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If $X$ is a topological space then the rational cohomology of $X$ carries a canonical $A_\infty$ structure (in fact $C_\infty$) with differential $m_1: H^\ast(X) \to H^{\ast+1}(X)$ vanishing and product $m_2: H^\ast(X)\otimes H^\ast(X) \to H^\ast(X)$ coinciding with the cup product.

If $X$ is a closed manifold then its cohomology algebra satisfies Poincare duality. This is a condition that refers only to the $m_2$ part of the $A_\infty$ structure. There are things more general than manifolds that satisfy rational Poincare duality, such as rational homology manifolds. In general, a space that satisfies rational Poincare duality is called \emph{rational Poincare duality space}.

Obviously, the statement that $X$ is a rational Poincare duality space places no additional restrictions on the $A_\infty$ structure beyond the condition on the product.

Question 1 Can one construct from the higher multiplications obstructions for a rational Poincare duality space to be rationally equivalent to a rational homology manifold, or a topological or smooth manifold?

Here is a second and somewhat related question. Poincare duality says that $H^\ast(X)$ is self-dual (with an appropriate degree shift). Thus the adjoints of the higher multiplication maps make the cohomology into an $A_\infty$ coalgebra (with appropriate adjustment of the grading).

Question 2 How does this $A_\infty$ coalgebra structure interact with the $A_\infty$ algebra structure on the cohomology?

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Jeffrey, you probably already know about math.purdue.edu/~nrounds/master.pdf also, take a look at rose-hulman.edu/~su/Zhixu_Su/Research_files/Thesis.pdf for examples. –  Paul Dec 23 '10 at 15:26
    
Paul (Kirk?), thanks for the links. I had seen the first but then forgetten about it, and I certainly wasn't aware of the second. It looks like they should have some of the information I want. –  Jeffrey Giansiracusa Dec 23 '10 at 16:37
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It's probably a bad place to ask, but what would be a good text for a low-dimensional topologist to learn about this kind of things (A_infty structures on homology and the like)? –  Maxime Bourrigan Dec 24 '10 at 0:36
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6 Answers 6

Jeff, your questions were in some sense the motivation for my thesis. Let me say a few things that you probably already know before I try and answer your questions.

The $E_\infty$ algebra structure on integral cochains of a topological space $X$ is a homotopy invariant of $X$. If $X$ is nilpotent and of finite type, then the quasi-isomorphism type of the cochain algebra is a complete homotopy invariant. This is a theorem of Mandell: http://arxiv.org/abs/math/0311016.

It is also true that the $C_\infty$ multiplication on cochains is a complete invariant of the rational homotopy type of a simply connected space, but I don't know of a place where this is written down explicitly. The problem is that Quillen and Sullivan wrote their papers before ideas like infinity algebras and Kozul duality were part of the general consciousness of topologists. However, Quillen shows in Rational Homotopy Theory that the (cocomutative) coalgebra on chains is a complete invariant of the rational homotopy type of a simply connected space. He goes to some trouble to construct a cocomutative coalgebra; nowadays we would say that he is just constructing a particular representative of the quasi-isomorphism type of cochain $C_\infty$ coalgebra which happens to be strictly associative.

Now, let me try and restate Jeff's question 1, which makes sense over Q or over Z. Fix a simply connected integral or rational PD space $X$. We know:

1) The cochain algebra $C^*(X)$, considered as an E or C infinity algebra on the integral or rational cochains, is a complete integral or rational homotopy invariant given restrictions on the fundamental group.

2) The homotopy class of the map $\mu_X: X \to BG$ which determines in the Spivak normal fibration is a homotopy invariant of $X$. (I don't actually know much about the rational version of this statement, but it looks like it's laid out in Su's thesis linked above.)

3) The topological (or smooth or PL) structure set of manifold structures in the homotopy type of $X$ is again a homotopy invariant of $X$.

I interpret Jeff's question 1 to be the following: the cochain algebra knows the all the homotopy invariant information about $X$, so how do we see the info of (2) and (3) as features of the cochain algebra? The problem is that the cochain algebra depends only on the homotopy type of $X$ as a $space$, not as a Poincare duality space. I don't think that that you can ever see, for example, the structure set from only the higher homotopies of the cup product. (Though I don't have a formal proof that it's impossible.)

