Does $H_*(A^-_0(K))=\mathbb{F}[U]$ imply that $K$ is an L-space knot?

Question

Let $K$ be a knot in the three-sphere. Let $A_s^-(K)$ be the Alexander filtrations of the knot Floer complex $CFK^{\infty}$. Would $A_0^-(K)$ has homology $\mathbb{F}[U]$ imply that K is an L-space knot? Are there counterexamples?

Thanks

Answer 1

The answer is no, and I claim that $6_2$ is a counterexample. (I will work over $\mathbb{F} = \mathbb{F}_2$ to avoid dealing with signs, but what I say holds for $\mathbb{Z}$-coefficients as well.)

In fact, according to knotinfo, $6_2$ is alternating, has genus 2, $\tau(6_2)=1$, and Alexander polynomial $t^2-3t+3-3t^{-1}+1$.

$6_2$ is not an L-space knot: its Alexander polynomial has some coefficients larger than 1.

It is commonly known that for alternating knot the Alexander polynomial determines the structure of $CFK^\infty$ (see, for example, Theorem 4 of this paper by Petkova).

In particular, for an alternating knot $K$ the complex $CFK^\infty(K)$ is a sum of "squares" and "staircases". Squares correspond to 4-dimensional subcomplexes with differentials $x\mapsto y+Uz$, $z\mapsto w$, $y\mapsto Uw$, while staircases corresponds to subcomplexes isomorphic to $CFK^\infty(T_{2,2k+1})$ for some positive integer $k$, and the "length" of the staircase is determined by $k$, which in turn is determined by $\tau(K)$.

In the case at hand, $K=6_2$ and $C=CFK^\infty(6_2)$ is a sum of two squares and a staircase corresponding to $3_2 = T_{2,3}$; moreover, the NE-corners of the two squares (i.e. $x$ in the notation above) sit in degree $(i,j) = (0,2)$ and $(0,0)$. It is also easy to see that squares contribute to the homology of the subcomplex $C\{i\le 0,j\le 0\}$ if and only if their SW-corner (i.e. $w$ in the identification above) is in degree $(i,j)=0$.

$A^-_0(6_2) = \mathbb{F}[U]$. In fact, the SW-corners of the two squares are in degrees $(0,1)$ and $(0,-1)$, so they give no contribution, so $A^-_0(6_2) = A^-_0(3_2) = \mathbb{F}[U]$.