---
tag: X126J76RH7
kind: definition
name: Knot Floer homology
msc: [57K18, 57R58]
defines: 5P608WRETM
uses: [QYEAZTYGPT, EDE9919B5R, 2WJJBN2T5P]
sources:
  - url: https://en.wikipedia.org/wiki/Floer_homology
    locator: Heegaard Floer homology
---

## Definition

Let $K \subset S^3$ be an oriented tame [[knot|QYEAZTYGPT]].
Let $(\Sigma, \boldsymbol{\alpha}, \boldsymbol{\beta}, w, z)$ be a doubly pointed Heegaard diagram for $(S^3, K)$, built from a [[Heegaard splitting|EDE9919B5R]] of genus $g$.
Here $\boldsymbol{\alpha} = (\alpha_1, \ldots, \alpha_g)$ and $\boldsymbol{\beta} = (\beta_1, \ldots, \beta_g)$ are attaching curves, and $w, z$ are basepoints off them.
Let $\mathbb{T}_\alpha = \alpha_1 \times \cdots \times \alpha_g$ and $\mathbb{T}_\beta = \beta_1 \times \cdots \times \beta_g$, viewed as tori in the symmetric product $\operatorname{Sym}^g(\Sigma)$.
Let $\widehat{CFK}(K)$ be the $\F_2$-vector space with basis $\mathbb{T}_\alpha \cap \mathbb{T}_\beta$.
Its differential is
$$\partial x = \sum_{y} \# \big( \mathcal{M}(\phi) / \R \big) \, y,$$
summed over $y$ and over Whitney disk classes $\phi$ from $x$ to $y$ with Maslov index $1$ and $n_w(\phi) = n_z(\phi) = 0$.
Here $\mathcal{M}(\phi)$ is the moduli space of holomorphic representatives of $\phi$, counted modulo $2$.
The *knot Floer homology* $\widehat{HFK}(K)$ is the homology of $(\widehat{CFK}(K), \partial)$.

## Remarks

It carries a Maslov grading $m$ and an Alexander grading $a$, and up to isomorphism it does not depend on the diagram.
The graded Euler characteristic is the Conway-normalized [[Alexander polynomial|JKA25KDPQP]]: $\sum_{m,a} (-1)^m \dim \widehat{HFK}_m(K, a) \, t^a = \Delta_K(t)$.
Example: the [[unknot|1TQ06R4Q66]] has a genus-$0$ diagram on $S^2$ with one generator, so $\widehat{HFK}$ is $\F_2$ in bidegree $(0, 0)$.
Non-example: if the condition $n_z(\phi) = 0$ is dropped, the complex computes $\widehat{HF}(S^3) \cong \F_2$, which forgets the knot.
Ozsváth and Szabó proved that the Seifert genus of $K$ is the largest $a$ with $\widehat{HFK}(K, a) \ne 0$.