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

Similarly to my previous question about direct limits, I have now basically the same question about inverse limits. It seems in fact, that I only need the result for products.

Question: Is there a natural smooth structure on $\prod \mathbb{R}$ such that $\mathcal{C}^\infty(U,\prod \mathbb{R}) = \prod\mathcal{C}^\infty(U,\mathbb{R})$?

share|cite|improve this question
Remark: $\hom(Y,\lim_i X_i) = \lim_i \hom(Y,X_i)$ holds in every category (and is trivial). So your question is: Does $\prod \mathbb{R}$ admit a smooth structure such that the resulting manifold is the product in the category of smooth manifolds? Well I don't think so, because $\prod \mathbb{R}$ is not finite dimensional locally. – Martin Brandenburg Apr 5 '12 at 6:56
We can consider larger category of smooth manifolds modeled on locally convex toplogical vector spaces. Or something like that. What I am basically asking is: For which definition of smooth function with values in real sequences we get that the space of all such smooth functions is just the space of sequences of smooth functions. (The topology on the target space can be medlled with but I would prefer it to be fixed.) If I didn't make pretty embarassing mistake, corresponding statement for continuous functions with values in $\prod\mathbb{R}$ is true. – Vít Tuček Apr 5 '12 at 13:06

This works in the Convenient Calculus of Kriegl and Michor ("A Convenient Setting for Global Analysis"). There, the linear category concerned is the space of bornological locally convex topological vector spaces and this is the linear subcategory of the category of Frolicher spaces. So $\prod \mathbb{R}$ being the bornological product is the categorical product as well and hence has the properties that you require.

share|cite|improve this answer

As Martin Brandenburg said, cannot exist a (locally euclidean) smooth variety structure, but in the more large category $(C^\infty-Ring)^{op}$ I think that the product exist (see "Models for Smooth Infinitesimal Analysis" Ieke Moerdijk, Gonzalo E. Reyes, see T. 2.8 pag. 30).

It is a funtor $P: C^\infty\to Set$ that map $\mathbb{R}^n$ on the set of funtion like $f\circ\pi_J:\prod_i\mathbb{R}_i \to \prod_J\mathbb{R}_j\to \mathbb{R}^n $ where $J\subset I$ is a finite subset, $\pi_J$ is the natural projection, and $f$ a smooth map (and by composition on morphisms). We have to show that $P$ is the sum of $I$ copies of $\mathbb{R}$ (where $\mathbb{R}$ is see as a $C^\infty$ spaces naturally). Give a manifold M, a morphism of $C^\infty$-spaces $\mathbb{R}\to M$ is uniquely represented by a smooth map $M\to \mathbb{R}$. Then give morphisms $g'_i: \mathbb{R}\to M$ i.e. smooth maps $g_i: M\to \mathbb{R}$ i.e. a map $g=\prod_i g_i: M\to \prod_I \mathbb{R}$ follow the morphism $g': P\to M$ such that on argument $g'_n: P(\mathbb{R}^n)\to M(\mathbb{R}^n)$ map $f\circ \pi_J $ (where $J=${$j_1,...,j_n$}) on $(f\circ g_{j_1}, ..., f\circ g_{j_n}): M\to \mathbb{R}^n$.

(I hope this work...)

share|cite|improve this answer

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.