trace((ad_x)(ad_y)) = 2n(trace(xy))

antiemptyv

1. The problem statement, all variables and given/known data

Let L be the Lie algebra [tex]sl(n, F)[/tex] and [tex]X = (x_{ij}, Y = (y_{ij}) \in L[/tex].

Prove

[tex]\kappa(X,Y) = 2n Tr(XY)[/tex],

where [tex]\kappa(,)[/tex] is the Killing form and [tex]Tr()[/tex] is the trace form.


2. Relevant equations

For any unit matrix [tex]E_{ij}[/tex] and any [tex]X \in L[/tex],

[tex]XE_{ij} = \sum_{m=1}^n x_{mi} E_{mj}[/tex] and
[tex]E_{ij}X = \sum_{m=1}^n x_{jm}E_{im}.[/tex]


3. The attempt at a solution

I have reduced this to the following:

[tex]Tr(XY) = \sum_{k=1}^n \sum_{m=1}^n x_{mk}y_{km}[/tex]

[tex]\kappa(X,Y) = Tr(ad_X ad_Y) = \sum_{k=1}^n (x_{ik}y_{ki} + y_{kj}x_{jk} ) - 2 \sum_{i=1}^n \sum_{j=1}^n x_{ii}y_{jj}[/tex]

Perhaps I have been at this for too long, but I don't see why exactly 2n times the first expression is equivalent to the second expression. Any guidance would be appreciated.