A question that's been on my mind for a while is whether any precise statement to the effect of "Heegaard Floer homology is a TQFT," for some reasonable definition of TQFT, can be made.

Of course, a lot of effort is being spent right now on HF as an extended TQFT (e.g. the bordered theory of Lipshitz/Ozsvath/Thurston and, more recently, the cornered theory of Douglas/Lipshitz/Manolescu). But right now I'm just wondering about the 3+1 structure.

The issue in 3+1 dimensions (leaving aside the mixed invariants and how to derive them from a TQFT framework) is that only cobordisms between connected 3-manifolds induce maps on HF. This is, in some sense, a fundamental feature of the theory, since the induced maps for closed 4-manifolds are zero. This was discussed in the MO question Seiberg-Witten theory on 4-manifolds with boundary.

So, what if one were to make this feature into a definition? "Some variant of TQFT" := a functor which only allows these cobordisms with connected inputs and outputs? Does this correspond to some definition that's already out there? Is it a reasonable thing to consider in the framework of, e.g., Lurie's classification of fully extended TQFTs? Or is there some other definition which could be used instead, more amenable to this framework?

I'm putting a "reference-request" tag on this question, because answering it as stated probably would consist of pointing out a relevant paper or two, but I'd be interested more generally in anything that continues the discussion from the MO question I linked above.


1 Answer 1


Katrin Wehrheim has this issue too for 2+1 dimensions; she's referred to it as "connected TFT" (at least in private communication). She and Chris Woodward are currently working on it (using Lagrangian correspondences and Cerf theory), and she has posted on her website the preprint: Floer Field Theory. Look at Definition 2.2.1. It is related to her other notes on the Symplectic 2-Category. In particular, because the surfaces are required to be connected, you don't have the product axiom.

Morphisms are assigned to the cobordisms through the Cerf relations. This is spelled out in their other note (available on her webiste): Connected Cerf Theory.

  • $\begingroup$ Thanks for pointing out those preprints- there's lots in there that seems directly relevant, and gives me hope that this type of TQFT is a natural thing to consider. I think the same issue with 2+1 dimensions appears in bordered HF too. If you have a 3d cobordism between two connected surfaces, with an arc inside connecting a basepoint on each, you get a bimodule. Somehow I think this is because you should think of the bordered surface algebra as being associated to a surface with one marked point, rather than just a "plain" surface. $\endgroup$ Oct 27, 2013 at 19:12
  • $\begingroup$ Then when you take Hochschild homology of this algebra, it gives you HFK of $pt \times S^1$ inside $F \times S^1$ (if $F$ is the surface), rather than HF of $F \times S^1$. In 4 dimensions, if you have a closed 3-manifold $Y$ and look at $Y \times S^1$, you could remove two balls and get a cobordism from $S^3$ to $S^3$. Then you might get confused, since this cobordism induces the zero map on HF, whereas in the usual TQFT picture it should be multiplication by the dimension of HF(Y). But in this perspective, maybe the more natural thing is remove the nbhd of two points from $Y$, rather (...) $\endgroup$ Oct 27, 2013 at 19:22
  • $\begingroup$ than $Y \times S^1$. Then, crossing with $S^1$, you get a cobordism from $S^2 \times S^1$ to itself, which can recover $Y \times S^1$ by gluing in two copies of $B^3 \times S^1$. The map induced on HF of $S^2 \times S^1$ is still zero, but by the previous philosophy, maybe one should consider ($pt \times S^1$ inside $S^2 \times S^1$) as a more natural thing here than just $S^2 \times S^1$. I'm not sure off the top of my head whether the induced map on HFK of $pt \times S^1$ is still zero. At least it's not equivalent to a closed surface invariant, because the $pt \times S^1$s are essential. $\endgroup$ Oct 27, 2013 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy