4
$\begingroup$

Suppose $X$ is a metric space with a free group action by a topological group $G$, which is also a metric space, such that $\pi\colon X \to X/G$ is a fiber bundle.

Does the quotient inherit the paracompactness of $X$? If this is not true in general I'm interested in the case, where $X$ is the space of embeddings of a closed surface $\Sigma$ into some closed manifold $M$ with the $C^k$ topology on which $ \text{Diff}(\Sigma)$ acts by precomposition.

$\endgroup$
3
$\begingroup$

For $k=\infty$ this is well-known: See 44.1 of here, or 13.14 of here, or here since the base of the principal fiber bundle is a smooth infinite dimensional manifold which is modeled on nuclear Frechet spaces, thus it is smoothly paracompact (there exist smooth partitions of unity). The modelling spaces are spaces of smooth sections of the normal bundle of $f(\Sigma)$ for a fixed embedding $f$.

If $k$ is finite, one has to adapt the proof: Use only smooth $C^\infty$ embeddings as bases for the trivializing charts, then the construction of the normal bundle does not loose differentiability, so one can still consider the space of $C^k$ sections of the normal bundle. Attention: the action of the group of $C^k$-diffeomorphisms (which is only a topological group: $k\ge 1$) is only continuous. The quotient is still a smooth manifold modelled on Banach spaces: Reading carefully through the proof one notices, that one can find explicit chart changes involving only smooth embeddings and smooth diffeomorphisms which act from the right, thus smoothly even on $C^k$ embeddings. Also smooth local fiber respecting diffeomorphisms act smoothly on on spaces of $C^k$-sections.

But the modelling Banach spaces of $C^k$-sections do not admit smooth partitions of unity. So the quotient space is only paracompact (since the base is second countable). See section 16 of here for this.

$\endgroup$
  • $\begingroup$ I'm trying to find the reference in "The convenient setting for global Analysis", where it is mentioned that locally nuclear Frechet space implies paracompact. Could you maybe help me out once more? $\endgroup$ – ThorbenK May 7 '18 at 17:42
  • 1
    $\begingroup$ @ThorbenK: 42.3 and the proof used there. $\endgroup$ – Peter Michor May 7 '18 at 17:49
  • $\begingroup$ I'm sorry if I'm a little bit slow on this but I can't patch all of this together. Proposition 42.3 says that $C^\infty(\Sigma,M)$ is metrizable, which of course also applies to the open subset $\text{Emb}(\Sigma,M)$ but how do you get from this to the fact that the base space of the described principal bundle is paracompact? Using 42.3 and the corollary in 27.4 I still need that the base space is Linderlöf. $\endgroup$ – ThorbenK May 7 '18 at 21:38
  • 1
    $\begingroup$ The base is Lindeloef, since it is the continuous image of Emb, which is Lindeloef, being second countable metrizable. $\endgroup$ – Peter Michor May 8 '18 at 6:29
  • 1
    $\begingroup$ @PeterMichor : I'm sorry to reopen an old comment thread, but I was also trying to follow through the reasoning, and I have a question about the corollary in section 27.4 of your book, which is used in the proof that $\mathrm{Emb}(L,M)/\mathrm{Diff}(L)$ is paracompact for finite-dimensional smooth manifolds $L$, $M$ with $L$ compact. If I understand correctly, the corollary says, roughly (omitting the word "smoothly" for brevity), that Hausdorff plus Lindelöf plus locally regular implies paracompact. $\endgroup$ – Martin Palmer Jan 28 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.