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 Answer 1


Two places in which this is discussed are Gompf and Stipsicz's book 4-manifolds and Kirby calculus (Sections 5.2 and 5.3) and Ozbagci and Stipsicz's Surgery on contact 3-manifolds and Stein surfaces (Chapter 2). The reason this is not discussed in many knot theory references is that this is more of a 3.5-dimensional issue (that is, in between dimension 3 and dimension 4) than a knot-theoretical one.

  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 $M$.

  • $\begingroup$ Since $q=1$, why we are free for the choice of longitude? $\endgroup$ Jan 25, 2022 at 16:37
  • $\begingroup$ 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. $\endgroup$ Jan 25, 2022 at 17:14

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