Proving the Cork Theorem

Question

I am reading Kirby's paper paper "Akbulut's corks and h-cobordisms of smooth simply connected 4-manifolds" and I have a question about how to actually prove the cork theorem from the results in the paper. Specifically, I want to show that any pair $M_0, M_1$ of homeomorphic but not diffeomorphic closed simply connected $4$-manifolds is related by a cork twist. A cork is a pair $(C,\tau)$ of a contractible $4$-manifold $C$ together with a diffeomorphism $\tau$ of its boundary (sometimes taken to be an involution) which does not extend as a diffeomorphism over all of $C$. The cork theorem then says that for any such $M_0, M_1$ we can find $C \subseteq M_0$ s.t $M_1 \underset{\text{sm}}{\cong} M_0 \setminus \mathrm{int}(C) \cup_\tau C$ (cutting out $C$ and gluing it back in along $\tau$ is the cork twist).

From Kirby's paper we know that any $h$-cobordism between $M_0$ and $M_1$ splits into a product $h$-cobordism and a contractible sub-$h$-cobordism $A$. We also know that the ends of $A$ can be taken to be diffeomorphic via an involution, and that $A$ itself is diffeomorphic to $B^5$.

One idea, then, is the following: $$M_1 = (M_1 \setminus \mathrm{int} A_1) \cup_\partial A_1 \underset{\text{sm}}{\cong} (M_0 \setminus \mathrm{int} A_0) \cup_\partial A_1 \underset{\text{sm}}{\cong} (M_0 \setminus \mathrm{int} A_0) \cup_\tau A_0$$

where the first diffeomorphism replaces $M_1 \setminus \mathrm{int} A_1$ with $M_0 \setminus \mathrm{int} A_0$ using the product $h$-cobordism, and the second diffeomorphism replaces $A_1$ with $A_0$ using the diffeomorphism between the two ends of $A$ that restricts to an involution.

The trouble is that the diffeomorphism restricted to the boundary (which we label $\tau$) clearly extends over the contractible itself, so it cannot be our cork diffeomorphism. But I don't see any other way to get an involution.

Answer 1

The proof starts with an $h$-cobordism $W:M_0\to M_1$ between two manifolds $M_0, M_1$ which are not diffeomorphic and produces a cork embedded in $M_0$ such that the cork twist yields $M_1$.

I'll borrow your notation. Using the flow of a Morse function adapted to the decomposition of the $h$-cobordism we obtain a diffeomorphism $$\hat \varphi: M_1 \setminus int(A_1)\to M_0\setminus int(A_0) $$

Denote by $\varphi = \hat \varphi|_\partial: \partial A_1\to \partial A_0$.

As you mentioned we have also a diffeomorphism $\hat \tau:A_1 \to A_0$ (which is constructed using an idea which goes back to Matveyev (R. Matveyev. A decomposition of smooth simply-connected h-cobordant 4- manifolds. Journal of Differential Geometry, 44:571–582, 1995), the point is that we can suppose that $\tau := \hat \tau|_{\partial A_1}\circ \varphi^{-1}:\partial A_0 \to \partial A_1$ is an involution.

Now it follows that we have a diffeomorphism: $$M_1 = M_1 \setminus int(A_1) \bigcup_{id} A_1 \simeq M_0\setminus int(A_0) \bigcup_{\varphi} A_1 \simeq M_0\setminus int(A_0) \bigcup_{\tau} A_0$$ where the first isomorphism comes from $\hat \varphi$ and the second follows from the diffeomorphism $\hat \tau$.

This exhibits $M_1$ as the result of a cork twist using $(A_0, \tau)$, the fact that $\tau$ does not extend to $Diff^+(A_0)$ follows from the fact that $M_1$ is not diffeomorphic to $M_0$.