I have been learning about Brieskorn homology 3-spheres $\Sigma(a_1,...,a_n)$ and Seifert manifolds. My reference is the first few pages of Saveliev's "Invariants of Homology 3-spheres."

If I understand correctly, it is explained that one can realize the homology three spheres by considering a trivial bundle $S^2 \times S^1$ and remove small discs $D_i$ about marked points $x_i ,..., x_n$ on the two sphere. One can then perform a rational Dehn surgery with surgery slopes $a_i/b_i$ along the links given by $\partial D_i \times pt$ to obtain the Brieskorn homology spheres.

More precisely Saveliev shows that there exist $b_i$ that make this Dehn surgery construction into a rational homology sphere and then states that it follows from Kirby calculus that the resulting 3 manifolds are independent of $b_i$. I should say that, unfortunately, I don't know anything about Kirby calculus. Clearly, in this construction the $a_i$ and $b_i$ do not play symmetric roles. Nevertheless, I am interested in the following:

A general question is: To what extent can we determine the homology three spheres from the $b_i$?

A specific question/motivation: I am not at all a low dimensional topologist. However, a Brieskorn homology sphere naturally arises as a surgery of the type above in a construction I am working on. I know that $n=4$ and all $b_i= \pm 1$, or $n=3$ and $b_1, b_2= \pm 1.$ It is natural to ask: what manifold am I possibly looking at?

  • $\begingroup$ You can readily compute the homology of such a manifold and check for yourself. $\endgroup$ Jul 18, 2014 at 22:36
  • $\begingroup$ I want to assume that the resulting manifold is a homology sphere, so I don't understand the comment. I thought this was clear, but I edited the last question to make sure. I don't know whether such a homology sphere exists so that is part of the question. $\endgroup$
    – user36931
    Jul 19, 2014 at 0:43
  • $\begingroup$ Thank you to both Bruno Martelli and Sam Nead for their very helpful explanations! $\endgroup$
    – user36931
    Jul 20, 2014 at 2:12

2 Answers 2


Seifert manifolds that are homology spheres can be determined as follows. Let $$ M =(S^2, (p_1,q_1), \ldots, (p_h,q_h)) $$ denote the Seifert manifold obtained by filling the $h$ tori in $D_h \times S^1$ (where $D_h$ is the 2-sphere minus $h$ discs) with coprime parameters $(p_i, q_i)$. Since we consider Seifert manifolds, we ask $p_i \neq 0$. We may say that $M$ fibers over the orbifold $$(S^2, p_1, \ldots, p_h).$$

The Euler number of $M$ is an important invariant, defined as: $$ e = \sum_{i=1}^h \frac {q_i}{p_i}.$$ A simple homology calculation shows that the integral homology $H_1(M)$ is finite if and only if $e\neq 0$, and in that case we have $$|H_1(M)| = ep_1\cdots p_h = \sum_{i=1}^h q_ip_1\cdots \widehat{p_i}\cdots p_h.$$ This implies that $M$ is an integral homology sphere if and only if $$\sum_{i=1}^h q_ip_1\cdots \widehat{p_i}\cdots p_h = \pm 1.$$ This in turn easily implies the following:

For every set $p_1, \ldots, p_h\geqslant 2$, of $h \geqslant 3$ pairwise coprime integers there is a unique homology sphere $\Sigma(p_1,\ldots,p_h)$, which fibers over the orbifold $(S^2,p_1, \ldots, p_h)$. Every Seifert homology sphere distinct from $S^3$ arises in this way.

Proof: it is easy to see that the $p_i$'s must be coprime. On the other hand, if they are, one can find easily $q_1,\ldots, q_h$ that give $|H_1(M)|=1$. Two different strings of $q_1,\ldots, q_h$ are related by ``moves'' that do not affect $M$. (The same $M$ can be described by different parameters.)

This theorem is probably written somewhere, like in Orlik's book or Seifert's original paper, but I can't remember.

  • 1
    $\begingroup$ @user36931 - the theorem that Bruno states can be found as Theorem 4.1 of "Seifert manifolds, plumbing, \mu-invariant and orientation reversing maps" by Neumann and Raymond. $\endgroup$
    – Sam Nead
    Jul 19, 2014 at 19:22

If you want to do lots of examples, to get a feel for how homology depends on the $a_i/b_i$, install Regina and use the following menu sequence.