If you want to detect manifold structures, you instead need to look at the Poincare duality map. In my thesis, I explain how you can write down Ranicki's total surgery obstruction -- which detects whether or not the structure set is empty -- as an obstruction to the existence of "local" inverse to the Poincare duality map. (This statement is over Z. I don't know of a rational version of Ranicki's total surgery obstruction, and Ranicki told me he doesn't either.)

Thus, as I understand it, the higher multiplications are not exactly the right place to look for obstructions to manifold structures; you need to look instead at the inverse of the Poincare duality map.

I know of a couple of different answers to Jeff's question 2 about the relation between the colagebra and the algebra structure on cochains.

1) There is the following paper of Tradler and Zeinalian: http://arxiv.org/pdf/math/0309455v2 One result of this paper is that the rational chains of a PD space form an $A_\infty$ coalgebra with an "$\infty$ duality". Presumably there is a dual statement for the cochains.

2) David Chataur has a result that for any PD space X, the PD map determines an equivalence of the cochains and chains of X as $E_\infty-C^*(X)$ modules. He sent me a sketch of the proof of this statement but I don't have his permission to disseminate it.

Morally, the answer should be that the PD map is an equivalence of "infinity Frobenius" algebras. Unfortunately there are many different definitions of Frobenius algebra, and there are technical problems with writing down the infinity versions of some these algebraic structures. (The ones that have a unit and a counit.) However see this paper of Scott Wilson's: http://arxiv.org/abs/0710.3550

That ended up being a long answer! Please ask if something isn't clear!

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Hi Nathaniel, thanks for the answer and welcome to MO! –  Jeffrey Giansiracusa Dec 24 '10 at 11:06
    
I suppose an $\infty$-Frobenius object is meant to be a fully dualizable object in a symmetric monoidal $(\infty,1)$-category, to use Lurie-esque language. –  Jeffrey Giansiracusa Dec 24 '10 at 11:47
    
Would such an object have a multiplication? It sounds like you giving an infinity version of a vector space with duality, not an infinity version of an algebra with duality. –  Nathaniel Rounds Dec 24 '10 at 20:15
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One natural definition of (noncommutative) $\infty$-separable algebra that Jeffrey is alluding to is as a fully dualizable object in the symmetric monoidal $(\infty,2)$ category of algebras, bimodules and maps of bimodules associated to the symmetric monoidal $(\infty,1)$-category $k-mod$ of dg vector spaces or simplicial k-modules for a ring k (which Jacob denotes $Alg_{(1)}(k-mod)$). The Frobenius condition is replaced by asking for an $SO(2)$-invariant such object, i.e. a Calabi-Yau algebra. –  David Ben-Zvi Dec 31 '10 at 20:05
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I don't know if asking this algebra to be a commutative algebra object in addition gives the right notion of commutative ∞-Frobenius algebra, which is presumably what is required here. My guess would be no, since the commutative algebra structure isn't really interacting with the Frobenius structure. Then again, Jacob recovers string topology exactly this way from cochains on a Poincare duality space.. –  David Ben-Zvi Dec 31 '10 at 20:12
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What Nathaniel wrote "This statement is over ${\mathbb Z}$. I don't know of a rational version of Ranicki's total surgery obstruction, and Ranicki told me he doesn't either." is strictly true in the sense that I had forgotten that I had a manifold interpretation of the vanishing of the rational total surgery obstruction in Proposition 7.7.5 (page 763) of my 1980 book on exact sequences in the algebraic theory of surgery (reference 2 below)! So I have now edited my original answer accordingly

The total surgery obstruction $s^R(X) \in {\mathbb S}_n(X;R)$ of an $R$-coefficient Poincare duality space $X$ can be defined for any ring $R$ with ${\mathbb Z} \subseteq R \subseteq {\mathbb Q}$. In the 1970's Quinn (and others) developed surgery obstruction theories for $R$-coefficient Poincar\'e duality spaces, but these theories have languished both for theoretical and practical reasons (e.g. a lack of examples). Proposition 7.7.5 of 2 does give a manifold interpretation of $s^R(X)=0 \in {\mathbb S}_n(X;R)$, but I am not sure if it is sufficiently geometric for practical purposes.

