Computation of killing form

Question

The killing form is denoted by $ B $. We know that for all $ X,Y \in gl\left(n,\mathbb{R}\right) $ $$ B\left(X,Y\right)=2n\ tr\left(XY\right)-2\ tr\left(X\right)tr\left(Y\right) $$ So for $ X,Y \in so\left(n,\mathbb{R}\right) $, $$ B\left(X,Y\right)=2n\ tr\left(XY\right) $$ since the traces of elements of $ so\left(n,\mathbb{R}\right) $ all are zero. However, the answer is $ \left(n-2\right)tr\left(XY\right) $... I can't find mistakes! Thanks in advance.

Answer 1

The Killing form on a Lie algebra doesn't restrict to the Killing form on its subalgebras. When you take the trace needed to compute the Killing form it's over a smaller and different vector space.

Answer 2

The Killing form of $\mathfrak{gl}_n(K)$ restricts to the Killing form on its ideals, but not on its subalgebras. Since $\mathfrak{sl}_n(K)$ is an ideal in $\mathfrak{gl}_n(K)$, the Killing form of $\mathfrak{sl}_n(K)$ really is $B(x,y)=2n\;tr(xy)$. The Killing form for $\mathfrak{sp}_{2n}(K)$ is given by $B(x,y)=(4n+2)\, tr(xy)$, and for $\mathfrak{so}_{n}(K)$ by $B(x,y)=(n-2)\, tr(xy)$; and this does not come from the Killing form of $\mathfrak{gl}_n(K)$.