Using Kirby calculus, Akbulut and Kirby first analyzed that the Brieskorn sphere $\Sigma(2,3,13)$ is diffeomorphic to the following link in their famous paper:

enter image description here

Then they switched the circles when they are symmetric. Next step is zero/dot change, which gives the desired Mazur type contractible 4-manifold.

What is this symmetry exactly? Does it preserve the isotopy type of the link? If yes, is this always true?

  • $\begingroup$ You probably mean to say the Mazur manifold has a presentation as surgery on the framed link in your diagram? I have not read the paper but the symmetry you are refering to probably is a reference to the link being Brunnian, i.e. you can simultaneously turn the 0-framed component into a circle, which knots the -1-framed component. $\endgroup$ Commented Dec 19, 2022 at 4:13
  • $\begingroup$ Yes, the picture describes the surgery diagram of the boundary of the Mazur manifold up to diffeomorphism. If we are using the symmetry of a Brunnian link, then our observation is independent of framings, right? The only required condition is that they are all unknots. $\endgroup$ Commented Dec 19, 2022 at 5:22
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    $\begingroup$ Also related mathoverflow.net/questions/372003/akbuluts-cork-involution $\endgroup$ Commented Dec 19, 2022 at 10:48

1 Answer 1


I don't know where you took the picture from, but that link is not symmetric (and it likely does not describe the correct manifold). Indeed, the 0-framed component of the link you drew is a trefoil: its diagram has three crossings and it is alternating. If you look at Figure 2 in Akbulut and Kirby's paper, the top crossing is the opposite than the one you drew here. EDIT: the diagram does represent the unknot... It's true that it's alternating, but it has a trivial Reidemeister-1 move available. Thanks Ryan for pointing this out. The rest of the answer is (as far as I can tell) still correct.

Once you switch those two crossings, Figure 2 in Kyle Hayden's paper displays the symmetry Akbulut and Kirby refer to: the link is isotopic to a link that has a symmetric diagram, so the two components can be swapped. Note that it's not symmetric as a framed link. That is to say: this is not an instance of a cork, at least not in an obvious way (if you wanted an obvious cork, just put framing 0 on both components and you get what in the literature is called the Mazur cork).

As for the question

is it always true?

I don't know what you mean exactly when you say "always". There is no reason why a link with two unknotted components should have an isotopy swapping the two components. The multi-variable Alexander polynomial, for instance, should be able to tell you that sometimes this cannot happen. It looks to me like the link L9n19 should give such an example.

  • $\begingroup$ Thanks! I used the original picture in their paper, it is the last diagram in Figure 23. To pass to the diagram appeared in Figure 2 for $\Sigma(2,3,13)$, they probably applied diffeomorphisms described in Proposition 2. $\endgroup$ Commented Dec 19, 2022 at 8:48
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    $\begingroup$ The 0-framed component is the unknot. $\endgroup$ Commented Dec 19, 2022 at 18:44
  • $\begingroup$ @RyanBudney: oops... this is a bit embarrassing :) I'll correct my answer. $\endgroup$ Commented Dec 19, 2022 at 20:17
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    $\begingroup$ No worries, mistakes happen. $\endgroup$ Commented Dec 20, 2022 at 2:21
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    $\begingroup$ @MarcoGolla The top crossing always makes me confused; this picture does not fit into the description in Figure 2 as you said. A similar but slightly different proof can be seen on pg. 27 in Akbulut's book "4-manifolds". $\endgroup$ Commented Dec 22, 2022 at 21:23

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