# Zero surgery on a Seifert fiber space

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).

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

• 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. – Stephen Dedalus Mar 23 at 20:13
• 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. – Stephen Dedalus Mar 23 at 20:31
• Other the usual convention, 0-Surgery corresponds to p=1,q=0 in her paper. – ThiKu Mar 23 at 22:01
• 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. – ThiKu Mar 23 at 22:25
• 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. – Stephen Dedalus Mar 23 at 23:36