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... and really without even the possibility of having objects, so it's not a matter of just finding the "correct" flavour of Fukaya category to use.

Question: Does there exist interesting symplectic manifolds $(M,\omega)$ without:

  • Any unobstructed Lagrangians (and therefore, with a $Ob(Fuk(M,\omega)) = \varnothing$)?
  • Any (smooth) Lagrangians?
  • Any (immersed) Lagrangians?
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    $\begingroup$ Should say something like "without any unobstructed Lagrangians": the unit circle in $\mathbb{R}^2$ is Lagrangian, hence so is the corresponding torus in $\mathbb{R}^{2n}$ hence any symplectic manifold has a Lagrangian in any neighborhood of any point. However this Lagrangian is "obstructed" (by the interior disk of the circle in $\mathbb{R}^2$ and its relatives) from participating in the Fukaya category as usually understood. $\endgroup$ Mar 6, 2017 at 19:16
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    $\begingroup$ Presumably, the empty Lagrangian is always there. So the best you can hope for is a Fukaya category whose set of objects contain exactly one element. $\endgroup$ Mar 6, 2017 at 22:56
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    $\begingroup$ It seems like a matter of convention. How does one make an "empty Lagrangian" into an object of the Fukaya category? What object of the derived category correspond to it under HMS? It seems to me like it should not count as Lagrangian (which should be a boundary condition for the Floer theory). I think that also makes sense with the usual convention for dimension (don't we assume that $dim(\varnothing)=-1$)? $\endgroup$
    – Nati
    Mar 6, 2017 at 23:47

1 Answer 1

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Let's assume that $M$ is an $n$-dimensional exact symplectic manifold and we are only interested in Fukaya categories of closed exact Lagrangian submanifolds. Then for any subcritical Weinstein manifold, $\mathcal{F}(M)$ is trivial. This follows from the fact that there is a well-defined open-closed string map

$\mathit{HH}_\ast(\mathcal{F}(M),\mathcal{F}(M))\rightarrow\mathit{SH}^{\ast+n}(M)$

and $\mathit{SH}^\ast(M)=0$, which is in fact nothing else but a modernization of a classical argument of Viterbo, which shows that any closed exact Lagrangian submanifold will contribute non-trivially to symplectic cohomology.

The above result can be generalized in both directions. Namely you can assume $M$ to be a subflexible Weinstein manifold rather than subcritical. For example, there are certain exotic cotangent bundles of spheres constructed by Maydanskiy and Seidel: https://arxiv.org/abs/0906.2230. On the other hand, you can also relax the assumption that the Lagrangian submanifolds under consideration are exact. For example, Seidel and Smith proved that for any Liouville 4-manifold $M$ which contains a weakly unobstructed Lagrangian torus $T$ with $\mathit{HF}^\ast(T,T)\neq0$, $\mathit{SH}^\ast(M)\neq0$. An application of this result is given in the paper of Lekili-Maydanskiy: https://arxiv.org/abs/1202.5625. They proved that for certain rational homology balls $B_{p,q}$, one has $\mathit{SH}^\ast(B_{p,q})\neq0$ but there is no exact Lagrangian submanifold in $B_{p,q}$. Note that this is an illustration of the fact that the above open-closed map is far from an isomorphism, it is an isomorphism only after certain non-compact exact Lagrangian submanifolds of $M$ are taken into consideration.

In the study of homological mirror symmetry, one usually defines the Fukaya category of an exact symplectic manifold by using only exact Lagrangian submanifolds. However, in view of the examples of Lekili-Maydanskiy, this does not seem to be a good definition. Since in this case the mirrors of $B_{p,q}$ are just certain finite coverings of Milnor fibers of $(A_{p-1})$ singularities with superpotantials, which admit non-trivial triangulated categories of matrix factorizations. It is believed by many people that a correct definition of $\mathcal{F}(M)$ should also involve certain immersed Lagrangian submanifolds as its objects. A good supporting evidence of this point of view seems to be the paper of Evans-Smith: https://arxiv.org/abs/1606.08656. Based on this viewpoint, I believe that there is always an immersed exact Lagrangian submanifold whose Floer cohomology is non-trivial in any Liouville domain with $\mathit{SH}^\ast(M)\neq0$.

Another typical example which is discussed in Seidel's lecture notes on categorical dynamics is $\mathbb{C}^2$ with a conic $\{xy=1\}$ removed. This manifold is algebraically and symplectically self-mirror. In this case, $M$ contains an exact Lagrangian torus $T$, and an immersed Lagrangian sphere $S\subset M$ can also be explicitly constructed. Computation shows that

$\mathit{HF}^\ast(S)\cong\mathit{HF}^\ast(T)$

Recall that the SYZ mirror construction of $M$ starts from a Lagrangian torus fibration on $M$ which has a unique singular fiber. Because of this, it's easy to see the algebraic counts of holomorphic discs only has to go through one wall and therefore the mirror $M^\vee\cong M$ consists of two $(\mathbb{C}^\ast)^2$ charts patched together using the wall-crossing formula of algebraic counts of holomorphic discs bounded by Lagrangian fibers on two different chambers. Under mirror symmetry, $T$ corresponds to the skyscraper sheaf of a point on one of the $(\mathbb{C}^\ast)^2$ chart, but $S$ corresponds to the skyscraper sheaf of the origin.

Another supporting evidence of my conjecture is the work of Alston, you can find many computations of Floer cohomologies of immersed Lagrangian spheres in his paper: https://arxiv.org/abs/1311.2327.

By the way, if you don't assume non-triviality of Floer cohomology or non-displaceability, then your second and third questions do not make sense.

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  • $\begingroup$ Could you please formulate the conjecture? Sorry, I could not extract it. $\endgroup$ Mar 7, 2017 at 6:07
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    $\begingroup$ @მამუკაჯიბლაძე Non-empty Stein manifolds contain Floer theoretically non-trivial immersed Lagrangian submanifolds. $\endgroup$
    – YHBKJ
    Mar 7, 2017 at 16:22
  • $\begingroup$ Hi YHBKJ, thanks for the answer! It looks very cool. I do have some things I don't understand though :-( $\endgroup$
    – Nati
    Mar 9, 2017 at 5:31
  • $\begingroup$ 1) How do you know the OC map is not zero? $\endgroup$
    – Nati
    Mar 9, 2017 at 5:31
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    $\begingroup$ @Nati Basically, this is because $\mathit{HF}^\ast(L,L)\cong H^\ast(L;\mathbb{K})$, then we get a map $H^{\ast+n}(M;\mathbb{K})\rightarrow\mathit{SH}^\ast(M)\rightarrow H^{\ast+n}(L;\mathbb{K})$ by composing Viterbo's map with the closed-open map, which is equal to the usual restriction map on cohomology induced by the inclusion. $\endgroup$
    – YHBKJ
    Mar 9, 2017 at 11:50

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