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Feb 9, 2016 at 11:07 vote accept Asaf Shachar
Jan 26, 2016 at 15:40 comment added Holonomia @Sebastian Goette: is the 'Borel construction' in your answer the 'fundamental construction' at page 12 of Donaldson's notes wwwf.imperial.ac.uk/~skdona/LIEGROUPSCONSOL.PDF ? Unfortunately, in Donaldson's fundamental construction there are no involved Riemannian metrics nor totally geodesic submanifolds. So would be nice if you provide a reference where all calculations can we followed.
Jan 26, 2016 at 15:19 comment added Sebastian Goette @Holonomia You are right. It suffices that $H$ acts isometrically on the typical fibre, which it does here. If I find some time, I will look up a reference for this construction (maybe Berline-Getzler-Vergne).
Jan 26, 2016 at 13:41 comment added Holonomia @Sebastian Goette: in the OP question the group $H$ is not assumed to be compact as you wrote. But perhaps this is not important for the Borel construction you have in mind, isn't it?
Jan 26, 2016 at 12:54 history edited Sebastian Goette CC BY-SA 3.0
added 319 characters in body
Jan 25, 2016 at 23:10 history edited Sebastian Goette CC BY-SA 3.0
Answered the right question now
Jan 25, 2016 at 20:32 comment added Asaf Shachar Indeed, I am assuming $g$ is right $H$-invariant over all $G$. That is, I meant for $d(R_h)_e:T_eG \to T_hG $ to be an isometry $\, \, \forall h \in H$.
Jan 25, 2016 at 19:30 comment added Holonomia is a Berger metric right $H$-invariant as in the OP question?. It is clear that the induced metric on $H$ is indeed bi-invariant. But it seems clear that the OP is assuming that the metric $g$ is right $H$-invariant over all the group $G$ and not just on itself. But perhaps I am misunderstanding the OP question.
Jan 25, 2016 at 19:07 history answered Sebastian Goette CC BY-SA 3.0