I am trying to understand how all the players in the title relate, but with all the grading shifts,and difficult isomorphisms involved in the subject I am having a hard time being sure that I have the picture right. I am going to write what I think is true, and if someone would confirm or deny it, that would be really nice.

The basic jumping off point is that if N is a simply connected manifold, symplectic cohomology of the cotangent disk bundle $D^*(N)$, symplectic cohomology $SH^*(D^*(N))$ as defined in Seidel's "A Biased View of Symplectic Cohomology" is naturally identified with isomorphic to $H_{n-*}(LN)$. And symplectic homology is isomorphic to $H^*(LN)$, these are the Hochschild cohomologies and homologies respectively of the algebra $C^*(N)$.

The reason is that the zero section generates the compact Fukaya category $Fuk^{cpt}(D^*(N))$ and $End(N_0,N_0) \cong C^*(N)$. There is expected to be a geometric Seidel map from $SH^*(D^*(N)) \to HH^*(Fuk^{cpt}(D^*(N))$ basically one considers cylinders in with a puncture which satisfy a deformed d-bar equation and which are asymptotic to periodic orbits of the Hamiltonian vector field and whose boundary lies on the zero section and is able to deform the compositions in the category to first order.

Question 1) Has anybody checked in this example that Seidel's map is an isomorphism?

Now we take equivariant versions of this, $SH^*_{eq}$ which should be identified with cyclic cohomology $CC^*(C^*(N))$ and we need to consider. Now we move to (linearized) contact homology, which Bourgeois and Oancea claim can be identified with $H_{eq}(LN,N)$(I give up on the gradings at this point :)) where N is included as the constant loops. Reasonable enough, since those are somehow the generators missing from contact homology. However, they also seem to be making an identification that $SH_* \cong H_*(LN)$ and not the cohomology... I get that it's not really a huge deal as vector spaces to identify a vector space and it's dual, but it adds to the confusion below.

With contact homology one can try a similar map to the Seidel map, namely work in the symplectic completion $T^*(N)$ and consider holomorphic disks with a puncture, boundary on the zero section, and as asymptotic now to a Reeb orbit as $|\rho|\to \infty$ `($|\rho|$ is some norm on $T^*(N)$). This gives a map from $CH_{*}\to CC^{*} $ which is mysterious because...(or maybe this is a map from "contact cohomology", I'm getting confused). Edit: A reference for this map is in a paper by Xiaojun Chen called "Lie Bialgebras and cyclic homology of A(\infty structures) in topology"

Question 2) We are supposed to somehow map $H_{*,eq}(LN,N) \to H_{*,eq}(LN)$. I'm not great with topology but this is a strange map, since it seems to go the wrong way. On the other hand one should expect it to be interesting by analogy with the Seidel map. What is this map at least conjecturally supposed to be? The only thing I could think of is the kernel of the map: $C_*(LN) \to C_*(N)$ induced by the map $LN \to N$ but this map doesn't even exist equivariantly.


2 Answers 2


Some blah on symplectic homology vs. cohomology. There's an invariant $SH(M)$ of Liouville domains $M$ which some people call symplectic homology and some symplectic cohomology. This is the direct limit of Hamiltonian Floer groups associated with functions of increasing eventual slope. The dual theory has two rather unpleasant features: it involves inverse limits, hence one must worry about $\varprojlim^1$-terms; and in general it's not countably generated. It's not often used.

Why the confusion about terminology? Well, depending on your convention for the sign of the symplectic action functional, you may regard this as Morse homology or compactly supported Morse cohomology of this function. From the perspective of Lagrangian Floer cohomology, consistency demands that one calls it symplectic cohomology; I do. However, symplectic field theorists (including Bourgeois-Oancea, I think) prefer the contrary convention.

Blah about grading. The integer grading on $SH^\ast(M)$ is defined when $c_1=0$, and is canonical when $H^1(M)=0$. One has Viterbo's map $H^\ast(M)\to SH^\ast(M)$, and one convention makes this preserve degree (I'll take that option), while another makes it shift degree by the complex dimension $n$ of $M$.

