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:
- Embed $M$ as a submanifold of $\mathbb{R}^n$.
- So the loop space $LM$ embeds as a submanifold of $L\mathbb{R}^n$.
- $L\mathbb{R}^n$ is metrisable.
- So $L M$ is metrisable.
- 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, Proc. London Math. Soc. 99 (2009) 195–216 (doi:10.1112/plms/pdn058, arXiv:math/0612096) 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 paper The Smooth Structure of the Space of Piecewise-Smooth Loops, Glasgow Mathematical Journal, 59(1) (2017) pp27-59. (arXiv:0803.0611, doi:10.1017/S0017089516000033) (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.)
Definition: ''A smooth manifold is a locally ringed space $(M;C^{\infty})$ which satisfies the conditions:
- Each $x \in M$ admits a neighborhood $U$, such that $(U,C^{\infty})$ is isomorphic to $(\mathbb{R}^n,C^{\infty})$ as a locally ringed space.
- The global sections of $C^{\infty}(M)$ separate points.
- The structure sheaf $C^{\infty}$ is fine as a sheaf of modules over itself.
- $C^{\infty}(M)$ has at most countably many indecomposable idempotents.''
Explanations:
1) is evident.
2) means that for $x \neq y \in M$, there exists $f \in C^{\infty}(M)$ with $f(x)=0$, $f(y) \neq 0$. This ensures the Hausdorff condition once we know that elements of $C^{\infty}(M)$ give rise to continuous maps $M \to \mathbb{R}$. This is as follows: $f \in C^{\infty}(M)$ given, $x \in M$. Pick a chart $h:U \to \mathbb{R}^n$; under this chart, $f|_U$ corresponds to a smooth function on $\mathbb{R}^n$, whose value at $h(x)$ does not depend on the choice of the chart. Call this value $f(x)$. Checking the continuity of $x \mapsto f(x)$ can be done in charts.
3.) By this I mean that for each open cover $(U_i)$, there is a partition of unity $\lambda_i$ with the usual properties and that $\lambda_i$ is a map of $C^{\infty} (M)$-modules. A standard argument shows that $\lambda_i$ is given by multiplication with a smooth function. Therefore, the underlying space $M$ is paracompact.
4.) An idempotent $p\neq 0 $ in a commutative ring $A$ is called indecomposable if
$$
p=q +r; r^2 =r; q^2 =q , q \neq 0 \Rightarrow p = q
$$
holds. Indecomposable idempotents in $C^{\infty}(M)$ correspond to connected components. Therefore condition 4 means that $M$ has only countably many connected components.
These conditions together imply that $M$ is Hausdorff and second countable, because a locally euclidean, connected and paracompact Hausdorff space is second countable, see Gauld, "Topological properties of manifolds", Theorem 7 (see
http://www.jstor.org/stable/2319220 ). Paracompactness alone does not guarantee second countability, see $\mathbb{R}$ with the discrete topology.
However, I think that sheaf theory and locally ringed spaces are the wrong software for differential geometry and differential topology.
Best Answer
Here's another proof, which shows that any connected paracompact locally Euclidean space X is second-countable. Cover X by Euclidean charts and take a locally finite refinement. Say an open set is good if it only intersects finitely many of the charts. Now take any point x and take a good neighborhood of it. The charts that intersect that good neighborhood can then themselves be covered by countably many good open sets. There are then only countably many charts intersecting those good open sets, and those charts can be covered by countably many good sets. Iterating this countably many times, you get an open set U associated to x which is covered by countably many charts such that if a chart intersects U, it is contained in U. It follows that the complement of U is also a union of charts, so by connectedness U is all of X. Thus X can be covered by countably many charts and is second-countable.