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I have an impression (but could be wrong) that I heard that for any semi-simple (real) Lie subalgebra $\mathfrak{k}$ of $\mathfrak{gl}(n,\mathbb{R})$ there exists a connected closed Lie subgroup $K\subset \mathrm{GL}(n,\mathbb{R})$ such that $\mathrm{Lie}(K)=\mathfrak{k}$.

Is it correct? A reference would be most helpful.

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    $\begingroup$ Yes: every perfect subalgebra of $\mathfrak{gl}_n(\mathbf{R})$ is even the Lie algebra of a Zariski-closed subgroup. I don't remember the reference. $\endgroup$
    – YCor
    Commented May 9, 2022 at 12:59
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    $\begingroup$ groupprops.subwiki.org/wiki/Perfect_Lie_algebra $\endgroup$
    – YCor
    Commented May 9, 2022 at 14:17
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    $\begingroup$ A reference for a comment of @YCor: For any Lie subalgebra ${\mathfrak g}\subset {\mathfrak{gl}}(n,{\mathbb C})$, its derived algebra $[{\mathfrak g},{\mathfrak g}]$ is algebraic, that is, it is the Lie algebra of an algebraic subgroup of ${\rm GL}(n,{\mathbb C})$. See Corollary 1 of Theorem 3 in Section 3.3.3 of the book: Onishchik, A. L.; Vinberg, È. B. Lie groups and algebraic groups. Springer-Verlag, Berlin, 1990. $\endgroup$
    – Mikhail Borovoi
    Commented May 10, 2022 at 5:28
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    $\begingroup$ If a perfect Lie subalgebra $\mathfrak{g}\subset \mathfrak{gl}(n,{\Bbb C})$ is defined over $\Bbb R$, then the corresponding algebraic subgroup is defined over $\Bbb R$ as well. $\endgroup$
    – Mikhail Borovoi
    Commented May 10, 2022 at 5:45
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    $\begingroup$ You should apply this result to the complexification of your Lie subalgebra $\mathfrak k$, and then take the closed Lie subgroup of the $\Bbb R$-points in the corresponding algebraic $\Bbb C$-group defined over $\Bbb R$. $\endgroup$
    – Mikhail Borovoi
    Commented May 10, 2022 at 13:27

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