This question is similar to 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?

This question is similar to 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?

This question is similar to 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 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 of the differential forms and of the cochains contain the same information about $M$.

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?

This question is similar to 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?