For simplicity let me work only with connected and simply connected spaces. "Space" will mean a space of this type.

A space is rational if its homotopy groups are rational vector spaces (equivalently, the integral homology groups are rational vector spaces). The rationalization of a space X is a rational space Y and a map $X \to Y$ with induces an isomorphism on homology groups with rational cohomology (a rational equivalence). Rational homotopy theory studies the homotopy theory of rational spaces (and rationalizations of spaces).

There are several equivalent ways to encode the rational homotopy type going back to work of Sullivan and Quillen. For example you can encode the entire rational homotopy type in terms of a commutative dga (Sullivan's rational forms). But this is a lot of structure to carry around, and I am wondering how to encode the rational homotopy type in terms of a smaller amount of data.

Specifically, I would like to know what additional information/structure you need to place on the rational cohomology groups in order to determine the rational homotopy type of a space. I have been told that "Massey products" suffice. But do you really need all Massey products? or can you get away with just some of them? Which ones? I would also be very interested in simple alternatives to Massey products which encode the rational homotopy type.


In this paper by Tornike Kadeishvili, it is shown that the rational cohomology of a simply connected space carries the structure of a $C_\infty$-algebra, and that the isomorphism type of this object determines the rational homotopy type of the space.

  • $\begingroup$ This is very helpful, thank you. The $C_\infty$-structure seems to be essentially encoding the Massey products. But I still wonder, is it not the case that some of the operations of $C_\infty$-algebras determine the others? What I am thinking here is how the induced quadratic form of a symmetric bilinear form actually determines the bilinear form (over the rationals at least). Could something similar happen here? $\endgroup$ – Chris Schommer-Pries Mar 30 '14 at 23:14
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    $\begingroup$ You're welcome. I don't understand these structures well enough to give a definitive answer to your question, but my suspicion is the answer is no. It is remarked on p12 of the linked preprint the relation of the operation $m_3$ with Massey triple products. Presumably the operation $m_n$ is related to Massey products of length $n$ in the same way. Now, I can imagine that there are spaces whose Massey products up to length $n$ all vanish, but which has a non-vanishing Massey product of length $(n+1)$. $\endgroup$ – Mark Grant Mar 31 '14 at 7:31

Let $X$ be a topological space, and let $A^*(X)$ be its CDGA of polynomial forms. As we work over a field you can produce a quasi-isomorphism (it is a chain homotopy equivalence): $$p:A^*(X)\rightarrow H^*(X,\mathbb{Q})$$ and transfer the commutative product from $A^*(X)$ to $H^*(X,\mathbb{Q})$. Then you a get a $C_{\infty}$-structure on the cohomology. As Mark writes, citing Kadesihvili, this determines the rational homotopy type of the space at least when $X$ is simply-connected and of finite type.

Just a few comments on this structure:

1) if you want to do some computations at some point you will need to have natural algebraic structures, a priori this transfer depends on choices. This transfered structure is not natural.

2) of course cohomology seems smaller and this statement is cool. But you will pay a price for this "smallness" in the complexity of the $C_{\infty}$-structure. And this is can be expensive.

3) You really need all Massey products, you can produce a lot of very funny examples playing with complement of links.

4) Dev Sinha and Ben Walter have two very nice papers on "Lie coalgebras and rational homotopy theory", the second one on Hopf invariants is related to my third remark. They explain how rational homotopy groups are related to higher linking numbers.

  • $\begingroup$ Regarding point (1). The Kadeishvili paper seems to show that the $C_\infty$ structure on the cohomology is unique. In particular it is natural and doesn't require any choices, unless I am reading the result the wrong way. Also, thanks for pointing out the examples of link compliments. It is always some much easier to get a handle on what is happening with a good supply of examples on hand! $\endgroup$ – Chris Schommer-Pries Mar 30 '14 at 23:18
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    $\begingroup$ Chris, it is unique upto equivalences of $\infty$-things, which is a rather ununique way of being unique. $\endgroup$ – Mariano Suárez-Álvarez Mar 31 '14 at 1:57
  • $\begingroup$ Thanks Mariano, you are absolutely right. The isomorphisms of $C_\infty$ objects include unspecified morphisms of higher degrees. I skimmed Kadeishvili's paper too quickly. But this observation just supports my gut feeling that the $C_\infty$ operations are carrying around irrelevant information and that there should be a smaller collection of data which determines the rational homotopy type. $\endgroup$ – Chris Schommer-Pries Mar 31 '14 at 5:49
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    $\begingroup$ "Regarding point (1). The Kadeishvili paper seems to show that the C∞ structure on the cohomology is unique. In particular it is natural and doesn't require any choices, unless I am reading the result the wrong way." No, it requires many choices, cycle choosing homomorphism for example, but the C∞ structure on the cohomology is unique up to isomorphism in the category of C∞ algebras. - Tornike Kadeishvili $\endgroup$ – Tornike Kadeishvili Jul 25 '14 at 12:07

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