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There is an algebraic theory that I'm thinking of trying to develop and I wanted to know if it had any real world prevalence --- I'd like to know an example of a generator L of a Fukaya category on a compact symplectic manifold where this Lagrangian has the properties that :

1) Simply connected,but not a sphere or a product of spheres 2) The homology of the Floer complex HF*(L,L) is isomorphic as an algebra to H*(L). (Edit: as an ordinary algebra, the higher operations induced by the perturbation lemma can be different.) Extra points if the Floer theory can in a reasonable sense be defined over C. I'd also be really interested to know why this situation can't happen.

Edit: I know it is kind of rare for a single Lagrangian(impossible?) to generate a Fukaya category. I'd be happy to have a bunch of such Lagrangians L_i where at least one of them was as above as long as Hom(L_i, L_j)=0 for i not equal to j...

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Regarding a single Lagrangian generating a Fukaya category, maybe this is relevant?: – Kevin H. Lin Sep 16 '10 at 8:55

Even the condition that you have a collection of Lagrangians which are categorically orthogonal and each with $HF^\ast(L)=H^\ast(L)$ as an $A_{\infty}$ algebra is unreasonable: There could a priori be symplectic manifolds with such Fukaya categories, but at the present state of knowledge, it is unlikely that we would be able to prove it since all methods for proving that a certain collection of Lagrangians generate the Fukaya category ultimately pass through a split-generation result for the diagonal (even the one used in Seidel's book can be interpreted in that language). On the other hand, the category you describe does not have such a resolution (you can see this by noting its Hochschild cohomology is a direct sum of homologies of free loop spaces and hence is of finite homological dimension).

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Hi Mohammed, It's great to have such an expert answering questions. One quick thing, I think maybe either I did not say what I meant or I am misinterpreting what you wrote, but when I say that I want the homology of HF*(X)=H*(L,L) I meant as an ordinary algebra(in other words agreeing up to m_2) there can be different higher operations. – Daniel Pomerleano Sep 16 '10 at 15:20
I intentionally answered the question for $A_{\infty}$ algebra structures as opposed to cohomology because I can't think of any reasonably geometric criterion which make the products agree, while the higher products to diverge. Moreover, I highly doubt that you can construct a smooth $A_\infty$ category by taking the $A_\infty$ structure on cohomology and changing the higher products. – Mohammed Abouzaid Sep 17 '10 at 1:05
Hum, that last sentence sounds like the kind of explanation I'd like to understand a little bit more, but I'm currently confused by it... isn't that exactly what happens in the Landau Ginzburg models (k[[x]],x^n) n>2 the Koszul dual to that is k[e]/e^2 in an odd variable, with a single higher operation e^(n-tensor power)-->1. Your claim is maybe that this can't happen except in the case of torus? Or is this an issue of Z/2Z grading versus Z grading? – Daniel Pomerleano Sep 17 '10 at 3:06
How about if I give you Z/2Z grading and I allow you to vary even $m_2$ but as you were suggesting I want $HF^*(L,L)$ to be smooth as a dg-algebra and I want it to make sense for some possibly small non-zero value of the parameter t? In other words, I want it to give me a non-formal deformation of the de Rham algebra to a smooth algebra? – Daniel Pomerleano Sep 17 '10 at 3:28
If you allow yourself to vary $m_2$, then you're essentially allowing an arbitrary proper Calabi-Yau algebra, and there are indeed going to be some that are smooth (though I'm not the expert on this subject). The question becomes too general at this stage to probably have a good answer. The example that you have in mind for the Landau-Ginzburg model requires changing $m_2$. – Mohammed Abouzaid Sep 17 '10 at 18:35

Even without the condition that $HF(L,L)$ is $A_\infty$-isomorphic to $H^\ast(L)$, your conditions are set up in such a way as to disable the standard tricks. This doesn't rule out the existence of examples, but it does make them hard to find.

One way to produce orthogonal objects in the Fukaya category is to take different spin structures on the same Lagrangian (this works for the Clifford torus in $\mathbb{CP}^{2n}$). Simple connectedness rules this out. So for orthogonality we need Lagrangians that are (at least Floer-theoretically) disjoint.

Free torus-orbits in toric symplectic manifolds give examples of interesting disjoint Lagrangians, but you've doubly disallowed tori! Thanks to Seidel, we know how to go about proving split generation by vanishing cycles of Lefschetz pencils, but these are spheres, and they tend to intersect one another.

We don't have any picture of the Fukaya category of a "generic" symplectic manifold, so it's hard to say how reasonable the conditions you impose really are. For all we know, it could be a common phenomenon that there are few unobstructed Lagrangians, or even none at all.

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Thanks Tim for taking me through the issues. This is kind of what I thought the theoretical status was... but it's great to hear an expert opinion. – Daniel Pomerleano Sep 17 '10 at 3:44
Mohammed is quite right (and I missed) that if $m_2$ is classical, that may well be for a geometric reason that also makes higher structure classical; e.g. look at the diagonal in $X_-\times X$, whose HF is $QH^\ast(X)$. The product is undeformed if $X$ is symplectically aspherical, say, but then there can be no non-classical higher structure either. And I don't know an example of a monotone Lagrangian torus whose primary deformation class (a quadratic form) is trivial but whose secondary class is not. – Tim Perutz Sep 17 '10 at 14:59
Ok I see what you are both saying about the product... But even if I lower my restrictions and ask for a Z/2Z graded Fukaya category which is generated by an orthogonal set of Lagrangians which are not spheres or products and keep the idea that HF(L,L) should be "defined over C" and admit vector space iso with cohomology, it sounds like even still we don't know such an example? – Daniel Pomerleano Sep 17 '10 at 15:29
I haven't thought of one, but the conditions look more realistic now. – Tim Perutz Sep 17 '10 at 16:46

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