Geometric Topology 22

The Fundamental Group of a Surface of Genus 3

Let $\mathscr{M}$ be an orientable surface of genus $3$ without boundary. It can be constructed as the quotient of a $12$-sided polygon $\mathscr{P}$, whose boundary code has the normal form:

Because all the vertices map to a unique point $x_0 \in \mathscr{M}$, each of the $12$ edges of $\mathscr{P}$ (suitably parametrized) maps to a loop in $\mathscr{M}$, and an element in $\pi_1(\mathscr{M},x_0)$.

To determine $\pi_1(\mathscr{M},x_0)$, we repeat the argument with $\mathscr{M}$ in place of $T$, but in more detail motivated by the image of $\mathscr{P}$ following, which $q$ maps onto $\mathscr{M}$. The proof divides into the following steps:

1. Take $U$ to be $\mathscr{M}$ minus the image of a small disk around the centre of the $12$-gon $\mathscr{P}$, and take $V$ to be the image of the larger disk centred at $\hat{o}$. Then $U$ can be deformed to the image of the boundary of $\mathscr{P}$, and is homotopic to a wedge of $6$ circles, whereas $U \cap V$ is an annulus homotopic to $S^1$:

   $$U \simeq \bigvee_6 S^1, \\ V \simeq \{\hat{o}\} \\ U \cap V \simeq S^1.$$

2. It follows that $\pi_1(U)$ is a free group on $6$ letters $a_1,a_2,a_3,b_1,b_2,b_3$, which we shall now (by abuse of notation) regard as path-homotopy classes (so $a_1$ really stands for $[a_1]$ etc.). On the other hand, $\pi_1(V)$ is trivial, and $\pi_1(U \cap V) \cong \mathbb{Z}$.

3. The homomorphism $i_* : \pi_1(U \cap V) \rightarrow \pi_1(U)$ maps the (clockwise) generator of $K = \pi_1(U \cap V)$ to $\mathbb{A}_3$, interpreted as a product of $12$ elements of $\pi_1(\mathscr{M},x_0)$.

4. Theorem vK2 implies that $\pi_1(\mathscr{M})$ is the quotient of the free product:

   $$\pi_1(U) {\Large*} \pi_1(V) \cong \pi_1(U)$$

   by the normal subgroup $N$ generated by $\mathbb{A}_3$. Thus:

   $$\pi_1(\mathscr{M}) \cong \pi_1(U) {\Large*}_K \pi_1(V) \\ \cong F_6 {\Large*}_K \{e\} \\ \cong F_6/N \\ \cong \langle a_1,...,b_3 | [a_1,b_1][a_2,b_2][a_3,b_3]\rangle.$$

Conclusion

Recall our notation for normal forms for the boundary codes. Each letter defines a loop in $\mathscr{M}$ (as all vertices map to the same point, call it $x_0$), and we use the same letter to denote an element of $\pi_1(\mathscr{M},x_0)$. In this way, the code becomes the relator:

If $\mathscr{M}$ is an orientable surface of genus $g$ and no boundary, then:

It is abelian if and only if $g =0$ or $g=1$, and $\mathscr{M}$ is the sphere $S = S^2$ or torus $T = T^2$.

If $\mathscr{N}$ is a non-orientable surface with $\chi = 2-h$ and no boundary, then:

It is abelian (and finite) if and only if $h=1$ and $\mathscr{N}$ is the projective plane $P$.

If $\mathscr{M}'$ is an orientable surface of genus $g$ with (for example) $r=1$ boundary component:

Knot Complements

Let $K$ be a knot in space. The aim of this final section is to describe the fundamental group $\pi_1(\mathbb{R}^3 - K)$ of the knot complement, which is a $3$-manifold.

Let $K$ be an unknot represented by $S^1 \subset \mathbb{R}^2 \subset \mathbb{R}^3$, and fix a basepoint $x_0$. An obvious element of $\pi_1(\mathbb{R}^3 - S^1,x_0)$ is $[\alpha_1]$ where $\alpha_1$ wraps once around $S^1$. It is intuitively clear that:

One can make this rigorous by first using Theorem vK1 to show that:

Then $S^3 - S_1$ can be retracted onto another circle (think Hopf link and Clifford torus). So $\pi_1(S^3 - S_1) \cong \pi_1(S^1)$.

The following result that we cite is deeper:

Theorem: If $K$ is a knot and $\pi_1(\mathbb{R}^3 - K) \cong F_1$ then $K$ is ambient isotopic to an unknot.

Arcs, Loops and Crossings

Motivated by the case of the unknot, let $K$ be a knot and 'flatten' it to resemble a diagram $D$, above which we place our basepoint $x_0$. Choose an orientation for $D$.

If $D$ has $c$ crossings, then it has $c$ arcs (connected strands). Place loops $\alpha_1,...,\alpha_c$ based at $x_0$ around each (think puppet strings), satisfying the right-hand rule.

Suppose that at a particular crossing with positive writhe sign $\alpha = \alpha_i$ overpasses $\beta = \alpha_j$ leading to $\gamma = \alpha_k$. Then the picture following shows that:

or:

At a negative writhe crossing:

The fundamental group $\pi_1(\mathbb{R}^3 - K)$ is then characterised by such 'conjugation' equations, one for each crossing. However it can be shown that one relation is always redundant, in an analogy with the colouring relations.

Final Remarks

Theorem vK2 can be used to prove that the knot group $\pi_1(\mathbb{R}^3 - K)$ is generated by the loops $\alpha_1,...,\alpha_c$ associated to the arcs of a knot diagram, subject to any $c-1$ of the crossing relations:

For a torus knot of type $(p,q)$ (with $p,q$ positive coprime integers), we have:

The nature of $\pi_1(\mathbb{R}^3 - K)$ permits one to construct finite covering spaces of $\mathbb{R}^3 - K$ (e.g. one based on the group $S_3$ if $K$ is a trefoil so $(p,q) = (3,2)$). This idea underlies work of William Thurston, which revolutionised the study of $3$-manifolds.

If $K = 3_1$, then $S^3 - K$ is $SL(2,\mathbb{R})/SL(2,\mathbb{Z})$, giving a link with the Lorenz attractor. If $K = 4_1$ then $S^3 - K$ can be obtained from two solid tetrahedra by gluing pairs of faces in a prescribed way (and removing the unique vertex in their quotient).