3
$\begingroup$

Now, we all know and love the quaternions, in which multiplication is not commutative, and the fact that, in all Clifford algebras, exponentiation is not commutative.

Having looked at the properties that we gain/lose going up the hypercomplex ladder (i.e. going from $\mathbb{C}$ to $\mathbb{H}$ to $\mathbb{O}$ to...), I am left wondering: if we keep extending these hypercomplex number systems, will there come a point where addition is no longer commutative?

If so, what is the dimension of this algebra?

And if you're still reading at this point, here's a bonus (sub-)question: What will be algebraic effect on this set, as a result of the lack of additive commutativity (does anything interesting result from this, e.g. an interesting property of the sedenions is the existence of zero divisors), apart from the ability to solve more equations?

$\endgroup$
  • 5
    $\begingroup$ Algebras have, by definition, commutative addition. $\endgroup$ – Zhen Lin Jun 11 '14 at 17:17
  • 4
    $\begingroup$ Moreover, if you want multiplication to distribute over addition, then addition must be commutative (exercise). What do you mean by "exponentiation is not commutative"? In any case, the point of these exotic number systems is not to solve more equations. Already this isn't the point of the quaternions. $\endgroup$ – Qiaochu Yuan Jun 11 '14 at 17:45
5
$\begingroup$

If by "going up the ladder" you mean, "continue to apply the Cayley-Dickson construction", then no.

The construction always produces a power-associative algebra (with commutative addition) all the way up. This is easy to see since the addition defined in the construction relies on that of the previous ring, which is commutative.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.