TITLE: How does the lie algebra capture compactness of the lie group?
QUESTION [8 upvotes]: This is a soft question.
Let $V,\rho$ be a representation of the lie algebra $\mathfrak{so}_3(\mathbb{R})$. Then if I understand everything right, $V$ is necessarily completely reducible, because the representation
$$\rho:\mathfrak{so}_3\rightarrow \mathfrak{gl}(V)$$
is the derivative of a representation
$$P: G\rightarrow GL(V)$$
where $G$ is either $SO_3$ or its universal cover $SU_2$. Because the latter is compact, there is a $G$-invariant inner product on $V$, obtained by beginning with an arbitrary inner product and averaging it over $G$ with respect to a Haar measure. Then the orthogonal complement of any $P$-subrepresentation of $V$ will also be a $P$-subrepresentation; thus $V$ is completely reducible as a representation of the lie group $G$. But a $P$-subrepresentation is also a $\rho$-subrepresentation and vice versa, so $V$ is also completely reducible as a representation of the lie algebra $\mathfrak{so}_3$.
Here's what's bugging me. In my mind, the fact that any representation of $SO_3$ is completely reducible is a consequence of the topological fact that its universal cover is compact, as in the above paragraph. But this same fact is reflected in the representations of $\mathfrak{so}_3$, which does not have any topological differences from any other three-dimensional lie algebra. So the topological information about $SO_3$ must be being captured in purely algebraic information about $\mathfrak{so}_3$.

What is that information? And how is it related to compactness?

Apologies for the softness of this question. I am unsatisfied but I am not sure in advance what kind of answer will satisfy me. Any thoughts you have for me are appreciated. Thanks so much.

REPLY [2 votes]: Let $\mathfrak g$ be a semisimple real Lie algebra. One can show that the following are equivalent:
i) For every $x \in \mathfrak g$, the linear map $ad(x) : \mathfrak g \rightarrow \mathfrak g$ is semisimple. ("Every element of $\mathfrak g$ is $ad$-semisimple.")
ii) The only $x \in \mathfrak g$ such that $ad(x)$ is a nilpotent endomorphism is $x=0$. ("$\mathfrak g$ has no nonzero $ad$-nilpotent elements.")
iii) The only $x \in \mathfrak g$ such that $ad(x)$ is diagonalisable (over $\mathbb R$) is $x=0$. ("$\mathfrak g$ has no nonzero $ad$-diagonalisable elements.")
iv) The only parabolic subalgebra of $\mathfrak g$ (in the sense of https://math.stackexchange.com/a/4135409/96384) is $\mathfrak g$ itself.
These equivalences actually hold over any field of characteristic $0$ and are part of Lemma 3.2.1. of my thesis. In the case of the field $\mathbb R$, one can further show equivalence to
v) The Killing form of $\mathfrak g$ is negative definite.
Combine this with Justin Campbell's answer, and i)-iv) give you a bunch of very algebraic criteria indeed. I like to advertise them because they seem to be not well-known enough. Look e.g. how easy my answer to Showing the Lie Algebras $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$ are not isomorphic. is if one knows criterion ii.