Nonassociative rings and algebras Definition

Structurable algebra

Builds on Field Nonassociative ring Ring with involution Jordan algebra

Definition

Let $F$ be a field of characteristic $0$ , and let $A$ be a finite-dimensional nonassociative algebra over $F$ with a unit element. An involution of $A$ is a linear map $x \mapsto \bar{x}$ with $\bar{\bar{x}} = x$ and $\overline{xy} = \bar{y}\, \bar{x}$ for all $x, y \in A$ . For $x, y, z \in A$ put $\set{x, y, z} = (x \bar{y}) z + (z \bar{y}) x - (z \bar{x}) y$ , and let $V_{x, y}$ be the map $z \mapsto \set{x, y, z}$ . The pair $(A, \bar{\ })$ is a structurable algebra if:

$x \mapsto \bar{x}$ is an involution of $A$ ;
$[V_{x, y}, V_{z, w}] = V_{\set{x, y, z}, w} - V_{z, \set{y, x, w}}$ for all $x, y, z, w \in A$ .

Here $[S, T] = ST - TS$ is the commutator of linear maps.

Remarks

Example: a unital associative algebra with an involution, such as $M_n(F)$ with the transpose. Then $\set{x, y, z} = x \bar{y} z + z \bar{y} x - z \bar{x} y$ , and clause 2 is checked by expanding both sides applied to a fifth element into associative words.

Example: a unital Jordan algebra with the identity map as involution. Clause 1 holds because the product is commutative, and $\set{x, y, z}$ becomes the Jordan triple product $(xy)z + (zy)x - (zx)y$ .

Non-example: the octonions with the identity map fail clause 1, since $\overline{xy} = xy$ differs from $\bar{y}\, \bar{x} = yx$ for noncommuting $x$ and $y$ . With the octonion conjugation instead, the octonions form a structurable algebra.

The rings with involution give the associative examples.

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Tag	`VAWQH650PF`
URI	`m2://VAWQH650PF`
Kind	definition
Writing	W0 stub
MSC	17A30, 17C50, 17B60
Defines	Structurable algebra
Uses	Field, Nonassociative ring, Ring with involution, Jordan algebra
Changed	3 November 2026

Structurable algebra

Definition

Remarks

Reading path

Further reading

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