Two-fold quasi-alternating links, Khovanov homology and instanton homology

Abstract

We introduce a class of links strictly containing quasi-alternating links for which mod 2 reduced Khovanov homology is always thin. We compute the framed instanton homology for double branched covers of such links. Aligning certain dotted markings on a link with bundle data over the branched cover, we also provide many computations of framed instanton homology in the presence of a non-trivial real 3-plane bundle. We discuss evidence for a spectral sequence from the twisted Khovanov homology of a link with mod 2 coefficients to the framed instanton homology of the double branched cover. We also discuss the relevant mod 4 gradings.

Figures (14)

• Figure 1: The $2^{3-1}=4$ distinct two-fold markings of the 3-component link L6n1.
• Figure 2: In these two figures we perform $0$- and $1$-smoothings on $(L,\omega)$ where $L$ is an unknot and $\omega$ is trivial. On the left, we represent $(L,\omega)$ by the two-fold marked diagram $(D,\check{\omega})$ in which $D$ is an unknot with a twist, and where $\check{\omega}$ is trivial. On the right, we represent $(L,\omega)$ by the two-fold marked diagram $(D,\check{\omega}')$, where $\check{\omega}'$ has a dot on each of the two arcs. The $0$-smoothing in either case is an unknot (which must have trivial marking data), and the $1$-smoothing is on the left an unlink with trivial marking data, and on the right an unlink split into two odd-marked unknots.
• Figure 3: Here we illustrate why L6n1 with the depicted non-trivial two-fold marking data (the left-most picture) is TQA. The middle picture is an unlink with non-trivial two-fold marking data, and thus is split into two odd-marked links. The right-most picture is a two-fold marked L4a1, which is alternating and hence TQA for any two-fold marking. Thus our two-fold marked L6n1 is TQA. Similar reasoning shows that L6n1 with the two other non-trivial two-fold markings are TQA.
• Figure 4: The Kanenobu knot $K_{p,q}$. There are $p$ and $q$ many half-twists where indicated. $K_{0,0}$ is a sum of two $4_1$ knots, which is alternating. Also, $\det(K_{p,q})=25$ for all $p,q$. Resolving the left (resp. right) encircled crossing in two ways, we obtain an unlink split into odd-marked unknots, and $K_{p\pm 1,q}$ (resp. $K_{p,q\pm 1}$), where the sign depends on the sign of $p$ (resp. $q$). Thus $K_{p,q}$ is TQA by induction.
• Figure 5: On the left, we provide the convention for 0- and 1-resolutions of crossings. On the right, we illustrate the (horizontal, unreduced) Khovanov differential for splits and merges. For a given complete resolution with circles $a_1,a_2,\ldots$, an element $a_1\wedge \cdots \wedge a_i$ corresponds to the picture in which $a_1,\ldots,a_i$ are solid and all other circles are dotted.

Theorems & Definitions (26)

• Definition 1
• Theorem 1
• Theorem 2
• Conjecture 1
• Corollary 1
• Corollary 2
• Definition 2
• Definition 3
• Remark 1
• Definition 4