HDB1 said:
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.
HDB1 said:
... 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.
HDB1 said:
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.