# What is the importance of Cartan decomposition of a semi-simple Lie algebra?

I just started learning about Cartan decomposition of semi-simple Lie algebras, and I'm curious to know what are their applications in studying semi-simple Lie algebras. My guess was that it might be helpful in their classification, but when I browsed a lecture notes on classification of semi-simple lie algebra, I didn't find any part which speak about Cartan decomposition, so I'm wondering why is it important to study Cartan decomposition of a semi-simple Lie algebra ?

## 1 Answer

So first things first the "semisimple Lie algebras" tag isn't very big so that might be why you are struggling to find people to answer your question (I came here from the Mathoverflow post but I think this is the right place for this question).

Secondly there are two decompositions often called the Cartan decomposition (you can see this in the comments on your mathoverflow post). One of these is the root space decomposition $$\mathfrak{g} = \mathfrak{h}\oplus \bigoplus_{\alpha\in \Delta}\mathfrak{g}^\alpha$$ but since this is very intimately linked to the classification via roots and Dynkin diagrams, I believe you must be referring to the other one.

The second one is the decomposition corresponding to a Cartan involution (I think this one is the better one to have the name Cartan decomposition). That is we decompose $$\mathfrak{g}$$ into a compact and noncompact part:

$$\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$$

The reason you might not see this at first when you glance through a treatment of semisimple Lie algebras is that they often focus on the complex Lie algebras first. The Cartan decomposition becomes more important when we start discussing real Lie algebras. They are really useful when it comes to classifying things about real Lie algebras.

For example, every Cartan subalgebra of a real Lie algebra is conjugate to one which intersects with the Cartan decomposition. i.e: $$\mathfrak{h} = (\mathfrak{h}\cap\mathfrak{k}) \oplus (\mathfrak{h}\cap\mathfrak{p})$$ Then two such $$\mathfrak{h}$$ are conjugate over $$G$$ if and only if they are conjugate over $$K$$ which simplifies understanding conjugacy classes of Cartan subalgebras in $$\mathfrak{g}$$ to understanding conjugacy classes of the "compact parts" of Cartan subalgebras in $$\mathfrak{k}$$.

• Thanks a lot @Callum for your helpful answer! Could you please elaborate a bit a further in why is the Cartan decomposition more important in studying real Lie algebras and not complex ones ?
– Mira
Commented Jun 17, 2022 at 17:42
• It is only defined for real Lie algebras. You can define it for a complex Lie algebra but only really by thinking of that complex Lie algebra as a real one. In that scenario it just picks out a compact real form: $\mathfrak{g} = \mathfrak{u} \oplus i\mathfrak{u}$ Commented Jun 18, 2022 at 6:25