Given an finite dimensional algebra $A$ over a field $k$ we define the left(right) regular representation of $A$ as the map $L:A\rightarrow\End_{k}A$ given by $L_{a}b:=ab$ ($R_{a}b:=ba$).

The trace form of $A$ is defined as $\langle,\rangle:T\times T\rightarrow k$:

The symmetric property can be interpreted as a weak form of commutativity of the product: $a,b\in A$ commute within their trace from. A more essential property arises for certain algebras and can be interpreted as “the product is associative within the trace” and written as

We shall call such an algebra weakly associative though the term is not standard.

This property is clear for all associative algebras as:

When we use a Lie algebra, the trace form is commonly called the Killing form which has property (1). A result of Koecher shows that Jordan algebras also have this property.

From this result many authors define an algebra to be semi-simple if its trace form is non-degenerate. In this way, $A/R$, $R$ the radical of $A$, is semi-simple. [Some variations on this definition are often required over small fields/characteristics, especially when characteristic is 2.]

More can be said when ideals are considered.

To proceed one factors out the radical so that $A$ is semisimple. Then given an ideal $I$ of $A$, if $I\cap I^{\perp}=0$ then as the trace form is a non-degenerate bilinear from, $A=I\oplus I^{\perp}$, and so by iterating we produce a decomposition of $A$ into minimal ideals:

Hence we arrive at the alternative definition of a semisimple algebra: that the algebra be a direct product of simple algebras. To obtain the property $I\cap I^{\perp}=0$ it is sufficient to assume $A$ has not ideal $I$ such that $I^{2}=0$. This is the content of the proof in

Alternatively any bilinear form with (1) can be used. However, the trace form is always definable and the desired properties are easily translated into implications about the multiplication of the algebra.