A recent paper proves the existence of homeomorphic but not diffeomorphic Mazur manifolds (see also examples of exotic pairs of contractible Stein manifolds).

Let's call them $M_1$ and $M_2$. If we glue $W= M_1 \cup_{\partial M_1=\partial M_2} M_2$, then we get a manifold homeomorphic to $S^4$ by Freedman's theorem. Doubling $M_1$ or $M_2$ yields the standard smooth $S^4$ by a theorem of Mazur. But I'm wondering if $W$ is diffeomorphic to $S^4$?

Actually, I don't know if twisting the double of an Akbulut cork can yield an exotic $S^4$? Any exotic pair of manifolds are related by twisting along a cork, so I guess I'm asking whether anything more is known when the complementary contractible manifolds are homeomorphic?

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    $\begingroup$ It can be worth saying that the double is guaranteed to be S^4 only if the Mazur manifold M you start with is made of 0-, 1-, and 2-handles along an Andrews-Curtis trivial presentation. This is not guaranteed for any contractible M, not even if made only of 0-, 1-, and 2-handles. $\endgroup$ Apr 14, 2020 at 14:25
  • $\begingroup$ @BrunoMartelli I see, yes I just had in mind the classic sort of example with one 0,1 & 2-handle like the examples in the paper. $\endgroup$
    – Ian Agol
    Apr 14, 2020 at 15:17

1 Answer 1


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It is diffeomorphic to $S^4$. I have drawn the Kirby picture, please let me know if it is not clear.

(Also I'm sorry that I don't know how to draw Kirby picture in computer so I do old fashioned drawing on notebook)

The Key ideas of the proof are,

1)When we upside down a compact 4 manifold $M$ with 0,1 and 2 handle, then 0 handle becomes 4 handle and 1 handle become 3 handle. Now when we upside down the 2 handle, the co-core become the new attaching circle, which is an unknotted meridian to the orginial attaching cirle of the 2 handle. And the new attaching framing is trivial,i.e, 0. [This is what we use to draw the kirby picture of $M_1\cup M_2$, using the fact that they have identical boundary]

2) 0 framed meridian of a knot always helps to resolve all the crossings of a knot (handle slides).[ This is what we use to simplify the Kirby picture]

3) If a 1 handle and a 2 handle (in the picture a dotted 1 handle and the attaching circle of the 2 handle) geometrically intersect at one point, then they cancel each other. Similarly a 0 framed unknotted 2 handle get canceled by a 3 handle. [this is what we used in last two steps to cancel all but only one 0 handle and one 4 handle and thus we get $S^4$].

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    $\begingroup$ Okay, thanks, I think I get the idea. $\endgroup$
    – Ian Agol
    Apr 12, 2020 at 19:05

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