3
$\begingroup$

We know that a Brieskorn homology 3-spheres $\Sigma(p,q,r)$ admit a free $S^1$-action, which makes it a Seifert fibered spaces with three singular fibers: $M(b;r_1,r_2,r_3)$. How should one get from $\Sigma$ to a surgery description of $M$?

$\endgroup$
5
$\begingroup$

(I'm adding another answer, in case *cough* you have no access *cough* to the book.)

In practice, you just need to crunch some numbers: you need to find integers $s_1, s_2, s_3$ such that $r_1r_2s_3 + r_2r_3s_1 + r_3r_1s_2 = b$; the surgery presentation is just obtained by doing 0-surgery on the unknot, and $r_i/s_i$-surgery on three (pairwise unlinked) meridians. (Maybe you can make a sanity check about the signs by looking at $\Sigma(2,3,5)$.)

This is also described (but not explained) in Casson and Harer's paper Some homology lens spaces that bound rational homology balls, which seems to be freely accessible.

$\endgroup$
2
$\begingroup$

One method is outlined in Montesinos' book "Classical tessellations and three-manifolds." Chapter 4 of that book deals with Seifert fibered spaces more generally, but the specifics for this question are covered in section 4.3 Constructing the manifold from the invariants and Figure 12 of that chapter.

Of course, this is not the only way to get surgery descriptions for Brieskorn homology 3-spheres and in fact for each $\Sigma(p,q,r)$, there are infinitely many hyperbolic manifolds which can be filled to give $\Sigma(p,q,r)$ which do not arise as part of the construction mentioned above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.