On Humphreys’ proof of conjugacy of Borel subalgebras

lie-algebras

I am reading Humphreys’ Introduction to Lie Algebras and Representation Theory and have trouble understanding the first step of the proof of conjugacy of Borel subalgebras (§16.4).

Hereinafter $ L $ is a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic $ 0 $.

Some Definitions and Facts

  • A Borel subalgebra of $ L $ is a maximal solvable subalgebra of $ L $.
  • Let $ H $ be a Cartan subalgebra of $ L $, $ \Phi $ be the root system of $ (L, H) $, and
    $$
    L = H \oplus \bigoplus_{\alpha \in \Phi} L_\alpha
    $$
    be the root space decomposition of $ L $. Fix a basis $ \Delta $ of $ \Phi $ and let
    \begin{align*}
    B(\Delta) &= H \oplus \bigoplus_{\alpha \succ 0} L_\alpha, \\
    N(\Delta) &= \bigoplus_{\alpha \succ 0} L_\alpha.
    \end{align*}
    $ B(\Delta) $ is called a standard Borel subalgebra of $ L $ (relative to $ H $).

Then, as stated in §16.3,

  1. $ N(\Delta) = [B(\Delta), B(\Delta)] $.
  2. $ B(\Delta) $ is a Borel subalgebra of $ L $, and $ N(\Delta) $ is a nilpotent subalgebra of $ L $. Moreover, we can see that $ \operatorname{ad}_L x $ is nilpotent for $ x \in N(\Delta) $.

Question

Fix a standard Borel subalgebra $ B $ of $ L $ and take an arbitrary Borel subalgebra $ B' $. Let
$$
N' = \{x \in B \cap B' \mid \text{$ \operatorname{ad}_L x $ is nilpotent}\},
$$
and consider the case that $ N' \neq 0 $. We can see that $ N' $ is an ideal in $ B \cap B' $ and that its normalizer $ K = N_L(N') $ is a proper subalgebra of $ L $ (p. 85, ll. 2–5). Next, Humphreys says that (p. 85, ll. 6–11. [[ ]] is by the quoter):

Next we show that $ B \cap B' $ is properly contained in both $ B \cap K $, $ B' \cap K $. For consider the action of $ N' $ on $ B/(B \cap B') $ induced by $ \operatorname{ad} $. Each $ x \in N' $ acts nilpotently on this vector space, so by Theorem 3.3 [[the main step of the proof of Engel’s Theorem]] there must exist nonzero $ y + (B \cap B') $ killed by all $ x \in N' $, i.e., such that $ [xy] \in B \cap B' $, $ y \notin B \cap B' $. But $ [xy] $ is also in $ [BB] $, so is nilpotent; this forces $ [xy] \in N' $, or $ y \in N_B(N') = B \cap K $, while $ y \notin B \cap B' $. Similarly, $ B \cap B' $ is properly contained in $ B' \cap K $.

I have trouble understanding the last sentence “Similarly, $ B \cap B' $ is properly contained in $ B' \cap K $”; it seems to me that we use the fact that $ B $ is standard in “$ [xy] $ is also in $ [BB] $, so is nilpotent” and that this argument is no longer valid when $ B $ is replaced by $ B' $.

Best Answer

I also thought that I need the condition that B is standard Borel for that part. However, it turned out that the only condition I need is that B is solvable.

Lie's Theorem (4.1 in Humphreys' book) says that a solvable Lie subalgebra of $\mathfrak{gl}(V)$ for a finite dimensional vector space $V$ must be contained in the Lie algebra of upper triangular matrices with respect to some basis of $V$. We can apply this on the solvable Lie algebra $\text{ad}_L B$. Then $[\text{ad}_L B, \text{ad}_L B]=\text{ad}_L [B,B]$ must be strictly upper triangular, so the elements of $[B,B]$ are nilpotent in $L$.

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