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Suppose a finite--dimensional Lie group $G$ is given. Does there exist a connected manifold $M$ and a Riemannian metric $g$, such that $G$ is the full isometry group of $(M,g)$?

For example if I try to do this for a connected $G$, then I often get a bigger group as the full isometry group, which includes e.g. the orientation reversing isometries. (Maybe one has to take a non--orientable space for that?)

Even if I try to realize $\mathbb R$ as a full isometry group, I fail. (One could take the full isometry group of $\mathbb R$ with the standard metric, which is given by $\mathbb R \ltimes \mathbb Z_2$ and divide out the $\mathbb Z_2$ action. But this leads to a fixpoint and the quotient is therefore not a manifold any more.)

There is an article of J. de Groot which proves that every abstract group can be realized as an isometry group of some metric space, but it is not clear to me, if this is true in the category of Lie groups and Riemannian manifolds.

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  • $\begingroup$ I expect that you assume that $M$ is connected, but not $G$? $\endgroup$ – YCor Oct 24 '17 at 7:50
  • $\begingroup$ Yes! edited the question. $\endgroup$ – Panagiotis Konstantis Oct 24 '17 at 8:46
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The article of de Groot is the one cited here: What kind group can be realized as a Isometry group of some space?

That every compact group is the full isometry group of a compact Riemannian manifold is shown in:

The isometry groups of manifolds and automorphism groups of domains, by Rita Saerens and William Zame (T. AMS, 1987). (such an isometry group must be compact priori).

Jorg Winkelmann (Comm. Math. Helv., 2005) shows that every connected real lie group is the full automorphism group of a complete, hyperbolic (in the sense of Kobayashi) complex manifold.

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  • $\begingroup$ The case of non-compact groups seems still to be unanswered. $\endgroup$ – Panagiotis Konstantis Oct 24 '17 at 9:21
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The group $\mathbb R$ can be realized as full isometry group of $(\mathbb R\times\mathbb S^1 ,g)$. Choose a generic periodic one parameter family of quadratic forms $h(t)$ on $\mathbb R^2$. Consider metric $g(x,y)=h(y)$ on $\mathbb R\times\mathbb S^1$.

Why: Note that each fiber $\mathbb R\times u$ maps to it-self. Note that orthogonal fibration $\mathcal{F}$ is preserved. Go along $\mathcal{F}$ once around $\mathbb S^1$. Since $h$ is generic you will not get to the same point. Therefore each isometry preserves the orientaion of $\mathbb R$-fibers.

This idea seems to work in general. Consider metric $g$ on $G\times \mathbb T^2$ which is invariant w.r.t. left $G$-translations and such that $g(e,t)=h(t)$ is a generic family of quadratic forms on $T_{(e,t)}$; here $t\in \mathbb T^2$.

Why: This way you get a holonomy map from $G\to G$ for any loop in $\mathbb T^2$. For generic $h(t)$ you may assume that there is no automorphism of $G$ which preserve this holonomy.

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  • $\begingroup$ Anton -- a "generic one parameter family of metrics" on what? $\endgroup$ – algori Jan 31 '12 at 0:21
  • $\begingroup$ @algori: hope it is clear now. $\endgroup$ – Anton Petrunin Jan 31 '12 at 0:39
  • $\begingroup$ Isn't the map $(x,y)\mapsto(-x,y)$ an isometry of your $\mathbb R^2$? $\endgroup$ – Mariano Suárez-Álvarez Jan 31 '12 at 1:42
  • $\begingroup$ @Mariano: formally NO, but you are right after changing coordinates --- I will update my answer. $\endgroup$ – Anton Petrunin Jan 31 '12 at 2:44
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    $\begingroup$ @Anton: Sorry, I don't understand two things concerning your example with $\mathbb R$. First I don't see why there shouldn't be any other (e.g. discrete) symmetries, despite the genericity of $h$. And second I don't get your argument about the isometries in $\mathbb R$-directions. Why is this proving that there is no orientation reversing symmetry? I guess I'm just blind not to see that. $\endgroup$ – Panagiotis Konstantis Jan 31 '12 at 11:25

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