In the proof of Lemma 3.2.2 in Chriss and Ginzburg's Representation Theory and Complex Geometry, the final step states that "the annihilator $\mathfrak{b}^\perp\subset \mathfrak{g}^\*$ gets identified with the annihilator of $\mathfrak{b}$ in $\mathfrak{g}$ with respect to the invariant form; the latter is equal to $\mathfrak{n}$, the nilradical of $\mathfrak{b}$". My question is, assuming that the invariant form is the Killing form, how to see that the annihilator of $\mathfrak{b}$ is precisely $\mathfrak{n}$?



The annihilator of a Borel subalgebra being its nilpotent radical
As in Chriss-Ginzburg p. 130, pick a regular semisimple $h\in\mathfrak h$ (Cartan subalgebra) and write $\mathfrak g\_a$ for the eigenspace of $\text{ad}\_h$ belonging to eigenvalue $a$. Invariance of the Killing form $\langle\cdot,\cdot\rangle$ implies $$0=\langle [h,X],Y\rangle+\langle X,[h,Y]\rangle=(a+b)\langle X,Y\rangle\qquad\text{for } (X,Y)\in\mathfrak g\_a\times\mathfrak g\_b.$$So
$$
\langle\cdot,\cdot\rangle\_{|\mathfrak g\_a\times\mathfrak g\_b}\text{ is }
\begin{cases}
\text{identically zero}&\text{if }a+b\ne0,\\
\text{nondegenerate}&\text{if }a+b=0,
\end{cases}
$$
and it follows readily that $\mathfrak b=\bigoplus\_{a\geqslant0}\mathfrak g\_a$ has annihilator $\mathfrak n=\bigoplus\_{a>0}\mathfrak g\_a$.


