# Homology groups of the complement of a knot in $S^3$

Let $$f: S^1 \times D^2 \to S^3$$ be an embedding, and let $$g=f|_{S^1 \times 0}$$ denote the induced embedding of a circle, and let $$K=g(S^1)$$. I want to compute the homology groups of $$A:=S^3-K$$ using the Mayer-Vietoris sequence.

Since $$A$$ is path-connected (by, for example, the Jordan curve theorem), $$H_0(A)=\Bbb Z$$.

Let $$B=f(S^1 \times D^2)$$. Since $$f$$ is an embedding, $$B$$ is homeomorphic to a solid torus, hence homotopy equivalent to a circle, so we know the homology groups of $$B$$. Also, we have $$A \cup B =S^3$$. Now we consider the Mayer-Vietoris sequence for the decomposition $$S^3 = A \cup B$$ (Actually, we have to consider an open neighborhoods of $$B$$ that deformation retracts onto $$B$$, instead of $$B$$). Note that $$A \cap B$$ is homotopy equivalent to a torus. Then it is easily shown that $$H_k (A)=0$$ for all $$k >3$$. So we only need to consider $$k=1,2,3$$.

For $$k=1$$, it is quite easy. We have:

$$H_2(S^3)=0 \to H_1(T^2)= \Bbb Z^2 \to H_1(A) \oplus H_1(S^1)=H_1(A) \oplus \Bbb Z \to H_1(S^3)=0.$$

Thus, $$H_1(A)=\Bbb Z$$.

For $$k=2,3$$ we need to consider the following:

$$H_3(T^2)=0 \to H_3(A) \oplus H_3(S^1)=H_3(A) \to H_3(S^3)=\Bbb Z \to H_2(T^2)=\Bbb Z \to H_2(A)\oplus H_2(S^1)=H_2(A) \to H_2(S^3)=0$$.

If we can examine the map $$H_3(S^3) \to H_2(T^2)$$, we will be done, but I have no idea here. How do I have to proceed?

On the other hand, is there another method, other than the Mayer-Vietoris, to compute the homology groups of $$A$$?

• Alexander duality? Dec 14, 2019 at 17:40
• @LordSharktheUnknown I don't know about it Dec 14, 2019 at 17:40
• en.wikipedia.org/wiki/Alexander_duality Dec 14, 2019 at 17:45
• @LordSharktheUnknown Wow That makes the question trivial Dec 14, 2019 at 17:49
• Also see Dold, Albrecht. "A Simple Proof of the Jordan-Alexander Complement Theorem." The American Mathematical Monthly 100, no. 9 (1993): 856-57. doi:10.2307/2324661. Dec 14, 2019 at 17:51

While it is indeed very quick to do this with Alexander duality, you can still do this with the Mayer-Vietoris sequence, and with a tiny bit of Poincare duality.

As you say, the problem is quickly finished if one can evaluate the connecting homomorphism
$$\underbrace{H_3(S^3)}_{\mathbb Z} \xrightarrow{\delta} \underbrace{H_2(T^2)}_{\mathbb Z}$$ This map $$\delta$$ is an isomorphism (from which it follows quickly that $$H_2(A) \approx 0$$).

To deduce this, you need to use the definition of $$\delta$$ in the setup of the Mayer-Vietoris sequence, and you need to know how the generators of $$H_3(S^3)$$ and $$H_2(T^2)$$ are defined in the early parts of the proof of Poincare duality.

Triangulate $$S^3$$ so that $$A$$, $$B$$ and $$T^2 = A \cap B$$ are subcomplexes.

Poincare duality tells you that the generator of $$H_3(S^3)$$ is a homology class $$[c]$$ represented by a 3-cycle $$c$$ that assigns constant coefficient $$+1$$ to each 3-simplex of $$S^3$$ (assuming that those 3-simplices are assigned orientations compatible with a global orientation of $$S^3$$).

To define $$\delta[c] \in H_2(T^2)$$, here's what you do. Restrict $$c$$ to either $$A$$ or to $$B$$ (it doesn't matter which, up to sign), let's say to $$A$$, and denote the restricted 3-cycle as $$c_A$$. Its boundary $$\partial c_A$$ is a 2-cycle supported on $$T^2$$. By definition $$\delta[c] = [\partial c_A]$$ From the construction it's pretty clear that $$\partial c_A$$ assigns constant coefficient $$+1$$ (or $$-1$$) to each 2-simplex on $$T^2$$. Hence, again by Poincare duality, $$[\partial c_A]$$ is a generator of $$H_2(T^2) \approx \mathbb Z$$.

• Thanks for the correction, it's nice to have old stupid errors fixed! May 16, 2021 at 14:34

As another method that uses only Mayer-Vietoris, we can use different subspaces.

Split $$K$$ into a pair of arcs $$a, b \subset S^3$$ that meet at a pair of points $$p_1,p_2$$. Then the open subspaces $$A = S^3 \setminus a$$ and $$B = S^3 \setminus b$$ cover the space $$X = S^3 \setminus \{p_1,p_2\}$$. In this case, the space you are interested in is $$S^3 \setminus K = A \cap B$$. Both $$A$$ and $$B$$ are contractible, and $$X$$ is a homotopy $$2$$-sphere, so the Mayer-Vietoris sequence gives

$$\widetilde H_{n+1}(S^2) \cong \widetilde H_n(S^3 \setminus K)$$

In other words, $$S^3 \setminus K$$ is a homology $$1$$-sphere.