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

[antiemptyv]

Homework Statement

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

Prove

$$\kappa(X,Y) = 2n Tr(XY)$$,

where $$\kappa(,)$$ is the Killing form and $$Tr()$$ is the trace form.

Homework Equations

For any unit matrix $$E_{ij}$$ and any $$X \in L$$,

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

The Attempt at a Solution

I have reduced this to the following:

$$Tr(XY) = \sum_{k=1}^n \sum_{m=1}^n x_{mk}y_{km}$$

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

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.

 

[mighty2000]

To prove that \kappa(X,Y) = 2n Tr(XY), we can use the fact that the Killing form is defined as \kappa(X,Y) = Tr(ad_X ad_Y), where ad_X and ad_Y are the adjoint actions of X and Y, respectively.

Using the given equations, we can rewrite the adjoint actions as:

ad_X(E_{ij}) = [X, E_{ij}] = \sum_{m=1}^n x_{mi} E_{mj} - \sum_{m=1}^n x_{mj} E_{im}

ad_Y(E_{ij}) = [Y, E_{ij}] = \sum_{m=1}^n y_{mi} E_{mj} - \sum_{m=1}^n y_{mj} E_{im}

Substituting these into the definition of the Killing form, we get:

\kappa(X,Y) = Tr(ad_X ad_Y) = \sum_{i=1}^n \sum_{j=1}^n (ad_X(E_{ij}))_{ij} (ad_Y(E_{ij}))_{ji}

= \sum_{i=1}^n \sum_{j=1}^n (\sum_{m=1}^n x_{mi} E_{mj} - \sum_{m=1}^n x_{mj} E_{im})_{ij} (\sum_{m=1}^n y_{mj} E_{mi} - \sum_{m=1}^n y_{mi} E_{mj})_{ji}

= \sum_{i=1}^n \sum_{j=1}^n (\sum_{m=1}^n x_{mi} y_{mj} - \sum_{m=1}^n x_{mj} y_{mi}) (\sum_{m=1}^n y_{mi} x_{mj} - \sum_{m=1}^n y_{mj} x_{mi})

= \sum_{i=1}^n \sum_{j=1}^n (\sum_{m=1}^n x_{mi} y_{mj} y_{mi} x_{mj}) - (\sum_{m=1}^n x_{mi} y_{mj} y_{mj} x_{mi})

= \sum_{