enter image description here

I know that surgery on the unlink with +0 slope gives $S^2 \times S^1$ (where all the links above are embedded in $S^3$). I figured (I think) that surgery on the hopf link (with +0 on both) returns $S^3$ by performing the surgery on one of the unknots and pulling the second unknot into the new torus (the 'slam dunk' trick) - but here is my first question:

In "On Kirby's Calculus of Links", Fenn and Rourke prove that you can get from any one integral surgery presentation of a 3-manifold to any other presentation through 'Kirby Moves' - where these are performing surgery on +1 unknots (or adding +1 unknots into the link) changing the surgery coefficients and linking of the components correctly. How do I perform these kirby moves to get that the surgery on the hopf link shown above equals $S^3$?

(Actually, I think this picture solves my first question:) enter image description here

I'm still not sure what 3-manifold the final surgery corresponds to - I couldn't simplify it.

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    $\begingroup$ The 3d one is the Euler class $-2$ circle bundle over $RP^2$. See eg chapter 4 of Gompf-Stipsicz which would make good reading for you. This is more of a Mathstackexchange level question; I'd suggest going there with similar such questions. $\endgroup$ May 1, 2019 at 11:33
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    $\begingroup$ Another good resource for these questions is Rolfsen's ``Knots and Links'' especially Section 9H. (Though Gompf-Stipsicz might be the more canonical choice.) Rolfsen's book should help you with your question, because it promotes the mentality that you can undo twist regions via 1/n Dehn surgery. $\endgroup$ May 1, 2019 at 14:22

1 Answer 1


The manifold obtained from the figure has homology $Z/2Z \times Z/2Z$ so I don't know that you will be able to simplify it more. At the very least, you will need a two component link in $S^3$ to define it and arguably by most measures of complexity, the picture above will be the simplest possible description.

As for what the manifold is it is the Seifert fibered space with base orbifold $S^2(2,2,2)$. Using Hatcher's description this should be the Seifert fibered space over $S^2$ with (2,1), (2,1) and (2,-1) as its exceptional fibers. The fundamental group of this orbifold is the $Q_8=\langle a, b | aba^{-1}b, aba^{-1}b\rangle.$ In particular, it is finite and the manifold is covered by $S^3$.


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