Instanton-phobia (part 2)

In this post, I'll try to describe a little bit about moduli spaces of ASD connections on closed four-manifolds.  This includes a very rough sketch of Donaldson's diagonalization theorem. Much of this is a summary of content in Saveliev's book "Invariants for Homology 3-Spheres".  

Four-manifolds: On three-manifolds, we were interested in flat connections, which roughly correspond with representations of the fundamental group.  Much of the excitement of four-manifold topology revolves around simply-connected four-manifolds, so this would not always be an effective invariant.   

Instead, we work instead with what is called the ASD moduli space.  Let $X$ be a closed, connected, oriented, smooth 4-manifold and $E$ a principal $SU(2)$ bundle (these are determined by $\langle c_2(E), [X] \rangle$, so you can just think of a choice of integer), we can consider the solutions to the so-called anti-self-dual (ASD) equation:  

$$F_A = - \star F_A.$$

I'll say what this means, but the equation and the terms aren't really necessary to understand the key ideas.  If you don't like this, skip to the next paragraph.  One can consider connections $A$ on the $SU(2)$ bundle over $X$.  They have a curvature, $F_A$, which you can think of as a kind of 2-form on $X$.  Also, $\star$ denotes the Hodge star operator (which turns a 2-form into a 2-form).  For context, a flat connection has $F_A = 0$, so any flat connection is an ASD connection (for some choice of bundle).  Hence, the space of ASD connections knows about representations of $\pi_1$, if there is any.  One works with these connections up to some kind of equivalence (called gauge equivalence).  

We are interested in studying the ASD moduli space, denoted $\mathcal{M}(X,E)$ which is the space of solutions to the ASD equation up to equivalence.  When I say we're interested, what I mean is that actually $\mathcal{M}(X,E)$ has interesting *topological* information about $X$.  I will try to explain how to extract this topological information via black boxes, hopefully without discussing what a connection is or much about the ASD equation.  

Recall that $b^+_2(X)$ is the positive-definite part of the intersection form of $X$.  ($b^+_2(X) >0$ if and only if there is a surface of positive self-intersection.)  If $b^+_2(X) > 0$, for generic metrics on $X$, the set of solutions to the ASD equations up to equivalence is a smooth, orientable manifold and $$\dim \mathcal{M}(X,E) = 8 \langle c_2(E), [X] \rangle - 3(1 - b_1(X) + b_2^+(X)).$$If $b^+_2(X) = 0$, then for generic metrics, $\mathcal{M}(X,E)$ has the same form, except there might be some singularities.  For example, when $\langle c_2(E), [X] \rangle = 1$ and $b_1 (X) = b^+_2(X) = 0$, then the singularities look locally like the cone of $\mathbb{C}P^2$ or $\overline{\mathbb{C}P^2}$.  In this case, the number of these singularities is equal to the half the number of square -1 elements in $H_2(X)$!     

Warning!!!!  ASD moduli spaces are not always compact.  This is something that is both a bug and a feature of the theory.

In general, while $\mathcal{M}(X,E)$ does depends on the metric, one can still extract topological invariants.  We now have enough to sketch the proof of Donaldson's diagonalizability theorem for the topology of simply-connected smooth four-manifolds.

Theorem (Donaldson): Let $X$ be a simply-connected, closed, smooth four-manifold, with $b^2_+(X) = 0$.  Then, the intersection form of $X$ is diagonalizable over the integers.  

By Freedman's theorem, this says that $X$ is homeomorphic to a connected sum of $\overline{\mathbb{C}P^2}$'s.     

The proof follows from studying the "right" ASD moduli space on $X$.  In particular, we choose $E$ such that $\langle c_2(E), [X] \rangle = 1$.  In this case, $\mathcal{M}(X,E)$ is 5-dimensional, and we may have some singular points modeled on cones of $\pm \mathbb{C}P^2$ depending on whether we have elements of self-intersection -1.  But that's not all.  It turns out that $\mathcal{M}(X,E)$ is not compact, but the "non-compact end" looks like $X \times (0,1)$, and adding $X \times \{1\}$ makes it compact.  That's right...our 4-manifold $X$ actually lives inside of the 5D ASD moduli space and after compactifying, the boundary is $X$ itself!  (This part is actually where the simply-connected assumption comes in.)  So, now, if we remove neighborhoods of the cone points, we have a cobordism from a connected sum of $\pm \mathbb{C}P^2$'s to $X$.  This is how we are going to get our intersection form information about $X$.  

Recall that if $M^5 : X_1 \to X_2$ is a cobordism between 4-manifolds, then the signatures of $X_1$ and $X_2$ are the same.  (The signature is $\sigma(X) = b^+_2(X) - b^-_2(X)$.)  We know $\sigma(X) = -b_2(X)$ and so the total number of $\pm \mathbb{C}P^2$'s is exactly $-b_2(X)$.  But, the number of $\pm \mathbb{C}P^2$'s is also supposed to be half the number of self-intersection -1 elements of $H_2(X)$.  Therefore, $-b_2(X)$ is half the number of self-intersection -1 elements of $H_2(X)$.  It is then an algebraic exercise to see that this implies that the intersection form is diagonalizable over the integers, and so we proved a theorem.  

In general, this is a common technique for proving topological theorems about four-manifolds using ASD moduli spaces.  Rather than knowing the entire ASD moduli space, one can try to make use of more rudimentary properties: what is the dimension, what are the singularities, what do the non-compact ends look like, etc.  In fact, 1-dimensional ASD moduli spaces can often be the right tool: to show a certain type of four-manifold can't exist, one builds a 1-dimensional (compactified) ASD moduli space for the purported four-manifold and shows that there an odd number of boundary components, which is obviously a contradiction.

In the next post, we will combine these ASD moduli space arguments with the Chern-Simons invariants to understand 4-manifolds with boundary.