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Floer homology associates a vector space $HF^\ast(L_1,L_2)$ to any pair of Lagrangian submanifolds $L_1,L_2$ inside a symplectic manifold $X$. By a categorification of Floer homology, I mean a category $\mathcal H\mathcal F(L_1,L_2)$ whose Hochschild homology is isomorphic to $HF^\ast(L_1,L_2)$.

Example: Let $M$ be a closed symplectic manifold, and consider the diagonal $\Delta\subseteq M\times M^-$, which is Lagrangian. It self-Floer homology $HF^\bullet(\Delta,\Delta)$ is (conjectured to be) isomorphic to the Hochschild homology of the Fukaya category of $M$. Thus $\mathcal F(M)$ is a categorification $\mathcal H\mathcal F(\Delta,\Delta)$ of $HF^\ast(\Delta,\Delta)$. Note $HF^\ast(\Delta,\Delta)$ is a unital algebra, and expressing this unit as a Hochschild cycle of $\mathcal F(M)$ is of fundamental importance in the study of the Fukaya category.

Are there any other known circumstances under which a categorification of Floer homology exists?

Floer homology also has a product $HF^\ast(L_1,L_2)\otimes HF^\ast(L_2,L_3)\to HF^\ast(L_1,L_3)$, so it is natural to further ask that a categorification $\mathcal H\mathcal F$ have natural functor $\mathcal H\mathcal F(L_1,L_2)\times\mathcal H\mathcal F(L_2,L_3)\to\mathcal H\mathcal F(L_1,L_3)$ which upon applying Hochschild homology recovers the product on Floer homology.

What is a natural geometric description of a functor $\mathcal F(M)\times\mathcal F(M)\to\mathcal F(M)$ corresponding to the product on $HF^\ast(\Delta,\Delta)$?

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    $\begingroup$ We now know that the map from the cohomology of the Thurston manifold to the Hochschild cohomology of the Fukaya category is not an isomorphism, because the mirror is not algebraic. $\endgroup$
    – Mohammed Abouzaid
    Apr 30, 2017 at 23:37
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    $\begingroup$ Regarding categorification: the most natural thing to look for is a mirror to the conjectures of Kapustin-Rozansky-Saulina on the B-side (see arXiv:0810.5415). $\endgroup$
    – Mohammed Abouzaid
    Apr 30, 2017 at 23:42
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    $\begingroup$ I also don't think it's possible to expect in general that the product on the Floer cohomology of a diagonal arises from a functor as you are asking: this is like asking for the (intersection) product on the homology of a manifold to arise from the structure of an H-space on the given manifold. In both cases, the degree of the product is given by the dimension, so it doesn't work out. The classical case of course offers a hint as to what one can hope for. $\endgroup$
    – Mohammed Abouzaid
    Apr 30, 2017 at 23:46
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    $\begingroup$ To elaborate on Mohammed's comment: an optimistic version of your question would be to seek a 3d lift of the A-model, just as Rozansky-Witten theory is a 3d lift of the B-model (ie recovers the B-model on compactifying on $S^1$, i.e. taking Hochschild homology of Hom categories). The latter only exists when the target is holomorphic symplectic. Indeed to get a 3d SUSY QFT giving a TFT the target needs to be hyperkahler, so a physicist would likely insist your $M$ be holomorphic symplectic. $\endgroup$
    – David Ben-Zvi
    May 1, 2017 at 16:27
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    $\begingroup$ There is a "3d A-model" studied in arxiv.org/abs/1002.4241 but it recovers the A-model (of a cotangent bundle) by reduction on an interval with suitable boundary conditions, not on a circle, so doesn't give what you want. Teleman and Dimofte have discussed informally the existence of a 3d A-model with properties like you want in the holomorphic symplectic setting, and loosely "3d mirror" to Rozansky-Witten theory, but nothing is written beyond Teleman's seminal work, cf his ICM, in the case M=pt/G.. $\endgroup$
    – David Ben-Zvi
    May 1, 2017 at 16:30

1 Answer 1

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I am sort of an armchair mathematician these days, but I suspect the answer to your first question is "no". At least in the hep-th literature categorification of Khovanov homology I have seen many instances of, Floer homology in general I know much less about. There are just so many:

  • Floer homology
  • Knot Floer homology
  • Bordered Floer homology
  • Instanton Floer homology
  • Monopole Floer homology
  • Heegard-Floer homolo
  • etc.

So the theory specializes in many different ways, and they all have really great properties. However, there are simply too many and I have lost track.

On the physics side, I know the way to understand all of these is Chern-Simons Theory and a small amount of google-fu returns

I am not an expert in 3-manifolds, so I can't say whether they fit into your question or not. I know a lot of the mathematical work is all about showing these functors map nicely onto the properties of 3-manifolds so that these define invariants.

Here I'm just being a librarian and not really opening up the papers in any way. I can read through Audin's Morse Theory and Floer Homology and notice complexes jump around as you change the choice of height function. There's a nice course on YouTube by Denis Auroux.

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    $\begingroup$ I don't understand what the point of listing all these Floer homologies is. Knot Floer homology is not a "Floer homology" as the OP defines it (it comes from a filtration induced by a knot on a Floer homology). But ultimately all of the rest of these Floer homologies do (or are expected to) come from a construction as the OP defines. Work of Manolescu-Woodward and Horton on symplectic instanton homology and the fact that Heegaard-Floer homology is "symplectic monopole Floer homology", together with the fact that bordered Floer homology is a computational scheme for Heegaard-Floer homology... $\endgroup$
    – dvitek
    May 1, 2017 at 4:11
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    $\begingroup$ ...(albeit one that comes from an interesting extension of HF to a 2+1+1-dim'l "TQFT"), means that all of the examples you've given are in fact examples of Lagrangian Floer homology. I also don't understand the relevance of the Chern-Simons shoutout - sure, some of the examples of Floer homologies you gave use the CS functional, but the OP's question is entirely contained within Lagrangian Floer homology (where you don't need a CS functional to define anything). $\endgroup$
    – dvitek
    May 1, 2017 at 4:16
  • $\begingroup$ @dvitek no idea $\endgroup$
    – john mangual
    May 1, 2017 at 6:51

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