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 $\langle,\rangle:T\times T\rightarrow k$ : $$ \langle a,b\rangle:=\tr(L_a L_b) $$

Proof. Given $a,b,x\in A$ and $l\in k$ then $L_{a+lb} x=(a+lb)x=ax+lbx=L_{a}x+lL_{b}x$ . So $L_{a+lb}=L_a+lL_b$ . So we have $$ \langle a+lb,x\rangle=\tr(L_{a+lb}L_x)=\tr(L_a L_x+lL_b L_x) =\tr(L_a L_x)+l\tr(L_b L_x)=\langle a,x\rangle +l\langle b,x\rangle $$ Furthermore, $\tr(fg)=\tr(gf)$ is general property of traces, thus $$ \langle a,b\rangle=\tr(L_a L_b)=\tr(L_bL_a)=\langle b,a\rangle $$ 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 \begin{equation}\label{eq:assoc} \langle ab,c\rangle=\langle a,bc\rangle. \end{equation}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 (). A result of Koecher shows that Jordan algebras also have this property.

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 $z\in A$ , $\langle xy,z\rangle=\langle x,yz\rangle=0$ . Thus $xy\in R$ . Likewise $yz\in R$ so $R$ is a two-sided ideal of $A$ .

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.

Proof. Given $a\in I^\perp$ , then for all $b\in A$ and $c\in I$ , then $bc\in I$ as $I$ is an ideal and so $\langle ab,c\rangle=\langle a,bc\rangle=0$ as $a\in I^\perp$ . This makes $ab\in I^\perp$ so $I$ is a right ideal. Likewise $\langle c,ba\rangle=\langle cb,a\rangle=0$ so $ba\in I^\perp$ and thus $I^\perp$ 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 $I\intersect I^\perp=0$ 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: $$ A=A_1\oplus \cdots\oplus A_s $$ 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 $I\intersect I^\perp=0$ 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 () 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.

Bibliography

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).