# 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 $L:A\rightarrow\operatorname{End}_{k}A$ given by $L_{a}b:=ab$ ($R_{a}b:=ba$).

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

 $\langle a,b\rangle:=\operatorname{tr~{}}(L_{a}L_{b}).$
###### 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

 $\langle a+lb,x\rangle=\operatorname{tr~{}}(L_{a+lb}L_{x})=\operatorname{tr~{}}% (L_{a}L_{x}+lL_{b}L_{x})=\operatorname{tr~{}}(L_{a}L_{x})+l\operatorname{tr~{}% }(L_{b}L_{x})=\langle a,x\rangle+l\langle b,x\rangle.$

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

 $\langle a,b\rangle=\operatorname{tr~{}}(L_{a}L_{b})=\operatorname{tr~{}}(L_{b}% L_{a})=\langle b,a\rangle.$

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

 $\langle ab,c\rangle=\langle a,bc\rangle.$ (1)

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

###### 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.

###### Proposition 4.

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

###### 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 $I\cap I^{\perp}=0$ 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:

 $A=A_{1}\oplus\cdots\oplus A_{s}.$

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 $I\cap I^{\perp}=0$ it is sufficient to assume $A$ has not ideal $I$ such that $I^{2}=0$. This is the content of the proof in

###### Theorem 5.

[1, Thm III.3] Let $A$ be a finite-dimensional weakly associative (trace) semisimple algebra over a field $k$ in which no ideal $I\neq 0$ of $A$ has $I^{2}=0$, 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 TraceFormsOnAlgebras 2013-03-22 16:28:01 2013-03-22 16:28:01 Algeboy (12884) Algeboy (12884) 4 Algeboy (12884) Topic msc 17A01 regular representation trace form