Are all manifolds affine? - MathOverflow

Are all manifolds affine?

2012-02-20T01:38:28Z

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!

Answer by Taladris for Are all manifolds affine?

2012-02-20T04:26:05Z

Hello,

I am not sure to fully understand the question. I understand it as: if two (possibly non paracompact or non Hausdorff) smooth manifolds have isomorphic algebra of smooth functions, are they isomorphic?

If I understand it well, then take the line with two origins (i.e. a quotient of the disjoint union of 2 real lines $\mathbb{R}\times\{0,1\}$ by the equivalence $(x,e)\sim(x',e') \Leftrightarrow (x=x' \text{ and } x\neq 0)$ ). Its algebra of smooth functions is isomorphic to $\mathcal{C}^\infty(\mathbb{R},\mathbb{R})$.

Answer by Todd Trimble for Are all manifolds affine?

2012-02-20T21:24:20Z

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.