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Let $G$ be a Lie group with a left invariant metric $g$.

Let $H$ be a (closed) Lie subgroup of $G$, and assume $g$ is right-$H$-invariant. (That is $d(R_h)_e:T_eG \to T_hG$ is an isometry for every $h \in H$). Note that this is equivalent to the statement, that for every $h \in H$ the map of right multiplication by $h$, viewed as a map $G \mapsto G$ is an isometry.

(This appears to be stronger thant requiring it will be an isometry as a map $H \mapsto H$).

Is it true that $H$ must be totally geodesic in $G$?

Since $g$ is bi-$H$-invariant, the geodesics of $H$ are the one-parameter subgroups of $H$. Hence, the question amounts to:

Are the one-parameter subgroups of $H$ geodesics in $G$?

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Edit. I totally misunderstood the question, maybe now its better. Assume that $g$ is left-$G$ and right-$H$-invariant. Then one can construct a Riemannian submersion $p\colon G\to G/H$. This is now a family of Riemannian manifolds with fibre $H$ and compact structure group $H$, because you could also write it as a Borel construction $G=G\times_HH\to G/H$. Here, you regard $G$ as a principal bundle over $G/H$ with fibre $H$. $G/H$ is equipped with the induced Riemannian metric, and $T^HG\subset TG$ is defined as $\ker(dp)^\perp$. Together with the induced metric on $H$, the Borel construction gives back the original metric $g$. But then its fibres are automatically totally geodesic, so now the answer is "yes".

Old answer to a different question :-) Consider Berger metrics on $SU(2)\cong S^3\subset\mathbb C^2$. Here the round metric is stretched or shrunk along the fibres of the Hopf fibration $S^3\to S^2$. These metrics are still left invariant, but not biinvariant. For $H$, take a one parameter subgroup, so $H\cong S^1$. The induced metric on $H$ is still left invariant, hence biinvariant, because $H$ is abelian.

If all one parameter subgroups were geodesics, then their left translates by elements of $SU(2)$ would be geodesics, too. So the Berger metrics would have the same geodesics as the round metric, but that is not the case. Hence, the answer in general is no.

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    $\begingroup$ 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. $\endgroup$
    – Holonomia
    Commented Jan 25, 2016 at 19:30
  • $\begingroup$ 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 $\, \, \forall h \in H$. $\endgroup$
    – Asaf Shachar
    Commented Jan 25, 2016 at 20:32
  • $\begingroup$ @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? $\endgroup$
    – Holonomia
    Commented Jan 26, 2016 at 13:41
  • $\begingroup$ @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). $\endgroup$
    – Sebastian Goette
    Commented Jan 26, 2016 at 15:19
  • $\begingroup$ @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. $\endgroup$
    – Holonomia
    Commented Jan 26, 2016 at 15:40

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