trace forms on algebras

Given an finite dimensional algebra $A$ over a field $k$ we define theleft(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 adjointrepresentation 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 ofleft/right adjoint representations.

The trace form of $A$ is defined as$⟨,⟩:T\times T\to k$:

$$⟨a,b⟩:=\mathrm{tr}(L_a{L}_b).$$

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

$$⟨a+lb,x⟩=\mathrm{tr}(L_{a+lb}{L}_x)=\mathrm{tr}(L_a{L}_x+l{L}_b{L}_x)=\mathrm{tr}(L_a{L}_x)+l\mathrm{tr}(L_b{L}_x)=⟨a,x⟩+l⟨b,x⟩.$$

Furthermore, $\mathrm{tr}(fg)=\mathrm{tr}(gf)$ is general property of traces, thus

$$⟨a,b⟩=\mathrm{tr}(L_a{L}_b)=\mathrm{tr}(L_b{L}_a)=⟨b,a⟩.$$

So the trace form is a symmetric bilinear form.∎

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

$$⟨ab,c⟩=⟨a,bc⟩.$$(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 formis 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 $z\in A$,$⟨xy,z⟩=⟨x,yz⟩=0$. 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 traceform 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 smallfields/characteristics, especially when characteristic is 2.]

More can be said when ideals are considered.

Proposition 4.

Given a weakly associative algebra $A$, thenif $I$ is an ideal of $A$ then so is $I^{\mathrm{⟂}}$.

Proof.

Given $a\in I^{⟂}$, then for all $b\in A$ and $c\in I$, then $bc\in I$as $I$ is an ideal and so $⟨ab,c⟩=⟨a,bc⟩=0$as $a\in I^{⟂}$. This makes $ab\in I^{⟂}$ so $I$ is a right ideal.Likewise $⟨c,ba⟩=⟨cb,a⟩=0$ so $ba\in I^{⟂}$ andthus $I^{⟂}$ is an ideal of $A$.∎

To proceed one factors out the radical so that $A$ is semisimple. Thengiven an ideal $I$ of $A$, if $I\cap I^{⟂}=0$ then as the trace formis a non-degenerate bilinear from, $A=I\oplus I^{⟂}$, and so by iteratingwe 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 thealgebra be a direct product of simple algebras. To obtain the property $I\cap I^{⟂}=0$ 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 finite-dimensional weakly associative (trace) semisimplealgebra 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 simplealgebras.

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.