Surgery exact triangles (side quest)

In a previous post I started to talk about how to prove the surgery exact triangle in Floer homology.  I hope to return to this series, but I thought I would give a simple application of the exact triangle to present an alternate proof of a recent result of Hom and Wan, where they show that the -1-traces on the knot $5_2$ and the pretzel knot $P(3,-3,-8)$ are non-diffeomorphic, even though these four-manifolds are homeomorphic.  (In other words, these are absolutely exotic four-manifolds.)  This is a very special case of their more interesting and general result relating the Heegaard Floer $\nu$-invariant to four-dimensional traces.  I thought I would instead present a goofy ad hoc argument, not for the purpose of shedding light on their work, but rather to give some more perspective on how the exact triangle can be useful.  (What is done below is secretly related to their more general work on the $\nu$-invariant, but not in an obvious way.)

We will start with a silly observation.  For notation, let $n$ be an integer and $K$ a knot in $S^3$.  Let $W_n(K)$ denote the cobordism from $S^3$ to $S^3_n(K)$ obtained by attaching an $n$-framed 2-handle along $K$.  Thus, the trace $X_n(K)$ is $B^4 \cup W_n(K)$.  It turns out we can use the information contained in surgeries other than $n$ to get invariants of the trace.  

Proposition:  The dimension of the Floer homology of $S^3_{n+1}(K)$ is an invariant of $X_n(K)$.  

Proof: We really just want to show that the dimension of the Floer homology of $S^3_{n+1}(K)$ is an invariant of $W_n(K)$. In the exact triangle in Floer homology $$\ldots \to F(S^3) \to F(S^3_n(K)) \to F(S^3_{n+1}(K)) \to \ldots $$ the map from $F(S^3)$ to $F(S^3_n(K))$ is the map induced by the cobordism $W_n(K)$.  Therefore, the dimension of $F(S^3_{n+1}(K))$ is completely determined by the rank of this cobordism map, which is hence determined by the diffeomorphism type of $W_n(K)$.   $\square$

Therefore, given $K_1$ and $K_2$, a way to distinguish the $n$-traces on $K_1$ and $K_2$ is to simply distinguish the size of the Floer homology of the $n+1$-surgeries.  (If one does Floer homology with coefficients in a field, and not an abelian group or module over a polynomial ring, then the Floer homology of $S^3_{n+1}(K)$ itself as a $\mathbb{Z}/2$-graded vector space is an invariant of $X_n(K)$.  With more general coefficients, one in principle runs into extension issues in the exact triangle.  Therefore, we just keep things simple and work with the dimension.)  

So, let's now try to show that the $-1$-traces on $5_2$ and $P(-3,3,-8)$ are not diffeomorphic.  By the proposition we want to distinguish the Floer homologies of the 0-surgery.  These knots are pretty simple and the 0-surgeries of their Heegaard Floer homologies have been calculated.  I'll describe at the end below how to calculate them, but the computation is that $$\dim \widehat{HF}(S^3_0(5_2)) = 4, \ \dim \widehat{HF}(S^3_0(P(3,-3,-8))) = 6.$$  Therefore, we see that the $-1$-traces on these knots are not diffeomorphic!  

Let's end this post with a brief discussion about how to compute the Floer homology of 0-surgery on $5_2$ and $P(3,-3,-8)$.  We begin with $5_2$.  This knot is alternating and has signature $+2$.  By Theorem 1.4 here applied to the mirror of $5_2$,  we see that $HF^+(S^3_0(-5_2)) \cong (\mathbb{F}[U,U^{-1}]/U \cdot F[U])^{\oplus 2} \oplus \mathbb{F}[U]/U$.  (The key point is that if you have a knot with "thin" knot Floer homology, then the Heegaard Floer homology of surgery is determined completely by the Alexander polynomial and signature.)  By the long exact sequence $$ \ldots \to HF^+(S^3_0(-5_2)) \to HF^+(S^3_0(-5_2)) \to \widehat{HF}(S^3_0(-5_2)) \to \ldots $$ we see that $\widehat{HF}(S^3_0(-5_2))$ is 4-dimensional (we get one for each $\mathbb{F}[U,U^{-1}]/U \cdot F[U]$ and 2 for each $\mathbb{F}[U]/U$).  But $S^3_0(-5_2) = -S^3_0(5_2)$ and the dimension of $\widehat{HF}$ is invariant under orientation reversal, so we get that $\dim \widehat{HF}(S^3_0(5_2)) = 4$.  

For $P(3,-3,-8)$ we have to do slightly more work.  Note that a different knot, $P(3,-3,0)$, is the square knot, i.e. $T_{2,3} \# -T_{2,3}$.  This knot is a connected sum of alternating knots, and so it has thin knot Floer homology.  The signature is 0 and Alexander polynomial is easy to compute.  Therefore, we can apply the same theorem from before - the Floer homology of surgery is determined by the signature and Alexander polynomial.  Using this formula, we get that $$HF^+(S^3_0(P(3,-3,0))) \cong (\mathbb{F}[U,U^{-1}]/U \cdot F[U])^{\oplus 2} \oplus (\mathbb{F}[U]/U)^2,$$ and hence $\widehat{HF}(S^3_0(P(3,-3,0)))$ is 6-dimensional.  However, conveniently $P(3,-3,-8)$ and $P(3,-3,0)$ have the same knot Floer homology (and signature and Alexander polynomial), so we also get that $\dim \widehat{HF}(S^3_0(P(3,-3,-8))) = 6$.  But how do we see that these two knots have the same knot Floer homology?  Well, the key thing is that $P(3,-3,0)$ is a ribbon knot built from attaching a band to two disks in $S^3$.  Here is a picture of $P(3,-3,0)$ where you can cut along the band:

 

Therefore, $P(3,-3,-8)$ is obtained from adding 4 twists to the ribbon band.  However, knot Floer homology is amazingly unchanged when you add twists to a ribbon disk built from 2 disks and a single band!  (See Theorem 1 here or Theorem 10 here.)  Hence $P(3,-3,-8)$ and $P(3,-3,0)$ have the same knot Floer homology.