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Suppose that $M$ is a finite-dimensional $C^{\infty}$-manifold, and let $\mathrm{Diff}\left(M\right)$ be the group of $C^{\infty}$-diffeomorphisms from $M$ to itself. When $M$ is compact, the usual compact-open (Whitney) topology turns $\mathrm{Diff}\left(M\right)$ into a Polish group (topological group whose topology is separable and completely-metrizable).

When $M$ is merely $\sigma$-compact, the phrase "diffeomorphism groups" commonly refers to the group $\mathrm{Diff}_{c}\left(M\right)$ of compactly-supported diffeomorphisms from $M$ to itself. It is now the question of what topology is appropriate here. When taking the local-compact-open topology (weak Whitney, uniform convergence on each compact set), it does not control the behaviour at infinity. On the other hand, the universal-compact-open topology (strong Whitney, uniform convergence everywhere) is non-metrizable. Thus, the working-topology is "specified" by the following notion of convergence:

$\left(\ast\right)$ $f_{n}\to f$ if all of $f,f_{1},f_{2},...$ are supported on the same compact set and, on this compact set, $f_{n}$ with their derivatives converge uniformly to $f$.

For instance, this seems to be the approach of [1, page 2] in the introduction of the subject. This is also the description of the toplogy in [2].

My question then is the following:

If $M$ is $\sigma$-compact and non-compact, is there a topology on $\mathrm{Diff}_{c}\left(M\right)$ that turns it into a Polish group, such that the convergent sequences are as in $\left(\ast\right)$?

Here is why I think that this question is non-trivial. On one hand, it is true that there exist topologies that induce this notion of convergence as was shown in [3] (see Lemma 1.7, Corollary 2.3, Remark 2.6). However, those topologies seem not to be metrizable. A different cnadidate one may suggest is the direct limit of topologies, when the limit is taken over compact sub-manifold (as stated shortly in [1, page 2]). However, in [4, Theorem 6.1] it was proved that this direct limit topology does not result a group-topology (!).

References

[1] Banyaga, Augustin, The structure of classical diffeomorphism groups, Mathematics and its Applications (Dordrecht). 400. Dordrecht: Kluwer Academic Publishers. xi, 197 p. (1997). ZBL0874.58005.

[2] Vershik, A. M.; Gel’fand, I. M.; Graev, M. I., Representations of the group of diffeomorphisms, Russ. Math. Surv. 30, No. 6, 1-50 (1975). ZBL0337.58003.

[3] Michor, P., Manifolds of smooth maps, Cah. Topol. Géom. Différ. 19, 47-78 (1978). ZBL0382.58009.

[4] Tatsuuma, Nobuhiko; Shimomura, Hiroaki; Hirai, Takeshi, On group topologies and unitary representations of inductive limits of topological groups and the case of the group of diffeomorphisms, J. Math. Kyoto Univ. 38, No. 3, 551-578 (1998). ZBL0930.22002.

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  • $\begingroup$ If $M$ is $\sigma$-compact (equiv. to second countable here) or even merely paracompact, then you might be able to get a stronger result. Not massively useful, but maybe one of these is assumed in your definition of finite-dimensional manifold. $\endgroup$
    – David Roberts
    Commented Nov 12, 2023 at 23:35
  • $\begingroup$ Thanks, I added this assumption. $\endgroup$
    – HUO
    Commented Nov 13, 2023 at 8:23
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    $\begingroup$ No. Write $M$ as increasing union of compact subsets $K_n$ so that the interiors cover too. Let $G_n$ be the subgroup of diffeos supported in $K_n$. It's closed under the given convergence, so for a metrizable topology with such converging sequences, is closed. It has empty interior. For a topological group, it just means it's not open. Indeed, just approximate the identity by diffeos outside $K_n$. So the resulting topology is not Baire. Conclusion: there is no Baire metrizable group topology for which these are the converging sequences. $\endgroup$
    – YCor
    Commented Nov 13, 2023 at 12:36
  • $\begingroup$ Thanks. Why not posting this as an answer? $\endgroup$
    – HUO
    Commented Nov 13, 2023 at 13:34

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