The annihilator of a Borel subalgebra being its nilpotent radical

Question

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 $\mathfrak{n}$?

Answer 1

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 $\mathfrak n=\bigoplus_{a>0}\mathfrak g_a$.

Answer 2

Say $\mathfrak{g}$ is a semisimple Lie algebra over an algebraically closed field of characteristic 0 such as $\mathbb{C}$. The nondegeneracy of the Killing form $\kappa$ of $\mathfrak{g}$ is one way to characterize semisimplicity. Note that $\kappa$ induces a canonical identification between $\mathfrak{g}$ and $\mathfrak{g}^*$.

The fact that the orthogonal complement (under the form $\kappa$) of a Borel subgalgebra $\mathfrak{b}$ of $\mathfrak{g}$ is just the nilradical $\mathfrak{n}$ of $\mathfrak{b}$ is quite old, though I'm not sure who first observed it. (I learned it from Dan Mostow in a seminar on Lie algebras at Yale in the early 1960s.) This is used for example in one approach to the more subtle fact that the intersection of any two Borel subalgebras in a semisimple Lie algebra contains a Cartan subalgebra. See the outline of an argument in my 1972 Springer text in Exercise 16.8.

To compute the orthogonal complement of $\mathfrak{b}$, note that the dimensions in the asserted answer are correct (by the nondegeneracy of the form), since $\dim \mathfrak{g} = \dim \mathfrak{b} + \dim \mathfrak{n}^-$ for the direct sum of negative root spaces $\mathfrak{n}^-$ of $\mathfrak{b}$. Thus it suffices to prove that any positive root space is orthogonal to $\mathfrak{b}$ under $\kappa$. For this the explicit computations in my earlier $\S8$ can be used.

I'm not sure what is written down explicitly in the literature, but I'd emphasize that the computation here is fairly straightforward and well-known.

Answer 3

This is a proof I came up myself. Since $\dim \mathfrak{b}+\dim\mathfrak{n}=\dim \mathfrak{g}$ and the Killing form $\kappa$ is non-degenerate, it suffices to prove that $\kappa(x,n)=0$ for any $x\in \mathfrak{b}$ and $n\in \mathfrak{n}$. Now by Jordan decomposition we may write $x=x_s+x_n$, and we just need to prove that $\kappa(x_s,n)=\kappa(x_n,n)=0$. On the one hand, since $x_s\in \mathfrak{h}$ for some Cartan subalgebra $\mathfrak{h}$ on which the restriction of $\kappa$ is non-degenerate, it follows that $\kappa(x_s,n)=0$; on the other hand, since $x_n$ and $n$ are in the same nilpotent subalgebra $\mathfrak{n}$, on which the restriction of $\kappa$ is null, it follows that $\kappa(x_n,n)=0$ as well.