Define the matrix commutator $\text{ad}_X$ as $$\text{ad}_XY=[X,Y]=XY-YX$$ where $X,Y\in\mathfrak{g}$ and $\mathfrak{g}$ is the Lie algebra associated to Lie group $G$.
Then on Lie group $G$, the Killing form is defined as $$B=trace(\text{ad}_X\cdot\text{ad}_Y)$$
What I don't understand is how to get the trace of $\text{ad}_X\cdot\text{ad}_Y$ . In my opinion, $\text{ad}_X$ and $\text{ad}_Y$ are just mappings, how can we get the trace of one mapping?
Could you please give some easy examples to explain this definition? For example, on Lie group $SE(2)$, how to get $B$?
Thanks.
The linear map $ad(x)\colon y\mapsto [x,y]$ has a matrix associated, with respect to the basis of the Lie algebra, as any linear map has a matrix associated with respect to the basis of the vector space. So we can multiply the two matrices $ad(x)$ and $ad(y)$ and take its trace.
For example, let $L=\mathfrak{sl}_2(K)$ with vector space basis $(e,f,h)$ and Lie brackets $[e,f]=h, [h,e]=2e, [h,f]=-2f$. Then the matrices for $ad(e)$, $ad(f)$ and $ad(h)$ are given as follows:
$$ ad(e)=\begin{pmatrix} 0 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\; ad(f)=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 2 \\ -1 & 0 & 0 \end{pmatrix},\; ad(h)=\begin{pmatrix} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$