# base point in Heegaard Floer homology

It is stated in Mcduff's overview paper "FLOER THEORY AND LOW DIMENSIONAL TOPOLOGY", end of page 9, that the Heegaard Floer homology without considering a base point depends only on the homology of the underlying 3-manifold.

Can anyone tell me in detail how it depends on the homology of the 3-manifold, or show me some references?

Thank you very much.

## 2 Answers

McDuff states that she's assuming throughout that she's dealing with rational homology spheres only (i.e. finite $H_1$), which simplifies the answer to your question a great deal. Also, for the sake of simplicity I will only consider coefficients in $\mathbb{F}=\mathbb{F}_2$ (the field with two elements).

"Ignoring the basepoint" can be translated into algebra by saying "set $U=1$" (in all theories except the hat, where there's no $U$): this has the effect of working on the chain complex $\widehat{CF}$ but with a lot more differentials than $\widehat{\partial}$ normally counts.

Except that now we have a nice algebraic version of dealing with this, and one just needs to take the tensor product with $R = \mathbb{F}[U]/(U-1)\cong \mathbb{F}$, and when we're working over $\mathbb{F}$ the algebra is much kinder. In particular, this new version of $HF$ will be $HF^+(Y)\otimes R = HF^-(Y)\otimes R$.

Since, for rational homology spheres, $HF^\pm$ is simply one tower plus torsion in each spin$^c$ structure, and since tensoring with $R$ kills the torsion, what you're left with is simply a copy of $\mathbb{F}$ for each spin$^c$ structure.

For 3-manifolds with $b_1>0$ the situation is slightly more complicated, and I think it's now proven (at least using the isomorphism with Seiberg-Witten) that this new version only sees (co)homological data, but you have to take into account the triple cup product $H^1\otimes H^1\otimes H^1 \to H^3$. E.g., for $T^3$ I would expect that ignoring the basepoint you get $\mathbb{F}^6$, while for $\#^3(S^1\times S^2)$ (which has the same $H_1$ as $T^3$) you get $\mathbb{F}^8$.

• Thank you Marco, your reply exactly answered my question. For the case that the first betti number is non-zero, Danny (see below) showed me the references. So for non-torsion spin-c structures, the infinity-version is zero, and for torsion ones, it is given by the "cup homology". – HuiRong May 23 '16 at 4:01

The Heegaard Floer version was carried out by Tye Lidman; see On the infinity flavor of Heegaard Floer homology and the integral cohomology ring. Comment. Math. Helv. 88 (2013), no. 4, 875–898, and Heegaard Floer Homology and Triple Cup Products. Lidman proves that $HF^\infty(Y,s)$ is given by the `cup homology' introduced by Tom Mark (Triple products and cohomological invariants for closed 3-manifolds. Michigan Math. J. 56 (2008), no. 2, 265–281).

• Danny, thank you for your references. – HuiRong May 23 '16 at 4:47