# Possible isometry groups of open manifolds

Consider a non-compact manifold $$M$$.

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

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...

• 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 $M$. – Peter Michor Jul 14 '19 at 8:55
• @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. – Misha Jul 14 '19 at 8:58
• 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 $g$. – Peter Michor Jul 14 '19 at 9:04
• @Ian Agol: I know: In the answer I said that I am restricting to orientation preserving maps to simplify the discussion. – Misha Jul 15 '19 at 3:09
• @AliTaghavi: $\chi$ is the alternating sum of Betti numbers assuming that they are all finite, like in our case. Otherwise, $\chi$ is undefined. – Misha Jul 17 '19 at 2:58

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.

• 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? – o r Jul 14 '19 at 7:53
• 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. – Misha Jul 14 '19 at 9:05
• 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 – Peter Michor Jul 14 '19 at 9:36
• 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 $(t, (x,y))\mapsto (e^tx, e^{-t}y)$. – Misha Jul 14 '19 at 10:35