I have a problem with understanding what is a neighbourhood of a singular fiber in a Seifert fibered space coming from the zero surgery. For me a 3-manifold $Y$ is a SFS if it has a decomposition into a disjoint sum of circles s.t. every circle has a neighbourhood of form $V(p,q)$, where $V(p,q)$ is obtained from $D^2 \times I$ by idenitying the top and the bottom disk by a rotation of angle $\frac{2\pi q}{p}$. Circles for which the neighbourhood is equal to $V(1, 0)$ are called regular fibers and others are called singular fibers.

Let's say that I have a Seifert fiber structure on an oriented three-manifold $Y$ with a Seifert invariant given by $\{b,g; \frac{a_1}{b_1}, \ldots, \frac{a_n}{b_n}\}$.

If I perform the zero surgery on a regular fiber of $Y$, then what is the Seifert fiber structure on the resulting space?

For a $\frac{p}{q}$-surgery it seems that the invariant should be given by $\{b,g; \frac{a_1}{b_1}, \ldots, \frac{a_n}{b_n}, \frac{p}{q}\}$ as the neighbourhood of the newly created singular fiber is isomorphic to $V(p,q)$, however for the zero surgery case $V(0,1)$ doesn't make any sense to me. My geometric understanding of the situation hints me that we should adapt the convention that $V(0,1)$ means the sum of concentric circles in each slice $(D^2 \setminus \{0\}) \times \{z\} \subset D^2 \times S^1$ and the core cicle $\{0\} \times S^1$. However, no literature I've found seem to corroborate that.

Let's say that I want to understand Seifert fiber structure on $0$-surgeries on torus knots.

Is there any canonical Seifert structure on these spaces somewhere in the literature?

The way I'd create a Seifert fibration on $S^3_0(T(p,q))$ is to start with $S^3$ and fiber it with $T(p,q)$'s in the standard way. That way I obtain a Seifert structure with two singular fibers with neighbourhoods $V(p,q), V(q,p)$. Then performing the $0$-surgery on any regular fiber should give us $S^3_0(T(p,q))$ with an invariant $\{b,0; \frac{p}{q}, \frac{q}{p}, \frac{0}{1}\}$. This doesn't seem to be correct though (for example the fundamental group of this presentation would be $\mathbb{Z}_p * \mathbb{Z}_q$ - see my comments under ThiKu's answer).

Edit: as it was pointed out by Bruno Martelli in his answer, I confused here accidentally the actual zero-surgery on a knot with so called fibre-parallel surgery. The subject is discussed in depth in his excellent book "An Introduction to Geometric Topology" in chapter 10.3.13.


What you call a 0-surgery on a Seifert manifold is a move that typically does not produce a Seifert manifold. It is a move that "kills the fiber" and it produces a graph manifold, homeomorphic to a connected sum of lens spaces.

Since this may be a potential source of confusion, let me mention that a a 0-surgery on the $(p,q)$-torus knot is not a "0-surgery" in the above sense, and it produces a Seifert manifold that fibers over the orbifold $(S^2, p, q, pq)$ with Euler number zero (since $H_1$ is infinite cyclic).

  • $\begingroup$ That clears things for me a bit. How to see the Seifert structure s.t. that the third singular fiber is of order $pq$? $\endgroup$ – Stephen Dedalus Mar 24 at 22:40
  • $\begingroup$ For instance, because it is the only way to get a Seifert manifold with infinite cyclic H1 $\endgroup$ – Bruno Martelli Mar 25 at 8:31

$0$-surgery at a torus knot yields a Seifert fibered space, an explicit construction is given in Louise Moser: Elementary Surgery along a Torus Knot (PJM, 1971). Be aware that a 0-Surgery is an (1,0)-Surgery in the notation of that paper.

  • $\begingroup$ Hey, thanks for the comment. I'm aware of this classic paper. Correct me if I'm wrong, but he assumes specifically that $p > 0$ on the second page (points (2)-(3)). Indeed, take a look at the presentation of the $\pi_1$ he gives on the third page. If everything worked well for the zero-surgery, $\pi_1(S^3_0(T(r, s)))$ would be equal $\langle q_1, q_2, q_3, f : q_1^rf^s, q_2^sf^r, q_3^0f^1, q_1q_2q_3 \rangle$ (and $f$ is central). This group can easily be seen to be $Z^r * Z^s$ which to the best of my knowledge is not $\pi_1(S^3_0(T(r,s)))$. That's one of the sources my confusion stems from. $\endgroup$ – Stephen Dedalus Mar 23 at 20:13
  • $\begingroup$ I'm not sure if my convention for role of the numerator $p$ and denominator $q$ is consistent with the paper's ordering, but if it is opposite we'd rather get instead the group $\langle q_1, q_2, q_3, f : q_1^sf^r, q_2^r f^s, q_3^1 f^0, q_1q_2q_3\rangle$. Let $a,b$ be s.t. $as + br=1$ then $q_1^{as}f^{ar}=1,q_2^{br} f^{bs}=\pm q_1^{br} f^{bs}$, hence $\pm q_1^{as+br}f^{ar+bs}=1$ and thus $q_1 = \pm f^{-ar-bs}$ and since $q_2=q_1^{-1}$, the group trivializes. Hence it's still not $\pi_1(S^3_0(T(r,s)))$. It's been a long day but I can't see a mistake now. $\endgroup$ – Stephen Dedalus Mar 23 at 20:31
  • $\begingroup$ Other the usual convention, 0-Surgery corresponds to p=1,q=0 in her paper. $\endgroup$ – ThiKu Mar 23 at 22:01
  • $\begingroup$ I do not understand the computation in your last comment. $q_2^{br}f^{bs}$ equals $1$ and it also equals $q_1^{-br}f^{bs}$. If you choose $a,b$ such that $as-br=1$, then you still get $q_1=f^{-ar-bs}$ and thus $1=f^{-ars-bs^2+r}=f^{-br^2-bs^2}$. On the other hand, $q_2=f^{ar+bs}$ and then $1=f^{ar^2+brs+s}=f^{ar^2+as^2}$. Recall that $a$ and $b$ are relatively prime, so the group is the cyclic group of order $r^2+s^2$ with $f$ as a generator. It is not the trivial group. $\endgroup$ – ThiKu Mar 23 at 22:25
  • $\begingroup$ Sorry, by trivial I meant indeed finite cyclic - I ran out of available characters. In any case, this group is different than expected $\pi_1(S^3_0(T(r,s)))$ and I can't see why given that we followed her reasoning. This doesn't seem to be the correct description of a Seifert fiber structure on the zero-surgery space despite following each step from the paper. $\endgroup$ – Stephen Dedalus Mar 23 at 23:36

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