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Reals, complex numbers, quaternions, octonions, etc are a hierarchy of algebras which can be constructed in a regular way.

One obvious property of this hierarchy is that each such algebra has $2^n$ natural basis elements.

Are there related algebras which have 3, 5, 7, or some other intermediate number of natrual basis elements?

If not, why not?

spraff
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    It's not too hard to prove that you can't have a 3- (or 5- or 7-) dimensional division algebra like Hamilton was originally looking for. You can use characteristic polynomials. – Kimball Dec 15 '16 at 16:15
  • Between the split quaternions and the split octonions is a six-dimensional algebra called the sextonions, see https://math.stackexchange.com/questions/484013/trinonions-quaternions-quinonions-sextonions-septonions-octonions – anon Jun 14 '22 at 19:57

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Well, you can always equip $\mathbb{R}^k$ with the product $(\mathbf{x}\ast \mathbf{y})(i)=\mathbf{x}(i)\cdot \mathbf{y}(i)$ to give you a unital associative commutative algebra, but for all $k\geq 2$ you won't have this be a division algebra.

The Frobenius Theorem says that the only real division algebras are $\mathbb{R},\mathbb{C},\mathbb{H}$ (up to isomorphism).

Other standard algebras include

  • group algebras (if $G$ is a group, consider the free $\mathbb{R}$-vector space generated by $G$, where we define the multiplication by the group law and extend linearly)
  • Clifford algebras which actually produces the examples of $\mathbb{R},\mathbb{C}$, and $\mathbb{H}$ (but not $\mathbb{O}$).
  • Field extensions (like $\mathbb{R}[x]$ or $\mathbb{R}(x)$ as $\mathbb{R}$-vector spaces, or $\mathbb{Q}[\sqrt{2}]$ as a $\mathbb{Q}$-vector space)
  • The algebra of $n\times n$ matrices $\mathbb{R}^{n\times n}$
  • The algebra of continuous functions $C(X)$ for any topological space $X$ (you can just consider $\mathbb{R}$ or $[0,1]$ if you're not comfortable with topological spaces)
  • Lie Algebras

All of these things are useful and some have entire fields of study devoted to them.

Hayden
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