Show that \(\pi_1(K)\), the Klein bottle, is given by pairs \((m,n)\) where \((m,n)\star (p,q) = (m+(-1)^np, n+q)\)
Show this is torsion-free
Show that \(T\) is a double cover of \(K\).
Solution
Goal: Prove that the fundamental group of the Klein bottle \(\pi_1(K)\) is isomorphic to the semidirect product \(G = \mathbb{Z} \rtimes \mathbb{Z}\) whose underlying set is \(\mathbb{Z} \times \mathbb{Z}\) with group operation \((m, n) \star (p, q) = (m + (-1)^n p, n + q)\). Prove that:
\(G\) is torsion-free.
The 2-torus \(T^2\) is a 2-fold covering space of the Klein bottle \(K\).
1. Realize \(\pi_1(K)\) as the group of deck transformations on \(\mathbb{R}^2\).
1.1. The Klein bottle \(K\) is the quotient of \(\mathbb{R}^2\) by the group of affine isometries \(G\) generated by the translations and glide reflections: \(a(x, y) = (x + 1, y)\) and \(b(x, y) = (-x, y + 1)\) (or \(a(x, y) = (x, y+1)\) and \(b(x, y) = (x+1, -y)\)).
1.2. Using the standard presentation \(K = \langle a, b \mid a b a b^{-1} = 1 \rangle \cong \langle a, b \mid b a b^{-1} = a^{-1} \rangle\): Every element of \(G\) can be written uniquely in normal form as \(a^m b^n\) for \(m, n \in \mathbb{Z}\).
1.3. Group multiplication in normal form: \[(a^m b^n)(a^p b^q) = a^m (b^n a^p b^{-n}) b^{n+q} = a^m (a^{(-1)^n p}) b^{n+q} = a^{m + (-1)^n p} b^{n+q}.\]
1.4. Under the bijection \((m, n) \leftrightarrow a^m b^n\), the group multiplication is: \[(m, n) \star (p, q) = (m + (-1)^n p, n + q).\]
1.5. The identity element is \((0, 0)\) and the inverse is \((m, n)^{-1} = (-(-1)^n m, -n)\).
1.6. The relation \(b a b^{-1} = a^{-1}\) means conjugation by \(b^n\) sends \(a^m\) to \(a^{(-1)^n m}\), so the group is the semidirect product \(\mathbb{Z} \rtimes_\theta \mathbb{Z}\) with \(\theta(n)(m) = (-1)^n m\); the normal form \(a^m b^n\) and the multiplication rule in <2>3–<2>4 are exactly this semidirect product structure.
2. Show that \(G = \pi_1(K)\) is torsion-free.
2.1. Let \((m, n) \in G\) and let \(k \ge 1\) be an integer such that \((m, n)^k = (0, 0)\).
2.2. Projecting \(G \to \mathbb{Z}\) onto the second coordinate via \((m, n) \mapsto n\) is a group homomorphism: The second coordinate of \((m, n)^k\) is \(k \cdot n\).
2.3. Since \((m, n)^k = (0, 0)\), we must have \(k \cdot n = 0\), which implies \(n = 0\) (since \(\mathbb{Z}\) is torsion-free).
2.4. For \(n = 0\), \((m, 0)^k = (k \cdot m, 0) = (0, 0)\), which implies \(k \cdot m = 0\), hence \(m = 0\).
2.5. Therefore, \((m, n) = (0, 0)\) is the only element of finite order in \(G\).
2.6. If \((m, n)^k = (0, 0)\), then <2>2–<2>3 force \(n = 0\), and 4 then forces \(m = 0\); hence the only element of finite order is the identity, so \(G\) is torsion-free.
3. Show that the torus \(T^2\) is a double cover of \(K\).
3.1. Define the homomorphism \(\phi \colon G \to \mathbb{Z}/2\mathbb{Z}\) by \(\phi(m, n) = n \bmod 2\).
3.2. \(\phi\) is a surjective group homomorphism because \(\phi((m, n) \star (p, q)) = (n + q) \bmod 2 = \phi(m, n) + \phi(p, q)\).
3.3. The kernel \(H = \ker(\phi) = \{(m, n) \in G \mid n \equiv 0 \pmod 2\} = \{(m, 2k) \mid m, k \in \mathbb{Z}\}\) is a normal subgroup of index 2 in \(G\).
3.4. For elements in \(H\), the multiplication rule simplifies because \((-1)^n = (-1)^{2k} = 1\): \[(m, 2k) \star (p, 2l) = (m + p, 2(k + l)).\]
3.5. Thus \(H\) is isomorphic to the free abelian group \(\mathbb{Z}^2\): \[H \cong \mathbb{Z} \times \mathbb{Z} = \mathbb{Z}^2 \cong \pi_1(T^2).\]
3.6. By covering space theory, the index 2 subgroup \(H \le \pi_1(K)\) corresponds to a connected 2-sheeted (double) covering space \(\widetilde{K} \to K\).
3.7. Since \(\pi_1(\widetilde{K}) \cong \mathbb{Z}^2\) and \(\widetilde{K}\) is a closed orientable 2-manifold (the orientable double cover of the non-orientable surface \(K\)), \(\widetilde{K}\) is homeomorphic to the 2-torus \(T^2\).
3.8. The Galois correspondence associates to the index-2 subgroup \(H\) a connected 2-sheeted cover \(\widetilde{K} \to K\) with \(\pi_1(\widetilde{K}) \cong H \cong \mathbb{Z}^2\); the only closed orientable surface with fundamental group \(\mathbb{Z}^2\) is the torus \(T^2\).
4. Q.E.D.
4.1. Steps <1>1–<1>3 complete the proof.