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I'm having trouble understanding the relation between Lie groups and Lie algebras. I'm studying on this article of Kostant, Section 2.1.

So $K$ is compact connected simply connected simple Lie group, $T\subset K$ is any maximal torus and t and k are their Lie algebras. So by general result on Lie groups we have that k is a semisimple Lie algebra and t $\subset$ k is a maximal abelian subalgebra. Now he considers the Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ to be the respective complexifications of t and k.

Now my questions are:

Why do you need to take complexifications?
What's the relation between the root system of the pair $(\mathfrak{g},\mathfrak{h})$ and the root system of the pair $(K,T)$?
What's the relation between the Weyl group of the pair $(\mathfrak{g},\mathfrak{h})$ and the Weyl group of the pair $(K,T)$?

I want to clarify that I know that I'm lacking in Lie groups representation theory. I need to have an idea of the concepts stated above because in the work I'm studying they rarely come out but I don't need to get in deep of Lie groups theory because my concern is about Lie algebras, and I what to treat those concepts as "facts" about Lie groups.

Thanks to everyone,
Bye.

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  • $\begingroup$ I don't have time to write a full answer now. You may find some parts of my earlier answer about what weights and root systems are useful: math.stackexchange.com/a/3949435/101420 $\endgroup$
    – Vincent
    Commented Sep 27, 2023 at 9:59
  • $\begingroup$ In short: Why do you need to take complexifications? Because the whole root/weight business is about eigenvalues/eigenvectors (see the other answer for a more precise description of what I mean by this). A real matrix/linear transformation can have non-real complex eigenvalues but then there are no corresponding eigenvectors in the real space it acts on. Passing to the complexification of that space we do get eigenvectors and working with these make computations a lot easier $\endgroup$
    – Vincent
    Commented Sep 27, 2023 at 10:01
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    $\begingroup$ No complexifications are more of a big deal when working with Lie algbras than with groups. But sometimes the complexification passes by unnoticed $\endgroup$
    – Vincent
    Commented Sep 27, 2023 at 10:24
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    $\begingroup$ The reason that the distinction between real Lie algebras and their complexifications comes up more explicitly when people are talking about groups is that the Lie algebre OF a group is by construction always real $\endgroup$
    – Vincent
    Commented Sep 27, 2023 at 10:28
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    $\begingroup$ Yes, I got it, when studying Lie algebras we always take complex Lie algebras, that's why I never needed complexification...from Lie groups arise real algebras... $\endgroup$ Commented Sep 27, 2023 at 13:15

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Learning Lie theory regarding root systems generally proceeds as follows:

You learn about complex Lie algebras first. As Vincent says this makes things easier, because you always have eigenvalues. For example this means that semisimple elements are exactly diagonalisable elements. So when you take a Cartan subalgebra you can find root spaces and they are all 1-dimensional, etc.

When you want to tackle real semisimple Lie algebras you run into a couple problems. For example, consider $\mathfrak{su}(n)$. All its elements are semisimple but none are diagonalisable, so what does it even mean to be a root of $\mathfrak{su}(n)$? One natural path here is to just complexify the real Lie algebra $\mathfrak{g}$ to $\mathfrak{g}^\mathbb{C}$ find the root system of $\mathfrak{g}^\mathbb{C}$ and then refer to that as the root system of $\mathfrak{g}$. Of course there are several different real Lie algebras which complexify to $\mathfrak{g}^\mathbb{C}$ so this idea doesn't distinguish between these "real forms" until we add the extra information of what complex conjugation does to the roots (adding that information is exactly what a Satake diagram is for).

At this point let's mention groups. The root system of a Lie group is just the root system of its Lie algebra. Again there are multiple groups with the same Lie algebra so a root system won't distinguish between them, but we have at least that all those (connected) groups are related by having the same universal cover.

So the root system of $(K,T)$ is the root system of $(\mathfrak{k},\mathfrak{t})$ which is the root system of $(\mathfrak{k}^\mathbb{C},\mathfrak{t}^\mathbb{C})= (\mathfrak{g},\mathfrak{h})$ basically by definition. Likewise the Weyl groups are the same.

An important side note to be aware of is that there is another way to talk about root systems for real Lie algebras, especially for non-compact ones. This is known as the restricted root system. If we take a real Lie algebra and choose a maximal split toral subalgebra $\mathfrak{t}$ i.e. a maximal abelian subalgebra containing only diagonalisable elements (note: this is a tighter restriction then maximal toral subalgebra as that only requires semisimple). For a compact Lie algebra there are no such elements so we don't use restricted root systems for them as they are trivial. Otherwise, we get, in effect, a smaller version of a Cartan subalgebra and we can do the same procedure as for the complex Lie algebras. We divide the Lie algebra into root spaces but now these may have dimension greater than 1. We get a root system this way although we may get a "non-reduced" one $BC_n$ alongside the usual possibilities and the rank of these root systems is the dimension of $\mathfrak{t}$ which is usually smaller than the rank of the root system of the complexified Lie algebra. Sometimes you will see the restricted root system referred to casually as the root system but in the compact case you know that isn't happening as the restricted root system is always trivial for compact Lie algebras.

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  • $\begingroup$ I got your points. But, in these definitions, what's the relation between the Weyl group $W$ of the root system of $(\mathfrak{g},\mathfrak{h})$ and the Weyl group $N(T)/T$? How does $W$ act on $T$? I think they are isomorphic, but where this isomorphism come from? $\endgroup$ Commented Sep 27, 2023 at 13:23
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    $\begingroup$ There are indeed isomorphic. The normaliser $N(T)$ preserves $\mathfrak{t}$ (or $\mathfrak{h}$) under the adjoint action and this descends to an action on the root system. Then you just have to show that it contains elements which act as the simple reflections. You can happily quotient out by $Z(T) = T$ as that acts trivially and finally show that the simple reflections generate the whole of $N(T)/T$. I believe there is a proof of these in Hall's book on Lie groups (prop 11.35 and theorem 11.36) but I don't have it in front of me. $\endgroup$
    – Callum
    Commented Sep 27, 2023 at 13:52
  • $\begingroup$ This is a very good answer! $\endgroup$
    – Vincent
    Commented Sep 27, 2023 at 14:18

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