$V = V(k)$ denotes a tubular neighbourhood of the knot $k$ and $C = S^{3}− V$ is called the complement of the knot. $H_{j}$ will denote the (singular) homology with coefficients in $\mathbb{Z}$
Choose the homology class of $k$ as a generator of $H_1(V)$ and represent it by a simple closed curve l on ∂V which is homologous to 0 in $H_1(C)$. These conditions determine the homology class of l in $∂V$ ; hence, l is unique up to isotopy on $∂V$ . A generator of $H_1(C)$ can be represented by a curve m on $∂V$ that is homologous to 0 in $V$ . The curves l and m determine a system of generators of $H_{1}(∂V ) $= $\mathbb{Z} ⊕ \mathbb{Z}$. By a well-known result, we may assume that m is simple and intersects l in one point. As m is homologous to 0 in $V$ it is nullhomotopic in $V$ , bounds a disk, and is a meridian of the solid torus $V $. These properties determine m up to an isotopy of $∂V$ . A consequence is that l and k bound an annulus $A ⊂ V $.
With respect to the complement $C$ of a knot the longitude l and the meridian m have quite different properties: The longitude l is determined up to isotopy and orientation by $C$ ; this follows from the fact that l is a simple closed curve on $∂C$ which is not homologous to 0 on $∂C$ but homologous to 0 in $C$. The meridian m is a simple closed curve on $∂C$ that intersects l in one point; hence, l and m represent generators of $H_{1}(C) $= $\mathbb{Z}^{2}$. The meridian is not determined by $C$ because simple closed curves on $∂C$ which are homologous to $m^{±1}l^{r}$, r ∈ $\mathbb{Z}$, have the same properties.

My Question is precisely the last statement, why the meridian isn't determined by $C$?


1 Answer 1


Yes, the homology groups $H_1(\partial V)$ and $H_1(C)$ along with the map $H_1(\partial V)\to H_1(C)$ only characterize the meridian slope up to some number of longitudes. There is a general principle called "half lives half dies," which is that for a compact oriented $3$-manifold $M$, the kernel of $H_1(\partial M;\mathbb{Q})\to H_1(M;\mathbb{Q})$ is half-dimensional. In the case of a torus boundary, like a knot complement, then the kernel is one-dimensional so is generated by a well-defined simple closed curve -- the longitude. There is ambiguity, however, in the meridian.

The second inclusion $H_1(\partial V) \to H_1(V)$ is what pins down the meridian, again by the generator of the kernel.

The meridian problem is this: given $C$, how do we glue in $V$ to yield $S^3$? The core of $V$ then corresponds to a knot, and the meridian of $V$ is the meridian. A priori, we also can't be sure that there aren't multiple ways to glue in $V$ that yield different knots, though.

The data of gluing in $V$ ends up, up to homeomorphism, only depending on where a meridian of $V$ is glued into the boundary of $C$ (a procedure called Dehn filling). This cannot be solved purely homologically, in the sense that for each possible meridian $m^{\pm 1}l^r$, if you glue in $V$ according to this loop you get a space whose homology is the same as that of $S^3$. That is to say, each of them gives you an integer homology $S^3$, often denoted as a $\mathbb{Z}HS^3$. For example, you can obtain the Poincaré homology sphere by using $m^{\pm 1}l^{\pm 1}$ (sorry, I forget which!) on a trefoil knot.

That's not to say the problem is not solvable. Gordon and Luecke proved that there is precisely one boundary slope whose Dehn filling yields $S^3$. Thus, for knots the meridian is completely determined by $C$. It is a very delicate argument, and my understanding is that through some careful combinatorial analysis, one shows that if there were two different slopes whose Dehn fillings yield $S^3$, then one could find that $S^3$ had a Lens space connect summand, which violates the fact that $H_1(S^3)=0$.

Gordon, C. McA.; Luecke, J., Knots are determined by their complements, Bull. Am. Math. Soc., New Ser. 20, No. 1, 83-87 (1989). ZBL0672.57009.

For links of more than one component, it turns out that $C$ does not determine the meridians, and there are infinitely many links with the same complement.

  • $\begingroup$ What is meridian slope? And if we demand that meridian intersects with longitude once, then why m±1lr will satisfy the said property? And why these elements are defined upto isotopy(What I mean is, we have something as generator in Homology, Why it has to be isotopic to other generators?) $\endgroup$ Mar 23, 2021 at 6:04
  • $\begingroup$ @RiyaSharma In 3-manifolds, homologically nontrivial closed curves in boundary tori (up to homotopy) are known as "slopes." A meridian slope is a slope that comes from a meridian of a knot (i.e., the boundary of a disk cross section of a tubular neighborhood of the knot). $\endgroup$ Apr 23, 2021 at 16:46
  • $\begingroup$ @RiyaSharma The $m^{\pm 1}l^{r}$ condition can come from using algebraic intersection numbers in $H_1(\partial V)$. Given a curve $m^ql^r$, the algebraic intersection number with $l$ is $q$ and with $m$ is $r$, so if it intersects $l$ only once it must be that $q=\pm 1$. $\endgroup$ Apr 23, 2021 at 16:48
  • $\begingroup$ @RiyaSharma You might take a look at Rolfsen's book "Knots and Links" to see why simple closed curves in tori up to isotopy are the same as up to homotopy. Since the first homotopy group is abelian, it's isomorphic to the first homology group, giving the correspondence to simple closed curves up to homology. $\endgroup$ Apr 23, 2021 at 16:50

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