# Integral surgeries on $3$-manifolds

Let $$K$$ be a knot in $$S^3$$. Let $$N(K)$$ be a tubular neighborhood of $$K$$, a solid torus. On $$\partial N(K)$$, we may specify a preferred longitude $$\lambda$$, i.e., a simple closed curve whose linking number with $$K$$ is $$0$$. Also, we can choose a canonical meridian $$\mu$$ whose linking number with $$K$$ is $$1$$.

A $$p/q$$-surgery on $$S^3$$ along $$K$$ is a closed oriented $$3$$-manifold given by $$S^3_{p/q} (K) = (S^3 - \mathrm{int}(N(K))) \cup_\varphi (D^2 \times S^1)$$ where $$\varphi: S^1 \times S^1 \to S^1 \times S^1$$ is a homeomorphism that sends $$\partial D^2 \times \{ 0 \}$$ to $$p \mu + q \ell$$.

I want to understand the generalization of this process to an arbitrary closed oriented $$3$$-manifold $$M$$.

1. The concept of integral surgery makes sense for any $$M$$? If yes, how?

2. Here, the choice of $$\lambda$$ is not obvious. How about the rational surgeries?

P.S. I checked several reference books about the knot theory. I couldn't find a precise approach for this generalization. Any reading advise will be appreciated.

1. Yes, this makes sense in arbitrary 3-manifolds, since the fact that $$q=1$$ is independent of your choice of longitude. The meridian is always well-defined, and any two longitudes differ by a multiple of the meridian, so having an expression of the form $$p\mu + \lambda$$ is independent of which $$\lambda$$ you choose. (The $$p$$ will vary, of course.) A different, arguably "better" perspective is that this corresponds to attaching a 4-dimensional 2-handle to $$M \times I$$ along $$K$$, which is what the two references I gave you are mostly about.
2. I'm not sure what you mean by "how about", so I'll at least tell you that the construction you have for knots in $$S^3$$ is completely general, and it goes by the name of "Dehn surgery". You can do it for any knot in any 3-manifold, and the datum you need is just the choice of (the homology class of) a simple closed curve in $$\partial N(K)$$, also called the slope (which is your $$(p,q)$$ or $$p/q$$). The only drawback is that if $$K$$ and $$M$$ are arbitrary you don't have a canonical way to translate this slope into a rational number, unless $$K$$ is null-homologous in $$M$$ (see below).
About 2., let me add that the choice of $$\lambda$$ is not only non-obvious, but also not possible in general. The only instance in which there is a canonical choice of $$\lambda$$ is when the knot $$K$$ is null-homologous in $$M$$, i.e. $$[K] \in H_1(M;\mathbb{Z})$$ vanishes. Then you have a preferred longitude (still called the Seifert longitude), which is the only curve on $$\partial N(K)$$ (up to isotopy) whose homology class dies in $$H_1(M\setminus K; \mathbb{Z})$$.
In general, whenever you have a knot $$K \subset M$$, you always have a rank-1 subgroup of $$H_1(\partial N(K); \mathbb{Z})$$ which dies in $$H_1(M \setminus K; \mathbb{Z})$$. This group needs not be generated by a primitive element, so there might be no simple closed curve on $$\partial N(K)$$ that bounds a surface in $$M \setminus {\rm int}(N(K))$$. It will be generated by a longitude if and only if $$K$$ is null-homologous in $$M$$.
• Since $q=1$, why we are free for the choice of longitude? Jan 25, 2022 at 16:37
• In fact, once you fix $q$ (whether it's 1 or 42 or 404), it's going to be the same $q$ for each longitude. This $q$ just measures how many times the slope you choose meets the meridian, so it is independent of the basis. Jan 25, 2022 at 17:14