MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 11 down vote accepted

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.)

share|cite|improve this answer
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 '10 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). – Loop Space Feb 23 '10 at 18:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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