# Fundamental group, homology of the complement of the “figure eight” knot in $S^3$?

The "figure eight" knot is the embedded circle in the three-sphere. What is the fundamental group and the homology of the complement of the knot in $S^3$?

• – Arthur Oct 15 '16 at 17:49
• Look up the Wirtinger Presentation. – N. Owad Oct 16 '16 at 4:10

First, the figure 8 knot is also called $4_1$ (the first knot with 4 crossings). On page 56 and 57 he describes the algorithm for writing down the Wirtinger Presentation of a knot diagram. And on 58 he gives the fundamental group of Euclidean space minus the knot, $\pi_1(\mathbb{R}^3-4_1)$, which is the same as $\pi_1(S^3-4_1)$, which answers your first question.
Then for the homology: It is well known the $H_1$ is the abelianization of $\pi_1$. So you will notice that when you let everything in $\pi_1(S^3-4_1)$ commute, you get just $\mathbb{Z}$. This is true for every knot.
Again, I will redirect you to Rolfsen, page 50 for the other homology groups. It happen $H_*(S^3-K)$ are are all isomorphic to $H_*(S^3-S^1)\cong H_*(S^2)$. Intuitively, you should think about homology as not being a fine enough invariant to capture knotting. It only sees an embedded circle as the standardly embedded unknot.