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Let $\mathfrak{g}$ be a real semisimple Lie algebra. Then there are simple ideals $\mathfrak{g}_1,\ldots,\mathfrak{g}_n \subseteq \mathfrak{g}$, unique up to order, such that $$ \mathfrak{g} = \mathfrak{g}_1 \oplus \ldots \oplus \mathfrak{g}_n $$ and every ideal of $\mathfrak{g}$ is a sum of some of these simple ideals.

Are the following two statements equivalent?

  1. each $\mathfrak{g}_i$ is noncompact
  2. $\mathfrak{g}$ does not contain a nontrivial compact ideal

Clearly 2. implies 1., but I'm not sure about the converse. Here, a Lie algebra $\mathfrak{g}$ is called compact if it is the Lie algebra of a compact Lie group or equivalently, if $\text{Int}(\mathfrak{g})$ is compact.

(Here, $\text{Int}(\mathfrak{g})$ denotes the adjoint group of $\mathfrak{g}$, that is, the unique Lie group whose Lie algebra is the image of the adjoint representation of $\mathfrak{g}$.)

abenthy
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    It would have been useful to specify that you are considering Lie algebras over the reals. – YCor Jan 09 '18 at 04:13
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    For the converse, what have you tried? What are the ideals of $\mathfrak{g}$? – YCor Jan 09 '18 at 04:13
  • @YCor Thanks I have now edited the question. For the converse, my idea is that for $n=2$, if $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are noncompact then they are both not the lie algebra of a compact group. But if $\mathfrak{g}_1 \oplus \mathfrak{g}_2$ were the lie algebra of a compact group $G$, then somehow that would contain compact groups $G_1$ and $G_2$ those lie algebras are $\mathfrak{g}_1$ and $\mathfrak{g}_2$, in contradiction to the assumption. – abenthy Jan 14 '18 at 09:48
  • Do you know if the statement is true? I couldn't find it written down anywhere. By an exercise in Helgason, a symmetric space is of noncompact type iff the lie algebra of its isometry group is semisimple and has no nontrivial compact ideals. So this would be reduced to "semisimple and noncompact". – abenthy Jan 14 '18 at 09:53
  • What is the meaning of $\text{Int}(\mathfrak{g})$ here? – No One Aug 10 '20 at 05:12
  • @NoOne $\text{Int}(\mathfrak{g})$ is the so-called adjoint group of $\mathfrak{g}$. This is the unique Lie group whose Lie algebra is the image of the adjoint representation of $\mathfrak{g}$ (see e.g. Wikipedia). I've now added an explanation to the question. – abenthy Aug 11 '20 at 07:20

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