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Let $(M,g)$ be a Riemannian manifold, let $\Omega^*(M)$ denote the cochain complex of differential forms on $M$ and $H^*(M)$ its cohomology considered as a chain complex with trivial differential. We have a map $f:\Omega^*(M) \to H^*(M)$ given by projecting a form to its harmonic part and taking its cohomology class and we have a map $h: H^*(M) \to \Omega^*(M)$ which sends a cohomology class to its harmonic representative. Then $fh$ is the identity and $hf$ is chain homotopic to the identity. Hence, we have the right setting to transfer the $C_{\infty}$-algebra structure of $\Omega^*(M)$ with structure maps $m_1=d$, $m_2=\wedge$, and $0=m_3=m_4=...$ to obtain an $C_{\infty}$-algebra structure on $H^*(M)$. We can get formulas for the structure maps of the transferred structure on $H^*(M)$ by taking sums over trees and putting the chain homotopies in the right internal edges, etc...

The topological information information obtained from this transferred structure is understood: up to homotopy, the transferred structure contains rational homotopy information.

However, my question is the following: What kind of geometric information is contained in the transferred structure (which depends on the metric $g$ and it involves choices) up to isomorphism ?

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  • $\begingroup$ This is an interesting question, but could do with being a bit more precise. For instance, could you say exactly what you mean by "up to homotopy" and "up to isomorphism" for $C_\infty$-algebras (or give a reference)? $\endgroup$
    – Mark Grant
    Dec 4, 2012 at 12:22
  • $\begingroup$ Let me see if I can make sense of my question. I agree that it is not clear. The transfer of structure depends on many choices. Suppose you have two metrics $g$ and $g'$ on $M$. Then these would lead to different $C_{\infty}$ structure maps $m^g_i$ and $m^{g'}_i$ on $H^*(M)$. However, when you take homology of these, i.e. when you form the bar construction and take homology of that you should get rational homotopy. My question is, do the transferred maps $m_i^g$ and $m_i^{g'}$ reflect any geometric information? $\endgroup$ Dec 5, 2012 at 1:32
  • $\begingroup$ Note that my question is metric dependent. $\endgroup$ Dec 5, 2012 at 2:54
  • $\begingroup$ @Manuel: Thanks for the clarification. I think that by an "isomorphism" of $C_\infty$-algebras $(A,\lbrace m_i\rbrace)$ and $(A',\lbrace m_i'\rbrace)$ you mean an isomorphism of vector spaces $A$ and $A'$ which throws the $m_i$ onto the $m_i'$. I think Kadeishvili's notion of isomorphism is strictly weaker than this, hence my confusion. $\endgroup$
    – Mark Grant
    Dec 5, 2012 at 15:40

2 Answers 2

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As far as I understand it, Theorem 7 of http://arxiv.org/abs/0811.1655 states that the $C_\infty$-algebra structures on $H^\ast(M)$ up to isomorphism correspond to rational homotopy types with the given cohomology algebra. This implies that the answer to your question is "any geometric information which can be read off from the rational homotopy type". This is a significant amount, see the book "Algebraic Models in Geometry" by Felix, Oprea and Tanré, or this Oberwolfach Report.

(Of course, it may just be that your interpretation of "isomorphism" of $C_\infty$-algebras is stricter than Kadeishvili's.)

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  • $\begingroup$ In fact you can put more structure on the transferred $C_{\infty}$-structure on cohomology. For example you can take into account Poincaré duality, this was studied by A. Hamilton and A. Lazarev. This additional data is enough to build string operations on loop homology of the manifold. $\endgroup$
    – David C
    Dec 4, 2012 at 7:56
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I am sorry but I will not answer your question rather it reminds me some beautiful results which go back to H. Whitney and is related to the transfer of the wedge product of forms.

If you pick a triangulation $T$ of your manifold $M$ and consider the rational cochain complex $C_T(M)$ associated to $T$. You have an integration map: $I:\Omega(M)\rightarrow C_T(M)$ and you H. Whitney has defined an inverse map $W:C_T(M)\rightarrow \Omega(M)$ (we have to use $L_2$-forms in fact). We have $IW=Id$ and $WI$ is homotopic to the identity. These maps are not multiplicative with respect to the cup product and the wedge product. But H. Whitney has given a very nice combinatorial formula for a product on $C_T(M)$, let call it the Whitney product and let denote it by $\bullet$, we have: $$c\bullet c'=I(Wc\wedge Wc').$$ What H. Whitney has done is to give a very nice and combinatorial formula for the product of the $C_{\infty}$-transfer of the algebra structure of differential forms on cochains.

But the story does not end here, if you want to add some geometrical data to the story you can build a combinatorial Hodge operator on cochains. And now look at the limit of all these structures when you refine the triangulation. In fact the combinatorial Hodge operator converges to the smooth Hodge operator and for the Whitney product all higher homotopy products will vanish in the limit. This was studied in details by S. O'Wilson in "Cochain algebra on manifolds and convergence under refinement", Topology and its Appl, 154 no. 9 (2007) 1898-1920.

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