5
$\begingroup$

$\newcommand{\wHFK}{\widehat{\mathrm{HFK}}}\newcommand{\wHFL}{\widehat{\mathrm{HFL}}}$Ni has shown that the knot Floer homology $\wHFK$ of an oriented link $L$ (in $S^3$ or more generally homology 3-spheres) detects $\chi(L)$, the maximal Euler characteristic of a Seifert surface $\Sigma$ of $L$. Here, by Seifert surface, we mean a compact oriented surface $\Sigma$ without closed components such that $\partial\Sigma=L$. In particular, we make no assumptions about the connectedness of $\Sigma$. So e.g. $\chi(U\sqcup U) = 2$ for $U\sqcup U$ for the 2-component unlink, and $\chi(T_{2,2}) = 0$ for $T_{2,2}$ the Hopf link.

Let us define $g(L)$ as the minimal genus of a Seifert surface of $L$ (where the genus of a disconnected surface is defined as the sum of the genera of the components). For example, $g(U\sqcup U) = g(T_{2,2}) = 0$. While there are some inequalities between $\chi(L)$ and $g(L)$, they do not generally determine each other.

Can one also read off $g(L)$ from $\wHFK(L)$ for all links $L$? How about link Floer homology $\wHFL(L)$?

More generally, we may consider a link $L$ with a fixed partition of its components into sublinks $L_1, \ldots, L_k$, and consider Seifert surfaces $\Sigma$ that have $k$ connected components $\Sigma_1, \ldots, \Sigma_k$, with $\Sigma_i$ a connected Seifert surface for $L_i$ for all $i\in\{1,\ldots, k\}$. Does $\wHFL(L)$ detect whether such a $\Sigma$ exists, and if so, what the maximal Euler characteristic of such a $\Sigma$ is?

A special case of this last question is: does $\wHFL(L)$ of a link $L$ detect whether $L$ is a boundary link? ($L$ is called a boundary link if $L$ is the boundary of a Seifert surface all of whose connected components have only one boundary component.)

$\endgroup$
3
  • $\begingroup$ Do you have a reference for "Ni has shown that..."? I believe you, I just want to look at the paper myself! $\endgroup$ Mar 17, 2022 at 13:53
  • 1
    $\begingroup$ Presumably, A note on knot Floer homology of links. Geom. Topol. 10 (2006), 695–713. $\endgroup$ Mar 17, 2022 at 16:12
  • $\begingroup$ @DannyRuberman Thanks! $\endgroup$ Mar 18, 2022 at 18:21

0

Your Answer

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