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

In summary: Thank you so much @fresh_42, please, is is the only cartan algebra in ,...Yes. It is the only CSA ( ... according to that basis; a different basis means a different representation. But is the standard basis, and the standard CSA.
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HDB1
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TL;DR Summary
Cartan subalgebra
Please, How we can solve this:



it is abelian, but how we can prove it is self-normalizer, please:Dear @fresh_42 , if you could help, :heart: 🥹
 
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As always: start with what we have.

The normalizer of is given as Now set and calculate

This gives you conditions for and and so for the form of
 
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fresh_42 said:
As always: start with what we have.

The normalizer of is given as Now set and calculate

This gives you conditions for and and so for the form of
Thank you so much @fresh_42, please, is is the only cartan algebra in , and does the characteristic of the field matter here?
I wonder why we chose ? Thank you in advance, I have to find other words to thank you, :heart:
 
  • #4
HDB1 said:
Thank you so much @fresh_42, please, is is the only cartan algebra in ,...
Yes. It is the only CSA ( ... according to that basis; a different basis means a different representation. But is the standard basis, and the standard CSA.
HDB1 said:
... and does the characteristic of the field matter here?
I don't think so, but as always: keep away from characteristic . I'm not quite sure how they are defined over characteristic fields.

HDB1 said:
I wonder why we chose ?
is only the linear span, hence the entire subalgebra. We do not choose . The procedure is as follows:

a) is not nilpotent, so it's no CSA of itself.
b) posseses no two-dimensional nilpotent subalgebras. Its two-dimensional subalgebras are solvable, but not nilpotent.
c) So any CSA of has to be one-dimensional.
d) is a one-dimensional CSA of

That's fine since it suffices for all our purposes. You could assume another one-dimensional CSA of say Maybe there is a solution with or 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, , is so convenient. It will be just another basis in the end that has a more complicated multiplication table.
 
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Dear, @fresh_42 , Thank for the clarification, but , please, I could not read the last comment, Thanks in advance, :heart:
 

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