Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I was wondering if it was possible to extend the De Rham theorem to differetiable spaces, as defined for instance in this paper of Chen. If it has been done, could anyone come up with a good reference ?

Alternatively, I could use a reference proving the "classical" De Rham theorem, by using explicitely the chain map $$ \alpha \mapsto \left[ \sigma \mapsto \int_\Delta \sigma^*\alpha \right] $$ so that I can extend the proof to differentiable spaces.

[edit] Seeing many of the aswers, there seem to be either another "de Rham theorem" or another way of seeing differential forms, so let me clarify: What I mean by "de Rham theorem" is the fact that the cohomology of differential forms is isomorphic to e.g. the singular cohomology, when describibg smooth manifolds. Now perhaps the answers given treated that, but it's not apparent to me, so perhaps one could elaborate on the link ?

share|improve this question
    
I havent the access to intere article, then I dont know exatly what a "differential space" should be. ANyway in this book: Moerdijk, I.; Reyes, G.E. (1991), Models for Smooth Infinitesimal Analysis THere is a specific chapter about DeRham cohomology . –  Buschi Sergio Apr 10 '12 at 13:02
    
@Buschi Sergio: Thank you, but I would like to avoid synthtic calculus and toposes, that weem to be of importance in this book. –  Samuel Tinguely Apr 10 '12 at 13:14

5 Answers 5

up vote 3 down vote accepted

You can also try Chapters 1 and 2 of

Johan L. Dupont: Curvature and Characteristic Classes, Lecture Notes in Mathematics, vol. 640, Springer Verlag, 1978.

share|improve this answer
    
The 2nd chapter in particular contains more or less everything I need, thank you ! –  Samuel Tinguely Apr 11 '12 at 13:35

How much do you want to stick to Chen's original definition of a "differentiable space"? I recommend to look for the very similar notion of a "diffeological space", for which more literature is available. The comments below should apply to either notion.

The book

gives a definition of the de Rham-homomorphism (§ 6.74). For a general diffeological space, there is no chance that it is an isomorphism.

Note that the usual proof one gives for smooth manifolds uses that smooth manifolds are paracompact. The topology of a general diffeological space, however, does not have to be paracompact. Already for infinte-dimensional manifolds (still a subclass of diffeological spaces) one has to require "smooth paracompactness" separately (see Theorem 34.7 in: Kriegl, Michor: The convenient setting of global Analysis).

share|improve this answer
    
Thank you, this will likely allow me to advance greatly. –  Samuel Tinguely Apr 11 '12 at 10:45

Under "alternatively", the following paper contains a lot of simplicial de Rham stuff that may or may not be useful to you: http://arxiv.org/abs/0810.1747. However, for your limited purposes there are probably simpler references.

share|improve this answer
    
Thank you for the reference. However, being rather new, I don't really see the differential forms in this article. Is there any hint as to where too look ? –  Samuel Tinguely Apr 10 '12 at 13:19

Let $X$ be a topological space, and $Symp_\bullet (X)$ the simplicial set with simplices $$Symp_m(X) = \{ \textrm{singular } m- \textrm{simplices } \sigma: \Delta^m\to X\}$$ (and usual faces and degeneracies). Let $A^\bullet(X)$ be the complex of rational polynomial differential forms on $Symp_\bullet(X)$. If $k$ is a $\mathbb{Q}$-algebra, denote $A^\bullet_k(X)=A^\bullet(X)\otimes_\mathbb{Q}k$. Dennis Sullivan proved that:

The integration map $$I : A^\bullet_k(X)\to S^\bullet(X,k)$$ induces a $k$-algebra isomorphism on cohomology.

Here $S^\bullet(X,k)$ is the complex of singular $k$-valued cochains.

The theorem is one of the versions of a more powerful result, see D.Sullivan, Infinitesimal Computations in Topology, Theorem 7.1. In this form the theorem is stated in R.Hain's paper The de Rham homotopy theory of complex algebraic varieties I, section 2 (K-theory, 1987).

share|improve this answer
    
I'm sorry I don't really se the link between this and differential forms. Is it just another way of seeing them ? Or is it something really different ? –  Samuel Tinguely Apr 11 '12 at 10:20
    
I am not sure I understand your question. Here $A^\bullet$ is the complex of differential forms on the simplex, which are polynomial in the ambient coordinates. This is "de Rham's theorem for topolgists": you have differential forms (on the simplex) on the left side, the singular cochains on the right, and integration gives an isomorphism on cohomologies. I though this is the kind of statement you are looking for. –  Peter Dalakov Apr 11 '12 at 14:45

I discover this question now, I would add some words to Konrad's answer. Actually there is a spectral sequence (as usual, in these cases, that means almost nothing) between the 2 cohomologies for diffeological spaces. I even had a preprint never published on that question (never published because there was no answer inside :-\ ) What I know is how to interpret the first obstruction (§ 8.30 of the manuscript): the cokernel of the first De Rham homomorphism is a class of $({\bf R},+)$ principal bundles over the space. For manifolds its gives no obstruction since every principal bundle with contractible fiber is trivial (which is a proof of the De Rham isomorphism in degree 1). For general diffeological spaces the set of such fiber bundles is some cohomology group, it's an invariant of the diffeological space (something like Picard group). I don't know much (to say nothing) about the other obstructions, I can say however something about the degree 2. Whatever... it would be very very interesting to understand geometrically these obstructions, and get from a geometrical point of view why/when, for manifolds (in particular), the De Rham homomorphism is an isomorphism. Actually it's a question I'm very interested in.

share|improve this answer

Your Answer

 
discard

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.