# Divisibility of Lee's class and its relation with Rasmussen's invariant

**Authors:** Taketo Sano

arXiv: 1812.10258 · 2025-10-08

## TL;DR

This paper explores the divisibility properties of Lee's classes in link homology over various rings, introduces a new invariant , and investigates its relation to Rasmussen's s-invariant, with implications for link concordance and the Milnor conjecture.

## Contribution

It defines the c-divisibility of Lee's classes over integral domains and introduces a new link invariant , connecting it to Rasmussen's s-invariant and link concordance.

## Key findings

-  is a link invariant derived from Lee's classes.
-  coincides with Rasmussen's s-invariant in certain cases.
-  can reprove the Milnor conjecture.

## Abstract

Lee homology (a variant of Khovanov homology) over $\mathbb{Q}$ possesses the "canonical generators" as its basis. The generators (Lee's classes) $[\alpha(D, o)]$ are constructed combinatorially from an oriented link diagram $D$, one for each alternative orientation $o$ on $D$. Let $R$ be an integral domain. There exists a family of link homology theory $\{ H_c(-; R) \}_{c \in R}$, where Khovanov's theory corresponds to $c = 0$ and Lee's theory corresponds to $c = 2$. For each $c \in R \setminus 0$, Lee's classes $[\alpha(D, o)]$ can be defined as elements in $H_c(D; R)$, but when $c$ is not invertible then they do not form a basis; in fact they are divisible by $c$-powers. We define the $c$-divisibility $k_c(D)$ of $[\alpha(D, o)]$ with $o$ the given orientation of $D$. For any link $L$ and its diagram $D$, we prove that $\bar{s}_c(L) := 2k_c(D) + w(D) - r(D) + 1$ is a link invariant, where $w$ is the writhe, and $r$ is the number of Seifert circles. We pose the question whether $\bar{s}_c$ coincides with Rasmussen's $s$-invariant. There are several evidences that support the affirmative answer. For instance, $\bar{s}_c$ is a link concordance invariant, and the Milnor conjecture can be reproved using $\bar{s}_c$. Also for the special case $(R, c) = (\mathbb{Q}[h], h)$, our $\bar{s}_c$ actually coincides with $s$ as knot invariants.

## Full text

_Full body text omitted from this summary view._ Fetch the complete paper as Markdown: https://tomesphere.com/paper/1812.10258/full.md

## Figures

25 figures with captions in the complete paper: https://tomesphere.com/paper/1812.10258/full.md

## References

31 references — full list in the complete paper: https://tomesphere.com/paper/1812.10258/full.md

---
Source: https://tomesphere.com/paper/1812.10258