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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

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  • $\begingroup$ Mathjax help. $\endgroup$ – user57432 May 16 '15 at 4:59
  • $\begingroup$ Welcome to MathOverflow! You can get math to show up just by putting it in dollar signs. I did that for you here. I might also mention something related to Heegaard Floer homology a little earlier in the question. $\endgroup$ – Dylan Thurston May 16 '15 at 12:04
  • $\begingroup$ As for the actual question, I have no idea. Can you say some more about why you believe it might be true? $\endgroup$ – Dylan Thurston May 16 '15 at 12:05
  • $\begingroup$ Thanks for editing the question! In fact, I don't think it is true. But I just wonder whether people have seen enough examples of computations of $A^{-}_{s}$. $\endgroup$ – Yajing Liu May 16 '15 at 18:04
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    $\begingroup$ Just to make sure, what do you mean exactly by $A_0^-$? Do you mean the subcomplex $C\{j\le 0\}\subset CFK^\infty$ with all differentials, or the graded object associated to this sublevel (i.e. with the horizontal differentials only)? My gut feeling is that, in the first interpretation every knot satisfies the assumption, while in the second only the unknot does. $\endgroup$ – Marco Golla May 17 '15 at 20:40
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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 $A^-_0(6_2) = A^-_0(3_2) = \mathbb{F}[U]$.

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