4
$\begingroup$

Let $p,q$ and $r$ be positive integers. A Brieskorn sphere is a closed oriented $3$-manifold defined by $$\Sigma(p,q,r) = \{ x^p+y^q+z^r=0 \} \cap S^5.$$

Its fundamental group is well-known due to Milnor. It is always a rational homology sphere. When $p,q$ and $r$ are further chosen pairwise coprime, then it is an integral homology sphere.

In this case, the plumbing graph of a Brieskorn sphere is well-understood, see for example Section 1 of Saveliev's book: Invariants of Homology 3-Spheres.

One needs to find unique integers $b,p',q',r'$ solving the equation \begin{equation} bpqr+p'qr+pq'r+pqr'=-1 \end{equation} where $1\leq p' \leq p-1$, $1\leq q' \leq q-1$ and $1\leq r' \leq r-1$. It is basically done by taking mod of these integers.

How about the rational case? Is it possible to find a unique representation for the plumbing graph associated to Brieskorn spheres?

Improve this question
$\endgroup$
3
  • $\begingroup$ Maybe the easy way out is to say that the (normal crossing divisor) resolution of the singularity gives you a plumbing. $\endgroup$
    – Marco Golla
    Commented Jan 29, 2020 at 0:04
  • $\begingroup$ Even if we assume that the plumbed $4$-manifold is negative definite, does it makes the problem easier? $\endgroup$
    – user150450
    Commented Jan 29, 2020 at 9:47
  • 2
    $\begingroup$ The (minimal) negative definite plumbing is unique; this is due to Neumann. (In particular, this tells you that topology determines the singularity, in a way.) $\endgroup$
    – Marco Golla
    Commented Jan 29, 2020 at 10:45

1 Answer 1

Reset to default
1
$\begingroup$

You may simply find the surgery diagram of Brieskorn spheres. This is from Özbağcı's lecture notes.

The small Seifert fibered $3$-manifold $M(r_1, r_2, r_3)$ is defined by the following rational surgery diagram.

enter image description here

Then

  1. $- \Sigma(2,3,5) \cong M \left(\frac{-1}{2}, \frac{1}{3},\frac{1}{5} \right) \cong \partial E_8 \cong S^3_1(3_1)$,

  2. $- \Sigma(2,3,4) \cong M \left(\frac{-1}{2}, \frac{1}{3},\frac{1}{4} \right) \cong \partial E_7 \cong S^3_2(3_1)$,

  3. $- \Sigma(2,3,3) \cong M \left(\frac{-1}{2}, \frac{1}{3},\frac{1}{3} \right) \cong \partial E_6 \cong S^3_3(3_1)$.

where $3_1$ denotes the right-handed trefoil, $S^3_n(K)$ denotes the $n$-surgery on the knot $K$ in $S^3$, and $E_k$ denotes the Dynkin diagram.

You may figure out (most of) these homemorphisms by using Kirby calculus.

For the plumbing graphs, you can read the lecture notes of Némethi.

Improve this answer
$\endgroup$