There is a classical result which says that the assignment $$M \mapsto C^{\infty}\left(M\right)$$ is an embedding of the category of (paracompact Hausdorff) smooth manifolds into the opposite category of $\mathbb{R}$-algebras. However, all of the proofs I have seen first establish this for manifolds of the form $\mathbb{R}^n$ and then use Whitney's embedding theorem. This of course uses the paracompact Hausdorff condtion in an essential way. My question is, is this functor still an embedding if we don't impose paracompactness conditions on our smooth manifolds? If so, is there a nice proof of this fact? If not, can someone provide a simple counterexample? Thanks!
[Math] Are all manifolds affine
ag.algebraic-geometryct.category-theorydg.differential-geometry
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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.)
Regarding question 1, yes you can always ensure the image is closed. You prove the strong Whitney by perturbing a generic map $M \to \mathbb R^{2m}$ to an immersion, and then doing a local double-point creation/destruction technique called the Whitney trick. So instead of using any smooth map $M \to \mathbb R^{2m}$, start with a proper map -- one where the pre-image of compact sets is compact. You can then inductively perturb the map on an exhausting collection of compact submanifolds of $M$, making the map into an immersion that is also proper.
Regarding question 2, generally speaking if a manifold is not compact the embedding problem is easier, not harder. Think of how your manifold is built via handle attachments. You can construct the embedding in $\mathbb R^4$ quite directly. Think of $\mathbb R^4$ with its standard height function $x \longmapsto |x|^2$, and assume the Morse function on $M$ is proper and takes values in $\{ x \in \mathbb R : x > 0 \}$. Then I claim you can embed $M$ in $\mathbb R^4$ so that the Morse function is the restriction of the standard Morse function. The idea is every $0$-handle corresponds to creating an split unknot component in the level-sets, etc.
edit: The level sets of the standard morse function on $\mathbb R^4$ consists of spheres of various radius. So when you pass through a critical point (as the radius increases) either you are creating an split unknot component, doing a connect-sum operation between components (or the reverse, or a self-connect-sum), or you are deleting a split unknot component. By a split unknot component, I'm referring to the situation where you have a link in the $3$-sphere. A component is split if there is an embedded 2-sphere that contains only that component, and no other components of the link. So a split unknot component means that component bounds an embedded disc that's disjoint from the other components.
Regarding your last question, the Whitney embedding theorem isn't written up in many places since all the key ideas appear in the proof of the h-cobordism theorem. So Milnor's notes are an archetypal source. But Adachi's Embeddings and Immersions in the Translations of the AMS series is one of the few places where it occurs in its original context. You can find the book on Ranicki's webpage.
Best Answer
The functor is not an embedding if we remove the paracompactness assumption.
I will need some preliminary definitions. Let $R$ be the long ray, i.e., the topological space given by $\omega_1 \times [0, 1)$ equipped with the order topology induced by lexicographic order, and let $L$ be the long line obtained by gluing together two long rays. Topologically, we can think of $L$ as the colimit of spaces $L_\alpha$ where $\alpha$ is a countable ordinal and $L_\alpha$ is obtained by gluing together two rays $R_\alpha = \alpha \times [0, 1)$.
According to the Wikipedia article, not only does there exist a smooth ($C^\infty$) manifold structure on $L$, there exist infinitely many smooth ($C^{\infty}$) manifold structures on $L$ which extend a given $C^1$ manifold structure. On the other hand, there is just one smooth structure on the ordinary line which extends a given $C^1$ structure. We exploit these facts to show that $M \mapsto C^\infty(M)$ is not an embedding.
Let $L$ and $L'$ be distinct smooth structures which restrict to the same $C^1$ structure. If $C^\infty(-) = \hom(-, \mathbb{R})$ were an embedding on general Hausdorff not necessarily paracompact manifolds, then $L$ and $L'$ would be isomorphic if their smooth algebras are isomorphic. So it is enough to show that $C^\infty(L)$ and $C^\infty(L')$ are isomorphic. Now it is well-known that every continuous function on the long line is eventually constant (constant outside some bounded neighborhood). In that case, consider the algebra $C_\alpha$ of smooth functions on $L_\alpha$ which are eventually constant. For $\alpha \leq \beta$ there is an obvious extension $C_\alpha \to C_\beta$, and we have
$$C^\infty(L) = colim_\alpha C_\alpha$$
Similarly, we may write $C^\infty(L') = colim_\alpha C^\prime_\alpha$. However, notice that the identity function $L \to L'$ restricts to a diffeomorphism $L_\alpha \to L^\prime_\alpha$, because each $L_\alpha$ is topologically an ordinary line, where we had observed there is just one smooth structure extending the given $C^1$ structure. The isomorphisms $C_\alpha \to C^\prime_\alpha$ induce an isomorphism $C^\infty(L) \to C^\infty(L')$, as desired.