If anyone is interested in reading more about the total surgery obstruction, here are some documents which are available online:

1 "The total surgery obstruction" 1978 paper

2 "Exact sequences in the algebraic theory of surgery" 1980 book

3 "Algebraic L-theory and topological manifolds" 1992 book

4 "The total surgery obstruction" 2010 MPIM lecture

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Jeffrey,

Nathaniel's 3) above is important to emphasize: a (homology) manifold is a space which satisfies local Poincare duality, i.e., it is some sort of sheaf of self dual complexes, whereas a Poincare space may have no such local structure. The total surgery obstruction measures this defect.

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Yes this is exactly the point. –  Nathaniel Rounds Dec 24 '10 at 5:16
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It was shown in the IHES paper in the 70's "Infinitesimal computations in Topology" that over Q homological Poincare duality is sufficient for a simply connected Q homotopy type to be realized by a smooth closed simply connected manifold in dimensions 5,6,7,9.10,11,13,... etc. In dim 8 ,12, 16, etc necessary and sufficient conditions to realize are given. These are tantamount to the Thom Hirzebruch signature formula holding for some choice of fundamental class where the quadratic form over Q is a sum of positive and negative squares. These statements imply the answer to question 1 is: There are no implications on the higher infinity structures implied by a closed manifold representative of a rational homotopy type.

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The homotopy Frobenius "algebra" mentioned by Nathaniel is actually a homotopy Frobenius bialgebra. By a Frobenius bialgebra, I mean here a vector space equipped with a binary commutative and associative product and a binary cocommutative and coassociative coproduct, which satisfy some compatibility condition. One can encode such an algebraic structure with an algebraic object called a properad, i.e. a multiple outputs generalization of the notion of operad. For properads, there also exist a bar construction and a cobar construction. The composite bar-cobar provides a cofibrant replacement functor for augmented properads. So Scott in his paper used this particular resolution (which is a W-construction à la Boardmann-Vogt, see Berger-Moerdijk). Bialgebras over this resolution are called "homotopy Frobenius bialgebras".

So far, the unit and the counit of Frobenius bialgebras are not encoded in the aforementioned properad. Ok, let us now consider the properad which encodes unital and counital Frobenius bialgebras. The problem is that it is no more augmented and we cannot use anymore the previous bar-cobar resolution. Joseph Hirsh and Joan Millès solved this in "Curved Koszul duality theory" http://arxiv.org/abs/1008.5368 . By applying their bar-cobar resolution to the properad encoding unital and counital Frobenius bialgebras, they get the proper notion of homotopy unital and counital Frobenius bialgebras, structure that should have the differential forms of a closed oriented manifold by Scott's method (obstruction theory).

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hi Bruno, can you explain exactly what the difference is between a Frobenius algebra and a Frobenius bialgebra? –  Jeffrey Giansiracusa Jan 20 '11 at 15:30
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Sure. A Frobenius algebra is classically defined as a finite dimensional unital associative algebra A equipped a nondegenerate bilinear form satisfying (a, b.c)=(a.b,c), see http://en.wikipedia.org/wiki/Frobenius_algebra. What I had in mind is a commutative algebra (not necessarily unital) equipped with an 'invariant' nondegenerate symmetric bilinear form, which means that (a.b,c)=(b.c,a)=(c.a,b). In other words, a cyclic commutative algebra. (I do not think that this is exactly the definition of a 'symmetric algebra' that one can find in the literature. Anyway.). In this case, the scalar product induces an isomorphism of A-modules $A\cong A^* $ between A and its linear dual $A^*$. So one can consider the transpose of the binary commutative product, which provides a binary cocommutative coproduct on $A$. The compatibility between these two says that the coproduct is a morphism of $A$-modules. Such an algebraic structure is called a Frobenius bialgebra (product, coproduct and compatibility condition). Therefore a 'Frobenius algebra' (or something close) induces a 'Frobenius bialgebra'. It does not work in the other way round. (Even if one can play with the unit and counit).

This is a general fact. An algebra over a cyclic operad is finite dimensional, because of the scalar product. But there is no such restriction for bialgebras over a properad. In the definition of a modular operad, one wants inputs and outputs to play the same role. So, they can only act on modules equipped with a nondegenerate bilinear form. The idea of properads was to break this and to split inputs and outputs. They can now act on any module. (And one can perform Koszul duality theory for properads). Notice that properads model higher genus phenomena.

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