trace forms on algebras
Given an finite dimensional algebra^{} $A$ over a field $k$ we define the left(right) regular representation^{} of $A$ as the map $L:A\to {\mathrm{End}}_{k}A$ given by ${L}_{a}b:=ab$ (${R}_{a}b:=ba$).
Example 1.
In a Lie algebra^{} the left representation is called the adjoint^{} representation and denoted $\mathrm{ad}\mathit{}x$ and defined $\mathrm{(}\mathrm{ad}\mathit{}x\mathrm{)}\mathit{}\mathrm{(}y\mathrm{)}\mathrm{=}\mathrm{[}x\mathrm{,}y\mathrm{]}$. Because $\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{=}\mathrm{}\mathrm{[}y\mathrm{,}x\mathrm{]}$ in characteristic^{} not 2, there is generally no distinction of left/right adjoint representations.
The trace form of $A$ is defined as $\u27e8,\u27e9:T\times T\to k$:
$$\u27e8a,b\u27e9:=\mathrm{tr}({L}_{a}{L}_{b}).$$ 
Proposition 2.
The trace form is a symmetric bilinear form^{}.
Proof.
Given $a,b,x\in A$ and $l\in k$ then ${L}_{a+lb}x=(a+lb)x=ax+lbx={L}_{a}x+l{L}_{b}x$. So ${L}_{a+lb}={L}_{a}+l{L}_{b}$. So we have
$$\u27e8a+lb,x\u27e9=\mathrm{tr}({L}_{a+lb}{L}_{x})=\mathrm{tr}({L}_{a}{L}_{x}+l{L}_{b}{L}_{x})=\mathrm{tr}({L}_{a}{L}_{x})+l\mathrm{tr}({L}_{b}{L}_{x})=\u27e8a,x\u27e9+l\u27e8b,x\u27e9.$$ 
Furthermore, $\mathrm{tr}(fg)=\mathrm{tr}(gf)$ is general property of traces, thus
$$\u27e8a,b\u27e9=\mathrm{tr}({L}_{a}{L}_{b})=\mathrm{tr}({L}_{b}{L}_{a})=\u27e8b,a\u27e9.$$ 
So the trace form is a symmetric bilinear form. ∎
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
$$\u27e8ab,c\u27e9=\u27e8a,bc\u27e9.$$  (1) 
We shall call such an algebra weakly associative though the term is not standard.
This property is clear for all associative algebras as:
$${L}_{ab}{L}_{c}(x)=((ab)c)x=(a(bc))x={L}_{a}{L}_{bc}x.$$ 
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.
Proposition 3.
Given a weakly associative algebra, then the radical^{} of the trace form is an ideal of the algebra.
Proof.
We know the radical of form $R$ is a subspace^{} so we must simply show that $R$ is an ideal. Given $x\in R$ and $y\in A$ then for all $z\in A$, $\u27e8xy,z\u27e9=\u27e8x,yz\u27e9=0$. Thus $xy\in R$. Likewise $yz\in R$ so $R$ is a twosided ideal^{} of $A$. ∎
From this result many authors define an algebra to be semisimple^{} if its trace form is nondegenerate. In this way, $A/R$, $R$ the radical of $A$, is semisimple. [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.
Proposition 4.
Given a weakly associative algebra $A$, then if $I$ is an ideal of $A$ then so is ${I}^{\mathrm{\u27c2}}$.
Proof.
Given $a\in {I}^{\u27c2}$, then for all $b\in A$ and $c\in I$, then $bc\in I$ as $I$ is an ideal and so $\u27e8ab,c\u27e9=\u27e8a,bc\u27e9=0$ as $a\in {I}^{\u27c2}$. This makes $ab\in {I}^{\u27c2}$ so $I$ is a right ideal^{}. Likewise $\u27e8c,ba\u27e9=\u27e8cb,a\u27e9=0$ so $ba\in {I}^{\u27c2}$ and thus ${I}^{\u27c2}$ is an ideal of $A$. ∎
To proceed one factors out the radical so that $A$ is semisimple^{}. Then given an ideal $I$ of $A$, if $I\cap {I}^{\u27c2}=0$ then as the trace form is a nondegenerate bilinear^{} from, $A=I\oplus {I}^{\u27c2}$, and so by iterating we produce a decomposition of $A$ into minimal ideals:
$$A={A}_{1}\oplus \mathrm{\cdots}\oplus {A}_{s}.$$ 
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 $I\cap {I}^{\u27c2}=0$ it is sufficient to assume $A$ has not ideal $I$ such that ${I}^{2}=0$. This is the content of the proof in
Theorem 5.
[1, Thm III.3] Let $A$ be a finitedimensional weakly associative (trace) semisimple algebra over a field $k$ in which no ideal $I\mathrm{\ne}\mathrm{0}$ of $A$ has ${I}^{\mathrm{2}}\mathrm{=}\mathrm{0}$, then $A$ is a direct product of minimal ideals, that is, of simple algebras.
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.
References

1
Jacobson, Nathan Lie Algebras, Interscience Publishers, New York, 1962.
 2 Koecher, Max, The Minnesota notes on Jordan algebras and their applications. Edited and annotated by Aloys Krieg and Sebastian Walcher. [B] Lecture Notes in Mathematics 1710. Berlin: Springer. (1999).
Title  trace forms on algebras 

Canonical name  TraceFormsOnAlgebras 
Date of creation  20130322 16:28:01 
Last modified on  20130322 16:28:01 
Owner  Algeboy (12884) 
Last modified by  Algeboy (12884) 
Numerical id  4 
Author  Algeboy (12884) 
Entry type  Topic 
Classification  msc 17A01 
Defines  regular representation 
Defines  trace form 