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Noah Schweber
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This depends on exactly what structure you’d like to preserve. If you ask that the multiplication is associative and commutative and all numbers have inverses, then the answer is no. The reals and the complex numbers are the only such things. However, if you drop commutativity there are the quaternions, a 4 dimensional “division” ring which is just like a field except not necessarily commutative. If you drop associativity, there are the octonions. These are an 8 dimensional nonassociative ring with inverses. This is a very strange object compared to the other three. However, there are even more things you can drop which give rise to the 16 dimensional sedonions.

It turns out all of these are important in topology because they give rise to “exceptional spheres”. Look at the vectors of magnitude 1 and they form a sphere which is closed under the operation. The operation can then be used to create something called a fiber bundle that relate the sphere of that dimension, the sphere of dimension one lower, and the sphere that is the sum of those two dimensions.

There are a variety of ways to prove that these are the only such things. I believe there is an elementary proof using linear algebra, as well as more complicated proofs using topology (along with less complicated, but only partial results using topology).

Connor Malin
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