## Infinity de Rham quasi-isomorphism

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This question is similar to http://mathoverflow.net/questions/9457/do-chains-and-cochains-know-the-same-thing-about-the-manifold in the sence that both deal with a natural "comparison" quasi-isomorphism that does not preserve the ring structure.

Let $M$ be a smooth manifold. There is a natural comparison map $Comp$ from the differential forms on $M$ to the smooth singular cochains of $M$ (i.e. the linear dual of the vector space spanned by smooth singular simplices). It is defined as follows: take a form $\omega$ of degree $p$ and set $Comp(\omega)$ to be the cochain $\sigma\mapsto \int_\triangle \sigma^*\omega$ where $\triangle$ is the standard $p$-dimensional simplex and $\sigma:\triangle\to M$ is a smooth singular simplex.

$Comp$ is a map of complexes (Stokes' theorem) and moreover, a quasi-isomorphism (the de Rham theorem). But as simple examples show, it does not preserve the ring structure. However it is natural to ask whether the ring structures, up to quasi-isomorphism, of the differential forms and of the cochains contain the same information about $M$. This translates into the following questions.

1. Can $Comp$ be completed to a morphism of $A_\infty$-algebras?

2. If the answer to 1. is positive (it presumably is), what about the $E_\infty$ case?

These questions also have natural rational versions. Namely, we can take an arbitrary polyhedron $X$ instead of $M$ and consider Sullivan's $\mathbf{Q}$-polynomial forms. There is a comparison quasi-isomorphism similar to the one above that will go from the $\mathbf{Q}$-polynomial forms of $X$ to the piecewise linear $\mathbf{Q}$-cochains. Can it be completed to a map of $A_\infty$ or $E_\infty$ algebras?

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## 2 Answers

This theorem was proven in 1977 in "V. K. A. M. Gugenheim, On Chen’s iterated integrals, Illinois J. Math. Volume 21, Issue 3 (1977), 703–715."

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Yes.

Here is one way to see it:

before passing to dg-algebras, let's look at cosimplicial algebras and then later apply the normalized cochain (Moore) complex functor.

Work in a smooth (oo,1)-topos, modeled by simplicial presheaves on a site of smooth loci. In there, we have for every manifold $X$

• the singular simplicial complex $X^{\Delta_{Diff}^\bullet}$ of smooth singular simplices on $X$,

• the infinitesimal singular simplicial complex $X^{(\Delta^\bullet_{inf})}$ of infinitesimal singular simplices.

There is a canonical injection $X^{(\Delta^\bullet_{inf})} \to X^{\Delta^\bullet_{Diff}}$. We may take degreewise (internally, i.e. smoothly) functions on these, to get the cosimplicial algebras $[X^{\Delta^\bullet_{inf}},R]$ and $[X^{\Delta^\bullet_{Diff}},R]$.

The normalized cochain complex of chains on $[X^{\Delta^\bullet_{Diff}},R]$ is the complex of smooth singular cochains.

The normalized cochain complex of chains on $[X^{\Delta^\bullet_{inf}},R]$ turns out to be, by some propositions by Anders Kock, to be the deRham algebra, as discussed a bit at differential forms in synthetic differential geometry.

Therefore under the ordinary Dold-Kan correspondence we have a canonical morphism

$$N^\bullet([X^{\Delta^\bullet_{Diff}},R] \to [X^{\Delta^\bullet_{inf}},R]) = C^\bullet_{smooth}(X) \to \Omega_{dR}^\bullet(X)$$

which is an equivalence of cochain complexes. But there is a refinement of the Dold-Kan correspondence the monoidal Dold-Kan correspondence. And this says that this functor is also a weak equivalence of oo-monoid objects.

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Thanks, Urs. Do I understand correctly that your answer concerns the  $A_\infty$ case? If so, what about the $E_\infty$ case? I think this would translate as whether or not the Moore complex functor is symmetric lax monoidal. Is this translation correct? – algori Jan 6 2010 at 0:23
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