Absolutely exotic contractible four-manifolds

Since exotic four-manifolds are always "in", I wanted to talk about a cool construction of exotica I like due to Akbulut-Ruberman.  In their paper, they construct compact contractible four-manifolds which are "absolutely exotic", which means they are homeomorphic, but there is no diffeomorphism between them.  To clarify the terminology, often times, for manifolds with boundary, one talks about "relatively exotic" manifolds, which means that you have a homeomorphism from one four-manifold to another, but the boundary homeomorphism doesn't extend to a diffeomorphism of the four-manifolds.  [For context, it's worth pointing out that long before, we know how to construct lots of non-compact, contractible exotica: namely exotic $\mathbb{R}^4$'s.  In fact, Taubes proved there are uncountably many exotic $\mathbb{R}^4$'s.]

I want to describe their work (but with a bit of a Floer homology lean).   For simplicity, let's construct a pair of non-diffeomorphic, compact contractible four-manifolds, instead of infinitely many distinct smooth structures.  If you're familiar with Floer homology for three- and four-manifolds, then skip down to Proof! below.  

We first need a couple facts about Floer homology.  Pick your favorite Floer homology theory $F$ (you don't really need to know any explicit Floer homologies to understand the proof).  

1) Given a four-manifold $X$ with boundary $N$, we get a relative invariant $F(X)$ which takes values in the Floer homology of $N$, $F(N)$.  This works essentially for any Floer-type theory of 3/4-manifolds.  This gives a powerful invariant of four-manifolds, because we can try to tell apart four-manifolds by their relative invariants.  Here, we need to be careful that thinking of $F(X)$ as an element of $F(N)$ implicitly depends on our identification of $N$ with $\partial X$.  Subtle but crucial point: If $X_1, X_2$ have boundary $N$ and $F(X_1) \neq F(X_2)$ it unfortunately doesn't mean that $X_1, X_2$ are non-diffeomorphic; it just means that there's no diffeomorphism from $X_1$ to $X_2$ that commutes with the identifications of $N$ with $\partial X_1$ and $\partial X_2$.  HOWEVER, if $N$ has no non-trivial symmetries up to isotopy, then that really does mean that $X_1$ and $X_2$ are not diffeomorphic.  

2) Given a cobordism $M : Y_1 \to Y_2$ we get an induced map $F(M) : F(Y_1) \to F(Y_2)$ which is a functor from the cobordism category to vector spaces.  In other words, $F(Y \times I) = id_{F(Y)}$ and $F(M \cup M') = F(M') \circ F(M)$.  In particular, if $M: Y \to Y'$ is an invertible cobordism, i.e. there exists $M'$ such that $M \cup M' = Y \times I$, then $F(M)$ is injective.  Another setting which guarantees injectivity of cobordism maps is when $M$ is a ribbon homology cobordism (i.e. a homology cobordism built out of only 1- and 2-handles).     

We are now ready for the Proof!

Step 1: Carefully pick a contractible four-manifold $W$.  Danny and Selman use this one (picture taken unethically from their paper):

 

If you haven't seen a Kirby diagram, start with $B^4$, remove the disk that the dotted circle bounds (pushed slightly into $B^4$), and then attach a 0-framed 2-handle along the knot with framing 0.  The boundary, denoted $Y$, can then be described as 0-surgery on this two-component link. 

Step 2: Find a self-diffeomorphism $\tau: Y \to Y$ such that $\tau_*(F(W)) \neq F(W)$.  This condition tells us that $\tau$ does not extend over $W$ as a self-diffeomorphism.  However, Freedman's work on homeomorphisms of simply-connected four-manifolds does tell us that $\tau$ extends over $W$ as a homeomorphism.   Our $\tau$ will simply be rotation about the vertical axis as in the figure above.  (The effect on Heegaard Floer homology is studied here for example.)   The pair $(W,\tau)$ is what's known as a cork - a contractible four-manifold with an involution on the boundary that does not extend smoothly.  (This manifold $W$ is particularly nice.  One can construct certain closed exotic four-manifolds by cutting out a copy of $W$ and regluing by a twist.  $W$ also can be equipped with a Stein structure.)

Step 3: Find a three-manifold $Y'$ and a cobordism $M: Y \to Y'$ such that i) $F(M)$ is injective, ii) $Y'$ has no non-trivial self-homeomorphisms up to isotopy, iii) $\pi_1(Y)$ normally generates $\pi_1(M)$, iv) $M$ is a homology cobordism.  (I'll give a bit more description on how to do this below.)

Step 4: Let $X_1 = W \cup M$ and $X_2 = M \cup_\tau M$.  We claim these are the desired compact, contractible, exotica.  That $X_1, X_2$ are contractible follows from $M$ being a homology cobordism and the $\pi_1$ condition on $M$ since $\pi_1(W) = 0$.  Further, $X_1$ and $X_2$ are homeomorphic because we can extend $\tau$ over $W$.  But why are $X_1$ and $X_2$ non-diffeomorphic?  Well, by construction $F(X_1) = F(M)(F(W))$ and $F(X_2) = F(M)(\tau_*(F(W))$.  Since $\tau_*(F(W)) \neq F(W)$ and $F(M)$ is injective, the resulting relative invariants in $F(Y')$ are different.  Since $Y'$ has no non-trivial symmetries, that means that $X_1$ and $X_2$ are indeed non-diffeomorphic!  Yey!

So, to finish this blog post, I should say a little more about constructing this mysterious homology cobordism $M$ to an equally mysterious symmetry-less three-manifold $Y'$.  In general, here's a simple way to build a homology cobordism from a homology sphere $N$ to a new homology sphere $N'$.  (This recipe basically works for more general 3-manifolds, but this is the simplest version to state.)  Attach a four-dimensional 1-handle to $N$.  This gives a cobordism from $N$ to $N \# S^2 \times S^1$.  Now attach a 2-handle along a knot $K$ which generates $H_1$; any framing is fine.  Call this cobordism $Q$.  It is straightforward to check that this is a ribbon homology cobordism.  Further, $$\pi_1(Q) = (\pi_1(N) * \mathbb{Z})/\langle \langle K \rangle \rangle$$ If we quotient by $\pi_1(N)$, then this kills the entire group, since $K$ is homologous to the generator of $\mathbb{Z}$.  Since there's no 3-handles, this is a ribbon homology cobordism, and hence is injective on Floer homology.  A similar thing also applies if we attach $k$ 1-handles and $k$ 2-handles, but we have to be a bit more careful with $\pi_1$ (basically, forgetting the letters of $\pi_1(N)$, the attaching curves have to generate the corresponding free group on $k$ elements).  The difficult part is carrying this out in a way that there's no symmetry of the resulting three-manifold.  (If you've never seen these arguments of building manifolds lacking certain symmetries, a nice example is this paper.  Often, the key to such an argument is analysis of the pieces of JSJ decomposition and possibly bringing in software like Snappy to deal with the hyperbolic pieces under the right circumstances.)  Rather than go through this in significant detail, I'll simply include the picture of the contractible four-manifold that Akbulut-Ruberman use. 

The black and rightmost blue circle describe the original contractible four-manifold $W$.  The homology cobordism $M$ is given by the two blue circles on the left (the new 1-handles) and the green and red curves (the two new 2-handles).  

Ok, now it's your turn to go out and find some exotica!  Note that even though we have exotic contractible four-manifolds (compact and non-compact), we still don't have any exotic four-spheres (or even exotic rational homology four-spheres)!