Recall from Sheet 2 that the torus \(T\) can be made from a solid hexagon with sides glued in pairs according to the labelled arrows. Use van Kampen’s theorem to compute a presentation of the fundamental group, and show that it is isomorphic to \(\mathbb{Z}^2\).
Solution
Goal: Compute a presentation of the fundamental group \(\pi_1(T^2, x_0)\) of the 2-dimensional torus \(T^2\) and prove that \(\pi_1(T^2, x_0) \cong \mathbb{Z}^2\).
1. Cellular / Polygonal decomposition of the torus \(T^2\).
1.1. \(T^2\) can be realized as a CW complex with:
One 0-cell: \(e^0 = \{v\}\),
Two 1-cells: \(a, b\) attached at \(v\), forming the 1-skeleton \(X^1 = S^1 \vee S^1\),
One 2-cell: \(e^2 = D^2\) attached along the boundary loop via the attaching map \(\phi \colon \partial D^2 \to X^1\) with word \(aba^{-1}b^{-1}\). <2>2. Proof: The standard square \([0, 1] \times [0, 1]\) with identifications \((x, 0) \sim (x, 1)\) and \((0, y) \sim (1, y)\) identifies all four vertices to a single point \(v\), the horizontal edges to \(a\), the vertical edges to \(b\), and the perimeter traversed counterclockwise gives the attaching loop \(aba^{-1}b^{-1}\).
2. Fundamental group of the 1-skeleton \(X^1\).
2.1. The 1-skeleton \(X^1 = S^1 \vee S^1\) is the wedge of two circles.
2.2. By the Seifert-van Kampen theorem, \(\pi_1(X^1, v) \cong \pi_1(S^1, v) * \pi_1(S^1, v) \cong \langle a, b \mid \emptyset \rangle = F(a, b)\), the free group on two generators.
2.3. The wedge \(S^1 \vee S^1\) is covered by two open sets each homotopy equivalent to \(S^1\) with contractible intersection, so van Kampen gives the free product \(\pi_1(S^1) * \pi_1(S^1) \cong \mathbb{Z} * \mathbb{Z} = F(a, b)\).
3. Effect of attaching the 2-cell on \(\pi_1\).
3.1. Attaching a 2-cell along a loop \(\gamma\) quotients the fundamental group of the 1-skeleton by the normal subgroup generated by the homotopy class \([\gamma] \in \pi_1(X^1, v)\).
3.2. Here the attaching map \(\phi\) represents the element \([\phi] = a b a^{-1} b^{-1} = [a, b]\) (the commutator of \(a\) and \(b\)).
3.3. Therefore, the fundamental group of \(T^2 = X^1 \cup_\phi e^2\) has presentation: \[\pi_1(T^2, v) \cong \langle a, b \mid a b a^{-1} b^{-1} = 1 \rangle.\]
3.4. Attaching a 2-cell along a loop \(\gamma\) kills the normal subgroup generated by \([\gamma]\); here \([\phi] = [a, b]\), so the cellular van Kampen theorem (Hatcher Proposition 1.26) gives the presentation \(\langle a, b \mid a b a^{-1} b^{-1} = 1 \rangle\).
4. Prove that the presented group \(G = \langle a, b \mid aba^{-1}b^{-1} = 1 \rangle\) is isomorphic to \(\mathbb{Z}^2\).
4.1. The relation \(aba^{-1}b^{-1} = 1\) is equivalent to \(ab = ba\), meaning \(a\) and \(b\) commute.
4.2. The group \(G\) is the free abelian group on two generators \(\{a, b\}\).
4.3. Define a homomorphism \(\Phi \colon G \to \mathbb{Z}^2\) by \(\Phi(a) = (1, 0)\) and \(\Phi(b) = (0, 1)\).
4.3.1. \(\Phi\) is well-defined because in \(\mathbb{Z}^2\), \((1, 0) + (0, 1) = (0, 1) + (1, 0) = (1, 1)\), so \(\Phi(aba^{-1}b^{-1}) = (1, 0) + (0, 1) - (1, 0) - (0, 1) = (0, 0)\).
4.3.2. By the universal property of group presentations, a map on generators extends to a well-defined homomorphism exactly when it sends the relator to the identity; the computation in 1 verifies this.
4.4. Define a homomorphism \(\Psi \colon \mathbb{Z}^2 \to G\) by \(\Psi(m, n) = a^m b^n\).
4.4.1. Since \(a\) and \(b\) commute in \(G\), \(a^{m_1+m_2} b^{n_1+n_2} = a^{m_1} b^{n_1} a^{m_2} b^{n_2}\), so \(\Psi\) is a group homomorphism.
4.4.2. The identity \(a^{m_1+m_2}b^{n_1+n_2} = a^{m_1}b^{n_1}a^{m_2}b^{n_2}\) follows from \(ab = ba\), which is the defining relation of \(G\); this is exactly the condition \(\Psi((m_1,n_1)+(m_2,n_2)) = \Psi(m_1,n_1)\Psi(m_2,n_2)\).
4.5. Check inverses:
4.5.1. For \((m, n) \in \mathbb{Z}^2\), \(\Phi(\Psi(m, n)) = \Phi(a^m b^n) = m(1, 0) + n(0, 1) = (m, n)\).
4.5.2. For any word \(w \in G\), using \(ab=ba\) reduces \(w\) to \(a^m b^n\).Then \(\Psi(\Phi(a^m b^n)) = \Psi(m, n) = a^m b^n\).
4.5.3. The two computations show \(\Phi \circ \Psi = \operatorname{id}_{\mathbb{Z}^2}\) and \(\Psi \circ \Phi = \operatorname{id}_G\).
4.6. Therefore, \(G \cong \mathbb{Z}^2 = \mathbb{Z} \times \mathbb{Z}\).
4.7. A homomorphism with a two-sided inverse is an isomorphism, so \(\Phi\) is an isomorphism.
5. Q.E.D.
5.1. <1>1–<1>4 gives the presentation \(\langle a, b \mid aba^{-1}b^{-1} = 1 \rangle\) and proves \(\pi_1(T) \cong \mathbb{Z}^2\).