# Proving the Cork Theorem

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.

• Corcaigh abú, boy! Commented Jul 31 at 21:02
• The cork twist not completely but a little similar to surgery(Some how). is not? Commented Jul 31 at 21:13

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 $$M_0$$.