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Let $\mathfrak g$ and $\mathfrak h$ be (finite-dimensional real) Lie algebras, and form their semidirect product $\mathfrak g \oplus_\pi \mathfrak h$ with respect to some homomorphism $\pi : \mathfrak g \to \text{Der}(\mathfrak h)$.

Suppose further that we have a linear representation of $\mathfrak g \oplus_\pi \mathfrak h$ (say as real matrices) so that $e_i$ and $f_j$ are bases of $\mathfrak g$ and $\mathfrak h$ respectively, and they satisfy the required commutator relations of $\mathfrak g \oplus_\pi \mathfrak h$. Then the canonical coordinates of the second kind $$ (x,y) \mapsto \prod_i \text{Exp}(x_ie_i) \prod_j \text{Exp}(y_jf_j) $$ represent (not necessarily faithfully) a connected Lie group with Lie algebra $\mathfrak g \oplus_\pi \mathfrak h$.

Question: Is this Lie group automatically a semidirect product of Lie groups in general?

More generally, given a Lie group whose Lie algebra is a semidirect product, is the group necessarily a semidirect product as well?

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  • $\begingroup$ What is the definition of the ”Lie algebra semidirect product”? I have never seen it before. Thanks! $\endgroup$ – DanielC Nov 11 '19 at 0:21
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    $\begingroup$ @DanielC One takes $\mathfrak g \oplus \mathfrak h$ as the underlying vector space and equips with with the Lie bracket such that $[X,Y] = \pi(X)Y$ when $X \in \mathfrak g$ and $Y \in \mathfrak h$. Of course one needs to show that such a Lie bracket exists and is unique for this definition to make sense, see e.g. Proposition 1.22 in Knapp's "Lie Groups, Beyond an Introduction". $\endgroup$ – MSDG Nov 11 '19 at 8:22
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I think the answer is NO.

The reason is simply that the Lie algebra does not change if you consider a Lie Group or its universal cover or some its quotients by discrete subgroups of $Z(G)$.

Probably your statement is true if you suppose that G and H are simply connected and consider the only simply connected Lie group with Lie algebra $\mathfrak{g} \otimes_\pi \mathfrak{h}$ but actually I have no references to suggest.

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    $\begingroup$ Thank you for the answer, I suspected that this might be the case. When the Lie groups are simply connected, then one can show that there is a unique action $\tau : G \to \text{Aut}(H)$ such that the simply connected semidirect product Lie group $G \ltimes_\tau H$ has Lie algebra $\mathfrak g \oplus_\pi \mathfrak h$, see e.g. Theorem 1.125 in Knapp's "Lie Groups, Beyond an Introduction". $\endgroup$ – MSDG Nov 11 '19 at 8:24

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