If $G$ is a real Lie group and $\mathfrak g$ is its Lie algebra. If $\hat{ \mathfrak g}=\mathfrak g+i\mathfrak g$ is the complexifiaction of the Lie algebra and let $\hat G$ be the associated complex Lie group.

Now is $G$ a real Lie subgroup of $\hat G$? If yes, why this is not a complexification (in general) of $G$? And can we say that every Lie group is embedded as a submanifold in a simply-connected complex Lie group?


No, $G$ doesn't have to be a real subgroup of $\hat G$. Let $G$ be the universal covering of $SL(2,\mathbb{R})$. Then $\mathfrak{g}$ is $\mathfrak{sl}(2,\mathbb{R})$ and $\hat{\mathfrak{g}}=\mathfrak{sl}(2,\mathbb{C})$. Therefore, $\hat G$ is $SL(2,\mathbb{C})$. If $G$ was a subgroup of $SL(2,\mathbb{C})$, that would make it a matrix group, but it is well-known that it is not.

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  • $\begingroup$ Thank you. But what about my second question. The universal complexification always exists so $\exists \hat G$ complex such that $G$ is embedded in $\hat G$ as a Lie subgroup? $\endgroup$ – Ronald Oct 3 '17 at 23:28
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    $\begingroup$ @Ronald No. $\endgroup$ – José Carlos Santos Oct 4 '17 at 6:28

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