Computing π₁

The circle is the only space whose fundamental group 7A.4 The Fundamental Group computed from scratch. Most spaces of interest are built by gluing simpler pieces, and the Seifert–van Kampen theorem computes the fundamental group of a union from the fundamental groups of the pieces and of their overlap. With it, every surface's fundamental group follows from a single picture, and so does the fundamental group of the complement of a knot. The results are given by presentations: generators, one for each independent loop, and relations, one for each way of filling a loop in.

The mechanism to remember is that gluing in a disc along a loop kills that loop, and gluing in higher-dimensional cells changes nothing. So the fundamental group of a cell complex is determined by its 1-cells and 2-cells. For three-manifolds this means it can be read off from a Heegaard diagram (10A.2 Building Three-Manifolds).

By the end of this chapter you will be able to:

• work with free groups, presentations and free products, and abelianise a presentation;
• state the Seifert–van Kampen theorem and apply it to wedges of circles and to attaching cells;
• compute $\pi_1$ of every closed surface, of $\mathbb{R}P^n$, and of the torus;
• write the Wirtinger presentation of a knot group, and show that the trefoil is knotted;
• explain why the fundamental group distinguishes all closed surfaces.

5.1Knotted DNA

5.2Free groups and presentations

The free group $F(S)$ on a set $S$ consists of reduced words in the letters $s^{\pm1}$, $s \in S$ (words with no adjacent $ss^{-1}$ or $s^{-1}s$), multiplied by concatenation and cancellation. The free group on one letter is $\mathbb{Z}$; on two letters $a$, $b$ it is non-abelian ($ab \neq ba$) and enormous. Every group generated by $S$ is a quotient of $F(S)$.

A presentation $\langle S \mid R\rangle$, with $R$ a set of words, is the quotient of $F(S)$ by the normal closure of $R$, the smallest normal subgroup containing $R$: the "largest" group generated by $S$ in which the relations $r = 1$ hold. For example $\langle a \mid a^n\rangle \cong \mathbb{Z}/n$ and $\langle a, b \mid aba^{-1}b^{-1}\rangle \cong \mathbb{Z}^2$. Deciding whether two presentations give isomorphic groups is in general undecidable (Adian and Rabin, 1955–58), which is part of why three-manifold topology is hard.

The abelianisation $G^{\mathrm{ab}} = G/[G, G]$ forces all generators to commute; for a finite presentation it is a finitely generated abelian group computed by integer linear algebra on the relations. It is the first homology group $H_1$ (7A.8 Homology in Brief).

The free product $G * H$ is the group of reduced words alternating between non-identity elements of $G$ and of $H$; if $G = \langle S \mid R\rangle$ and $H = \langle S' \mid R'\rangle$, then $G * H = \langle S\cup S' \mid R\cup R'\rangle$. Given homomorphisms $i : K \to G$ and $j : K \to H$, the amalgamated free product $G *_K H$ adds the relations $i(k) = j(k)$ for all $k \in K$.

5.3The Seifert–van Kampen theorem

Wedges of circles. The wedge $S^1\vee S^1$ (two circles joined at a point, a figure eight) is covered by two open sets, each a circle with a short whisker of the other, whose intersection is a contractible cross. So $\pi_1(S^1\vee S^1) \cong \mathbb{Z} * \mathbb{Z} = F(a, b)$, free on the two loops. The same works for any number of circles, and for any connected graph: its fundamental group is free, on as many generators as edges outside a spanning tree.

5.4Surfaces, projective spaces and the torus

A closed surface described by a polygon word (7A.3 Manifolds and Surfaces) is a cell complex: after gluing, the boundary of the polygon becomes a wedge of circles, one for each letter (all corners are identified in the normal forms), and the polygon is a 2-cell attached along the word. So:

surface	presentation of $\pi_1$	abelianisation
$S^2$	trivial	$0$
torus $T^2$	$\langle a, b \mid aba^{-1}b^{-1}\rangle \cong \mathbb{Z}^2$	$\mathbb{Z}^2$
genus $g$, $\Sigma_g$	$\langle a_1, b_1, \dots, a_g, b_g \mid [a_1, b_1]\cdots[a_g, b_g]\rangle$	$\mathbb{Z}^{2g}$
$\mathbb{R}P^2$	$\langle a \mid a^2\rangle \cong \mathbb{Z}/2$	$\mathbb{Z}/2$
$N_k = \#^k\mathbb{R}P^2$	$\langle a_1, \dots, a_k \mid a_1^2\cdots a_k^2\rangle$	$\mathbb{Z}^{k-1}\oplus\mathbb{Z}/2$

Here $[a, b] = aba^{-1}b^{-1}$ is the commutator. The abelianisations are all different, so the fundamental group distinguishes every closed surface from every other, which completes the proof that the surfaces in the classification theorem are pairwise distinct (7A.3 Manifolds and Surfaces). And only $S^2$ has trivial fundamental group: a simply connected closed surface is a sphere.

