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

[HDB1]

TL;DR Summary

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,