Proving Cartan Subalgebra $\mathbb{K} H$ is Self-Normalizer

[HDB1]

TL;DR

Cartan subalgebra

Please, How we can solve this:

$$
\mathfrak{h}=\mathbb{K} H \text { is a Cartan subalgebra of } \mathfrak{s l}_2 \text {. }
$$

it is abelian, but how we can prove it is self-normalizer, please:Dear @fresh_42 , if you could help,

 

[fresh_42]

As always: start with what we have.

The normalizer of $H\in \mathfrak{sl}(2)$ is given as $\{X\in \mathfrak{sl}(2)\,|\,[X,\mathbb{K}H]\subseteq \mathbb{K}H\}.$ Now set $X=eE+hH+fF$ and calculate
$$
[X,H]=[eE+hH+fF,H] \in \mathbb{K}\cdot H
$$
This gives you conditions for $e$ and $f$ and so for the form of $X.$

 

[HDB1]

As always: start with what we have.

The normalizer of $H\in \mathfrak{sl}(2)$ is given as $\{X\in \mathfrak{sl}(2)\,|\,[X,\mathbb{K}H]\subseteq \mathbb{K}H\}.$ Now set $X=eE+hH+fF$ and calculate
$$
[X,H]=[eE+hH+fF,H] \in \mathbb{K}\cdot H
$$
This gives you conditions for $e$ and $f$ and so for the form of $X.$

Thank you so much @fresh_42, please, is $\mathbb{K}{H}$ is the only cartan algebra in $\mathfrak{sl}{2}$, and does the characteristic of the field matter here?
I wonder why we chose $\mathbb{K}{H}$? Thank you in advance, I have to find other words to thank you,

 

[fresh_42]

Thank you so much @fresh_42, please, is $\mathbb{K}{H}$ is the only cartan algebra in $\mathfrak{sl}{2}$,...

Yes. It is the only CSA ( ... according to that basis; a different basis means a different representation. But $E,H,F$ is the standard basis, and $\mathfrak{K}H$ the standard CSA.

... and does the characteristic of the field matter here?

I don't think so, but as always: keep away from characteristic $2$. I'm not quite sure how they are defined over characteristic $2$ fields.

I wonder why we chose $\mathbb{K}{H}$?

$\mathbb{K}H$ is only the linear span, hence the entire subalgebra. We do not choose $H$. The procedure is as follows:

a) $\mathfrak{sl}(2)$ is not nilpotent, so it's no CSA of itself.
b) $\mathfrak{sl}(2)$ posseses no two-dimensional nilpotent subalgebras. Its two-dimensional subalgebras are solvable, but not nilpotent.
c) So any CSA of $\mathfrak{sl}(2)$ has to be one-dimensional.
d) $\mathfrak{K}\cdot H$ is a one-dimensional CSA of $\mathfrak{sl}(2).$

That's fine since it suffices for all our purposes. You could assume another one-dimensional CSA of $\mathfrak{sl}(2),$ say $\mathfrak{K}\cdot (eE+fF+hH).$ Maybe there is a solution with $e\neq 0$ or $f\neq 0.$ You could calculate whether this is possible or not, but with regard to chapter 15.3, it doesn't really make sense to search for another one if the first one, $\mathbb{K}\cdot H$, is so convenient. It will be just another basis in the end that has a more complicated multiplication table.

 

[HDB1]

Dear, @fresh_42 , Thank for the clarification, but , please, I could not read the last comment, Thanks in advance,