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I am reading B.Hall's book "Lie groups, Lie algebras, and Representations". You can find there a definition for semisimple Lie algebra.

DEF1:

A complex Lie algebra $\mathfrak{g}$ is reductive if there exists a compact matrix Lie group K such that: $$\mathfrak{g} \cong \mathfrak{t}_{\mathbb{C}}. $$ Where $\mathfrak{t}$ is the Lie algebra of K. Lie algebra $\mathfrak{g}$ is semisimple if it is reductive and its centre is trivial.

$\square$

And the definition for compact real form.

DEF2:

A real subalgebra $\mathfrak{t}$ of complex semisimple Lie algebra $\mathfrak{g}$ is a compact real form of $\mathfrak{g}$ if:

  • $\mathfrak{t}$ isomorphic to Lie algebra of some compact Lie group.
  • Any element $Z \in \mathfrak{g}$ can be written uniquely as $Z=X+iY$ with $X,Y \in \mathfrak{t}$

$\square$

The question is, how can I see that any semisimple Lie algebra has such a compact real form? As far as I understand it is not true that for any $\mathfrak{t}$ elements of $\mathfrak{t}_{\mathbb{C}}$ can be uniquely written as $X+iY, \ X,Y\in \mathfrak{t}$. As an example if some $X$ belongs to $\mathfrak{t}$ together with $iX$ then $X$ as an element of $\mathfrak{t}_{\mathbb{C}}$ can be written as $X=X$ or as $X=i(-iX)$.

It seems like using the Killing form, one can show that $X$ and $iX$ can't belong to the algebra of compact group simultaneously. It would contradict the fact that for a compact group, the Killing form is negative definite. However, is there a method without using those?

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Hall's definition of semisimple appears to be precisely that it has a compact real form as well as trivial centre so this is automatic. He doesn't prove that this is identical to the usual definition of semisimple so I assume what you're asking is why his definition is equivalent to the usual one.

To address your points at the end it is not true that $X$ and $iX$ are both in $\mathfrak{t}$. This is a real subalgebra so it is not closed under multiplication by $i$, indeed $\mathfrak{g} = \mathfrak{t}^\mathbb{C} = \mathfrak{t} \oplus i\mathfrak{t}$. Using the Killing form we can confirm this since $\mathfrak{t}$ is negative definite so $(X,X) \leq 0$ and $(iX,iX) = -(X,X) \geq 0$ so $iX$ cannot be in $\mathfrak{t}$.

To show that any semisimple Lie algebra (from the usual definition) admits a compact real form we can proceed as follows. Take any real form $\mathfrak{g}_0 \leq \mathfrak{g}$ and pick a Cartan decomposition: $\mathfrak{g}_0 = \mathfrak{k}_0 \oplus\mathfrak{p}_0$ i.e. $\mathfrak{k}_0$ is a maximal compact subalgebra and $\mathfrak{p}_0$ is its orthocomplement. Then $\mathfrak{u}_0 = \mathfrak{k}_0 \oplus i\mathfrak{p}_0$ is a compact real form of $\mathfrak{g}$. Obviously you need to see that we can always find at least one real form and that such a thing admits a Cartan decomposition but this isn't too hard.

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    $\begingroup$ Hi! My question wasn't about equivalence to the standard definition. Correct me if I am wrong, but the fact that some subalgebra is real doesn't necessary means that it can't be complex. In fact, you can think about any complex algebra as about real one. As I stated at the end of the question, it is not the case when we talk about compact groups due to argument with the Killing form. I wonder whether it is possible to show it without using the Killing form because this way, though it is not hard, demands a few additional steps to show that the Killing form itself is negative definite. Thanks! $\endgroup$ Commented Jul 19, 2021 at 18:06
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    $\begingroup$ @NikolayEbel That is true but we have made a much stronger assumption here. We have already assumed that $\mathfrak{g} = \mathfrak{t}^\mathbb{C} = \mathfrak{t} \oplus i\mathfrak{t}$ for some compact, real $\mathfrak{t}$ in Def 1. By Def 2 this immediately means $\mathfrak{t}$ is a compact real form (although we can make others). Note that by writing $\mathfrak{t}^\mathbb{C}$ we are assuming that $\mathfrak{t}$ is a real form (and thus $i\mathfrak{t}$ is complementary to $\mathfrak{t}$) although we can deduce it must be via the Killing form as well. $\endgroup$
    – Callum
    Commented Jul 20, 2021 at 9:22
  • $\begingroup$ $\mathfrak{t}^\mathbb{C}$ precisely consists of unique linear combinations of the form $X = iY$ for $X,Y \in \mathfrak{t}$. $X$ and $i(-iX)$ are not different combinations since $i(-i) = 1$ $\endgroup$
    – Callum
    Commented Jul 20, 2021 at 9:28

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