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This question is motivated by the answer in this question (you dont have to read it to understand the following.... you may have to read other things though)following).

I am not that proficient in calculating Floer homology, and I held back on answering and just commenting on the question because of the following.

Let $L$ be the equator on $S^2$ then it is obvious that no "Hamiltonian" symplectomorphism can take $L$ to another Lagrangian not intersecting $L$. Indeed, $L$ cuts $S^2$ into two equally sized pieces and it is not difficult from there.

It is quite a different thing to assert that the intersection FLoer homology $FH(L,L)$ is non-trivial, which is used in the solution to the problem (together with a kunneth formula).

The thing that initially bugged me about this was that arbitrarily close to $L$ there are Langrangians whose Floer homology intersection is $0$ both with $L$ and themselves. I have later come to realize that I myself when considering exact Lagrangians in cotangent bundles have encountered this discontinuity frequently, and should really not be surprised. Here the exactness is very important for non-triviallity. However, for me this comes up in a different way because I am considering finite reductions of the loop spaces using Chaperons broken geodesics (as Viterbo does).

I, however, realized that the discontinuity does not seem that bad and this leads me to my question:

Question: Is there a natural Floer homology with real coefficients, which resolves these discontinuities in the sense that the differentials are continuous.

A little more motivation: In the above case it seems that with coefficients in the novikov ring $\mathbb{Z}[t]$ we can realize the Floer complex with two generators, and then for all Lagrangians not dividing $S^2$ into two equally volumed parts the differential is an isomorphism. But if the Lagrangians are both of that particular type the differential is $0$. This could be described continuously with real coefficients as:

$d \colon \mathbb{R}[t] \to \mathbb{R}[t]$

where $d$ is the differential from the $\mathbb{Z}[t]$ coefficient complex multiplied by the scalar $(A_1-A_2)^2+(A_1'-A_2')^2$ where $A_i$ are the two volumes for the first Langrangian and $A_i'$ are the volumes for the second Lagrangian.

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# Real interpretations of Discontinuities in Floer homology

This question is motivated by the answer in this question (you dont have to read it to understand the following.... you may have to read other things though).

I am not that proficient in calculating Floer homology, and I held back on answering and just commenting on the question because of the following.

Let $L$ be the equator on $S^2$ then it is obvious that no "Hamiltonian" symplectomorphism can take $L$ to another Lagrangian not intersecting $L$. Indeed, $L$ cuts $S^2$ into two equally sized pieces and it is not difficult from there.

It is quite a different thing to assert that the intersection FLoer homology $FH(L,L)$ is non-trivial, which is used in the solution to the problem (together with a kunneth formula).

The thing that initially bugged me about this was that arbitrarily close to $L$ there are Langrangians whose Floer homology intersection is $0$ both with $L$ and themselves. I have later come to realize that I myself when considering exact Lagrangians in cotangent bundles have encountered this discontinuity frequently, and should really not be surprised. Here the exactness is very important for non-triviallity. However, for me this comes up in a different way because I am considering finite reductions of the loop spaces using Chaperons broken geodesics (as Viterbo does).

I, however, realized that the discontinuity does not seem that bad and this leads me to my question:

Question: Is there a natural Floer homology with real coefficients, which resolves these discontinuities in the sense that the differentials are continuous.

A little more motivation: In the above case it seems that with coefficients in the novikov ring $\mathbb{Z}[t]$ we can realize the Floer complex with two generators, and then for all Lagrangians not dividing $S^2$ into two equally volumed parts the differential is an isomorphism. But if the Lagrangians are both of that particular type the differential is $0$. This could be described continuously with real coefficients as:

$d \colon \mathbb{R}[t] \to \mathbb{R}[t]$

where $d$ is the differential from the $\mathbb{Z}[t]$ coefficient complex multiplied by the scalar $(A_1-A_2)^2+(A_1'-A_2')^2$ where $A_i$ are the two volumes for the first Langrangian and $A_i'$ are the volumes for the second Lagrangian.