In this exercise we describe an algorithm for computing a presentation of the fundamental group of the complement of a smooth or piecewise linear knot \(K\) in \(\mathbb{R}^3\), called the Wirtinger presentation. To begin, we position the knot to lie almost flat on a table, so that \(K\) consists of finitely many disjoint arcs \(\alpha_i\) where it intersects the table top together with finitely many disjoint arcs \(\beta_\ell\) where \(K\) crosses over itself. We build a 2 dimensional complex \(X\) that is a deformation retract of \(\mathbb{R}^3 - K\) by the following three steps. First, start with the rectangle \(T\) formed by the table top. Next, just above each arc \(\alpha_i\) place a long, thin rectangular strip \(R_i\), curved to run parallel to \(\alpha_i\) along the full length of \(\alpha_i\) and arched so that the two long edges of \(R_i\) are identified with points of \(T\). Any arcs \(\beta_\ell\) that cross over \(\alpha_i\) are positioned to lie in \(R_i\). Finally, over each arc \(\beta_\ell\) put a square \(S_\ell\), bent downward along its four edges so that these edges are identified with points of three strips \(R_i\), \(R_j\), and \(R_k\); namely, two opposite edges of \(S_\ell\) are identified with short edges of \(R_j\) and \(R_k\) and the other two opposite edges of \(S_\ell\) are identified with two arcs crossing the interior of \(R_i\).
Assuming this bit of geometry, show that \(\pi_1(\mathbb{R}^3 - K)\) has a presentation with one generator \(x_i\) for each strip \(R_i\) and one relation of the form \(x_i x_j x_i^{-1} = x_k\) for each square \(S_\ell\), where the indices are as in the figures above. [To get the correct signs it is helpful to use an orientation of \(K\).]
Use this presentation to show that the abelianization of \(\pi_1(\mathbb{R}^3 - K)\) is \(\mathbb{Z}\).
Solution
(a).
1. The \(1\)-skeleton of \(X\) is the union of the strips \(R_i\), each of which is a rectangle with its two long edges glued to the table \(T\).
by construction, each \(R_i\) is a disk attached along two arcs to \(T\).
2. Each strip \(R_i\) contributes a generator \(x_i\) to \(\pi_1(X)\).
attaching a \(2\)-cell (the strip) along two arcs to the contractible table \(T\) creates a loop; the generator \(x_i\) is the loop that runs along \(R_i\) from one long edge to the other.
3. Each square \(S_\ell\) is a \(2\)-cell attached along a loop that runs through three strips \(R_i, R_j, R_k\).
by construction, the four edges of \(S_\ell\) are glued to \(R_j\), \(R_k\), and two arcs crossing \(R_i\).
4. The attaching loop of \(S_\ell\) is \(x_i x_j x_i^{-1} x_k^{-1}\).
traversing the boundary of \(S_\ell\): along \(R_j\) (giving \(x_j\)), across \(R_i\) (giving \(x_i\)), back along \(R_k\) (giving \(x_k^{-1}\)), and back across \(R_i\) (giving \(x_i^{-1}\)); with the orientation convention this reads \(x_i x_j x_i^{-1} = x_k\).
5. Hence \(\pi_1(\mathbb{R}^3 - K) = \langle x_1, \ldots, x_m \mid x_i x_j x_i^{-1} = x_k \text{ for each } S_\ell \rangle\).
2 and 4, by van Kampen (the \(2\)-cells impose exactly these relations).
(b).
1. Abelianizing, each relation \(x_i x_j x_i^{-1} = x_k\) becomes \(x_j = x_k\).
in an abelian group, \(x_i x_j x_i^{-1} = x_j\).
2. Hence the abelianization is generated by \(x_1, \ldots, x_m\) with relations \(x_j = x_k\) for each crossing.
1.
3. The crossing graph (vertices = strips, edges = crossings) is connected, since \(K\) is a single knot.
the knot is connected, so its diagram is connected.
4. Hence all generators \(x_i\) are identified, so the abelianization is \(\langle x \rangle \cong \ZZ\).
2 and 3.
5. Q.E.D.
5 (a) and 4 (b).