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Consider a non-compact manifold $M$.

Does there always exist a Riemannian metric on $M$ such that the isometry group is non-compact?

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Let $M$ be the triply punctured 2-sphere (i.e. $M=S^2-\{p_1, p_2, p_3\})$); one can also take any noncompact connected surface of finite topological type as long as $\chi(M)<0$ but the proof is a bit more involved.

Suppose that $g$ is a Riemannian metric on $M$. To simplify matters, I consider only orientation-preserving isometries.

Then $\mathrm{Isom}(M,g)< \mathrm{Conf}(M,g)$. I claim that $\mathrm{Conf}(M,g)$, the group of conformal automorphisms of $(M,g)$, is finite. The mapping class group of $M$ (the group of self-diffeomorphisms of $M$ modulo isotopy) is finite (isomorphic to the dihedral group of order $6$); hence, it suffices to prove that if $f: (M,g)\to (M,g)$ is a conformal automorphism isotopic to the identity then $f=\mathrm{id}$. This is quite standard (most likely, you will find it in Farb-Margalit's book on the mapping class group).

Note that $f$ is isotopic to the identity if and only if it induces an inner automorphism of $\pi_1(M)$.

By the uniformization theorem, $(M,g)$ is conformal to the quotient of the hyperbolic plane $H^2$ by a discrete group of isometries $\Gamma$ (isomorphic to $F_2$, the free group on two generators). The map $f$ then lifts to an isometry $F$ of $H^2$ (a linear-fractional transformation in, say, the Poincare disk model of $H^2$) which commutes with every element of $\Gamma$ (since $f$ induces an inner automorphism of $\pi_1(M)\cong \Gamma$). In particular, $F$ acts on the boundary circle $S^1$ of $H^2$ fixing fixed points of all elements of $\Gamma$. Since $\Gamma$ is free of rank 2, the set of fixed points of its nontrivial elements is infinite. Hence, $F$ fixes at least three points of $S^1$. Hence (being a linear-fractional transformation), $F=\mathrm{id}$; thus, $f=\mathrm{id}$.

To conclude, every Riemannian metric on $M$ has finite group of isometries.

Edit: I can prove the same claim for each noncompact complete hyperbolic $n$-manifold of finite volume, but a proof is more difficult.

In general, the question can be reformulated in purely topological terms: Which smooth noncompact connected manifolds admit smooth proper actions of noncompact Lie groups? I suspect, this was studied in 1960s...

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  • $\begingroup$ Counterexample: Represent $M$ as the open disk in $\mathbb R^2$ with two smaller closed disks omitted, and consider the vector field $\partial_x$ restricted to $M$. Then apply the construction of my answer. I think that your answer is correct, if you also ask for a complete Riemannian metric on $M$. $\endgroup$ – Peter Michor Jul 14 '19 at 8:55
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    $\begingroup$ @PeterMichor: I am not using completeness of the metric $g$: Uniformization theorem does not require it. As for your example of a vector field, it will be "incomplete" and will not define an $R$-action. I think you will have this problem in general. The proof that I wrote is quite standard in hyperbolic geometry. $\endgroup$ – Misha Jul 14 '19 at 8:58
  • $\begingroup$ Aha. Now I am confused. I think that my construction goes through in this case. I gave an argument for making the vector field complete by multiplying it with a function (for example $1/\|X\|^2_{g}$ with respect to a complete auxiliary metric $g$. $\endgroup$ – Peter Michor Jul 14 '19 at 9:04
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    $\begingroup$ @Ian Agol: I know: In the answer I said that I am restricting to orientation preserving maps to simplify the discussion. $\endgroup$ – Misha Jul 15 '19 at 3:09
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    $\begingroup$ @AliTaghavi: $\chi$ is the alternating sum of Betti numbers assuming that they are all finite, like in our case. Otherwise, $\chi$ is undefined. $\endgroup$ – Misha Jul 17 '19 at 2:58
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The answer is sometimes yes:

If $M$ admits a complete vector field $X$, such that its flow $\operatorname{Fl}^X: \mathbb R\times M \to M$ is a proper action, then there exists a Riemannian metric which is invariant under the flow, by the following

Theorem. If M is a proper G-manifold, then there is a G-invariant Riemann metric on M.

which is due to Palais if I remember correctly. A proof of this theorem can be found in 6.30 of Topics in Differential Geometry. AMS, 2008.

Finally, the isometry group then contains a non-compact 1-parameter group (this is not enough: it might be dense) which moves each point towards infinity for $t\to\infty$. So the isometry group cannot be compact (in the compact-open topology).

The existence of such a vector field is a nontrivial condition, since then $M$ is the total space of real line bundle, as pointed out by Misha Kapovich. This is actually proved in 29.21 of 1.

See his answer for an example where the answer is no.

ADDED:

The converse is also true: If the (connected component of the) isometry group is not compact, it contains a closed 1-parameter group whose generating vector field then has proper flow.

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  • $\begingroup$ If I look at the vector field $y\partial_x - x\partial_y$ then it should be complete on $\Bbb R^2-\{0\}$, where it has no zeros. But I think the flow is not proper, as for any point $x$ we have $\varphi_X^{2\pi n}(x)=x$. Is that correct? $\endgroup$ – o r Jul 14 '19 at 7:53
  • $\begingroup$ True, one needs a vectorfield without periodic orbit. $\endgroup$ – Peter Michor Jul 14 '19 at 8:11
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    $\begingroup$ If you have a proper $R$-action then $M$ is diffeomorphic to the total space of an $R$-bundle; of course, many noncompact manifolds do not admit such. $\endgroup$ – Misha Jul 14 '19 at 9:05
  • $\begingroup$ But the $\mathbb R$-bundle is over a non-Hausdorff space, in general. In particular, in the 3 punctured sphere: Where 1 orbit becomes two, the two cannot be separated. See www.mat.univie.ac.at/~michor/vect-mf.pdf $\endgroup$ – Peter Michor Jul 14 '19 at 9:36
  • $\begingroup$ Proper actions have Hausdorff quotient spaces. What you explained nicely in your linked note is that given an incomplete vector field on a manifold, you can extend it to a complete vector field on a (potentially) non-Hausdorff manifold. What happens in the 3-holed sphere example is that the quotient is obviously non-Hausdorff and, hence, you cannot get a proper $R$-action (even though, individual trajectories are proper, of course). What you have here is a variation on the standard example of a non-proper $R$-action on the punctured affine plane, given by $(t, (x,y))\mapsto (e^tx, e^{-t}y)$. $\endgroup$ – Misha Jul 14 '19 at 10:35

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