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Aug 29, 2023 at 15:09 comment added Ben McKay For a counterexample, you can take $G$ a torus, $H$ a subtorus, and $\mathfrak{n}\subset\mathfrak{g}$ a linear subspace, complementary to the Lie algebra $\mathfrak{h}$ of $H$, but so that $\mathfrak{n}$ is the tangent space at $1\in G$ of a dense subgroup of $G$.
Aug 29, 2023 at 10:53 comment added A. J. Pan-Collantes By the way, I guess that there are examples when this semidirect sum is not coming from the existence of a semidirect product structure in $G$...
Aug 29, 2023 at 10:49 comment added A. J. Pan-Collantes @BenMcKay I see, my requirement can be weakened to: $(G,H)$ is such that $\mathfrak{g}$ is decomposed as the semidirect sum of a "given" Lie subalgebra $\mathfrak{n}$ and the Lie subalgebra $\mathfrak{h}$. Anyway, I prefer my formulation, because (for me) the idea of semidirect product is nearer to intuition.
Aug 29, 2023 at 9:03 comment added Ben McKay $G$ doesn't have to be a semidirect product. The Lie algebra is a semidirect sum.
Aug 25, 2023 at 6:35 history answered A. J. Pan-Collantes CC BY-SA 4.0