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I'm trying to find an example of a Lie group $G$ which admits a bi-invariant Riemannian metric, and which has a closed subgroup $H$ such that the manifold $G/H$ does not admit a $G$-invariant Riemannian metric.

I know so far that if $G/H$ admits an invariant Riemannian metric, then $H$ has to be compact. So I'm searching for a Lie group $G$ which admits a bi-invariant Riemannian metric and a closed subgroup $H$ which is not compact, since this will solve the problem. Does anyone know such group?

Thank you for any help!

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    $\begingroup$ If $G$ admits a biinvariant Riemannian metric $ds^2$, then projection fo $ds^2$ to $G/H$ is left-invariant under the action of $G$. As for your other question, think about abelian groups. $\endgroup$
    – Misha
    Commented Nov 19, 2013 at 22:33
  • $\begingroup$ So the first claim is true even if H is not compact @Misha ? $\endgroup$
    – user42999
    Commented Nov 19, 2013 at 22:51
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    $\begingroup$ That's what I wrote. On the other hand, very few connected Lie groups admit biinvariant metrics: products of abelian and compact groups. $\endgroup$
    – Misha
    Commented Nov 19, 2013 at 23:03
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    $\begingroup$ @user42999: your statement "if $G/H$ has invariant Riem. metric then $H$ is compact" is false. 2 counterexamples: (1) $G$ not compact, $H=G$; (2) $G$ arbitrary, $H$ any infinite discrete subgroup: this second example shows that your statement is false even if $G$ acts faithfully on $G/H$. $\endgroup$
    – YCor
    Commented Nov 20, 2013 at 8:38

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