Seidel's map for cotangent bundles. (Edited: my first version was not correct.) There are actually two versions of Seidel's open-closed string map, derived from the same moduli spaces: $$ \kappa: SH^\ast(M) \to HH^\ast(F(M),F(M))$$ and $$ \lambda : HH_\ast(F(M),F(M)) \to SH^{n+\ast}(M). $$ Here $F(M)$ is the Fukaya category of exact, compact Lagrangians. Moreover, there are extensions of these maps to the wrapped Fukaya category $W(M)$.

The absence of a dualisation - hence the connection of symplectic cohomology to both Hochschild homology and cohomology - looks strange. Mohammed Abouzaid points out in his comment below that this is a manifestation of a self-duality property for the wrapped category. He shows in http://arxiv.org/abs/1003.4449 that for $M=T^\ast L$ and the wrapped category, the map $\lambda$ is an isomorphism.

Cyclic version. My expectations are slightly different from those stated in the question. I'd guess that $\lambda$ extends to a map from cyclic homology to circle-equivariant symplectic cohomology, $$ HC_\ast(F(M)) \to SH^{n+\ast}_{S^1}(M) $$ and that this should be an isomorphism when $\lambda$ is.

For cotangent bundles $T^\ast L$ of simply connected, spin manifolds, one has $SH^\ast_{S^1}(M) \cong H_{n-*}^{S^1}(\mathcal{L}L)$.

Linearised contact (co)homology is, according to Bourgeois-Oancea (if I have it right), the mapping cone of the (cochain level) Viterbo map $H^\ast(M; H^\ast_{S^1}(pt.)) \to SH^\ast_{S^1}(M)$. For cotangent bundles as before, Viterbo's map should be identifiable with the map induced by the equivariant inclusion of constant loops: $H^\ast(T^\ast L)[u] = H^\ast (L)[u] = H_{n-\ast}(L)[u] \to H_{n-\ast}^{S^1}(\mathcal{L}L)$.

  • $\begingroup$ Mohammed, thanks for the corrections to my rather careless comments. $\endgroup$
    – Tim Perutz
    Jul 4, 2011 at 23:21
  • 1
    $\begingroup$ It is only $HH^∗$ of the wrapped Fukaya category which has a chance to be isomorphic to symplectic cohomology (which is verified for cotangent bundles). There are two versions of the Calabi-Yau property: the one most people are familiar with concerns the existence of a pairing, and the fact that it holds for compact Lagrangians implies that $HH^∗$ of the ordinary Fukaya category is always, up to shift, dual to $HH_∗$. There is another version (due to Kontsevich) which is a statement about duality in a category of bimodules, and which implies an isomorphism $HH^∗ \cong HH_∗$ up to shift. $\endgroup$ Jul 5, 2011 at 1:06

Let me rephrase: we consider the map from the symplectic cohomology of D*N, which (Viterbo 97, et al) is the homology of the free loop space of N, to the Hochschild cohomology of the Fukaya category consisting of compact Lagrangian submanifolds. If we allowed only the zero-section N, that Hochschild cohomology would be the Hochschild cohomology of the cochain algebra C*(N). (For the definition of the map itself, see Seidel 02).

Clearly, this doesn't work out to be an isomorphism for N = T^n: the symplectic cohomology is an exterior algebra tensor a Laurent polynomial algebra, whereas the Hochschild cohomology is an exterior algebra tensor a power series algebra (by the "odd" analogue of Hochschild-Kostant-Rosenberg).

The obvious rejoinder is that one should allow Lagrangians which are N equipped with a nontrivial flat vector bundle. However, if one takes this to mean ordinary finite-dimensional flat vector bundles, that doesn't cure the problem. There are more interesting versions with certain infinite-dimensional vector bundles, for which see Abouzaid 09.

  • $\begingroup$ This is a really interesting comment and I'll have a look for sure...in the question I was looking at simply connected manifolds, when (I think) we know that isomorphisms exist, but the question is about specific maps. $\endgroup$ Jun 26, 2011 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.