I have a subalgebra $B = b(n,F)$ of L.
$b(n,F)$ is the subalgebra consisting of all upper triangular matrices. Then $b(n,F)$ is clear NOT a cartan subalgebra since it is not nilpotent.
But I think that the relation $N_L(B)=B$ still holds...
Can someone please help me to show this? Or to tell me I am not correct in my assertion.
Thanks in advance!
Every Borel subalgebra $B$ of a semisimple Lie algebra $L$ is self-normalizing (the letter $B$ stands for Borel subalgebra), because it is a maximal solvable subalgebra. The proof goes as follows:
If $x \in L$ normalizes $B$, we may create $B + Fx$, which is certainly a subalgebra of $L$. Here $F$ is the field.
Clearly
$[B + Fx,B + Fx] ⊂ B$, so $B + Fx$ is solvable. Now B is Borel (maximal solvable) so that $x$ must be an element of $B$. Hence we obtain $N_L(B)=B$. Clearly $B$ is not nilpotent, but solvable.
In your example, where $B=\mathfrak{b}_n(\mathbb{C})$, the Lie algebra $L$ is $\mathfrak{sl}_n(\mathbb{C})$, and its Cartan subalgebra $H$ consists of the diagonal matrices of trace zero, so it is abelian.
