Nonassociative algebras

Let $A$ be an algebra with identity 1 over $\mathbb {R}$. Let a conjugation operation $x \mapsto \bar{x}$ be given on $A$, possessing the properties $\overline{\bar{x}}=x, \overline{x y}=\overline{y x}$. Let us equip the space $A \oplus A=\left\{ (x, y) \mid x, y \in A\right\}$ with the bilinear multiplication operation

One obtains an algebra called the doubling of the algebra $A$.

Verify that $\mathbb {C}$ is the doubling of the algebra $\mathbb {R}$, and $\mathbb {H}$ the doubling of the algebra $\mathbb {C}$. The doubling of the algebra $\mathbb {H}$ is called the Cayley algebra $\mathrm{Ca}$.

Verify that $\mathrm{Ca}$ is a noncommutative and nonassociative algebra. Express explicitly the conjugation operation on $\mathrm{Ca}$.

Consider $\mathbb {F}_{2^{n}}$ as a vector space $V$ of dimension $n$ over $\mathbb {F}_{2}$. Alongside the addition operation inherited from $\mathbb {F}_{2^{n}}$, introduce on $V$ the multiplication operation $(x, y) \mapsto x \circ y=\sqrt{x y}$. Here $x \mapsto \sqrt{x}$ is the automorphism of $\mathbb {F}_{2^{n}}$ inverse to $x \mapsto x^{2}$, so that $\sqrt{x+y}=\sqrt{x}+\sqrt{y}$. Show that $(V,+, \circ )$ is a commutative (nonassociative) algebra over $\mathbb {F}_{2}$, possessing the properties:

In any algebra the identity for associators holds

Verify this by direct checking, and show that if, in an algebra $A$ with identity 1 over a field $P$, the inclusion $(x, y, z) \in P \cdot 1$ holds for all associators, then $A$ is an associative algebra.

The octonion algebra $\mathcal{O}$ is an 8-dimensional non-associative algebra. It is defined in terms of the basis elements $e_{\mu }$ ($\mu = 0, 1, \ldots , 7$) and their multiplication table. $e_{0}$ is the unit element. We use Greek indices $(\mu , \nu , \ldots )$ to include the 0 and latin indices $(i, j, k, \ldots )$ when we exclude the 0. We define

The multiplication rules among the basis elements of octonions $e_{\mu }$ are given by

and

where $\delta_{ij}$ is the Kronecker delta and $\epsilon_{ijk}$ is $+1$ if $(ijk)$ is an even permutation of $(123)$, $-1$ if $(ijk)$ is an odd permutation of $(123)$ and 0 otherwise. We can formally summarize the multiplications as

where

with $\mu , \nu = 0, 1, \ldots , 7$, and $i, j, k = 1, 2, \ldots , 7$.

Consider the vector space $\mathbb {R}^{3}$ and the vector product $\times$. The vector product is not associative. The associator of three vectors $\overrightarrow {u}$, $\overrightarrow {v}$, $\overrightarrow {w}$ is defined by

The associator measures the failure of associativity.