Excision (part 1)

In algebraic topology, the excision theorem is an extremely useful tool for computing homology.  In case your algebraic topology class is ancient history, let's recall the statement: if $U \subset V \subset X$ are a sequence of spaces with $\overline{U} \subset V^\circ$, then $$H_*(X - U, A - U) \cong H_*(X, A).$$In other words, relative homology is unaffected by "cutting" out $U$.  In Floer homology, we also have excision, which involves a different kind of cutting.  In this post I want to discuss some aspects of Floer homology in excision.  In the next post, I'll give proof of a similar statement in annular Khovanov homology.  

 

Simplest version of excision: 

First I want to state and prove an easy version of Floer excision.  I'm going to write this in a way that applies in most Floer theories, so I'll use $F$ to denote an arbitrary Floer homology.  Let $Y$ be a closed connected oriented three-manifold.  Suppose that $x$ is a primitive element of $H_2(Y)$ which can be represented by a connected surface $S$.  Note that this requires $b_1(Y) > 0$ and that $S$ is non-separating.  For simplicity, let's assume that $S$ has genus $g \geq 2$.  (There's an analogous story for $g = 0, 1$ but it's more annoying to write down.  Sorry not sorry.  If you want some good deets see this paper of Bagherifard for example.)  Then, $S$ picks out a special summand of $F(Y)$, denoted $F(Y | S)$.  Heegaard / monopole Floer homologists should think of $F(Y | S)$ as summing over spin$^c$ structures whose $c_1$'s evaluate to $2g-2$ on $S$ in the $+$ / "to" theories.  In instanton Floer homology, this is the $2g-2$-eigenspace of the surface operator $\mu(S)$.  

The group $F(Y | S)$ is very well-behaved.  It is non-zero precisely when $S$ is Thurston norm minimizing.  (Recall that the Thurston norm of a closed possibly disconnected surface is the negative of the Euler characteristic after throwing out the spherical components.)  This is part of the broader statement that Floer homology detects the Thurston norm.  In general, it can be small (e.g. $F(S \times S^1 | S)$ is 1-dimensional) or can be very large (e.g. $F(S^3_0(K) | S)$ becomes arbitrarily large where $K$ is a complicated twist knot and $S$ is a capped-off Seifert surface.)  

Theorem (Excision, v1):  Let $Y'$ be the three-manifold resulting from cutting $Y$ along $S$ and regluing.  Then $F(Y | S) \cong F(Y' | S)$.  

Usually $F(Y | S)$ doesn't have a very refined grading, but it should have at least a relative $\mathbb{Z}/2$-grading.  I'll let the reader deduce from the proof we give that this preserves the relative $\mathbb{Z}/2$-grading. 

Why might you care?  Here's an example.  Let $Y$ be a fibered three-manifold with fiber $S$.  Then $Y$ is obtained from cutting and regluing along $S$ in $S \times S^1$.  That means that if $Y$ is fibered then $F(Y | S)$ is 1-dimensional.  Wow, that was fast!  (There's also the awesome converse that if $Y$ is an irreducible 3-manifold, then $F(Y | S)$ is 1-dimensional if and only if $Y$ is fibered.  But you can't prove it in a simple way like this.)

Let's talk about the proof.  The input is going to be some gluing map for how to reglue $S$ to get $Y'$.  Every (orientation-preserving) diffeomorphism of $S$, up to isotopy, can be factored into a product of Dehn twists by some old work of Dehn (also attributed to Lickorish).  So, actually, we just need to prove that cutting and regluing along a Dehn twist along $S$ does not change $F( \cdot | S)$.  Of course, we can restrict to positive Dehn twists.  But how can we understand Dehn twist information with Floer homology??? Dehn surgery baby!  And how can we do that?  Surgery exact triangle! 

The key input here is the Lickorish trick, which we need a quick aside on.  In general, suppose that $\Sigma$ is a surface in a three-manifold $M$ and let $\gamma$ be a simple-closed curve on $\Sigma$.  Then $\gamma$ inherits a natural framing from $\Sigma$, called the $\Sigma$-framing.  This is the framing curve that is a parallel copy of $\gamma$ that sits on $\Sigma$.  Let's call it $f_{\Sigma,\gamma}$.  Then Lickorish proved that if we do $f_{\Sigma,\gamma} - 1$-surgery on $\gamma$, that's the same as cutting $M$ along $\Sigma$ and regluing by a positive Dehn twist.  (If you haven't seen it before, this is a good route to prove that all 3-manifolds are surgery on a link in $S^3$.)

Back to our case.  Recall we would like to prove that if $Y'$ is obtained by cutting along the non-separating surface $S$ in $Y$ and regluing by a positive Dehn twist along $\gamma$, then $F(Y' | S) \cong F(Y | S)$.  Therefore, we would like to show that doing a $f_{S,\gamma} - 1$-surgery on $\gamma$ does not change $F( \cdot | S)$.  So, let's now consider the surgery exact triangle applied to $\gamma$.  We get an exact sequence that respects the $S$-decomposition:

$$ \ldots \to F(Y | S) \to F(Y' | S) \to F(Y_{f_{S,\gamma}}(\gamma) | S) \to \ldots $$ 

We'd like to show that $F(Y_{f_{S,\gamma}}(\gamma) | S)) = 0$.  Now, in $Y_{f_{S,\gamma}}(\gamma)$, we have to be a little careful.  We can take an isotopic copy of $S$ away from where the surgery happens, and that's the $S$ we mean here.  However, this $S$ will no longer minimize Thurston-norm.  Why not?  Well, when we do $f_{S,\gamma}$-surgery on $\gamma$, this has the effect of making $\gamma$ bound a disk that is loosely disjoint from $S$.  That means that in the same homology class of $S$, there is a surface, possibly disconnected, obtained from surgering $S$ along $\gamma$, which necessarily has lower Thurston norm.  This means that $F(Y_{f_{S,\gamma}}(\gamma) | S) = 0$.  This completes our proof of excision!!!

 

A  more general excision statement:

There is a more general version of excision that I think is really cool.  This is described well in this paper of Kronheimer-Mrowka.  Suppose that $S = S_1 \coprod S_2$ consists of two disjoint surfaces of the same genus (at least 2) in $Y$. Here, $Y$ could be connected or two components $Y_1 \coprod Y_2$ with $S_i \subset Y_i$.  In the connected case, $F(Y|S) = F(Y|S_1) \cap F(Y|S_2)$ and in the disconnected case $F(Y|S)  = F(Y_1 | S_1) \otimes_{\mathbb{F}} F(Y_2|S_2)$ if we work with coefficients in a field $\mathbb{F}$.  We can cut along both $S_1$ and $S_2$ and reglue to get $Y'$.  Here, when we cut along $S_1$ and $S_2$ there are four surfaces, say $S_1, -S_1, S_2, -S_2$ if we track the orientations, and we need to glue $S_1$ to $-S_2$ and $S_2$ to $-S_1$.  The excision theorem then says that $F(Y|S) \cong F(Y'|S)$.  

As a warm-up exercise, prove that this general excision statement implies the previous version we discussed.  For a more fun exercise, prove that if $\widetilde{Y}$ is obtained from $Y$ by taking an $n$-fold cyclic cover corresponding to a connected non-separating surface $S$ then $F(\widetilde{Y} | S)$ is the tensor product of $n$ copies of $F(Y | S)$.