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Let $G$ be a simply-connected nilpotent Lie group and $H$ be a closed simply-connected nilpotent subgroup of $G$.

Why $G/H$ is diffeomorphic to some $\mathbb R^n$?

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  • $\begingroup$ I'm not entirely sure what you're looking for. One equivalent criterion is that the maximal compact subgroup of H is also a maximal compact subgroup of G. $\endgroup$
    – user98602
    Commented Apr 15, 2017 at 4:05
  • $\begingroup$ I am trying to understand why (in the case of complex Lie groups) why the bundle $G/\Gamma\rightarrow G/G_\Gamma$ is holomorphically trivial? where $\Gamma$ is discrete and $G_\Gamma$ is the group associated to $\mathfrak g_\Gamma=\mathfrak g+i\frak g$ $\endgroup$
    – Ronald
    Commented Apr 15, 2017 at 4:19

1 Answer 1

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$p:G\rightarrow G/H$ is an $H$-principal bundle over $G/H$, since $H$ is contractible, the bundle is trivial, we deduce that $G$ is diffeomorphic to $G/H\times H$, since $H$ is contractible, $G$ retracts to $G/H$ and $G/H$ is contractible.

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  • $\begingroup$ Thanks Tsemo, did you also see my comment to the question? I mean in the case of complex groups $\endgroup$
    – Ronald
    Commented Apr 15, 2017 at 4:25

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