# The set of smooth maps from exotic smooth manifolds to the reals

Here a $M,N$ are topological manifolds and $\mathcal{A}$ and $\mathcal{B}$ are atlases. The brackets $[]$ denote the formation of the equivalence class of atlases.

Let $(M,[\mathcal{A}])$ and $(N,[\mathcal{B}])$ smooth manifolds exotic to each other ($M$ and $N$ homeomorphic, lets say $h(M)=N$, but not diffeomorphic with the smooth structures).

I wondered if the following statements are true.

1. $f\in C^\infty (M,[\mathcal{A}])$ is not equivalent to $f\circ h \in C^\infty (N,[\mathcal{B}])$
2. There exists an $f\in C^\infty (M,[\mathcal{A}])$ such that there is no $g\in C^\infty (N,[\mathcal{B}])$ such that $f=g\circ h$ and the other way around: There exists an $g\in C^\infty (N,[\mathcal{B}])$ such that there is no $f\in C^\infty (M,[\mathcal{A}])$ such that $g=f\circ h^{-1}$.
3. For all $f\in C^\infty (M,[\mathcal{A}])$ there is no $g\in C^\infty (N,[\mathcal{B}])$ such that $f=g\circ h$.

Or in more transparent version with atlases dropped from notation and $M=N$ as topological spaces, but still not diffeomorphic.

1. $C^\infty M\neq C^\infty N$
2. $C^\infty M\not\subset C^\infty N$ and $C^\infty N\not\subset C^\infty M$
3. $C^\infty M\cap C^\infty N=\emptyset$

Thanks in advance and maybe the diffoelogy characterization of smoothness is helpful.

Kind regards

Mar

(corrected the error)

• I'm guessing it's supposed to be an intersection in the second 3. – Glougloubarbaki Feb 10 '14 at 18:31
• 1. is true, so we cannot have both $C^\infty M \subset C^\infty N$ and $C^\infty N \subset C^\infty M$, so by symmetry we have 2. 3. is wrong because constant functions are always smooth – Glougloubarbaki Feb 10 '14 at 18:33
• That was actually my intuition: The constant functions are always smooth. but how can one proof it? – w_w Feb 11 '14 at 13:09
• @user127698 Work in coordinates. – Neal Feb 11 '14 at 13:19
• Thanks. My question bordered on a stupid question in this case. But as long as you do not know it is better to ask. – w_w Feb 12 '14 at 6:48

1. True, essentially by the definition.

2. False in general. Edit: There are examples of homeomorphic but not diffeomorphic manifolds $M, N$ so that there exists a smooth homeomorphism $$h: M\to N.$$ (Milnor's exotic spheres satisfy this.) Note that, of course, the inverse of such $h$ is not smooth. Therefore, the pull-back via $h$ defines an embedding $$h^*: C^\infty(N)\to C^\infty(M), \quad h^*(\varphi)= \varphi \circ h.$$ Alternatively, if you take a homeomorphism $h$ whose inverse is smooth, you obtain the embedding
$$h_*: C^\infty(M)\to C^\infty(N).$$

3. Always false, take $f$ which is constant.

Note that I was addressing here question in the first three numbered items, the last 3 numbered items make no sense to me.

2nd Edit: One more thing, which answers a separate question by Jason (in comments). An isomorphism of ${\mathbb R}$-algebras $$C^\infty(N)\to C^\infty(M)$$ will imply diffeomorphism of the corresponding manifolds $M$ and $N$ (no need to assume a priori that they are homeomorphic). This is because there is a natural bijection $$C^\infty(M,N)\to Hom(C^\infty(N), C^\infty(M))$$ see theorem 2.3 in the book

Navarro Gonzalez and Sancho de Salas, "$C^\infty$-differentiable spaces", Springer Lecture Notes in Math. vol. 1824.

• Can you give an example where the sets of smooth functions on $M$ is contained in the smooth functions on $N$, that is $C^{\infty}M\subset C^{\infty}N$? – w_w Feb 11 '14 at 13:08
• While 1. is true by definition, I think a much more interesting question is whether or not $C^\infty(M)$ and $C^\infty(N)$ can be isomorphic (as $\mathbb{R}$-algebra). I know for compact manifolds, if $C^\infty(M)\cong C^\infty(N)$, then any isomorphism can actually be used to construct a diffeomorphism between $M$ and $N$ (this is an exercise in Milnor/Stasheff's Characteristic classes book), but I actually don't know what happens when $M$ and $N$ are both non-compact. – Jason DeVito Feb 11 '14 at 14:10
• @user127698: See the edit. – Moishe Kohan Feb 11 '14 at 14:12
• @JasonDeVito: This is a good question, see an edit to my answer. – Moishe Kohan Feb 11 '14 at 15:36
• @user127698: This is definitely true for compact manifolds, it is most likely to be true for noncompact manifolds too, but I have to check this. – Moishe Kohan Feb 13 '14 at 19:48