## Which Fréchet manifolds have a smooth partition of unity?

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A classical theorem is saying that every smooth, finite-dimensional manifold has a smooth partition of unity. My question is:

1. Which Fréchet manifolds have a smooth partition of unity?

2. How is the existence of smooth partitions of unity on Fréchet manifolds related to paracompactness of the underlying topology?

From some remarks in some literature, I got the impression that not all Fréchet manifolds have smooth partitions of unity, but some have, e.g. the loop space $LM$ of a finite-dimensional smooth manifold $M$.

For $LM$, the proof seems to be that $LM$ is Lindelöf, hence paracompact. Is this true for all mapping spaces of the form $C^\infty (K,M)$ for $K$ compact?

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## 2 Answers

Use the source, Luke.

Specifically, chapters 14 (Smooth Bump Functions) to 16 (Smooth Partitions of Unity and Smooth Normality). You may be particularly interested in:

Theorem 16.10 If $X$ is Lindelof and $\mathcal{S}$-regular, then $X$ is $\mathcal{S}$-paracompact. In particular, nuclear Frechet spaces are $C^\infty$-paracompact.

For loop spaces (and other mapping spaces with compact source), the simplest argument for Lindelof/paracompactness that I know of goes as follows:

1. Embed $M$ as a submanifold of $\mathbb{R}^n$.
2. So the loop space $LM$ embeds as a submanifold of $L\mathbb{R}^n$.
3. $L\mathbb{R}^n$ is metrisable.
4. So $L M$ is metrisable.
5. Hence $L M$ is paracompact.

(Paracompactness isn't inheritable by all subsets. Of course, if you can embed your manifold as a closed subspace then you can inherit the paracompactness directly.)

I use this argument in my paper on Constructing smooth manifolds of loop spaces to show that most "nice" properties devolve from the model space to the loop space for "nice" model spaces (smooth, continuous, and others). See corollary C in the introduction of the published version.

(I should note that the full statement of Theorem 16.10 (which I did not quote above) is not quite correct (at least in the book version, it may have been corrected online) in that the proof of the claim for strict inductive sequences is not complete. I needed a specific instance of this in my preprint The Smooth Structure of Piecewise-Smooth Loops (see section 5.4.2) which wasn't covered by 16.10 but fortunately I could hack together bits of 16.6 with 16.10 to get it to work. This, however, is outside the remit of this question as it deals with spaces more general than Frechet spaces.)

On the opposite side of the equation, we have the following after 16.10:

open problem ... Is every paracompact $\mathcal{S}$-regular space $\mathcal{S}$-paracompact?

So the general case is not (at time of publishing) known. But for manifolds, the case is somewhat better:

Ch 27 If a smooth manifold (which is smoothly Hausdorff) is Lindelof, and if all modelling vector spaces are smoothly regular, then it is smoothly paracompact. If a smooth manifold is metrisable and smoothly normal then it is smoothly paracompact.

Since Banach spaces are Frechet spaces, any Banach space that is not $C^\infty$-paracompact provides a counterexample for Frechet spaces as well. The comment after 14.9 provides the examples of $\ell^1$ and $C([0,1])$.

So, putting it all together: nuclear Frechet spaces are good, so Lindelof manifolds modelled on them are smoothly paracompact. Smooth mapping spaces (with compact source) are Lindelof manifolds with nuclear model spaces, hence smoothly paracompact.

(Recall that smooth mapping spaces without compact source aren't even close to being manifolds. I know that Konrad knows this, I merely put this here so that others will know it too.)

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Hi Andrew, I totally forgot about the source. Many thanks for your answer! When you are saying "smoothly paracompact" you mean exactly "has a smooth partition of unity", right?  – Konrad Waldorf Feb 23 2010 at 17:55
Yes.  I mean that every open cover has a smooth partition of unity subordinate to it (the closures of the supports are a locally finite cover refining the initial one). – Andrew Stacey Feb 23 2010 at 18:37

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One also has to assume that M is compact for LM to be paracompact, as well as finite dimensional. A question that should have a slightly easier answer is this: which Frechet spaces have partitions of unity?

(Dang, I just registered and lost all my rep from when I was an unregistered user. I would leave a comment otherwise)

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If you email the mods or leave a message on meta they can combine the accounts for you. – jc  Feb 23 2010 at 1:45
can you explain why M needs to be compact? or give a reference or something? Thanks! (I know the moderators would be happy to combine your accounts).  – Chris Schommer-Pries Feb 23 2010 at 4:10
You should now own all your posts and have all your rep. Let me know if you have any other account problems. – Anton Geraschenko♦ Feb 23 2010 at 4:52
Hi David, M does not have to be compact to make LM paracompact. See Brylinski's book, Prop. 3.1.2 (citing Pressley-Segal). [I should have written Lindelöf instead of metrizable above] – Konrad Waldorf Feb 23 2010 at 6:15
Whoops! Probably getting confused with needing compact source. I considered referring to my  esteemed colleague above, but I'm not sure about the propriety of saying 'so-and-so knows this', since it is a cheap way of answering without answering. – David Roberts Feb 24 2010 at 22:56
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