Alexander Duality and Homology of Knot Complement

Poincare duality with boundary

Theorem. Let $M$ be an $n$-dimensional compact, oriented manifold with boundary, and $[M]\in H_n(M,\partial M)$ be its natural orientation class. Then

$$H^i(M,\partial M)\overset{\cdot\cap[M]}{\longrightarrow} H_{n-i}(M)$$

and

$$H^i(M)\overset{\cdot\cap[M]}{\longrightarrow} H_{n-i}(M,\partial M)$$

are both isomorphisms.

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Alexander duality

Corollary. (Alexander duality in $S^n$) Let $K\subset S^n$ be a subcomplex of $S^n$ under some finite cell structure. Then there is a canonical isomorphism

$$\tilde{H}^q(K)\cong\tilde{H}_{n-q-1}(S^n-K)$$

Proof. Suppose that $N(K)$ is a regular neighborhood of $K$. Then $\forall q>0$, using Poincare duality with boundary:

$$\begin{aligned} \tilde{H}^q(K) &\cong H^q(K)\\ &\cong H^q(N(K))\\ &\cong H_{n-q}(N(K),\partial N(K))\\ &\cong H_{n-q}(S^n,S^n-K)\quad\\ &\cong H_{n-q-1}(S^n-K) =\tilde{H}_{n-q-1}(S^n-K). \end{aligned}$$

The last equality is because of the long exact sequence of homology induced by the pair $(S^n,S^n-K)$, and for $q\neq 0$, we have $H_{n-q}(S^n)=0$. Therefore, the case for $q\neq 0$ has been established. For $q=0$, notice that

$$H_{n-1}(S^n-K)\oplus\mathbb{Z}\cong H_n(S^n,S^n-K)$$

and

$$\tilde{H}^0(K)\oplus\mathbb{Z}\cong H^0(K)$$

Still using the above result: $H^0(K)\cong H_n(S^n,S^n-K)$, we have

$$\tilde{H}_{n-1}(S^n-K)\cong \tilde{H}^0(K).$$

This completes the case for $q=0$. $\square$

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Knot complement

Using Alexander duality, the computation of homology groups of knot complement in $S^3$ is simple.

e.g. Suppose that $K\subset \mathbb{R}^3$ is a knot. Then by Alexander duality in $S^3$,

$$H_1(S^3-K)\cong H^1(K)\cong H^1(S^1)\cong\mathbb{Z}$$

and

$$H_2(S^3-K)\cong \tilde{H}^0(K)=0.$$