Having two closed exact Lagrangian submanifolds $L_1$ and $L_2$ that intersect cleanly inside a Liouville manifold $M$ with $c_1(M)=0,$ is there (with possibly some other conditions) any relation between $$H^*(L_1 \cap L_2)$$ and $$HF^*(L_1,L_2)?$$ In particular, (when) are they necessarily isomorphic?
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$\begingroup$ The intersection $L_1\cap L_2$ is zero dimensional so there is no cohomology in positive degrees. $\endgroup$– Liviu NicolaescuCommented Nov 11, 2019 at 16:05
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$\begingroup$ Not necessarily. I didn't say they intersect transversely. $\endgroup$– FilipCommented Nov 11, 2019 at 17:01
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1$\begingroup$ You can see work of Abouzaid arxiv.org/abs/0904.1474 for more refined statements about the Fukaya category generated by L_1,L_2. $\endgroup$– Daniel PomerleanoCommented Nov 11, 2019 at 18:58
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$\begingroup$ Great, thanks ! $\endgroup$– FilipCommented Nov 11, 2019 at 20:40
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1 Answer
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There is a spectral sequence from the cohomology of the intersection to the Floer cohomology. This is explained (and used) in Seidel's paper "Lagrangian 2-spheres can be symplectically knotted"
https://arxiv.org/abs/math/9803083
and is based on the thesis of Pozniak, available here:
http://www.math.ethz.ch/~salamon/PREPRINTS/Pozniak.pdf
(As far as I remember, Pozniak didn't mention the spectral sequence, but it's one way of understanding his work).
Edit: In particular, to answer your last question, if the Lagrangians are exact and the intersection is connected then the spectral sequence degenerates at $E_2$ and the Floer cohomology is isomorphic to the cohomology of the intersection. If there are J-holomorphic discs (e.g. potentially when the Lagrangians are non-exact) then this might not be true. The example to bear in mind is $L_1=L_2$: the intersection is connected, and we have $HF(L,L)=H(L)$ whenever L is exact, but not in general.
answered Nov 11, 2019 at 18:50
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$\begingroup$ Thanks! And just to clarify: for this spectral sequence to exist, you need $L_1$ and $L_2$ to intersect cleanly? $\endgroup$– FilipCommented Nov 14, 2019 at 14:59
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1$\begingroup$ Yes, sorry: I assumed you wanted that from the title of the question, but it's definitely worth stressing! The columns of the E_2 page are the cohomology groups of the components of the intersection (shifted vertically in some way that I can't recall). $\endgroup$ Commented Nov 15, 2019 at 1:52
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$\begingroup$ In addition, as Seidel explains in that paper, when $H^1(L_1)=H^1(L_2)=0,$ the Floer cohomology $HF^*(L_1,L_2)$ is $\mathbb{Z}$-graded, where the grading is defined up to a shift. Is there any canonical way to choose this shift, under some special conditions on the ambient and Lagrangians? Ideally, getting the isomorphism $H^*(L_1,L_2) \cong H^*(L_1 \cap L_2)$ without a shift? $\endgroup$– FilipCommented Feb 27, 2021 at 23:35
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1$\begingroup$ If you pick gradings on L_1 and L_2 independently then you can just adjust the grading on one of them to shift the grading on HF(L_1,L_2) however you like. $\endgroup$ Commented Feb 28, 2021 at 7:39
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1$\begingroup$ It's true that the grading on HF doesn't make sense unless L is graded... But then again, if you use the pearl model for HF, the Floer differential equals the Morse differential (there are no discs), so in some sense there is a Z grading (all pearly trajectories increase Morse index by 1). $\endgroup$ Commented Mar 4, 2021 at 20:09