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Let $G$ be a connected Lie group and suppose that its Lie algebra ${\frak g}$ splits into a direct sum of ideals $${\frak g} = {\frak a}\oplus{\frak b}.$$ Let $A$ be the connected Lie subgroup of $G$ with Lie algebra ${\frak a}$ and $B$ the one with Lie algebra ${\frak b}$. Let $\tilde{A}$ and $\tilde{B}$ be the universal covering groups of $A$ and $B$, respectively.

Then, $\tilde{A}\times\tilde{B}$ is a simply connected Lie group with Lie algebra ${\frak g}$, so it is isomorphic to the universal covering group $\tilde{G}$ of $G$. Hence, we have a covering homomorphism $$\varphi:\tilde{A}\times\tilde{B}\longrightarrow G.$$

Question: Do we have that $\varphi(\tilde{A}\times\{1\})=A$ and $\varphi(\{1\}\times B)=B$?

This would imply that $G=AB$ since $\varphi$ is a surjective group homomorphism. Hence, if this is true we would have that:

Proposition. If ${\frak g}={\frak a}\oplus{\frak b}$ then $G=AB$.

Is that true?

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  • $\begingroup$ When you say, let $A$ be $the$ connected Lie group, what do you mean exactly ? $\endgroup$ – Thomas Feb 11 '16 at 13:55
  • $\begingroup$ @Thomas You can find the following result in any introductory textbook on Lie groups: Theorem: Let $G$ be a Lie group and ${\frak h}$ a subalgebra of its Lie algebra ${\frak g}$. Then, there is a unique connected Lie subgroup $H$ of $G$ with Lie algebra ${\frak h}$. $\endgroup$ – SHP Feb 11 '16 at 14:10
  • $\begingroup$ Yes, the question has a positive answer. On the other hand, beware that $A$ and $B$ are not necessarily closed and that $A\cap B$ is a possibly nontrivial discrete subgroup. $\endgroup$ – YCor Feb 12 '16 at 0:14
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What about $G=(S^1)^2$, $S^1$ the Lie group of complex number of modulus 1. Let $A=(u,1)$ with $u\in S^1$, $B= (v^n,v^m)$ where $(n,m)=1$, $v\in S^1$. The kernel of the map $A\times B$ to $G$ is the finite group $\bf Z/mZ$ $(r^{-n}, (r^n,1))$, $r$ a $m-$th root of $1$.

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