nLab
Cayley-Dickson construction

Idea

Consider how the complex numbers are formed from the real numbers. If you generalise this carefully, then you can perform this operation again to get the quaternions, and so on. This operation is the Cayley–Dickson construction.

Definitions

Let $A$ be an possibly nonassociative $*$-algebra over the field $ℝ$ of real numbers: an algebra equipped with an involution $x\mapsto \overline{x}$ which is an antiautomorphism.

Then one defines a new algebra, the Cayley–Dickson double $A^2$ of $A$ which is the direct sum $A\oplus A$ as an $ℝ$-vector space, and with the multiplication rule given by

$$(a,b)\cdot (u,v)=(au-\overline{v}b,b\overline{u}+va).$$(a, b)\cdot(u,v) = (a u - \bar{v} b, b \bar{u} + v a).

The map $a\mapsto (a,0)$ is a monomorphism $A\to A^2$. If $A$ is unital with unit $1$ then $A^2$ is unital with unit $(1,0)$. In the unital case, the element $\mathrm{i}≔(0,1)$ has the property ${\mathrm{i}}^2=-1≔(-1,0)$. Then the formula

$$\overline{a+b\mathrm{i}}:=\overline{a}-b\mathrm{i}$$\widebar{a + b \mathrm{i}} := \bar{a} - b \mathrm{i}

defines an involutive antiautomorphism on $A^2$ and the doubling procedure can be iterated.

Regarding that $\mathrm{i}a=\overline{a}\mathrm{i}$, if the involution in $A$ is nontrivial then the double $A^2$ is non-commutative. If $A$ is not commutative then the double is not even associative. However, if $A$ is associative then $A^2$ is still an alternative algebra.

The standard example is the sequence of consecutive doubles starting with $ℝ$ (with the identity map as involution): the real numbers $ℝ$, the complex numbers $ℂ$, the quaternions $ℍ$, the octonions or Cayley numbers $𝕆$, the sedenions $𝕊$, etc. These are the normed division algebras, followed by further algebras which are not division algebras.

References

• M M Postnikov, Lectures on geometry, Semester V: Lie groups and Lie algebras, Lec. 14 (russian and english editions)
• John Baez, The Cayley–Dickson construction
• John Baez, This Week’s Finds — Week 59