Calculating homology groups of a generic knot

Question

I am trying to calculate the homology groups of a knot embedded in $S^3$. This is what I have so far: $$H_0(S^3-K)=\mathbb{Z}$$ since $S^3-K$ is path connected; from Alexander duality I have that: $$H_n(S^3-K)=0 \qquad n\geq3$$ Again from Alexander duality I think it should be $$H_2(S^3-K)\simeq H^0(K)=\mathbb{Z}$$ I need to find $H_1(S^3-K)$, but I do not know how. I was thinking to use Mayer-Vetoris method as follows: $$0=H_1(S^3)\rightarrow H_1(S^3,S^3-K)\rightarrow H_1(S^3-K)\rightarrow H_1(S^3)=0$$ but I do not know how to find $H_1(S^3,S^3-K)$

EDIT I just thought I have to use Hurevicz theorem, so that $H_1(S^3-K)\simeq \pi_1(S^3-K)/[\pi_1,\pi_1]$

you don't know $H^1(K)$? Remember that, because $K$ is an embedded knot, $K$ is homeomorphic to $\Bbb S^1$ — Aitor Iribar Lopez, Mar 24 '21 at 10:49
I wouldn't do this by Hurewicz since it's hard to get a handle on the fundamental group. Try $H^1(S^1)=\mathrm{Hom}(H_1(S^1),\mathbb Z)=\mathbb Z$ by universal coefficients, or even Poincare duality. — Andres Mejia, Mar 24 '21 at 22:04
How do you know that $S^3-K$ is path-connected? This is true but not at all obvious. — Eric Wofsey, Mar 27 '21 at 00:56

Paul Frost · Accepted Answer · 2021-03-27T00:18:52.137

I am surprised that you know Alexander duality but not $H^1(S^1)$. However, you are right that Alexander duality shows $$\tilde H_i(S^3 \setminus K) \approx \tilde H^{2-i}(K) \approx \tilde H^{2-i}(S^1). $$ Here $\tilde H_j$ and $\tilde H^j$ denote the reduced homology and cohomology groups. Except for $j = 0$ they agree with the unreduced groups. Moreover, if the reduced group in dimension $0$ is $0$, the unreduced group in dimension $0$ is $\mathbb Z$.

You correctly conclude that $H_i(S^3 \setminus K) = 0$ for $i \ge 3$.

The (reduced) cohomology groups of $S^1$ can be computed by exactly the same methods as the homology groups of $S^1$ which gives $\tilde H^1(S^1) = \mathbb Z$ and $\tilde H^i(S^1) = 0$ for $i \ne 1$. Let us come back to this point later. Anyway, we get $$\tilde H_0(S^3 \setminus K) \approx H^2(S^1) = 0, $$ $$H_1(S^3 \setminus K) \approx H^1(S^1) = \mathbb Z, $$ $$H_2(S^3 \setminus K) \approx \tilde H^0(S^1) = 0 . $$ Concerning the cohomology of $S^1$: We can either compute it as indicated above or use Alexander duality. In fact, consider $S^1$ embedded as the set $S = \{(x_1,x_2,0,0) \mid (x_1,x_2) \in S^1\}$. It is well-known that in this special case $S^3 \setminus S \simeq S^1$. Thus you know the homology of $S^3 \setminus S$ and via Alexander duality the cohomology of $S = S^1$.

score 0 · Answer 2 · answered Apr 10 '21 at 18:34

Classical knots in 3-space have no interesting homology. Instead, Alexander looked at the homology of their 2-fold covering spaces (an easy to see invariant that distinguishes a lot of knots). Reidemeister showed, shortly thereafter, that linking numbers in non-cyclic coverings (just a bit harder to see) fill in all the known gaps (as revealed now through 19 crossings). -- Ken Perko, lbrtpl@gmail.com

Calculating homology groups of a generic knot

2 Answers2