Excision (part 2)

In Excision part 1, we studied how Floer homology changes when we cut and reglue along a surface.  In this post, I wanted to discuss an analogue for annular Khovanov homology.  (As a warning, I am not an expert in Khovanov homology, and definitely not knowledgeable about annular Khovanov homology.  Make sure to really fact check this!)  

First, recall that given an oriented link $L$ in $S^3$, Khovanov homology $Kh(L)$ is a bigraded vector space which categorifies the Jones polynomial.  Annular Khovanov homology, $AKh$, is a *triply* graded invariant of oriented links in $D^2 \times S^1$, developed by Asaeda-Przytycki-Sikora.  The third grading is called the annular grading, and we write $AKh(L,f)$ to mean the summand in annular grading $f$.  It's an important point that this is an invariant of the link up to isotopy in $D^2 \times S^1$ and not up to diffeomorphism.  In particular, you can cut along a meridional disk and reglue by a full twist to (almost always) change the isotopy type of the link.  Annular Khovanov homology can distinguish these!  Analogous to the fact that Floer homology can detect fibered 3-manifolds, annular Khovanov homology can detect if a link in $D^2 \times S^1$ is a braid closure.  Some quick notes by Melissa Zhang I like on the definition of annular Khovanov homology can be found here.

We now state the excision statement we are interested in.  Let $k$ denote the minimal geometric intersection number of $L$ with some $D^2 \times pt$.  Then, $AKh(L, k) = AKh(L',k)$, where $L'$ is obtained from $L$ by cutting along any $D^2 \times pt$ that intersects $L$ in $k$ points and inserting any $k$-strand braid.  

The proof will be very similar to the proof of excision for Floer homology discussed in this post.  

In order to prove this excision theorem we need to talk a bit about the annular gradings in $AKh$.  Consider the projection of $D^2 \times S^1$ onto $[0,1] \times S^1$ by forgetting the second coordinate in $D^2$, which provides a link diagram in the annulus.  For each complete resolution of this diagram (i.e. choice of smoothing at each crossing), choose an orientation of this smoothed link.  This gives a generator of the annular Khovanov chain complex and the annular grading is the associated element in $H_1([0,1] \times S^1) \cong \mathbb{Z}$.  It then follows from the definition that if $L$ is a link which intersects each $D^2 \times pt$ at most $k$ times geometrically, then $AKh(L,n) = 0$ for $n > k$.  (For an exercise, verify that if $L$ is a $k$-strand braid closure, then $AKh(L,k)$ is 1-dimensional.  This does not require understanding the differential, other than knowing that it preserves the annular grading.)      

Ok, we're ready to prove the excision theorem!  Suppose that $\sigma$ is an element of the $k$-strand braid group and we are going to insert $\sigma$ into our link $L$ in $D^2 \times S^1$.  We'd like to see that $AKh(\cdot, k)$ does not change.  We can factor $\sigma$ as a product of the Artin generators $\sigma_1,\ldots,\sigma_{k-1}$ and their inverses.  Therefore, it suffices to show that applying an Artin generator $\sigma_i$ does not change $AKh(\cdot, k)$.  Let $L'$ be the result of inserting $\sigma_i$ into $L$ at $D^2 \times pt$.  Note that $L$ is a smoothing of $L'$ at the new crossing induced by $\sigma_i$.  There is a skein exact sequence in annular Khovanov homology which respects the annular grading, which gives an exact sequence relating $AKh(L, f)$, $AKh(L', f)$, and $AKh(L_0, f)$, where $L_0$ is the other smoothing at the crossing induced by $\sigma_i$.  However, note that the geometric intersection number of $L_0$ with $D^2 \times pt$ is now $k-2$.  Therefore $AKh(L_0, k) = 0$ and we see $AKh(L,k) \cong AKh(L',k)$.  Easy peasy!