Alexander Duality and Homology of Knot Complement

Poincare duality with boundary

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

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


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

are both isomorphisms.

Alexander duality

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


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

H~q(K)Hq(K)Hq(N(K))Hnq(N(K),N(K))Hnq(Sn,SnK)Hnq1(SnK)=H~nq1(SnK).\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 (Sn,SnK)(S^n,S^n-K), and for q0q\neq 0, we have Hnq(Sn)=0H_{n-q}(S^n)=0. Therefore, the case for q0q\neq 0 has been established. For q=0q=0, notice that

Hn1(SnK)ZHn(Sn,SnK)H_{n-1}(S^n-K)\oplus\mathbb{Z}\cong H_n(S^n,S^n-K)


H~0(K)ZH0(K)\tilde{H}^0(K)\oplus\mathbb{Z}\cong H^0(K)

Still using the above result: H0(K)Hn(Sn,SnK)H^0(K)\cong H_n(S^n,S^n-K), we have

H~n1(SnK)H~0(K).\tilde{H}_{n-1}(S^n-K)\cong \tilde{H}^0(K).

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

Knot complement

Using Alexander duality, the computation of homology groups of knot complement in S3S^3 is simple.

e.g. Suppose that KR3K\subset \mathbb{R}^3 is a knot. Then by Alexander duality in S3S^3,

H1(S3K)H1(K)H1(S1)ZH_1(S^3-K)\cong H^1(K)\cong H^1(S^1)\cong\mathbb{Z}


H2(S3K)H~0(K)=0.H_2(S^3-K)\cong \tilde{H}^0(K)=0.