# Conjugacy of Borel Subalgebras: Proof in Humphreys' Introduction to Lie Algebras and Representation Theory

In the title referenced above a proof of the conjugacy of Borel subalgebras is given on page 84: We assume $$L$$ semisimple and let $$B$$ be a standard Borel subalgebra and $$B'$$ any other Borel subalgebra. We set $$N'$$ equal to the set of all nilpotent elements of $$B \cap B'$$ and assume $$N'\neq 0$$ for case 1. (Note that N' is an ideal of $$B \cap B'$$). If $$x \in N'$$ then $$ad x$$ acts nilpotently on the vector space $$B/B \cap B'$$ whence there exists nonzero $$y$$ in this vector space that is annihilated by all of $$N'$$ (by a linear algebra thm). In other words, there is a $$y$$ in $$B$$ outside of $$B'$$ that is sent into $$B \cap B'$$ by any $$ad x$$ for $$x \in N'$$. But since $$[x,y] \in [B,B]$$ then $$ad [x,y]$$ is nilpotent on $$L$$ (since $$B$$ is standard) and hence in $$N'$$. Therefore there is a $$y$$ in $$B$$ outside of $$B'$$ that is in the normalizer, $$K$$, of $$N'$$. Humphreys then wants to make a symmetric conclusion: That there is a $$y'$$ in $$B'$$ outside of $$B$$ that is in the normalizer, $$K$$, of $$N'$$. But how is this possible? Since $$[x,y'] \in [B',B']$$ we may only conclude $$ad [x,y']$$ is nilpotent on $$B'$$ and not on all of $$L$$ since we do not know that $$B'$$ is standard. Thus we can't know for sure if $$[x,y] \in N'$$ and can't conclude $$y'\in K$$. Thoughts?

As mentioned in the question, we have some $$y'\in B'$$ outside of $$B \cap B'$$ such that for all $$x \in N'$$ we have $$[x,y']$$ in $$B \cap B'$$. If all such $$[x,y']$$ lie in $$N'$$ there is no problem. So assume there is some $$x' \in N'$$ with $$[x'y']:=z \notin N'$$. Then we may Jordan-decompose within $$B \cap B'$$ as $$z=z_s+z_n$$ with $$z_s \neq 0$$, so by virtue of $$z$$ and $$z_n$$ commuting and being nilpotent on $$B'$$ we see $$z_s$$ is also nilpotent on $$B'$$. Since $$z_s$$ is semisimple this implies that on $$B'$$ we have $$ad (z_s) = 0$$. So the centralizer of $$z_s$$ in $$L$$, say $$C$$, contains $$B'$$. Now find a CSA for $$L$$, say $$H'$$, including $$z_s$$. Clearly $$H'$$ must also must be contained in $$C$$. Since $$z_s \in B' \subseteq C$$ we see that $$C$$ is an ideal in $$L$$ and therefore semisimple. As a CSA of $$L$$, $$H'$$ is also a CSA of $$C$$ and so there is a BSA $$B'' \subseteq C$$ containing $$H'$$. Now $$C$$ cannot be all of $$L$$ so by the inductive hypothesis we have $$f \in \varepsilon(C)$$ sending $$B''$$ to $$B'$$ and so also an $$F \in \varepsilon(L)$$ sending $$B''$$ to $$B'$$ . Hence $$B'$$ contains a conjugate of the CSA $$H'$$ of $$L$$, which is of course also a CSA of $$L$$. But it is not difficult to show that any BSA of $$L$$ which contains a CSA of $$L$$ is standard. Thus $$B'$$ must be standard, whence $$[B',B']$$ is nilpotent on all of $$L$$, but of course this contradicts $$z \notin N'$$.