Projective spaces. $\mathbb{R}P^n$ has a cell structure with one cell in each dimension $0, 1, \dots, n$; the 2-cell is attached to the circle $\mathbb{R}P^1$ by a map of degree $2$. So $\pi_1(\mathbb{R}P^n) = \langle a \mid a^2\rangle = \mathbb{Z}/2$ for every $n \geq 2$, by Corollary 5.2. In particular $\pi_1(\mathbb{R}P^3) = \mathbb{Z}/2$, a Path milestone that 7A.6 Covering Spaces proves again with covering spaces and interprets physically.

5.5Knot groups

A knot is an embedded circle $K \subset \mathbb{R}^3$ (or $S^3$), and its knot group is $\pi_1(\mathbb{R}^3\setminus K)$. Equivalent knots, those that can be deformed into each other through embeddings, have homeomorphic complements and isomorphic groups. The unknot's complement deformation retracts onto a circle linking it, so its group is $\mathbb{Z}$.

The Wirtinger presentation. Draw the knot as a diagram in the plane with breaks at undercrossings, so that it falls into arcs. Give the knot an orientation. Each arc $k$ contributes a generator $x_k$: a loop from a basepoint above the plane that dips down under the arc and back, going around it in the right-handed sense. At each crossing, where an overcrossing arc $x_k$ separates two undercrossing arcs $x_i$ and $x_j$, there is one relation,

depending on the sign of the crossing. (Van Kampen, applied to the complement cut along the plane of the diagram, proves this.) One of the relations always follows from the others.

For the trefoil, with three arcs $x$, $y$, $z$ and three crossings (Figure 5.2), the Wirtinger presentation reduces to

the second form by $a = xyx$, $b = xy$ (Exercise 5.5). This group is non-abelian: sending $x \mapsto (1\ 2)$ and $y \mapsto (2\ 3)$ in the symmetric group $S_3$ respects the relation and is onto a non-abelian group (Exercise 5.6). Since the unknot's group is $\mathbb{Z}$, abelian, the trefoil is knotted: no deformation through embeddings untangles it. Its abelianisation is $\mathbb{Z}$, the same as the unknot's; every knot group has abelianisation $\mathbb{Z}$, so the non-abelian structure is essential.

5.6History

Walther Dyck introduced group presentations in 1882 and Heinrich Tietze developed them in 1908. Max Dehn studied knot groups and proved in 1910 that the trefoil is knotted and distinct from its mirror image (the latter in 1914). Wilhelm Wirtinger described his presentation in lectures in 1905; it was published by Emil Artin in 1925. Herbert Seifert (1931) and Egbert van Kampen (1933) proved the theorem that bears their names. Sergei Adian (1955–57) and Michael Rabin (1958) proved that most properties of finitely presented groups, including triviality, are undecidable. Stasiak and colleagues' work on DNA knot mobility appeared in 1996 and Taylor's knotted protein in 2000.

5.7Exercises

Solution

Send $a \mapsto (1\ 2)$, $b \mapsto (2\ 3)$; since the group is free, any assignment extends to a homomorphism. The image is generated by two transpositions, hence all of $S_3$, which is non-abelian, so $ab \neq ba$ in the free group. Since transpositions are their own inverses, $aba^{-1}b^{-1} \mapsto \big((1\ 2)(2\ 3)\big)^2$; the product of two transpositions sharing one point is a 3-cycle, and the square of a 3-cycle is a 3-cycle, so the image is non-trivial.

Solution

(a) $b^3 = xyxyxy = (xyx)(yxy) = (xyx)(xyx) = a^2$, using $yxy = xyx$. (b) $b^{-1}a = y^{-1}x^{-1}xyx = x$, and $a^{-1}b^2 = x^{-1}y^{-1}x^{-1}xyxy = x^{-1}xy = y$. Conversely, with $x = b^{-1}a$, $y = a^{-1}b^2$: $xyx = b^{-1}a\,a^{-1}b^2\,b^{-1}a = a$ and $yxy = a^{-1}b^2b^{-1}aa^{-1}b^2 = a^{-1}b^3 = a^{-1}a^2 = a$.

Solution

$(1\ 2)(2\ 3)(1\ 2) = (1\ 3) = (2\ 3)(1\ 2)(2\ 3)$. In the abelianisation, each relation $x_j = x_kx_ix_k^{-1}$ becomes $x_j = x_i$; the arcs of a connected diagram are linked by crossings, so all generators become equal, leaving one generator and no relation: $\mathbb{Z}$.

Solution

(a) $V$ deformation retracts onto its core circle, so $\pi_1(V) = \mathbb{Z}$, generated by the core. Attaching the 2-cell along $\alpha$ kills $[\alpha] = n$ times the generator, giving $\mathbb{Z}/n$ ($\mathbb{Z}$ if $n = 0$); the 3-cell changes nothing. (b)–(d) follow by substituting $n = 1$, $0$, $p$.