Let g\mathfrak g be a finite-dimensional over a field k\Bbbk. Its Killing form is the symmetric bilinear form

B:g×gk,B(x,y)=tr(adxady).B:\mathfrak g\times \mathfrak g\to \Bbbk,\qquad B(x,y)=\mathrm{tr}(\mathrm{ad}_x\circ \mathrm{ad}_y).

Here ad\operatorname{ad} is the . The form is .

Examples

For sl2(C)\mathfrak{sl}_2(\mathbb C), with basis

H=(1001),E=(0100),F=(0010),H=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad E=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad F=\begin{pmatrix}0&0\\1&0\end{pmatrix},

one computes, using adX(Y)=[X,Y]\operatorname{ad}_X(Y)=[X,Y], that

B(H,H)=8,B(E,F)=4,B(H,E)=B(H,F)=B(E,E)=B(F,F)=0.B(H,H)=8,\qquad B(E,F)=4,\qquad B(H,E)=B(H,F)=B(E,E)=B(F,F)=0.

For X,Ysln(C)X,Y\in\mathfrak{sl}_n(\mathbb C), the Killing form is

B(X,Y)=2ntr(XY).B(X,Y)=2n\,\mathrm{tr}(XY).
Remarks

Over a field of characteristic 00, nondegeneracy of BB characterizes .