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Lie's Third Theorem says that every finite-dimensional real Lie algebra is (up to isomorphism) the Lie algebra of some Lie group. We can improve the statement in two different directions:

  1. Every finite-dimensional real Lie algebra is (up to isomorphism) the Lie algebra of some matrix Lie group. This is explained in Lie Groups, Lie Algebras, and Representations by Brian C. Hall, section 5.10. (By definition, a matrix Lie group is a closed subgroup of $\operatorname{GL}(n;\mathbb{C})$. It follows from the closed subgroup theorem that matrix Lie groups are indeed Lie groups. See Corollary 3.45 of Hall's book.)
  2. Every finite-dimensional real Lie algebra is (up to isomorphism) the Lie algebra of some simply connected Lie group. This is proved from Lie's Third Theorem, using the construction of the universal covering group of a Lie group $G$, which is shown to have same Lie algebra as $G$.

I was wondering if we could mix 1 and 2. I was trying to decide if the following is true or false:

Conjecture: Every finite-dimensional real Lie algebra is the Lie algebra of some simply connected matrix Lie group.

It happens that the universal covering group of a matrix Lie group is, in general, nonmatrix (i.e., non-isomorphic to any matrix Lie group. See Sect. 5.8 of Hall's book for an example of this phenomenon). We cannot apply then this strategy to prove that the conjecture is true. Also, two simply connected Lie groups are isomorphic if and only if their Lie algebras are isomorphic. So if the conjecture were false, we could try to find a simply connected Lie group which is nonmatrix. The Lie algebra of such a group would be a counterexample to the conjecture. But the only example I know of a connected Lie group which is nonmatrix is one that is not simply connected (namely, that of Sect. 4.8, A Nonmatrix Lie group, from Hall's book).

Since I have studied almost exclusively matrix Lie groups (reading Hall's book; they are the ones that the book studies), I don't have more ideas about where to look for counterexamples or positive arguments for the conjecture. All I know from Lie group theory is the five chapters of Hall's book.

Any help will be appreciated :)

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No, that's not true. The only (up to isomorphism) simply connected Lie group whose Lie algebra is isomorphic to $\mathfrak{sl}(2,\Bbb R)$ is the metaplectic group, which is not a matrix group.

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  • $\begingroup$ I think you gave an even stronger counterexample to what I was originally asking to. According to the wikipedia page you linked, the metapletic group $\operatorname{Mp}_2(\mathbb{R})$ has no faithful finite-dimensional representations. In particular, it neither has faithful "closed" finite-dimensional representations (that is, such that the image of the representation $\operatorname{Mp}_2(\mathbb{R})\to\operatorname{GL}(n;\mathbb{C})$ is closed). Am I right? $\endgroup$ Commented Jun 28, 2021 at 9:54
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    $\begingroup$ Yes, but note that asserting that a Lie group has no finite-dimensional faithful representations is equivalent to asserting that it is not isomorphic to a closed subgroup of some $GL(n,\Bbb C)$ $\endgroup$ Commented Jun 28, 2021 at 9:56

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