Surgery, Seiberg-Witten invariants, and cobordisms

Here's a quick comment which is orthogonal/complementary to a recent paper, which is also related to some of the results in this paper.  What I'm going to say below has an analogue for Donaldson invariants, but I thought I'd talk about Seiberg-Witten theory today to switch things up. 

I want to talk about how certain five-dimensional homology cobordisms govern the Seiberg-Witten invariants of four-manifolds. 

The Seiberg-Witten invariants are a powerful gauge theoretic invariant of smooth four-manifolds (and three-manifolds) that gives lots of exciting topological and geometric information.  

At a first pass, the Seiberg-Witten invariants take in a four-manifold (closed, oriented, $b^+ > 1$) and can be thought of as vanishing (such as for symplectic manifolds) or non-vanishing (manifolds with positive scalar curvature).  For the second pass, think of it as a function $SW: Spin^c(X) \to \mathbb{Z}/2$.  Per usual, if you don't like spin$^c$ structures, then just think of $H^2(X)$.  For example, if you look at the $K3$ surface, the Seiberg-Witten invariants are trivial except for the unique self-conjugate spin$^c$ structure $\mathfrak{s}_0$ (i.e. the one whose first Chern class is trivial).  For a third pass, we think of this as $SW: Spin^c(X) \times \Lambda^*H^1(X) \to \mathbb{Z}/2$.  The rough idea is that the Seiberg-Witten invariants count solutions to the Seiberg-Witten equations, and the $\Lambda^*H^1(X)$ is part of the cohomology of the space the solution space lives in; choosing an element of $\Lambda^*H^1(X)$ corresponds to cap product of your moduli space of solutions with a certain homology class.  As a concrete example to remember, the Seiberg-Witten invariants of $K3 \# S^3 \times S^1$ are non-vanishing on $(\mathfrak{s}_0, pt \times S^1)$ but vanishes for any input of the form $(\mathfrak{s}, 1)$.  Note also that if the input from $\Lambda^*H^1(X)$ is 0, then the associated Seiberg-Witten invariant is trivial.  In general, $SW_{X \# S^3 \times S^1}(\mathfrak{s} \# \mathfrak{t}, (pt \times S^1) \wedge \eta) = SW_X(\mathfrak{s}, \eta)$. 

Something cool about the Seiberg-Witten invariants is that they are well-behaved under surgery.  For example, if $\gamma$ is a loop in $X$, let $X_\gamma$ denote the result of removing a neighborhood of $\gamma$, i.e. an $S^1 \times D^3$, and gluing in an $D^2 \times S^2$.  (There are two ways to glue in, but let's ignore that discrepancy.)  There's a cool fact which is that, modulo care with spin$^c$ structures, the Seiberg-Witten invariant of $X_\gamma$ for $\eta \in \Lambda^*H^1(X_\gamma)$ are basically those of $X$ for $\gamma \wedge \eta$. 

There are two cool things to extract from this: 

1) The Seiberg-Witten invariants of $X_\gamma$ only depend on the homology class of $\gamma$. 

2) If $\gamma$ is nullhomologous, then $SW_{X_\gamma}(\mathfrak{s}_\gamma, \eta) = SW_X(\mathfrak{s}, 0 \wedge \eta) = 0$.  Because $\gamma$ is nullhomologous, any of the spheres from $S^2 \times D^2$ we glued in will actually be (rationally) homologically essential.  Conversely, given a (rationally) homologically essential sphere with trivial normal bundle, it came from surgering a (rationally) nullhomologous loop in another manifold.  In conclusion, any manifold with a square zero rationally homologically essential sphere has trivial Seiberg-Witten invariants!

Here's a neat observation as a result. Suppose that $W^5:X \to X'$ is a smooth homology cobordism built out of only 1- and 2-handles.  For simplicity, $H_1(X) = H_1(X') = 0$, but maybe there is some $\pi_1$.  That means we attach $k$ 1-handles to $X$, and $k$ 2-handles which algebraically cancel the 1-handles.  After handleslides, this means that $X'$ can be obtained by surgering $k$ loops where the $i$th loop is homologous to the $i$th copy of $S^1 \times pt$ in $X \#_k S^1 \times S^3$.  Since surgery only depends on the homology class of the loops, we see that $SW_X = SW_{X'}$.  

This means that any five-dimensional homology cobordism between four-manifolds with different Seiberg-Witten invariants (e.g. $K3 \# \overline{\mathbb{C}P^2}$ and $\#_3 \mathbb{C}P^2 \#_{20} \overline{\mathbb{C}P^2}$ are related by a homology cobordism) must contain 2- and 3-handles.  (If it had no 3-handles, then it couldn't have 4-handles, and would just be 1- and 2-handles.  Also, if it just has 3- and 4-handles, we could flip it upside down and have only 1- and 2-handles.)  Note that this is stronger than saying that any h-cobordism needs 2- and 3-handles.  Kind of weird, right???  Anyway, that's my thought for the day.