Surgery exact triangles (part 1)

This is a series of posts outlining a proof of the surgery exact triangle in Floer homology.

One of the most important features of Floer homology theories for 3-manifolds is the surgery exact triangle.  Let me state this loosely for an arbitrary Floer homology theory $F$.  Suppose that $K$ is a knot in a 3-manifold $Y$ with framing $\lambda$.  Then, there is an exact triangle 

$$ \ldots \to F(Y) \to F(Y_\lambda(K)) \to F(Y_{\lambda + \mu}(K)) \to F(Y) \to \ldots   $$

Here, $Y_{\lambda}(K)$ means surgery on $K$ corresponding to the framing $\lambda$, and $\lambda + \mu$ is the framing on $K$ which wraps one more time meridionally than $\lambda$.  (This exact triangle exists as stated, for example, in Heegaard Floer homology, monopole Floer homology, and framed instanton Floer homology.  There are lots of variants of the surgery exact triangle in a bunch of Floer theories.)  The surgery exact triangle has a bazillion applications in low-dimensional topology, including a proof of Property P and the Dehn surgery characterization of the unknot, and the majority of my papers at least implicitly depend on it some way or another.  

I think there's a lot of expository literature on how to use the surgery exact triangle, but I don't see as much on the structure of the proof.  So, I wanted to write a couple posts going more into the details of how to prove the surgery exact triangle.  (You can see Saveliev's book "Invariants of Homology 3-Spheres" for a good exposition also.)  There are different proofs in different settings, but I want to focus on an approach taken by Kronheimer-Mrowka-Ozsvath-Szabo in their proof of the Dehn surgery characterization of the unknot.  This is also explained well in Scaduto's paper here.

As in any good exact sequence, the first thing to understand is what the maps are.  These are very easy to describe.  There are three associated 2-handle cobordisms, $W_{01}, W_{12}, W_{23}$ 

$$ W_{01}: Y \to Y_\lambda(K), \ W_{12}: Y_\lambda(K) \to Y_{\lambda + \mu}(K), \ W_{23}: Y_{\lambda+\mu}(K) \to Y,  $$

shown in the following pictures

The maps in the exact triangle are just the Floer homology maps associated to these cobordisms!  So, if we write 

$$ Y = Y_0, \ Y_\lambda(K) = Y_1, \ Y_{\lambda + \mu}(K) = Y_2 $$

and track all our indices mod 3, the surgery exact triangle becomes 

$$  \ldots \to F(Y_0) \xrightarrow{F(W_{01})} F(Y_1) \xrightarrow{F(W_{12})}  F(Y_2) \xrightarrow{F(W_{23})}  F(Y_0) \to \ldots  $$ In Heegaard Floer homology or monopole Floer homology, one sums over all spin$^c$ structures on these cobordisms (and three-manifolds), and that makes the gradings a bit more complicated.  In instanton Floer homology, it's a bit more subtle - one has to choose bundle data on each cobordism, and only certain choices will produce exact triangles.  (In some ways, this is a feature - one can produce different exact triangles with different bundle data, and the different exact triangles carry slightly different information.)  For the purposes of this post, we are going to suppress all of that extra data and math mostly on vibes.  Definitely, I will not be keeping track of signs in this argument.  

Our goal is to show that this data really is an exact sequence, which is going to require some serious business.  But, as a warm-up, we'll begin with half of exactness, namely, why the composition of two maps in the exact triangle is 0: $F(W_{i,i+1}) \circ F(W_{i-1,i}) = 0$ for each $i$.  The key thing is to look at the topology of these cobordisms.  

Lemma: $W_{i,i+1} \circ W_{i-1,i}$ is diffeomorphic to $Z_{i-1,i+1} \# \overline{CP^2}$ for some four-manifold $Z_{i-1,i+1}$.  

Proof:  This can be seen by taking the core disk of the 2-handle in $W_{i-1,i}$ and gluing it onto the co-core of the 2-handle in $W_{i,i+1}$.  These glue up to give a sphere with Euler number $-1$, and hence a $\overline{CP^2}$-summand.  For a more visual proof, we show this through some Kirby calculus.  For simplicity, suppose $i = 1$.  Then, we have 

The isolated $-1$-framed unknot demonstrates the $\overline{CP^2}$ summand.  

Ok, so how does this help us show $F(W_{i,i+1}) \circ F(W_{i-1,i}) = 0$.   Let's recall the composition law in Floer theory, which says that for any two cobordisms $N_1, N_2$, we have $$F(N_1 \cup N_2) = F(N_2) \circ F(N_1).$$ Therefore, we just want to see that $F(W_{i,i+1} \cup W_{i-1,i})$ is zero.  However, in our setting, we're going to have $F(Z \# \overline{CP^2}) = 0$, for any cobordism $Z$.*  

Let me comment a little more on why this cobordism map is zero.  If your brain naturally goes to Heegaard Floer homology or monopole Floer homology, you might be a little grumpy - you are going to tell me that you think that the cobordism map for $Z \# \overline{CP^2}$ should be the same as that for $Z$.  That's true and not true - for each spin$^c$ structure $\mathfrak{s}$ on $Z$ there are two different spin$^c$ structures $\mathfrak{t}_\pm$ on $\overline{CP^2}$ so that $F(Z,\mathfrak{s}) = F(Z \# \overline{CP^2}, \mathfrak{s} \# \mathfrak{t}_\pm)$.  But these two spin$^c$ structures are going to cancel out when we sum over all spin$^c$ structures.  (More generally, the contribution of a spin$^c$ structure $\mathfrak{s} \# \mathfrak{t}$ on $Z \# \overline{CP^2}$ will cancel out with that of $\mathfrak{s} \# \overline{\mathfrak{t}}$.) If your brain naturally goes to instanton Floer homology, then the point is that the bundle data will be such that the bundle is non-trivial on the $\overline{CP^2}$-summand.  Choose the metric on $Z \# \overline{CP^2}$ to be stretched along the neck of this connected sum, and an index computation in that case will show that there are simply no index 0 instantons on $Z \# \overline{CP^2}$.  (In fact, such instantons always have index at least 2.  The key point is that an instanton $A$ on $\overline{CP^2} - B^4$ equipped with a non-trivial bundle can't be a central reducible, and so its index is at least -1.  An irreducible instanton $A'$ over $Z - B^4$ necessarily has index at least 0, so if we glue $A$ and $A'$ together along a reducible over $S^3$, we get $ind(A*A') = ind(A) + ind(A') + 3 \geq 2$.)    

Hopefully this gives some insight into how the 4D topology can have some influence on the maps in the exact triangle.  In the next post, we'll set up more of the ideas that we need to prove exactness.  This will involve looking at some gauge-theoretic moduli spaces indexed by higher-dimensional families of metrics, similar to how we defined invariants of diffeomorphisms of four-manifolds from this post.  

*In instanton Floer homology, if you have a trivial bundle on the $\overline{CP^2}$, then it will be the case that $F(Z \# \overline{CP^2}) = F(Z)$.  In every exact triangle I know in instanton Floer homology, the double compositions will have non-trivial bundles on the $\overline{CP^2}$-summands.