Packet Tree > New 3D Triangulation > Type of triangulation > Seifert fibered space

$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\cross}{\times}\newcommand{\bdy}{\partial}$You should also read the Wikipedia page on Seifert fibered spaces. To now answer your questions:

1) The $b_i$ have a great deal to do with the topology of the Seifert fibered space - they are part of the Seifert invariants and they also determine the Euler class. This measures how much the circle fibers are "twisting". For example, if the Euler class is zero then the SFS has product geometry.

2) The manifold you are looking at is determined by its Seifert invariants, Euler class, and various choices of orientability. All of this can be deduced from the information you have.

Here is a general discussion, written at a lower level, which may be helpful to you.

If you take $S^2 \cross S^1$ and remove four vertical disjoint solid tori $D_i \cross S^1$ then the resulting manifold $M$ has integral homology $H_1(M) = \ZZ^4 = \langle x, y, z, t \rangle$. Here $t$ goes around the fiber and $x, y, z$ lie in the planar surface $P = S^2 - \cup D_i$. Each of $x, y, z$ goes around one of the first three components of $\bdy P$, so the last component of $\bdy P$ lies in the class $-(x + y+ z)$.

Note that $\bdy M = \bdy P \cross S^1$ is a union of four tori, $T_i$. So, for example, the first homology of $T_1$ is generated by $x$ and $t$. If we Dehn fill $T_1$ along the slope $a_1/b_1$ that is the same as killing the element $a_1x + b_1t$ in $H_1(T_1)$. Thus the new homology is $\ZZ^4 / \langle a_1x + b_1t \rangle$. If you do four fillings, then you are killing four elements of $\ZZ^4$. It is now a matter of linear algebra to decide if a four-tuple yields a homology sphere.

Computing a determinant shows that the filling $((a,1),(b,1),(c,1),(d,1))$ yields an integral homology sphere if and only if $abc + abd + acd + bcd = \pm 1$. For example $(a,b,c,d) = (2,3,5,-1)$ or $(2,3,7,-1)$ give homology spheres.

Brute force computer search finds more examples: $(-19, 21, 23, -26)$, $(3, -4, -23, -25)$, $(5, -9, -22, -23)$, $(-5, 6, 13, -23)$, $(-3, 4, 7, -17)$, $(2, -3, -11, -13)$, $(-8, 9, 11, -13)$, $(3, -7, -10, -11)$.

  • $\begingroup$ Thank you @Sam Nead. I have two questions 1) the two manifolds you find are not homeomorphic? 2) Do you think it is hopeless to give a simple description up to homeomorphism of all manifolds in the class that I am looking at ? $\endgroup$
    – user36931
    Jul 19, 2014 at 9:33
  • $\begingroup$ 1) They are not homeomorphic - they have different Seifert invariants. 2) I think that you are asking for an explicit enumeration of all of these manifolds? I think doing that is equivalent to listing all solutions to the quadratic form I gave. (Well, there is an issue of sign, as well.) $\endgroup$
    – Sam Nead
    Jul 19, 2014 at 9:52
  • $\begingroup$ To be precise - there is more than one form. The forms are $\epsilon_d abc + \epsilon_a bcd + \epsilon_b cda + \epsilon_c dab$ and we are asked to find all ways to represent $\pm1$. The signs $\epsilon_w$ are the same as the $b_i$. $\endgroup$
    – Sam Nead
    Jul 19, 2014 at 9:58
  • $\begingroup$ @ Sam Nead thank you for your answer. I think it gets to the heart of what is confusing me. Your answers seem to indicate that the Seifert invariants capture the homology sphere uniquely. However, from what I understand in Saveliev, there are relations e.g. $ M(e,(a_i,b_i)) = M(e',(a_i,b'_i))$ iff $b_i=b'_i.$ This is the part of the story I can't really absorb. Are you saying there is no such relation if instead of fixing $a_i$, we fix $b_i$? What is a reference for this? Sorry for asking so many questions. $\endgroup$
    – user36931
    Jul 19, 2014 at 14:05
  • $\begingroup$ That should read $b_i=b'_i$ (mod $a_i$). $\endgroup$
    – user36931
    Jul 19, 2014 at 14:10

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